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35
F62
1766
ARTES
1817)
SCIENTIA
VERITAS
LIBRARY
OF THE
UNIVERSITY OF MICHIGAN
PLUATE OF UNYN
THEBOR
SI QUÆRIS PENINSULAM-AMŒNAM
CIRCUMSPICE
I'm Drenn
THE
UNIVERSAL
MEASURER,
AND
MECHANIC,
In THREE PART S.
A WORK equally uſeful to the GENTLEMAN,
TRADESMAN, and MECHANIC.
With eleven NEAT COPPER-PLATES.
The SECOND EDITION.
>
By A. FLETCHER Philomath.
LONDO N:
Printed for G. ROBINSON and J. ROBERTS, in Pater-nofter Row,
MDCCLXVI.
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япадёт
10
44-45
[iii]
53448
3
GENERAL CONTENTS.
PART FIRST.
Of practical geometry, fhewing how to conftruct, reduce, divide, in-
Scribe, circumfcribe, &c. figures in plain and folid geometry; plain
trigonometry, wrought geometrically, arithmetically, and infiru-
mentally; heights and diſtances; levelling; furveying of mines;
with a great many practical problems.
PART SECOND.
C
The theory of geometry; menſuration; conic fections, and mechanics
demonftrated in an eaſy, new, and univerſal method, independent
of fluxions. To which are prefixed, the elements of algebra; infinite
feries; progreffions; fumming of feries; logarithms; &c.
PART THIRD.
Decimals; feet and inches; fliding-rules; menfurations of planes,
and folids of all kinds, both by the pen and fliding-rule; furvey-
ing, plotting, and dividing of lands; gauging, inching, ullaging,
&c. with a large and curious collection of questions and folutions
relating to meaſuring; guaging, mechanics, as pendulums, pumps,
barometers, mills, engines, wheel-carriages, ftrength of walls,
beams, &c. gunnery; circular motion, proving the true figure of
the earth, &c.
ADVERTISEMENT.
BROUGHTON, Jan. 1766.
The AUTHOR continues, as ufual, to teach
all the Branches of the MATHEMATICS,
according to the beſt and lateſt Improvements.
PEEEEEEEREREKERERPRETPIL
1
[ v ]
THE
PRE FAC E.
SINCE trade and business must neceffarily engage the care and
attention of the greater part of mankind; it is no wonder that
the claffics and study of dead languages, ſhould be lefs cultivated
than the mathematics, which have gained the preference, and of
late years been much improved: tho' (like all other fiiences) the
mathematics are but in their progreffive state towards perfection;
certainly then, every endeavour to improve them, or to render the
Study thereof eafy and pleaſant, deferves encouragement.
Meaſuring and mechanics are branches of mathematics,
which come fo often in ufe, that few men who live in the world,
but have occafion for their affiftance one time or other. The gen-
tlemen may be agreeably employed in meafuring the works of the
feveral artificers he may have occaſion to employ, in furveying, ſet-
ting out, and dividing lands; and thereby not only prevent frauds,
but be enabled to form a better judgment of its value.
The mechanic is nearly concerned to employ fome of his lifure
hours in acquiring a competent knowledge of the principles, where-
on his labour depends; that he may be capable not only to ſet ajuſt
value on his workmanſhip, but to contrive the nearest way to work;
and to improve the invention of his implements, for expedition as
well as eafe.
The favourable reception this book has met with, encouraged me
to employ my utmost care and attention in improving this fecond
1
vi
The PRE FACE.
edition, in order to make it more compleat; with the addition of
mechanics: a ſubject of the greatest fervice in the affairs of
life.
It would ill become the author, to ſay more in behalf of theſe
ſheets ; let it fuffice, therefore, to give a brief account of this work,
which is divided into three parts.
{
•
The first is a compleat body of practical geometry, with the
demonftration of every difficult problem.
Problem ift to 40th the construction of figures in plane geo÷
metry; conic fections; and architecture; with the method of fix-
ing butments to bridges. From prob. 40 to 52, the rules of pro-
portion; extraction of the Square and cube roots geometrically.
From 52 to 68 problems concerning the circle. From 68 to 87,
circumfcription, and infcription of figures. From 87 to 92, re-
duction of figures, and plots. From 92 to 119, divifion of figures
all
manner of ways. From 119 to 128, plane trigonometry,
right and oblique angled, ſolved both geometrically, and arithmetic-
ally. From 128 to 142, heights at diftances; with the defcrip-
tion and ufe of the inflruments for that purpose. From 142 to
145, levelling, and furveying of mines. From 145 to 152, cu-
ricus questions in navigation, mechanics, &c. From 152 to 164,
the moſt uſeful problems in ſolid geometry; the construction of the
five regular bodies; how to take dimenfions of folids, as caſks, fruf-
tums, &c.
1
PART 2d, is altogether theory; containing from problem 164 to
178, algebraic definitions, characters; addition, fubtraction, mul-
tiplication and divifion, of whole quantities; as alſo of fractions,
and furds; involution, evolution, proportion, progreffion; infinite
feries; folution of equations, by converging, and infinite feries,
reverſions of ſeries; an eafy way to fum up any ferics; with many
new inveſtigations, logarithms, &c. Prob. 178. The chief theo-
rems in Euclid, with many now theorems and their demonflrations,
The PREFACE.
vii
191,
Prob. 179, of fines and tangents; with the axioms of trigono-
metry. From 179 to 182, of conic fections. From 182 to
the theory of menfurations; with the demonftration of theo-
rems, for meaſuring, and gauging, all kinds of planes, and folids,
both by particular rules, and by one general method, as alſo by a
new and univerſal feries; with theorems for knowing the forms
of cafks, inching, ullaging, &c. Prob. 191, a general method of
maxima, and minima. From 191 to 201, the theory of mecha-
nics fhewing the general laws of motion fimple and compound, e-
laftic and non-elaftic, of gravity, vibrations of pendulums, and
mufical ftrings; mechanic powers, wheel carriages, centers of gra-
vity, ofcillation, percuffion and gyration, preſſure and ſtrength of
walls, beams, &c. motion of projectiles, hydrostatics, pneumatics,
hydraulics, &c. Prob. 201, the principles of fluxions and fluents,
with their application to many parts of mechanics.
PART 3d. Divided into three fections. Sect. 1ft, Decimals
made plain and caly, with feveral uſeful contractions, and the ex-
traction of roots. This fection (after common arithmetic) ſhould
be learned, then practical geometry. By obferving this method,
you will be better enabled to fluly menfurations to advantage.
Sect. 2d. Construction, deſcription, and uſe of Cogglefhal's flid-
ing-rule. Sect. 3. Multiplication of feet and inches; called cross
multiplication. Sect. 4. Superficial mcafure of planes and of folids ;
with the methods of taking dimenſions with, or without inches.
Sect. 5. The ſeveral artificers work. Sect. 6. How to meaſure all
forts of folids. Sect. 7. Surveying, plotting and dividing of lands,
&c. Sect. 8. Gauging in all its parts; with the deſcription and
uſe of inftruments for that purpoſe. Each example in theſe
three fections is wrought by pen and fliding-rule. Sect. 9.
Questions with their folutions. 1. From queſt. 1 to 40, exerciſes
in menſurations, furveying, and gauging. 2. From 40 to 50, men-
furations of folids, by their centers of gravity.. 3. From 50 to 85,
fbewing the nature and uſe of the mechanic powers. 4. From 85
f
viii
The PREFACE.
to 115, of wheel carriages, ftrength and firefs of walls, beams, &c.
5, From 115 to 140, pendulums, founds, muſical inſtruments.
6. From 140 to 166, the maxima of bodies moving in and by fluids;
the forces of moving bodies, and the refiftance they meet with; the
perfections of mills, engines, &c. 7. From 166 to 192, the ſpe-
cific gravities of bodies, with a table; the admirable properties of
the air, weather-glaffes, pumps, and fire-engines. 8. From 192 to
231, the practice of gunnery in refifting and non-refifting mediums
9. From 231 to 261, of friction, fails of ſhips, ſpouting-fluids,
waves, whirling bodies, &c. 10. From 261 to 294, laws of
central forces, circular motion, true figure of the earth; gravities,
denfities, &c. of the planets and tides.
To conclude, I have been careful to omit no article that ſeemed
useful or neceffary, and as cautious to avoid things perplexing or fu-
perfluous: and to make the whole more familiar and eaſy to the
learner, I have kept as much as poffible within the limits of algebra;
altho' by this work may be acquired, a clear and diſtinct notion of
fluxions and fluents.
THE
UNIVERSAL
MEASURER
AND
MECHANIC.
PART FIRST.
3
A Complete BODY of PRACTICAL
I
•
GEOMETRY.
I
POSTULATE S.
T may be taken for granted, that a right line may
be drawn between any two given points,
2. That a given right line may be produced at
pleafure.
3. That from a given point a line may be drawn
either parallel to another, or in fuch a manner as
to make any given angle with any other given line.
W
4. That a circle or any arch thereof may be defcribed upon a plane
with a pair of compaffes opened to any extent.
5. That a line may be defcribed by the motion of a point.
6. That a fpace, plane, or area, may be defcribed by a line moving
parallel to itſelf, or about any point as a center.
7. That a folid may be defcribed by a plane moving parallel to its felf,
or about any part thereof as an axis.
A
2.
THE UNIVERSAL MEASURER
1
It is to be obferved, that in working geometrically, all points and lines
fhould be as fmall as poffible, (the former being without dimenfions, and
the latter only one dimenfion, viz. length) and the figures will be more
exact. Alfo, when you begin to work any problem, obferve carefully what
things are given, and with theſe work exactly as the reading directs, ma-
king all lines of illuſtration pricked or dotted, and all lines given or re-
quired, black; be careful to draw all the lines as directed whether they
are in the figure or not.
い
​{
PROBLEM
I.
To draw a line CD parallel to a given line AB, and at the distance
of the line e f from it. Fig. 1.
1. With the given line e f in the compaffes, and one foot on any two
different points in AB, as at E and F, ftrike two arches n and s.
2. Lay a ruler to touch the tops of theſe two arches, and draw the
line CD, which is the parallel required.
Note. Black lines or arches commonly reprefent things given and re-
quired, and thofe pricked or dotted, fhew how the problem is worked.
In geometry, as in arithmetic, there is always fomething given and re-
quired and here obferve, the things given you may prick down as you
pleafe, then work by the directions for thofe required.
PROBLEM II.
To draw a line GH parallel to a given line IK, and to pass thro
a given point P. Fig. 2.
1. With one foot in any part of the given line I K,
Ealf circle to pafs through P, 2. make the arch I M
as at Q, fweep a
to the arch P K
3. thro' M and P draw the line G H, and its done. See Problem 42.
}
PROBLEM III.
To bifect or divide a given line A B into two-equal parts. Fig. 3.
1. With any radius greater than half the faid line, and one footfirft
on one end and then on the other, ftrike two arches, croffing each other
in C and D, 2. thro' the points C and D draw the line CED; then
will AEE B.
AND MECHANIC.
3
PROBLEM
IV.
To erect a perpendicular BD, upon the end B, of any given line
A B. Fig. 4.
1. Lay a ruler along A B, and produce it to C, then with one foot
on B fweep the pricked half circle, cutting A BC in n and m, 2. upon
n and m ſtrike two arches croffing each other in D, draw DB and its
done.
PROBLEM V.
To raiſe a perpendicular DB upon any given point B, in a given
line A C. Fig. 4.
. As in the laſt problem, take any two points n and m, equally diſtant
from B, upon which as centers, ftrike two arches crolling each other in
D, thro' D and B draw DB, and its done.
PROBLEM
VI.
From a given point P upon a given line AB, to let fall a perpen-
dicular. Fig. 5.
1. With one foot in the given point P, ftrike an arch to cut the given
line A B in two points n and m, 2. upon n and m with any radius ftrike
two arches croffing each other in C, 3. thro'P and C draw PC E, which
is the
required.
PROBLEM
VII.
Given any angle BAC, to make another angle bac equal to it.
Fig. 6.
1. Draw a line ac, 2. with any radius and one foot on A, the given
angular point, ftrike an àrch B C to cut each fide, with the fame radius on
a, ſtrike the arch bc, take the arch BC in your compaffes and lay it from
c tob, 3. thro'a and b draw ab, and its done.
PROBLEM VIII.
To divide a given angle BAC, into two equal angles, DAC
-DAB. Fig. 7.
1. With any radius and one foot on A, fweep the arch nm, 2. with
any radius upon n and m ftrike two arches crofling each other in D, diaw
AD, and its done.
THE UNIVERSAL MEASURER
NOTE. When three letters B A C expreſs an L, the middle letter A
flands at the angular point.
PROBLEM
IX.
How to make a line of Chords. Fig. 8.
1. Draw a line A m, which make radius, and upon Afweep the arch m
90. 2. Frommupon this arch, lay the radius Am to 60. 3. Take half the
arch m 60 and lay it from 60 to go. 4. Divide the arch 60, 90, into three
= parts, which wil lalfo divide m 60 into fix = parts. 5. Set one foot of
your compaſſes in m, and ſtrike arches from each of theſe parts to the
line A m, and its done.
NOTE. Each of thefe equal parts in the arch fhould be divided into
ten more.
PROBLEM
X.
To make any acute angle; ſuppoſe one of 35° Fig. 10.
1. Draw a line AC, with the chord of 60° upon A fweep an arch C
40. 2. From the fame chords take 350, and lay it from C upon the
arch, thro' which draw the line A B; fo is the angle D A C one of 35º.
PROBLEM
XI.
Upon the point A, and with the line AC, to make a right angle,
i. e. one of 90°. Fig. 10.
This is the fame as erceling a ≈.
1. Upon A with the chord of 600 ftrike the arch C 90°, from the
fame fcale of chords take 90º, and lay it upon the arch from C to 90.
2. Thro' A and go draw AD; fo is the angle DAC one of 90º, as
required.
PROBLEM
XII.
{
Upon the point A, and with the line AB, to make an obtuſe L,
BAC, fuppofe of 95°.
Fig. 9.
1. Upon A with the chord of 600, fweep the arch B C, then, becauſe
the arc 1 B C is to be above goº, which is the whole length of the line
of choice, you may take any two chords whofe fum is 95°, as 60° and
350, and lay them upon the arch B C one after the other, from Bto C,
then draw A C, and its done.
}
AND MECHANIC.
5
PROBLEM
XIII.
Upon any given line AB, to make an equilateral triangle. Fig. 11.
1. Make the given line radius, and upon A fweep the arch B C, alfo
upon B fweep the arch A C, croffing each other in C. 2. Join A C and
B with lines, and you'll have A B C, the triangle required.
PROBLEM XIV.
Given three lines AB, AD, AC, to make a triangle. Fig. 12.
1. If A B be one of the given lines, take either of the other two in
your compaffes as A D, and with one foot on Bftrike an arch, then with
A E in your compaffes and one foot on A, ftrike another arch crofing
the former arch in C. 2. Join CA and CB, and its done.
NOTE. One of the
given lincs muſt be fhorter than the fum of the
other two, or it cannot poffibly be a triangle.
PROBLEM
XV.
Required the greatest triangle that it is poffible to make with a
given bafe AB, and fum of the other two fides DE. Fig. 13.
1. This muſt be when the perpendicular CP is the longeſt poffible;
therefore with F E FD-half D E in your compaffes, and one foot
feverally on A and B, fweep two arches crolling each other in C. 2. Join
AC and CB, and its donc. For it is plain, if DECA+CB be a
thread or cord faftened to two pins A B, and a needle put in C, the
double thereof and moved any way, keeping the cord at a conſtant
fretch, the CPby fuch motion will be shortened, whence the ifcofceles
triangle ABC is that 'required.
PROBLEM
XVI.
Given the diagonal A B, and the four fides AC, AD, AE, AF,
f a trapezia AIGH, to make the trapezia. Fig. 14.
1. Make AG A B. 2. With any of the given lines in your com-
pafics as A C, and one foot on either A or G, fuppofe in A, flrike an
arch H, alfo with fome other of the lines, as A E, and one foot in G,
fweep an arch, crolling the laft arch in H. 3. Join H A and HG, and
6
THE UNIVERSAL MEASURER
one of the triangles is made; then with the other two lines A D and A
F, form the triangle AIG, below the diagonal A G, and its done. It
is but making two triangles by problem
PROBLEM XVII.
Given any right lined figure abcde, to make another equal and
fimilar to it. Fig. 15.
1. Divide the given figure a b c d e into As, by drawing the diago-
nals a c and a d, 2. by problem 14, with the three lines a c, a b, and
bc, make the ▲ A B C, alfo upon A C, with the two lines a d and d c,
make the A A D C, laftly, upon A D, with the two lines a e and d e,
nake the A AED, and its done.
PROBLEM
XVIII.
Given any right lined figure a b c d, to make another figure A B
CD fimilar to it, whofe fides may be twice as large, or in any
other proportion. Fig. 16.
1. This may be done by the laft problem, by doubling every line as
you lay it down : or thus, by problem 7. make the L BAD = L bad,
continuing A D, until it be twice a d, then upon D make an LAD C
twice L ad c, making DC
CL dc b, making C
-
twice dc, then make an L D C B upon
B = twice cb, which in this cafe will meet
A B in B, making A B twice a b, and the L A B C = La bc, if it is
truly drawn.
*
PROBLEM
XIX.
Upon any given line A B to make a fquare ABCD. Fig.17.
1. By problem 5, upon B one end of the given line, raiſe a LB D,
making it equal to A B, 2. with AB in your compaffes and one foot
ſeverally on D and A, ſtrike two arches crofling each other in C, 3.
join CD and CA, and its done.
PROBLEM
XX.
Given two lines A B and B C, to make a right angled parallelo-
gram, or rectangle ABCD. Fig. 18.
1. By problem 5, upon A raiſe the LA D, making it —B C, 2. with
A B in your compaffes and one foot on D, fweep an arch, 3. with B C
in your compaſſes and one foot on B, croſs the laſt arch in C, join D C
and B C, and its done.
t
AND MECHANIC.
7
NOTE. If the dimenfions of figures be given in numbers, they may be
made by a ſcale of equal parts, or a diagonal fcale, the fame way as if
the dimenfions were given in lines; as in fome of the following figures.
PROBLEM XXI.
To divide a given line AB into any number of equal parts, ſup-
pofe into five. Fig. 19.
1. At each end of the given line, by problem 7, make an L, viz. the L
BA5=LAB5, 2. take any ſmall diſtance in your compaffes, and run
it along the line A 5, dividing it into 5 = parts, run the fame diſtance
from Bto 5, dividing B 5 into 5 = parts, 3. thro' theſe divifions draw
lines, as 1-4, 2-3, &c. and they will divide A B as required. See
theorem I.
PROBLEM
XXII.
How to make a ſcale of equal parts, as AB. Fig. 20.
1. Having drawn a line A B, take any diſtance in your compaffes and
run it along the line 11 times, as from A to o, from o to 10, &c. to
100, 2. divide the diſtance A o into 10-parts, which ferves the whole
fcale for units; ſo that e 20 is 25, the diſtance e 70 is 75, &c.
PROBLEM
XXIII.
How to confiruct and uſe a diagonal ſcale. Fig. 22.
1. A diagonal fcale is divided into 10 large equal divifions, as the
diſtance e q. or e 100, is one of theſe divifions, and theſe fignify hun.
dreds, the diviſions 10, 20, 30, &c. between e and 100, along the fide
of the end divifion, denote tens, and thofe on the end, as 1, 2, 3, &c.
fignify units, being all equal divifions. Now to find any number upon
this ſcale, ſuppoſe I chain 43 links, or 143 of any equal parts, as inches,
feet, yards, &c. firft fet one foot on x which ſtands over againſt 40,
and under 3, is the place of 43, and extend the other foot along that
line to y, fo have you 143 in your compaffes; in like manner, the dif-
tance w u is 185, tu 85, &c. which will become eafy with practice.
THE UNIVERSAL MEASURER
PROBLEM
XXIV.
To make a rhombus ABCD, whofe height B P may be 20, and
each fide A B = BD, &c. 30. Fig. 23.
AB
1. From the ſcale of equal parts take 30, and lay it from A to B.
2. Make AB radius, and with one foot on A, fweep the arch C, and
on B fweep the arch D, from the fame ſcale take 20, with which, by
problem 1, in any two different points on A B, fweep two arches over
whofe tops draw CD, cutting the first two arches in C and D. 3. Join
BD and CA, and its done.
PROBLEM
XXV.
To make a rhomboides AB CD, whofe length A BCD may
be 143, fide or end AD-BC85, and height DP 80. Fig. 21.
1. From the diagonal fcale, with the length 143 and height 80, by
problem 20, make a rectangle DCE P, producing EP towards A. 2.
From the fame fcale, with the breadth 85, and one foot on C and D
ſeverally, the other will croſs EP in A and B, join CB and D A, and
its done.
PROBLEM XXVI.
To make a rhomboides A B CD equal to a rectangle DCEP,
whoſe acute angles ſhall be equal to a given angle z. Fig. 21.
1. Make A B P E, and upon A and B, by problem 7, make the
Ls EB C, and P A D, each the given angle z, drawing A D and BC,
till they are each the given breadth 85. 2. Join D C and its done.
But if the height 80 were given, and not the breadth 85, you muſt con-
tinue A D and B C till they meet D C, drawn parallel to A B at the
diſtance of So from it. By problem 1. For demonftration fee the-
orem 1, and 2.
In fome of theſe figures you'll find more lines, arches and letters than
is mentioned in the reading for the problem or figure, fuch figures ferve
more problems than one: thefe lines, &c. need not be drawn or regard-
ed, till you come to the problem that refers to them: but I would ad-
viſe a learner that is defirous to underſtand geometry, to draw a figurė
for each problem.
AND MECHANIC.
PROBLEM XXVII.
To make any regular polygon, fuppofe a pentagon. Fig. 24.
I.
Deſcribe a circle and croſs it with a diameter A 5, which divide
into as many equal parts as the polygon hath fides, fo a pentagon having
five fides, the diameter muſt be divided into 5 equal parts. 2. With the
whole diameter A 5 in your compaffes, and one foot on A and 5 feverally,
ſweep two arches croffing each other in D. 3. Lay a ruler to D and the fe-
cond divifion on A 5 and draw a line DE, cutting the circle in E. 4 Join
A F., which will be a fide of the required pentagon, and will divide the
periphery of the circle into 5 equal parts as required,
PROBLEM
XXVIII.
Upon a given line CD to make any regular polygon, fuppoſe a
pentagon. Fig. 25.
1. As directed in the laſt problem, find A E, the fide of a pentagon of
any fize, then upon each end C and D of the given line C D, makes an L
to the LEA 5, in figure 24, and at Q_where the lines meet, is the
center of a circle, which being deſcribed with the radius QDQC,
the given fide CD will divide it into 5 equal parts as required.
PROBLEM XXIX.
To make an oval or ellipfis, TISG. Fig. 26,
1. Deſcribe two equal circles, paffing thro' each others centers B and
C, and croffing each other in a and e. 2. With the diameter S Cor BT
of either circle, and one foot on e fweep the arch I, alfo with one foot
on a, fweep the arch G, and its done.
PROBLEM XXX.
Another way to make an oval or ellipfis. Fig. 26,
1. Upon any plane where you would make an oval, ſtrike in two pins,
ſuppoſe at B and C, and to theſe pins faſten the ends of a thread, as B
PC. 2. With a pin extend the thread as far as poffible, and by moving
the pin about, you may deſcribe the periphery of an ellipfis,
B
TO
THE UNIVERSAL MEASURER
XXXI.
PROBLEM
Given the axis eu and ordinate Ą B, of a common parabola A u B,
to draw the parabola. Fig. 27.
1. A B being to u e, produce u e towards C. 2. Upon u e defcribe a
circle u A CB, to pass thro' the three points u, A, B, cutting ue pro-
duced) in C. 3. Upon ue, defc:ibe as many more circles as you pleafe,
each paffing thro' u, and cutting the axis u e in a, b, d, &c. 4. Take Cẹ
and lay it from A ton, from b to m, from d to o, &c. 5. Thro' theſe
points n, m, o, &c. draw lines or chords each parallel to AB, and where
thefe chords cut their respective circles, are the points thro' whichthe
parabolic curve muſt paſs, and may be drawn with a iteady hand. See
theorem 60.
PROBLEM
XXXII.
Given the tranfverfe axis u T of an hyperbola A B u, and u K =
TH the distance of the focus from the ends u and T of the axis,
to draw the hyperbola. Fig. 28.
น
1. Produce u K any length as to C for the abfciffa. 2. Divideu Cinto
as many parts (or not) as you pleaſe, as in the points m, n, a, &c.
3. With the radii Tm, Tn, Ta, &c. and one foot on H, ſtrike the
arches ee, ee, ee, &c. 4. With the radii u m, un, ua, &c. and one
foot on K, cross the former arches, with the arches 00, 00, &c. 5. Thro'
Note. The more inter-
thefe interfections draw the curve A C B.
fections you make uſe of the better, in both theſe figures, as they will
not only help you in the drawing, but will alfo render them more exact,
See theorem 62,
?
PROBLEM
XXXIII.
Upon a given line A С to make a gothic arch ATC. Fig. 29,
1. Divide the baſe or ordinate A C, into 3, 4, or 5 = parts, fuppofe
into 3 parts, CD DEEA, then with A D two of thefe parts,
and one foot in D and E feverally, fweep two arches A B and G C,
crofling each other in T, and its done; ATC being a gothic arch of
the third point; becauſe A C was divided into 3 parts, but if
divide A Cinto 4 or 5, &c. parts, the arch is faid to be of the 4th,
$th, &c. points,
you
AND MECHANIC.
PROBLEM
XXXIV.
Upon a given line or baſe AC, to make a low arch. Fig. 30.
1. Upon A C by problem 13, make an equilateral triangle AC F;
2. Upon F, with the radius F AF C, fweep the arch A E C; now if
the LFB be produced, it will bifect the arch in E, whereof E B being
about of AC, will give the height or thickness of the freight arch for
fquaring or cutting the ftones, &c. and by the figure may be ſeen the
different forms of the itones, &c.
PROBLEM XXXV:
To defcribe a Catenaria TIS whofe height CI may be 55 feets
and its breadth TS 200 feet. Fig. 31.
1. The geometrical conftruation of this curve being tedious, it will
be eaſieſt done thus; on an upright wall or plane lay the diſtance 200;
from any fcale of equal parts, parallel to the horizon, as from T to S,
there ſtrike in two pins, upon which fufpend a flexible line, (a chain of
very fmall links is beft) as TGS, till its middle at G be 55 of the fame
= parts below T S, fo will T G S form the catenaria required.
PROBLÈM XXXVI.
To take the plan of any curve, arch, bended hedge, &t. end lay it
down on paper. Fig. 31.
1. Meaſure ftreight from S one end of the curve SG T, towards T
the other end, fetting down the diftances Sa, Sa, &c. where you meas
fure perpendicularly from S T to the curve, fet down alſo theſe Ls a e,
ae. &c. called offsets, and always be careful to meaſure the offsets
from the ranging line TS to the extremities of the faid curve, which
done you may have a tree plan by drawing a line T S, upon which
from a fcale of parts, lay down all the diſtances S a, Sa, &c. upon
which points erect the Ls a n, an, &c. and from the fame fcale lay on
thefe offsets their reſpective diſtances, as from a to n, &c. thro' thefe
points draw the curve.
THE UNIVERSAL MEASURER
PROBLEM
XXXVII.
The arches uſed in architecture, are as follow;
1. Semi-circular arches, being half-circles. 2. Scheme arches are
leſs than a femi-circle, and commonly contain 70 or 90°. 3. Elliptical
atches, are fuch as confift of a femi-ellipfis, formerly uſed over chim
heys for mantle trees. 4 Gothic arches. 5. Streight arches, are fome-
times over windows, doors, galleries, cielings, &c. becauſe they take up
but little room. 6. Theſe are the arches which were formerly uſed in
vaults, bridges, and other buildings, amongſt which the femi-circular
arch was moſt eſteemed both for beauty and ſtrength. But the ſtrongeſt
arch poſſible is the catenaria; whoſe ſtrength is fuch, that a number of
globes, or a heavy flexible line, being put into that form, will not fall,
but fupport themſelves; this appears from the formation of that curve,
(Fig. 31) thoſe above the line TS, muſt ſupport with the ſame force,
as they gravitate below it.
PROBLEM
XXXVIII.
To find the puſh or butment of an arch AE M. Fig. 32.
1. Let E be the middle of the underfide of the bridge, &c A E M.
2. Join A E, and let A D be L to the horizon A M. 3.Upon A, with the
radius A E, draw the quadrant B E D, ſo is A B = AE = AD. 4. Join
DB, which cuts A Ein I. 5. From I upon A M let fall the LIL. Laſt-
ly, make A C— A L, then A C will be the length of the butment re-
quired, fit to hold the pufh of the arch A E, being fufpended upon the
point A as a center. For if A E, (the chord of the arch A E) become
A B parallel to the horizon, it is plain, it muſt have its own length in
the fame parallel on the other fide of A to ballance it with, which fup-
pofe 20; but when the fame A F is raiſed into A D to the horizon,
upon the fixed point A, it will then require nothing to ballance it with.
Whence it follows, that if 20 ballance A E when in a parallel pofition,
and o when in a Lone, that half the fum of 20 and o, will ballance it
when in a mean between A B and A D, i. e. when it makes an Lwith
the horizon of 45°.
PROBLEM XXXIX.
Given two lines BA and BC, to find fuch a third line A C, as
that the triangle ABC may be the greatest poffible. Fig. 33.
1
AND MECHANIC.
13
1. With the two given lines make a right LC B A, and join the ends
Ĉ and A, ſo is C A the line fought, and A.BC the A. For if BD-BC,
the As A B C and ABD, ſtand upon the fame baſe A B, and therefore
are (by theorem 33) as their heights. It is plain by the figure A B C,
that BC has the greateſt height, and confequently it is the greateft tri.
angle poffible. By the fame method it may be proved, that if any num-
ber of lines be given to find another line fo as to make the greateſt fi-
gure poffible, the line required muſt be the diameter of a circle, and
the given lines fo connected, as to be inſcribed in the half periphery.
PROBLEM XL.
Given the hypothenufe AB of a right angled A, to find the legs,
fo that the A may be the greateft poffible. Fig. 34.
1. Upon the given hypothenuſe A B, ſweep a femi-circle whofe mid-
dle is at C, join C A and C B and its done, A B C being the triangle
fought, and A CBC the two legs. For, as in the last problem the
▲ A B C, is greater than the ▲ A B D.
:
Note All right-lined figures infcribed in circles, are the greateſt
that can poffibly be made for a circle being the greateſt figure under
the ſame periphery, it follows, that the nearer any figure approaches
to it, the greater it will be.
***********************************00
The RULES of PROPORTION,
Extraction of the Square and Cube Roots, Geometrically:
IN TWELVE PROBLEMS.
PROBLEM XLI.
To divide a line A B into fuch proportion, as the line Cu is to the
line Du. Fig. 35.
1. Upon either end of the given line A B, as on A, make an L by
drawing A w. 2. Make AFC u, by taking Cu in your compaffes
' and laying it from A to F; alſo, lay Du from F to w, join w B, and
parallel to w B, thro' F, by problem 2, draw F G and its done. For
AF is to F w, as Cu is to Du. And if thefe lines be laid down from
a diagonal ſcale, as marked in the figure, you'll find G B = 58,7 and
AG= 134,3 from the fame fcale. See theorem 9.
14
THE UNIVERSAL MEASURER
PROBLEM XLII.
Given two lines A B and Du, to find a third line in proportion
to them. Fig. 36.
1. Make an angle DA C, lay Du from A to E, aud A B from E to
D, join E B. 2. Thro' D, and parallel to E B, by problem 2, draw DC,
or a parallel may be drawn by making Ls thus, upon E fweep an
arch to cut the two containing fides E A and E B of the LA EB, with
the ſame radius defcribe an arch upon D, take the arch deſcribed upon
E, and lay it upon that defcribed on D, from the line E 1), and where
it cuts draw the line DC, which (per theorem) will be parallel to E B,
whence you'll have B C for the line required. For as A E (Du70)
is to AB 124, fo is ED (AB 124) to BC 219,6. See theorem,
and problem 45.
PROBLEM
XLIII.
Given three lines, AB, Cu, and Du, to find a fourth propor-
tional. Fig. 37.
1. With any two of the given lines make an L BA E, making A B
← A B, A E ≈ Du, and ED Ca, join B E, and thro' D paralley
to BE (by the laſt problem, or by problem 7) draw D C, cutting A B
produced in C; fo is B C the fourth proportional line required. For,
(by theorem 9,) as A E (D u) is to A B, fo is ED (Cu) to B C.
PROBLEM XLIV.
If 180 Labourers do a piece of Work in 115 days, in what timè
will 106 labourers do it. Fig. 37.
1. Make any L CAD, and from a diagonal ſcale lay 106 from A tơ
B, 115 from A to E, and 180 from E to D, join E B, and thro' D,
parallel to EB, draw DC cutting A B C in C, fo is C B = 186, from
the fame ſcale, as required.
By theſe two laſt problems you may work any proportion by lines
or fcales of equal parts, whether direct or inverſe.
PROBLEM XLV.
Between two lines ED 36, and CD 100, to find a mean propor
tional. Fig. 38.
AND MECHANIC.
1. Having made one right line E C of the two given ones DE and
D C, upon È C as a diameter, deſcribe a femicircle EP C. 2. Upon
E
D, where the two lines meet, raiſe the LDP, cutting the arch in P,
fo is DP the mean geometrical proportion required. For as CD: DP
:: DP: DE, by theorem 14: whence CDXDE DP 3600,
fo DP 60, as will alſo be found if you lay DP on the fame ſcale of
equal parts that you took DE and DC from.
Note. By this figure you may alfo work problem 42.
PROBLEM
XLVI.
To extract the fquare root of any number, ſuppoſe 3600, by lines,
or a dia ona! fiule.
RULE. A geometrical mean proportional between any
two lines or numbers, is the ſquare root of their product.
EXAMPLE. 100 x 36 3600, and a mean proportional between
100 and 36 is by the laſt problem found 60, whence 60 is the fquare
rout of 3600 Alfo, 10x 360 = 3600, and a mean proportion between
10 and 360, will alfo be found 60. Likewife 40 × 90 = 3600, and
30 X 120 = 3600, and a mean proportional between 40 and 90, or be-
tween 30 and 120, will alfo be found 60, the fquare root of
3600,
PROBLEM XLVII.
Between two given lines A and B, to find two mean proportional
lines. Fig. 39.
1. Make a right LK CH, drawing the fides C H and CK at large.
2. Lay the line B from C to E, and the line A from C to D, and join
ED, find its middle, which is at F, then with F EFD, upon F
defcribe the femicircle EGD. 3. With the leffer line B in your com-
paffes and one foot on D, crofs the femicircle in G. 4. Upon the point
G, move a ruler till it cut the lines CH and C M in H and K, at an
equal diſtance from F, fo fhall E H and DK be the lines fought. For
as EC (B): DK:: DK to EH, and as E C: DK::EH:CD,
(A), 13. E. 6,
16
THE UNIVERSAL MEASURER
PROBLEM XLVIII,
To extract the cube root geometrically. Fig. 39.
RULE. The cube root of any number is the firſt of two
mean proportionals between unity and that number. Or,
which is the fame, and eaſier to do, Take any number which
is less than the cube root of the given number, fquare this
number, and divide the given number thereby, then the
firft of two mean proportionals, between this quotient and
the aboveſaid taken number, will be the cube root of the
given number.
1
EXAMPLE. What's the cube root of 67584? Take 32, which
fquared is 1024, then 67584 102466; fo the firſt of two mean
proportionals between 32 and 66, will be the cube root of 67584;
therefore, if from the diagonal ſcale you make CE=32, and CD=66,
and work as in the laft problem, you'll have D K 403 nearly, from
the fame fcale, for the cube root of 67584, or the first of two mean
proportionals between 32 and 66. By this, and problem 46, with a
good diagonal ſcale, you may readily find the cube and fquare root of
a number.
PROBLEM
XLIX.
To find out two lines E F and F C, which ſhall have fuch propor.
tion to each other, as the fquare of a given line A hath to the
Square of another given line B. Fig. 40.
1. Make a right LED C, (i. e. CDL to DE) then lay the line
B from D to C, and A from D to E, join C E. 2. From Dupon CE,
let fall the L D F, and its done. For, by theorem 15, as □ A : —B,
::EF: FC.
PROBLEM L..
To divide a line CD in power, as the line A is to the line B.
Fig. 41.
1. By problem 41, divide C D into fuch proportion as A to B, i. e.
as B: A::CE: ED. 2. Upon CD, as a diameter, deſcribe the femi-
circle CFD. 3. Upon E raiſe the LEF, cutting the femicircle in F
AND MECHANIC.
27
3. Join F C and FD, which are the two lines required.
orem 16, as B: A::CFDF, and as A+ B:
:CF, and :: A: FD.
For, by theo-
CD :: B
PROBLEM
LI.
To enlarge any line C E in power, according to any proportion ;
fuppofe as the line A to the line B. Fig. 41.
}
1. By problem 41, make it as A: B::CE: CD; i. e. make C B
= A, and C A = B, join E B, and thro' A parallel to E B, draw AD,
cutting CE produced in D. 2. Upon CD defcribe the femicircle C F D.
3. Upon E erect the EF, cutting the femicircle in F, join CF for
the line required. For, by theorem 15, as A: B :: O. CE: O
CF.
PROBLEM
LII.
To cut a line AD in extreme and mean proportion. Fig. 42.
1. Upon A raiſe the L AF, making it AD. 2. Biſect A F in G
and join G D. 3. produce F A, making GIGD. 4. Make A C—
A I, and its done. For as AD: AC:: AC: CD. 30. E. 6.
DEMONSTRATION. Put a AD AF, then A Gaby con-
ftruction, and put e AIA C, then a e-CD, and the laft
proportion will be, a:e::eae, therefore, ee-aa-ae, or,
a a ➡ee+ a e, which quadratic equation folved (by article 81, part ·
2.) gives eaaaax:
2
i
: now,
GD=Q
AG+AD=‡a a + aa — a a, and G D - GA — A C—e,
i.e.aa.ae as before; fo the figure is truly contructed.
Note. Becauſe the fquare root of 5 cannot be exactly had, it fhews
that no number can be exactly divided in extreme and mean proportion,
but any line may be fo divided geometrically, as above.
( 18 )
The chief PROBLEMS belonging to the CIRCLE
are fifteen, as follow:
PROBLEM
LIII.
Given any Circle A B C D, to find E its center. Fig. 43.
1. Any where in the circle draw a chord O O, which bifect with
the line BD. 2. Bifect B D with the line A C, fo is E, the croffing
of thefe two lines, the center of the circle; and the lines themſelves are
two diameters, at right angles to each other.
PROBLEM LIV.
Given any arch of a circle G C, to find its tangent DC, fine GB,
fecant A D, &c. Fig. 44.
Thro'
1. Make A EL to A C, fo is the arch E G Ca quadrant. 2.
A the center of the given arch, and over G its top, draw the line AD.
3. Draw CDL to A C, or parallel to A E; alſo GB parallel to A E,
and EF and G1 both parallel to A C, fo is CD the tangent, DA the
fecant, G B the fine, GIA B the co-fine, B C the verfed fine, IE the
co-verfed fine, A F the co-fecant, and EF the co-tangent of the arch
GC, each of which may be meaſured on the diagonal ſcale that A C was
taken from. In like manner, becauſe one of theſe arches E G and GC is
the complement of the other, therefore, E F is the tangent, and CD the
co-tangent of the arch EG, &c.
PROBLEM LV.
To defcribe a circle thro' any three points, as the three angular
points of a triangle ABC. Fig. 45. Or, to find a point D
equally diftant from the three points A B C, not lying in a freight
line.
1. By problem 3, bifect the diſtance between any two of the three
points, as A and 3; again bifect the diſtance between the third point and
either of theſe, as B and C, and at D where theſe two bifecting lines meet,
AND MECHANIC.
19
is the center of the circle required, which fweep with the radius DA, and
its done.
PROBLEM LVI.
Given any arch of a circle as A B C, to find the whole circle
Fig. 45.
1. Take any three points in the given fegment, as AB C, and by the
laſt problem find D, the center of a circle to pafs thro' theſe three points.
2. With the radius DA DB-D C, defcribe the circle.
If the chord A C, and verfed fine or height D B (fig. 46.) were given,
to find the diameter of the whole circle, it is the fame, for the circle re-
quired muſt paſs thro' the three points A B C.
To do this arithmetically.Let AC120, and D B 36, then,
as E D 36: DP 60 :: D P 60: DC (fig. 38.) 100, which added to
ED 36 gives 136 E C, for the whole diameter.
PROBLEM LVII.
Given D B the diameter, and IB the verfed fine of a fegment
O BO, to find ⱭI half the chord of that ſegment. Fig. 43.
This is but to find a mean proportional between DI and B I,
2s is done by problem 45. Theſe two problems are of great ufe in
meaſuring ſegments of circles, globes or ſpheres.
PROBLEM LVIII.
To find the length of any circular arch as A B C. Fig. 46.
1. Divide the chord A Cinto four parts, and fet one of thefe parts
on the arch from C to G. 2. From G to E the third of theſe parts,
draw G E, which is half the length A B C nearly.
PROBLEM LIX.
To find the length of the abovefaid arch A BC nearly by arithme-
tis. Fig. 46.
1. Multiply A B the chord of half the fegment by 8, and from that
product take A C the chord of the whole fegment. 2. One third of the
remainder is nearly equal to the length of the arch.
THE UNIVERSAL MEASURER
60 A B.
8
}
480
100 = A C.
3)380
126.66 the anſwer.
Note. This method gives the length of the arch ſomewhat too little,
and the greater the arch the greater the error. There can be no rule
given for finding the length of an arch exactly, till fòme method be found
for fquaring the circle truly; yet this way of finding an arch, will ferve
in common practice.
If you know what degrees are contained in any fegment's arch, you
may find the length of the arch truly by this
RULE. As the periphery of any circle in degrees is to its periphery in
equal parts, fo is any arch in degrees and decimal parts of a degree, to
the fame arch in equal parts.
EXAMPLE. Suppofe the circumference of a circle to be 71, and the
arch to contain 52° 15′, or 52.250, what is the length of that arch?
As 3600: 71 :: 52.250: 10.304 the anſwer.
PROBLEM LX.
From any given circle AO FC, to take away a ſegment C F
O A, containing an LCF A to a given L z. Fig. 47.
1. Draw the line B ALL to the diameter A F. 2. Make the LC
AB the given L z. drawing A C till it cut the circle in C, and it is
done. For the LC A B taken from 90° leaves the L F A C, and F A C
taken from 90º leaves the L A F C, (becauſe the LA CF = 90°) con-
ſequently the L CFA LCA B.
—
PROBLEM LXI.
Upon a given line BF, to make a fegment of a circle FLOB, in
which the L F L B ſhall be to a given L ABF. Fig. 48.
1. Upon the angular point B, erect the 1 B L. 2. Upon F make the
LBFI the compliment of the given L, viz. L F B L, cutting BL
in I, fo is I the center of the fegment required, which deſcribe, join
FL, and it is done,
*
AND MECHANIC.
23
PROBLEM LXII.
Given any circle A to find how many circles of equal magnitude will
touch it and one another. Fig. 49.
1. Becauſe the three fides of an equilateral ▲ are equal, if with half
of any of thefe fides in your compaffes, and one foot on each angular
point, you ſweep a circle, it is plain theſe three circles will touch one
another, as the circles A B C, deſcribed upon the angular points of the
equilateral ▲ A B C; now, becaufe any circle may be divided into fix
equilateral As, therefore divide the periphery of the given circle A into
fix parts. 2. Thro' A its center, and each of thefe parts, draw lines
A B, A C, &c. each twice its radius beyond its periphery, which will
be the diameters of the circles required, in number fix.
PROBLEM LXIII.
To find D E the radius of a circle = to the fum of three or more
given circles, whofe femi-diameters are AB, A C, and A D.
Fig. 50.
1. Make a right LDA E. 2. Lay A B from A to B, and A C,
from A to C, join B C, fo is C B the femi-diameter of a circle equal to
the two given ones AB and A C.
from A to D, join E D and its done.
3. Lay B C from A to E, and AD
See theorem 13.
PROBLEM
LXIV.
To find AB the femidiameter of a circle, to the difference between
two given circles, whofe femidiameters are AC and C B.
Fig. 50.
1. Make a right LEA D, in which make A C—A C. 2. With BC
in your compaffes and one foot in C, cross the LA E in B; fo. is BA
the femi-diameter required. See theorem 13.
PROBLEM LXV.
To draw a tangent T P, to any aſſigned point P, in a given curve
APD. Fig. 51.
1. A tangent to any point in a circle, is at right angles with a diameter,
from that point; therefore find C the center of a circle (upon AB the
22
THE UNIVERSAL MEASURER
axis of the given curve) to touch the given point P. 2. Join P C, and L
to PC draw P T, which is the tangent to the point P, both in the curve
and circle.
PROBLEM LXVI.
To draw a line CP perpendicular to any given curve or arch, A
PD. Fig. 51.
"
This is already done in the laſt problem. For CP is to the tan
gent T P, and confequently to the curve in the point P.
PROBLEM LXVII.
To make a ſpiral upon a given line E e.
Fig. 52.
1. Divide the given line Ee into parts ed, dc, cb, bB, BC, &c.
two of which neareſt the center, bifect in the point A. 2. Upon A, with
the radiufes AB, A C, AD, &c. defcribe the feveral femicircles above
the line e E. 3. With the radiufes g b, gc, gd, &c. and one foot in
g, ſweep the femicircles below the line e E, and it is done.
CIRCUMSCRIPTION and INSCRIPTION
of FIGURES, in twelve Problems.
DEFINITION. A figure is faid to be inſcribed in
another figure, when all the angles of the former are in the
periphery of the latter. Or, to inſcribe a figure A within a
figure B, is to cut the greateſt figure A that can be out of
the figure B.
PROBLEM
LXVIII.
In a given circle A EBD to infcribe a ſquare. Fig. 53.
1. Draw the diameter D E, and bifect it with the diameter A B.
2. Join the points A E, E B, BD, and D A, with right lines, and it is
done.
Note. By this method you may make a fquare,
AND MECHANIC.
23
PROBLEM LXIX.
About a given circle to circumfcribe a Square, a b d e. Fig. 53.
I. Crofs the circle with two diameters DE and AB. 2. Take the
radius D C CA, &c. of the given circle, and one foot in A, D, B, E,
feverally, defcribe the pricked arches, croffing in the points a, b, d, e,
which points join with right lines, and its done.
PROBLEM LXX.
i
a, and
Out of a given circle to cut the greatest rhomboides poffible, whofe
length may be to its breadth, as the line b is to the line
each of its acute Ls to a given L z. Fig. 54.
1. Draw the diameter A C, and produce it if neceffary. 2. Lay the
line a from the center D to C, and make the LECP - the given L
z, making CP to the line b. 3. Draw DP cutting the circle in u. 4.
Draw u z parallel to CP. 5. Produce u z till z nz u. 6. Make Dy
- Dz, and with zu in your compaffes and one foot in y, croſs the circle
in a. 7. Join Y a, and produce it till ye ya, join a n and eu, and it
is done. For the As Dzu and DCP are alike; therefore as a : b
DC: CP:: Dz:zu::an: au, by taking an = z y = 2 Dz,
and au 2 z u.
PROBLEM
LXXI.
Out of a given circle to cut the greatest rectangle poſſible, whoſe
length may be to its breadth as 3 to 2, or 30 to 20. Fig. 55.
1. Draw the diameter A B, produced if neceffary. 2. Take 30 from
a fcale of parts and lay it from the center C to a. 3. Upon a, erect
the Lan, making it 20 from the fame fcale. 4. Draw Cn cutting the
circle in b. 5. Make A e, Bq, and B feach A b, and join fq, qe, eb,
and bf, and its done. For e b being parallel to n a, the As n Ca, and
b Co are alike; and therefore, as in the laft problem, as 20: 30: ǹ à
:aC::bo:o C::eb:e
:eq.
PROBLEM
LXXII.
Within a triangle A B C, to infcribe a circle. Fig. 56.
1. By problem 8, bifect any two of the Ls, fuppofe the Ls B and C,
and where the biſecting lines meet, as at D, is the center of the circle.
24
THE UNIVERSAL MEASURER
PROBLEM
LXXIII.
Given DR the radius of a circle, or the circle itfelf L G F, to be
cut into a triangle, whofe fides may be in a given ratio, as the
lines a, b, c. Fig. 57.
1. With the three given lines a, b, c, or any other three lines in the
fame ratio, make a ▲, a b d; and by problem 55, find E the center of
its circumfcribing circle. 2. Upon E, with the given radius DR, defcribe
the circle, and thro' the angular points a, b, d, draw the femidiameters
Ea, Eb, and E d, which join with right lines, and you'll have F L G
for the A required.
Note. By this method, if any figure be inſcribed in a circle, you may
infcribe another fimilar figure in any other circle, whether greater or
lefs, by defcribing one circle upon the center of another, and drawing
femi-diameters thro' every L, &c.
PROBLEM LXXIV.
To infcribe in a circle FLG, a AFG L, having = Ls with a
given A, a, b, d. Fig. 57.
1. Draw the line GA touching the circle in any point G. 2. Upon G,
make the LF GA – Lab d, or any 4 of the given ▲ drawing G F till it
L
cut the circle in F. 2. Upon F make the LG F L L bad, or any
other of the given Ls; join G L, and G L F is the A required. 2. E. 4.
PROBLEM
LXXV.
To circumfcribe about a given circle F G H, a AL.m n, having
Ls with thofe in a given A A B C. Fig. 58.
1. Extend the fide A C, making the two external Ls DCB and
E A B. 2. At E the center of the circle, make the LF EHLDCB,
and the LHE GLEA B. 3. Join thefe points F, G, H, with right
lines to touch the circle, and you'll have Lm n for the triangle required.
3. E. 4. Theſe two problems may be done alfo by problem 73.
PROBLEM LXXVI.
To inſcribe and circumfcribe a circle about any regular figure, Sup→
pofe a hexagon. Fig. 59.
AND MECHANIC.
25
1
1. Bifect any two fides EE and ED, of the given polygon, and at C,
where theſe biſection lines meet, is the center of the polygon. 2. From
C draw the line CE, fo is Cd the radius of the infcribed circle, and CE
that of the circumfcribing circle. 41. E. 4.
PROBLEM LXXVII.
To inſcribe and circumfcribe about any circle any regülar figure,
ſuppoſe a_hexagon. Fig. 59.
1. By problem 27, in the given circle infcribe the figure. 2. Bifect
the fides of this infcribed figure with the lines Cd, Cd, &c. and where
theſe bifecting lines touch the circle's periphery as in e, e, &c. draw
lines parallel to EE, ED, &c, which will cut one another; and form
the figure required.
PROBLEM LXXVIII:
In a given circle HIK, to deſcribe any number of equal circles,
Suppose three. Fig. 60.
1. Divide the given circle's periphery into as many parts as are the
number of inſcribed circles, H, I, K, and from the center C, draw lines
CH, CI, &c. to each of theſe parts. 2. Draw a line I K between any
two of thefe parts. 3. On C K produced, lay K LIK. 4. Par
allel to IL (having joined I L) draw È F, cutting C I in F, fo is FÈ
the radius of one of the circles required, = EHGK.
EHGK. Otherwife,
by problem 73, make a regular polygon of as many fides as you want
circles, upon each L of this polygon fweep an circle to touch each o-
ther, and the center of the polygon will be the center of the circumfcris
bed circle; then it will be, as the radius of this circle is to the radius of
one of its infcribed ones, fo is the given circle's radius, to the radius of
one of the required ones.
PROBLEM LXXIX.
About a ABCD, to deſcribe a ▲ cf g, having – Ls with
a given AEF G. Fig. 61.
1. Upon any fide A B of the given, make an L A Bf-LG,
and the L BALE, continuing f B and f A till they meet CD
produced in g and e, and it is done; for A B is parallel to C D; the L
J = LfBAG, and the Le = LfAB¤ LE, &c.
D
26-
THE UNIVERSAL MEASURER
1
PROBLEM LXXX.
About any ▲ A B C to deſcribe a□. Fig. 62.
1. Produce A B, or any fide of the given A, and from the third an-
gular point C, (by problem 4) let fall the CD, making DE — AD:
2. With the radius A D or D E and one foot on A and E feverally, de-
fcribe two arches croffing each other in F, join E F and A F, fo is
ADEF the
required.
PROBLEM LXXXI.
Within any ▲ EPF, to infcribe a□LLSK. Fig. 63.
1. By problem 6, let fall the L P O. 2. Parallel to P O draw F Q¿
making it the baſe EF. 3. Join OQ cutting P F in S, from S draw
S K parallel to PO, and S L parallel to E F, alfo, L L parallel to PO,
and it is done. For by fimilar As, as FO: EF (FQ) :: KO (eS)
:SK, and, FO:FE:: eS: LS, therefore, K SSL, confequently,
KL LS (being a rectangle) is a .
PROBLEM LXXXII.
To augment a☐ A B C D to any proportion, ſuppoſe to an
octuple proportion. Fig. 64.
1. Produce A B and A D, any two fides of the given, as alfo the
diagonal A C. 2. Make A L and A H each AC, and complete the
AHEL, which is double the given A B C D. 3. Make A M and
A I each = A E, and complete the AIF M, which is double the □
AHEL, or 4 times ABC D. 4. Make A K and A N each AF,
and complete the AKG N, and it is done. For, by theorem 13, the
=
AK G N is 8 times the ABC D. To diminiſh ais but juſt the
converfe work, as is evident by the figure. What is faid here offquares,
will alſo ſerve in all like figures, as is plain by theorem 36.
PROBLEM LXXXIII.
Within any ▲ A B C, to infcribe a rectangle, whofe length may
be to its breadth as Eto F. Fig. 65.
1. By problem 43, make it as E:CA:: F: CF, then make CQ
CF and parallel to the L BO. 2. Join OQ cutting C Bin a, draw
a d and be, each parallel to BO,,and join a b, fo is a b c d the rec-
AND MECHANIC.
27
tangle required; for, (by fimilar triangles, as in the laſt problem) as
OC: Od::CF (CQ): da, and as OC: Od (a e) :: AC: ba,
fo by equality of ratios, CF: AC:: da:ba; but by conftruction,
AC::da:
as E: F:: AC: CF, therefore, as E: F::ba: da. Q. E. D.
PROBLEM LXXXIV.
Required the greatest rectangle that can be cut out of any given ▲
A B C. Fig. 65.
I
. Bifect B O the L with the line a b. 2. Parallel to B O draw ad
and bc, ſo is a b c d the rectangle required. By theorem 142.
PROBLEM
LXXXV.
Within any regular polygon Aqnom, to infcribe an equilateral
A B D. Fig. 66.
I. Thro' A, (or any L of the polygon,) and C the center of the cir-
cumſcribing circle, draw the line A C G. 2. With the radius C A, upon
A, defcribe the arch C E, whofe middle is at F, fo draw A F, cutting the
fide of the polygon in B. 3. With AB in your compaffes, and one foot
on A, croſs q n in D, join A D and D B, and it is done; for join A E which
will be the fide of a hexagon, alſo AC by conftruction, now LCA E,
the L of a hexagon is double the L CA F, that of an equilateral A,
and by conſtruction, arch CF = F E, Q, E, I, or, the LCA E being
=60°, the L CAFLCAD will be
it will.
=
30°, be the polygon what
PROBLEM LXXXVI.
In a given rectangle A BEF; to inſcribe a ▲ A P B, whoſe two
fides A P.and P B may be as a to e.
Fig. 67.
I. By problem 41, make it as a : e: : A C : C B, (i. e. lay a from
A to a, and e from a to e, join e B, and thro' a draw a C parallel to
e B) alſo make it as A C — B C : B C
BC: BC:: AC: Ce. 2. With the ra-
dius Ce defcribe the femicircle C RD, and draw the radius OR to
C D, on which lay A E B F from O to n. 3. Thro' n and parallel
to CD, draw the chord P P, cutting the femicircle in two points P and
P, (which fhews that the problem admits of two different folutions)
join PA and P B, and it is done. See theorem 20.
ཀ
28
THE UNIVERSAL MEASURER
REDUCTION of FIGURES
Of one form to another, keeping their areas or magnitudes
equal. In feven problems,
•
PROBLEM LXXXVII.
To reduce a rectangle A B C D to a CEF G. Fig. 68.
If you confider what two things multiplied together will give the area
of any figure, then a mean geometrical proportion between theſe two
things is the fide of a☐ in area to that figure; fo produce CD a
fide of the given rectangle, till C L AD, or C B its breadth, upon
LD as a diameter, deferibe a femicircle cutting C B produced in G, fa
is CG (by problem 45,) a fide of the required. In this manner
you may reduce any number of rectangles to a , for it is but reducing
every fingle &c. to a, and then adding all thefe is together,
as taught in problem 63.
!
PROBLEM LXXXVIII.
To reduce any trapezia Ą B C D to qAAB C. Fig. 69.
1. Draw the diagonal A C, and parallel thereunto draw B E, cutting
BC produced in E, join A E, and it is done for the As B A C and
EA C both ſtand upon A C, and have each an L in the parallel B E,
and fo (by theorem 31) are equal.
PROBLEM LXXXIX.
From a given point P in any right lined figure A BCDE,
duce it to a APHG. Fig. 70, See the last problem.
to re-
1. Jain B D, and parallel thercunto thro' C draw a e, and join a B,
fo is the CD reduced to the ▲ Ba D, for they both ſtand upon
▷ D, and have each an L in the parallel a e. 2. Join P B and parallel
to it draw a H, fo is the ABP AA PHB, for they both ftand
pon B P, and have each an L in the parallel a H; whence the AP
HB is the ſpace O Da P. 3. Join A P, and draw E G parallel to
AND MECHANIC,
29
AP, then join E G, fo is the APG A=APEA, for they both ſtand
upon A P, and have each an L in the parallel E G, fo P H G is the ▲
required.
PROBLEM XC.
Given any number of different As, ſuppoſe two AB,C and C D E,
to make one ▲NFO-to them all, whofe or height Fm fall
be to a given line a b. Fig. 71.
1. On the fame line N E, join A C and C E, or any two fides of the
given As. 2. By problem 1, at the diſtance of a b draw G Q parallel
to N E, cutting the fides B A and B C, of the AA BC, in I and F.
3. Join F A, and parallel thereunto draw B N, cutting C A extended
in N, fo is the FNCA ABC, (having joined F N). 4. Produce
C D till it cut F QinQ, join Q E, and parallel thereun to draw DO,
cutting A E in O, then join OQ, fo is the AO QC=AEDC.
5. Join Q N or F O, and it is done. For draw the LS Bn and F m¸
now becauſe B N is parallel to F A, the As NB C and AFC are fimi-
lar (having the fame L at C ;) therefore, as BN: NC:: Fm; A C,
ergo, NC × Fm ACxBn, i. c. ▲ A B C - ANF C; after the
fame manner it may be proved, that the ACDE A COO, alfo,
(by theorem 31.) AFOC.
=
PROBLEM XCI.
s
To find the fum and difference of any right lined figure A B C D E.
Fig. 70. and a given parallelogram ABCD. Fig. 72.
I.
=
By problem 89, the irregular figure A B C D E, is reduced to
the AG PH, whofe bafe G H, bifect with the line QM, and draw
RG parallel to QM. 2. Thro P and parallel to G H, draw R T, fo
is the AG PH, reduced to the rectangle G B QM. 3. Make the L
TMN LB or C of the given A B C D, and parallel to M T
draw GS, fo is the rectangle G R QM=□SGMT, extend G H
on both fides, making M NA B, or C D, and make MUA C or
BD, thro' U draw F Z parallel to G H, draw Z N parallel to U M,
fo is the M UZN the given one A B C D, then (by problem 90,)
draw U G, and parallel thereto draw T L, cutting HG extended in L,
parallel to U M draw L F; laftly, lay A B from L to w and from F to
30
THE UNIVERSAL MEASURER
1
LFUM
GSTM,
LFZN, and the difference is
t, and its done. For, (by problem 90,)
therefore the fum required is the
the
Ft w L; whence, (by problem 8,) you may reduce any irre-
gular plot to a ; and (by problem 90,) any number of As ors to
one A or, and by this problem find the fum of fuchs and s,
and fo perform menfurations geometrically!
PROBLEM
XCII.
Toreduce any plot ABCDE, according to any proportion, as a to b.
Fig. 73.
1. By problem 50, take D E, or any of the fides, and make it as
bab::DE:e E, and draw the diagonals E C, E B, &c.
2. Thro' e and parallel to D C, draw e f, cutting EC in f, parallel to
BC and B A draw f g and g h, and its done. For by theorem 15, as
a: b: A B C D E efgh E.
PROBLEM
XCIII.
To enlarge a plot e f g h E according to any proportion, as b to a.
Fig. 73.
1. By problem 51, enlarge any fide Ee in power as b to a, i, e. make
it as b: ba:: Ee: ED, draw the diagonals Ef, Eg, &c. at large,
2. Thro' D and parallel to e f draw DC, cutting E f in C, thro' C and
parallel to f g draw C 3, cutting E g in B, &c. and its done. For as b
a plot e f g h E: plot A B CD E, by theorem 16.
AND MECHANIC.
3I
DIVISION of FIGURES or PLOTS.
In twenty-fix PROBLEMS.
This depends chiefly on Theorems 14, 15, 16, 31.
PROBLEM XCIV.
To divide a circle EFDG into any proportion, fuppefe as the
line De is to the line e A, by another circle a de b defcribed upon
C the fame center. Fig. 74.
1. Having drawn the diameters DE and F G at right angles to
each other, then (by problem 41.) divide the femi-diameter CD, as
De:eA::Dn: n C, and upon n E as a diameter, defcribe the femi-
circle n a E, cutting F G in a. 2. Upon C with the radius Ca, def-
cribe the circle a d e b, and its done; for, as De: e A: the area
of the ring D d Faeb GE: to the area of the circle a de b; for
Dee a is or proportional to the area of the circle FFDG, and
e A to the circle a de b by conſtruction, fo by theorem 16, as D e +
e A: DC, or □ DE :: e À
:e Ad Cord b, for Ca is a mean
proportional between CD and C n, or — C D divided, as by problem
50, for let a = C D C e, and e = Cn, figure 74, then : a+e:xe
(ECX Cn) ae+ee a C; alfo, let a D e, and e E e,
figure 73, then axe (Dexe E) aeee, and a e + a a =
(Cee+eE) Ea e, whence it is the fame whether you find a
mean proportional between a whole line C D and C n one of its feg-
ments, or divide the faid whole line in power, as by problem 50.
=
=
PROBLEM XCV.
IfE FDG be a round table, whofe diameter ED or F G, is
100 inches, let it be required to divide it into two unequal parts,
OEGO and OFO, as 10 to 1, by a chord line OO. Fig. 74.
1. This is eaſieſt done by the table of arcas of fegments, thus, the
diameter of the circle being 100, its area will be 7854, then as II
32
THE UNIVERSAL MEASURER
(viz. 10+ 1): 1 :: 7854: 714, the area of O F O ; but the greateſt
area in the table of fegments is 0,7854, (viz. the area of a circle whoſe
diameter is unity) the 11th part of which is ,0714, the neareſt to which
in the table under fegment arca, is,071741, and the verfed fine againſt
this,071741 is 0,147; then fay as I: 0,147 :: 100:14,7=Fz, the
height or verſed fine of O F O, which taken from F G 100, leaves G z
8593
the verfed fine of OG O, as required.
PROBLEM
XCVI.
To divide a□A Gfe into any poſſible proportion, as á to b, by
a line drawn from any point, as at P.
Fig. 75.
1. It is plain by the figure, that the line of divifion muſt cut the fides
e A and fG, fo thro'g and n the middles of the other two fides, draw
the line ng, which muft divide the given AG fe into 2 = parts, which,
by problem 41, make as a : b::n L: Lg. 2. Thro' L and the given
point
P draw U L P, and it will be as a : b:: UAEGP: feUP as
required. For if tw be parallel to e for A G, then becauſe as a : b::
n L: LG; and alls or As ſtanding upon the fame or equal bafes e f
A G, are as their heights L g and L n, therefore, as a : b::n L:
LG::AG wt: wfet; and becauſe LtLw, and e A,
gn and fG are all parallel, it follows that ALPwA Lut, confe-
quently the line u P may turn upon L as a center, and will ſtill keep the
tw,
AG fe divided in the fame proportion as long as it cuts the fides e A
and G f, but if it vary fo as to cut a fide G f and an end A G, the pro-
portion cannot hold; for then the ALP w will not be▲ Lut, and
this you are to underſtand in the following problems where mention is
made of poffible proportion.
PROBLEM XCVII.
To divide a plot A B CD E, having two oppofite fides A B and
CD parallel, into any poffible proportion as a to b, by a line
drawn from a point P any where given, and paſſing thro' theſe
two parallel fides. Fig. 75.
1. By problem 89, the figure is reduced to the trapezia G AB D,
thro' the middle of whofe fides G A and B D, draw the line ng, which
by problem 41, make asa:b::nL: Lg. 2. Thro' L and the given
AND MECHANIC.
33
point P, draw the line U L P, and its done. For, as a : b :: UA E
CP (U AnGP) : DBUP(UPfe:) for draw f e parallel to AG,
fo the s Bge and D g fare equal, as by the laft problem.
PROBLEM
XCVIII.
To divide a▲ A B C into any proportion as a to b, by a line drawn
from an La to the oppofite fide BC. Fig. 76.
1. By problem 41, make it as a : b: CD: DB, join A D, and
it is done. For the As CAD and CA B and D A B, all ſtanding on
one line B C, and meeting in one point A, have the fame height, and
fo, by therem 31, are as thefe lines on which they ſtand, i. e. as a b
ACAD; ADA B.
PROBLEM XCIX.
To divide a plot A B C D into any proportion that is poffible as
a : b, by a line n A drawn from an LA to its oppoſite fide DC.
Fig. 77.
1. From the L A, by problem 89, reduce the plot to a ▲ AED,
whofe bafe D E divide, by problem 4, as a : b:: Enton D. 2. Join
n A, and its done. For, by the laſt problem, as A: b:: ▲ An E (=
AnCB) A An D.
PROBLEM C.
To divide a ▲ A B C into any proportion as a : b, by a line drawn
from a given point E in one of its fides B C. Fig. 78.
1. From the given point E to its oppofite L A, draw the line EA,
then, by problem 41, divide BC, i. e. as a : b.: : BD : D.C. 2. Thro'
D and parallel to E A, draw D N, join E N, and it is done, i. e. as
BENA: ENC,:: a to B. For join A D, and it will be, by problem
98, as a : b:: ABAD: ACA D; and by the figure, ADAN+
ANDCHAD A C; alfo, A CDN+ANDE, but ▲ DAN
AED N, for they both ftand upon N D, and have each an L (A and
E) in the parallel A E, confequently A ENCLDA C.
See this done by a different method, by problem 104.
E
34
THE UNIVERSAL MEASURER
PROBLEM CI,
From any affigned point P in the fide of any plot or right-lined fi-
gure A B C D E, to divide the fame into any proportion that
is poffible, as a is to b. Fig. 79.
1. By problem 89, the figure is reduced to the ▲ GPH, whoſe
baſe G H, make, by problem 41, as a :b:: HS: S G. 2. Join PS,
and it is done. That is, as a :b:: SPDCB (SPH): PEAS(PGS)
by problem 98. For, A PE=APAG, by problem 88 and by the
figure, APGA+APAS PGS PEAS. Q.E. D.
=
PROBLEM CII.
u,
To divide a A B C into any proportion as a to b, by a line F
making given Ls B FU and BUF, with the fides of the ▲ which
it cuts, or (which is the fame thing) to be parallel to a line D
any where given. Fig. 80.
1. By problem 41, make one of the fides, which the divifional line
FU is to cut, as a to b, viz. as a: b:: Bg:g C, join A g, and thro'
A parallel to D draw A e; then, by problem 50, make it as e B:
B:: Bg: BF, viz. B f, as by the figure. Laftly, thro' F and par
allel to D or e A, draw F u, and it is done. That is, as a: b :: B U
F: UACF. For the As B A G and B A e, having the fame height,
are as their bafes B g and e B, and by the problem ▲ Bu F muſt be
ABAg; therefore, by theorem 15, as e B (A cВA):□e B::
Bg (ABg A) ABU F): BF. Q. E. D.
Note. If the line of divifion F U were to be the fhorteft poffible,
(a: b continuing the fame) then, by problem 15, it muſt cut the ſides
a = Ls, viz. L B FUL BUF; alſo, if F U were to be parallel
fide A C, it is but drawing A e, fo as to make L Be A➡L BCA,
viz. A e would become A C.
સ
to any
PROBLEM CIII.
To divide a plot A B C D E into any proportion that is poſſible,
as a to b, by a line 2 d drawn parallel to any given line R.
Fig. 81.
AND MECHANIC.
35
.
1. It appears by the figure that the line z d of feparation muſt cut the
fides A E and C D, therefore produce thofe fides till they meet in H,
and then the figure reduced to a ▲ is G A F, whoſe baſe G F, make
as ab: Ga to F a, join A a, and draw A Q parallel to the given
line R. 2. Make it, (by problem 50,) as HQ: HQ::Ha:
2.,Make
Hd, fo thro' d and parallel to R, or A Q draw z d, and it is done.
That is, as a : b:: z ABCd:d DE z. For, as in the laſt problem,
the As Ha A and H QA, are as their bafes H a and HQ; and z d
is parallel to A Q. Therefore, by theorsm 15, as HQ:□HQ::
Ha (▲ Ha A = ^HDZ); □Нd. In the fame manner, you may
divide a plot by the ſhorteſt line poffible, of by a line drawn parallel to
one of its fides. See the laft problem.
PROBLEM CIV.
To divide a ▲ A B C into any proportion as a to b, by a line PM
drawn from a point P without the A. Fig. 82.
1. Thro' the given point P, and parallel to A Cdraw P G, cutting
B Aextended in G. 2. By problem 41, make it as a : b:: AH; Hb;
alfo, by problem 43, make it as GP: AC:: AH: HI, 3. By
problem 45, find A La mean proportional between A G and A 1,
which A L lay from H to M, join P M and its done. That is, as a : b
:: A MD: M DC B. For, by problem 100, figure 78, it is evident,
that, as CE: CD :: CA: C N, fo by joining NE that problem is
done; whence, if the point R were in the line A C, as at D, it would
be as AD: AC:: AH: a fourth term, which laid from A towards B,
will give the point required. But, P is the given point, and the As A
MD and G M P are fimilar, becauſe of their two parallel fides A D and
GP; fo, by theorem 15, as AG MP: AAMD:: □ GM:□A
M. QE. D.
PROBLEM
CV.
To divide a plot, ABCDEF, according to any proportion that
is poſſible, as a to b, by a line SP, drawn from a given point with-
out the plot, as at P. Fig. 83.
36
THE UNIVERSAL MEASURER
1. It appears by the figure, that the divifional line muſt cut the fides
AF and CD, ſo produce theſe fides till they meet in K, then from the
LC oppoſite to the given point P, the figure, by problem 89, is reduced
to the AMCN, whoſe baſe MN, make as a b:: MO: ON,
2. Thro' P, and parallel to M N, draw PL, cutting C D extended in
L, then make it, as LP: KO:: KC: KX; find K q a mean pro-
portional between L K and K X. 3. Always divide or find R the mid-
dle between K and X, and join R q, which Rq fet from R to S, and draw
S P, fo it will be as a : b:: AB CSP: PSDEF as required. See the
laft problem.
PROBLEM CVI.
To divide a ▲ A B C into any proportion, as a to b, by a line RQ
drawn from a given point within the ▲ as P. Fig. 84.
1. The dividing line being to cut A C and BC,
thefe lines, as A C, by problem 41, thus, as a
AC,
thro' the given point P, draw M L parallel to A
divide either of
b:: Aaà C, and
C, and upon A C let
ऊ
fall the LS UP and e B, then make it as UP:e B:: Ca:CN.
2. Thro' N and parallel to B C draw N M meeting LM in M, from
the of M P take the of L P, and a fide of that which remains
is QN, ſo thro Q and P draw the line of ſeparation Q_PR, and it
will be as a b::AQ BR: CR Q. For by conſtruction, U P:e B
::±Ca:CN, ergo U P XNC= e BX †a C, i e. ANLC=A
CQR = QRL N; fo that the line QR is to pafs thro' P in fuch a
manner as to make the ſpace QRLN ANL C, &c.
Note. P may be fo far within the A and the proportion fo unequal,
that this, and problem 119 may each be impoffible.
PROBLEM CVII.
To divide a plot A B C D E F into any proportion that is poffible,
as a b, by a line W U drawn thro' a given point P, within
the plot. Fig. 85.
AND MECHANIC.
37
1. It appears that the line of divifion will cut the fides AF and B C,
fo produce theſe two fides till they meet in K. 2. By probleni 89, the
plot is reduced to the ▲ LCM, whoſe baſe L M make as a : b:: LN
:N M, then, as a from P to A F: a from G to L M: : KN:
KO. 3, Thro' 0, and parallel to K C draw R O, alfo thro' P and pa-
rallel to A F draw q R meeting RO in R. 4. From the of PR take
the of q P, and a ſide of that □which remains is — OU, ſo thro' U
and P draw the line UP W, and its done; that is, as a : b: ; ABWU
: UWCDEF. See the laft problem.
PROBLEM
CVIII.
To divide a A B C D into four right angled ▲s and a□ in the
middle, whofe areas may be all equal, viz. each = a fifth part
of the given. Fig. 86.
1. Make D E—C F, cach two fifths of AD a fide of the given,
and draw C F which will be parallel to D C and A B. 2. Upon C D
as a diameter defcribe the femicircle cutting E F in G, join G D and G
C, fo is G D C one of the As required. The other three As may be
found in the fame manner; or thus, which is eafier, draw B GL to G
C and AHL to B I, and G D produced will cut A H perpendicularly
in L, fois DGC, CIB, B H A, and A L D the four As, and GIHL
theas required. For G P being DA, it is plain, by theorem
24, the DGC is of the A B C D, which with the other three
As, make of the faid, ſo confequently there mult remain for
the little
=
fo
GIHL. QE. D.
13
PROBLEM
CIX.
To divide a right angled ▲ A B C into four right angled ones
and AD E, to each other, and
CDF, BFD, B D E,
BDE,
fimilar to the given one A B C.
Fig. 87.
1. Ftom D the middle of the hypothenufe draw D B, alfo D E and
D F, parallel to B C and A B, and it is done. For, becauſe D E is pa-
rallel to BC, and D F to A B, and D C D A, therefore, E A=
EBDF, and FB FCE D. QE, D.
38
THE UNIVERSAL MEASURER
PROBLEM
CX.
To find a point
in a given ▲ A B C, from which if three lines
be drawn to the three As A, B, C, they fhall divide the whole
▲ A B C into three other As A OB, BOC, and C O A,
!
whofe areas ſhall be as m, n, and u respectively. Fig. 88.
1. In A C and AB produced if neceflary, make C E and A F each
n-fn-
nt nu, joining E B and C F, make C em, Ad=n, and
draw e b and d f parallel to E B and C F. 2. Thro' b and f draw be
and f P parallel to AC and A B, and at O where thefe two lines inter-
fect each other is the point required. For draw bH and BD Ls to
A C, then the As C B E and C be, as alfo CBD and C b H are fimi-
lar; therefore, it will be as m (Ce): m+nXu (CE) :: Ce: CB
::bH: BD:AAOC:▲ A B C, by theorem 31, for the As
AOC and A B C have one bafe A C; in the fame manner it may be
proved, that as n:m+n+u::AAOB;▲ABC, confequently,
as u :m+n+u::ABOÇ: A A B C. Q. E. D.
PROBLEM
CXI.
By a line from an LA in a given ▲ A B C, to cut off a part
towards the LB that ſhall be to a given □, whose fide is a b.
Fig. 89.
1. Take A h
fite fide B C, (viz.
the neareſt diſtance between the LA and its oppo-
a 1,) and by problem 42, make it as A h (Cb)
: a b (Cd): : ab: dq, which dq lay from B to P, join A P, and it
is done. For the A AP Bis
becaufe, by conftruction, A h (
the area of a
whofe fide is a b,
=
Ad) xBPabx a b. This may
be done differently by the following problem.
PROBLEM
CXII.
By a line A G drawn from a given L A, in a plot A B C D E,
to cut off a part towards the L B to a given rectangle a ba b.
Fig. 90.
1. Ifthe line of divifion cut the fide CD, then extend CD and A E
till they meet, but the A A B C is greater than the
abab, fo AG
*
AND MECHANIC.
39
muft cut B C, whence, if by problem 87, you reduce the rectangle a b
a b to a, the work will be the very fame as in the laft problem.
Or thus, thro' A and parallel to B C draw L S, and upon C B raiſe
the LB L, then, by problem go, make the rectangle BLSO to the
given one a b a b, make O GO B and draw A G, fo is the ▲ ABG
BLSO ab ab as required; for BLS O and AABG
have the fame height B L, and B G the baſe of the A is double B O
the bafe of the rectangle. Q. E. D.
PROBLEM CXIII.
From a given point P in the fide of a ▲ A B C to cut off a part
BGP to a given rectangle b c f e. Fig. 91.
1. Thro' P and parallel to its oppoſite ſide A Edraw L3. 2. Upon
Braife the BL meeting S L in L. 3. Upon BL, by problem go,
make the rectangle B LSO the given rectangle b c fe, or make it
as BL: eb:: bc: LS-BO, which B O lay from Q to G, join
PG and it is done, as is evident by the laſt problem.
=
PROBLEM CXIV.
To cut off from a plot A B C D E F á part A B P G F, = to a
given rectangle a b c d, by a line P G paſſing thro' a given point
P in one of its fides BC Fig. 92.
1. Becauſe the dividing line is to cut E F and BC, produce theſe fides
till they meet which is at H. 2. By problem 91, add the figure HBA F
to the given rectangle a b c d, and you'll have the be fc, to be cut
off from the AHCE thro' P, fo work as in the laſt problem, and you'll
get A BP G F the given a b c d, or the AHPG = the
befc.
PROBLEM
CXV.
To cut off from a ▲ A B C a part A n P to a given □ Abc D,
by a line n P making an L n P A with the fide AC of the A
which it cuts to a given LE b A. Fig. 93.
40
THE UNIVERSAL MEASURER
1. From E upon A C let fall the LE F, and make b Q= 2 E F.
2. Upon A Qas a diameter, defcribe the femicircle cutting c b in m,
thro' m and parallel to A C draw m n meeting A B in n. Laſtly, thro'
n and parallel to Eb draw n P and it is done. For the As A E b and
A n P are fimilar, becaufe E b in parallel to n P, and L A common;
fo as E F AB: nS (mb): A P. But by problem 45, m b=
A b xbQ = Ab x 2 E F, by conſtruction; whence, as EF:: A b
Abx: Ab x 2 EF:
:: Abx2 EF
EF
=AP, and APX
Abx: ABX2EF:
n SAP X √: Abж2 E F
EF
Abx Abx 2 EF
Abx2 EF:
Abarea ▲ An P, which
2 EF
is alfo the given A b c D. QE. D.
PROBLEM
CXVI,
From a given plot z TBDCN, to cut off a part n Tz N P
— to a given rectangle A e u b, by a linez P making given Ls with
the fides C N and B F, viz, ↳ NP n =LEbF. Fig. 94.
1. Let the fides B T and CN, by which the dividing line is termi-
nated, be produced till they meet in A; then it is plain, that if to the
given rectangle A e ub, by problem 91, you add the ſpace A T z N,
and of that fum make a Ab c D, the work will be the very fame as
in the last problem, and the directions for it will ferve for this alfo.
Note. In this manner you may cut off a part by a line drawn parallel,
to any
of the fides fuppofe D B, it is out to draw E b parallel to DB,
&c. or alfo by the ſhorteſt line poflible. This is but to make A E
A b and then join G B, and let fall the L E F, &c.
PROBLEM
CXVII.
=
To cut off from a ▲ B C, a part EAD to a given □whoſe
fide is a b, by a line PD drawn from a point P without the A.
Fig. 95.
:
AND MECHANIC.
1. Thro' P and parallel to A B, a fide of the given A oppoſite to P,
draw QP meeting the fide A C in R, draw A QL to A B, and upon
it make the AQ_SO = the given □. 2. With the radius A G —
2 AO — RP, and one foot on A, deſcribe the arch G, with the radius
OA and one foot on O, crofs it with another arch in G, draw A G
produced to F, making GFR P, join F O, and make O DF 0;
thro' D and P draw DP and its done. For, draw R H parallel to E D,
upon O with the radius O A defcribe the circle, and extend F O tom,
making the diameter n O m. Then, by conſtruction, G FRP,
A FAT and FOOD, whence A Dm F and D T■F,
and, by theorem 11, A DXDT=mFxnFAFX FG AT
XRP; whence, AD × DT=AT XD H, becauſe DHR P,
(for D His parallel to R P, and R H to P D) now if to each of theſe
s be added the ADX AT we ſhall have A D×DT+AT
ATXDH+A D, or A DAT XAH; now, let bRr
=AQ = 0 S, a = A0=0T, r=RP DH, then becauſe the
As E A D and R AD are funilar, it will be as 2 a +r (A H): b
by: 4aa+zar:
=
(R):4aa+2ar:(:ATXAH):-
OE,
2a+r
b:4aa+2гa:
whence O EXAD=
·× √: 4a a + 2 ar:
4 a +25
A..
bx: 4aa+2 ar
4a+2r
a = area ▲ A DE, which is alſo ares
=
AQSO. Q. E. D.
PROBLEM CXVIII.
To cut off from a given plot I H L CE B, a part LEDIH
= to a given rectangle A B a O, by a line E D drawn from a
given point P without the figure. Fig. 96.
F
42
THE UNIVERSAL MEASURER
1. Becauſe the dividing line feems to cut the fides C L and BI, pro-
duce thofe fides till they meet in A. 2. Thro' P and parallel to A B
draw R Q meeting the AQ in Q and the fide A Cin R. Now it is
plain by the figure, that if to the given rectangle A B a O you add the
fpace A LHI, and of that fum upon A Qmake the reangle AQS O,
the work will be the fame as in the laft problem, and performed by
the fame directions and letters.
PROBLEM
CXIX.
From a given point P within a A A BC, to cut off a part AED,
to a given rectangle A QSO. Fig. 97.
1. Thro' the affigned point P and parallel to A B, draw a line QS
meeting A C in R. 2. Upon A B erect the LA Q meeting SQ in
3. Upon A Qmake the AQSO to the given one A QSO.
4. With the radius A GAO 42 R P and one foot on A deſcribe
the arch G, and with the radius O A upon O crofs it with another arch
în G, join A G, make A FA O, and thro' O and F draw O F, make
OD FO, ſo thro' D and F draw D E and its done. For, upon O,
with the radius O A deſcribe the circle, and extend F O till it become
the diameter n o`m, parallel to E D thro' R draw RH, fois HD≈RP,
becaufe Q´S is parallel to A B; alfo A D = m F, becauſe A F≈ AO,
FG 2 R P and DOFO by conftruction, whence, DT
Now, by theorem 10, A F XF GnFxmFADXDT=AQ
X2RP (2 AO) TAXRP, whence A DX DT-ATX HD the
cry fame as in problem 117, this being the fame ftep in the demonftra-
tion there.
In F.
Alfo, as problem 1 18 is done by problem 117, it will be as cafy by
this problem to cut off a part from a plot by a line drawn thro' a point
within the plot, &c.
N. B. Theſe problems for cutting areas, and the foregoing ones far
dividing areas in proportion, may both be wrought one way; for ex-
ample, if it were required to divide 54 in fuch proportion as 5 to 1,
firſt, 5+1=6, then fay, as 6:54 : : 1 : 9 or 5 5 45,
1 : 9 or 5 45, fo that to di-
vide 54 as 5 to 1, is the fame as to cut 54 into 2 parts,
the greater
and 9 the lefs, be the figure what it will. Alfo if you would cut 9 acres
AND MECHANIC.
43
from fome figure or plot whoſe area is 54 acres and chooſe to do it by
proportion, then 54945, fo that proportion will be as 9: 45 or
lower as 1 to 5.
Thus have I explained two ways of performing thofe difficult prob
lems; but the eaſieſt and moſt practicable method for dividing, &c.
fuch furfaces, you'll find in furveying, yet the foregoing methods are
geometrically true, and may ferve as good excercife for young ftudents
in geometry.
PLANE TRIGONOMETRY,
In Nine PROBLEMS.
Shewing the Rules, Axioms, &c. with the Geo-
metrical, Inftrumental, and Logarithmetical
Solutions of all the Cafes of Right and Ob-
lique Triangles.
PROBLEM CXX.
DEFINITIONS.
1. Trigonometry implies the meafuring of As, and is here under-
ftood to point out the relation or proportion that is between the fides
and Ls of a A, from the refpective data in the various cafes of right and
oblique angled As.
2. Every ▲ conſiſts of 6 parts, viz. 3 fides and 3 Ls, any three of
which, if one fide be known, the reſt may be found; but from the 3 Ls
only no determined ▲ can be obtained, for the problem admits of in-
finite anſwers, fince any fide may be fuppofed at pleaſure, and with it
and the Ls may fides be found correfponding; the fame reafoning holds
1
ᏎᏎ .
THE UNIVERSAL MEASURER
when only two of the fix parts are given. In taking heights, diſtan
cês, &c. your only buſineſs will be to fecure three parts of a ▲, and
one of them muſt be a fide, to find the reft, or elſe all your labour will
be to no purpoſe.
3. A diagonal ſcale is a fcale of = parts diagonally divided, from
which larger numbers may be taken than from a ſcale of parts only.
Thoſe ſcales are for meaſuring and laying down right lines, as the fides
of As, &c. but the Ls are meaſured and laid down from a ſcale of
chords.
4. All the parts of the fame figure whether fides, Ls, baſes or dia-
gonals, muſt be taken from the ſame ſcale of parts, otherwiſe your
work will be faulty; but an L may be meaſured from any ſcale of
chords.
if
=
5. If your numbers to lay down be too large or too fmall for your
diagonal ſcale, they may be diminished by divifion or augmented by
multiplication thus, 25205 divided by 100 makes 252.05, ſo you may
take 252 and a little more from your diagonal ſcale; then fuppofe a
fide of the fame figure make 750 on your ſcale you muſt multiply by
100, the reaſon of which is evident, viz. becauſe you divided by roo
in laying down, fo that your 750 becomes 75050; the contrary holds
your numbers be too little; this alfo muſt be obferved on Gunter's
fcale or line of numbers on all ſcales, rules, &c. fo that if you multi-
ply or divide any right-lined part or number, in order to increaſe or di-
miniſh that part or number, you muſt do ſo with all the reft. Six Ls
will not admit of fuch multiplication or diviſion, fo that whether they
are great or fmall they muſt be projected or meaſured in their original
ftate. But the radius of the ſcale of chords may be increaſed and dimi-
niſhed at pleaſure. And as 600 is the radius of a circle whofe periphery
is 360°, you muſt always with the chord of 60 and one foot on the
angular point, deſcribe an arch to cut the two fides fubtending an L,
produced if neceſſary, and then this arch or fubtenſe meaſured on the
fcale of chords you took the radius from fhews the quantity of that L if
not above 900.00, but if more you muſt meaſure it at twice, and take
AND MECHANIC.
45
the two meaſurements in one fum. This meaſure of the arch is to be
underſtood when an Lis mentioned. See problems 10, 11, 12.
7. You muſt meaſure all Ls or project from the fame fcale of chords
you took the radius 60° from. Hence it appears that the fides of an L
have no relation or connection with the quantity of that L; for let the
fides of an L be ever fo long or ſhort the meaſure of the L is fill the
fame invariable quantity.
8. A plane L is made by the meeting of two right lines called its
fides, and if its meafure be 900 it is called a right L, if lefs than 90 it
is called an acute L, if more than 90° it is called an obtufe L; but the
two laft are generally called obtufe Ls.
9. A plane A is made by the interfection of three right lines called
its fides, and if any two of theſe fides make an L of 90° it is called a
right angled A, and thefe two fides are together called legs, the longer
of which is commonly called the bafe, and the other the Lor cathetus,
and the longeſt ſide which is always oppofite to the right L is called the
hypothenufe.
10. If each Lofa A be lefs than 90° it is an acute angled A, but if
one L be more than 900 it is an obtufe angled A.
II, If the 3 fides of a A be the A is equilateral, if two fides are
equal it is called iſofcelar, but if all the fides are unequal it is a fca-
jene A.
12. Since, by theorem 4,the fum of the 3 Ls of every plane A is 1800,
it follows, that if any two of thefe Ls be given the third L is found by
ſubtracting the ſum of thofe that are given from 180°; and in a right
angled ▲, becauſe the right L is 90°, if one of the acute Ls be given,
the other is found by fubtracting the given one from 90º.
13. Complement is what an arch or L wants of 90°, and fuplement
is what an arch or L wants of 180°.
14. In right angled plane As, are 7 cafes, and in oblique angled ones.
5, all folved arithmetically by thefe 4 axioms.
4.6
THE UNIVERSAL MEASURER
AXIOM I.
In all right angled plane As, as the greater leg the leffer leg
:: radius: the tangent of the lefs acute L, or as the leffer leg: the
greater :: radius: tangent of the greater L, (fee theorem 47) this is
when the legs are given to find an L, or a leg and an L to find the
other leg.
:
AXIOM II.
In all plane As, as any fide: the fine of its oppofite L:: any other.
fide the fine of its oppofite L. And the contrary. See theorem 48.
This is when oppofite fides and Ls are given and required. And hence
we learn, that the greateſt fide fubtends or oppofite to the greateſt L,
the leaſt fide to the leaſt 4, = fides to Ls, and the contrary. Theſe
L,
two axioms will folve all the 7 cafes of right angled plane As, and by
them alfo may all the 5 cafes in oblique angled As be folved, by divis
ding the oblique angled A into two right ones with a L.
AXIOM III.
I
I
As the fum of any two fides: their difference: : the co-tangent of
their included L, viz. ¦ the contained Lor the L made by theſe two
fides, or the tangent of the fum of the other two Ls: tangent of
their difference; then the fum of the unknown Ls + their differ-
ence is the greater L, and the faid fum- the ſaid difference is the
lefs L. See theorem 50. This is when two fides and the L made by
2
them are given to find the other two Ls.
AXIOM IV.
When the three fides are given, as the longeſt fide: the fum of the
other two the difference of thefe two fides: the difference of the
fegments of the baſe or a fourth number, which taken from the great-
eft fide or baſe, the falls on the middle of the remainder, (fee the-
orem 49) of which is the leſs ſegment of the bafe, and this added
to the 4th number gives the greater fegment of the baſe, and then the
Ls are found by axiom 2.
AND MECHANIC.
47
I
Or an L may be found in this cafe without letting fall a 1, thus,
take each ſide ſeparately from the fum of all the fides, but firſt that
fide oppofite to the L you want to find, noting their remainder; then,
as the product of half the fum of the fides and firſt remainder : the pro-
duct of the other two remainders :: of radius: the of the tan-
gent of the L required. Thus when any three parts of a ▲are gi-
ven, to find a fourth part conſider to which axiom it belongs, by which
you'll be able to form the proportion; where obſerve, that if fines or
tangents be mentioned in the proportion, you muſt uſe the numbers be-
longing to fuch fines or tangents, but thofe terms which mention nei-
ther are natural numbers.
15. On the ſcale for this purpofe are three lines the length of the
fcale, the first marked at the end Num. is a line of numbers, and is the
fame with that on the flider in the common fliding rule, the next marked
Sin. is a line of fines marked 10, 20, 50, &c. to çc at the end; then
the line of tangents ending at 45 marked Tan. this line is alfo numbered
back again to 800 cr 890. The fame lines are fet on fliding fcales which
are made to work with a flider inſtead of compaffes. This line of tan-
gents both increafes and decreafes from 45 the end towards the left
hand, fo as any number of degrees increafing is their compliment to
90 decreafing,
Note. In all fcales you muſt reckon the divifions the fame way that
the numbers or figures are reckoned. In extending on this line between
any two tangents both greater or lefs than 45, the companies cannot
fall off the ſcale, but if one is greater and the other lefs than 45, the
compaffes will go beyond the line; in fuch cafes let the foot beyond
45 reft, and bring up the other to 45, there reftit, and turn the other
foot upon the line within 45, and it points the anfwer. This can only
happen in the uſe of axicm the third, for in axiom firft one turn is al-
ways radius 45.
16. The radius of the line of fines is 90, radius of tangents 45, each
at the lines end.
17. On this ſide of the fcale are other lines, as T R being the tan-
gent-line divided into S points, with halves and quarters of the feaman's
48 THE UNIVERSAL MEASURER
1
compafs, to be uſed by axiom 1; when the Ls are required or given in
points of the compafs, called rhumbs. SR a line of fines divided into
rhumbs, to be uſed by axiom fecond, when the Ls are in rhumbs.
VS verfed fines, of uſe in ſpherical trigonometry. Mer. meridional
parts, which with E P joining it, a line of parts, is uſed in Mercator's
failing. Thefe lines are all conſtructed by the rules given in part third,
for the fliding gunter, &c. viz. only laying down the, logarithms of
natural fines, tangents, &c. from the fame fcale the line of numbers is
from.
18. On the other fide of this fcale are the fcales for projection; and
is a ſcale of 24 inches or two feet, two diagonal ſcales, one double the
other; Rum. a ſcale of rhumbs ; Cho. a ſcale of chords made by prob-
lem 9 ; Sin. a ſcale of fines, with Sec. a fcale of fecants following, both
beginning at one point Sin. Tan. and S T a fcale of tangents and femi
or half tangents, theſe may be made by problem 54; for if G C be an
arch of 35°, then G B is fine of 35°, A D the fecant, CD the tangent
of 350, or femi-tangent of 70°. Fig. 44.
19. Rules of proportion on Gunter's ſcale, extend from the firſt term
to the ſecond or third of the fame denomination, and that extent will
reach from the third or fecond term of another kind to the answer or
fourth term of the fame kind, viz. extend from L to L whether fine or
tangent, or from fide to fide, fo will your extent th us taken reach in the
firft cafe from fide to fide, and in the latter from L to .
20. Rules of proportion on the fliding fcale called fliding Gunter.
Set the firſt on the flide either of the two means on the rule, then a-
gainst the other mean on the rule ftands the fourth term on the rule.
Or, fet the first term on the rule to either of the means on the flider,
then againſt the other mean on the rule ftands the fourth term or an
fwer on the flider.
21. Rules of proportion by the logarithms, add the logarithms of the
two means, viz. of the ſecond and third terms together, from that fum
take the logarithm of the firſt term, and the remainder is the logarithm of
AND MECHANI C.
49
the fourth term. Or as you write the logarithm of the firſt term out of the
table of logarithms, begin at the left hand fide, and take the refidue of
each figure to 9, and the laft to 10, this is called the complement arith-
metical marked C. Ar. which with the logarithms of the two means be-
ing added into one fum, rejecting 10 in the index, (or number pricked
off to the left hand fide) gives the logarithm of the fourth term with-
out fubtraction.
22. In the following problems you have the proportions all fet down
which you may work as you pleaſe, by ſcale and compaffes, fliding
fcale, or by logarithms; which laſt is moſt exact, becauſe the logarith-
metical tables give the anſwer to moſt places of figures.
PROBLEM
CXXI.
Cafes firſt and ſecond. In a right angled plane ▲ A B C, there is.
given the two acute Ls B 33°.45′ and C 560.15', with the
leg A B 121.394, to find the hypothenufe B C, and the other
leg AC. Fig. 98.
1. Make the line A B
121.394 from a diagonal ſcale. 2. Upon
A, by problem 11, make the right Ln A m. 3. Upon B, by problem
10, the Ld Ba= 33°45′, theſe laſt two lines meeting in C, form the
▲ required. Then the leg A C laid on the fame diagonal fcale is found
≈81.1, and B. C = 146.
Arithmetically, by Axiom 1.
As tang. LC 560.15
10.1751074
9.8248926
: tangent radius 450 10.0000000
10.0000000
or by taking
:: A B 121.394
2.0841692
2.0841692
the com.ar.
: AC 81.113
1.9090618)
L 1.9090618
Becauſe the log. index is 10, you muſt take it from 19, and reject 20
in the fum.
In finding the logarithmetical fines and tangents, if the degrees be
under 45, find it at the top of the table, and the minutes in the little
1
G
!
50
THE UNIVERSAL MEASURER
column on the left hand ſide; but if the degrees be above 45, find them
at the bottom of the table and the minutes in the little column on the
right hand ſide: fo above 560 and againſt 15' in the tangent column
ftands 10.1751074 the logarithmetical tangent of 56° 15′ and in the
fide column ſtands 9.9198464, the logarithmetical fine of 56°.15. If
you have the log. fine given to find the degrees and minutes, it is the
reverſe of the former, for with the given log you enter its column whe-
ther fine or tangent, and having found the degrees, you take the minutes
which come neareſt. The log. of a number is had by finding your num-
ber under minutes in the table, and againſt it ſtands the logarithm rẹ-
quired; always obferving that the index muſt be 1 leſs than the number
of places of whole numbers in the given number.
If the logarithm of a number be given to find the number, it is juſt the
converſe work; fo 1.9090918 being given I look for it amongſt the
logarithms, and finds .9090744 to be the neareſt, and againſt it under
numbers ftands 8111, fo the anſwer will be 81.11, becauſe the given
index is 1, &c. for others.
By fcale and compaffes. The extent from 560.15' to 450 in the
tangents reaches in the numbers from 121.394 to a little above 81.
By the fliding fcale. Set tangent 560.15' to tangent 450, then a
gainſt number 121.394 ftands better than number 81, for the anfwer.
Note. Log. radius is always 10.000 &c. Laftly, B C is had by ax-
iom 2d, and fo may A C too; thus, as fine LC 560,15': radius fine
90 LA:: A B 121.394 to B C 146; and as fine L C 560,15': fine
▲ B-330,45' :: BA 122.394: A C 81,11.
PROBLEM CXXII.
Cafe 3, Given the two acute Ls B 33°45′ and C 560,15,
with the hypothenuſe BC 136, to find the legs A B and A C.
Fig. 98.
AND MECHANIC.
5º
י
1. Draw the line Ba, and on B, by problem 10, make the Ld Ba
33°45′. 2. From a diagonal ſcale take 146 and lay from B to C,
then at C make the L BC A 560,15′; or, from Cupon B a let fall
the LC A, and the figure is made; then A B and A Claid on the fame
diagonal ſcale are found 121 and 81. Arithmetically, by axiom 2d,
as fine L A 90° : fine L B 33°45′ :: BC 146: A C 81,11, and fo
is fine LC 560,15′ : AB 121,39.
PROBLEM CXXIII.
Cafes 4th and 6th.
Given the legs A B 121,4 and A C 80, to
find the oblique Ls B and C, and the hypothenufe BC. Fig. 96.
1. By a diagonal fcale make B A 121,4. 2. Upon A, by problem
11, make the L B A C 90°, and from the fame diagonal ſcale take
80, and lay from A to C, join B C and the figure is protracted. Then
BC laid on the diagonal fcale is found 145. For the Ls, with the
chord of 60° and one foot in either B or C fuppofe C, deſcribe an arch
dm, which laid on the fame chords gives 560,30′ for the LC, whofe
complementis 330,30' for the L B. Arithmetically, by axiom 1, as B A
121,4: A C 80:: tangent 45°: tangent L B 330,30'. And by axi-
om 2, as fine LB 330,30′ : fine L A 90º :: A C 80 : B C 145
nearly.
PROBLEM
CXXIV.
Cafes 5th and 7th. Given the hypothenuſe BC 146, and a leg
BA 121,39, to find the other leg AC and Ls B and C.
Fig. 99.
1. Draw the line B A, on which from A to B by a diagonal fcale lay
121,39. 2. Upon A, by problem 11, make a right L BAD. 3. From
the fame diagonal ſcale take 1 46, and with one foot in B deſcribe an arch
e a croffing A D in C, join B C and the figure is made.
chord of 60° and one foot in either B or C, ſuppoſe B,
Then with the
deſcribe an arch
for the L B, the
d n, which laid on the fame chords gives 330,45
complement to which is 56°15′ the C, and the leg A G laid on the
52 THE UNIVERSAL MEASURER
diagonal ſcale is found
81,1. Arithmetically, by axiom 2, as B C
146 : A B 121,39 : : radius (fíne L A 900) : fine LC 56º,15′, and as
radius: fine LB 330,45': B C 146: A C81,11.
PROBLEM
CXXV.
Cafe ift. Of oblique angled plain As given two Ls A 620,30′ and
B 37º,30′, and, by definition 9th, the third LC is alſo known
to be = 80°, with a fide A C 350, to find either of the other,
fuppofe BC. Fig. 100.
1. By a diagonal ſcale make A C= 350. 2. Upon A, by problem 10,
make an L B A C 80°, and the two lines A B and C B meeting in
B form the required. Then B C and A B laid on the fame diagonal
ſcale are found 510 and 566. Arithmetically, by axiom 2, as fine.
LB : fine L A 62°º,30′ :: AC 350: BC 50 9,976; alſo, as fine L B
37°,30′ : fine L C 80º :: A C 350: A B 566,2.
PROBLEM
CX XVI.
Given the two fides B C 509,976, and A C 350, with the L B
37°,30′ oppofite to one of them A C, to find one of the other Ls
A or C. Fig 100.
1. By the diagonal ſcale make B C509,976. 2. Upon B, by prob-
lem 10, make the LE B C = 37°,30′. 3. With 350 from the diago-
nal ſcale and one foot in C defcribe an arch to cut E B in the points À
and F, which fhews that this cafe is doubtful, whether both the un-
known Ls be acute as A and C, or the one F C B acute and the other
CF Bobtufe; for from the things given either B C F or B C A may
be the required, by joining C A and C F, fo let A B C be the A re-
quired, then with the chord of 60 and one foot on C defcribe the arch
m n, which laid on the fame chords gives 800 for the LA C B; then,
by definition 9, LC 800+ LB 370,30' gives 117°,30', which taken
from 180 leaves LA 62,30′ the arch a e being ftruck upon A with
the chord of 60. Arithmetically, by axiom 2, as A C=CF 350:
CB 509,976: fine L B 37,30': fine L A 62,30', if both the Ls be
acute; but if C F B be the L required, then this 620,30' taken from
1800 leaves 117°,30' for the C F B.
1
AND MECHANIC.
93
Note. When you are to take the fine or tangent of any L above ço,
you muſt firſt take fuch an L from 180, and take the fine or tangent of
the remainder. For it is plain by figure 44, that B G is the fine of the
arch HEG as well as of the arch C G, and that thefe two arches to-
gether make a femicircle. See it otherwiſe proved in theorem 48.
PROBLEM CXXVII.
Cafes 3d and 4th. Given the two fides A C 350, and C B 510,
with the LC 80° comprehended between them, to find the other
two Ls A and B, and the third fide AB. Fig: ioo.
=
1. By problem io, make an Ln Cm 80°, extending the fides
Cn and C m, till by the diagonal fcale C A = 340 and C B = 510,
join A B and the A is made. 2. With the chord of 60 and one foot in
either A or B, fuppofe in A, defcribe the arch e a, which laid on the
fame chords gives 600,30' for the LA, and by definition 9, you'll find
it 566. Arithmetically, by axiom 3. firft 350 added to and taken
from 510 gives 860 and 260 for the fum and difference of the two fides,
and 80 taken from 180 (definition 9) leaves 100 half of which is 500,
the half contained Ls complement or half fum of the oppofite Ls A and
B; Then, as 860: 260:: tang. 50: tang. 12.30' half the difference
of the faid Ls, fo 500+ 12,30′ =62,30′ the greater L, and 50
12,30′ 37,30′ the leffer L; laftly, as fine 62,30 fine 80:: CB
510: A B 566,3. By axiom 2.
PROBLEM CXXVIII.
Cafe 5. Given the three fides À B 213,5, A C 107,5, and BC
250,2, to find the Ls. Fig. 101.
1, To lay down this figure is already taught in problem 14, and to
meafure the Ls you muſt have recourſe to the foregoing problems or the
general rule laid down in definition 6 and problem 120; therefore to
do it arithmetically by axiom 4, firft the bafe is B C 250,2, then
213,5 + 107,5 = 321 the fum, and 213,5 — 107,5 = 106 the differ-
ence of the other two fides; then as 250,2: 321 :: 106: 136 near-
ly, the fourth number which is BF, how BC 250,2 —BF 1
136—
1
54
THE UNIVERSAL MEASURER
FC 114, half of which is 57 CGFG, and FG 57+ FB 136=
BG193; then in either of the right angled As A B G or AC G there
is given the hypothenufe A B or A C, and a leg B G or C G, fo by
axiom 2, as A B 213,5 : B G 193 :: radius: fine LG A B 64,45',
the complement to which is 25,15′ = L A BG. Again, as A C 107,5
: fine L B 25,15' :: AB 213,5 : fine L C 57,55',
: fine L A 96,50′, or having any two of theſe three
found by definition 12.
*****
and
:
:: BC 250,2
Ls the third is
茶​米糕​糕
​The Application of
TRIGONOMETRY
To the meaſuring Heights and Diſtances;
with the Deſcription and Ufe of the common
Inſtruments for that purpoſe.
PROBLEM CXXIX.
DEFINITIONS continued from Problem 120.
23. Of the Chain. Mr. Gunter's chain of 100 links each 7,92
inches is now the only chain in ufe among furveyors, whofe length is
22 yards or 66 feet or 4 poles, perches, rods, &c. and this divifion of
the chain into parts, makes links the decimal parts of a chain, con-
fequently chains and links work together in all refpects as decimal frac-
tions.
24. At first fetting forward to meaſure, to prevent miſtakes and for the
fake ofexpedition, let the perfon going before with the end of the chain,
have tenſmall ſticks, each about two feet long cut to a point at one end
AND MECHANIC.
55
to ſtick down into the ground with more eaſe at the end of every chain,
which muſt be carefully taken up by the perfon following with the other
end of the chain, and thus proceed till all the ſticks are taken up by
the perfon following the chain, if the line be fo long, being always
careful to move in a ſtreight line from mark to mark. If the diſtance be
not furveyed when all the ſticks are out, let them be returned to the
leading affiftant, and fo proceed till the whole diftance is meafured,
obferving to fet down how often the ſticks have been changed, that no
error be made in that cafe.
24. At the end of every 10 links there is commonly a piece of brafs
a ring or fome other mark for the more ready reckoning the links.
26. Chains and links may be fet down thus, 7ch. 40l. or 7,40 ch.
or 7,4 ch, or 740 links, which laſt is beſt fignifying 7 chains and 40 links
or 740 links.
27. If the chain be too long for ufe in meaſuring fmall parcels, you
may take dimenfions with the half chain, and fet down half the number,
as for example, fuppofe you have 11 fticks and 25 odd links, it muſt
be fet down 5 chains 75 links or 575 links, and then in plotting or
finding the area it will be the fame as if you had meaſured with the
whole chain.
28. For taking angles in the field are various inftruments contrived,
but all that can be propofed from them is to meaſure an ▲ included
between two lines or hedges that may be taken as freight lines, the
chief of which are, the plain table for ſmall incloſures, the theodolite,
circumferentor, and femicircle for champaign grounds, the three lall are
nearly of one form, being only a circle or femicircle divided into de-
grees and minutes, if the inftrument be large enough to admit of fuch
fubdivifions.
29. The plain table is a board or rather three pieces put together,
forming a parallelogram fit to hold a fneet of paper, with a frame round
it to keep the table together and to hold the paper faft, with a looſe
index with fights to lie upon the table &c. its ufe is chiefly to take the
lot of a field. See furveying.
56 THE UNIVERSAL MEASURER
30. The femicircle or the odolite is half a circle or two quadrants
whofe limb is divided into degrees, &c. as by problem 9, beginning on
the limb at one end of the diameter and ending in 180 at the other, and
beginning again at 181 and ending at 360, fo that as the index goes off
at one end at 180, it comes on the other at 181, confequently it an-
fwers all the purpoſes of a whole circle divided into 360°, in taking
angles it muſt be placed flat with its face horizontal, on which the in-
dex turns on the center, on the index are two upright pins or fights to
fpy the object thro' in the time of obſervation.
31 There is a line and plumet put thro' the central hole when al
titudes are taken therewith, and the object is ſeen thro' two fights
fixed in the diameter, the limb hanging downwards, the line and plumet
cuts the angle of the altitude; if you count the degrees from the mid-
dle of the limb but the co-altitude, if the degrees are numbred from
the corner next your eye, the fame is true in a quadrant which is half
a femicircle with a line and plumet ſuſpended at the center of the arch
a (in fig. 102) the fights upon the edge as a B, being directed to the
object C, the line hanging at liberty will be parallel to CD, if CD is
upright whence its plain (by fig. 102) that the Lu a BL B C D
and the arch e uLA BC, as by laſt definition.
32. Circumferentor is a circle upon a card divided into degrees &c.
and fixed on a board or in a box, a needle touched upon a loadſtone
moving flat upon its center, this is ufed in furveying mines &c.
34. Protractor a ſmall inftrument in form of a half circle, for laying
down Ls readier than a ſcale of chords, thus lay its diameter along
the line on your paper with its center on the angular point, then a-
gainſt the degrees of the L on the protractors limb or arch make a mark
on the paper, thro' which and the angular point draw a line and the
L is made.
33. Level, fill a cylinderic glafs tube with water excepting a ſmall
ſpace which cloſe at both ends, and when the tube is laid in a level
1
AND MECHANIC.
57
horizontal fituation the vacuity will fhew its felf in the middle, but will
vary there-from on the leaſt inclination,
34. Some note down the dimenſions of a ſurvey into a field book di-
vided into columns, but I would recommend the following, draw an
eye draught of the plot as near as you can, then when you have mea-
fured
any line, angle or offset, &c. write down its length or quantity,
along the fide or in the angle correfponding in your eye draught; fo fhall
you have a rough plan to make a true one by, which is both eaſier and
fafer to plan by than any table by what form or method foever it be
contrived.
PROBLEM CXXX.
To find the altitude C D of any object, as a tower, &c. by a quad-
rant B e a, being an inacceffible altitude. Fig. 102.
1. The method of obferving by a quadrant is plain by the figure
compared with definition 31; thus, at any convenient place B, view C
the top of your object along the edge B a of the quadrant, and ſuppoſe
the plumet a u hanging at liberty to cut 52,5' from e, viz. arch e`u
52,5' which, by definition 31, is the meaſure of the LC B A. 2.
Meaſure in a ſtreight line from your eye at B to the tower, which fup-
pofe B A = &5 feet, and your obſervation is finiſhed.
To lay this down. You have given the leg A B 85, and the Ls
B 52,30′ and C its complement, fo the figure is laid down by problem
121, and the required height is C A laid on the diagonal ſcale is found
105,8, to which add A D ſuppoſe 5 feet the height of your eye a-
bove the level of the bottom D F (as you muſt always do infuch caſes)
and it gives C D 110,8 feet. Arithmetically, by axiom 2, as fine
LC 37,30': fine LB 52,30':: BA 85: CA 105,8 feet.
PROBLEM
CXXXI.
To find A B the height of any object DHC, as a freeple, &c.
and come no nearer its bottom than S, being an inacceffible alti-
tude. Fig. 103.
H
THE UNIVERSAL MEASURER
1
1. Upon S obferve B the top of the object, along the edge of a quad-
rant, or by a femicircle, the plumet u a hanging at liberty, as by the
faſt problein, and ſuppoſe the LB'S "A to be found 51;30′, 2. Move
"back´any diſtance in a line with SA, fuppoſe to F75 yards, and upon
-Fmake obfervation to B as before, fuppofing the LB'FA to be 26,30'.
3. Now to lay this down. Make S F-45 from a diagonal ſcale, “and
upon S and F, by problem 10, make the Ls BS A51,30' and BFA
26,30' and at B where thefe two lines meet let fall the BA cutting
-F S produced in A, and by the diagonal fcale you'll find the height
BA 62 yards the anſwer; alfo by the fame ſcale you may find
SA &c. Arithmetically, In the oblique A B F S is given S F= 75,
L BFS 26,30', and L BSF (complement L BSA 51,30' to 180)
128,30', whence LB FS250 (definition 12) fo, by axiom 2, as
fine LB F S 250 fine L-BFS26,30′ :: FS75 yards; fide F B 79,18,
then as fine L BA F 90: fine L A F B 26,30ì :: F B 79,18: A B
61,97 the height required.
:
Note. If from a point Fyou obferve a point B, it is plain that you
feeit in a freight line F B, in like manner a point B'is feen in a ftreight
"Ime from a point S, whence it is plain, that theſe two lines muſt meet
in the point B, fo if you meaſure the diftance S F 75 yards, feet, or
chains, &c. and lay 75 from any ſmall parts on paper from S to F to
repreſent theſe 75 yards, &c, and draw the lines F B and S B to make
the fame Ls (F and S) as you took by obfervation, it is evident, theſe
two lines FB and SB muſt meet in B, forming a true plan of your ob-
fervations, in which plan all the unknown lines, &c. will be jaft fo ma-
ny of the faid parts as they would be yards, &c. in the thing itſelf.
This holds true in all heights, diftances, &c. whatever, which being
confidered, theſe things may be eafily done.
PROBLEM
CXXXII.
To find the heights C z and B'A≈e z of two inacceffible objects
Standing upon one another. Fig. 103.
AND MECHANIC
59
1. Upon any place where both points B and C may be feen, as at F,
obferve theſe two points with your inftrument as before directed, and
let the Ls, CFA44°, and B F A 26,30, 2. Move directly for
wards to the bottom of the object any diftance, fuppofe from F to S 75
chains there obſerve B and C as before, and let the Ls CSA be
67,50 and B.S.A=51,30. To lay down this, draw a line F SA at
pleafure, make FS75 from a diagonal fcale or fcale of parts,
and upon F and S, by problem 10, make the Ls as you found them by
obfervation, and C and B the meeting of theſe two lines will be the tops
of the objects, from which let fall the 4s BA and Ce, then by the
fſame ſcale of parts you'll find A B=62 and C z=50 chains the
heights required. Arithmetically, as by the laft problem, you found
A B, fo find Ce, and the difference is Cz.
PROBLEM CXXXIII.
To find m F the perpendicular height of a hill. Fig. 104.
1. As in problem 131, chufe any two convenient places on the fame
plane or level as n and´s which let be 100 chains diftance, and from
each place s and n obferve T the top of the mountain, and let the Ls
be s 300 and n 42°, by which the figure may be laid down exactly as
in problem 131, and by the diagonal fcale you'll find m T=160 chains
the required height; the arithmetical folution is alfo the very fame.
with that in problem 131.
PROBLEM CXXXIV.
To find AB the height of a tower, &c. by making two obfervati-
ons on the fide of a hill ſome diſtance from it. Fig. 105.
1. On any part D of the hill's fide obferve a point E, where you in-
tend your fecond ſtation to be, and let the L BDE = 13° made be-
tween looking at E and at B the bottom of the tower, and the L ADE
48°, being the degrees cut between obferving E and A the top of
the tower. 2. At D view fome other mark beyond E, as F in the fame
line D E, meaſure the diſtance D E which fuppofe 87 yards, then let
60
THE UNIVERSAL MEASURER
the degrees cut between looking at F and B be 180, and between Fand
A 680 when you ſtand at E. 3. To lay this down, draw the line
D E F at pleaſure, make D E87, on D make the Ls BDF 130
A D F 480; alfo, upon E make the Ls BEF=180, AEF = 680,
and at A and B where thefe lines meet will be the top and bottom of
the tower, which diſtance A B is found = 155 yards by the diagonal
ſcale. The arithmetical calculation may eaſily be deduced from what
is done in problem 131.
PROBLEM
CXXXV.
To find A B the height of any tree, tower, &c. by a fquare a b E,
or two ftreight flicks fastened together at right Ls. Fig. 106.
1. Let the ſticks be each of the fame length, and faſtened fo as to
form a true fquare, their length may be about four or five inches;
this done, fet an end E to your eye holding the fide E b level, move
back or forward till over the top of the other ſtick you can juſt ſee the
top A of your object, and at the fame time obferve fome point B to-
wards the foot of it along E b, then meaſure the diſtance E Bandit will
be to A B. For the As E ba and E B A are fimilar, and Eb is =
ba, fo E B will be = B A.
PROBLEM CXXXVI.
To find A B the height of any object by the ſhade of the fun.
Fig. 106.
1. Near the bottom of the object fet up a ſtreight ſtick a b parallel to
the faid object, then it is plain, if B E be the length of the object's fha-
dow, that b E will be the length of the ſtick's fhadow; therefore, as
Eb the ſtaffs ſhadow is to ba the ſtaffs length, fo is E B the length of
the object's fhadow to B A its height. For b a being parallel to B A,
the As E a b and E A B are fimilar.
!
$
AND MECHANIC.
619
#00000000000000
The 5 following problems belong to
DISTANCE S.
PROBLEM CXXXVII.
To find the distance of any object C, from a place A. Fig. 107.
1 Having A for one ſtation, or place of obfervation, you müſt make
choice of another which let be B, and the diftance A B 145 feet,
place your theodolite at A with its face parallel to the horizon and that
end of the diameter where the degrees begin towards B, then along
the faid diameter, or thro' the fights on it view B, ther e fcrew the in-
ſtrument faſt, and turn the index till through the ſights thereon you
ſee the object C, and fuppoſe the index then cut 80, (viz.) tó move
over the arch a e. 3. This done come to B, and obferve your firſt ſta-
tion A along the diameter, then turn the index to C, and as before
let the L A B C be 44° and you have done your obfervation. To lay
down which is the very fame with problem 125, for in the ▲ A
B C is given the fide A B 145 and the Ls CAB 78º, and A BC 44°
which laid down (as by fig.) you'll find by the diagonal feale A C=
120 the anſwer. Ar. as fine L B C A 58° : fine L A B C 44° :: A
B 145: AC 120 feet.
PROBLEM CXXXVIII.
To find the diſtances of feveral places fuppɔſe two C and D, from
any propoſed ſtation A, by any inftrument that can take an L.
Fig. 107.
1. You may work this problem like the laſt only becauſe here are
two objects, you muſt take two Ls at A which let be D A B = 520,
and D A C= 26°. 2. Meaſure from A to your ſecond ſtation B which
fuppofe 145 yards, there take two Ls again (viz.) L A B C= 44° and
LCBD 38°, then to lay this down on paper, make A B 145
62
THE UNIVERSAL MEASURER
=
from a diagonal fcale, (by problem 10) upon B make the Ls A B G
44° and C B D = 38°, alfo, upon A make the Ls D A B = 52° draw-
ing A D till it cut B D in D, and L D A G≈ 26° continuing A È till
it cut B C in C, fo is D and C the required object, and by the fame
diagonal fcale will be found AC 120, and. A D 200 yards the dif-
tance between A and the two objects C and D; alfo, by the fame fcale
you may find B C and B D their diſtances from B and C D their dif-
tance afunder. The arithmetical calculation, is cafy from the laft
problem.
PROBLEM CXXXIX.
To meaſure any level field a b c d, which can be all feen from t
different ftations, and not come near it. Fig. 108.
1. This is but to take two convenient ſtations F and S, and on each
of them make obfervations to every L or corner in the field &c. (by
laft problem) thus placing your theodolite at F, look for S and every
corner, and let the degrees cut by the index between looking at S, and
d be 26°, between d and c 34°, c and b 25° b and a 16°. 2. Measure
the diſtance F S which let be 120 chains, and placing your inftrument
on S look for F and every corner of the field, and let the degrees cut
by the index between looking at F and at a, b, c, d, be 40, 18, 24, 36º.
To plan this, draw a line FS 120, upon F make all the Ls as you
took them at your firſt ſtation, and upon S as you took them at the fe-
cond, drawing Sa till it meet F a, S b'till it meet F b, Sc till it meet
Fc, and S d till it meet F d. Join thefe points a, b, c, d, and you'll
ave a true plan of the required field, which by the fame ſcale you took
F from, you may find the diſtance of every corner a, b, c, d, from
each ſtation F and S, as alſo their diſtance from each other, and alſo
meaſure a diagonal and two´Ls, and fo find its area.
Note. In the fame manner you may take the plan of a field at two
ftations within the field, and ftill mind to take an L between each ſta-
tion and the next corner when you are upon the other ſtation, for its
plain the figure could not be laid down if you wanted the L F Sa and
!
AND MECHANIC.
·63
FS; alfo, let your ftations be as far diftant as you conveniently can,
then the points a, b, c, &c. where the lines meet will not be ſo acute,
for in fuch ſmall Ls it is not eaſy to fee where the lines interfect one an-
other, and confequently the plan cannot be true.
PROBLEM
CXL.
$
To find the diftance of fome place as A by three flicks. Fig. 36.
A
1. Set up a ſtreight flick any where as at C, from which move in a
right line towards the object A any diſtance fuppofe to B 200 yards,
there ſet up another ſtick, and move from B any diſtance in a right line
› acroſs E B as at E 100 yards, there fet up another ſtick; then come
back to C and move in a line parallel to E B, till over the ſtick you
can fee the object A, and fuppofe C D'the distance moved to be 300
yards, then it will be as DF-200 the difference between CD and BE
EF=BC 200 ; : CD 300 to C A 300, or :: EB 100: BA 100,
the object's distance from C and B, for all the As are fimilar.
PROBLEM CXLI.
Fo take a distance A B as the breadth of a river &c. by a carpen-
ter's or maſon's ſquare E D F. Fig. 109.
1. At A the rivers fide ftick down a ftreight ſtaff L to the horizon,
upon its top D hang a fquare F D E, fo that along the fide D E you
can juſt fee B the farther fide of the river, and at the fame time mind Ĉ, a
- point on the ground where D F the other fide of the fquare points, and
let AC be two feet and A D 8 feet; then by problem 45, it will be as
CĄ 2: AD8AD8: AB 32 feet the anfwer.
Note. If you flit the head D of an upright Raff A D, and in that put
a ſmall ſtick D E, bending it up or down till along the fide D E you can
just fee B the farther fide of the river, then turn the ſtaff D A about and
obſerve where D E points on the land, which point meaſure from A, and
you'll have the diſtance required.
GA
THE UNIVERSAL MEASURER
The two following Problems include
The Art of LE VELLING.
PROBLEM
CXLII.
3
To find the level between two places, fuppoſe A and I. Fig. 110.
1. For this purpoſe you may make an inftrument by faſtening a fquare
ea upon the head of a ſtreight ſtaff A B, and at the LB faften a thread
and plumet c a e; now it is evident, when this ſtaff A. Bis upright,
the thread ce will (hanging at liberty) juſt touch the fide ca of the
fquare. 2. Having thus placed the ſtaff at A in a poſition, obſerve
Calong the fide Bn of the fquare, or, which is better, thro' two fights
placed thereon; then meaſure the diſtance B C 10,6 feet for the level
between A and C, alfo meaſure the height A B 5,5 feet for the height
of C above the level of A. 3. Fix your inftrument at C and do exactly
as you did at A, and you'll fee the point E, and thus do till you come
to fee I, the top of the hill, and let the dimenſions be as here fẹt down,
BC = 10,6
The feveral DE = 10,8
levels are FG = .5
The feveral
heights are
A B = 5,5
CD=2
EF=
2,6
GH = 1,7
The fum is IK = 11,8
1
HI = 2
and A K = 28,4
Hence it appears, that if you dig a pit at
that if you dig a pit at I 11,8 feet deep, there will
be level for the water, and that level will be 28,4 feet long. But that
water in ſuch caſes may have a little current there is an allowance made
of 4 or five inches for every mile or 1760 yards; i. e. if a level be 1760.
yards long, one end of that level muſt be 4 or 5 inches lower than the
other, that the water may run off; and ſo in proportion for any other
length.
PROBLEM
CXLIII, ..
It is required, to know whether water can be carried from 1 to A
(by the level, mentioned in definition 33.) Fig. 110.
1. Take two freight poles, of any proper length each divided into
inches and decimal parts of an inch. 2. Set one of thefe poles upright
AND MECHANIC.
at A, and the other any where elfe conveniently, as at E. 3.
3. Fix your
inftrument at fuch a point C as you may therefrom fee both theſe poles
then look thro' the fights on the level for the pole fet up at A, cauſing
fome affiſtant to move a piece of white paper up or down the faid pole
L A, that you may juſt ſee the upper edge of it which in this figure i
at B, ſee what inches the faid edge cuts, which fuppofe 90 inches from
A to B; obferve the fame method in looking for a paper fo moved upon
the ſtaff at E, and you'll fee the point E, fa is A B the height, and EB
the level of the point E. Then remove the ſtaff at A to another point
as H, letting the ſtaff ſtand at E, and on fome point between E and H
do as before, and thus going on you may foon level any fuch place, as
is plain by the figure.
This is the apparent level and will do in ſhort diſtances without cor-
rection, but if the diftance be long there must be made an allowance,
on account of the earth's ſpherical figure. Thus, let A be the center
of the earth (fig. 44) A E and A G two half diameters, Ed a tangent
to the furface at E, now if the level be placed at E it will give the ap-
parent level E d, between the places E and G, whereas the true level
is the arch EG, fo that the apparent level exceeds the true one, E
G by the line Gd, for which reafon this line G d must be found and
taken from the apparent lev el to get the true level between the places
E and G; thus, (by theorem) 2 A G+GdxdG=□ Ed, or, be-
caufe A G is great in refpect to Gd, at any diftance under 30000 yards,
we may fafely take 2 A GxdG□Ed, and then Gd=
Ed
2 Å G
!
6,222 parts of a yard, when Ed is taken - 1760 yards or one
mile, (the radius AG of the earth being about 6966412 yards) now,
0,222 × 36 = 8 inches nearly. Hence the difference between the true
and apparent level is 8 inches per mile, and fo in proportion for any
other diſtances.
2
Note. By this method you may eafily calculate a table of corrections
for diſtances.
L
5+
THE UNIVERSAL MEASURER
PROBLEM CXLIV.
Ede.
To furvey mines, . Fig. 111.
7. Over the top of the pit lay a piece of wood, from the middle of
which let down a line and plumet to the bottom, by this you'll know
how to place your inftrument fo as its center may be in the fame 1 line
both at top and bottom, on this inftrument fee definition 32) the de-
grees begin at north and are numbered eastward to 900, fouth 1800,
welt 270°, and north 360°, where they again begin; wherever you
place
this inftrument for obfervation be careful to place it horizontally
on a board, &c. fo that it be perfectly level, and let the north end of
the needle point north where the degrees begin; having thus placed it
at the bottom of the pit, hold the end of a fmall line over its center,
and let an affiſtant go with the other end as far as he can go in a direct
line, obferve what degrees are cut by this line, fuppofe 200 or 200
weftward of ſouth, meaſure the line which fuppofe 7 fathoms, fet theſe
down, and remove your inftrument to where your affiftant ſtood, and
there work as before; and thus go on as far as is neceffary, and let
the dimenfions be as follow;
7
Distances
{}}
!
5
11.5
2007
$20 weitward of fouth.
degrees cut 180 or, due fouth.
335
(25° weft. of north,
This done it is common to place the inftrument with its center on the
top of the aforefaid line, and fo by the degrees meafure out the de-
grees exactly as they were taken at the bottom, and inſtead of meafur-
ing thefe diftances in fathoms, &e. they have fmall pieces of wood
numbered 1, 2, 3, 4, &c. put into the twine of the line at the
ends of the 1ſt, 2d, 3d, and 4th, &c. diſtances, which marks fhew how
far you are to go above ground in every turn; bat this method may
fometimes create great if not impracticable labour, in going thro' woods,
water, &c. to prevent which it will be beft to lay down a figure care-
fully from the dimenfions taken, and fo lay off the thing at once, thus
draw a line N. S. which make at large, N reprefenting north and S
fouth, on N make the L$ NA 20°, and from a diagonal feale lay
AND MECHANIC
ก
7 og N A, from N to A, then thro' A and parallel to N S, draw o s, foë
another north and fouth line, and lay 5 from A to B (becaufe the fe-
cond diſtance was fouth) on B make an & A B C =25ª, laying 11,5
from B to C, join C N, which is found by the fame diagonal ſcale =&
fathoms, alfo thro' C draw the north and fouth line d Sparallel to пS,
and meaſure the L NC d, which is found about 560 from the north
eaftward; therefore, if from the top of the pit you meafure out 8 fathoms
over 2360 or fouth 560 weſtward, it will point out the place at top
where you meafared to below.
The 7 following problems will exercife the fkill of the young ftu
dent in
G
M E. T
E. TR Y.
PROBLEM CXLV.
Three ſhips A, B, and C, fail from one port P; A fails due fouth
24 miles, B ſouth weſterly and C south easterly, each a diſtance.
unknown, but B's is greater, B and C are 78 miles afunder,
and the fum of their courfes is 98°, the three ships are all in the
fame east and west line. Quere, the courses and distances failed
of B and C. Fig. 112.
1. Draw È W for the eat and weft line making it 78 miles.
2. Make the Ls C WE and CE W each 8°, and with the radius
CE = C W upon C defcribe the circle. 3. Parallel to EW and at
24 diſtant from-it draw QP, cutting the circle in P and Q either of
which may be the port according as B's or C's diftance is greater, join
PE and PW, draw P A to E \, fo doth P W reprefent B's diſt-
ance failed, the ↳ A P V her courſe, PE C's difiance failed, the £
A PE her courſe, PA 24, A's diítance failed; then by the fime
parts that EW and A F were taken from, you'll find P W = 69,3
68
THE UNIVERSAL MEASURER
27,7, and by the chords, LWP A is found
miles, P E
69,45'
and LE PA 28,15′ as required. If the L WP E≈ 980, then (by
theorem 6) the arch W D E muſt be twice 98 = 196°. Then the
arch W PE muſt be (360 - 1960)
=
GE, the LC WELWEC, each
from 180 and divided equally.
PROBLEM
164º, and becauſe C W =
80 WCE 164° taken
—L
CXLVI.
A ſhip at S obferves three ports A, B, and C, whofe distances are
known to be A B = 99, A C = 71, B C 86, and the angles
of obſervation are AS B = 127°, ASC = 128°, ſo the L
BSC must be 105°, (for they all muſt be = 360°) how far
is fhe diftant from each port. Fig. 114.
1. With the three given diſtances 71, 86, and 99make a ▲ ABC,
now becauſe L ASB 1 270 is oppofite to A B, therefore, as in the
laſt problem, make L E ABLEB A each = 37°, and upon E
where the fides meet with the radius E A E B defcribe the arch
A SB, alſo L BS C being 105° you muſt make L D C B = LD B C
cach 15°, (ſee the laſt problem) and upon D with the radius DC or
DB deſcribe the arch D S B, croffing the laſt arch in S the place of
the ſhip, whoſe diſtance from the three ports (by the diagonal ſcale)
is found S A= 40,5, S B=67, and S C = 42.
**
PROBLEM CXLVII.
A ſhip at S obſerves three ports A, B, C, whoſe diſtances àr
known to be A B = 80, AC=72, and BC= 120, the an-
gles under which ſhe obſerves theſe ports are CSA = 25°,
B S A = 19º, ſo CS B muſt be 44°. Quere, her diſtance
from each port. Fig. 113.
1. With the three fides make the ▲ A B C. 2. Upon B make the
LD B C = 25º, and upon C make the L B CD 19° drawing C D
and B D till they meet in D. 3. About the ▲ BDC defcribe a circle,
lay a ruler to D and A, and draw the line A D s cutting the circle in
AND MECHANIC.
69
$ the place of the fhip, join S Band SC, which by the diagonal ſcale
are found SA = 164, SB = 95, and S C = 158, the diftances fought,
if the port A was fartheft from the fhip, but if it was neareft, then the
diſtances are P a 96, PB 167, and P C 149. For, by theorem
, the Ls DSB and D C F ſtanding both on the fame chord &c. are =,
and for the fame reafon LCS DLCBD, the fame holds when a
is one of the 3 ports. Theſe 3 laft problems fhew how a figure may
be conſtructed, when an angular point or points is required to be found,
&c. The 3 following ones are on the fame fubject.
PROBLEM
CXLVIII.
In'a plane ▲ A B C (Fig. 115.) is given the bafe A C, with its
oppofite L A B C, and length of a line B D drawn from the ſaid
L to bifect the faid fide A C, to conftruct the ▲.
1. On the given fide A C (by problem 61) defcribe the fegment
A B C of a circle fo as to contain an L A BC the given one. 2. With
the line B D in your compaffes, and one foot in D the middle of A C,
fweep an arch to cross the fegment in B. 3. Join B A and B C, fo is
A B C the A required. If the line B D were to bifect the given L
A B C inſtead of the bafe A C, then in practice, you may make the L
ABC the given one, which bifect with B D, making it the given
length, then prick A C on the edge of the ruler, which apply to the
point D, fo as theſe two pricks may fall on the lines B A and BC, and
its evident the A will be formed.
a
PROBLEM CXLIX.
In à ▲ is given an L, the ratio of the containing fides, and the ra-
tio of the fegments of the baſe, to make the A. Fig. 116.
1. Draw a line P C at pleaſure, in which take it as A E: E B ::
greater fide : leffer ſide; and as AF: FB:: the greater fegment of
the bafe: the leffer fegment. 2. Upon A B (by problem 61) make a
fegment of a circle to contain an L AR B given one P QT. 3. Make
it as E C : E A :: E B ; E ALEB, and with the radius F. C fweep an
70 THE UNIVERSAL MEASURËR
يع
arch E QR, cutting the laft arch in R, join R B and B A, fo is ARB
the ▲ required, for (by theorem 20) as AR: BR:: A E: B E, if
the 1 were F Qgiven initead of the L, then upon E raiſe the E
and thro' parallel to R A and R B draw QP atid QT, ſo will P
QT be the required A. For by reafon of thefe parallel lines it will
be every way fimilar to the AAR B.
PROBLEM
CL.
Given the 3 Ls 45°, 56° 15', and 78° 45' with the 3 fides in one
= 100, to find the fides feverally. Fig. 117.
1. From a fcale of equal parts make K Proo the given fum.
2. Make the L9 P K p= 45° and p P K 56° 15'. 3. Bifect theſe
two Ls and at D where the bifecting lines meet, draw D B and D C
parallel to K pand p P, fo is B CD the A required.
fcale of equal parts, you'll find DB 33,01, DG
And by the fame
28,06, and BC
what num-
= 38,93. Arith. you may fuppofe any one of the fides
ber you pleafe, and then by it and the Ls (by axiom 2.) find the other
two fides, and it will be (by fimilar As): fum of the 3 fides thus had
: any one of them :: 100 the given fum: the true fide, and fo may you
find them all three.
Demonſtration. The LDB C=p KP and L PKD = L K D B
LpK P, whence, D B-KB; in like manner it may be proved,
that DC-P C, therefore DC+DB+B C=100 K P. Q. E.D.
PROBLEM CLI.
Let A BCD repreſent a rectangular billiard table; required in
what direction a ball from a given point P muſt be ſtruck fa that
after three reflections it may fall into a purfe at the LB. Fig. 118.
I.
Parallel to B D draw P K, making H K twice B D, upon K
raiſe the LK M making it = A H+C D, join M P fo is d PM the
direction required. For if the ball P be ftruck in the oblique direction
ds it will ſtrike A B in E, and from thence reflected back to F, from
F to G, and from G to B, ftill making the Ls of reflection and incidence
#
AND MECHANIC.
71
(fee theorem 164) whence the Ls H, A, C, and D being right ones,
the As PHE, F E A, F C G and B G D are all fimilar, and per fig.
PH+AF+FC+ B D is given PH + 2 BD and E H +
EA+GC + G D is alſo given AH+CD, therefore, as PH +
2 BD:AH+CD:: PH: HE, which gives the above conſtruction.
000000000
000000000000
The 12 following problems are the moſt uſeful in
SOLID GEOMETRY,
For fhewing the nature of folids, &c.
PROBLEM
CLII.
From a given point in any plane to erect a to that plane. rr. E. 12.
In common practice you may take a carpenters fquarc, which ap-
plied to the given point will direct the L.
PROBLEM CLIII.
To know if any wall, &c. be to the horizon.
Stick one end of a ſtaff into the ground with a line and plumet faſt-
ened to the other, then if the wall, &c. be L you will by ſtanding at a
fmall diſtance from it obferve it to be parallel to the line and plumet.
1
PROBLEM
CLIV.
To make a pyramid or folid angle. Fig. 119.
1. Suppoſe you would have a fquare pyramid, then make 4 = ifo-
celes As, join their vertexes together laying their fides along each o-
ther, and you'll have the form of a fquare pyramid A v BCD made
of the 4 = As A B v, ADv, CBv, and CDv; after the fame manner
may a pyramid of any number of fides be made, for every fide of the py-
ramid has a ▲, be its fides more or lefs. But here it is to be obſerved,
that the fum of the vertical 4s of the plane Ls which conffitute a ſolid L
72
THE UNIVERSAL MEASURER
muſt be less than 360°, otherwife the angular points when connected
will be all in one plane, as in a circle's center. Alfo, any two of the
faid Ls however taken muſt be greater than any one of the other Ls,
or they cannot poffibly form a folid L.
PROBLEM CLV.
How to make a cone. Fig. 120.
1, With any convenient radiús deſcribe a fector A v B. upon paſte-
board, ſtiff paper, or the like, then cut it out cloſe by the edges A ▼
and Bv, as alſo by the arch A B, then this turned together fo as v A
lies upon v B will form a cone. By the fame method you may make a
cylinder with a rectangle,
PROBLEM
CLVI.
Given do the radius of a sphere, to find the fide of any of the five
regular bodies infcribed therein. Fig. 121.
1. Having defcribed the femicircle divide the diameter into 3 = parts,
viz. da bab r, erect the Ls a e and c f, and draw e r, ed, df,
cut e d in extreme and mean proportion in h, upon r raiſe the Lr G
making it ➡rd, join c G cutting the circle in n, join rn, now if we take
rd the axis of the ſphere 2, then by the nature of a circle we have
1. re= 1,62299)
1,15479
[Tetrahedron 1
for a fide | Hexahedron
2. df
infcribed in
3. de=1,41421
4. dh0,71364
of the
greateſt
Octahedron
Dodecahedron
that ſphere.
5. rn=1,05146)
Icofehedron
See the five regular bodies in the explanations, as well as the five fol-
lowing problems.
PROBLEM CLVII. CLVIII. CLIX. CLX.
CLXI.
To repreſent all the five regular bodies.
AND MECHANIC.
73-
1. A tetrahedron is a folid contained under four equilateral As and
is no more but a triangular pyramid whoſe baſe and three fides are all
equilateral As, and is formed by turning up the three As A B and E
upon the fides of the AD. Fig. 122.
2. A hexahedron, or cube, is form'd by turning up the fours
mark'd 2, 4, 5, 6, upon the fides of the
the fides of the mark'd 3, over which
as a cover bend the
mark'd 1. Fig. 123.
3. An octahedron, is a folid contained under 8 equilateral ▲s, and
is formed by folding the As in Fig. 124.
=
4. A dodecahedron, is a folid contained under 12 equilateral and
equiangular pentagons, and may be formed by folding up the 12 regu-
lar pentagons, in Fig. 125.
5. An icofehedron, is a folid confifting of 20 triangular pyramids,
whofe vertexes meet in the center of a ſphere, which is imagined to
circumfcribe it, and may be form'd by folding &c. the Ls in Fig 126,
Note. Theſe figures fhould be drawn on ſtiff paper or paſte-board,
and then in the paſte-board cut the lines half way thro', and bend or
fold in the feveral planes of which the body confifts, and they will clofe
together in fuch fort as to form the body deſigned.
PROBLEM
CLXII.
To find the axis of any cone or pyramid, or of any fruftum thereof.
Fig. 127.
1. Meaſure any length from B the bafe as B C, with a ſtreight ſtaff
or line. 2. Meaſure from C to u, (in a direction parallel to the baſe
AB) to the fide u B of the folid, then it is evident, that B C is LD
— u o, the axis of the fruftum A B C G, and Cu (Bo) taken from
BD leaves Lu (-Do) which doubled gives Gu the breadth &c. of
the leffer bafe of the fruftum, make A a G F and Ae Gu, fo will
As A v B, a E B, and eu B
E a and ue be parallel to A v, whence the
will be alike, and it will be as e B (A B — G u) : o u, or: e a :: A B
:Dvor: A v, and :: a B (AB - EF): ǹ E, or : D F, &c. whence we
have a rule to take the dimenfions of a growing tree, &c. thus, ſuppoſe
a tree be 20 feet high, 40 inches girt at bottom, and 38 inches 3 feet
K
74
THE UNIVERSAL MEASURER *
*
.
above the bottom, it will be as 3 feet is to 1 inches (40 — 381) fois 20
feet to 10 inches, which taken from 40 leaves 30 inches the girth at top.
The fame holds true in any fruftum of a pyramid, by ufing the peri-
pheries at the baſe and what diſtance from it you pleaſe; or inſtead of ·
the peripheries you may take two fimilar fides, and ſo find a like fide
at the top,
PROBLEM CLXIII.
To take the dimenſions of any folid, ſuppoſe of a caſk, HB H L
dDdE, Fig. 128.
1. Cloſe to each end of the caſk ſet a ftreight ruler as A P and CQ
to EL its axis. 2. Upon the bung B lay another ftreight ruler A C,
parallel to the faid axis EL, and meeting the former two rulers in A
and C, then it is evident that C A will be HHEL the cafk's
length, and that twice A H added to Hdis BD the bung diameter.
Alfo, twice Nm taken from BD leaves a diameter m D, as taken atm,
between the bung and the head.
Note. The thickneſs of the cafk and what the ftaves overshoot the
heads muſt be confidered. Alſo if the heads of the cafk be, then A C
will be parallel to E L when A His CH, but if the ruler A G is not
parallel to E L, it is plain by the figure, that the ſum of the parts CH
and A H added to Hd will give BD.
If the caſk ſtand upright, as fig. 129, you may do it eaſier thus, lay
a ftreight ruler d Q over the top of the cafk with a line and plumet QP
at the end Q; now it is plain that if the cafk beQP will be paralleito
EL its axis, and that twice HQ added to H d gives the diameter B D,
the line touching the cafk in the point B, alfo twice n m taken from
BD leaves the diameter n D, &c. after the fame manner you may take
the dimenſions of other folids as occafion may require.
The End of the First Part
!
X
淡菜​糕
​礼
​THE UNIVERSAL
MEASURER
AND
MECHANIC.
*************************************
PART SECOND.
CONTAINING
The Principles of ALGEBRA, GEOMETRY, TRIGONOME
TRY, MENSURATIONS, CONIC SECTIONS, SURVEYING,
GAUGING, MECHANICS, PROJECTILES, HYDROSTATICS,
HYDRAULICS, MECHANIC POWERS, PENDULUMS, CEN-
TERS OF GRAVITY, PERCUSSION, OSCILLATION, LAWS
of MOTION, DESCENT OF HEAVY BODIES, PNEUMATICS,
PUMPS, BAROMETERS, STRESS and SRENGTH of WALLS,
BEAMS, &C. WHEEL-CARRIAGES, RESISTANCE of FLU-
IDS, &c. &c.
I
T is comn:on for writers on algebra and geometry, to add a collec-
tion of queftions by way of illuftration; however I have ventured
to deviate from the common method, by fupplying this fpeculative part
with uſeful articles and general theorems analytically demonſtrated
which I hope will be thought as good excrcife as the folutions of ma-
ny queſtions, and in the mean time furnish a book with a large collection
of principles applicable to the follations of queftions in many branches
of the mathematics and natural philofophy: fuch a collection and fo
digeſted, I prefume cannot be found in any other book, which I hope
will be thought worthy of my reader's notice.
76 THE UNIVERSAL MEASURER
DEFINITION. An algebraic procefs is no more than a
comparative way of reafoning, and is founded on theſe fix
fundamental axioms, viz..
1. Quantities that are equal to one and the fame thing, are equal
themſelves.
2. If quantities laid on each other agree in each part, they muſt be
equal. By this axiom many of the propofitions in Euclid may be de-
monftrated.
3. If equal quantities are added to or taken from equal ones their
fums or differences will be equal.
and
5.
4. The whole is equal to all its parts taken together; and if
quantities are equal, their like parts, as halvės, doubles, trebles, &c.
will be equal.
6. But if quantities are unequal theſe parts or fums, &c. will be un-
equal, and that which was greateſt will remain fo, &c.
PROBLEM
CLXIV.
Algebra is an art by which the moſt difficult problems in arithmetic
and geometry are folved, by affuming fymbols or letters for the things
required, and thofe given alſo if you chuſe, for this purpoſe the initial
letters in the alphabet are uſed, and the final ones for unknown quan-
tities. Theſe letters or quantities must be connected together by addi-
tion, fubtraction, multiplication, divifion, &c. as the nature of the prob-
lem requires; which done, obferve if the unknown quantities exceed
the known ones in number, the problem is often impoffible, but if con-
trary it is capable of many anſwers; and when the number of known
quantities is equal to that of the known ones, the problem can have but
one anſwer, which anfwer or anſwers are had by working out all the
unknown quantities except one', and this one being on one fide of the
fign or equation by itſelf, and the known ones on the other fide there-
of, the value of this unknown one is thereby known, by taking the va-
lues of the known ones in numbers and working with thefe numbers as
the connections of their refpective numbers direct. Any letters may be
taken, or repreſent any numbers univerfally; ſo that in folving any
queſtion algebraically you'll have a rule or theorem for all queſtions
of that nature. Obferve carefully that whatever you do on one fide of
AND MECHANIC.
77
the equation, you muſt with the fame value do the like on the other fide,
otherwife, it is plain, the thing-will lofe its equality. In vanishing the
unknown letters, you muſt uſe ſuch means as you think will beft and
caſieſt do it, whether by addition, fubtraction, &c. Alſo, for every
different value in your problem or queftion, make choice of different
letters to prevent miſtakes.
Of SIGNS and CHARACTERS,
their names and fignifications in algebra and geometry.
= Equal to. 10=10, a=e, i. e. 10 is equal 10, or a equal e.
+Plus or more.
Minus or lefs.
aez, i. e. a more or added to cis equal z.
a-e-z, i. e. e taken from a leaves z.
× Multiplied. axe or ae, either of which denotes the prod. of a and e.
Divided by a÷e, or
or Square root.
a
fhews that a is divided by e
e
9, or 93, the fquare root of 9 is
equal 3.
or Square. 9 or the fquare of p is to be taken.
:,::, :
19 ¯¯¯|²
Proportional. 2:4:: 6 : 12, as 2 is to 4 fo is 6 to 12.
Lefs than. 46, 4 is lefs than 6.
Greater than. 64, 6 is greater than 4.
... Ergo, or therefore.
L Angle.
RL or
Right angle.
Parallelogram.
Δ Triangle.
Circle, or the fun.
Parallel.
Minus and Plus.
Plus and Minus.
Involution.
to Evolution.
C Complete fquare.
axexu, or a.e.u, or a eu, denotes the product of the quan-
tities a e and u.
(47 E. 1) the 47 propofition of Euclid's first book.
Ls Right angles.
Ls Angles.
s Parallelograms.
. Parallel to.
Thenth root. ———
As Triangles.
s Squares.
Ls Perpendiculars
Proportional to.
√ Cube root.
3
The 'n power.
-³ Cubed power.
✔ate or √: a te,
n
or a+e all thew or expreís the root,
whofe index is n is to be taken of a + e, alfo a+e" fhews the fum
of a and e is to be raiſed to the power denoted by n, and a +exu,
or : a+e: xu denotes the product a and e multiplied u.
denote 2 degrees 5 minutes 6 feconds.
2°, 5′,6"
2
78
THE UNIVERSAL MEASURER
Numbers prefixed to quantities are called coefficients, and ſhew how
often the quantity is to be taken, as 5 a denotes 5 times a, 6 a e, 6
times the product of a and e,
by twice e. When letters are joined together without the figns + or —
ча
2 e
or ja ÷ 2e is 7 times a divided
e
among them they are called fimple quantities, as a, 2 á e,—
ie, 2 other
wife compound quantities fuch as a † e. a
a
etcd, a e ghe.
Quantities that have the fign + or no fign before them are poſitive, or
—
a_cd
affirmative quantities, and thoſe that have before them as
are negative quantities. Every quantity has its fign on the left hand
fide implying its pofitive or negative relation. Multiplication joins the
quantities together, and quantities divided as vulgar fractions, the di-
vidend above and the divifor under a line drawn between them.
N. B. Quantities may be ranged in any order without altering their
value, as 5 a b—c, or c+5 ab, or 5 ba — c, &c. are all the fame,
for c is ſtill negative and 5 a into b poſitive.
Figures following quantities are called indexes or indices, or expo-
nents; thus, a is
m
√aaa, :a+e:implies a +e is to be involved
to the m power, and evolved to the n power, fuch are called furd quari-
tities. To illuftrate theſe things more clearly;
Let a = 6, b = 5 and c = 4, then will
aa+3abcca+3b ac² 36+ 90 16110
—
2 aaa-3 aab+ccc2a³-3 a¹b+c³-432-540+64-44
ax:a+b:-caa+ba-c36+30-462
√2ac+cc = 2 a c + c² | 2 = √: c² + 2 ac := √ 64 = 8
ac | +2ac: √64
ааа
૨૩
2+ c 2 + c
216
10
=21.6
a3
તે
216
C C =
16 = 21,6 — 16 = 5,6
2+ c
Na
1Q
№6
a
6
but √
=√²= √1,5. And ſo on.
C
4
4
2
If no figure be before a quantity, you are to ſuppoſe an unit before
it, ſo 1 a, ¡c, are the the fame as a, e.
AND MECHANIC.
79
Iate=s
1 + 2
3
WN
2
a
- d
2a=s+d
3 X 5
}
55+5 d
4 10 a = 5s +5
On the left hand ſide of algebraic operations is ruled two lines with
figures between them called fteps, and thoſe on the left hand ſide of the
ſteps are to fhew how thefe operations are performed; ſo here againſt
the third flep we have 1+2, i. e. the quantitics in the 1ft and 2d ftep
are added together and their fum placed in the 3d ftep; alfo againſt the
4th ſtep 3 X5, i. e. 5 times the quantities in the 3d ſtep are placed in
the 4th ſtep. Obſerve, where a daſh is over a figure as 5, it fhews that
5 denotes the number 5, without which it would denote the 5th ſtep.
Some authors do not make uſe of ſteps, tho' I think them much eafier,
as they fhew how every ſtep is wrought. Quantities as a and a, or
2 b c and c b, or 8 da e and + 9 d a e, &c. confifting of the fame
letters are called like quantities, let the figns and coefficients be what
they will, and vice verfa,
PROBLEM CXCV,
To add algebraic quantities.
Rule. When the quantities are unlike they can be no way added but
by ſetting them down with their reſpective ſigns before them; but if they
are alike and have the fame figns, the coefficients are added together
and to their fum the quantities are annexed and figns prefixed, but if
the figns are unlike, you must take the difference instead of the fum of
the coefficients, and fo prefix to it the fign of the greater cocfficient,
for a negative quantity is always leſs than nothing, fuppofe a man is only
worth 10051 and owes 2000, the true fum or ballance will be
his real worth, or 995! worfe than nothing.
I + 2
I
1 + 2
I
EXAMPLES.
I. 2. 3. 4.
5.
7.
1 aaa17a-6b| 5a+bd bd-cc-/a+d
a
-9951
a a | 12a-b12bd-ga la+db-10ce—e
2 a|—2a| o |29a −7b|3vd+2a 1200—1 rec+d—√o
I
2.3
8.
9.
a+via-e:+5 latviate: +5
2ata—e: -10|2a1/:a~e:
10
3a+21/:a~e:—5/3a−5+√:a+e:+v/:â—e:
80
THE UNIVERSAL MEASURER
10.
-Jo√:ctu: +ve
u:
u:
2
√:etu:
I + 2:
3/9 √e + u :
1.1.
I
aa aa a
a
3
"
1+2+3
2 3
4
√ aaa+ ³√ aa√ a
✔aaaaa:
3√ a
3
aaa+aa—a+√a³+a³|³—va+v:a³+a²: -a]
If the indices of furds be unlike, as a and³a, or the quantities
within the furd fign be unlike, as: a +e: and
a +ee: or have.
unlike figns as: a-e: and :e
e: and √: e—a: it is evident they have dif-
erent values, and fo muſt be added as if they were different fimple
quantities, as in the foregoing examples.
PROBLEM
CLXVI.
Subtraction of algebraic quantities.
Rule. Subtraction being the reverſe of addition. If therefore you
ſuppoſe all the figns in the fubtrahend to be changed, viz. — into + and
-, you may work as in addition.
-+ into
12. 13.
14.
15.
16.
1 | 2a | -2 al
2 a 1
a |
0 129a-7b13bd+2a
+ a | 12 a − b | 2 b d − 3 a
·3a-e
I-2
31 a 1
a |
a | 17a-6b| bd + 5a + e
17.
18.
19.
I
2bd-11c-e
a+va aaa+a³
2
e11c-2bd
1/a-a
a3_ aaa
I-2
3
о
O
2+21/d 2√ ada
.
/:a
e + 2:
20.
a + c
a -e+2:
a + c
et?:
+
a + ċ
દી
I-2
3√ √
a
: a
e+ 2:
a + c
AND MECHANIC.
81
PROBLEM
Multiplication.
CLXVII.
Rule. Multiply the coefficients of fimple quantities together, and to
the product annex the quantities, and prefix the fign + if both figns
were or both
but if one was + and the other you mu't prefix
←, i, e. † into + or - into gives + in the product, but — into
See examples 86 and 87.
22. 23. 24.
26.
2:a-ee:
+
gives
21.
a
a
16 a
25.
+180e
2
a
IY2 2 aa
a a
-aa
a+a+2 a a
3 2 aaa
2 a b
360abe 14 aaec:a-
7aae c
-ee:
By ex. 21 and 22, it appears, that to multiply two different powers
a² and a5 &c. of the fame quantity a, is but to take the fum of their in-
n m n+m
dexes 2 and 5, i. e. a* xa5a7, and univerfally, (ex. 27) a xa a
V
e v. e vm+en
Alfo, (ex. 28)*+a_ant, and, an+aman
十​日
​=a
m
By
reducing the indices to a common denominator and adding them toge-
ther, (ex. 29) If any number be fquared or cubed, &c. and the furd
fign of that power prefixed to it, fuch a furd will be to the first taken
number, i. e. 3×3=9 and /9=3, alfo 3X3X3-27, and 3/27=3&c.
whence a"™ or a" "+babba alfo, aa—e—a³—àªè &c.
If the multiplier be a compound quantity you must work with every
term therein as a fimple quantity, and the fum of all theſe fimple prod-
ucts will be the product fought.
IX a
1 x b
I
- 2
34
30.
a+b
a-b
a a+ba
-ab-bb
5 ] aa—bb
3 + 4 15 laa
31.
ate—ue+
+2
a e
2 a
e
Ixa
3 a a ta e
aaae- -uea
IX-e4ae-ee Luee-
a a e
C
aee
C
aae acc
C
3+45 Jaa-ee-uea+uee+
L
82...
THE UNIVERSAL MEASURER
I
32.
Vaa -ee
a
e
I
2
33.
a e
a
دم
aa+ae"
2
I Xa
3
24
a
3
Ix-e
4
4
— ae — ee
3.+4
5
Vä
a e
ત
-- Va¹e¹—e*
5
a a
e ef
If furds to be multiplied have the fame index (as in ex. 33) you
may multiply them as whole quantities, and to the product join the
furd fign.
PROBLEM CLXVIII.
Divifion.
Rule. Infimple quantities caft fuch quantities out of the dividend as
you find in the divifor, and what remains will be the quotient; obferv-
ing that +by+or — by — gives +, but + by or by + gives
-in the quotient: that is, divifion is just the reverfe of multiplication.
See example 51.
2 a
u
34.
35.
36.
37.
38.
I
a a
a a
a a
32 a3
7 a e
2
a
a
a
e u
1-2
3
a
a
a
16a a
7 a
40.
41.
ba
m
ban
3
m
ર
e a
n- -11
ન
e
-I
2ab
I S
70
39.
- 8 a
ae
2 u e
4 a
u
2 ban
1÷2 31a
42.
Sabay
43.
15 × €³+y³]"×:a+b:
16 abbay 7 cx e³+y³]" ×:atb:
In any operation where the quantities are much compounded as in
ex. 43, it will be beit to put a fingle letter to fuch compounded parts,
(called ſubſtitution) and work with this letter as the nature of the op-
eration directs, and then in the anſwer you may write fuch powers &c,
of the compound quantity as thofe of the ſubſtituted one directs, (this.
is called reftitution;) thus, ex. 44, put A eee+yy yn and B =
a+b, then ex. 43 will ftand+15 BA÷7c BA≈
15 AB
7 CBA
where by
the rule it plainly appears that is the quotient required. If the di-
7c
AND MECHANIC.
83
1
for and dividend be both com pound quantities, let each be ranged in
one order that the greateſt term may ftand firft &c. in order, or they may
ſtand in any order to be moſt convenient for divifion, then take fuch a
term for the firſt term in the quotient as that the first term in the di-
vifor being multiplied thereby, the product may be to the first term
in the dividend, multiply all the divifors by this quotient term, and
make ſubtraction as in common diviſion, and look upon the remainder
as a new dividend, then find a quotient term as before; and thus go on
antil diviſion is finiſhed, and if at laft there be a remainder you may pro-
ceed on and fo get a feries, which is the method of putting a fraction
into a feries by divifion. It is to be obferved, that after a few of the
firſt terms are diſcovered by divifion, the law of continuation will ap-
pear, and may be carried as far as you pleaſe without dividing.
Ex. 45 and 46. Divide aa-ee by afe, and a3+5a³e+5ae²+e³ by ate.
ae)aa ee(a-e
ae)a³+5a²e+5ae³te³ (a²+4ªetes
a3 a¹e
aa+ae
-ae-ee
-ae-ee
Ex. 47. Divide e e by e… a.
a² 3
a
c—a) e e (etat—++&c.
ee-ae
તે
Lae
+ae-a*
+il
૨૩
+a² _ a³
a
€
ઘે
e
+
e
remainder +
3
ત
3
+4a³e+5ae¹
+4a²e+4ãe²
+aete³
+ae²+e³
the quotient required. In which you
fee the figns are all pofitive, and e-
very following term = the next
foregoing one multiplied by
a
whence the terms may be continu-
ed at pleaſure, which appears to be
a decreafing geometrical feries,
whofe firft term is e, fecond a, and
3
common ratio -ore toa. And
t
becauſe every whole (e-a) ee or
e e is to the fum of all its parts,
e-a
(e+a+=+=_&c.) we have this gencral rule for the fam of ſuch a
e
e
feries; viz. divide the fquare of the firſt or greateſt term by the differ-
ence of the firſt and fecond terms, and the quotient is the fum of all
the terms in the feries, the terms continued until the laſt term be in-
finitely ſmall; as for inftance, if a body move 12 the firſt hour, 10 the
84 THE UNIVERSAL MEASURER
fecond
the third, 1000 the fourth &c, here e 12 and a 10,
2008
if the body move for ever at that rate it cannot exceed.
ee
e-a
=72.
Ex. 48. Divide ey by e- 1, or which is the fame, ey
an infinite feries
e_1)ey (y+2+2 +
ey-y
+y
y-
+
y
e
y
e
Ꭹ .
e
+
remains
y
ee
У
ee
y
ее
put
into
e-I
y&c. where it appears that each followa
eee Sing term is had by multiplying its fore-
y
e3
going term by e which being the com-
mon factor or ratio, fhews the feries to
be a geometrical diverging one, ift term
y, common ratio or divifor e; hence as
before we have another general rule,
divide the product of the greateſt term
and common ratio by the difference
between the ſaid ratio and unity, and
the quotient will be the fum of all the
terms in the feries, the laſt or leaſt
term being indefinitely fmall or in ef-
fect = 6, fő in the laſt queſtion,
I whence the common
ΙΟ
3
a
€ 12
1.2
14,4
ratio is = 1,2 me, therefore ey 12X1,2
=
e--I 1,2-[
if inſtead of the laſt or leaſt term being taken
affignable quantity as a, then it is evident that
a
much by
e
I
0,2
72 as before, but
o, we take it fome
ey
e… I
ey-a
-
will be the fum too
and therefore will be the fum of
ey
a
e-I e-I e-I
thoſe terms between y and a; and if inſtead of ſuppoſing the ſeries to
begin at y the greateſt term and decreaſe to a, we ſuppoſe them to be-
y
=}{
༨.
gin at a and increaſe to y, i, e. iffory+++ &c. we take
i.
eee
a+ae+ace+ae³+&c. we fhall have ya for the fum of any geome-
e-I
trical increaſing ſeries, whoſe firſt term is a, common ratio or multiplier
e, which is a new, fhort and eaſy inveſtigation of geometrical progref-
fions.
4.
AND MECHANIC.
85
#
Ex. 49 and 50. If
a-elaae
aae-aae
aae
a-e
and
૨૩
z6-23
6
20 +223 +1
be the fums of two feries,
what are the terms of theſe two feries?
(ae +ee+
+ae²
ae¹ - e³
дез
e3
e4
Let
دم
a
a
et es
a
a a
remains +
aa
eee
+
a
e4
a á
+&c. where the law of continuation
appears.
2°+22³+1)26—2³ (1--23 + 25-2, &c.
zo+zz³+i
Z
-323-1
~32³—6— 3
+5+ 3
Z3
IO
23
+s+ 1/8 + 2560
23
Z
5
7
z3
Z
zo &c.
Z
PROBLEM
CLXIX.
Of algebraic fractions, or broken quantities.
The rules in vulgar fractions hold true in algebraic fractions. The
moſt uſeful cafes in reduction of broken quantities are thefe 4. 1ſt, To
reduce fractions to their loweſt term; 2, to a common denominator;
3, a mixed expreffion to a fraction; 4, and the contrary.
Lemma. If the numerator and denominator of any fraction be each
multiplied or divided by the fame quantity, the new fraction will be —
a
a
în value to the old one ; ex. 5, fo 2=2 X
e
e
u au
a ZZ
X
u
eu
e
ZZ
auzz &c. which is plain from the nature of divifion in common arith-
Cuzz
zo 5
metic. Alfo ex. 52, ex· =aa, and z³ × 3 = 520 =
aa
e
eaa
e
Z
Z
=5, &c.
whence to reduce a fraction to its lowest terms, we have this general
rule, caſt like quantities out of both numerator and denominator.
Ex. 53.
fo
eaau ea
auc C
and
ęzzbd zzb aa-ee
alfo,
defc fc
ate
a-e, alſo,
zz-ee z-e &c. Hence alfo it appears, the method of reducing
bz-be
b
fractions to common denominators, for if you can multiply the deno-
minator of a fraction by fuch a quantity ſo that it may be the deno-
minator of another fraction, the numerator of that fraction multiplied
by the ſame quantity, the fraction will (per lemma) keep its value, and
have a denominator or common with the other fraction.
86
THE UNIVERSAL MEASURER
a
nm
-therefore, and when reduced to
Ex. 54. fo
a
W
au
X
u
eu
e
au
eu
nm alfo 2 and
a and iD
eu
I denom. will be and alſo
ad
cb
b
eu
in a common denomina-
tor will be and which gives the common rute, viz. multiply the
bd bd'
denominators continually for a common denominator, and every nu-
merator into all the denominators except its own denominator for a
new numerator. Fractions thus reduced to a common denominator may
be added by taking the fum of the numerators, or fubtracted by taking
their difference and writing it above the common denominator; thus,
aum, alfo, the difference
a
nm
Ex. 55. b+ added to gives b+
e
eu
eu
ad-bc bc-ad
between and
a
b
с
IS=
ΟΙ
bd
bd
Ex. 56. Becauſe (per lemma) a =-
ae
and
cde
de, &c. therefore,
e
C
bt
a
eb+a
whence b+2
deubau+de
+
&c. Mixt quan-
e
a
u
eu
tities being thus ordered, their product is had by taking the product
of the numerators and placing it over that of the denominators.
Ex. 57. fo b + 2:x-beta; x neb+na
e
n
m
e
n
m
em
ae::a+e:_aa-ee &c. If fractions have a common
Ex. 58.
e
C
ec
denominator, and you divide the numerators of two fuch fractions the
quotient will be the quotient of theſe two fractions.
a
C da bc da
db db bc
alfo ate aa-ee
cd
ate
dc
Ex. 59. fo÷1
I
b
وت
aa-ee
= -in its loweſt terms, otherwife the numerator of the dividend
a-e
and denominator of the divifor is a new denominator.
a
c
dá
Ex. 60. So.
= is === and
ears
-
cea
&c. Thefe few
b d =cb b C brs
examples with the reafons and plaineſs of the rules, are fufficient for a
perfect knowledge of algebraic fractions.
PROBLEM
CLXX.
Of involution, evolution, and infinite feries.
Rule. Involution is the raifing of quantities to any propofed power,
and is nothing but a continual multiplication of the given quantities or
root into itſelf.
AND MECHANIC.
87
61.
62. 63.
64.
65.
66.
n
a
-a
-ab
2e
Va
ате
H
Q
2 2
aa
+aa aabb
4ee
G.
2
1 & 3 3
Q
3 14
aaa
23
-a³ a³b3
8e3
लोल
a
6
að
6
+20 +2ºº
64e6
a 2·3
ayaa apel
ate
3 n
6n
2n
a l-e
!-el
9n
3
35
ta⁹
-ab⁹
51269
a
ate
5 &
2
2
9
6 an
a
-ab
5.20
n
AJ
a
926
2n
ate
Ex. 67, and 68. Let it be required to raiſe the compound quanti-
ties ae called a binomial, and a — e called a refidual to the fourth
power.
I 1a Le root or fingle power.
•
fae fee
aa42ac+ce fquare, or fecond power of a † e
ate
2
аже
IX a
3
aa+ae
Ixe
4
3 +4
5
6
5 X a
7
5 x e
7 +8
8
9
IO
9 × a
9 x e
I I
a+
12
12 + 11
Again
IX a
aaa2aaeace
aae+2aeeeee
aaa +3aae+3aeeeee *cube or 3d power of a te
ale
† za³e ¦-3aaee+e³a
a³e+3aaee+3ae3-e+
13 | a++4å³e+6ª²e²+4ªe³+e¹ biquadrat of a e
I a-e root
ae
dd-de
-ac+ee
aa—zae + ce
a-e
aaa-2aaeeea
2
3
I
e
4
3 +4
5
6
5 X a
5 Xe
7
8
7÷8
9
a 3
-3aae3aee-e³ cube
ΙΟ
a-e
9 X a
II
at
9x-e
12
11+ 12
13
a
-aae2eea-es
3a3e + зааес-ае з
- a³e +3aaee-3ae³ +e+
-4a³e6a e-4ae3e4 biquadrat
6a0² + 4.00
+4004
23
Q
+ 4ão?
4 ad
32
3
1000
+ rav4.
f
aoq
420 +62
+10ão²
Errão +10 ad
3
+5æð +10 åðˆ + 10 ad + 500²+
a
-
a
a
2 a
+ 20 ad² + 10094
a.
ato.
atult and
درقه
88
THE UNIVERSAL MEASURER
n 2
From hence it appears, that if any binomial a+e, be raiſed to
power whofe index is n, the terms without their coefficients will ſtand
a" ea"-
n-1eea
I
+eeea¹³ +e¹a "4+ &c. until the index
of e ben, where it is evident, that as the index of a decreaſes that
of e increaſes in arithmetical progreffion, common difference —
If the index of the power be n, it alſo appears that the coefficients will
2 n-3
be 1, n, nx", nxx, xxx=3&c. until
X-
3 4
2
I
n. I n 2
3
2
nx
n I
2
unity.
the number taken from n leave o; i. e. every coefficient is had by mul-
tiplying the foregoing one by n, leffened and divided by an unit more
than in that before it. Ex. 69. Theſe coefficients prefixt to their refpec-
tive terms will ftand thus a"
n-3+
&c. or taking A
ece2
n-I
n I n-2
nea + n eea + n
2
n-in-2
2 3
the 2d coefficient, B- the 3d, C=
I
the 4th, D=the 5th, &c. we ſhall have A=n, B=Ax", C=BX
2
n-², D=Gx¹²3, &c. fo a² + Ae a¹¹+ Beean + Cege an−3
3
4
2
an
3
3
4
4
I
+De¹an―4+ &c.atel" by which any binomial may be raiſed
to any power without the trouble of multiplication, fo if n=4, then
A=4, B=4x4=6, C=B ×¹-2=6x-2=4, D=Cx3=4x+3=1
which put in the laſt expreffion for A, B, C, &c. and 4 for n, we ſhall
have a+el+4ea3+6eeaa + 4e3ae4, the fame with the ftep 13.
in ex. 67. And if for ae we takea-e, it will be the fame, only e-
very other fign will be negative, as appears by ex. 68; as for Ex, 70.
Let it be required to raiſe a―e to the 6th power. Here n—6—A, then
B=Ax=6x 15, and C=Bx-2=15x-2=20, and D-C
n-༢
2.
6-1
2
CBX!
3
3
{
..
X 1-320 × 6-3-15, fo the coefficients will be 1, 6, 15, 20, 15, 6, 1,
y.
X
4
for the first half of the coefficients being found, the fecond is alfo
found, the latter decreaſing as the other increaſes, (fee ex. 67 and 68)
now n being=6, the terms will be aº—a³e — aªe² —a³e³ † a³eª—àè§
+e, to which prefix their coefficients, then a°-6a5e+15a4e-
6
20a³é³15a³e¹-6ae5eae
3 م3
6
+
2
Ex. 71. Raife atu+z
to the ath power. This is the fame as ex. 69, by writing u + z inſtead
n
of e, fo we fhall have a "+ Au† za
n I
1-2
+ Bu÷z |³a +c
u+z/³an-3+ &c. = a+u+zn In this manner may any expref-
fon be raiſed to any power, or put into an infinite feries by involution.
AND MECHANIC.
89
Evolution is the reverſe of involution; i. e. as the one raifes roots to
powers, fo the other finds roots to powers. But to put any furd into an
infinite feries by evolution is the fame as the laſt example or ex.
I
69, if
for the index you take inſtead ofn. Exam. 72. Find the nth root of
ate, or put a el into an infinite feries. Here, by ex. 69, we ſhall
I
+
I.
have a "Aea¹
I
1
2
1
+ Beea + Ceeean + &c.—a en for
it is evident the method of involution makes no difference whether the
1
index be a whole or a fractional quantity, as n or Ex. 73 What's the ▷
root of a + e, or put a +
1 I
I
n
2
풀 ​X
n
I
into an infinite feries. Here A=1, B—A‚×
CBX
I
a
I
2
2
I
=
D=CX
1
n
3
3
=‚׳=-&c.
X
Ti &c. which fubſtituted in the laſt ſeries, for A,
T28
4
B, C, &c. and
for "gives a
+ea
I
2
—
- Leea
+ feeca- &c
=ae
a te But becauſe by the nature of divifion, am is the fame as
I
112
(ſee ex: 40, 41.) viz. a
a
I
a
m
m-
m+alfo,
ลูก
a
a
an
}
n-n
a za°, but
an
ал
is known to be 1, confequently aº1 or uni-
ty, let the quantity a be what it will. Likewife an÷
I
an
n
a
aan, which fhews that quantities with negative indexes become
£
2
I
divifors, and therefore a + €] = a² +
e
e
a²+2√√ a
ee
8a7 +
eee
&c.
164
But the above coefficient may be taken otherwiſe; thus, A=1, B—A
x = = =1, c =
C- + I_ + 13, D~~5-135, which ſhews the
128 2,4,6
X
16 2,46
2 2,4
numerators to be a feries of odd numbers, and the denominators a fe-
ries of even ones, obferved with the indices of a, &c. in a few of the
firſt terms, the law of continuation (as in ex. 47) is evident, and may
be continued to what number of terms you pleaſe.
Ex. 74. What is the fquare root of a
the fquare root of a -e?
it will be found√ a
1.3.5.70,5
2,4,6,8,10√a
e
I,ee
1,3,eee
By the above work
2√24a³ 2,4,64/as
13.5eeee
2,4,6,84a7
c. here all the ſigns are negative; by negative
figns of the terms, (fee ex. 70) and pofitive figns of the coefficients
(fee ex. 73.)
M
90 THE UNIVERSAL MEASURER
Ex. 75. Put aaa-eee
I
B = AX ============
2
1
into an infinite feries. Here a A, and
a =
I 2 -2
---
1,2
5
**
3
3
6 3
3,0
9
1.2,5,8
3,6,9,12
438
1,2.5, D-1.2,5
D= 1.2,5 × 1-3
-X
3,6,9 3,6,9 4
I
&c. now by writing aaà—eee
for a +e, and 3 for in the general feries in ex. 72, with theſe values
of A, B, C, &c. we ſhall have a
1,2,5.8,e**
3,6,9,12a
II
I,eee
1,2.e6
3,aa 3,6,45
1.2,5e9
3,6,9,28
&c. cube root of a³-e. By this method we may al-
fo put any fraction into an infinite feries. As for
I
Ex. 76.
e e
e-
(by
(by ex. 73)=eex e-
Here A-]=
I,
I I
2
B == 1, ==== 1, C = 1, —=—=—=—1, D=—1:===²=&c.
2
fo by putting thefe values of A, B, C, &c. with
I
4
1 for ", and e- ą
for ae in the faid feries in ex. 72, we fhall have e tae
3
+aaae
a a
a
.4
+ &c, = e
3
e e
+ &c. =
Ex. 77. Put Va Le
I
e e
d
4aac
which multiplied by ee gives e Le
£, as by ex. 47.
e-a
√ia-e:
Firſt, a+ €² =✔a+
T²=√a+
adly, a el
Now by multi-
plying thefe
terms together
and taking
the fum
viz. ✔:ate: ×´a—e]
e
into a feries,
e e
+ &c. by ex. 73-
zee
2√ª 8√ä³
I
2
1
e
√a^2₁/a+as+&c. by ex. 74, — for +
22 8aa
e
2â
+
zee
e
ce
I
+&c.
&c.
8aa
зее
+&c.
8aa
We get
1+1+
ee
3
2aa
+&c. = a+eß²÷ a—ep
AND MECHANIC
PROBLEM
CLXXI.
Of equations, converging feries, &c.
An equation is when two different expreffions equal in value are fet
down with the fign = between them, as 10 5 + 2 = 20 ΙΟ - 30
In ftating algebraic queftions, you muſt be careful to connect the fuppo-
fed letters as the nature of the question requires, viz. fo as the verbal
and ſymbolical expreffions be exactly the fame in effect, which done,
the next and most difficult part is to clear the equations of all the un-
known quantities except one (fee prob. 164) which muſt be done by a
perfect knowledge of the foregoing work.
Ex. 78. If the product of two numbers be 35, and their quotient 1,4
what are theſe numbers? Firſt put a the greater number fought,
e the leffer, p = their product: and
their product: and q = their quotient.
Note. It is beft if there be fractions in the terms, to clear them out
by multiplying each term in the equation by the denominator of the
fraction. Alfo, if there be furd terms, clear them out by involution;
evolution, &c.
Then
J
I
ae=p=35
2
by the queſtion.
2
a
=9=1,4
1.
хе
I e
÷
3 = 4
5 x e
69
e
3
a = qe
P
4
a
5 qe=
a
P
e
6qee
qe e = p
7 | ee=1/2 =
7 w 2
8
9
3
= 25
૧
Р
= 5 for the leffer number
!
५
a = qe= 1,4 × 57 for the greater number.
When there are more unknown quantities in the queſtion than one,
fome of them may be left out in ftating, by fubftituting their values,
and the work much fhortened; fo here, becaufe- =q, or a=qe,
a
C
by writing q e for a, we fhall have qexeqeep, and dividing
each fide by q we get e ep÷q, the fame with ftep 7.
Ex. 79. If a +es 12 be the fum of two numbers a and e,
and a a+ee⇒ z=74 be the fum of their fquares, what are theſe
numbers? Here, becauſe a + e=s, or s
ea, by taking e from
each fide
92
THE UNIVERSAL MEASURER
I
S
Z
S S
Z
SS
3
e e
we ſhall have
I
ཡ
SS
1
22
1 2
ë| +ee=z=ss-2se+ee+ee per
2 ee
- 2 se
[queftion.
se
2
Z
3 c [
4
=ee—se+4s s (ſee ex. 80)
2
Z
4 ww 2
5. №:
2
5+5 S
S
=e=6+✔/:74—7²:=6+√1
2
2
=7, one of the required numbers, and s—e—a, viz. 12 — 7 = 5 tho
other of them.
Otherwiſe. If we take u = 6, half the fum, and d- half the differ-
ence; then will au+d, and eu-d, (a denoting the greater
and e the leffer number,
Then, per
queftion
I
{
1
234
2
Z
3 u u
5
4 w 2
whence
aa+ee=”u+d]²+u—d)²=uu+2ud+dd+uu —
2uu+2dd z. By addition.
uu+dd=z
dd=1z-uu=37—36 = 1
d = √:iz-uu:=√1=I
[2ud+dd-z
6 | a=u+d=6+1-7 and e—u—d—6—1—5 as before
By the foregoing examples it is obfervable, that algebraic queſtions
may be performed feveral ways, but that which is ſhorteſt and eaſieſt is
to be preferred.
Ex. 80. In ſtep 3d foregoing we have ee - s e, where if for s we
write 2a, then it is plain, (by ſtep 5th in ex, 68) there wants but a a
to be added ee 2 ae to make it a complete fquare, viz. aa— 22e
+ee, whoſe ſquare root is a e, (as in ftep 5th foregoing) which
gives this general rule for compleating the fquare, and fo folving this
fort of equations, viz. fquare half the coefficient of the unknown quan-
tity, and add it to each ſide of the equation, &c. (as per ſteps 45 and
46 above) which fort of equations admits of three cafes, all folvable
this way, viz.
to which
1. aa+2aes)+ce Caa+2ae+ee-stee
2. aa-2ae-s
2ae-aa-S
we
have
-2ae+ees+ee
aa—2aefee—ee—s
whence
S
ate stee:
a_e=1/:stee:
a-e=√:ee-s:
Ex. 81. But to exprefs thefe equations univerfally, let us fuppofe
J
a" = a, then
AND MECHANIC.
93
1. a²+2a" e +ees
which by
2. aan
- 2 ea"+ee=s
2e
3. a¹ª — 2תº tee__s
2eaees
Cand
tranfpofing
e gives
a=√:+e±ee
I
D
a=√:—e+stee
2 = √ ÷ + e + s + eel:
In the 2d and 3d cafes you may take either the fign- or before se ejn
and ſo get two different roots: either of which muſt be taken as the na-
ture of the problem requires. The reaſon of two roots is, becauſe a —e
or e —a fquared gives the fame thing, viz. aa— 2aeee. But in
cafe 2d, one of theſe roots will be negative or less than nothing, foit
can have but one pofitive root; but in cafe 3d, both the roots are po-
fitive; theſe are called adfected quadratic equations; becauſe there are
but two dimenſions of the unknown quantity, and the index 2 n of the
one double to n that of the other; and if an be taken out, it is called
a fimple quadratic equation. Alfo, fuch as add a-cas, wherein
the unknown quantity a is only of the firſt power, or index unity, are
called fimple equations; all other forms of equations are called adfected
equations, and cannot be folved by compleating the fquare; but all
kinds of equations may be folved by the following general method,
called converging feries.
n
I
Ex. 82. Required the value of e, in a e¹+ be n-¹ +cen-²+
den-3
+ &c. Q; wherein n, a, b, c, d, &c. reprefent any quan-
tities pofitive, or negative; to folve which put r+z=e,
Then
per exam.
69.
Ier" + nг"
Ι en
I
2
e
I
nr^ z+ &c.
n
I
+:n
I :r
z+ &c.
rn
+:n~2:rn —
3
z + &c.
J 3 Len
continuing the terms no further than the firſt power of z.
Now thefe values of e", e^—¹, &c. fubſtituted in the given equation,
we have a r+anr
cr
2
I
2
I
n-
z+br" —¹ +in- 1: bra z f
A
n +:02:cr" —
- 2: cr³ — ³ + &c. =Q; now, by tranſpoſing all
the terms wherein z is not concerned, and dividing by the coefficient of
z, we have
2 =
Q-ar" - bra
I
narn +:n 1: br
3
+:0-2: crn
n
&c.
3+ &c.
Ex. 83. As an uſe of this general expreffion, let it be required to
find the value of e in this equation -eee +300 e 1000.
=
Firſt, by ſuppoſition and trial get a number to come as near the value
of e as you can, and call that number r, and what r differs frome call it
z; then if r is greater than e it will be r - ze, otherwife r+ze
t
94
THE UNIVERSAL MEASURER
fo here by trial, eis found to be fomewhat more than 3, ſo take r = 3
and r+ z = e, then in this ex. n is = 3, a = I, b=0, c = 300,
and Q = 1000, d, &c. being not in this ex. becauſe there are no e
terms for them, they are taken o, and fo the general equation be-
comes
etar
naran
+:n~2: c1
Q+arn -crn
1000 + 27 900_ 127
27 + 300
273
2
2: CIA-3
2
1000 + r 3
300 r
3x² + 300
=0,5=z, whence, r+z = 3,5 = c.
But this method cannot at one operation give thie juft value of e, becauſe
in the general equation (ex. 82) all the powers of z above z' are left
out, yet by making this value of e, 3,5=r, and repeating the operation
z will be had to two or three more places of figures; and by making a
third operation, you'll find z or e to 4 or 5 places more, and thus pro-
ceeding e may be found to any exactnefs required, every operation
doubling at leaſt the places of figures; fo by writing 3,5 for r in the
1000+42,875-1950_-7,125
laft equation, we have z =
263,25
3,473
36.75 + 300
0,027, fo (r+ z) 3,5 - 0,027=3,473
0,027=3,473 = e nearly; and if this
➡e
be taken for a new rand proceed as before, you'll find e=3,4729635 i
and fo on if neceffary.
Ex. 84. If the root of a fimple power be required, then a, b, c,
d, &c, are each o, for en Qin this cafe, fo we fhall have
e-ro
z=
n г"
I
A general method for extracting roots by converging ſeries.
Ex 85. Every adfected equation hath as many roots pofitive and
negative as it confifts of dimenfions; thus, if eb, and ec, then
b-e=oc
ос eo, and c-ex: b e =ee
c b = o, or e e — 2 ze+bc=0 (by putting — 2 z =
fame with cafe ad in ex. 81.
ce - eb +
-ċ — b) the
e et
Again, e - bx c —e—eb —ee-cb+ce = o, or
2ze cb, by putting 2 z = b+c; this is cafe 3d in ex. 81. Alfo,
: e
C:
a e ³ — be³ — ce² ÷
a:x:e -b:x: e -c:= ele e
abe +ace + b c e — a b c o, an adfected cubic equation. Such
equations are fet down by fome writers thus,
cect
a
b
}
ab
eet ac
b c
{
e — a b c = •
J
AND MECHANIC,
95
Here e has three values or roots, viz. a, b, &c. and fo for other
higher equations, it appears that a - b
c, the coefficient of the
fecond term is fum of all the roots with a contrary ſign; that of the
third term is ab+ac+bc, the fum of all their products, that can
be made by two at a time, the laſt term of all is the product of all
the roots, viz. a, b, c, +. From a due confideration of which, rules
might be invented for finding each of theſe roots; but as one pofitive
root commonly anſwers the end, and that may be had univerfally by
the rules in ex. 83, it is thought fufficient. However, it is eaſy to fee
that if any fuch equation be divided into fo many roots or parts,
a - e=o, e − b = o, &c. as being multiplied together will produce
the given equation, that you will have (a, b, c, &c.) all the required
roots.
Ex. 86. That
multiplied by +, or + by-produces -, may
be thus proved, viz. if + bx-c, I fay p-bc,
put
I - C
2 x b
that is
I x b
5 - b c
4=6
7
2
3
- am FKON∞
7
ba
8
b + c = a
b
R
C
bb=ba+b× — ¢
bb = ba+P
b b + b c
bbba
ba
- be
ba+p babc
+p=-bc, Q. E. D.
Ex. 87, That multiplied by-produces + may be thus proved,
viz. If-bx-c, the product p is+bc;
*
}
са
ca=
-C C
C=
-ca+-bx- c. By cx. 86.
ca+ P
cb-cc
put
I
- b
? X
that is
IX
- C
5 + cb
4=6
7 + ca
Ham tnx no
I
2
a
a = b + c
b
3
7
b c са
8
са
b c - ca-
CC
ca+p
+bc=+p. Q. E. D.
96
THE UNIVERSAL MEASURER
PROBLEM CLXXII.
Ex. 88. To find s, the fum of any feries a+b+c+d+e+&c. by means of their differences.
Here we
have
-་
1 2 3 4 5
86 9
a+b+c+d+e&c. = s
— a + b,
a+b,
1
A
B -
b+c,
a 2 b + c; b
2 + 3 b
c+d, —
2. c + d, c
- 3 c + d,
b+3c-
4
b + 6 c
4 d te, &c.
a
ہے
a+b
2 b+c
a + 3 b
d+e,
2 d + e
3d+e
By taking the firſt term from
that which follows it, firft in
one row and then in another.
the first arifing!
23
3 c + d
4
b + 6c
difference or re-
mainder in the
4
4 d te
5
ſtep, or row.
put
C
9
D
then
IO
aa, the first term in the feries.
per 6,
I I
b
a + A
per 7, 10,
12
a
2 A + B
8, and 11,
13
d
a
3 A + 3 B + C
9, and 12,
14
e = a +
fum of the
Jaft
15
Ave steps
4 A + 6 B + 4 C + D
From the 14 ſtep the law of continuation is
evident, and fhews the coeff, of A, B, C, &c.
to be thoſe of a binomial raiſed to the ſecond,
third, &c. powers.
a + b + c + d + c = 5 2 + 10 A † 10 B + 5 C + D, &c. = s.
AND MECHANIC.
97
AN
Hence, Ifn the number of terms in the propofed feries a+b+
*
n
d+&c. we ſhall have s±na+:n×
Π
A:+:n x
2
2
B::nx X
n:- I
2
2
X
3C:+&c
3
4
3
Ex. 89. Required the fum of the feries of fquare 9 +16+25 +36
+&c. to n terms in number.
Here a 9 the firſt or leaſt term, A = 7
the firſt term of the firſt differences, B = 2
that of the fecond differences, Co; fo, na+h
n
I
X A, &c. becomes±9n+nx²====7+¤×
2
-7n,2nnn
2 =S=9ń +7,¹² = 7"+
2
2 n n n + 17 n n + 3 5 a_nan
6
2
400+20
79
2
2
n
2
3
35"=nan + 17nn + 3.5".
3
6
6
Ex. 9o. Required the fum of the feries 2+6+12+20+30†
&c; to n terms:
Here a=2, A= 4, B = 2, and C = 0,
4, 6, 8, io,
fb,na+nx
- I
A+nX
X
B, &c.
2
2; 2, 2,
2
3
=2n+ 4 ¤n — 4n
+
2 nnn+3nn+2n
S
2
6
3
Ex. 91. Required the fum of the rectangles m x n,+m-ixn-I,
+m−2xn~2, +m−3xn−3 +&c. Theſe terms actually mul-
tiplied ftand
nm, nm — m
m—n+1, mn— 2 m—2n+4, mn÷3m—3n+9&c..
mn+1,
m- — " + 3,
3;
,2
-m—n† 5, &c.
›2
,0
Here an m, A=~m~+1, B2, C, D, &c. each, whence
I
we ſhall have n² m×:—m-n+1:פ×+ 2n x
n
X
3
2
n
2
n
Σ
2
2 3 m n ² + 3 m n − n 3 +" for the required fum, continu-
6
ed to n terms. As an inſtance of which, let it be required to find the
number of ſhot in an oblong pile, whoſe length at the baſe is 46, and
breadth 15 fhot.
N
2
THE UNIVERSAL MEASURER
Here m
46 and n = 15, fo 3 mnn+3mn-—n³+n
6
= 4960,
the number of ſhot in fuch a pile. And if m = n = 15, the pile will be
n15,
a ſquare pyramid, and we ſhall have 2 n 3 + 3nn+n
=1240 for
1, and working as before, we ſhall have
for the number of fhot.
Ex. 92. By taking a
1 + 2 + 3 + 4 + 5 + &c.
9
+ n
n n - n
+n=
2
=S
1² + 2² + 3² +4²+5²+ &c.
nnn
nn
+n
3
+ +
2 6
n
S
1 ³ + 2³ + 3³ +4³ +5³ +&c.
3
+ n³
n
n
3
n
=
+
+
S
4
2 4
ª + 2ª + 3* + 4ª +54 +&c. +nª
n's
5
n 4
n 3
n
+-+
11
5
2
3
30
15
1³ +25 +35 +45 +55 +&c. +n³="6+
n
5 n*
n
S
12
12
6
1° +2°+3°+4° +5° +&c. +ns= + +
n
n
n
5 n3
n
ク ​2
+
Ú
2 6 42
93. Hence the firſt term of the ſum of 1d +2 + 3 + 4ª+&c.
+nd is nd +_ and if n be indefinitely great, viz. divided into an
======
T
I
indefinite number of equal parts, then this n+1 will be the true
I
n+ I
fum of 1ª + 2ª &c. n, for an indefinitely great number or
quantity cannot be increaſed or diminiſhed by addition or fubtraction,
and therefore all the following terms in the faid fum may be rejected.
See theorem 78.
Ex. 94. Req. the fum of the arith. feries a+e+a† 2e +&c,
Here aa, Ae, B, C, &c. eacho±e
+e, &c.
o, &c,
n n
n
n
T
whence na nx
A &c. n a +
: ±e:
2
༡
2 nan²e Ine
S.
See ex. 103.
2
AND MECHANIC.
.99
PROBLEM
CLXXIII.
Of proportional and progreffional quantities, analogies, &c.
Ex. 96. If the quantities a, b, c, d, are in direct proportion, viz.
a: b:: cd, the product of the two means b and c is the product
of the two extreams a and d; i. e. bca d, fora: b::c:
bc
it
ed,
į, e, the product of the ſecond and third terms divided by the firft term
is the fourth or required term, by the nature of the golden rule;
bc
whence, if we multiply each fide of the equation :d, by a, we'll
abc
have = ad, viz. b c = ad, (by ex. 53.)
a
a
a
Ex. 97. Becauſe theſe two terms of the proportion become a fraction
it is eafy (by ex. 53) to reduce compounded ratios into more
fimple ones. As for
abc:
Ex. 98. If it be as a b c : adc :: A: A²; it will be the fame thing
if we fay asab a d, or as b: d :: A: A². Alſo aa—ee::a—c;
or ae: the fame ratio; becaufe each term is divifible by a
without any remainder, &c.
e,
Ex, 99. Alfo we fee how to turn equations into analogies or propor-
tions; for it is but to divide each ſide of the equation into two factors,
one fide for the two means, and the other for the two extreams; fo,
if a ad b, then a : b
:d:a, ord: a:: a: b, for a xażbyd.
Ex. 100. If abd+cd, then, d: a::a: bc, or,
va
d: 1 :: a b + c, &c.
Ex. 101. If
Then
and
I + d b
I + c d
I - bd
I
c d
Itac
Laſtly
2
I ad-bc, then a:b::c:d, in direct proportion.
2 a:c:: bd, called alternate proportion
3b: ad c, called inverfe proportion.
d:
4 da+db_bc+db, then a+b:b::cd:d, comp. pr.
5a+c:c:.b+d; d, alternately compounded.
6 }`ad-bd-bc-bd, then a—b : b::c--d:d, divided
7 ad-cd-bc-cd, then a-c:c::b-d:d, alter. div.
8 adac-bcac, then a: ba::c:dc convert.
9a+bab::c+d:cd, mixtly.
| Iolac+adcbbd=ac+bcadbd, viz. da=be
100 THE UNIVERSAL MEASURER
€
Thus you fee how many ways it is poffible to vary four proportional
quantities; which is the fubftance of Euclid's fifth book.
102. If a quantity a, be continually increaſed or diminiſhed by a quan-
tity e, the reſults a, a te, a +2e, a + 3e, &c. are called arithmetical
feries or progreffionals increafing, and a, a
are called arithmetical feries or progreffionals decre afing.
Let a a
laſt term,
e, a — 2 e, a
3 e, &c.
the first term, e the common excefs or difference, y=the
the number of terms, and s = the fum of all the terms
in the feries; then y will be
will be a: n Te, or, a- n - Te, as the
feries increaſes or decreaſes, as is plain from the feries themſelves;
y a
=2, and 3. D=
whence we get, firſt ay- ne+e; 2. e =
}
y—a te
the feries being in creafing. Then
Example 102
2
2 nann-n:e
per Ex. 94.
4
and alfo
n I
-s, or 2na +:n n-
(n:e=2s, when n, a and e are given.
5nne+na+a—y=2s+ e, fory = a+
(ne-e, as above.
:a+v: xn =s, when a, y, and n are given,
2
=s, as per ſtep 6.
6
per 5 and 2.
per 2 and
7 reduced
4:
ク
​2 na
(to find s.
y
- a
:= 2 S
n-
I
8 na
na+nv
2
a+y
per 3 and 8.
:
:X:
9
2
n:y-nete:+nv_2ny
per 1 and 8
IO
2
y — a +e:_s_yv—aa+aetye
e
2 e
(when a, e, andy are given.
nne+ne
2
=s, when n, e, and y are given,
The 4th, 6th, (ar 8th) 9th or 10th fteps, contain the principal theo-
rems in arithmetical progreffion; by which any of the other letters, as
well as s, may be found.
103. Geometrical progreffion is had by multiplying and dividing, as
that of arithmetical is by adding and fubtracting. Thus,
a, ae, aee, a cee, &c. increaſing,
î
C
a
૧
её
ece deoreafing,
geometrical feries
AND MECHANIC.
ΙΟΥ
Whence it appears, that if a the firft term, e
common ratio, n=
number of terms, s the fum of all the terms in the ferie, the laft term
y will be a e
n
whence, and by ex. 48, we have
ep-
a e ae
S-
I
e-
I
e
V
2
a=
n
e
1 and 2.
3
J
s=ey
ת
a
a en
e I
when
( a, e, and n are given.
fory—a eª—¹, by the above.
y
ye" - y
I
e
n
e
when e, n, and
(y are given, as alfo e, y, and a.
Σ
———, for y = a en —¹, as before.
a
I
I
4
en
Aww:n-1:
5
y
e=
a
or
log.e=log.
log.
y ÷ a
n I
I and 6
6
s = y +
y
a
n
y
I,
a
when n, y, and a, are
(given.
Theſe are the principal theorems in geometrical progreffion; from
which any of the other letters may be found if neceſſary.
.
104. Muſical, or harmonical proportion, is that between thofe numb-
ers, which affign the lengths of mufical intervals; or the lengths of
ftrings founding mufical notes. Thus, if a ring be divided in propor
tion as 3, 4, 6, thefe lengths will found an octavo 3 to 6, a fifth 4 to
6, a fourth 3 to 4: for thefe numbers, 3, 4, 6, are as in theorem 178.
Alfo, 5, 6, 8, 10, are 4 numbers in harmonical proportion; for
ftrings of fuch lengths will found an octavo 5 to 10, a 6th greater 6 to 10,
a 3d greater 8 to 10, a 3d leffer 5 to 6, a 6th leffer 5 to 8, a 4th 6 to 8 ;
whence, if for 3, 4, 6, we take a, b, c, we ſhall have, as a:c::a
-b:b-c, ergo, a b
ac-ca
ca-cb. Any two of theſe three let-
ters being given, the third may be had thus, a➡
e are given, and c=
a b
2 a
bc
when b and
2ac
>
2c-b
(=6) when a and b are given. Alſo b=-
a+c
= 4, a muſical mean proportional between the whole ſtring c = 6 and
a=3 the octavo; which 4 is the length anſwering to the fifth, viz.
4 to 6. Again, if for 5, 6, 8, 10, we take a, b, c, d, by the nature
of theſe numbers we fhall have, as a :d :: a -b: c — d, ergo,
ac-ad-da db: any three of theſe 4, a, b, c, d, being given,
the fourth is alfo had as before.
ૐ
+
102 THE UNIVERSAL MEASURER
105. From a view of thefe numbers, it appears, that a feries of
numbers in harmonical proportion, are inverſely as another feries in
arithmetical proportion: thus,
Harmonical
Arithmetical
60.
IO, 12, 15, 20, 30,
6, 5, 4, 3, 2, I.
That is, as 10: 12::5;6, and as 12:15:4:5, &c.
106. Hence, Ifa, b, c, d, e, f, &c. denote an harmonic feries, and
m=b-
a, we fhall have a =
ab, b=
a b
b -m
a b
C
b
a b
d
a b
b
3 m
b_
4 m
&c. and the laſt term =
n being the number of terms in the feries. Theſe terms divided by
2 m
a b
:n
1: m
a b gives the feries.
I
I
I
bbm'b-2m' b-3 m
&c. to
I
b—3`m’
I
b. n.
i : m'
in harmonic proportion,
PROBLEM
CLVI.
To find a vulgar fraction in few figures, equal to a given decimal
fraction q; or, (which is the fame thing) to find the ratio of
the decimal's denominator to its numerator in fewer figures.
Take the following folution, as a general one.
Ex. 107. Required the ratio of 1 to 3,14159265, (in fewer figures)
being the ratio of a circle's diameter to its periphery.
Here the neareſt ratios in the leaft whole numbers are 1 to 4, but
a
nearer as 1 to 3, which in fractions are 4 and 4. Let 2 = 4, the
nearer ratio, and, the other ratio, viz. a 3, b = 1, c = 4,
d=1, and fuppofe
or=
C qd
qb-a'
a e + c
eb + d
=9=3,14159265, whence e-9d.
e=
C
q b — a
having thus found e, fubſtitute it in the fuppofed fraction
ae+, ſo you'll have the firſt approximated value of q, which call
ab+d
a
a
a fecond and for the firſt taken take, and proceed on as
b
before, for a fecond approximated value ofq; and thus repeat the ope-
AND MECHANIC.
103
racon with the two laſt neareſt values of q, you may come to an approx-
e
imation as near q's value as you pleaſe; ſo here, e =
q
q b
a
4-3,1415
3,1415-3
,8575
,1415
=6 nearly, whence,
a e + c
=
3×6+4
eb+d IX 6+ I
22
a
7
e will be =
q
==q nearly. Again, Let
(954)
a e + c
whence, q=c+d
eb+
imate which call
3
3,14159X I
3,14159X7 — 22
7 × 16, + I
355, the ſecond approx-
22
7
즉​, and 등
​3. Then
d
1
,1415
16 nearly
,00887
22 X 16, +3
=
=
113
a
22
C
and
=
and we ſhall have e
c - q d
b
7
q b
a
,00885145
—,00003055
102928
=9=3,14159265, &c.
32763
29 nearly, whence ae+c
10295022
eb+d
32770 +7
Note. The value of e is ftill to be had only in the neareſt whole
numbers, the ſecond and third approximations are the ratios in com-
mon practice, viz. 2, or as 7 to 22, called Archimede's ratio, or near-
er, as 113 to 355, called Metius's ratio; this ratio is truer than the
common factor, I to 3,1416, for 355÷113=3,141592. Therefore
in uſing this ratio, the error cannot exceed,0000003, but ¹3 =3,142
&c. fo this ratio may err, nearly ,002, tho' it is the moſt common one
in practice and indeed may ferve for ordinary ufes well enough.
z
Whence, as I to 4, or nearer, as I to 3, or nearer, as 7 to 22, or
nearer as 113 to 355, or nearer as 32763 to 102928, fo is any circles
diameter to its periphery, &c. for any other.
PROBLEM CLXXIV.
How to compute logarithms.
108. What natural numbers do, multiplication, diviſion, involu-
tion, and evolution, logarithms, or artificial numbers, performs by
addition, fubtraction, multiplication, and diviſion, i. e.
109. The fum of the log. of any two numbers, is to the log. of
the product of thoſe two numbers.
110. The difference between the logarithms of any two numbers, is
to the log. of the quotient of the one number divided by the other.
111. If you mult. the log. of any number by the index of any power,
the product will be the log. of that number involved to that power.
Ex. 112. If you divide the log. of any number, by the index of any
root, the quotient will be the log. of that root. See ex. 72.
104 THE UNIVERSAL MEASURER
113. Lemma. If from each of the indefinite roots of two numbers;
an unit be taken, the ſum of the two remainders will be to the pro-.
duct of the faid two indefinite roots minus unity; or which is the fame, if
from the indefinite root of any number you ſubtract 1, the double of
the remainder will be to the fquare of the faid indefinite root leffen-
+ e, be any finite number, and n an indefinite
!
ed by unity, i. e. if
1
number, then we are to prove that I + e n − 1 × 2 is = 1 +e|
L
en
× 1 + è ª – i, viż. that 2:1 +e
e
OK. 72 and 71, &c.
L
Π
1. Firſt by
2=1+
I
2: 1 + e | ¹² = + 2 e + 2x:
2.
X:
| -
n
2 n
I
we ſhall
have
2
n
=+
X:
I
(ee + 2x: 12-1x12-2:eee,
2 n
=2+ㅁ​+금
​1 + ep³ = 2 + 2 e +
I
n
n
3 n
금​x:
I 2
X:
I: ee +
2 n
I
I X
2:eee, &c.
3 n
(2 × 12 - 1×12-2
2 n
2: 1 Tel — 2 = 2e +
e + 2 x : 12-1:ee+;
2 n
x: 1 2-1: ee+,
1
2
3
n
2
2
4
Ite
n
1=
2-T
et
X:
n
5
2
I
3 or 4
e+
n
n
2 n
Lee +
X
2
2.
X
X
n
2
(를
​2 eee + &c↓
3
5 reduced
2 e-
6
2 ee + 23 n
eee,
&c.
n
2 n
2
6 reduced
7
x: e
n
2
Lee+ 1/
I..
&c!
3
Note. Log. or 1. ſtands for logarithm, and logs. or l's. for logarithms:
Here the ſeries in the third and fourth ſteps, are exactly the fame.
Q. E. D. But becauſe n, by fuppofition is indefinitely great, it is evi-
dent that 2 muſt be indefinitely fmall, therefore, 2 added to, or
n
n
taken from any number, can make no fenfible alteration in the faid
AND MECHANIC.
105
2
number, fo in the co-efficients
2
I
2, &c. this
2
n
n
may be
n
omitted, and then the feries will be as in the 5th ſtep, which contract-
ed, gives that in the 7th ſtep.
n
114. From hence it appears that if one be taken from Ie the
indefinite root of 1 +e, of any number, the remainder-
3
4
I
n
Ze*
е
+ } è ³ — — e ¹ +&c. would be the log. of 1 +2, were it not inde-
finitely ſmall, viz. multiplied by, but if we put Q = 10000&c.
n
D
(where roooo &c. is fuppofed to confift of as many places of figures
as n doth) the ſeries will then become finite, and we ſhall have Qx
&c. for the log. of 1 +e.
: €
3
e + + } e
4 3 es
е
115. It is further evident, that there may be as many forms of logs.
as n can be ſuppoſed different indefinite numbers. Thus, if Q= 1,
then n = 10000 &c. and the above feries will become e
&c. which is called the hyperbolical log. of 1+ e. Again, if
n = 2,30251 &c. then Qwill be 0.434291 &c. and Q_x: e-
1 e ² + + e ³ 3 &c. is the tabular or Brigg's log. of 1+e, but this fe-
ries tho' it be the most natural one, yet it converges fo flowly that it
is of little ufe in the conftruction of logs. fo to find a feries that will
converge ſwifter, let us fuppofe___ to denote the number whofe log.
is req. then by ordering
I
I
HE
e
I
I
n
I
e
I
n
I
e n
— 1 as you did 1 + el I viz. by taking the nth root of
and fubtracting I from that root, you'll have QX: e+že²+že
+ 1 e¹ + &c. for the log. of
ব
3
4
êt
I
n
I
.>
to which add Q×:e — ÷
I
e
e² + + e ³ — — е++ &c. the log. of 1+e, and you'll have 2Qx:
7
e+÷e³+} e³+e+ &c. for the log. of
I te
I
C
ལ་
I
× Ite, or
Ι
and thus you have 3 different feries viz. one for the log. of
I
Ie, one for the log. of ________ converging faſter, and 1 for the log.
I - e
I te converging yet faſter.
I
I
106 THE UNIVERSAL MEASURER
116. Ex. required the log. of any number N; firſt, let N
and then we ſhall have e
N I
N+I
I te
I - e
which being ſubſtituted in the laſt
feries for e, will give the logarithm of the number N, as required.
N
117. Required the log. of a fraction as Here, as before, we
D
muſt put
N
Ite
D I
N-D
and then we'll have e=
which put in
e
N+D
the laſt feries for e, gives the anſwer.
୨
118. What is the hyperbolical, or Neper's log. of 2? Here N=2,
N I 2
I
I I
13
=, therefore, e+že³+3e5+ e² + 3/
IS
IT
17
19
e7
7
2 I
fo e =
N+I 2+ I
e' + vie" + re¹³ + + e + " +е²+++е²=
IT tre
+e tzte
ZT
0,346573590280, which multiplied by 2 Q2 gives 0,693147180560
for Neper's log. of 2, true to the laſt figure, by thefe 11 terms of the
feries only.
119. Required the tabular or Briggs' log. of 2. Here, if inftead of
multiplying 0,346573590280 by 2 2 Q you multiply it by
0,868588963926 = 2 Q (by 115.) you'll have the tabular log of 2;
or, which is eaſier, Let 2 Q=R=0,8685889, &c. then the foregoing
feries becomes Rx:e+
=, we ſhall have,
e e e e
5
c
+ +
5 7
7.
; nowe being before found
3
0,8685889
Re=
= .2895296 +
Re³
3
0,8685289
=,0107233 +
3
Re5
81
0,8685889
=,0007149 –
5
Re
1215
0,8685889
=,0000568-
7
15309
Re⁹
0,8685889
=,0000049 +
9
ΙΣ
177147
Re
Į I
0,8685889
=,0000005
1948617
Sum,3010300 for the tabular log.
of 2, as required. Having thus found the log. of 2, you may from it
AND MECHANIC.
107
་
find the logs. of 4, 8, 16, 32, 64, &c. for twice the log. of 2 gives
the log. of 4, (becauſe 2× 2 =4, fee ex. 109.) to which and the log.
of 2 and you'll have the log. of 8, &c. and by the above ſeries you may
find the log. of the prime numbers, 3, 7, 11, 17, &c. But as we go
to higher numbers, the feries converges flower, therefore take the fol→
lowing method for fuch cafes.
120. In this manner the logs. of fmall numbers may be readily
found. But it will be more expeditious to find the logs, of large num
bers, from thoſe of ſmall ones already found. Thus, fuppofe
I a = b
b- I a, b, &c. being three numbers in
arithmetical progreffion whoſe com-
mon difference is unity.
c=b+ I
{
2
I X 2
3 + I
3
ac-bb I
4
ac+1=bb
ac+I
bb
4 ÷ a c
5
ac
a c
put
N D
by ex. 117
7
=e=
N + ·D
6ac1N, and ac D, then a c + 1 _ N
2 ac+I
per laſt ſtep
a c
D
I
,
hence
Co
8
log.
ac+I
Re+
ac
Re 3
+
3
Res
5
+&c. which
feries call S.
ac+I
then per 5
tranfpo.
9th ftep
we get,
{
9 | 2 log.b-log. a-log.c-S, viz.=log.
IG
II
log. a + log. c + Slog. b,
ас
2 log blog. c flog. a. Hence, if any tw
of thefe 3 logs. be given, the third may be found,
Ex. 121. Required the tabular logarithm of 7, having found the log.
of 2 and 3, as directed in ex. 119, you may from them have the logs.
of 8 and 9, viz. 3 times log. of 2=log. of 8, and twice log. of
log. of 9, and per 11th ftep, log. of 7 2 log. 8-log. 9
log. 9 — S, (a
being7, b8 and c=9) now to find S, we have per ftep 7, e
I
foS Re+
2ac+1 127
=
Re 3 Re5
3
+ +, &c.log.
5
log. 84, whence, by refolving the feries, we fhall have
ठउ
ac + 1
ac
Re=
,868588963:
=,006839283
127
Re 3
,868588963
,000000141
3
Res
LA
5
5145149
,868588963.
165191847035
,000000000
here S, or fum,006839424 tabular log.
108 THE UNIVERSAL MEASURER
of 8. Whence 2 log. 8-log. 9-S= 0,845098040, equal tabular
log. of 7, true to the laft figure, and only 3 terms of the feries uſed,
and alfo turns out the tabular log. of the fraction 64 or 3, and if we
have greater numbers fewer terms will do; thus, if the logs. of 2, 3
and 5 be known, then thoſe of 48 and 50may be known, from whence
the log. of 7 (= log. 49) may be alfo found, for let a 48, b49
I
&c. 50, then e =
2ac+I
1
ठ उ
ठ
>
and Re=
,868588 &c.
4801
4801
,0001809 &c. = S, then per ftep 10, we have log. 48+ log. 50 +S
=log. 49, or log. 48+ log. 50+S
2
log. 48 + log. 50+$ =,845098040=log. 7 as be-
4
fore, and takes in only the firſt term of the feries, which would hold
true to a place or two more if continued, and greater the numbers are,
the log. by only the firſt term of the feries will ſtill be truer to more
places of figures, by which a table of logarithms may readily be made,
or examined.
PROBLEM
CLXXV.
Given any hyperbolical logarithm L, to find its natural number
I + e.
122. If you revert the feries found in ex 114, you'll have the re-
quired value of 1+e.
and as n may be
I
n
I = L'
Otherwife, fince we have 1 +
any indefinite number, we may take r:
I + er r for 1+ el =I
That is
2 Q. r
r
I
1+er=1+
I r: 1+ er
r = L
2
2/1
r
L
3 1 + c = 1 + 2 + "x: 1-1: LL
ry:r. - I: X
+
2
rr
2
r
2 L L L
+ &c.
3
rrr
3 reduced
4I+e=1+L+ +
LL LLL
L 4
+-
+
2
2,3 2,3,4
L 5
+ &e.
2,3,4,5
Here (as in ex. 113) r being fuppofed indefinitely great, the numbers
I,
J
2, - 3, &c. in ſtep 3, are left out in the 4th ſtep, which gives
the anſwer.
AND MECHANIC.
100
PROBLEM
CLXXVI.
Reverſion of infinite feries.
122. Before we do this it will be needful to know how to put, or
m
raiſe an univerſal infinite feries, hz +bhz
+dh z
m + 3n
m
m+I
m + 2 n
+ch z
n
+ &c. = hz x: 1+bz" + cz 2 " + d z 3
+ &c. to an infinite power whofe index is v, or to find the value o
h z
my
2 n
3 n
f
x: 1+bznfcz +dz + &c. in fimple terms. To
make this plain let us put p=1+bz" + cz z n
л
V v m
fz 5n+ &c. then neglectingh z
per ex. 69, we'll have 1 + pl
+ &c. whence
V
n
Z
3 n
4 n
dz³ " + ez¹¹+
+dz
till afterwards, and working as
1+AP+Bpp+CPPP+DP4
BPP
Bpp=
CPPP=
Dp+=
Ep5=
I here break off the feries at z 5 n,
1+Ap=1+Abz"+Acz³"+Adz³n + Aez 40+ Afz 5+ &c.
5
+Bb¹z¹"+2 Bbc z 3n+ B cc z4n+2 Bbdz 4+2 Bebz 5+2 Bedz sa
+ Cb ³ z³n + 3C b*cz 4n+3Cbccz5n+ 3 Cbbdz 5n+ &c.
3
+ Db4z4n+4Db³cz5n+ &c.
Eb 5 z 5n+ &c.
i. e. I take in all the terms in p, p, p 3, p4 and p 5, which come below
Z
6
n
›, you may break off ſuch a ſeries where you pleaſe, but the more terms you take,
the nearer the truth
your anſwer will be, and nearer the law of continuation you'll approach, if not quite.
124. Now the like terms of z collected together and multiplied by hˇz vm gives hˇz
v m
V
+ Abh
N
v mt n
2
+ A + B b² × hˇ z vm + 2n +
vm+5n+ &c.
V
A e + 2 Bbd + B c c + Db 4 + h
y
3 Cb2 + 4 Db3c + Eb 5 Xh z
Z
+ A f + 2 B e b + 2 B c d + 3 C b c c +
-hzm
To illuſtrate this theorem by one particular cafe.
125. Let it be required to raiſe 2 z
6
42 -62 9 — 8 z
12
8
2 — 625 +82 12 Z
I I
n
&c. — 2 z² X : I −3 Z4 +
+ 16 z 14 = 22
&c. to the 5th power. Here we have h2, m = 2, b =
2, m = 2, b =
I
&c. n=3, v=5= A, B VX.
2
3, c = 4, d = 6, e = 8,
= 10, &c. and ſubſtituting theſe values in the above general theorem,
it will become, 32z*°—:32:X:15z¹³: +:32x20 + 90z¹: +32x−30-240-270z¹ &c. =32z¹°-32:15213
+32:1102¹º:—540:32z¹ &c. the required feries; or, eaſier thus, 32X:z¹³°—15z¹³ + 110z¹6—540¹⁹ &c,
10
19
Ad + 2 Bbc + Cb3 X hz
vm + 4 n
V
vm + 3 n
+
_hzx1+b+cz²+dz³¹+ &c.
× 3
2
3 n
110 THE UNIVERSAL MEASURER
126.
If the fum of a ſeries of terms conſiſting of the feveral
o, the fum of the terms, where
powers of a variable quantity be
any one and the fame power of that variable quantity is involved, will
Lemma.
alfo be = 0,
AND MECHANIC.
2
III
That is, If
I
2
then 1-2"
I÷zm
IZ
r
confequen.
n
Л
r
az +bz+cz +dz +ez'+fz'+gz'+=0,
о
a+b+&c.= o, Ergo, a+b=0,
z n
30+o+c+d+&c.— =0,whence c+d=0
4
z m
{
0,+e+f+8=-=-=-=0, fo, e+f+g=0,
r
5a+b=o, c+d=o, e+f+g=0, &c.
127. Therefore, if a ſum of a ſeries of terms confifting of the ſeveral
powers of a variable quantity beo, there must be at leaſt two terms
wherein each of the faid powers are involved. Otherwiſe, the values
of the co-efficients in fuch an affum'd feries cannot be determin'd, now
to do this univerfally.
128. Let it be required to revert the univerſal infinite feries, e
m
m+n
m+2n
+ be
+ce
m+ 3n
m+4n
de
+Ee
+ fe
I n
in
+ Cz
m+50
+&c.—z, or which is the fame thing, to put z into fuch a
feries, as may exprefs the value of e, to effect which, affume the infi-
L
111
nite feries z +B z
e, then by what is faid in ex. 122, we'll have,
!
I + 2n
1+ 3 n
11
+Dz
+ &c.
m
I + n
I+2n
1+30
נת
m
I. e
=z+mBz
m
+m Cz
m
+m Dz
+
I
B3z
: m
m
m
3
2
2. be
B² z
1+20
I - 2n
m
Ι
1+30
m
I
+ m
B Cz
m
m
+ m
2
2
+ &c.
נת
m+n
I
I+2n
=bz
+m+nBb z
n
גנן
+ m + n
+3n
1+ 3 n
Cb z
+m+n
m
2
B2bz
+ &c.
3. ce
·4. de
&c.
m+ 2n
-Cz
m+ 30
=dz
&c.
1+30
+m+2nBcz
m
+ &c.
1+ 30
m
I +30
&c. breaking the ſeries off at zi
112 THE UNIVERSAL MEASURER
mtn m+2n
, се
Here its plain, if theſe values of e ™, be
&c. be
ſubſtituted in the affum'd ſeries, it will exprefs the value of e, but then
the co-efficients B, C, D, &c. are not known, ſo to find them we have,
i+n
in
', are, b and m B, ſo (per lemma) b÷mB=0,
firſt, thoſe of z
whence, B
ות
1+20
Secondly, thofe of z
m
, are, c+mc+
m-I
m
B² +m+nBb, which (per faid lemma) areo, whence
2
I
m+1+2 nbb,
2 me
b
b b
C=
,
(for B=-
and B 2
as juſt
2 m m
m
mm
now found). Thirdly, the fum of the co-efficients of z
1+ 3n
are
m
m
I
m
I m
d+mD+m
BC+ m
2, B³ +, m+n C b
2
、 2
3
+m+n
mt n
+
n — I
2
B² b+m+2n B c, which alfo, are =o (per
lemma) now by reduction and ſubſtituting the values of B and C, as
:2mm +9inn + 9n²+3m+6n+1:—b³
above found, we have D=
+:1+ 3n+m:bc
Z
m
6 m 3
d
">
now fubftituting thefe values of B, C, D,
m
It n
+
b z
m
m
m
in the affum'd feries, it will become ez
:m+1+2n:b2:
ננ
: 2m² + 9mn + 9n² + 3 m
6 m 3
2 m c
I+ 20
Z
+
2 m m
+60+1:
-b
b 3
1+ 3n+m: b c d
I
1+ 3n
Z
A general
m m
m
m
cafe
may
in e +be+ce³+de¹ + &c.
be readily found. Thus, Ex.
theorem whereby a feries reprefenting the value of e in any particular
129. required the value of e
mt n
z.
Here e
m =e,
bzi+n
=e²,
I
fo m=1, and n=1; therefore, e=z m
m
&c. z
T
bz2+:2b2-c:z3:5bc5b3d: z 4 &c.
130. Required the value of e, in e-be-ces-de- &c.
mtn e3, fom I, and n=2, b, c, d, &c.
➡z, here eme, e
bzi + n
cach negative. Therefore, ez
in
m
מן
+36²+c:zst: 8bc12b3+d: z7 &c.
&c.=z+bz3
AND MECHANIC.
113
130. Required the value of e, in e-be 3-ce-de- &c.
in
-z. Here, e :e, e
m - a
each negative; therefore, ez
e 3 fom 1 and n=2, b, c, d, &c.
I bz
m
In
m
7° &C.
+c:25 +: 8 b c + 12 b 3 + d: z
131. Required e, in e²+be+ce°+de
u+w
&c.=z+bz3+:36*
&
+&c. =Z.
I
=e+, fom 2 and n=2, whence e-zi
Here, em=
m_e¹e
I + n
b z
I
&c.-22
b z
N4
2
b
S
4 C
I
+:
: Z
+
m
8
4
3363
16
글
​:2
ź &c.
m
9 b c - 2 d
I
I
132. Required e, in ✔e+bet+ ce¹² + de¹¹+ &c. — z.
Here, em
2bz9+: 17b2
2)
mfn
е
2 C: Z
e4 whence mand n=1,
and n=1, ſo ez
+:48 bc - 200 b ³ 2d: z
16
=
2
23 &C.
133. Let it be required the fquare (e) root out of this equation,
3 e- 6e ² + 8 € 3
e 13 e + + &c. = 15 y.
Note. When the firſt or leaſt term has a coefficient, it is beſt to di-
vide the whole equation by it.
So this equation divided in this manner is e
+ &c. = 5 y z whence, b = -
2, c = 1, d
ftituting theſe values of b, c, and d, in ex. 129,
of e there and here are alike) it will become e
+ 5 3 2
उ
4 &C.
2 e² + — e³ —² 3 c²
13, now by fub-
(becaufe the powers
= : 2 2 2 + 1 / 1 2
z 윽고
​z
3
134. If (in Ex. 128.) z were another univerfal infinite feries,
P
Qv + R v
P+ t
P + 2 1
+PE
+Tv
P 3 t
+ &c.QR,S,T,&c.
being known coefficients, and it were required to find e in terms of v,
it is evident, the laſt general theorem will anfwer this end, by ſubſti-
tuting this feries in it, for z and the powers of z, i. e. raiſing the ſaid
feries, to the powers of
I I + n
m
m
&c.
135. If you would have more terms in value of e, than what is in
theſe examples, it will be eafy in moft numerical cafes, from theſe four
terms to give the law of continuation, but if in any caſe they do not,
you may either continue the general theorem further by taking in terms
before you break off, or you may from what is here
beyond z
I+
m
3 n
done, eafily revert any particular cafe by itself. See the following ex.
P
114 THE UNIVERSAL MEASURER
ое
136. Required e ine-be3+ ce5-de'+oe- &c. —Z.
Here, as in ex. 130, m=1, and n≈2, but every fecond coefficient
negative, whence, (per 128, general theorem,) ez bz 3
++; 3 b² z 5_cz5: + : 12 b³ +d8bc: z7&c. here, if we
2
I
I
2, 3, 4, 5 120
d:
I
2,3,4,5,6,7
I
take - b=
C=
2,3
I
臼
​we'll have e−z +
23
5040,
1
+
2,3
3z5
36
Z
+
120
"827
720
12 2
7
>>
5040
325 52
+-
40 112
B = 2/23
2
&c. which terms duly reduced, gives e =z +
+
216
Z
3
f
6
+&c. where it plainly appears that if we take A = Z,
A, C-
→ & C + ÷ D +
ZZ
32Z B, D = 5ZZ C, &c. we'll have e=A+‡ B
4
E +7 F +&c. every of thefe denominators in-
fhews the law of continuation.
I
greafing 2, which evidently
137. Now if we take z
PC, ET D, F,
A = 5
B =,0625
C=,01171875
1;
0,5A, then will BA, C=τ B,
E, GF &c. that is
,50000000= A
2083334B
234375
34877 D
5934 = E
1093F
212 = G
D=,00244141
E=,00053406
F
=00012016
G
=,00002754
H=,00000639
43H
I=,00000149
9=x+1
可
​fum,52359877ė,
T
which collected gives, fum =,52359877
and by
aking in more terms you may get e to what exactneſs, or places o
figures you pleaſe,
138. If the radius of a circle be 1, the fine of 30° will be or 0,5,
and then e,52359877, will be the length of the arch 30°, which
multiplied by 6 (becaufe 30° is of the periphery) gives 3,14159262,
true to the laſt figure, which fhould be 5, inſtead of 2, for the peri-
phery of a circle whofe diameter or 1, or of that half circle whofe ra-
dias is 1. See art. 173.
AND MECHANIC.
115
PROBLEM CLXXVII.
The ufe of infinite feries in the reduction of equations.
139. When an equation is given in two variable or unknown quan-
tities, e, and y, to find (y) one of them in terms of (e) the other, and
can be done no other way, we must have recourfe to infinite feries, now
to find the indices of fuch an affum'd feries as will beft answer this end,
viz. to make the found feries converge the fafteft &c. take theſe rules.
140. Rule 1. Make the given equation =0, then for y in the faid
equation fubftitute A e ", if fluctional letters be the fame in every term,
they may be rejected as conſtant ones.
141. Rule 2. In this new equation make two of the leaſt indices of
e, which have contrary figas each other, when e is ſmall in refpect
of a, otherwiſe the two greatest powers of e with contrary figas muſt
be made each other, by which we find n, the index of the first termi
of the required feries.
142. By making the co-efficients
were made
0, of thoſe terms whoſe indexes
each other, we'll have the value of A, the co-efficient of
the firft term of the required feries..
143. Write the value of n found as above, in each of the indices of
e, and fet down the difference, that there is between one of the equal
ones above-named, and every one of the different indexes of e, which
call a feries of differences.
143. To this ſeries of differences fet down all the leaft different num-
bers, that can be made, by doubling, trebling, adding together &c. the
terms of the faid feries, with thoſe that ſhall be thus produced, till you
have thereby got as many different terms, as you defign the required
feries to confiſt of.
145. Rule 6. Add each of theſe terms to the value of n, if the leaft
indices were made equal, otherwife they must be taken from the fame,
and the ſum or difference, are the indexes of the terms of the required
feries.
146. If A, have two or more equal values (per 142) the feries of
differences found per 143, must be divided by the number of thofe
equal values before they are added to, or taken from the value of n,
But if A have another poflible value befides thefe equal ones this other
value is uſed, and thefe equal ones not regarded.
;
147. If in the given equation after the ſubſtitution of A e n for Ys
there be any term that hath no power of e in it, then the index ofe is
fuch a term is always ➡ ✪.
116 THE UNIVERSAL MEASURER
148. If there be any fraction, or furd in the given equation, it muſt
be cleared thereof by divifion, or involution &c. if it can, if not, it
muſt be cleared, by putting that part where the furd &c. is into a
feries.
n
149. The value, or values of A, found by (142), Rule 3. or by the
comparing of like terms as by Art. 126, are the fame, fo having found
A, as per art. 142, you may either uſe this found value, or A it's ſelf
in the affum'd feries, for if A eº — the firſt term of any ſeries, and S
the reft, i. e. if A en sy in any equation, then becauſe the
first term of any feries is either the greateft, or the leaſt of all the terms,
confequently the terms in that equation, which have the leaſt, or great-
elt indices of e, may be made per 126, for if the fum of a ſeries
be ſuppoſed = 0, the co-efficients are to be fuch as will actually make
that feries 0.
3
3
et
150. Example. Given a 3 e a e³ — a ³ y + y 4, viz. a
3
a e 5 — a ³ y — y 4 — o, to find y in terms of e, in a feries converging
fafter as e is greater, or which is the fame, as is fuppofed to be ſmall
in respect of e..
3
- A¹ e 4 n + a³ ef
I and 3, the
By putting A eny, we'll have A e¹ a ³
a e 3 =0, per rule 1, whence the indices of e are n, 4 ",
two greateſt of which with contrary figns (per rule 2,) are 3 and 4",
L
fo n=
4
, and their co-efficients are A4 and a, ſo (per 142) A = a
and theſe indices of e will be 4, 3, 3 and 1, (per 1'43) each of which
differs from (3 or 4 ") one of the above equal ones by 0,2, 24, the
feries of difference (per 143) now, let the required affum'd ſeries con-
fift of 5026 terms, then the leaft different numbers that can be pro-
duced by doubling &c. and adding together of o, 2, 2 will be 0, 2, 24
44, 41 &c. Then (per rules 2 and 6,) each of theſe taken from
(n) leaves, — — — 4,4 &c. the indices fought;
2
whence, the affum'd feries is Ae+Be+Ce+ De
I 4
+ Ee
will become
5
,
6
-1
&c.y, which being put in the given equation for y it
A¹e³+4A³Be+4A³Ce & +44³De+4A³Ee+4A³ Fe
-ae 3
-a € ³ — a ³
3 e
}
+a³ Ae
+6A²Be+12A*BCe ++6A²C²e
он
&c.=0.
+a³ Be
$4
3
+ a³ Ce !
Now (per art. 126 and 127) by making thefe terms equal o, that
have the fame power of e in them, as they here ſtand one below ano-
ther, we'll have Aa (as found before) B=4a4, C=-1a4,
A
IO
AND MECHANIC.
117
I
mkt
D
a
3
a, E
E = ÷ a
I
I
', Fa, &c. whence, y=ae
I 7
I
a
3 a
4
T
a
a
+
&c.
5
4 e
4 e 2
8 e 13 8
+
e ¹4
151. Required the value of y, in a ay + y 3
3
3
32e
e3o, e being
ſmall in reſpect of a, here, by putting A e n for y, we have a a A e¹
+ A³ e ³n — ė :o, and (per rule 2) the indices of e are n, 3" and 3,
the two leaſt of which with contrary ſigns are n and 3, whence, n=3,
and Aay (per art. 142) and the indices of e, in numbers are
3, 9 and 3, each differing from n=3 (143) by 0,6, for the ſeries of
differences, then by doubling, trebling &c. and adding together of 0,6,
we get 0,6, 12, 18, &c. (per rule 6) each of which added to n = 3,
gives 3, 9, 15, 21, &c. for the indices of the required affum'd (uni-
verfal infinite) feries (Ae+ Be
+ &c.) whence
n + 2 m
+Ce
n+ 3 m
+De
n +4 m
y= Ae³ + Be⁹+ Ce¹5 + De¹¹ + Сe¹7 + &c. Then will
y³ A³ e⁹ +3 A Be's+3 A Ce¹+3 AD e27 +3 A B² e²*
(+ 6 A B C e²7 + B³ e²7 &c.
aa y➡aa A e³+a a Be² + a aСe¹5+aa De¹¹ +aa Ee²7 &c.
y³ — e³ = — e³ + A³ e⁹ +3 A² Be¹5 +3 A² С e²¹ + 3 A B² e²¹
(+ 3 A² De ²² + 6 A B C e²7 +B³ e²7 &c.
Now by equating the like terms, &c, (as per 126) we fhall have
A
I
aa4}=0,
aaB} =
}=0,
A=aa, B =
whence, {c=
+
e s
a
I of
3
a
5 5
B = as
aaD
3 A2 C
3 AB2
=0,
a a E
3 A2D
6 A B C
BB B
o, &c.
I 2
Therefore, y=
e
aa
14, D= 20
E=6, &c.
12 e 21
220
I
+ 55 e 27
&c.
226
152. Given a³ y3 —
*
4 a² e² y²+5 a e¹ y-2 e- e³ y³ — y6
o, to find y, when e is fmall in refpect of a. Here by putting
n
2
2. 21 + 2
n+
+5a Ae
A e y we have a A e -4a A e
33 +3
n
6
2e
A e
3 3 3 m
6 6 n
A e =0: now by making any two of
the leaft indices with contrary figns (3n, 2 n + 2, n + 4, 6)
equal each other, (becauſe no fuch two of thefe can be lefs than the
—
118 THE UNIVERSAL MEASURER
reſt and yet equal each other) we have n 2, and the coefficients be
longing them are A3 a³.
being divifible by A a
laſt quotient is A a
I
A
2 2
·4 A² a²+5 A a
-4.
20 (per 146,) which
I, and the quotient again by A a — 1, the
2; whence, A has two equal values, cach
=—, and one poffible odd value, fo the equal ones are not regard-
a
a
ed, whence our indices will be 2, 5, 8, 11; ſo y
I I
تم
2 C
c² (or A)
8e5
+ Be³ + Ce³ + De¹¹+ E e ¹¹ &c. which being involved, &c.
as in the two laſt examples, we fhall have y =
2
2 e
+
a
32 e
8
+
3 4 X 3 2 e
I I
+ 24 × 64e¹4
a7
ΙΟ
+ &c.
a
13
a
a 4
153. If in any fuch equations e fhould differ but little from a, (or
the equation be of the logarithmetical kind, where it is fometimes im-
poffible without a different fubflitution) it will be beft to work as fol-
lows.
154. Given y³a a ye³, to find y; e being fuppofed
nearlya, ora―ze. Then e³
8 3
2
4 a a z
+
2 azz
27
3
8 a
3
4 a az
+
27
3
- z zz, whence, y³ a à y — e³ — y³ fa a y
3
2 a zz + z ³, and putting A z¹ = y, the indices will be, 3 n, n,
O, 1, 2, 3, the two leaft of which with contrary figns (becaufe e is
leſs than a) are n and o, viz. no, and becauſe no,
3 n is alfo
8a3
27
O;
o, and the co-efficients belonging them are A³ + Á a a
from which the value of A, may be had by folving an adfected equa-
tion (fee ex. 83) &c. as before, you'll have B-
42 a
3a² + 9 A²
C
a =
-B3-I &c. the affumed feries being A +Bz
2a-3AB2
>
D=
aa + 3 A
2
+ 3 A²
+3
+ Cz² + D z³ &c. = y.
3
zñ
155. If e be greater than a, or a-ze, then by writing A zn
for y, our given equation y³ +a² y — e³
+aa Azn
27 a
3
2
+
27 a
8
4
e3o will become A3 z 3 n
Z 9azz
+z30, the two great-
2
eſt of theſe indices with contrary figns (becaufe e is greater than a),
AND MECHANIC.
19
$
are gn and 2, ſo n = , their co-efficients are A 3
9 a fo A
=
II
2
I
9413, and thefe indices in numbers will be 2, 4, 0, 1, 2, 3, each of
2
भु
which differs from n, or by o, 3, 11, 21, or 0, 3, 4 7, the feries
of differences, which by doubling &c. gives, o, 3, 3, 3, 4;, each of
theſe taken from n,, leaves,, 1, 0,
I
3
I
3>
मु
3
A z³ + B z³ + Cz˚, viz. +C+D z ¯ + Eż
4, fo y =
+ &c.
for the affum'd feries.
PROBLEM
CLXXVHI.
CONTAINING, The principals of Geometry, or chief
theorems in Euclid compendiously demonftrated.
Before
you enter upon the following theorems, make your felf maf-
ter of what is contained in prob. 120.
156. If two right lines C G and D E (Fig. 130,) any way croſs each
other, they will make four angles, the oppofite of which are equal, viz.
LEBC = L G B D, and L EBGLCBD, this is fo plain it needs
no demonſtration. Theo. 1.
157. If a line CG, be cut by two or more parallel lines D E, and
A C, it will make equal angles with all theſe parallel lines, viz. L A
La, and LCL c, &c. this is plain per figure. Theo. 2.
158. If a line. B G meet another line D E in any point B, it will
make two angles therewith whofe fum is 180°, for (fig. 130) upon
Rfweep the femicircle, fo are the arches G D and E F G, the mea-
fures of the angles G BD and G B E, which two arches are per fig.
180° or a half circle. Theo. 3,
159. the fam of three angles of any plane triangle are equal to 180°
(fig. 130) for, thro' any angular point B, draw a line D E parallel to
the oppofite fide AC, and produce the other two fides A B and C B,
then per theorem . angle bL B, and by theo. 2. La = L A., and
LCLC, which three angles a, b, and c, are manifeftly — a ſemi-
circle, or 1800, i.e. a+b+c=180° = A+B+C. Theo. 4.
❖
16. If any fide CA of a triangle, A B C, be produced, the out
ward angle B A I is equal to the fum of the two inward oppofite Ls
B and C (fig. 130) for by theorem 3. 1800 BA CLB+LC,
confequently, L.BAI-LB+LC. Theo. 5.
1
120 THE UNIVERSAL MEASURER
161. In a circle LO BF (fig. 48) any angle O I F at the center I,
is double of the angle O B F at the circumference, when the fame arch
OLF, or fame chord FO is the bafe of the angles. For by theo. 5,
LLIF LIBF LFF B, but I BIF (being femi-diameters)
ſo their oppoſite angles I BF and IF B are equal, whence, LLIF
=2 LIBF=2LIFB in like manner it will be prov'd that L LIO
2 LIBO, confequently, LOIF2LOBF. Theo. 6.
This way of reaſoning holds true, let the point B be where it will in
the arch O B F, which fhews that all angles are equal which are in the
fame fegment of a circle i. e. L BOFLBLF &c.
Theo. 7.
Becauſe the diameter of a circle fubtends 180°, viz. half the peri-
phery, its evident by theorem 6, that an angle C F D (fig. 41) in a
femi-circle is a right angle, Theorem 8.
From whence it follows, that an angle in a ſegment O B QF (fig.
48) greater than a half circle is acute, and an angle B QF in a fegment
lefs than a femi-circle is obtufe, confequently, if any trapezia BQFO
be inſcribed in a circle, the fum of any two oppofite angles B Q +
BOF, or OFQ + 0 BQ, is to 180º, and therefore theſe forts
of trapezia's can only be infcrib'd in circles.
162. If any two figures are alike, their like fides are proportional,
that is, thoſe fides which fubtend the equal angles, as alfo thofe fides
which are about the equal angles, are proportional, and confequently
if any two or more figures have all their fides proportional, their angles
are equal. Theorem 9.
For fuppofe the fides A B and A C (fig. 36) of a triangle A E B, to
be produc'd by the motion of the fide or line E B, parallel to itſelf, and
in it's motion to increaſe in length, ſo that the points E and B may al-
ways be found in the produced fides A E and A B, then it's evident
that when E B becomes CD, that A E will become A D, and A B
will be A C, i. e. as EB: DC:: AE: AD, and :: A B : A C, if
we make C F E B and join E F, then by the above faid motion E B,
has increaſed or gained in length, DF, and A E has gained E D, alſo
A B, has gained B C, i. e. as A E:AB::ED: BC
e. as A E: A B ::ED: BC= E F and ſo on,
for other like figures, which are but fome number of triangles put to-
gether, for all plane figures may be fuppofed to be made up of triangles
and therefore what holds in fimilar triangles, muft alfo hold in all fimi-
lar planes; that the triangles A E B and A CD are fimilar, it's plain,
for E F being parallel to `A C, the LFED LEA B, (per theo. 2)
as alfo LEFD LAC D, becauſe CF is parallel to E B, and the
LD, is common to both triangles, in like manner it may be proved
that all the 3 triangles, A E B, AD C, and ED F, are alike; there
are other ways of demonftrating this famous theorem, whofe ufes are
AND MECHANIC.
121
fo many, that few problems can be folved, or theorems demonſtrated
without it, as appears by what follows.
163. If two right lines F E and G I, croſs each other within a cir
cle they will form two fimilar triangles y EI and y GF. Fig. 131.
For by theorem 7, Ly IE➡ Ly F G, becauſe they both ftand on
the fame chord E G, and for the fame reafon, Ly E Ly GF, fo
by the last theorem, it will be as y IyEyFy G.
xyEyIxy G. Theo. 10.
YEI
Ergo, y F
164. If two right lines A G and A F, meet each other without a
circle, (the other ends terminating in the periphery) the angles A IG
and A E F, will be fimilar, for theorem 7, LA FELAGI, and
La common, fo it will be as A G; AF: ; AI: A E. Ergo, A G
X AE = AFX AI. Theo. 11.
165. If a triangle G I F, inſcribed in a circle have one of its angles
GIF bifected, with the line ID, which produced meets the circle in
e, there will be two fimilar triangles IG e and I FD, for by hypotheſis,
L DI F = L DIG, and by theo. 7, L I F D = L I e G, both ftand-
ing on the fame chord I G (fig. 131) and therefore as I D+ De (le)
: IG :: IF: ID, ergo, ID + De × ID=IGIF, but by
theo. 10, FDxD G = De + ID, whence ID F D X D G
IGXIF, or by tranpofition, IDIGXIF-FDXDG.
Theorem 12.
If we fuppofe the triangle GIF to be ifofceles, viz. IF IG, then
its manifeit the bifecting line I D, will alfo bifect the bafe ( F and be
a perpendicular thereunto, i. e. DIF and DIG, will be two equal
right angled triangles, and then the laft theorem will become D
=QFI - OF D, or = GI - GD. Theo. 13, which is the fa-
mous.theorem in 47, E. 1, offo much fervice in all parts of mathema-
tics and in words is, thus; In any right angled plane triangle, the
fquare of the hypothenufe is equal to the fum of the fquares of the two
legs. Now if (in fig. 49) we put u = DE, c = DC, p = D F, a ≈
EF and eCF,
CF, b the hypothenufe C E = a + e, we may prove
this theorem another way; Thus, the 3 As CDE, CDF, and EDF,
are all fimilar. for (by problem 129) LCFD 90° - L F C D = 4
CDF, and L C D E 90º — LF C D = LCED, whence LCD F
LCFD, in like manner it is proved that L F D E = LFCD, the 3
angles CDE, CFD, and EFD, being each 90°,
122
THE UNIVERSAL MEASURER
Then by
theo. 9
2+3
2
1 1a:p::p:e, fo, aepp, whence/aep, Theo. 14
b:cc:e. Ergo, cc-eb exa+e=ac+cc
3 b:n::u:a. Ergo, uuab=axa+aafae
4 ccuuaa+2ae+eea+ejbb QE,D,
cc:un :: ae +ec:aa+ae::e: a, 1hco. 15
6a+e(b):aa+ae (uu) ::a:
5
per 2
and 3
이
​aaa aae
a fe
=aa, Theo. 16
166. (Fig. 131) If e I be the diameter of a circle, and n I a per-
pendicular in the triangle IG F, then the As Le F, and IGn arc a-
like, for by theorem 7) IeFLIGF, and by fuppofition ↳
In G900 and (by theo. 8) LIFe=90°, fo it will be (by theo. 9)
as nI IG::IF: Ie.
Theo. 17.
L
167. (Fig. 131) If any trapezia, IG e F, be infcribed in a circle,.
and the diagonals I e and F G be drawn, and LG In be made
e.I F, the triangles A e F and I Gn will be fimilar, for L nIF = 4
Gle by the figure, and LIe FLIGn, by theo. 7, as alío, LIFn
ELIe G, whence the triangles IFn and Ie G are alfo alike. So
per theorem 9,
I '.*
Į + 2
viz.
I
Ie:eF:: IG : Gn, and as IF: Fn
Fn::el: eG
lex GneFIG, and lexFn= eG XI F
3 Iex: Gn+Fn: eGXIF+eFxIG
Iex GF GXIF+eFxIG, for Fn+ Gn-GF, perig,
2 50 4
Theorem 18. This is called Ptolemy's theorem, faid to be found
by him.
If any two fides of the abovefaid trapezia be parallel, fup,
pofe e F |, to IG, then the other two fides e G and F I will be equal
as will alfo the diagonals, viz. F G e I, and then the last theorem.
will become Ie=□e GeFXIG, or le
le
Ge=ef
XIG. Theo. 19.
□
168. (Fig. 67) If any point C be taken in a right line A B, and it
be made as AC-BC: BC:: AC:C O, and with the radius CO
a circle be defcribed upon O, and lines drawn from A and B.to any
point in the periphery, it will be as BC: AC:: BP: AP. For
P
IAC-BC: BC:: AC: C O, ergo, BC XAC=
COX; AC - B C
Since
+
2 in ::
}
23
fo per 3
4 +
4
5
3 and 5
CO,
:AC+CO:X BCCO X AC
AOX BC, per fig,
*
CO: AO:: BC: AC, but per fig. CO-BO BC,
and A O-CO=AC.
CO:AO::CO_BỘ:AO CO.
CO:AO::BO;CQ, or PO:BO::A0;
BP.
PO, :: AP
6 BC: AC: BP; AP, Q, E, D. Theo. 20.
AND MECHANIC.
123
169. In theſe theorems, when I have found any two angles of one
triangle reſpectively equal to two angles in any other A, I ſay theſe
two triangles are fimilar or equiangular which is evident from theorem
4, and becauſe the fides oppofite to fuch equal angles are proportional
(by theo. 9) it is manifeft that in all As, equal fides fubtends equal
Ls, the greateft fide fubtends the greateft L, the leaft fide, the leaft L,
and contrary. Theo. 21.
170. In any rectangle (A B C D), fig. 132, it's area is had by tak
ing the product of the length and breadth, viz. A BRC area
ABCD, this will appear if you number the little parallelograms in
the figure for you'll find their number equal 66 A B II X B C 6.
Theo. 22.
If DE CF, then becauſe A D and B C, as alío A B and DC, are
parallel, the fides and angles (by theo. 21 and 22) in the As À DE
and BCF, are rofpe&ively equal, confequently their areas will be
equal. Hence, if from the end BC of the ABCD, we take A
BCF (= A A E D) and add it to the end A D thereof, it's plain the
□s A B C D'and A EFB are equal, whence A B x AD — area of
the rhomboides AEF B. Theo. 23. By the fame way of reaſon-
ing it will appear that ABX e F - (ABX BC) area triangle AFB.
2
2
Theorem 24.
Whence if d = the diagonal of any trapezia, e and n two As, fall-
ing thereon, then half de area one of the As, and half d n = area
de+dn :e+n: xd
of the other A, fo half de + halfd n
area trapezia. Theo. 25.
2
But if a trapezia E A er, have two parallel fides e A and E F, then
half ADX DE (E F — DF) — AADE, and A D X D F =
AD Fe, and the ſum of theſe two is, half A DX: EF - DF: +
ADX:EFDF (Ae):
area of the trapezia
ADXDF =
E A e F.. Theo. 26.
2
If S the fide of any regular polygon, Rthe radius of its in-
fcribed circle, and n the number of its fides, then half R S - the
area of one of the triangles of which the polygon is compofed, which
triangles being all equal, we have
it's area. Theo. 27.
RS n
2
If the fides be infinitely fall fo as to lie in the periphery of a cir-
éle, then R will be the rading of that circle, and S being infinitely finall
may be rejected, then will n the periphery of the faid circle, whence
the laft theo. will become half Rn a circles area.
Theo. 28.
"
124 THE UNIVERSAL MEASURER
If inſtead of taking n equal the whole periphery, we take it — any
part thereof, we'll have half R n = area of a ſector, whofe arch is n,
and radius R. Theo. 29.
171. From theo. 21, it is evident that any parallelogram G Hm Q
is divided into two equal parts or triangles H G m and Q G m, by a
diagonal m G, as alfo are the is SGTR and LR y m, by the
diagonals G R and R m, whence, it is manifeft, that from the
G Hm Q, you take As Lm R—Lmy, there will leave □SGQy
= □ G T L H, from each of which, take away the As GTR
and GSR, and there leaves TR yQ=□ SRL H, &c. Theo.
30. Fig 133.
P
If I B be parallel to A D, (fig. 133) and ICBD = p, be to
them, and a➡ A C, e CD, then half p a =
CD, then half pa▲ AIC, and half
: X÷a+e:= A ABD, the former taken from the latter leaves
ре
pe=AABD
▲ A B D – ▲ A B C, which fhews that triangle ABC is
half ap=AAIC, confequently, all triangles (and alfo s, they
being the halves of As) which ftand on the fame bafe or fide and have
the fame height, are equal. Theo. 31.
From this lalt method of reafoning it appears, that there is no dif-
ference whether a and e, be taken as lines or as areas, whence it fol-
lows that all prifms and cylinders, or all pyramids and cones, which
ftand upon the fame or equal bafe A C, and between the fame parallel
planes I B and A D, are equal, viz. folid A IC folid A B C (fig.
133) Theo. 32.
172. If B and b, be the bafes, and p the perpendiculars, A and
a the areas of two triangles, then by theo. 24, half P B = A and half
pba, fo
a, ſo it will be as PB: pb: A: a, (for the being conſtant
may be rejected) See prob. 173. If A = a, then P B
as B: b::p: P, if P= p, then
A
B
름​,
2
pb, whence
or Aba B, whence as
:P,
A:B::a:b. If Bb, then A pa P, whence as A:a:: P:
i. e all As, or alls, are in proportion as the products of their baies
and heights, and if the areas are equal, the bafes and heights, are in-
verfely proportional, but if the bafes or heights are equal, then the
areas are as the heights or bafes. Theo. 33.
The very
fame things hold in prifms or cylinders, cones or pyra-
mids, by looking upon B and b, as the areas of their bafes. Theo. 34.
If theſe figures, are fimilar, viz. their fides &c. proportional, thu,
Pb
if it be as BP::b:
=P,
p, then ab
B
Pbb
B
fo as BP: A ::
Pbb
A Pbb
a. Ergo, a BP
===
that's a BBPAPbb, or a BB
B
B
AND MECHANIC.
125
en 4 bb, whence, as A: a :: BB: b b, and becauſe all planes are
made up of triangles it follows from this procefs, that all plane figures
are as the fquares of their like fides. Theo. 36.
If D, E, F, and d, e, f, be the dimenfions of two folids, and if it
bd
be
be as a :
b
·
: d:
D, and
e:
; b = E, and :: f :
b f
= F,
a
a
a
b
a
or putting r
we'll have rd D, re E, and r f F, then
theſe two folids are fimilar becauſe their dimenfions are proportional,
now if p def the one folid, pdefrrrpDEF muft by the fame
rule, be equal the other folid and therefore as pde f: pdefrrr::
ddd; Pd e f r r r ddd
pdef
=rrrddd = DDD, &c. for E and F,
whence all like folids, are in the triplicate ratio of their homologous,
or like fides, i. c. as any folid is to the cube of any part of it fo is any
other like folid, to the cube of it's like part, or fide. Theo. 37.
PROBLEM CLXXIX.
Of fines tangents &c. with the axioms &c. in plane trigono-
metry.
173 If there be three arches A F, A N, A P, (fig. 134) in arith-
metical progreffion, viz. if A F = A, then A N≈ 2 A, and A P =3 A
draw the lines as you ſee in the figure, then the As z C N and C O B,
COX Nz
being fimilar it will be, as CN: CO :: Nz:
CN
:: Cz :
- O B and
C 7 × CO - CE, but fince F N = NP, FO will be =
CN
OP, whence it's plain O B will be an arithmetical mean between PG
and FH, and is therefore
PG+FH
(half their fum) and for the
2
fame reafon C B
-
COX NZ
CN
NZX2O C
CG + CH
2
CG+CH
confequently,
PG+FH
=0B
2
and
Cz x Co
whence, P G =
2
CN
FH and C G
CN
Cz x 20 C_CH, and if radius
CN be taken equal unity, then PG-Nzx2OC-F H, and C G
= Ċ z × 2 O C — CH, i. e. If N the fine of the mean arch A N,
be multiplied by 2 O C, twice the co-fine of the (arch N F
common difference, and from that product, the fine F H of either ex-
NP)
CN
126 THE UNIVERSAL MEASURER
treme (arch as A H) be taken, there will leave P G, the fine of the
other extreme arch A P. Theo. 38.
Alfo, If Cz the co-fine of the mean A N, be multiplied by twice
the co-fine of the common difference, and CH the co-fine of either
extreme A F, be taken from that product, there will leave CC, the
co-fine of AP the other extreme.
Theo. 39,
Now if e denote the fine of any arch A, and hrlf y it's co-fine, we
may find the fine and co-fine of n A, or n times that arch, thus let A
be confidered as an arithmetical mean between 0 and 2 A. Then by
fine 2 Aey fine Axy-fine o. Theo. 40. radius=1.
fine 3 A-cyy-e-fine 2 Axy-SA
Thco. 38
we have
fine 4 A
fine 4 Aeyyy-2ey finc 3 A xy-S: 2 A
fine 5 Aeyyyy-3eyy+e- fine 4 A xy-S: 3 A,
From whence it appears that the fine of n A will be =ex:y
n
2
n
yn - 3+
-3x
n
X
4
y
口
​5
&c.
I
Ι
2
In like manner by theo. 39, we have co-fine 2 A
co-fine A xy- co fine o, viz. Axy-1 (for radius
y y
I =
2
V+2
2
=
co-fine of o,) co-fine 3 A =
yyy-3y
=
co-fine 2 A X Y
CO.
2
n
fine A, and fo on as before, we'll get co-fine n A=
y n
2
ny +
n
X
n 3
n
n 4
n - 5
6
y
n
X
X
y n
2
2
2
2
3
+ &c.
Theorem 41. which feries must be continued till the indices of y bé-
come = o, if n A be given ſuppoſe = S, then by folving the laſt equa-
tion, we'll have half y the co-fine of and by the former equation
I
n
you may alſo find the fine of, this is evident, for half y being the
.n
co-fine of A, the feries expreffes the co-fine of n A, &c. Alſo, becauſe
the fine of any arch is half the chord of double that arch, it follows
that if 2 e be written for e, we may find the fide of any regular poly-
gon infcribed in a circle, or if we double the fide when found by the
faid feries, of the multiple or fub-multiple arches n A or it will
I
n
be the fide of a regular polygon, or twice fine of the arch, &c. be-
i
}
AND MECHANIC.
127
4
caufe by the nature of the circle, ee+ y y = 1 (viz. fquare radius)
therefore, if n be equal any odd whole pofitive number, we may have
the fine of n A in terms of e only, (becauſe the fine of 3 A, 5 A, &c.
turns out in even powers of y) without furds, thus let 4
put in thefe terms initead of yy it's equal, fo we'll have fine
3e-4eec, line 5 A = 5 è
5 è — 20 e
20 e³ + 16 es fine 7 A
3
56e3+112e5+64e7 &c. whence fine n Ane-
n
4 e e, be
3
A =
7e —
2
n
n
I
X
[e³
I
2
T
n
X
2,3
n² - 25 e?
2,3-
4,5
6,7
it's
2
+ 1/ X
n
I
n
X
-9.25
D:
X
I
I
4,5
2,3
&c. now if the arch A be taken very fmall it
will be equal e,
fine very nearly, then by writing A inſtead of e, we'll have fine n A
-
2
2
n
n
11
n
I n
=n4
A
X
A 3 +
X
X
9 A5
n
X
I
I
I
2,3
2,3
4,5
n
I n 9
X
X
253
4.5
6,7
n`-25 A7+&c. now if n be fuppofed inde-
finitely great, then n A will be equal fome afignable quantity fuppofe
equal z, and the numbers 1, 9, 25, &c. taken from. n may be rejeccd
as fmall in refpect of n, and then the laft feries will become z -
2 5
Z
3
Z
++
2,3
2.3,4,5
Z
7
2, 3, 4, 5, 6, 7
+ &c. (the fame feries with that
given in art. 136) which feries gives the fine of any arch z. Theo. 42.
In like manner you'll get I
8
2
2, 3, 4, 5, 6, 7, 8
2
4
6
Z
Z
Z
+
+
2
2, 3, 4
2, 3, 4, 5, 6
&c. for the co-fine of any arch z.
Theo. 43.
174. (In fig. 134) The triangles CNz, CO B and P N v are alike,
C z X P O - P v, then P. G ≈ (OB +P v)
fo, as CN: CO:: Nz:
COX NZ
GN
OB, and as CN: PO::
CO xNz + Cz×PO
Cz:
CN
CN
COXNZ
CZY PO
CN
and F H (OB-v I or it's equal P v)
that is, the fine (PG) of the fum (PA) of any two arches (A N and
NP) is equal fum of the products of the fine of the one into the co-
fine of the other (radius = 1). Theo. 44.
And the fine (F H) of their difference (FA) is the difference of
the fame products (radius C N being=1). Theo. 45.
128 THE UNIVERSAL MEASURER
175. To find the fine of any arch, ſuppoſe of 50. Firt as 10800 the
minutes in 1800 is to 3,14159265358 &c. (ſee article 136 to is 300
the minutes in 50 to ,08726646, for z, or the length of the given
arch, radius being equal 1, then for the fine, (by theo. 42) for the
co-fine (by theo. 43.)
z=+,08726646
2
3
=,00011076
6
+
+,00000004
I 20
+ 1,
3
2
,00380771
2
Z
+
=+,00000241
24
fo fine 50,08715574, and it's co-fine,9961947
Thus may the fine and co-fine of any arch be found, but greater
the arch is, the flower the feries will converge, and therefore a great-
er number of terms muſt be taken, but having thus found a few fines,
you may by the foregoing theorems, foon find as many as you pleaſe,
and ſo conſtruct new tables of fines, or prove old ones, and from a ta-
ble of logs. you may have the log. fines of theſe natural fines fo found.
Thus, to find the log. fine and log. co-fine of 50, firſt, becauſe the
natural numbers in the common log. tables are but to 4 places of fig-
ures, take the log. of 8715 940296, and of 9962 (near=9961947
the co-fine of 50) = 998344, then becauſe the co-fine has one place
of figures more in it than the fine has, (in this caſe) and the log ra-
dius of theſe tables is fet at 10,000cco = log. fine of 90°
log, tan-
gent of 450, the indices of theſe logs. muſt be 8 and 9, ſo 8.940296
is the log. fine of 5º and 9,998344 it's log. co-fine.
176. Let r radius, s = fine, c co-fine, t tangent, z = co、
tangent, n= fecant, e = co-ſecant, v = verſed fine, then by theo. 9,
and 13, any two of theſe letters being given, any of the next may be
found, as appears by the right angled fimilar triangle in figure 44•
prob. 54.
I. S = √ Iг-cc =
=cn_tz=1
e
tr
tr
r c
rr
n
Z
tt
rr+zz
√nn—rr = 21V
e
n
e
— vv|* =√v |³, V being = verſ-
-- VV
V
ed-fine fupplement.
rs
2. c=r_v=V — r = √πT SS ====√
12
Sz
r z
rz
√
rr + zz
rr
t z
n
n
rr+tt
r
ce-rr,
0
Tr, and fo on for any of
AND MECHANIC.
129
the reft.
Whence
177. It follows, that if log. radius
rro, slog. fine, clog.
co-fine, t = log. tangent, z = log, co-tangent, n = log. fecant, and e
log. co-fecant, then by the nature of logs. we'll have
s=c+t_r=1+r 0,=c+r-z=2r
—z — 2 r—e, and t-s+r-c
(= 21-Z,
z,
=s+r—t=20—0—6+2—r=z+r—e, alſo, z = c +
0 = c + r s=3r-s z=etr-z=2r
(~~c=35
C
1)
(rs2r- t,
c, and ezt î
c_t=nr—t=2r+s,
fo that if all the log. fines or log. co-fines &c. be found firſt, all the
reft are eaſily found by them.
r
178. By the fimilar As (fig. 134) as 2 O B, or P G +FH:20D,
or PI: OB: OD::OL: OF
OL: OF NM: NK, that is, as the fum
of the extream arches is to their difference (fine greater fine leffer)
fo is tangent mean arch to tangent common difference of the arches,
and becauſe the mean arch fum of the extreams, therefore, in any
two arches (A F and A P) as fum of their fines is to their difference
(s greater - s leffer): tangent their fum to tangent half their dif-
ference. Theo. 46.
2
179. In the right angled ▲ A B F (fig. 135) HI is the tangent a e
the fine and e BIB the radius of the arch or L A B F (by prob. 54)
then per fimilar As, BIH and B F A, as BI (rad.): I H (tangent L
ABF) BF: A F. Theo. 47.
Alfo, as e B (rad.): BA:: ea (fine L A B F): F B, whence in the
A BAG, (divided into the two right angled ones A B F and AG F
by the perpendicular A F) it will be as R: A G :: fine LAG F:
A GX SLAG F A F, and by the laſt proportion, as R:: AB
R = radius
:: SLABF:
XS LABF
ABX SLABF
R
A F, as before, whence, A B
AGXSLAGF, in like manner, by drawing G K L
AGXSLGAK
R
B A produced, it will be as R: A G:: SLGAK:
GRY SLABG
GK, and alfo, as R: GB:: SLABG:
A
the fame G K, whence A G ×SLG AK (= comp. L G A B to 180°)
GBXSLAB G. Theo. 48.
R
130 THE UNIVERSAL MEASURER.
:
.ee
BG and eG F; then by theo. 13,
a a
ee = (AB BF)
40+2e4-ee, that is, cc-
2 eu, which turned to an
Let a AB, cAG, u
сс -еe=A Faa uu + que
A F, therefore c c
2eu
eu - uu+a a, or a a сс
analogy gives as u: a + c :: à-c: u
J
a
u u
2 c. Theo. 49.
In any triangle ABC (fig. 135) make BD) – B A, and join A D,
which bifect in F with the line B G, draw F E, parallel to A C, then
will AF FD, DE EC, the
BACHA CB
2
BAF =
BAD + B D A
2
BAC - ACB)
and LG AF BAC-BAF-
2
alſo A B+ AC≈ 2 E B and B C — A B ≈ 2 DE, then the As BEF
and B G C being fimilar it will be, as BE: EC:: BF: FG, but
(by theo. 47) as AF: R:: BF: tangent BAF :: F G : tangent
FAG, or as BF: FG:: tangent BAF: tangent F A G, confequent-
ly, as BE E C (DE) :: tangent B A F: tangent FAG.
Other
wife, As BC: BA:: fine A: fine LC (by theo. 48) and as B C +
BA: BC BA::SLA+ SLC: SLA-SEC:: (by theo. 46,
tangent
A+ C
2
A C
tangent
2
Theo. 50.
Thefe 4 laft theorems demonftrate the 4 axioms in plane trigono-
cB =
metry.
=
180. Given, I a≈e, cВ ! Br, the radius, and e the verſed-
fine of an arch e I (fig.135) z, to find the length thereof. Suppofe the
given verfed fine I a, to be divided into an infinite number of parts
and let a q be one of theſe parts, which put 1, draw v q parallel to
which becaufe qa, is very fmall, the fmall arch ve may be look-
ed upon as a ftreight line, and then the As Bae and ven are ſimilar,
ea,
(vn being = and || qa) fo as en (2 re-ecl): e B (r) : : vn (1) :
foas (2ře
ve, now becauſe there are as many fuch parts ve
r
V:2 re .ee:
in the arch eĬ, as there are units in I a, e, its plain that ve xe=z
then by comparing, this equation with ex. 77, that
re
V:2re-ec:
2,2re, 2 re
is putting it into a feries, we'll have z =
2 re
a re, e e
+
24/2re
2.3 re.ee &c.=√25×
2,82 re
rel
2
1+
2 e 3
+
2,4 r
e //
to
2,3 + &c. which
2,8,4rr
by dividing each term in the feries by the index of e in that term (ſee
AND MECHANIC
izt
rule to prob. 185,) gives the true value of z
2
2
2
3 e e
+
2
3
+
3, 5 eee
35
,4,555
e
2, 3 d
+
3 ee
2², 4, 5 d d
2+, 4, 6, 7 r 3
2 rex: 1 +
+ &c. ot, z=dex: 1+
+ &c. by putting d = 2r. Theo. 51.
Becauſe e the verfed fine in arches under 40° or 350 is ſmall in rea
ſpect of the arch, or of its radius &c. its plain this feries in fuch a caſe
will
converge very faft, and that the two firit terms thereof will give
the length of the arch near enough for moſt uſes, that is z = √2 rë
evare
at
12 r
te:
2
I 2 r
√2re
nearly, or ź=
I 2 r
X: 121
PROBLEM
CLXXX.
Fig. 136.
Given any conic ſection &c. to find its property.
Definition. If a cone U E F (formed as in prob. 155) be cut by a
plane paffing thro' the vertex u and center of the bafe dividing the
cone into two equal parts of fections, the plane of each fection will be
an ifofeles triangle. (182) If a cone be any where cut by a plane
CP parallel to E F its baſe, the plane of that fection will be a circle.
183. If a cone (fig. 136) be cut by a plane TS paffing from T the
extremity of the bate to u F the oppoſite fide, the plane of this fection
will be an ellipfis, whofe tranſverſe axis or diameter is TS, being the
longet diameter, and the fhorteft diameter B A B cutting TS at right
angles in A its middle, is called the conjugate diameter, any part S a
or Ta of the axis of any conick fection, is called an abfciffa, and the
line ba b, which cuts off that abfciffa is called an ordinate, which being
at right angles to each other, the ordinate is faid to be rightly applied.
(184) If the cutting plane SA (fig. 137) paſs any how ſo as A S pro◄
duced will meet the other fide Eu of the cone produced as in T, the
plane of this fection is an hyperbola, whofe tranfverfe axis is TS,
B A B and b a b, ordinates to the abfciffa's S A and Sa. (185) If the
cutting plane S A (fig. 138) be parallel to u E, a fide of the cone,
then the plane of this fection is a parabola whofe axis is S A, b a b an
ordinate to the abfciffa S a, (186) of theſe five fections, viz. triangle,
circle, ellipfis, parabola, and hyperbola, the 3 laft are only called co-
nick fections becaufe the properties of the triangle and circle, may be
known without help of the cone. Thus, in the triangle C S M (fig.
134) RL being parallel to S M, we have per fimilar As, as CN:SM
132 THE UNIVERSAL MEASURER
: co:
S M
CN
XCO = RL, whence, if we put p=
S M
and
C N
e any abfciffa C O, then, e p R L, which equation is called, the
nature, property, or equation of the figure, (in this cafe of a plane
triangle.) Theo. 52.
I
e
Again, let a = CH (fig. 44) the diameter of any circle, z = AB,
BC any abfciffa, to the chord, or femi-ordinate B G y, then
by theorem 10, we will have, a exe=yxy,
=yxy, viz. ae—ee—yy,
which is an equation of the circle. Alfo, becauſe by the figure, I a
—e— z, therefore, ½ a — z—e, and fo, a a
is, AGA B = □ BG), a fecond equation.
4
zzy y, (that
Theo. 53.
187. It is eaſy to underſtand, that the ordinates B A B and b a b,
are chords of circles of the cone whofe diameters are EF and C D,
therefore, by
Théo. 53
By
theo. 92
2 X 3
I and
but
4
I
23
4
44
BAAF XA E, and
Sa: aD::SA: AF
Ta:aC::TA:AE
}
ba=aDx aC.
for the As Sa D and SAF
are fimilar as alfo are the As T A C and TA E.
Sax Ta: aDxaC:: SAX TA: AFX AE.
5 Sax Taba :: SAX TA: BA, in fig.136, 137.
6iSa:□ba:: SA: BA, in (fig. 138) where S A
is parallel to u E, for, then CaE A, and fo in ſtep 3, T a becomes
=TA. Whence, if we put 2 a the tranfverfe, and 2 c = the con-
jugate diameters of the ellipfis, and hyperbola, but in the parabola,
(fig. 138) a =SA, the axis, and 2 c B A B the greateſt ordinate,
then, in any of these three fections, if E S a any abſciſſa, and 2
bab, an ordinate to that abfciffa, we by ftep 5th have, aa:cc::
CC
2a-exe:
exe: X: 2ae
a a
ee:=yy. Theo. 54, for the ellipfis,
alfo, a acc:: 2a+exe:
cc
×:zae+eeyy, for the
äа
=yy for the
hyperbola. Theo. 55, and by ſtep 6, a:cc::e:
parabola. Theo. 56.
ecc
a
Thefe are the chief properties of the three conick fections, from
which other equations relating to them are deduced, if we take z =
A a (fig. 136) = a e, then we'll have X:aa zz:=yy for
сс
a a
another equation of the ellipfis, and when the ellipfis becomes a circle
then a = c, and then thefe equations are the fance with thoſe of the
AND MECHANIC.
133
circle, which fhews that the difference between the equations of the
circle, and ellipfis is only the factor, if TS, be taken for the con-
a a
jugate diameter of the ellipfis, and B B, for the tranſverſe, the method
of demonftration will be the very fame as before, which fhews that
the property of the ellipfis is the fame in refpect of each diameter and
its ordinate. Theo. 57.
PROBLEM CLXXXI.
Given the property of a conick fection, to find its latus, re&tum,
or parameter, and its focus.
188. The letters a, e, y, z, repreſenting the fame things as in the
laſt prob. if we take =p in the parabola, or
CC
a
4CC
2 a
=p in the
ellipfis and hyperbola, then this p will be a conſtant factor, to any va-
riable abſciſſa e, by which its ordinate is found, fo this p being made
as an ordinate is called the latus rectum, or right-parameter, and that
point in the axis, or diameter of the fection thro' which it paſſes, is
called the focus, node, or burning point of the fection, of which there
are one in the parabola, one in the hyperbola, being the point K, in
fig. 28, where H is the focus of the oppofite hyperbola, which being
taken in with K, in the conſtruction, the hyperbola is faid to have two
focuſes, in the ellipfis are two focuſes, which are the points B and C
in fig. 26. Now by theo. 54, 55, and 56, we have
2ae-ee
p
2a:p:2a-exe:
=
: 2ª—exe: PX: 2 ae
P
2 a
2 a
ee:
X:2ae- ee:
yy, for the ellipfis. Theo. 58.
2a:p :: 2a+exe: x2aeeeyy, for the hyperbola.
Theo. 59.
2 2
py::ye, ergo peyy, or univerfally peny, for the para-
bola. Theo. 60.
189. Let the ordinate bab (fig. 138) be the parameter of the para-
bola, given to find Sa (e) the diſtance of a, the focus from S, the
vertex of the parabola, here by theo. 60, pe=yy=PP, (becauſe
per laſt art. =p) hence, e = Pagain, for the ellipfis and hy-
сс
a
perbola, it appears by the equations of theſe two curves, that the dif-
ference is only in the figns and †, viz. —, in the ellipfis and +,
134 THE UNIVERSAL MEASURER
in the hyperbola, whence, being in the equation will ferve both
curves, ſo putting z 2ae the diſtance between any ordinate
y, and the center of the figure we ſhall (by theo.
54,
and 55)
Р
a a
X:
aa Izzyy, for another equation of theſe two curves, now if
ž
P
a a
2 y be the parameter, or y = p, then,
X:aa+zz:=yy=
z: a apa: and
inſtead of its p, we'll have zaa — cc in the cl-
App, hence a azzp a, and fo,
4 cc
сс
writing
2 a
lipfis. Theo. 61, and z = √: a acc: in the hyperbola. Theo.
62, which fhews that PC + PBTS (in fig. 26) &c. for the pàra-
bola and hyperbola. Fig. 27, and 28.
PROBLEM CLXXXII.
Shewing ſome other uſeful properties of conick ſections.
190. By the fimilar As T A E and TS m, as alfo, SAF and STn
(fig. 136) it is as TS: Sm:: TS (TA):
A E, and, as ST:Tn::ST (SA):
Sm x rs
2 TS
TnxST
2 ST
=
S m
2
In
=AF,
2
but by theo. 53, AEX AF□BA =
S m
2
X x In
or 4 BA
2
Smx Tn, that is the conjugate diameter of any ellipfis, is a mean
proportional between the diameters of the cone at each end of the
tranſverſe.
Theo. 63.
191. Becauſe caſks are fuppofed to reſemble theſe conick ſections,
now to find the true form of a cafk &c. let mm D fig. 128) a
diameter taken in the middle between the head Hd and bung B D,
k = Hd, the head, b BD the bung diameters, a =
the caſks length, e — v E.
EL, half
}
AND MECHANIC.
135
2
aab-haa:
a abb
3
Then by the
property of
the parabola
by proper ellip.
I
ين
ce-hh, cx:e+a:=mm, cx:e+a:
bb, c being parameter.
b - h
mD-Hd.
4
=tt, and
Laabb
tt, tbeing half
www
bb-hh
bb_mm
tranfverfe.
from 1st step
bb+hh
4
mm=
Theo. 64.
2
3b+h
per 2
5
m=
Theo. 65.
4
3 bb + hh
per 3
6
m m
Theo. 66.
4
bb - hh
alfo by writ- [
7
mm: hh+
:Xz. Theo. 67
a
ing z in the
3 first steps
b — h
8
m :: b.
: X zz. Theo. 68.
fora, they
a a
become.
9
mm:bb
bb-hh
:X z z. Theo. 69.
a a
By the three firſt of thefe theorems, you may find a diameter in the
middle between the bung by calculation, which compared, with one
taken by your gauge rod &c. will fhew what form the caſk is neareſt,
or becauſe diameters are as their peripheries, you may uſe the girths
inſtead of the diameters, the 3 laft of thefe theorems are for ullaging
ftanding cafks.
Notc. The first of the above ſteps, is for the fruftum of a parabolic
conoid, the ſecond ttep for that of a parabolic fpindle, and the third
for the zone of a ſpheroid.
192, If a folid U FF (fig. 139) formed by the revolution of the
femi curve U E P, about the axis u P, be cut by a plane S A parallel
to the faid axis U P, its manifeft, that B F B and BE B, the bafes of
the two parts fo cut, are ſegments of the circle of the baſe of the whole
folid UE F, whofe diameter E F, call 2 a, chord B A B 2 b, verfed-
fine E A = e, then by theo. 53, we'll have 2 aeeebb, then
by the nature of the generating curve U E P, we may find the equation
of the ſection BS B, made by the faid cutting plane S A. Thus, let
UE P be a half parabola, whoſe parameter, let beq, it's axis u P
c, the radius of its bafe E PFPa, the height S A of the cut-
ing planeh, and SIA P, its diftance from the axis (u P) — z⋅
136 THE UNIVERSAL MEASURER
Then by theo. 60, q c = a a, and q×: c—h:
J
:=zz, whofe differ.
ence is q ha a zz=BA by theo. 53, hence the lection made
by this cutting plane, is a parabola, having the fame parameter with
the generating parabola. Theo. 70.
Again, if Eu F, be a femi fpheroid, whofe axis of revolution is u P
c and E F 2 a the other diameter of the generating ellipfis, E A
сс
e, then by the property of the ellipfis as x: 2a e-ee:
CC
a a
CC
a a
X
BA): 2 ae ee, or lower, as : I, or as cc:aa::SA:
a a
BA, whence, as ca::SA: BA, hence, SBB the plane of the
fection cut off, is an ellipfis fimilar to the generating one U E F. Theo.
71.
193. If from the vertex U of a cone (fig. 141) a perpendicular UQ
let fall upon SA (produc'd) the axis of the conick fection, this per-
pendicular, as alſo U p, the axis of the cone may be found by having
given the dimenfions of the fruftum, ME FS, thus, make SI = Sm,
thro' I draw IG and I H parallel to EF and EU, then draw IT,
HP and SR, Ls to FU, IG and E F, and the right angled As HPG
and I T G will be fimilar, having the common L at G, as alfo wili ITS
and USQ, becaufe of the equal Ls at S, likewife, the As HIG and
Um S, are fimilar becaufe H I is by conftruction parallel to U m.
by theo. 9.
and
I
I = 2
by theo. 9.
by 3 and 4.
2 3 4 in
GH: GI:: HP: IT,
GH:GI::SU: SmSI, by conſtr.
HPIT:: SU: SI,
IT : IS:
S :: UQ:US,
UQ:US:: HP: US, fo, UQ=HP.
GP SR
5
by theo. 9.
6
FR:SR:GP:
FR
=HP=UQ&÷
as alfo
7
FR: SR:: Fp
FPX SR
= UP
FR
The 7 following problems contain the theory of menfurations, or
the investigation of theorems, for meaſuring all kinds of planes and
folids.
PROBLEM CLXXXIII.
Having given the 3 fides of any plane ▲, to find its area.
Let the ſide be denoted by the fame letters, a, c, and u, as in theo.
49. where, we'll have e
cca a
2 a
u u
which fquared, gives
AND MECHANIC.
137
c² + aª + uª + 2 a² c
4 a a
-
2 c² u² - 2 a² u²
which ta-
ken (by theorem 13) from cc, will leave the fquare of the perpendi-
cular of the A, whofe fquare root (fee theo. 24) multiplied by ½ a,
half the baſe, gives the area of the A√: - C
2
2 a²c² + 2 c² u² + 2 u a
a
u + c a+c+".
2
X
2
:= √:
✔
Theorem 73-
u¹ +
- c 4
ctatu
· a 4
X
2
a+c+u
2
194. It appears from fig. 58, that if lines be drawn from each an-
gular point, L, m, n, to E the center of the infcrib'd circle F G H,
that the ▲ Lmn, will be divided into 3 As, ELm, Emn, and En L,
whofe Is are each the radius (r) of the ſaid inſcribed circle, and
therefore by theo. 24, ra+ru+½rc= the area of the ▲
Lm n, whence, if the radius of the infcrib'd, and
area is eaſily found, or if the three
radius by the laſt theorem is cafily had, for
=√:=c+a+u
2
fides
u + a cfu a
2
by divifion gives r = √: Tx
Theorem 74.
X
2
I
3 fides be given the
be given the faid
rx: a+u+c:
X
a + c + u
2
which
c+a+uxc-a+u×c÷a—u
afcfu
PROBLEM CLXXXIV.
Given the 3 angles A, B, and C, of any ſphericaltriangle ABC,
to find its area. Fig. 142.
195. If the circle AR D, be moved equally round the globe, upon
the two poles A and D, it will defcribe equal ſpaces in equal times,
and therefore, thofe fpaces muit be as the angles at the faid poles,
that is, putting D = the axis of the fphere, and p 3,1416, we'll
have (by theo. — ) p DD = the ſurface of the whole ſphere, then,
as 360°: pDD::LA,
LA, or LD: ſpace A Q_D, &c. put G area
A ABC. Alfo, ADEF H (becauſe the fides and Ls in the
one A are refpectively thofe in the other ▲) Rfpace CBQ_D,
TCED, and SA CEP.
138 THE UNIVERSAL MEASURER
1
Then by
what is faid
before
1 + 2 + 3
4
5 ÷ 3
per figure
3
4
2 m & not∞ a
7
8
180 pDD | 9
9
6 and 7
8
9 ÷ 720
о
360: p DD:: A: G+ R
360: p DD:: B : G +
G+
S
R
36: pDD:: C: G+T=H+T
:p
3X360: 3 PDD :: A+B+C: 3G+R+S+T
3×360×:3G-+R+S | T:=3pDDx: A+B+C;
360X 3G+R+S+T=pDDx: A+B+C:
R+S+T÷ ÷ p DD — G
360 xpDD+2G:=pDDx: A+B+C
720 GPDDX: A+B+C-130:
PDD×:A+B+C
A+B+C-180:xPDD. Theo. 75.
G=:
720
196. Becauſe all polygons are made up of triangles, let Sfum
and n the number of angles of any fpherical polygon, of what
ordinate, or inordinate figure foever it be, we'll have PD
+S— 180n, for its area. Theorem 76.
PROBLEM CLXXXV.
720
X:360
Given a e+bee + ceee+de+&c. A, the area of
any curve-lined Space TGC (fig. 31) e being the abfciffa
TC, a, b, c, d, &c. being known, or unknown coefficients, to
find y the ordinate G C to the faid abfciffa, and from the equa-
tion of the cure thus found, to determine its area.
197. Let e z, be drawn very near, and parallel to G C, ſo may the
fmall part e G of the curve be taken as a ftreight line, or the ſpace
GCze, as a Trapezia, with two parallel fides ez and G C, whofe
half fum will be qd, an ordinate in the middle between G Candez,
let z, the abfciffa Tz, then by the above given area, a z + 3 b z z
+ + c z ³ + 4 dz&c. will be the area of the fpace, Tez, then,
+3
its plain that if from the ſpace TG C, be taken the ſpace T ez,
there leaves the trapezia G C ze, which divided by C'z (e-z) gives
(by theo. 26) the ordinate d q (q). thus
3
4
From ae+be² + ‡ce³ +‡de* + &c. = ſpace TGC,
3
take az÷4bz² + ‡cz³ +źdz¹ + &c. = ſpace T e z,
3
e—z) ae—az +žbe²—bz² +7ce³—cz³ + 4de¹—‡dz* (a+be+
(bz+cee + &c. =9.
ae-az
be²-1bz2
be²-4bez
bez-1bzz
bez-1bzz
AND MECHANIC.
139
3
4
The required ordinate or equation of the curve, but when z be-
comese, then q becomes y, and this equation or quotient be-
comes, a+be+cée + de³ + &c. y, for the equation of that
curve whofe area is aebee +÷ce³+de+ &c. A, viż.
if the equation be a y, the area will be a e, if the equation, be a t
bez
y the area will be a e, bee if the equation be bey, the
area will be bee, and univerfally, if peny, the area will be
pentin
this laft equation is (by theo. ¿o) that of a parabola, and if
n + i
I
工
​it be a common parabola, then n, in which cafe Pen+ I
3
n + I
be-
comes pe
(becauſe pe³ y) + ey, for the area of a common
pey)
parabola. Theorem 77.
Ifn=1, the (by theo. 52) pė≈y is the equation of a plane A,
in this cafe Perf
becomes pee
n+ I
(becaufe pe=y) ey, the
fame with theo. 24, from theſe things we may deduce the following.
general rule, viz.
the
198. Having the ordinate (y) clear on one fide of the equation,
if the other fide be in furds or tractions &c. it must be reduced into
a feries, &c. Then multiply each term by the abfciffa (e), which done
divide each term by the index of e in that term, fo have you
or the fum of an infinite number of ordinates which compofe the ſpace,
or figure, the reverfe of this rule gives the curve's equation, this rule,
is alfo fairly proved in art. 93 in prob. 172. Theo. 78.
area
199. In thefe theorems &c. I make no difference between the or-
dinate, ſemi-ordinate for (fig. 31) if you uſe the ſemi-ordinate, G C
(y) with the abfciffa TC (e) you'll have the femi-fegment TCG,
but if I G (y) be taken, you will have the whole fegment IT G, for
GI=2CG, &c. for others.
200. From the above rule, arifes the method of fluxions and fluents
for if e be a variable quantity, the fluent of ve" è
n+1
ve
is =
&c.
" + I
Let z or e always
201. But to apply the above rule to practice. Let z or e
denote an abfciffa, to an ordinate y, A the area of the ſpace, and S its
folidity, then firſt let the area of that ſpace or curve be required whofe
equation is X:
a a
C
Z z: yy, or
V: aa ZZ := Ya
a
140 THE UNIVERSAL MEASURER
which put into a feries (by art. 73, prob. 170) will be y=
y=x: a
ZZ
4
N
6
22
2,44
3
3 Z
2,4,6 as
3,528
2,4,6,8 a7
. &c. each term in
which multiplied by the abfciffa z, gives zy=
C
z 3
X: za
a
N
2 5
2 a
327
8 a 3
48 a
5
of
Z 3
5
Xza-
2
a
6 a
40 a
3
&c. this feries will (by theo.
&c. and dividing each &c. as by the laſt theorem A=
327
336 a 5
57) give the area of an elliptical ſpace intercepted between the dia-
meter c and ordinate
y.
Theo. 79.
If z be taken a, we'll have Acax: I
1 5
3456
TTG
&c. which feries continued will turn out ca × 0,78539 &c.
A, for the area of a whole or femi-ellipfis (as you take ċ and a for
the whole or half diameters &c.) Theo. 8o.
C
If from 0,78539 &c. a c, you take the feries in theo. 79, its mani-
felt there will leave the area of an elliptical fegment, or by ordering
—√2ae-cey, as you did✔a
C
a
a -zzy (ſee theo. 54)
you'll have a feries for the fame fegment, cut off parallel to the tranf-
verſe, when a is taken for the conjugate, but cut off parallel to the cons
jugate when a is taken for the tranfverfe.
202. If a be taken c, the ellipfis becomes a circle, and then thefe
theorems holds true for the like parts of a circle. Theo. 81
•
If we take the diameter of a circle = 1, its periphery (by art. 136)
be 3,1415926, &c. and (by theorem 28) half of 1 (0,5) multiplied by
half of 3,1415926 (1,5707963) gives 0,78539815, for the area of a
circle whofe diameter is unity, or 0,7854 may ferve as a conftant fac-
tor, the very fame will turn out by theo. So, whence, 0,7854 dd =
area of any circle. Theo. 82.
as the fquare
1, viz. I to
Theo. 83.
we take the
Becauſe circles are like figures, therefore (by theo. 36)
of 3,1416, the periphery is to 0,7854 its area fo is
0795 &c. the area of a circle whofe periphery is unity.
It is plain (fig. 134) that if from the fector CPN F,
ACPF, there will leave the fegment P N F, that is, if C P =r, oÑ
the verfed fine e, Po half chord
PNFN, the half arch, then ra
the faid fegments area, but (by theo, 51) a=
a=
C, Co=d=r
e, and a
de (by theo. 29 and 24) = A,
√2re
X: 12гte:
12 r
AND MECHANIC
141
nearly, therefore ra =
√2re
X: 12re: the area of the faid
12
fector, from which take dc the ▲, and there leaves the ſegment
√2re
1601
T
x: 12r+e:—d c, Theo. 84. nearly, or
39 e
X
12
32r-3e
4º √2re, nearer, and if t and q, be the diameters of an ellipfis,
15
we'll (by theo. 81) have:
160г — 39 e
: X
32r-3e
4 et
15 g
2re, for the
area of an elliptical fegment. Theo. 85.
203. Thus you may proceed to find the area of any space &c. if
its equation be given, and having fo found the areas, you may com•
pare the figures or find what proportion one plane bears to another,
as for inftance, if a circle be circumfcrib'd about an ellipfis and ano-
ther circle inſcrib'd therein, let p=0,785.4, a and c, the diameters of
the ellipfis, which are alfo the diameters of the faid circles, then
paa and pcc, and p c a, are by the foregoing theorems, the areas of
thefe 3 figure, which are, as pa a: pcc,::aa : cc, alfo, as paa:
paa::aa: ac, likewife, pa a Xpcc=□pca, which fhews that
every ellipfis is a mean proportional, between its circumference and
infcrib'd circle.
PROBLEM CLXXXVI.
To find the contents of folids, when the equations of their generating
planes are given.
204. As a plane is compofed of an infinite number of lines, or ordi-
hates, fo is a folid of an infinite number of planes or areas, whence it
follows, that if one of theſe areas be multiplied by their number, the
product (if the areas are equal) will be the content of the folid, there-
fore as an area, divided by its length gives a breadth (q) ſo a ſolid
divided by its length gives an area (pqq), whence, if inſtead of taking
a+be+cee+ de ³ + &c. = y, we take it = y y, and work as in
the laft problem, we'll find that when the equation of the generating
curve is ve¹y, or the equation of the folid generated thereby is
3
vvenyy, or cemy y, the folidity will be
m+ I Ι
Pce which
m+I
gives this general rule, viz. if the ordinate (y) or radius of the baſe
be clear on one fide of the fign, then fquare the whole equation,
and multiply each term therein by the abfciffa (e ar z) then divide
each term by the index of the faid abſciſſa or axis in that term, which
142 THE UNIVERSAL MEASURER
!
multiplied by p gives the folidity, p being 3,1416, if y = radius of
the baſe, but p=0,7854, if y = the diameter of the bafe for
and 3,1416 ÷ 4 = 0,7854) Theo. 84. or writing y y for ce m
its equal we'll have Pyve
S, which is a general theorem for all
m÷÷÷ 1
cones, pyramids and parabolic conoids, thus ifce
yy, then PVye
MT I
I
2
pyyes, for a common parabolic conoid.
Theo. 87.
then Py Ve
pyye
2.)
If qe y (the equation of a A) or ceeyy,
m I
pyye S, for the folidity of a cone (m in this cafe being
Theo. 88.
If y be the fide of any regular polygon, and p fuch a number as
that pyy the area of that polygon, ethe axis of a pyramid whofe
bafe is the faid polygon, then pyy e S, its folidity. Theo. 89.
сс
аа
205. In the ellipfis (ſee theo. 54) X: 2ae-ec := -yy, which
multiplied by ep, and each term divided by the index of e im that
term, Theo. 86, gives Xpace - peee S, for the folidity of
cc
a a
3
CC
a ſpheroid's fegment, Q_Mm (fig. 144) = (becauſe yy:2
CC
I
—ee) pe × & Cee + yy. Theo. 90.
x a a
a a
с
If C×:aa-zz=yy, thea (by theo. 86)
C C
px:aaz-
a a
Z Z Z =
S, the folids of any half zone, if z,
Cn, or of the whole
zone MM mm (fig. 140) if z = nn, of a spheroid pccz-
p ccz z z
3 a a
сс
whence, becauſe x: aa—Zz: yy, therefore a a
a a
ZZCC which put in the laſt equation for aa gives after reduction
cc-yy
==
Pzx: 2cc tyy: S, Theo. 91, for the zone of a fpheroid as
alfo of a globe, as is evident becaufe a, the axis of the fpheroid is not
in the expreffion, if e be taken
a, then
C C
a a
PX: aée
S, (fee theo. 90) becomes pcca S. Theo. 92.
eee=
AND MECHANIC.
143
If in theſe theorems, you write a for c, and c for a, the fame theo-
rems will hold true in the fame parts of an oblate fpheroid, as is plain
from theo. 57. Alfo, if the ellipfis become a circle viz. if c 2,
then thefe theorems hold true in a globe, or fphere and its parts, viz.
蛋 ​paaa
folidity of a half, globe, pex: ae—ee= folidity of a
2
globe's fegment
pex:ee +yy: &c. Theo. 93.
206. If QV Qv be a fpindle formed by the revolution of the ellip-
tic arch QVQ about QQ (fig. 140) a chord parallel to TT (a)
one of the ellipfis's diameters, whofe other diameter is 2 OV (c) let
Cd, the diſtance between the center O, of the ellipfis and the
axis of rotation QQ, e = Cn, Aarea of the elliptic fegment.
MV d, b Vv, and y Mm, then (by theo. 57) X:aa-4ee:
dy=
=
c c
а а
√: aa—4ee: — 2 d, fo
=2d+y=Mm, whence y=
a
CC
yy=
4ee:+4dd-4d
a a
Vaa-4ee
-4ee which mul-
a
tiplied by pe, and each term divided by its index of e gives pc ce
+4ddep8dpxfpace MqOV, which ſpace (fee theo.
4 cce³ p
3 a a
79) is manifeftly
and c
2
ed+ey+A which being put for MqO V,
2dy: for 4ccee
2d12dd12dy:
a a
its equal, we'll have 2 ca
I
pe-8 Ad, S, the folidity of
V M m v a fruftum or half zone of the fpindle QV Qv, but d +
fo 2 cc8dd8db + 2 bb, which put for 2 c c in the
b
= c, fo
laft equation gives 2 bbyy+8d: x by
3xpe=
A:
C
S, the fame folidity. Theo. 94. If ac, or the ellipfis become a
circle, then the fegment M V d will be a circular fegment whofe area
is A, and the last theorem will become S
2bbyy+8d:xb-y-3A: pe, for the fruftum of a
e
circular fpindle, Theo. 95. Ifin thea. 94. we write
d for + d,
then that theo. will become S: 2bb+yy-8dx:b-y-3A;
e
Xpe, for the fruftum or zone of an hyperbolic fpindle, Theo. 96.
A being area of a hyperbolic fegment M V d, if yo, the C n =
e becomes CQ, half the fpindle's length, then the 3 laft theorems
144
THE UNIVERSAL MEASURER
become S-2bb +8dx:b
3 A
Xpe, for the folidity of an
e
Theo. 97.
elliptic, circular, and hyperbolic fpindle.
If do, then theorems 94, 95, and 96, becomes, S: 2bb +
yy: Xpe, for the zone of a globe or of a ſpheroid, or of an hy-
perbolic conoid. Theo. 98. And if y = 0, then S = 3 pebb, for a
globe, fpheroid, and hyperbolic conoid. Theo. 99. Theſe two laft
theorems are the fame with theorems 91 and 92, which ſee.
C b
2
a+b
↓ 2
If you would have the above theorems without d, you may put in
its equal
in the ellipfis, or
in the hyperbola's fruftum
and by the equations of theſe two curves, work out a and c, but then
the theorems will be very complex. Alfo, you may aproximate the
area of the fegment M V d, (fee theo. 84 and 85,) and put it in theſe
theorems for its equal A, but they are eaſieſt as they now ftand.
207. If QV Qv (fig. 140) be a parabolic fpindle, form'd by turn-
ing the parabola Q_V Q, round its double ordinate QQ, let M m
y (being | Vv) 2 CV = V vb, the double axis of the generat-
ing parabola, or diameter of the ſpindles greateſt circle, Cne, and
c=the parameter of the faid parabola, then (by theo. 60) c x Vd
ee, and therefore, yb
= Md, or cx:
b - y
2
fo y y=bb-
4 bee
C
2 ee
C
+4eeee, whence (by theo. 86) Spex:
CC
bb- 4 bee+ 4e+: but 2-b-y, whence pexbb-
3c
3c
5 cc
C
4 eeee: will become epX:bb-2bx: b—y² +
4 bee+
5 cc
4
|
5
by := {pex:
8bb+4by +-3ÿÿ
5
3
10bb+3yy—2bb+4by.
5
xpe: 2bb+yy-b-yl: x+pe, which are 4 thco-
rems, for the fruftum of a parabolic fpindle. Theo. 100.
If yo, the fruftum becomes a whole ſpindle, and then we'll have
: 2 bb - 3 bb : pe=}pebbS, for the whole fpindle (ife
QQ). Theo. 101.
If U EF (fig. 139) be a conoid form'd by the revolution of the
parabola E SU about its axis u P, let u Pa, PEPFb, the
greateftordinate SI=y another ordinate, I Pe, and c para-
I
=
AND MECHANIC.
145
Spe:x:ca+‡yy−
: 2
2
meter, then (by theo. 60) c x: a e: yy, whence (by theo. 86)
pc ex: a eS, but ca ceyy, folyy-ca———ce,
whence Spe:X: ca+yy-ca: ca+yy: alfo,
cabb, fo Spex:bb+yy: the folidity of a parabolic co-
noid's fruftum, Theo. 102. And if yo, then Spa bb, for the
whole conoid, as by theo. 87.
207. If A BFE (fig. 127) be the fruftum of a cone, let its height
n Ee, diameter of the greater baſe A Bb, that of the leffer baſe
EF
y, put c the parameter of the A B v A, and d=b
then (by theo. 52), cx:a—e:=y (a being
fo cc: aa
ولا
the axis or D v)
2ae+ee:—yy, whence (by theo. 86) Spcc ex
: aa aeee: but becauſe the As a EB and A v B are fimilar,
ce will be d, focceed d, alfo, cab, or ccaabb, and
cexcaccae= db, which put in the equal of S, gives Spex
:bb-db+dd: the folidity of the fruftum. Theo. 103.
If the fruftum become a cone, then y = 0, d = b, e = a, and the
laſt theo. will become Spa bb, the fame with theo. 88, but if
y= b then d = o, in which cafe S pebb, the folidity of a cylinder
Theo. IO4.
If ba fide of a fquare prifm's bafe, then p= 1, fo Sebb.
Theo. Io5.
If in theo. 103, you write b y for its equal d, you'll have S≈
pex:bb-bb+yb÷÷dd:=pex:yb+dd: Theo. 106.
If in this laſt theo. you put bb - 2 by + y y, for its equal d d,
bb-2yb + y v
you'll have Spex:yb+
= 13/pex:bb+
yb+yy. Theo. 107.
3
And becauſe circles are as the fquares of their diameters, therefore
if A and a, denote the areas of the two bafes, then 3 Spex: A+
√ Aa+a, a general theorem for all fuch fruftums. Theo. 108.
If A B C be a cone or pyramid (fig. 133) with an oblique baſe A C
(whofe area let bea) where the axis BD (e) cuts the plane. A C of
the baſe produced, then by what is faid in this article, with theo. 34,
we'll have Se a, for the folidity of this oblique folid. Theo. 109.
Or if A B C be an oblique fruftum whofe bafes A C (a) and B (A)
are parallel, then for the fame reafons Sex: A+√Aa+a:
Theo. iio.
208. If the fruftum F Em'S (fig. 141) of a pyramid, or cone, Eu F,
be obliquely cut by a plane S A, the parts FSA and ASm E fo cut
T
146 THE UNIVERSAL MEASURER
are called hoofs, or for diftinction F S A is called hoof, and ASm E
the complement hoof, and if the cutting plane S A, be an ellipfis,
parabola or hyperbola, the hoof is called an elliptical, parabolic, or
hyperbolic hoof; in any of which it is plain by the figure, that the
hoof FSA, is— AUF — AUS, thatis put Aarea of the fection made
by the cutting plane S A, a area of the circular fegment whofe ver-
fed fine is A F then by theo. 109, up a conical part A U F
and UQXA = oblique cone A U S, ſo ♣
FSA, of any kind or form
(p being up and q = Qu)
3
pa
?
q A = hoof
whatever either conical of pyramidical,
Theo. 111.
III
And this hoof taken from the fruſtum F Em S, leaves A Sm E, or
taken from Sʼn A F, leaves comp. hoof Sn A Theo. 113.
But by theo. 72, you may exterminate p and q,
hoofs in terms of the fruftum &c. thus by writing
and fo have the
GPX SR
FR
for q
and
FpXS R
FR
a
for P, theo. 111, will become
x:FPXSR:
3 FR
A
PR
P, the
3 FR
X GPX SR:: Fpxa-GPXA:
3 FR
fame folidity.
Theo. 113.
This theorem is general, whether the cone &c. be right viz. its
axis fall in the middle of its bafe, or oblique, viz. when the axis is
not in the middle of its bafe, by fome called a pyramidic cone, fuch
an one is the cone, A u F, or A US, but in theſe theorems following,
I mean the right cone, made by prob. 155, where the laft theorem
will become more fimple, for then Ep Fp, and G PIP, and by
FAXSm
fimilar triangles, as SA: FA::Sm(=SI):
let DEF, dSm, and h
2 SA
D-d
2
dx FA
and GP=
2 SA
=GP,
SR, the height then F Fp - 1 D, FR
whence the laſt theorem will become
dx Fax A
h
: DX a
X
=S,
S, Theo. 114.
SA
3D-3d
Let Tm Sn (fig. 136) be the fruftum of a ſquare pyramid, cut
diagonally by a plane S T, each fide of the greater baſe being — D,
and of the lefferd, and height SRh, here a≈DDA=
D+d
2
XTS SAFA (n T). D, which put in the last theorem gives (for
¡
AND MECHANIC.
147
a, A, and FA=D=Ta): DDD -: dxDxD + dx
2hDDD - d DD h –– D d d h
6 D-6 d
h
2 3:D+d:
=2DD+ Dd: xh, S,
which taken from: DD+ Dd+dd: (fee theo. 107) xh, leaves
: 2dd +D d: xh, S, for theſe two hoofs, STn, and ST m.
Theo. II5.
Let Tm. S (fig. 136) be an elliptical hoof of TmSn, the fruftum
of a right cone, with circular bafes, whofe diameters are S m≈d,
Tn D, and height SR h, put p=0,7854, now by theo. 63,
Dd=BAB, the conjugate diameter of the ellipfis, fop Ddx
STA, and a = p DD, which put in theo. 114, for A and a,
TS, for S A as alfo Tn for F A, we'll have S=:
h
3: D-di
= :DD
DDD P
I
and
: D D — dy Dd :x
dxDxDdxpxTS
TS
: X
ph
3: D― di
X
=S, the comp. hoof T m S.. Theo. 116.
the whole fruftum Tm Sn, leaves : DDd-dd:
d ph
3: D-d:
In the parabolic hoof S AF (fig. 138) we have (by theo. 77) SA
* BB
area parabola BS B = SAX: ✔d x D
d: = A,
alfo, a area of a circular fegment whofe chord is B AB, and height
FA, which FA is D-d (becauſe S A is || E m) thefe put in theo.
114, for A and F A gives S: Da4dxD-dxSAx:vdxD--d
SA
h
D
3: D-d d: D-d
ха
4 d
3
✔:dxD-d x h = hoof
†
SAF which taken from the fruftum (EFSm): DD+dxD+d:
Xph, (ſee theo. ro7) leaves, S:DD+dxD÷d : Xp
Da
4 d
+ √id xD −d : ×h for the upper, or comp.
3
D-d
hoof S A Em.
:
Theo. 117.
If the cutting plane be an hyperbola (fig. 137), in this cafe, F A
and SA muſt be had before the hoofs can be meaſured, and then D,
d, A, a and h, being as before, we'll (by theo. 114) have SD a ~
dAX FA
for the lower hoof S AF. Theo. 118.
SA
:X
h
3:D-d
143 THE UNIVERSAL MEASURER
But if the cutting S A, be L E F, then (per fig.) S A=h, and F A
Dha d A
by which the laſt theorem becomes, S =
D-d
2
the fame hoof. Theo. 119.
D-d
But as
2
و.
3:D-d:
6
is commonly ſmall in refpect to D, we may in fuch a
h
8D+6d
8 d + 6 D
:X· √DD-dd
45
cafe, take: 47DD+23 Dd_ 47 dd + 23 D D-
S, hoof SAF nearly. Theo. 120.
PROBLEM CLXXXVII.
To find multipliers for reducing caſks to cylinders, &c. with feve
ral other useful theorems in menfuration, and gauging.
209. Having in the laſt prob. fhewn how to inveſtigate theorems
for the moſt uſeful folids, I fhall here fhew how a multiplier may be
found for reducing a folid to a cylinder &c. which is of great ufe in
the gueffing way of cafk gauging. Theſe caſks are faid to be of four
forms or varieties, viz.
variety the
I two equal
fruftums
2
3
or zones
of a
fpheroid
parabolic spindle
parabolic conoid
cone
curving
moſt
lefs
leaft
nothing
between
the head
and bung.
With theſe four varieties fome mix two more, firſt, the fruftum of a
circular ſpindle after ſpheroid, and the fruftum of an equilateral hy-
perbolic ſpindle, before parabolic conoid, but theſe four above are ſuf-
ficient.
Now, let e the cafk's length, b-bung, and h-head diameters,
d-b-h, their difference, m the required multiplier, then m d, or
mx: b h added to h muſt be the mean diameter, viz. that of a
cylinder to the caſk and of the fame length, whence pe x h+md]
=S, the cafk's content,: 2bbhh:Xpe, by theo. 91,
2bbhh-2 dd :x+pe, (by theo. 101,): bb+hh: xipe,
(by theo. 102):bb÷hh+bh: Xpe, (by theo. 107), divide
cach of theſe by p e, and you'll have
va.
I
2
Z
for i | 2bb+hh-3x: htdm: 3hh+6hdm+3ddmm,
2bb÷hh-0,4dd3hh÷6dhm+3ddmm,
bb÷hh=2hh+4dhm+2ddmm,
bb+hh+bh3hh+6dhm+3 ddmm.
riety
3
4
18
And by tranfpofing 2 h h, in the third variety, and 3 h h in the other
three varieties,
AND MECHANIC.
149
2bb2hh6dhm+3dd m,
22bb-2hh-0.4dd6dhm+3ddmm;
6 dhm+3 dd mm,
We'll
have
4
bb-2hhbh
3
bb-hh=4dhm+2ddm m.
In theſe 4 equations; for bb write its equal h÷d, viz. hh +2
hd+1dbb, and in variety 4th, forh b, write its equal: h+d:
Xh, and you'll have,
I 4dh2dd
24dhi,6 dd6dhm+3 ddmm,
413 dh+dd
32dh+dd4dhm+2ddmm.
Now by tranfpofition, and diviſion, theſe will become,
:6m-4:×h
I.
-d
2 3 mm
2.
: 6m 74:xh
=d
- 1,6 + 3 mm
:4m−2:Xh
3.
=d
4.
-d
I - 2 m
:6m−3:xh
I- 3 m m
leaft limit
Now to have d the greateft poffible, its
plain the denominator 2
3 m m muft
be:
= 0, or 2 = 3 m m, and to have d
the leaft poffible, the numerator 6 mi
- 4 = 0 &c. or 6m 4 &c. by which
the value of m, in each variety is,
greateſt limit
m
m
m
m
|| ||
variety.
m
✔
,8165
I
m
NIS
›7303
m
,7071
m
123+
4
√,5774
Thus you ſee the leaſt and greateſt value that m can poffibly have,
in theſe 4 varieties, and that there can be no one multiplier fet to any
one of the faid varieties, but to find (m) a multiplier for any particu-
lar cafe, let us fuppofe a caſk of the most common form, bung dia-
meter b
32, head diameter h=26, the b — h=d=6, then
by what goes before we'll have
:—h: =0,6888 &c. Theo. 121.
2bb‡h h—0,4dd;—h:—0,6753 The. 122
I 2bbhh
I
d:v
3
for
2 m=
V:
m
V:
va-
riety 3
4
m=
3
bb+hh: — h:= 0,5259 &c. Theo.
h:=0,5259 123.
2
bb+hb+hh
:-h:=0,5087 &c. The. 124.
3
150
THE UNIVERSAL MEASURER
By putting fquare of h+md
2bb+hh
3.
=2bb+hh.
d d = &c. &c.
EF
210. Let there be given a trapezia A BEF (fig. 127) with two
parallel fides A B and EF, to find the length ou= LD, of a given
part to be cut off, by a line G u parallel to A B or EF; let b
1=n E, a = Eu, Sarea part Gu E F to be cut off, and 2 d
a B. Then by fimilar triangles, as 1:d::a:
AB-EF
daa
uzbutuz xuE=
-
d a
T
AEzu, which added to FG z E
I.
a=
2 d
(b×a) gives daª +ba=S, whenee a√: ydls+bb : — Ib
Theo. 125.
1
211. If A BEF (fig. 127) be the fruftum of a pyramid &c. and it
be required to cut it into two other fruftums A BuG and Gu EF,
let h En, the height b = a ſide of the leffer bafe EF, d dif.
ference between a fide of each baſe, a B, c = a factor by which if
you multiply the fquare of a fide of the pyramids bafe &c. and that
product by its axis, may give the pyramids folidity, a = Eu, the re-
quired length, and z the folidity of the part GuEF to be cut off,
then by fimilar As, as d:h:: b::b:
bh
hb³c-
d
yv, fo
folidi-
d
:
chbi
b3h3
ty top pyramid E F V, then (by theo. 37) as
or in
d
d3
hh
lower terms, as c:
::
dd
chb3
d
ch 3 h 3 + z d h h
dh
+z:
-cube
cddd
•
of LV, fo
+
a,
d
cdd
d
Theo. 126. or putting q
Lit will be
-
bh a.
C
d
3: h3 b3 zhh bh
3: h3b3qzhhd:
212. If A B C D (fig. 142) be a ftreight fided fruftum with parallel
bafes A B and CD, and there be given the folidities of the three parts
or fruftums ED, DG, and DI, A, B, C, refpectively, to find the
folidity of a fourth fruftum DI=D, the perpendicular diſtances be-
ing equal to each other fuppofe a (viz) a = La= 2a, a 2 a,
3 a =&c. let e TL, then per fimilar As, as AB: PT:: CD:
PTXCD
A B
TL=e, let
PT
AB
I, then eCD, and for the
AND MECHANIC.
151
reafon eta EF, e+2a≈GH, e÷3a
e+3a=IK, e+4a=
Lm, CD, EF, GH &c. being parallel to one another and may be
taken for the ſquare roots of the areas at the ſeveral diſtances TK,
OCD XTL
Ta, Ţ 2a, &c. and therefore (by theo. 89) eee=
е
2
folidity of TCD, likewife +al x:e+a:
folidity TE F, now
two folidities is manifeftly
3
3aae+зaee+aaa
3
3
DEFXTa
3
the difference of the
the fruftum CDFE, in which expref-
fion, by writing 2a, 3a, 4a, feverally for a we ſhall have the required
values of A, B, C, D.
viz.
I
3aae3aee+aaa
3
6aee+12aae+8aaa
L2a
A,
= B
2
and
3
4
I a
2 ÷ 2 a
6
Qaеe+27aae +27aaa
12aee48aae +64 a aa
53ee3ea taa:=
3ee6ea4aa:=
3
3
3
= C
D
The height of the
fruftum being
·L3a
L42
A
a
B
2 a
C
3÷3a
73ee9ea +9 aa:
3 a
C
8
: 3ea5aa :=
3 a
22
C
B
8 X 3.
9+5
10 X 4 a
9:9ea15 aa: =3×:
1
3 a
2 a
10:3ee12 ea +16aa:—
II 12eea + 48 ea a + 64 aa a
A +3x:
C
B
cal
a
3 a
2 a
4a :X*
A
C B
-+3×:
:= D
a
3- a
2 a
as appears by the fourth ftep.
Theo. 127.
152 THE UNIVERSAL MEASURER
Again, if a La the height dd the area at C D, alſo, DD—
the area at E F, and e difference between the fquare roots of theſe
two area's viz. e — D — d, then ed― D, which put in theorem
106 inſtead of b, we'll have: dd +de+ee: ×a= fruftum
CDFE, now if za be put any depth as L 2a, L3a, &c. then
by the property of the A it will be, as a: e:: za: ze, by which
we have dd:d+ze:ze × z a, — that fruſtum whoſe depth is
Theo. 128.
-
za.
PROBLEM
CLXXXVIII,
To investigate theorems for the furfaces of folids.
213. To underſtand the furface, or fuperficial content of a folid,
imagine it to be fewed in cloth &c. and then this cloth to be ripped
open and ſpread out flat, and meaſured in this polture as a plane, will
give the fuperficial content of the folid, from which it plainly appears
that if a=
the periphery of the leſſer baſe, A — that of the greater
baſe, and e➡ the flant length of any ftreight fided fruft
theo. 26) have, :a+A:×÷e=C, the curve, or cor
Theo. I29.
And if a
cylinder, &c.
we'll (by
Jurface.
A, then e AC, the convex furface of any priſm,
A, then e A:
Theo. 130.
If a
o, thene AC, the curve furface of any cone, &c.
Theo. 131. Convex, or curve ſurface, means the ſuperficial content
of the folid without its baſes or ends.
214. If the femi-fector, BI e, be turned about the dius Bi fig.
135) its manifeft, the arch ei will defcribe the fegment of a glo or
ſphere, every point in the ſaid arch, as v, e, &c. defcribing the peri-
phery of a circle the fum of which peripheries are the convex fur-
face of the faid fegment; therefore, any part of this arch, as ev may
be taken as an abfciffa, and that the periphery deſcrib'd be v or e,
as an ordinate to that abſciſſa, let the arch ve be very ſmall, ſo as to
be eſteemed a right line, which will be (by the nature of the circle)
at right angles to the radius e B, fo will the As Bae, and v ne be
fimilar (vn being || BI) and therefore as e a (y) : e B (r) : : v n (e) :
➡ve the abfciffa, let c = periphery to the radius r, then as r:c
y
:: y: cy
er
y
r
periphery to the radius y, which multiplied by the abfciffa
(ſee art. 198) gives cyer
ce=C, the curve ſurface deſcribed
Y
by the arch ve. Theo. 132.
AND
153
MECHANIC.
And if p3,1416, and d—e— 2 r, the diameter of the whole
circle, then c = pd, and ce= C, becomes = pdd = C, the ſurface
of a globe or ſphere. Theo. 133.
3
=
Becauſe pdddS, (fee theo. 93) and pdd C, therefore,
CdS, for the folidity of a globe or fphere.
It appears that e=vn=qa, may be taken
Theo. 134.
any part of the radius
IB whence cepde=C
= C = curve ſurface of any fegment, fruftum
or part of a fruftum whofe height is e, and therefore, as d: pdd ::
=pde, that is, as the axis of a globe is to its ſurface ſo
e:
pd de
d
is the height of any fegment &c. to its curve furface. Theo. 135.
The furfaces of the fpheroids &c. being tedious to turn without
fluxions, fee them &c. in prob. 190.
Shewing
PROBLEM
CLXXXIX.
inveſtigation of a new and ſwift converging feries,
areas and folidities of curve-lined ſpaces, &c.
1
215. Let ean abfciffa to the ordinate y, and de p¤¹ x²+fe" |
y the equation of the curve, firſt involve z+fe" to the mth power
and it will be z
+mfe"zm-1+mxm-1: ffe 2 "z +
m
m 2
&c. which multiplied by de Pn-I and by the abſciſſa e, and then
>
eac
each term divi
by the index of e
in that term (fee art. 198) we
de po
Z
m
mfe" z
n m I
m
m
I
X:
+
+ X
P
P+I
2
will have
วก
n
m-2
ffe Z +&c.
P + 2
A the required area, but to make the fc-
ries converge fafter, let v
put in the last feries inſtead
the mth power we will have,
z+fen then z=v-fe, which being
of of z, and then each term involved to
m
m
V
my
ift
P
P
The
2d
3d
terms =
I.
fen
mv
m-I fe
P+1
+
m I
m-220
m
V
&c.
2
P
m-iv
m
I V
2
m
X
I
2
El
I
X.
M-IV
m
2
U
m
ffe 2n
P+I
2ffe 20
P + 2
+ &c,
&c.
154
THE UNIVERSAL MEASURER
1
m.
V
mi
m-I fe
n
Now by collecting thefe 3 laft fteps we have
+
P
PP + I,
m-2
2ffe
20
&c. where the law of continuation is plain
m,m-r,v
P,P+P+2,
/
now this multiplied by
+
m'm-1.ffe 2n
V
v²'p+1°p+2°
de Pa
becomes
de
pn
m
n
P11
n
mfe
WP-k I'
&c. A, by which the area of any known fi-
gure may be had nearly, in a few terms, as for example.
сс
Let x:ae Fee:= yy, or
2.C
a
a a
2 C
a
vaeec = 2y (lee art. 189)
x√exv:a Fe: which compared with the general equa-
tion de P´n
I
m
1xz+fe"=y, it appears that 2d, ee, n= 1,
a
pn-1, ergo, p, az, f=1, m=1, vae, which
1=1,
put in the laſt feries, we will have A = 4yX;
e
e e
I.3
1.3.5 V
eee
e
+
4
3.5.7v²
57.9v³
-&c. for the area of an elliptical fegment if you
write for, but for the area of an elliptical fegment if you write
for. Theo. 136.
And if a c, the elliptical fegment becomes a circular one, viz.
A = 4√dvx:
V
V V
+
1.3 3.5 d
V V V
V
4
+
3 5.7dd
5.7.9d3
&c.the arca
of a circular fegment whoſe verſed fine is v, diameter of the circle D,
and d D — V.
Theo. 137.
Again, for folids, let q=3,1416, then 3, 1416 y y=qae‡qee
аа
gex: a Fe: which compared with the general equation de Pn - I
&c. we'll have q=d, eże, n≈i 10 p=2, z=a, f=41 and m≈
1, by which the general feries becomes qee
x: a + e +
7
1
2
3
0 := qae e qeees, the fame in effect with theorems 98,
and 90.
AND MECHANIC.
155
PROBLEM CXC.
To measure any plane or folid without its equation given.
216. If instead of taking a, b, c, d, &c. (fee prob. 172) for a feries
of numbers, we take them for as many ordinates at right angles to an
abfciffa e SC (fig. 31) and v diftant from each other, viz. a a, aa,
&c. v, now if y - any of theſe ordinates, a
n the number of ordinates between a and y,
number of ordinates, it is manifeft, from fteps
the firft of them, and
or n+1 = the whole
10, 11, 12, of prob.
ก
I
n
I
n
2
172, that ya+n A÷x
B+nx
X
C
2
2
3
+ &c. now ife the abfciffa, or diftance between the ordinates a
and y, and v = the diſtance between each ordinate betwixt a and y,
its plain that nvme, therefore, n= which put in the laſt equa-
e
tion inſtead of n, and reduced to fimple terms will be y=a+A+
св
+
e3 C
ee C
e C
+
3
&c. which multiplied by e, and
2 v v
3 V
2 V V
ee B
2 V
eA
CA + еe B
3.2Vv
each term divided by its index of e (fee art. 198) gives A=ex:a+
2 V
e B
-+
eee C
2.2V 4 V V V
ee C
e C
-+
for the area
3.2 vv
2.3 v
of that ſpace whofe abfciffa is e, and ordinates, a, b, c, d, &c. to y,
let the equation of the ſaid ſpace be what it will, hence, if we take
two ordinates, then ev, and B, C, &c. (as appears from ſteps 7, 8,
9, prob. 172) areo, therefore, area ex: a A: but Ab
b - a
a (by ſtep 6, prob. 172), therefore, area=e×: a+
2
:
e×:a+b: again, if three equi-diftant ordinates be taken then (by
ſteps 8, 9, prob. 172) C=0, &c. and B = a
2b+c, and here
e B
ev, whence area
ex: a + + CA +
e
ee B
3,2 V V
V
3.2 V
:=ex: a
+ A + B÷B:ex:a + 4b+c: and thus going on by
taking ev, ev, ev &c. you'll get the following tables,
giving the area when 2, 3, 4, 5, &c. ordinates are taken, in which
(for brevities fake) I put A the fum of the two extream ordinates.
B the fum of the next two, C the fum of the next two &c. viz.
if there be taken five equi-diſtant ordinates a, b, c, d, f, I take A — 2
+f, Bbd, and c=d+o.
156 THE UNIVERSAL MEASURER
So,
now
(Axe=a+b:xe,
A+4B: × ==a+b+c:מe. Theo. 138.
4 equi-:A+3B:X
diftant |
ordi-
A+ 3B :X = &c.
nates
are ta-
the a-
rea is
when
ken,
6
7
་
8
3
:7 A + 32 B+ 12 C : ×
01:
90
: 19 A + 75 B + 50 C : x288
:41 A+216 B+27 C+ 272 D:X
{ :751A+3577B+1323C+2989D:×
e
840
e
17280
&c.
It is proved in what follows that three ordinates being taken, one
at the greater end, one at the leffer end, and one in the middle between
the two ends of any ſpace to be meaſured, will by this method give
the content near enough for common ufe in any figure whatever, for
which reaſon the table of three ordinates is noted with theorem 138,
if ae A, b E F and e AD, (fig. 132,) then per table of two
ordinates, we have area: a+b: xe, the fame with theo. 26, and
if ao, then eb area, the fame with theo. 24, &c.
=
217. The fame things hold alfo true in folidities, for its plain by
the procefs, that there is no difference whether a, b, c, d, &c. be ta-
ken as areas or as ordinates, therefore, if they be taken as ſpaces, or
areas, then the above tables give folidities inſtead of areas, fo that if
a and c the areas of the two bafes of any ftreight fided fruftum whofe
length is e, and b an area in the middle between a and c, then a and
✔ca fide of a fquare at each end, and by the property of fuch a
Na+vc
folid
✔b, a fide of a fquare in the middle, whence, 4 b
=a+2√ ac+c, which put in theo. 138 for 4b, we get: 2a +
2 √ ac+2c: xe, the fame with theo. 108. If there be fuch a
prifmoid as that Tx Da, and tx dc, the areas of its two ends
or baſes, and if D and d be parallel as alſo T and t, then by the pro-
D+d
perty of the folid
b, the area in the middle be-
2
T + t
2
2
tween the two baſes a and c, ſo 4b=T+tx D + d = DT +
DtdTdt, which values of a, 4 b, and c fubſtituted in theo. 138
AND MECHANIC.
157
gives, S = : 2 DT + D t + dt +2dt: x the ſolidity of the
prifmoid. Theo. 139.
Again, for the hoofs of a fquare pyramid, let Da fide of the
greater baſe, d = that of the leffer, and e the hoofs height, then a
=DD,b=D+0xD+d-4: DD+ Dd, and c = 0, ſo by
Ꭰ
2
2
theo. 138, S: DD+ Dd: × the fame with theo. 115.
3
= nì
!
If h the head, b the bung diameters, ni a diameter in the
middle between the head and bung, and ethe length of any caík,
whatever, then p b ba, 4 pmm4b, and hhc, fo: a + 4 b†
:bb+4mm+hh: x pe= the content of that caſk
e
either true or very near, or the ullage of it, if b, h and m be verfed
fines of theſe diameters. Theo. 140.
For (by theo. 66) 4mm =3bb+hh, which put in the laſt theo.
for 4m² gives : 4 bb+2hh: × pe S, the fame with theo. 91,
alfo, (by theo. 64) 4 mm =2bb2hh, which put in theo. 140,
for 4 mm gives 3bb3hh: xpe S, the very fame with theo.
3b+h=9bb6bh+
2
102. Again (by theo. 65) 4 m m =
hh: by which theo. 140, becomes : 13 bb+6 bh + 5 h ²
8bb4bh+3 hh,
eps nearly, but (by theo. 100) Spex:
5
=
'pe
I 20
2
: X 1
×:64 bb + 32bh+24h h: this taken from pex: 13 bb
+6bh5hh, viz. from pex: 65 bb +30bh+25hh:
leaves pex: bb-2hb+hh:pexb-h,too much
which is all that any of theſe theorems differs from truth, and is fo
fmall, that in gauging, the error cannot exceed part of a gallon,
but if inſtead of taking h, m, and b, as three equi-diftant ſpaces, we
work with them as 5, viz. h+h A, the fum of the two extream
ones, m+m B, the fum of the next two, and b=C, the middle
ΤΟ
one, then by table 5, we have: 7 A+ 32 B +12 C: x CPS,
wherein A phh+phh2phh, B-pm'+pm
=px:36+b
8
X
90
2pm m
>
and C pbb, which put for A, B and C, in the faid
158 THE UNIVERSAL MEASURER
32
ер
table, becomes:14hh+ ³½ 3b+h) ‍+ 12 bb : × CD =:
90
e p
५०
X: 48
bb +24 b h +18hh: S, the very fame with theorem 100.
Ifenne (fig. 31) be a part of a ſpheroid's frullum, let c = CI, a =
CT, bna, the radius of the greater bale, hna, that of the leffer
and eaa, the fruftum's length, put z— Ca, then by the property
of the ellipfis, (putting n=
I 2n xa a
2nX: aa
cc
C_C).
a a
zz:= 2bb
2
2
4nX: aa (z+ c ) ) zz — ze —
ze-ee:=2bb2hh
we
have
2
enx:aa — (z+e) zz
p
2ze - ee :
ee: ahh
3 +
ee:=4mm
1+2
3-4
4
4nX: aa ZZ: 4
5
but
6
5:6
nee = 4 mm
lı
- 2 bb
2 h h, or 4 mm nee+2bb2 h.
:bb+hh+4 mm: xep S by theo. 138,
7:3bb+3hh+nee: xep
nee
6
Or putting h and b
: Xpe. Theo. I4I.
bb + hh
S=;
2
whole diameters, m in the middle be
I
tween them, then p muſt be ,7854 inſtead of 3,1416, and then S
bb +hh
2 nee
:
+
2
other methods.
3
: Xpe, for the fame folidity true as found by
218. By this method we may alſo approximate the center of any
curve, or ſpace of a given equation and that in a few terms, by what
any common ſeries can do, as for example. Required A the area of
a curve whofe equation is =y, let the abfciffa e be divided
I
Ite
into 6parts, and then we will have theſe 6 equi-diſtant ordinates,
I
VIZ.
I
I
I
}
I
>
which
5
I
1 + 0/0
1 + 2/2/ I + 종 ​1 + // 1 + 3/
66 , fo by table 7, we have A1+
reduced are 1., g. 5, TO THI TI,
6
6
४
+9 = 1,5, B = ++= 1,4025974, C=8+18=1,35, D=8
유아 ​TI
+0=0,6666666, whence 41 A + 216B + 27 °C + 272 D =
e
I
582,2443717, which multiplied by ‚viz.by, gives 0,6931479
840
A, which by article 113, is the hyperbolic logarithm of 2, (for
-e-ee &c. the fame ferics as you'll fee there) and would
I
Ite
AND MECHANIC.
159
require 1000000 terms of that feries, to get 6 places of figures true
as is here done by only 7 terms, and by taking e = 2, 3, 4, &c, you
may have the hyperbolic log. of 3, 4, 5, &c.
་
Note. What is here faid of areas, is true of convex ſurfaces.
PROBLEM
CXCI.
An equation being given, to find the greatest or leaſt of its values
poffible, called maxima and minima.
219. If a be an indefinitely fmall quantity, and b+a7e, as alfo
b-ae, I fay that be, fór by tranfpofition, b7c — a, and b ▲
ea, but a being indefinitely fmall, can by adding or by fubtracting
make no alteration, and therefore, in ſuch caſes may fland as o, viz.
b7e-o, and b Leto, confequence be.
220. In the equation p q
t+be" _e▾
.e =m, wherein p, q, t, n,
b, v, m, are known quantities, and e a variable quantity, let the value
of e, be required when m is a maximum, or the greateft poflible ſo-
lution, let a—an indefinitely fmall quantity, and put f= p q — t, and
z "be" then the equation will be f+za
e, each have a added, then m =f + z + a|
m, and let z and
V
e
=m,
n
e+
+ al
(by
n
I
n
I
+ nx
aaz
+ &c.
2
ev
F I
V
I
_vae
V
aae
▼— 2 ___ &c. but the former part
2
evolution) f+za+naz
of this equation being a maximum is greater than the latter part there-
of, that is f+za
V
€
n
I
7f+z+naz +0x
n
I
aaz
2
n
2
+ &c. — e vae
I
2
-VX
aae▾ -&c. (which
2
V-I
Π
+x
by tranfpofition and divifion becomes) ve 7nz
n
I
a &c. -VX
J
2
V
2
a &c. Again, if you take z a ande-
a, for z✦a and e+a, and proceed in the fame manner, you will find
n
V
I
n-I
V
ve
n z
n X
a &c. + v·X
a &c. but
2
2
a being fmall, all the terms multiplied thereby, may be rejected, as
being=0, and then we'll have v e ˇ-17n z
+o, and ve
160 THE UNIVERSAL MEASURER
1
t
1
Loz
n-r -ve'
or n z
whence we have.
n-1
o, whence by art. 219, ve
V-I
n I
n z
VII
=0, viz.nbe
n-I
n
I
- ve
= ♡,
221. This rule viz. having got m the maxima &c. on one fide of
the equation, and the other fide confifting of one variable quantity e,
and its powers, in at leaſt two terms, then multiply each term by its
index ofe, and from the faid index of e in every term take unity,
which done for m write o, fo you'll find e, and if there be any terms
in the equation not engaged with e, fuch terms vaniſh, as is plain
by the above procefs; by the fame rule you find a minimum, which
is only known from a maximum by the nature of the queſtion, the
one being contrary to the other. The above proceſs alfo demonftrates
the method of fluxions, and by the rule fuch expreffions may be flux-
ed, thus, the fluxion of fe is
аё
21/ae
V
-V I
=ae
I —
I
for vac
=
ve~~I è, that of a e, is
— — — a÷√ae, to
1-4
= ae
which join e, and you have the fluxion of a e, hence, the fluxion,
of
any expreſſion madeo, gives the maximum or minimum, of that
expreffion.
-ee
222. Let it be required, to divide the quantity b into two fuch parts
as being multiplied together, may produce a maxima, let e one of
the parts, then be muſt be the other part, fɔex:b e: eb
eem and per rule, b 20=0, whence, eb. Theo. 142.
Alſo, if ex: b-e: were to be a minima, we would by the fame
proceſs have e=1b, but then its plain, the lefs we take e, the lefs
will be the product, fo in this cafe e muft be indefinitely fmall, and
thus it may fometimes happen that a maximum or minimum, may be
had by reafoning from the nature of the problem. Again, if b is to
be divided into 3 fuch parts as that their product may be the greateſt
poſſible, let b — ¤,
e, = one of theſe parts, then will efum of the
other two parts, but (per laſt theo.) the greatest product that can be
made out of the fum of any two parts of e, is exže, fo
e
e
X
2
2
x:b-e:
e:=
bee
eee
=m, a max,
fo
per rule,
2be - zee
4
4
= 0, or 2 be — 3 e e = 0, or 2 be = 3 ee, or 2b=3e, whence, e=
b. Theo. 143.
Required e, the axis of the greateft cone, that can be cut out of
the fpheroid whofe axis is a, and radius of its greateſt circle c.
AND MECHANIC.
161
CC
1. By the property of the ellipfis we have 4 cc
a a
Хае
ce: =yy,
which multiplied by pe gives 4CCP X: aee-eee:=peyy, the
conem, ſo
4ccp
per rule,
a a
a a
X:2ae-3ee:≈o, whence e = +
a, but if e the axis of the greateft infcribed cylinder, then per faid
cc
property, Xaa -ee:=yy=☐ radius cylinder's baſe, which
4 aa
multiplied by pe gives CCP x: a ae-eee: peyy its folidity
=m,
fo
per rule,
ccp
4 aa
4 aa
X:22 - 3ee:0, whence ea√, thefe
alſo hold true in a globe, becauſe c is out of the equation. Theo. 144.
Required En (h) the axis of the greateft cylinder, that can be cut
out of a given cone Av B (fig. 127) whoſe axis v D — a, diameter
A B of its baſed, let e EF the diameter of the cylinders baſe,
then
per
a
d
fimilar As, as a :
I
2
d::h: e
, ergo, h=
ad.
a e
2
d
Let ==c, then cx: d
d
then cx:d-e:h, then peehcpx:dee-
eeeS the cylinders folidity, which ordered by the rule, will give
e=÷d, whence —×:d—e:=xe—e=ae÷‡e=
* anh.
Theo. 145.
ае
:
Required e, the axis of the greateſt cylinder, whofe diameter is d,
and diagonal a, here, a a -ee-dd, and pedd-px: aae-eee;
m, fo per rule, a a
3ee0, whence ea, which is alfo
the axis of the greateſt cone that can be made out of a ſlant ſide a,
=m,
and baſes diameter d. Theo. 146.
Required e the diameter of the greateft cylinder (open at one end)
made out of a given fuperficies c, let S its folidity, a=,7854, then
4 S — curve ſuperficies, and ace
area baſe, ſo 45
e
+ace=48c-1
+aecc,m, a minimum, or the leaſt ſurface out of the great-
eft folidity, fo per rule, -4 Se
= 45, whence c =
ee
e
4S
+nae=0, or 2 a e
=0, or 2aę
ce
2
Theo. 147,
X
2.
162 THE UNIVERSAL MEASURER
n
223. If q be a variable quantity and q x3 = 9
then per rule, n + vxq
V
then n + xq
Theo. 148.
V
v+n
v+n
= m,
=0, and if for qª we take za
D-I
V
Enz
q+vq
ZAO.
Again, if d+zm, first (by art. 220) d+z", becomes o †
+Az n-1 which call t, alfo d+z" becomes v xdz"|" I
n -1
V
V I
V
which call w, then twnz xvd+z" =o. Theo. 149.
By theſe two laft theorems you may find the maximum of a furd
or of a fraction &c. thus if
ę s Q C
:e+Cxe+Q:
2
m,
e s Q C x
e e + e C + e Q + QC. Here t2e+C+Q, and w
S
ee + e C + e Q + Q C-x-1 fo (by theo. 148) tw xes QC
I
SCQ
whence
+SQCX:ee+eQ+eC+Q_C {
zee+ec+eQ: x − s Q C +
sac +
ee teC+eQ + Qc ²
2ee-eC-eQ=ee+eC+eQ+QC, and fo QC = ee•
-
Again, if ee-eee
I + el
cc +eQ FeC+Q C
I
=m,
m, =: ee
eee:X ITE here in
,to+1, and w: i+e X-1, fo (per theo.148)
twx:ee-eee:+: ze -3ee: × 1tel
1+e:x:2e3ee: - ee+eee
ee eee
I
1+el
28
26€
2 e -3ee
of
−3ee:
Ite
´Ï+ef²
2eee0, whence eete
1, fo (by art. 81) e =
I + V
2
=0,6180 &c. for any fuch like.
PROBLEM CXCII.
This and the 8 following problems, contain the theory of mechanics,
but first of definitions.
224. By mechanics, is meant the forces, velocities, motions, actions
&c. of bodies one upon another.
225. Motion is a reftlefs ftate, or continual change of place of a body.
226. Uniform motion, is when a body moves over equal fpaces in
equal time.
{
AND MECHANIC.
163
227. Velocity, or fwiftnefs, is that affection of motion by which a
moving body runs over more or lefs fpace in equal times.
228. Accelerated motion, is when the velocity continually increaſes
but if the velocity continually decreafe, its called retarded motion, if it
increaſes or decreafes uniformly, it is equally accelerated or retarded.
Alfo, if its motion be confidered in regard to fome other body at reſt,
its called abfolure motion, but if the motion of a body be confidered
with refpect to fome other bodies alſo in motion, its called relative
motion, and that way it moves is its direction.
229. Momentum, quantity of motion, or impetus, is the power or
force, with which a moving body ftrikes another body at reft, or is
the motion a body has both in respect of its velocity and matter.
230. Celerity of motion is that affection by which a body paffes
over a given ſpace in a given time, or what is called the ſwiftneſs or
flowness of motion.
231. Compound motion, is that which is produced from different
powers acting in different directions, Sir I. NEWTON has eſtabliſhed
and divided motion into theſe three general laws, viz.
232. Law 1. All bodies continue their ſtate of reſt, or of uniform
motion in a right line, till they are made to change that fate by fome
external force impreffed on them.
233. Law 2. The change of motion produced in any body is always
proportional to the force, whereby it is affected, and in the fame di-
rection wherein that force acts.
234. Law 3. Re-action is always equal and contrary to action, or
the action of two bodies upon each other are equal, and in contrary
directions.
235. Body is the mafs or quantity of matter, if a body yields to a
ftroke and recovers its former figure again, its called an elaftic body,
if not its in-elaſtic, or a non-elaſtic body.
236. Denfity of a body, is the ratio of the quantity of matter it
contains, to that in another body of the fame bulk.
237. Force is a power exerted on a body to move it, if it act but
for a moment it is called the force of percaffion, or impulfe, if it act
conftantly, its called the accelerative force, if conftantly and equally,
it is called a uniform accelerative force.
238. Gravity, is that force wherewith a body endeavours to def
cend towards the earth's center. Abfo ute gravity, is when the body
falls in free ſpace, and relative gravity when it defcends in a fluid.
Alfo, fpecific gravity, is the greater or leffer weight of bodies of the
fame bulk.
164 THE UNIVERSAL MEASURER
239. Center of gravity of a body is fuch a point in it, as that the
body being freely ſuſpended on that point, would reft in any poſition.
240. Center of motion of a body is a fixt point about which a body
is moved, and axis of motion is a fixt line about which a body moves.
241. Center of ofcillation, is that point in any pendulum into which
if its whole weight or gravity be collected, the time of its vibrating.
will not be altered thereby, and is the fame with the center of percuf-
fion in bodies that move about a fixt point, but when the body moves
in a parallel direction, the center of percuffion is the ſame with the
center of gravity, and in either cafe, is that point in which the force
of the ftroke is greateſt.
242, Friction, is the reſiſtance arifing from the rabbing of bodies a-
gainst one another.
243. The length of pendulums &c. are meaſured between the cen
ter of ofcillation, and the axis of motion, or pin on which they hang,
the pin being very fmall.
244. It is manifeſt from art. 239, that the whole weight or force
of a body may be confidered, as acting, or contracted into its center
of gravity, and though lines and ſurfaces are confidered, as having no
weight, yet if they are taken with any thickneſs as the baſes of folids
&c. they must have weight and confequently centers of gravity.
245. If right line be fo drawn thro' any plane or folid, as to biſect
all the ordinates or fpaces which compofe that plane or folid, this line
is called the diameter of gravity, becauſe the center of gravity is al-
ways in fome point thereof, fo if any rectangle have drawn within two
diagonals, the interfection of thefe diagonals is its center of gravity &c.
246. Axis of fufpenfion, a line at right angles to the axis of motion.
247. Equilibrium, the equality of weight, of bodies keeping one
another at reſt.
248. In the following theory, all planes are fuppofed to be perfect-
ly even and fmooth, all bodies perfectly fmooth, except it be mention-
ed otherwife.
249. All lines ftreight, levers, inflexible, chords or ſtrings, pliable
and without weight, unleſs expreffed to the contrary.
250. Rare, or lightneſs, little matter under much bulk, and is op-
poſite to denfe.
1
AND MECHANIC.
165
=
PROBLEM
CXCIII.
To demonftrate the univerſal laws of motion.
251. Let q=the body, or quantity of matter in the body to be
moved, f force acting on the body q, m = momentum, generated
in q, v velocity, generated in q, Space defcribed by the body q.
in the time t, with the velocity v. Then, first in all kinds of motion
m is proportional to q v,
for its plain, if the velocity be double, the
quantity of motion m will be as 2 q, if the velocity be treble it will be
as 3 q &c. confequently, if the velocity be v, m will be as qv; other-
wife, fince v acts equally on every particle of matter in q, its plain
that m will be as 2 v, 3 v, 4 v, &c. if the particles of matter be 2, 3,
4, &c. confequently, if the particles of matter be q, m will be as qv,
Secondly, in all kinds of motion S is, as tv (252) for if the velocity be
uniform, viz. if a body moves v miles every hour, its manifelt, that
t hours, it will move IV, 2v, 3v, - t v miles, or the
fpace S, but if the velocity be uniformly accelerated viz. be 1, 2, 3,
t, in the ift. 2d. 3d. t th equal parts of time, then S
in 1, 2, 3,
•
1+2+3
+t, by theo. 24tt, but, as the time increafes, the velo-
city alſo increaſes, therefore, t is as v, fo Stv, or S is as tv, as
before. Allo, becauſe v is as t, therefore, in this cafe Stt,
becomes, Sv v. or S v.v, (253) hence it appears that in uni-
form motion, if the time be given, or conftant, S∞ v and if the velo-
city be conftant, then St, but in uniformly accelerated motion, S
vv, when the time is given, and Sott, if the velocity is given,
but in both motions we have S ∞ tv.
254. In any uniform motion, m is as f, for if no force act on the
body to put it in motion, it will have no momentum, therefore the great-
er the quantity of motion, and becauſe the velocity is ſtill the fame,
we must always have m ∞ f, let the time be what it will, but in uni-
formly accelerated motion, where the velocity, and confequently the
force, increafes with the time, we muſt have m tf.
255. From theſe articles, by uſing the fign
we'll bave the following theorems, from m,
and m, ∞ tf, viz.
90
as it were the fign=,
qv, Sα tv, fx m,
mxxmxSt. Theo. 150, q ∞, &c. as before.
txxx. Theo. 151, t
Sq
m
996
m
+ x & x √ Bα = .
t
να
x Samα1.
Theo. 152, v
ft
fs
166 THE UNIVERSAL MEASURER
tf Theo. 153, Sqft
Sxtvx™m x 1.
9 9
m∞ f¤vq× 15. Theo. 154, m∞ft.
gigi
१.
fα V
fom o vq « ? Theo. 155, fơ quan qu c
C x c
t
τι S
t
Thefe on the left fide of the theorems, are for uniform motion, and
thofe on the right for accelerated motion, and it m∞ f, orm OC if,
were alike, (fee art. 254) thefe theorems would be the fame in boch
motions, and therefore, if what arifes from art. 254., be entirely left
out, the proportions on the left hand fide of the theorems, will hold
true in both motions.
256. Of compound motion, let us fuppofe a body to be moving from
A towards B, (fig. 143) in the direction A B. but by fome force act-
ing thereon, is compelied to move in the direction A D, or which is
the fame thing, if while a body would move from A to B, the
ABDF, would move from A B to D F in the direction A F = BD,
its plain the body by this compound motion, would in the fame time,
defcribe the diagonal A D, that it would defcribe the line A B, in by
its own motion, that is, while the body by its own force, would be
Carried from A to D. Hence, if AB exprefs the force of the body,
then BD AF, will exprefs that of the impreffing body, and AD,
the joint force or effect of both A B and B D (= A F) viz. force A D
force A B+ force B D. Theo. 156.
Hence, if a body at D, acts on an obtacle, or plane A B, in the di-
rection DB, and with the force DB, let fall the LDC, and com-
pleat the rectangle B CD E, fo is force DB - force DE— force E B,
(by the laft theo.) whereof force E BDC, acts perpendicularly a-
gain't the plane A B, but force D E acts parallel thereunto, and fo can
avail nothing, fo in this cafe force DEo, and fo force DC the
whole, or greateſt force poffible of DB, againſt the plane whence the
greatest force imprefs'd, is in a line DC, perpendicular to the plane
of the obftacle. Hence appears, the method of dividing one force into
two or more forces, for if D A exprefs the force of a body at A, in
direction DA, and it be required, what part of that force acts in any
other direction BA, then if from D you let fail the LDC upon the
faid B A, you'll have CA for the required force, viz. as DA to CA
fo is radius to co-fine L AD C fo is force in direction D A to force in
direction D A to force in direction B A.
Theo. 157.
Alfo. if a body at D acts on a plane B A in direction DB, and it be
required what impreffion it makes on the faid plane, firft let fall the
DC, then (per laſt theo.) as DB: the bodies whole force :: DC a
AND MECHANIC.
167
that part of its force which acts on the plane B A, and is as radius:
the fine of the incident CBD Theo. 158.
If there be any number of forces, fuppofe 3, A, B C, (fig. 144)
acting against one another in the point D, fo as to keep one another
in equilibrio, viz. force of C both the forces of A and B, firit produce
AD and BD, and compleat the DICH, of which DC is a dia-
gonal.
Now its plain, if a body be kept in equilibrio, the contrary forces in
any one line of direction muſt be equal, viz. force in direction AD –
that in direction ID, &c. or the body cannot be in equilibrio but will
move, if theſe contrary or oppofite forces do not deſtroy each other,
which they cannot do unleſs they be equal, whence if DI-CH, ex-
prefs the force of A, then D H = CI, will exprefs the force of B,
which two forces (by theo. 156) are CD; but the force of C is
alſo theſe two forces, fo CD is alfo = the force of C, whence the
forces of A, B. C, are as DI, IC, CD, or before the ſides of As, are
as the fines of their oppofite Ls, there alfo holds true viz as DICI
(A : B) : : S. L D CÏ, or CDB : S. LCDI, or CD A, and as C I
: CD (B:C) : : S LCDI or CDA : S. LC HD, or HDI, or AD B.
If there be ever fo many forces acting againſt a point D, and keep on
another in equilibrio, they may be reduced to the action of two equal
and oppoſite forces by the fame method, for if HD and I D, be two
forces they are the ſingle force D C, &c.
Theo. 159.
+
257. The meeting of hard non-elaftic bodies, if a moving body A,
whofe velocity is V, momentum M, and weight or quantity of mart
Q, ftrike another body B, whofe velocity is v, momentum m, and
weight q. and ife, be put
and ife, be put the velocity of both bodies after the ſtroke
then (by art. 234) we'll have fee
Theo. 154.
1. QV+qv=Qe+qe, fo e
=
move both the fame way.
QV+q when the bodies
Qv v
Q+q
2. QV - q r = Qet qe, foe when the bodies
move to meet each other.
Whence in both cafes it will be e —
Q+q
QV9V. Theo. 169.
етя
For its plain if the bodies both move the fame way, B will gain what
A lofes by the ftroke, but if they move the contrary way, or meet
each other, then the greater momentum over powers the leffer, and
the bodies muft both move in direction of that which had the greater
momentum and with the difference of their momentums, if B be at
168 THE UNIVERSAL MEASURER
Qv
reft before the ſtroke, then =0, and foe=
q + q
Theo. tốt,
Again, becauſe the quantity of matter multiplied by the velocity
(by theo. 154) gives the momentum, we'll have Qe=
i
for the momentum of A after the ftroke, and q e =-
for the mom. of B, after the ſtroke, whence QV
Qq
Q+q
QQ V+Q q v
тя
Q qvqq v
Q79
Q Q v + Q q v
Q+9q
the momentum of A loft in the ftroke, and
confequently to that gained by B, but this lofs or change of motion
in either body meaſures the magnitude of the ſtroke, wherefore, A
and B ftrike each other with a ſtroke always=
Qq × V÷v, or
етя
proportional to V+v if they meet, but to V.v, if they move the
fame way, and if B be at reſt before the ſtroke, then y = 0, and the
magnitude of the ftroke will be
conftant). Theo. 162.
9 V
Q+q
or as V (becauſe
ая
is
Q + q
If we put z the velocityof A, loft in the ftroke, and y that of B,
gained in the ftroke, then, becaufe (by theo. 162)
Q v
V
=e,
Q+q
the common velocity after the ftroke, we'll have V
Qv ± qu
Q+q
q V + q v
Qutq
— Z, and
Q V + q v
Qv=qv=v=
итя
Q VFQ v
=y, which
етя
two equations turned into analogies gives, Q+q: q:: V TV: Z₂
and Q+q: Q :: V-v: y, in which for v write o, if B be at reſt
before the ftroke. Theo. 163.
T
258. Of elastic bodies, or fuch as give way when preffed by the
ftroke &c. but after the removal thereof, reftore their form again
pow if this reftoring force is equal to the force of compreffion, they are
faid to be perfectly elaſtic, but if thefe forces are unequal, their ratio
is called the elaftic force, and obferve, that all homogenious bodies of
the fame kind have their elaftic ratio invariable, for it muſt be the fame.
in all bodies whofe conftituent parts are the fame, now in theſe fort of
bodies the velocity loft and gained mult confilt of two parts, viz. one
from the force of compreffion, and the other from that with which
AND MECHANIC.
169
the parts reftore themfelves, which parts are equal in perfect elaſtic
bodies, and in a given ratio in non-perfect elaſtic ones, now if a and b
expreſs the velocities of A and B after the ftroke, and ir as the
compreffing force is to the reftoring force, then by the laſt theo. we
will have
I.
9X: VIV
ётя
2. Q v : V = v
Qv: VI
Q+9
the velocity of A, loft by the compreffing force,
= the velocity of B, gained by that force.
the velocity loft by A in the reſtor-
qrx: v=v:
IX!
3
9-9
Qrx: VI v:
2 X r
4
९:- १
v
I+3
5
Q+q
2 + 4
6
:
:+r:X9X:V
I Lr: ~Q X÷V Tthe total velocity gain'd by B
ing force,
the velocity of B, gained by that
force,
the total velocity loft by A,
1
V - 57
}
7
V
Q+9
: 1+ r : xq x : V?
Q+q
v:
=a=
-2qx:V+v:
Q+q
+V
= V = − b = ±v+
6±88v+
2
+ r : ~ Q v : V
: 1+r: ~Q Y : I v:
· V
VI
етя
when I or the bodies are per-
fectly elallic.
you muſt take the upper
When the bodies move the fame way,
fign before v, but if they tend to meet, you muſt take the lower fign,
and if B be at reft before the ſtroke, then v=o, alfo, if we fuppofe
the bodies void of fpring, or elafticity, theſe two laſt ſteps will be the
fame with theo. 160, whence they are a general theorem in all cafes
that can happen in the collifion of elaſtic or non-elaſtic bodies. Again
if A ſtrikes B at reft, and a cauſes B to ftrike another body C at reft
with the velocity b, and ifc exprefs the velocity of C acquired by the
ftroke we'll by theo. 161) c=
Sq b
(putting r+S) or becauſe
9+ C
(by foregoing 6th ſtep) b=
S Q V
Q+ q
it will be c =
S Sq Q V
:q+C:x:Q+q
which multiplied by C gives
SS qQ VC
:q+C:x:Q+9:
-
the momentum of
Y
170 THE UNIVERSAL MEASURER
Gafter the ftroke, and if this momentum be taken a maxima, wheres.
in q is variable, we'll (by theo. 149) have QC qq, viz. AC-BB
hence as A: B: B: C, that is B must be a mean proportion between
A and C, that the velocity or momentum of C after the ftroke may
be the greateft poffible. Whence, if it be required to find (n) any
number of bodies fuch, that by the firft ftriking the fecond, the fecond
ftrikes the third, the third the fourth, and fo on to the laft which be-
ing ftruck by the laft but one, may have the greateſt velocity, and
(confequently the greatest momentum becaufe the weights are conſtant)
¡t is manifeſt theſe bodies muſt be in geometrical progreffion, they be-
ing all at reft before the ftroke of the firft body) whofe ratio let be e,
and Q the first body V its velocity, then the fecond body will be eQ
and
S Q V
Q+ q
its velocity will become SQV_Sv whence.
Q+eQ 17e
SQ &V
ite
will be its momentum, confequently the lalt body will be e n − 1
2
·
n I
n [
S
es
S
its velocity
V, and its momentum
I + e
I te
1
Qv.
In all thefe expreffions, if the bodies are perfectly elaític, then r I,
or r + 1 = S = 2, and if they are hard non-elaſtic bodies S goes out
of the expreffion.
259. In all theſe cafes the bodies are fuppofed to impinge on, or
ftrike one another perpendicularly, but if they ftrike one another ob-
liquely the fame things may be found after the manner of theo. 157,
and 158. Thus fuppofe the body A, (fig. 145) to be moving in the
line AC, with the velocity A C and the body B, to be moving in the
line BC, with the velocity BC, and let the plane or line EFC, be
that where the bodies touch each other when they ftrike, or meet in
the point C, on which EFC, let fall the perpendiculars A E and BF,
which (by theo. 158) expreffes the velocities wherewith A and B ap-
proach each other, compleat the rectangles E G and F H, fo A C the
motion of A divided into two others A G and A E, to which the mo-
tion in A C, is as. A C to A G and A E refpectively, in like manner the
motion of the body B is refolved into two other motions B F and B H,
to which the motion B C is as BC to B F and B H, refpectively. But
fince AG and B H are parallel, the bodies by the velocities in theſe
directions cannot ſtrike each other, fo that velocity with which A,
comes directly againſt Bis-G C, and that wherewith B comes againſt
A, is HC, now let C L, be the velocity of the body A, from. C to-
wards L after the ſtroke (which G L may be found by the foregoing
theorems whether the bodies be claftic or non-elaftic,) and becauſe the
2
AND MECHANIC.
171
velocity of A in 'direction A G, is not altered by the ſtroke (as afore
faid, it must therefore be the fame both before and after the ftroke,
fo make CM= A G, and compleat the rectangle CLMN, and draw
the diagonal C N, which CN will (by theo. 158) exprefs the velo-
city of A, after the ftroke, in like manner you may find that of B after
the ftroke.
260. In like manner it may be proved, that if a perfect elastic body
u, be thrown obliquely in the direction and with the velocity u Cagainit
a firm obftacle G, the body will be fo reflected from the obftacle as
to make the angle of reflection n CL- the angle of incidence u C L,
and the velocity Cn the velocity n C, for by letting fall the Lu L,
the motion u C is refolved into two motions u L directly against the
obitacle, and LC parallel thereunto, which LC being parallel can a
vail nothing, nor be any way obſtructed by the ſtroke, and will be the
fame after as before the ſtroke, fo make C L = LC, and compleat thẻ
rectangle Cn, then becaufe by the nature of elaftic bodies, if the body
is thrown with the L velocity u L, it will be reflected with the fanie or
an equal velocity Ln, whence the reflected velocities are C L and
Ln: (=LC and u L) the one velocity Cn (fee theo. 258) — u C;
but fince no bodies in nature are perfectly elaftic, they must be fome-
thing longer in reftoring their forms, and therefore, thenCL will
bc fomewhat lefs than UCL. Theo. 164.
י
PROBLEM
CXCIV.
Of the defcent of heavy bodies.
bo
261. If from any part in or on the earth, a heavy body be thrown -
upwards it will fall down in direction to the carths center, as is feen
by trial from whence it appears that this center is the center of gra-
vitation, and that if a heavy body could arrive there it would be at
reft. Alfo, the nearer this center any body is let fall, the greater
muſt be its velocity, but the height to which we can project any,
dies being but ſmall in reſpect of the earths axis, the force of gravity
may any where on the earths furface be looked upon as equal at
equal heights, and to defcend in parallel direction with uniformly,
equally accelerated velocities, this kind of motion is treated of in the
laſt problem which may be done by lines thus, let A E (fig. 145) de-
note time add E C, velocity acquired by falling in that time draw Dv
and v F parallel to EC and AE, then by fimilar As, as AD: Dv ::
À E EC, viz. as any time A D is to Dv the velocity required in
that time, fo is any other time A E, to E C, the velocity acquired by
falling in that time, which times and velocities conftitute the fimilar
fpaces or As A Dv and AE C, which ſpaces are as ADxD v to
172 THE UNIVERSAL MEASURER
AEXEC, and fo is the diftance defcended in the time A D to that
defcended in the time A E. Hence, the diftances defcended, or fallen
through, are as the times and velocities conjointly. Theo. 165.
But (by theo. 36) as ▲ ADv: A AEC ::□AD: □ AE, or::
Dv: EC, that is the diſtances defcended. are as the fquares of
the times, or as the fquares of the velocities. Theo. 166.
If you compleat the parallelogram AECG, uniformly defcribed
with the velocity AG EC in the time A E, to acquire the velocity
A G
EC. Theo. 167.
If a body in defcending the time A E, acquire the velocity E C, it
muſt be thrown up with the fame velocity E C, to come to the place
from whence it began to fall in the fame time A E, by the third law
of motion. Theo. 168.
At the point D, the rifing body will have the velocity D▾ equal to
that which a body would acquire by falling in the time A D, the dif-
ference between the times of the defcent A E, and afcent E D.
Theo. 169.
All bodies of the fame kind whether great or fmall falling from a
ftate of reft, will acquire equal velocities in equal times, for its evident
that gravity acts equally on every particle of matter, and were it not
for the refiftance of the air, which acts on all bodies according to their
denfities, all bodies whether light or heavy, would from a ftate of reſt
fall equal ſpaces in equal times, in which they would acquire equal
velocities, from what is faid in this prob. it is evident, that the theo-
rems hold true, whether the body defcend perpendicularly or ob-
liquely, by uſing the oblique defcent inſtead of the defcent.
PROBLEM
CXCV.
The theory of pendulums and vibrating chords.
that at B, then
AD: √ AB::
262. Let A HB (fig. 146) be a half circle, with its diameter A B
perpendicular to the horizon, and fuppofe a heavy body falling along
A B, put v the velocity acquired at D, and V
(by theo. 166) as AD: vv :: AB: VV, or a
v: V, and per fimilar As, as AD:AB::AE: AC, whence, as
AE: AC: v: V, that is, if the heights be equal, the ac-
quired velocities will alſo be equal, whether the body deſcend per-
pendicularly down A B, or obliquely along A E C. Theo. 170.
Alfo, if T the time of defcent from A to B, and t that from
A to D, then becauſe the velocities are as the times, it will be as AG
✅AE :: Tat, or as AC : AE :: TT : tt, alſo if inftead of an
inclined plane A C, you take AC to be any curve, this and the laſt
theo. will hold true, as will appear by dividing the faid curve AC into
-
AND MECHANIC.
173
an infinite number of equal parts which equal parts may be taken as
right lines, and the velocities will be equal at equal heights &c. Thco.
171.
Again, becauſe an angle A HB, in a femicircle is a right angle, we'll
by fimilar As, have as AB: AH:: AH:□AH
A I, whence
AH::AH:
A B
A H
WAB
curve or inclined plane or chord A H, and becauſe (by theo. 165) the
distance fallen thro' divided by the velocity gives the time, therefore
A B the time of falling from A to H. Whence
✔AI, the velocity acquired at H, by falling thro' the
AH÷
A H
✔AB
the times of defcent along the diameter of a circle perpendicular to
the horizon, and any choid A H or HB, of the fame circle are equal-
Theo. 172.
=
Let d the length of a pendulum, or 2d the diameter of a cir
ele, cany chord thereof, t = time of a heavy body freely falling
thro' the chord c, vthe velocity acquired at the end of that fall,
and let D, C, T, V, be the like things of another circle &c. then per
laft theo. as √2 D:2d:: Tt, or as 2 D:2d: ; TT: (t,
and if the choids be very fmall, Cand c, may exprefs the arches
themſelves, and then, the times of vibrations of pendulums in very
ſmail arches are as the fquare roots of their lengths. Theo. 173.
C
V2 D
Alfo, by theo. 170,
C
=V, and
=v, whence as
√2d
· 2D
c c
:
ad
: : V V : v v, and ifC = c, then as
T
I
:
or as 2d: 2 Dr
2 D 2 d
or as d :D :: VV: vv, that is the lengths of pendulums are inverfe
ly as the fquares of their vibrations. Theo. 174.
If p3,1416, then in article 424, it is proved that P√2d:
a
+
3,3 aa
2, 2, 4, 4, 4 dd
+
2, 2, 2d
3, 3, 5, 5 a a a
2. 2, 4, 4, 6, 6, 8 ddd
= Σ
+ &c.
it
the time of falling thro' an arch whofe verfed fine is a, diameter d,
which when the arch is very fmall, may be taken p2.d, and
(by theo. 172) 2d, is as the time of falling down the diameter, as
alfo down the chord, fo as 2d: p2d, or as i :p, or as 2: p
the time of falling thro' the chord to the time of falling thro' the
the diameter to half the periphery, whence, as 2: p:::
the time of vibrating in a circle, down whoſe diameter 2d, a
arch,
SP
}
*
174 THE UNIVERSAL MEASURER
2
Ž
heavy body would fall in the time t, the arch of the circle in which the
pendulum (whofe length is d) vibrates being fmall then as dt pl
ttppD
:D: "PPD
PPD
=tt, whence d
or as I :ipp:: D:d,
4 d
that is, as the fquare of a circles diameter is to the ſquare of half its
periphery fo is the length of a pendulum vibrating in a mall arch, to
the diftance which a heavy body will perpendicularly defcend in the
time of one ofcillation. Theo. 175.
Again"(by theo 173) as
pendulum d, be made to
:
4
dit :: √D: T=t√, but if the
d
deſcribe a greater arch or the fame circle,
then from the above feries, by putting n =), we'll have T =t
D
n
་
3. 3 nn
X: 1+ +
+
if t
2,2
= one fecond of time,
n
T=1+
༢, ༢. ༦, ༼nn rn
2, 2, 4, 4, 4
2, 2, 4, 4, 0, 0, 8
and d = D, then the last feries becomes,
3, 3nn
3, 3, 5, 5 nnn
+
+
2,2 2, 2, 4. 4,42, 2. 4, 4, 6, 6, 8
&c. and
&c. that is, if a
pendulum ofcillate once in one fecond of time in a very ſmall arch, it
will ofcillate once in T feconds in a greater arch of that circle. Theo.
176.
It may alſo be proved, the vibrations being ſmall, that if b the
degrees in an arch in which a pendulum meaſures equal time, the
number of ſeconds loft per day in another arch whole degrees = c,
will be: cc - bb: nearly. Alfo, if n = the number of threads
in an inch of the fcrew at the lower end of the pendulum, y = the
time in minutes that the clock gains or lofes in 24 hours, then the
number of threads that the bob is to be let down, or raiſed up to beat
feconds will be ny. Theo. 177.
262. To find the vibrations of an elaſtic ſtring or muſical chord,
AB (fig. 147) whofe length A B, let be 1, and its diameter d, to
make this plain, let us fuppofe the given chord A B to be ſtretch'd by
a weight f, put over a pulley at B, and to be ſtruck in the middle C,
by another force or weight e, which puts the chord A B into the ob-
lique pofition A D B, now its plain the whole force of f upon the ftring
A B, is in the direction B D, which force may be expreffed by B D,
and divided into the two forces BC, tending to pull the ſtring ftreight
and DC tending to pull it directly upwards from D to C, whence as
fx CD
DB: f::CD:
the force of f in direction DC, which muft
DB
AND MECHANIC
175
bee, becauſe they ballance each other and becauſe C D is but very
ſmall we may take CBD B, and then
fxCD fxC D
= e, =
D B
C B
fyCD fxCD
or
it
21
∞ e, now becauſe the ſpace A D B, is but ſmall
may be taken as a plane A, and then its CD will always be pro-
fs
portional to it, viz. C D α S, whence ∞e, theſe ſpaces A DB
&c. made by the vibrations of the ſtring A B, being as aforefaid, but
ſmall, may be looked upon as uniform, and then by theo. 154) m ∞
et, or
9
m x
m
t
fs
o e, whence, or Sft ml, but (by theo. 154
t
v and (by theo. 153) Sα tv, whence fvtt x 1qv, affitt
1q lidd, becaufe Id d is as q, the ftrings weight, or folidity,
whence t∞
Id
that is, the time of one vibration is as the diameter
and length of the fring directly, and as the fquare root of the tenfion
(f) inverlely, and becauſe C D, nor S is not in the expreffion, it fol-
lows, that the time of a vibration is the fame, whether the cord vi
brate thro' a greater or leffer ſpace. .Theo. 178.
264. Let a the fpace perpendicularly defcended by a heavy bo-
dy in the firft fecond of time, and w the cords weight, then w oldd
folw ftt, and becauſe the ſpaces fallen thro' are as the fquares of
the times, and t being as the time of one vibration and alſo as
I w
t
its plain, that to cauſe a vibration of the ftring A B, the point D, muſt
I w
| W
fall thro' a fpace ∞
tt, whence as a : 1 fecond ::
2 f
2 f
1 w
2dt
the fquare of the time of one vibration in feconds,
fo
W
2af
t, the time of one vibration in feconds, whence ✅✅✅
2 af
Tw
number of vi-
brations in one fecond. Theo. 179.
PROBLEM CXCVI.
The theory of wheel carriages, and of the mechanic powers, viz:-
the wheel, pully, ferew, ballance, leauer and wedge,
265. Let A PFM, be a wheel whofe weight is w, radius CA =
r (fig. 147) and EF h, the perpendicular height of an obftacle E
176 THE UNIVERSAL MEASURER
to the horizon ND. over which the wheel is to be drawn, let OK be
2 tangent to the wheel at the point E the top of the obitacle meeting
the diameter AG, produced in O, then the wheel coming to the point
E, ftands upon G, preing there with its whole weight, let therefore,
CO, the direction in which it gravitates exprefs its whole weight
which (by theo. 158 is refolved into the two forces CE and O F., of
which CE preffes directly against the top E of the obftacle and fo
is destroyed by the re-action of the laid point E, therefore, the only
force to be overcome is the other force O E, whence becaule wheels
are drawn by axle-trees thro' their centers, it follows that a force =
O C. in direction C M parallel to F. K, will hold the wheel in equili•
brio on E, fo by fimilar As, as EO: CO::HE: EC, whence E O
W
I
XHE, but by the property of the circle HE=AH×HG
=2rhhh, therefore E O =
√2 0-0, the wheel in direction
وا
CM parallel to o E K, is drawn with the greatest eafe, for fuppofe it
drawn in any other direction CM, then this force CM being
= the
two forces Cr and mr, of which Cr draws the wheel directly againit
E, and fo is deſtroyed by equal re-action of the point E, what there-
fore remains to draw it up in direction parallel to o K is mr, now if
S= fine Lm Cr, (which Cm the direction of the force called the line
of traction makes with E C) and radius=1, then as 1:r::S:rm,
XIm=1, víz. whenrm becomes xrm,
or as Sr::rm:
г
then CM will become XCM=xx√2rh-bh :=
A
2rh-hhf, the force fit to fuftain the wheel on E, in any direc-
tion CM of the force. (Cm being to r mas r to S). Theo. 180.
If EF (h) be very fmall, then hh may be rejected and then f=
=
W
Varh I
√2rb, = (if w be given)
I
f
六
​(2h being conftant) that
js in rough uneven furfaces, the force to draw the wheel will be in-
verfely as the fquare root of the wheels radius or diameter, if h=0,
then fo, which fhews that no force is required to move a heavy
body on an horizontal plane perfectly fmouth. Theo. 181.
Ifr ∞ h, and the line of traction parallel to OK, then f=
W
✔:2rh—bb:=y: 21-
y: 251-r1:= =
VII=W, that is,
AND MECHANIC,
177
the force will be proportional to the weight of the wheel, if Cm the
line of traction be parallel to the horizon ND, then LmCE-LCEH
form-CH,S=r-h, whence, f
W
W
s – b
W2rh-h-
V: 2 r I: when his conftant. Theo. 182.
266. If the obſtacle EF (h) be fuch, as that it may be depreffed or
broken down by the wheel, then CE denotes the whole force with
which the wheel bears upon the obſtacle, and (by theo. 158) is the
two forces H £, parallel to the horizon which must tend to drive the
obitacle before it, and C H perpendicular to the horizon and fo tends
to prefs it down, but CH-r-h, the preffing force, whence the
larger the wheel, the greater will be the force to preſs down the ob-
ftacle.
267. If the obſtacle be driven forward before the wheel, then HE
=V:2rh hh: the driving force, whence the leffer (r) the wheel
the greater the advantage, in poching the obstacle forward.
268. The line of traction C M being parallel to OK, its plain that
if E Fo. the line of traction must be Cm parallel to the horizon
ND, for if any other line CE, be this line of traction, it is refolvable
into two other forces C H, drawing the wheel downwards, and ſo in-
creafes the loads weight, and HE parallel to ND, whence HE, is
the only force that tends to draw the wheel forwards; again, if M
be the line of traction, it is the two forces, CB, drawing the wheel
upwards, and fo diminishes the loads weight, and B MND, foBM
is the only force that tends to draw the wheel forwards. Hence, it
follows, that if the radius of a carriage wheel be the height of a
horfe's breaft: or the traces parallel to the plane on which he draws,
the carriage will go with the moit eaſe, and wheels whole radius ex-
ceed that, are worſe than thoſe which are lefs, in which the traces
are a little elevated, in which pofition of the traces, the horſe on his
account draws eafielt, as well as on account of the wheels.
269. Of leavers there are 4 kinds, fift, is when the prop, of
fulcrum C (fig. 148 is between the power P and weight w. Second
kind, is when the weight C is between the prop w and the power P.
Third kind, is when the power C is between the weight and prop w
and P, the fourth kind is a beaded leaver as BCw, or B C A&c.
270. I wCP(fig. 148) be a freight line or leaver at reft parallet.
to the horizon with a weight wat on end, and another Pat the other
24
178 THE UNIVERSAL MEASURER
end the lever being at reft over a prop C, if it be made to move into
the pofition BCD, the weight w, will defcribe the arch w D, and P,
the arch Pe, which arches are plainly equal the velocities of theſe
weights, and becauſe theſe weights or bodies w and P are in equilibrio,
the force on each ſide of C, muſt be equal, that is, (by theo. 155,)
WXWD
P xe P, (for f ∞ qv) but the arches W D and Pe, are
as their radii C w and CP, fo wxw C - PX PC. From the first
of the equations, it follows that the product of the power and its dif-
tance moved is the product of the weight and its diftance moved in
the fame time, fo if a weight be raited by any engine whatever you
may by oblerving the diſtance that it and the power moves in the fame
time, know the ratio of their weight, which is of vaft ufe in mechanics,
as you'll find the queftions. Theo. 183.
271. Since weights hang perpendicular to the horizon, and BP is
▲ CP as alſo AwCw, it muſt be the fame thing whether P act at
P, or at B. and w at w, or at A, it will fill be wx w CP PC,
and it LA Cw LPCB, it will be wx AC-PXBC, or wx Aw
PXPB. Alſo it is the fame whether you fuppofe P to preſs or lift
and therefore, if w be the prop, it will alſo be wxwC=PxPC. or
as PC : w C :: weight upon w, to the weight upon P. Theo. 184.
The other mechanic powers are dependant on theo. 183.
PROBLEM CXCVII.
Of the centers of gravity, percuſſion, ofcillation, and gyration.
Fig. 149.
272. If there be any number of bodies a, b, c, d, ſuſpended on the
fame line or freight lever A B at the points A, C, D, B, and be in e-
quilibrio upon a point G, this point G, is the center of gravity of all the
bodies, then by theo. 184, we'll have CG xb+AGxa≈ DGX
A C, and D GĦAD—AG,
c+BG xd, but per fig. CG AG
alfo, BG-AB-AG, fo AG-ACxb+AGxa AD-AG x
¤† AB-AG xd, which by tranfpofition and divifion gives A G=
ADXc+ACxb+ABxd+ax the diftance of G the com--
a + b + c + d
mon center of gravity from A. Alſo, by reduction c =
AGx: 2+b+d:— ACxb-AB Xd, any of the bodies as c,
AD-AG
in like manner, we may get AC =
AGX:a+b+c+d:- ADXC-ABxd-a, the distance
}
AND MECHANIC.
179
:
between a and b any two of the bodies, but if the lever A B have any
weight, fuppofe that of the body c, and D its center of gravity,
thefe equations will be the fame, becauſe the weight of every body
acts in its center of gravity. Theo, 185.
If the freight beam or ballance A B (fig. 149) fufpended on G, be
in equilibrio with the bodies a, and d, (b and c beingo) but if the
body a be removed to B, it will be in equilibrio with a body w, fuf-
pended at A, then (by the laſt theo.) A G =
BGxa
BG xd
a
and A G
whence BGxd xw = BG xa xa, that is, dwaa, fo
✅ dwa, which put in the laſt equation for a, gives BG xd w
BG xdw, or,
AG XW BG xd
(each a) ſo as w:d :: □ BG:
A G
DAG.
BG
Theo. 186.
If there be any irregular folid Ac GB, whoſe weight is w, (fig.150)
and its center of gravity at G, and if by a
the folid be in equilibrio upon the prop D,
upon q, it be in equilibrio upon the prop C,
w acts at G, we'll (by theo. 185) have w —
body c, laid on the point d
and alfo by a body a, laid
then becaufe (by art. 244)
Dd XC and w
q Cx a
DGA
C G
whence, DdxcxCGgGax DG, but CGCD+DG, fo
DdXcx CD+DG = q CX3XDG, and therefore
DG. Theo. 187.
D d x c x C D
qCxa- Dd xc
273. To find the center of gravy of any furface or folid, as ABC
(fig 150), let Gits celler of gravity A B, the diameter of gravity
bilecting all the particles of matter b, c, d, g, h, i, 1, &c. in the faid
body, and then by theorem 185, we have A G
Ahxb 19c Lào với Ahxh &
C
+xc.
L &C.
now if any one of
thefe variable quantities A b, c, &c. fuppofe A b, be called e, and
the body Sand if aebeec+eeed=y, or ⇒ yy, the equa-
tion of the body &c. then (by prob. 185, or 186) b+c+d+g+&c.
the fum of the particles of matter in the body will be #ae + ce b
+ feeec + ƒ e ¹d, and A b x b + Acxc+Agxg + &c. =
ace+beee+ce+des, whence A G
4
tie bee reee + det Theo. 188.
*+toe+cee+deee
{
180 THE UNIVERSAL MEASURER
This general theorem may be laid down in a rule, and applied uni-
verfally to the centers of gravity, as thoſe in prob. 185, and 186, aré
to contents.
274. If e be the diameter of gravity, y an ordinate &c. bifected
thereby, and if it be cey or yy, then by the last theo. A G =
ce" +2 cc"+I
in + 2
се
n + I
ly, we'll have
n -- I e, and by taking n = 2, 1, 2,
n+ 2
In the common parabola
in the conoid of this parabola
in the cone or pyramid
in a plane triangle
ni
2
A G
e-
enfu ni fm makt ei fen
!
refpective..
e,
e,
e,
In the A A CD, (fig. 151) if you draw a line from any of the Ls,
A, C, D, to the middle of its oppoſite fide, of that line by what s
done above muſt fall at G, the center of gravity of the A. whence it
follows, that if the 3 fides of any plane A be biſected by lines drawn
from their oppofite angles, that any two of thefe lines wil meet in the
As center of gravity, which will be of any bifecting line diſtant from
its angular point, a hollow cone has the fame equation with a ▲, and
therefore its center of gravity will be at of its flant length, diſtant
from its vertex, in like manner you may compute the center of gravity
of any other hollow folid.
↓
275. If G, be the center of gravity of the fector SA B, let S A
the radius a the chord A B, e any abfciffa Sa, c=the arch AQ B
(fig. 151) then as r:e::a:
a e
1
chord a b, and as r; e :: c:
се
T
=arch a q b, now (per theo. 188) ex, and divided by the index
cè
of e plus 1, gives for the dividend, and, divided by the in-
dex of e+1, gives
I
역
​ае
Γ
e ea
3r
се
21
for a divifor fo
ee a ce
* 2 ae
(ware)
3 r
I
3c
2 ra
2 ra
a a
SG, and a 2 r, then
3¢
20
4rr
11
the diſtance of
30 30
3c
the center of gravity of a femi-circle from the circles center.
e a
Note. The reaſon for uſing two equations = ab and
r
ce
aq b, in this art. is clear from the demonſtration of theo. 188, where
all the particles of matter are fuppofed to be reduced to the diameter
AND MECHANIC.
181
of gravity in ftreight or L lines, but the area of the fector is compofed
of an infinite number of arches fimilar and parallel to A QB &c. this
obſerve in any figure whoſe baſe is a curve.
1
yy=
276 If the femi-parabola SAD fig. 152) be fufpended upon the
femi-ordinate A D≈a, da—y, d D—a B—´e, and parameter=
unity, then per conicks, aa SD and y y =Sd. but perfig. SD_Sử
—ak, i e. aa—yye, which (by theo. 188) multiplied by y, and
divided by the index of y plus 1, gives 2
and divided by the index of y plus 1,
for, whence 2ayyyyy
4
=
ay - y Vy, for a dividend,
4
VY, for a divi-
3 aa — yy
gives
νυ
3
3
6 a ay – 3 \ ♥ ♥ _ GD,
Y Y V
12 aa
4YY
the center of gravity of the parabolic trapezia a d DA, and becomes
= }} y, when a = y, the center of gravity of the parameter S A D.
1
277. If A Q_B, the fegment of a circle, be fufpended upon its verf-
ed fine QC, you may find G its center of gravity, by having given
R and C, thele of the fector S AQ E, and ▲ SA B, (fig. 153), thus,
let D
the area or weight of the ▲, E that of the ſector, F
that of the fegment, c=arch AQ_B, a chord A B, r = radius S A,
LSG of the A, z = GC, the diſtance between the centers of
gravity of the lector and fegment, then by art. 244, and théo. 184,
DXCR
DX CR
FXCG, or z =
DXER
>
but E — D F,
-DF, foz=
F
E-D
alfo, Dae, Ercand 2ra
274, which put in z =
2 raae
3C
DX CR
E-D
e=CR, by art. 275, and
inſtead of D, E and CR, gives z=
zace e from which take the L (SG) e, and there leaves
3 rcc3aec
2 rá a
3rce+aee, for the diſtance of G, from the middle of the
3rc-3ae
chord AB. After the fame method, may be had the center of gra-
vity of other compound figures, or of parts of figures.
278. If we take ban ordinate or ſpace in the middle of the body,
a one at the leffer, and cone at the greater bafe, each equally
diſtant from b, and all 3 bifected by e, the diameter of gravity, we'll
have the diſtance of the center of gravity from the leffer baſe equat
ec + 2e
theo. 189, as will appear by applying prob. 190 to theo.
+46
b
182 THE UNIVERSAL MEASURER
188. This theo. 189 anfwers for the centers of gravity as theo. 138,
doth for contents, as for example. Let the center of gravity of the
folid in theo. 139, be required, when it is fufpended upon e its axis.
Here adt, b = 4 : TD + d T+dt+ Dt: and c TD, which
fubftituted in theo. 189, fora, band c, gives: 3 TD +Td+td+tD
2 TD + 1d + 2 dt - 1 D
Xe AG (fig. 150) and if this folid becomes the fruftum of a cone
or pyramid, then TD and td, and then AG:
3RD+2Dd+dd
DD + D d +dd
Xe, and if a whole cone or pyramid then do, and then A G =
=e, if there be a plank of equal thickneſs length e, breadths
3 D De
4 DD
at the two ends b = a and c, then b
:
for b, gives A G=
2ae + ce
તે
3d+30
a c
which put in theo. 189
2
which is the fame thing as if e be the
diameter of gravity, and flant length of a hollow fruftum of a cone or
pyramid, whole peripheries at its two parallel bafes is a and c, and if
co, it becomes A G
=e, for the hollow cone or pyramid
}
&c. for other figures.
2 ae
за
279. If the diameter of gravity be the fame with the axis of the fo
lid viz. each = e, then by theo. 189, AG X 4b+c+a: ec
+2eb, whence : a+4b+c=x -/-/-
theo. 138) :a+4b+c: x =-=-=-
is
= :
c+2b: xee
6AG
: but (by
the body, fo that if either the
body or center of gravity be given, the other of them may be found.
Theo. 190.
280. If P C (fig. 148) one part of a lever, fixt in a wall at C, the
force to move the faid part will be PX P C, by theo, 184.
281.`It is known, that a line and plumet is only at reft, when the
line is to the horizon, whence, and by art 244, if the beam B C
(fig. 154) be fufpended by a cord F H, thro' its center of gravity G,
the beam will reit in any pofition when the cord F H, is perpendicular
to the horizon, therefore, if the chord F H be taken away
and the beam hung by the cords F C and FB, or by any two cords
EB and IC, in the fame directions, it muſt alſo be at reft, the fame
holds if it be hung by the cords EB and h C, becauſe theſe cords meet
at Hin the faid line, whence, and by art. 234, whether a body be
수
​AND MECHANIC.
183
1
+
uftained by two ropes, F. B and IC, or F B and FC, or e B and h C,
or by two poſts A B, and D, or q 8 and p G, or by two planes L
to A B and D C, the body can only be at reſt, when thele interfect in
the plumb line FH, paffing thro' the center of gravity G of the body.
if from any point Hin the line F H, we draw HI parallel to F B then
the whole weight the preffure at C, the thraft or preffure at B, are
reſpectively as FH, FI and IH, and in thefe directions, for the body
may be fuppofed to be fupported by two planes q B´and pCL to BA
and CD, which planes re-act again(t the body in theſe directions, and
fo is the fame thing as if the body was fuftained by the two ropes B F
and CF, and in either cafe, the faid ratio is true, by art 256.
282. From what is faid above, it follows, that a body fo fuftained
cannot be at reſt, if the ropes F B and F C, &c. do not interfect in the
faid line FH, and that a body will always be at reft, when hung or
ſupported by a line the horizon, palling thro' its center of gravity.
283. If there be any number of beams AB, BC, CD, DE, &c. t
be kept in equilibrio by their own weight, when fet one to the end of
another upon the horizon A C, as you ſee in fig. 155, whoſe centers
of gravity are at the points a, b, c, d, e, &c. and weights q, r, s, t, &c.
the weights upon the angles, B, C, D, with which theſe beams prefs
bC XI_B, C =
Bbxr c D x $
+
BC
each other, are A =
D=
Aaxq
A B
+
BC
CD
cCxSdExt, &c. for (by theo. 184) as AB: A a::q
CD
DE
AaX9, the weight of the beam A B, againſt the LB, and as CB:
A B
Cb::r:
Cbyr
C B
the weight of the beam CB, againſt the ſaid L B,
—
&c. whence, B, C, D, &c. are the weights againſt their reſpective
angles, thro' thefe Ls, let r1, Sm, tp &c. be drawn perpendicular to
the horizon A F, and produce DC, to meet Ir in r, then (by art. 256)
as SLA BC: SLABr:: weight B: force in direction B C equal
B × S. A Br
in like manner force in direction C3 CXS, DCS.
S. ABC
but to preferve the equilibrium, the force in direction C B, muſt be
that in direction B C, viz.
=
BXS A Br
S. A B C
S. B C D
whence, as
C cvs.ncs
S. BUD
S. ABC S. FCD
B: G&
and by the fame way of reafoning as C
S. ABr S.DCS
104
THE UNIVERSAL MEASURER
407
រ
$. BCD S.CDE
D::
S. BCS S.EDI
S. A B C
S. A Brx S. BCS
S. C D E
S. CDPXS.EDP
therefore, as weight B weight D : ;
S CD F
s Du s xs ED
ა SX SE DE
S. A B C
S. ABIX S. C BI
that is, if the beams be in equilibrio, the weight
upon any angle G, must be as
S. m C
S. B C D
B × S. m C D› fo if the po
fitions of all the beams be given, the ratios of the weights upon the Ls
may be found, and if one weight be given, all the other weights may
be found.
284 Alſo, (by art 256) the weight C, the forces in directions CB
and CD, are as r B, C B, and C r ieſpectively, but as C B to Cr::
S. L Cr B, or SCD, or r C m: S. rCB:: co fine elevation of C D
co-fine elevation of CB, :: fecant elevation of CB: fecant elevation
of CD, becauſe the fecants are inverfely as the co fines, whence, the
force or thruſt at C in direction C B, or at B in direction B C. is as the
fecant of the elevation of the beam or line RC above the horizon.
285 Draw Cq and Dn parallels to DE and CB, and let C D ex-
preſs the force in direction CD, or DC, then C q is the force in di-
rection D E, and Dn, the force in direction CB, but by theo. 156,
Dq, or the weight D, is the two forces D C and Cq, alfo Cn, or
the weight C is the two forces CD and Dn i. e. the weights on C
and D to preferve the equilibrium will be as Cn to Dq, whence if all
the weights are given with the pofition of two lines, (CD and DE)
the poſition of all the other lines C B, B A, &c may be found.
286. If the weights were to act upwards in the direction m C, p D,
&c. or the figure A B C D E F, turned upfide down, and the weights
remain the fame, and the points A and F be fixt as before, its plain
(by art. 281) that the whole figure will be exactly as before it was fo
turned, and that whether the lines A B, BC, CD, &c. be flexible,
or inflexible chords or beams.
287. Whence it follows, that if any chords of lengths be ſtretch-
ed to the fame degree of curvature, the firetching forces as the
-weights of the chords.
288. If Ba be an inclined beam or wall fig. 156) fupported by a
prop A a the horizon, let C be the beams center of gravity, w➡
its weight, e
fine ↳ B a P, or co-fine La BP, y = the weight borne
y
by a P, then w
theo. 184) m Pxy
the weight at B, draw Cm M, a P, now (by
1!
Bmxw-y: fo: mB+mP: (BP) xy=
'
}
€
}
t
AND MECHANIC.
185
Bmxw, but by fimilar As, as B m: BP:: BC: Ba, whence B È
Xw= Baxy, and (by theo. 48) as radius 1: Ba :: e: BP, ergo,
BCxwxe which
Ba=
Ва
B P
e
whence xy=BCxw, ory=
BP
e
1
В Р
will be the leaſt, when the divifor BP is greateſt, and becauſe B P is
always the prop it muſt (by prob 15) be greateſt when ↳ Ba A =
LBA a, and then it becomes BQL the prop a A.
Theo. I9I.
Again, let cB C, v prop a A, then becaufe B Qis La A, and
Ba= BA, therefore, a A and L a B A are each bifected by BQ,
now, ſuppoſe m Am B, then LamA
As, as BP: Pa:: Bm: Cm (viz.) as
VI
Z z
Z
I
Z
=V:
:= Cm,
I + Z
1 + 2
2 L a BA, and per fimilar
2 La BA,
I
+z: √:1—zz
::I:
tangent LCB m, (z being =
m Pco-fine La m P, radius = 1) fo if eco-fine of
tangent of that is
I
:: fo (by theo. 47) as
L
Ite
::: 1 :
√:
2
for BQ, we get y
I. e
I
=BQ, which written in
2 cew
V
√:
any L, the
I
e
:-c:
I te
BCXwxe
B Q
: now if y be a maximum
1 + e
2 C W
X
V
e-eee
Ite
yy, will alſo be one two, viz. yy =
See-Seee
I te
2 C W
m, (putting S=0
V
is already found = 0,6180 e. Theo. 192.
IL
which (by theo. 149)
288. If CDE (fig. 157) be an upright wall whofe weight is y,
and A BC, another wall &c. of the fame height A B, with a flope
fide B C, next to the wall of equal thickneſs C D E, and the vacant
A B C D between theſe two walls be filled with ftones, earth, fand &c.
of w weight, let P be that part of w which is fuftained by the wali
CD C, then its evident the ABC D, as a folid body would ſlide down
the inclined and upright planes B C and D C in the fame time, fo (by ·
DCXW
theo. 183) as BC: w::DC:
B C
by BC, and therefore, w —Q= w
W-
Q, the weight ſuſtained
DCXW-P, that ſuſtained
B C
+
A a
186 THE UNIVERSAL MEASURER
by DC, but as BC:
DCcw
B C
(the force in direction B C) :: A C:
ACY DC W
□ B C
the force in direction ACL to the wall CDE and
may be confidered as acting against the point L in direction G L || A C
becauſe G is the As center of gravity (C L being = CD and G L
DB) now the fides of the wall CDE, being parallel its center
of gravity is at R in the middle of C E, make RFCL, then its
plain, that a power lifting at F, the end of the lever FR, upon the
prop R is the fame force acting at L in the direction G L. to over-
turn the wall EDC upon E, fo ER x y, the force with which the
wall refifts is ACXDXW÷BC, the force of the A B C D
DC XW
against the wall CDF, but Pw
B C
is the force or weight
fuftained by the wall CDE in direction D C, that is y is the preffure
upon R and P that upon C, let the point e, be the center of gravity of
theſe two preffures, then, E e x y +P = AC× D C x w÷□ B C
in a cafe of equilibrium. Theo. 193.
But if the wall be only preffed in direction GL, then R Exy=
ACX DC ✓ w
B C
289. To find Q, the center of gyration of any fyftem of bodies
(fig. 158) A, B, C, connected together, and revolving about a center
S. 1. Let m the force or momentum of the whole fyftem, at any
point O, or Q, vthe velocity of any point in it at 1 diftant from S,
and A, B, C, the weights of their refpective bodies, A, B, C, now its
evident the velocity of any point in the fyftem, will be as the diſtance
of that point from the point S, fo the velocities of the points A, B, C,
will be v XS A, v x S 3, or y a, v b, vc (putting a = SA, b SB,
c = SC), but c muſt be negative, becauſe it and the common center
of gravity of the fyftem are on different fides of S, then (by theo. 155)
ya A, vb B, vc C, will be the forces of thefe bodies acting at A,
B, C, and (by art. 280) as SQ; a :: va A:
vaa A
SQ
the force of
A acting at Q, and for the fame reaſon the forces of B and C there,
v b b B
will be
>
and (
сх
vca, vec C
C сса
now its plain the fum
sa
sa
s Q
of all thefe forces muſt bem, i. e. X÷Aaa+Bbb+Ccc :
SQ
m.
AND MECHANIC.
189
If we ſuppoſe all the bodies placed at Q, their joint force, there
will be vxSQX: A+B+C, which taken
V
: A+B+C:=
vụ:Aaa+ Bbb + Ccc:
se
m, we get vxSQX
whence, SQ
Aaa Rbb +Ccc: fhewing how far Q, the center of gyra
A+ DJ C
/:
tion is from S, the point of fufpenfion, at which point the fame
force can ftop, or produce, the motion of the whole fyftem, as can ftop
or produce the motion of a fingle body (= A+B+C) placed at Q.
290. If the ſyſtem be at reft its force will be A+B+C, preffing
with that weight on its common center of gravity which let be at the
point G in the ſyſtem, whofe velocity being v XS G, this force will
then be v XS X: A+B+C: put this,m, and take the point O
SG
inftead of Q. then vxSGx: A+B+C:=
whence, SO-
A aa Bbb + Ccc
SGX: A+B+C:
vx: Aaa+Fbb +Ccc:
SO
the diſtance between S the point
offuſpenſion and O the center of percuffion, or point in the ſyſtem where
the ftroke is greateſt poflible, and appears to be a third proportional
to S G the center of gravity, and S Q the center of gyration, that is,
SG:SQ::SQ:SO, fince the force divided by the weight, gives
the velocity, if we divide X: Aaa+Bbb+Ccc:
and put the quote
we'll have S G =
SO
by A+B+C,
XSG, the velocity of the center of gravity,
Aaa+Bbb +Ccc, whence, SO=
SOX:A+B+C
Aaa Bbb + С c c
SGX: A+B+C :
the diſtance between S, and O the center of
ofcillation, and appears to be every way the fame with the center of
percuffion laft found.
It is the fame thing whether the bodies be all in one right line
CS A B, or out of it in the fame plane CCA BB, as is plain by what
goes before, fuppofe them all joined by the line CSA B, and put p⇒
GA, q=GB, r=GC, g=GS, and then a, b,,c, being as before
we'll have,
Aaa=□: ġ—p: × A=ggA—2gpA+PPÅ
Bbbg+g: XBgg B÷2gqB+qqB
Gcc=r-g:XCggG-2 gr C+C.
188 THE UNIVERSAL MEASURER
1
In theſe 3 equations, if G be the common center of gravity, B will
be in equilib. with C and A, or q
B PA+r C, fo the three terms
2gp A, +2gq B and 2 gr C, deftroy each other, and the ſum
of the reft will be, Aaa + Bbb + Ccc =gg×: A + B + C:
PPA + qq Brr C, divide each fide by g X, A+B+C: and
it will be Aaa+Bbb+Ccc_ App + B qq + Cr r
:A+B+C:xg
fore) SO, whence,
+g(as be-
A+B+C:Xg
SO — g=OG, now
App+Bqq+Crr
: A+B+C:xg
its plain per fig. that if OA=p, OBq, OC=r, OS must be =g,
for GSGA — O'S OA = SA
SA a, &c. fo if g in the laſt e-
quation ſtand for SO, we'll get SO APP÷B q q+C r r
A+B+C:XUG
whence
it is evident, that if the plane of the motion be the fame, and either O,
or S, be made the point of fufperfion, the other of them will be the
center of oſcillation, put SOS, OG, or S G — g, then S
App÷Bqq+C rr, we'll have Sg-gg=
: A+B+C: xg
=
g, then S-g being
App + Bq q + Cr r
A+B+C
૧
and if S g - g g be made a maximum, we'll find 2 g = S, and that
the oſcillations are made in the leaſt time poffible when OG =SG=
App+Bqq+Crr
N:
A÷B÷G
: where p GA, q = GB, r=GC.
221. Since the velocity of the fyftem when the point G moves in a
right line SO, is to its velocity when it revolves about the
point S, as SG to SO (the momentum in each cafe being the fame) it
follows that OG the difference of velocities is as the force preffing on'
S in direction to SO, and confequently the other part 3 G is ſpent
in accelerating the motion of the fyftem, that is, as SO :OG::force
of gravity in direction LSO: force acting at S in the like direction,
this alfo holds true when the axis of fufpenfion has a motion or when
the fyftem or body has a progreffive motion as well as an angular one,
as a cord wrapped about any body, or fyftem of bodies, and one end
of the cord made faft, then the body let go it will defcend rolling,
ftretching the cord with a force as O G, and its defcent to that of
falling freely as SG to SO, i. e. as SO: SG:: falling freely to roll--
ing, and :: OʻG : weight the cord fuftains.
292. Becauſe (by art. 272) SG X: A+B+C:
Aaa
Cc, our equation for S O will become SO =
Aa+Bb-
Bbb + C c c
: A+B+C: XSG
AND MECHANIC.
189
Aaa+Bbb+Ccc
Aa+Bb TUC
Theo. 194.
A=
Theo. 194. Where if A = o, and Bb➡
Cc, it will be SO –
Bbb Ccc
Bb TC c
Bbb +Ccc
infinite, viz.
о
the pendulum will not vibrate but reſt in equilibrio upon S as a center.
о
Bb Cc
Alfo, in this cafe S G
C + B
B÷C
thatis, the center of
gravity will be at S, and fo produce an equilibrium.
By comparing the laſt theorem with theo. 188, and art. 273,
n÷2
n+ 3
e, inſtead of A G =
we'll
e, ſo SO=4 e, in
1+2
e nearly, in the
have S O
the common parabola
cone or pyramid,
e in the plane triangle,
the radius in the plane of a circle. = e in a
2rr.
right line, rectangle, ſmall prifm, or cylinder, d + 2
5d
in afphere,
where r its radius, d diſtance between its center and point of fuf-.
penfion, e being the axis of the body which is fufpended at the vertex.
PROBLEM
CXCVIII.
To determine the ftrefs and ftrength of beams &c.
293. Strefs, any force acting against a beam to prefs or break it,
ftrength, is the force or power with which the beam refilts the firefs.
294. The beams &c. here meant are fuppofed to be homogeneous,
and placed parallel to the horizon, whether they be fixt at one end
as into a wall &c. or ſupported at both ends as upon props &c. except
it be faid to the contrary.
Theſe beams may be of wood, ſtone,
iron, &c.
295. Let BCDQ (fig. 159) be a beam with one end B C fixt into
a wall &c. and a weight w fufpended at D Q the other end of it, then
(by art. 280) AQxw, is as the ſtreſs at any point A, that is, the
ſtreſs on this account is as the length, but if e denote the weight of the
beam AD, AQx e, will be as the ftrefs at the fame point A, (for
the weight of every body) acts in its center of gravity (and this cen-
ter is in the middle, becauſe the beam is of equal breadth, and of =
depth throughout.
L
2
296. In fuch beams, the length AQ, will be as the weight e, and
fo AQ may alſo be pute, and then if we, we'll have e e for
AQX w, and ee, for AQx e, hence the ftrefs on A when the
weight is equally diffuſed over all the beam, is to the ftrefs there when
190 THE UNIVERSAL MEASURER
the weight is fufpended at the end D Q, as ee to `e e, or as 2 to 1,
and in general, if g be the diſtance between BC, the fixt end, and
the beam's center of gravity, orgne, then negew, is the
whole ftrefs of the point C, on both accounts, e being CQ, the
axis of the beam, let it be of what form it will, and q its weight, alfo
w the weight fufpended at DQ.
297 It is manifeft that the force of w upon the point B, is the fame
if there be another beam to the beam CD, with one end fixt, and
the other end touching D'Q, but if it be join'd to DQ. that fo both
beams bear the weight w, then the ftrefs at B will be but half of what
it was before, hence, the ftreſs ou a beam fixt at one end is equal the
ſtreſs on one of a double length when fupported at both ends; the fame
is true if the weight be equally diffufed over all the beam, as is the
cafe when it bears only its own weight, and for the fame reafons, the
ftrefs on a fquare plate &c. fupported by two fides, is the ftrefs on
one of a double fide or length, when fupported by all its four fides.
298. if A a a B (fig. 159) be a beam fupported at each end, by two
props a, a, and bearing a weight at any point P, proportionable to its
length, then B P and A P will be as the weights on the fegments B P
and A P, fo (art. 295) weight A P x diſtance BP, or weight BP X
diſtance A P, viz. APX BP ſtreſs on the point P, but if the weight
be not fufpended, but equally diffufed over all the beam then (art.297)
the ſtreſs at any point P will be as
ལ་བ
A PX B P
2.
the breadth of the beam B Q
continue the fame, while b va-
299. If e BD the length, and b
(fig. 159) and if the weight and length
ries, its plain the ſtreſs at B, will be inverfely as b, but (art. 295) it is
directly as e, ſo the ſtreſs, is as, and if there be another beam of
the fame depth, length
be, then Ebe B, or
E and breadth = B, and if it be as B: E::
e
음​=
==. Hence, the ftrefs on any two
B
Gmilar fections of the fame thickneſs, and matter, is equal, if they are
alike ſupported..
300. Since the ſtreſs is as the length when the beam is placed hori-
zontally, therefore, if the beam a B (fig. 156) ftands a flope, let fall
the La P upon the horizontal line AB, then the ftrefs on the beam
Ba in this poſition, muſt be as B P, but by trigonometry, as radius (1)
Ba fine L Ba P (c) viz. co-fine L of inclination a BP: Ba× C,
i.e. the ſtreſs on any beam &c. horizontally placed, multiplied by the
AND MECHANIC.
191
co fine of any inclination, give the ftrefs on it when it hath that in-
clination, and fo much for ftrefs.
301. To find the lateral ſtrength, in any part of any beam B CQ_D
(fig. 160) as at G. Let Gn A, be the fection of the beam there, fup-
pofing it to be cut right a cross, GA being perpendicular to the hori-
zon, or the beam parallel thereunto, put G Ad, the depth or abfcif-
ſa of the ſection b An, its ordinate or breadth.
Note. That fide of the beam perpendicular to the horizon is called
d. the depth, and that fide of it parallel thereunto is called b, the
breadth, fuppofe the depth G A to be divided into an infinite number
of equal parts, by ſmall threads (n A) paffing thro' it parallel to the
horizon, fuppofe a y any one of thefe threads or fibres, as an ordinate
to the imali variable diftance Gae, and en =ya, let the ſtrength
or refiílance of a fibre at A be 1, then, as d : 1 :: e: ftrength
e
I'
of a fibre at a; for when the beam is broken, the parts E F (fig. 161)
which before were together, are feperated at the diftance E F, and the
parts at D not parted at all, now if this diſtance E F, or ſtrength of a
fibre at that place be put 1, then its plain per figure, as D E (d) :
EF (1) :: De (e) : e f, the ftrength of a fibre at f, but the number
of fibres between A and a, aree, and the equation of the fection is
I
n+2
e
e
n
c=y, the product of thefe 3, is en=
which by theo.
d
n+3
78, is
e
db b
(when ed and y b) T, the total
d.n+3, n T3
ftrength of the fection, G An, whence, in the common parabola, rec-
tangle, triangle, n =†, no, n = 1, then T = b dd T = ÷ bdd,
T=4bdd, if the fection be an ellipfis, or circle, a =AG, that dia-
meter of it perpendicular to the horizon, cthe other diameter pa-
rallel thereunto, — D, eA a, the depth of any fegment, y=ya,
с
a
:
its femi-chord, then y =D: ae-ee: foxe x D √: ae-
ee is as the quantity generating the ſtrength of the ſection. this com-
pared with theo. 78, or the general feries in art. 215, we ll have T=
0,2451 Daaa, in the ellipfis, and T=0,2451 a a a in the circle, &c.
Note. Its plain the feparating parts, are at the upperfide of the beam
when fixt at one end,
192 THE UNIVERSAL MEASURER
302. Let g and S be the diſtances of the centers of gravity and of
cillation of the fection G An from G, a its area, d≈ A G, then
d, alfo, (theo. 78) 2 =
(art. 292) g
d, and S=
n+2
n+2
n÷3
d b
>
bddd
the product of thefe 3 equation is g Sa =
>
nti
nT 3
go
Sa
d
bd d
n+3
then Gaxa =
d
whence,
T, as in the laft article, now if we take Ga
gSa
T, the total strength of the fection G A n,
in which the point a is that where all the fibres being collected, do
act with their whole ftrength; this point a in a rectangle viz. GA =
1 d, for g = ½d, and Sd, fog Sdd, in a circle whofe
diameter A G, g
s
2d, S = &d, fo Ga =
=AG,
fere, in the periphery of a circle (the beam being a hollow cylinder)
g=d, s=d, fo Ga 3dd nearly, as in the two former,
in a fquare ſuſpended by one of its angles, diagonald, g={d,
Sd nearly, fo Gad.
8
g
S d
303. Hence if a power v, be applied to G, as the end of a lever G a,
upon the prop a, then G'a xv, is as the force to pull the beam afun-
der lengthwife, which let be GD (e) xw, the force to break it
laterally, i. e. (becauſe G a = gS÷d) g S ÷ d
gSd) gSdew, whence v
ewd
But if z the weight of the beam, G
S g
the diſtance be-
tween its fixt end and its center of gravity, then Gzewg Sv
÷d, or v=
d G z te dw
g S
>
and if y the weight of the beam be-
low where it is pulled afunder, or that part of its weight which any
ways affifts in pulling it, then d Gz+edw=:v
:v -y: xgs.
Note. If the beam be fixt at one end, g and S muſt be meaſured
from the under fide CQ, but if fupported at both ends they muſt be
meaſured from the upper fide thereof B D, and then e taken for half
the length, and w half the weight &c. (art. 291) alſo, if the prop or
fulcrum, dent into the beam, it will diminiſh its ftrength by fo much
as d is fhortened, by that denting.
304. If BG AC (fig. 160) be a cylinder twined, or twiſted about
its axis Te T, let e Pe Rd, periphery of the fection P R R R
b, ep any ſmall variable radius e, then if the power of a particle
at P to twine about the ſection be I as eP: 1 :: ep: P==
e
eP
the
AND MECHANIC.
193
power of a particle at p, the number of fuch particles between e and p
eb the periphery of, or particles in prpr,
aree, and as d: be:
e
d
whence - xex
eb
d
eeeb
dd
ee e eb
(by theo. 78)=
= (when e
d
4dd
d4ddb, the total twilting ſtrength of the whole fection PRPR,
and (art. 301) is equal the lateral ftrength of a triangle 1 — d, and
bafe | horizon b.
=
Note. This article does not agree fo well with wood as with metal,
for the texture of wood is not the fame in length and breadth as it is in
metal, where the power of ſpliting and breaking is the fame. Theſe are
the principles of ſtrength.
305. If a beam &c, is to be equally strong throughout, or the
ſtrength in any part of it proportional to the ftrefs on that part. First,
if it be fixt at one end, and a weight fufpended at e diftance from that
end, the trefs there (art. 295) is as w, and (art. 301) the ſtrength as
bđu÷ n73, therefore e ∞ bdd (w and n +3 being conſtant)
and if the upper, or under fides of the beam be a rectangle, or bevery
where the fame, then e o dd, the equation of the common parabola,
ifbd, then e ∞ d dd, that of the cubic parabola, but if the weight
prefs uniformly over all the beam then (art. 296) e e ∞ bdd, and if
b be every where the fame, then ee ∞ dd, ore ∞ d, the equation of
the plane triangle if bd, then e e ddd, the equation of the femi-
cubic parabola.
306. If A B (fig. 159) be a beam fupported at each end, and bear-
ing a weight on any variable point, or uniformly on all parts of it then
(art. 298) the ſtreſs at P is A Px BP, fo APX BP x bdd ∞ (if
b=d) ddd ∞ (if b be conftant) d d, the equation of the ellipfis, but..
if a weight be fixt at any point E, then B P ob dd, ∞ (when bd)
ddd, then A QE and B QE are two cubic parabola's, but if b is
conftant, then BP ∞ dd, then A QE and B QE are two common.
parabola's.
307. Let there be two beams q and Qof the fame matter, lengths
e and E, each fixt at one end, depths there d and D, breadths b and
B, weights of the beams w and z, weights borne at their unfixt
ends v and y, the diftance from their fixt ends to there
centers of gravity g and G, then their total ftreffes (art. 296)
will be g w + ev and GzEy, and ftrengths (art. 301) bdd÷:
n+3: and BDD÷: m+3: now if the ſtrength muſt be propor
tional to the ſtreſs, then (putting p=1÷n+3 and P
and P='1
m3 as pb dd: gw+ev:: PBDD: Gz + Ey, but becauſe
B b
I
194 THE UNIVERSAL MEASURER
folidity is as weight, as cbde: w::CBDE:
A
WCBDE
cb de
=z, which
put in the laft proportion for z &c. we get p CBD d G Ew +
pcbddEey PCBDx: Degw+ Deev. Theo. 195.
308. If the fections are alike, viz. mn, then pP, c=C, SA a and
their areas then acpbd and A≈p C B D = P c BD, then
the laft equation will be Ad G Ew+ad Eey A Degw+ ADeev,
if yo,
o, then Ad GEwA DegwA Deev, if the beam q and
Qbe priſms, or cylinders, then GE, and g=e, then AdEEW
A Deew + 2 A eev D, and if they be fimilar folids, then d E
De, and fo Ewe w 2 ev, alfo if the beams are fimilar, cylinders
or fquare prifms, then, we may write E and e for 2 G and 2 g, DD
and d d for A and a, dE for De, fo the last theorem will become
EEEw+2eeey=EEex: w + 2 v :, whence y : =
FEEw + FEex: W + 2 v: in which if e be made variable
2e ee
and the equation a maximum, we'll (art. 221) find E-2X: +2v:
3 W
309. It is prov'd by experiments, that the bending of a beam is as
the weights laid on it, alfo by weights hung to the ends of elaftic firings
as wires, hairs &c their lengths by firetching is found proportional to
the weights hung at them, except when they are going to break and
then this increaſe is f.mewhat more, by taking the weights off none of
the bodies are formed to regain their former figures, except well tem-
pered fprings, tho they by often bending grow weaker, hence there are
no natural bodies perfectly claític. Thefe things are alfo proved in
art. 263.
b
310. Let eĄB (fig. 161) the length, d D E, the depth, and
the breadth of the ftreight beam A B, fupported at both ends, if
weights be laid on it till it break the parts E F, which are together at
the beginning of the bending, will gradually feperate till the beam
breaks, at which time they will be all parted, as in the figure, and
while they are thus parting the bending or curvature is increafing, but
the greater DF. is (all elfe being the fame), the fewer parts will be
ſeperated, fo the curvature is as, and the longer the beam, the leſs
weight it will bear, therefore, bdd being as the ftrength, the weight
it will bear is as bdd ÷ e, which put for the greateſt weight, and
for the curvature when breaking, w any other weight and C the cor-
d
AND MECHANIC.
195
*
bdd
I
e w
refponding curvature, then it will be as
: W::
:C,
ف
b d d d
311. When CD is very fmall, A D B is very near a circular arch,
whofe radius let be r, then ee (□ AD) = [] C D + □ С A = 2 r
XCD
i
CD+□ CD=2rXCD, whence
CD
but the lefs r, the greater the curvature, therefore
the laſt art.) C∞, whence CD ∞
e W
bdd
eee w
b d d d
ee
I
I
or
or ∞
81
∞ ( by
ee Γ
CD I
and it was the fame
po.
in reſpect of both curvature C, and deflection CD, from the firſt
fition of the beam whether it be freight or bended. Alfo in the out-
moft ftrength of beams, or their breaking pofition b d d xew (art.
305) then C ∞ ———, and the deflection C D ∞ ee÷dd,
d
If the length of a beam be e, fixt at one end, and at the other a
weight proportional to e, the ftrefs by art. 296, is as e xe, or ee,
but if it be fupported at both ends, and the weight in the middle,
then the ftrefs by art. 297, is as exe, oree, hence, the ſtreſs
when fixt at one end, is to the ftrefs when fupported at both ends, as
4 to 1, and therefore, if C D ∞ eeee÷bddd (e being as w) when
the beam is fupported at both ends, then 16eeee÷bddd ∞ CD,
when it is fixt at one end, (taking 2 e for e) if a beam be fupported
by two props within the ends, as at E at F (fig. 204) let a equal the
whole length D HD, c — ED, and z = FD, then FHE➡ a
e, and therefore G H will be as the biquadrate, of a Z -e, and
CD, as 16eeee, alfo d D as 16zzzz, the fum of theſe
will be
3
as all the deflections of the beam, which call S, and if we take z —e,
4
-
e
S
Z
then a 2 + 32 eeee S, its plain S will be greatest when eo',
and if the equation be ordered as a minimum (art. 221) e variable and
a = 1, we'll have
8 + 48e 96ee + 192eeea, whence
tranfpofing 8, and dividing by 48, we get e 2 ce + 4eee,
which folved gives e nearly, when the bending is the leaft, if the
beam is fupported at each end, and other two props to be fet each at
the distance e from the end, then each deflection for the length e will
be as eeee, fo their fum will be as 2eeee, and therefore, a- -2c
+2eeee S, ſo inſtead of the above equation we'll have e― 2 ee
+1,5eee = ½, where e, and thus may any number of props be
fet fo as that the beam may have the leaſt bending. -
4
196 THE UNIVERSAL MEASURER
Laft'y. Suppofe the beam or bar A C B (fig. 161) to bend thro' the
fpace Ca, before it begin to break, let cthe weight laid foftly up-
on a, which breaks it when fo bent, w = a weight falling freely upon
C, from the diſtance or height d, juſt to break it or bend it thro' the
dillance Ca, put b = Ca, e = C n, any ſmall variable diſtance, now
(art. 309) the beam bent into the pofition A a B, exerts a force C
diſtance Cab, therefore, as bc::e: e c -force at n,
1, foce -
w force at n acted on w, but (theo. 152) v v =
fe
W
= (becauſe f=ec
e is variable)
f S
१
b
= (in this cafe)
ес
w)
D
eec-ebw
b w
= (fee theo. 78, becauſe
eec
2eb w
2 eb w
2 b w
but (theo. 166) v v ∞ d, whence, d =
and when eb, then 2 b w dbbc — 2 bbw, fo
eec
2b w
b c
W=
2 d + 2 b
b c
2 d'
nearly, becauſe b in moſt ſuch caſes is very
fmall.
1. Hence, when do, w
bc
26
=c, that is, half the weight will
break the bar, when bent to its breaking poſition, that will break it
when unbent.
2. The weight that by falling a given height will break any beam
is nearly as the ſpace (b) thro' which it bends before it breaks, hence
brittle bodies break fooner by percuflion than others of equal ſtrength.
Note. Here is no notice taken of the bar's weight.
3. If gravity have no concern, or the weight be thrown horizontally
againſt the beam, then -w, will only be, fo proceeding as
ес
before, we'll find 2 dwbc, whence w
b c
2 d
PROBLEM CXCIX.
Of Hydroftatics, hydraulics, and pneumatics.
312. Hydroſtatics, is a fcience that treats of the properties of fluids.
313. Hydraulics, is the art of raifing, carrying &c. of water as by
pumps, &c.
314. Pneumatics, is a fcience that treats of the air's properties.
AND MECHANIC.
197
315. The motion of a fluid is accounted for as that of a heavy body,
but the motion or preffure in a fluid is equally diffufed all around in
all manner of directions, and can only be at rett when its furface is
parallel to the horizon.
316. If a homogeneous body be immerfed in a fluid of the fame den-
fity with itfelf, it will remain at reft in any place, and in any poûtion,
but a body of greater,denfity than the fluid will link to the bottom,
and a body of leffer denfity will rife to the top and fwim, hence the
body of greater density lofes fo much weight as that of an equal quan-
tity of the fluid, and fo tends downwards only with the difference of
theſe weights, and this is the relative gravity of the body in the fluid,
but if the body is fpecifically lighter than the fluid it feems to lofe more
weight than it has, and hence the body will tend upwards with the
difference of theſe weights, and this is the relative levity of the body
in the fluid, and the weight of this body is equal to the weight of a
quantity of the fluid as big as the immersed part of the body.
317. If a plane furface be perpendicular to the stream of any fluid,
the tream ftrikes the plane with the fquare of its velocity, for its
plain, that with n times the velocity the force of each particle will be
n, and n times the number of particles will ſtrike the plane in the fame
time, confequently the plane is ftruck by n.n, and becauſe the whole
plane is thus ftruck, it follows that n n a is as the impreffion on the
plane (a being the area of the plane.
318. Hence (and by theo. 158) as DB: DC (fig. 143 ) : :
the abfolute velocity of the fluid to that part of it which impreffes the
plane, and fo is fquare radius: fquare incident angle CBD,
when the plane A B is oblique to DB the direction of the ftream, ſo
that if a area plane A B, r radius, S fine LCBD, we'll have
as rra: SSa :: the impreffion in a direction to that in an oblique
one.
319. If a plane in motion be ftruck perpendicularly by a fluid, it is
plain, that each particle ftrikes that plane with a velocity equal to the
difference between the velocities of the fluid and plane, whence, the
impreffion of the fluid on the plane will be as the fquare of the differ-
ence between their velocities, ſo if v = the velocity of the fluid, and ▾
-e that of the plane, their difference is e, fo: v -e:xeevee
eee, the impreffion, and in cafe of a maximum e = ÷ v.
320. If there be two planes whofe areas are A and a, velocities of
the fluids V and v denfities of the fluids D and d, fines of incidence S
and s, then (by art. 318) the impreffions on the planes are as A VV :
a vv, on account of the velocities, which velocities (per laſt art.) are
198 THE UNIVERSAL MEASURÈR
as SS ss, and the denfities being as D to d, the products of theſe given
as A D VVSS adv vss, the ratio of the impreffions.
I
321. If a fluid falls from any height h it acquires a velocity, with
which it moves uniformly, and fo (by theo. 167) defcribes a double
ſpace in that time, in which it fell to gain that velocity. Let S the
height in feet fallen by a heavy body in one ſecond of time, v the
velocity of the fluid, or ſpace it defcribes in 1 fecond, a the area of
the plane, or bafe of a column of the fluid, h the height of this
column, fit to acquire the velocity v, then (by theo. 167) 28 = ve-
locity generated by gravity in falling thro' S, therefore (by theo. 166)
h, the height fallen thro' to gain the velo-
as 4SS: S :: V V :
V V
4 S
city v feet per ſecond, fo=2h, and Va
2S
2S
2 ha a column
I
of twice the height h, and (by theo 155) 2×2ha 4Shavva
the motion generated by the weight of the column in 1 fecond, viz. the
body x velocity 2 S, this v v a'is the fame with nna in art. 317,
which fhews that the force of any fluid againſt a plane is equal to
the weight of a column of that fluid the bafe of the column being to the
area of the plane, and its beight twice the height defcended by a
falling body to acquire the velocity of the fluid.
v v da
322. If d =
denfity of the fluid, then d ah
F, the
2 S
v v da
F, the
S
force of the fluid againſt the plane, or dah
fame force, if the fluid ftrike the plane obliquely, let z = fine of
incidence thendahzz=
v v dazz
2 S
F (fee'art. 320) if the plane
be alſo in motion, the relative velocity of the fluid againſt the plane,
must be taken inſtead of the abfolute velocity.
323. If inſtead of the fluid triking the plane, we fuppofe the plane
to move in the fluid, or a cylinder to move in it in direction of its axis
b, the area of its baſe its velocity or the ſpace it defcribes in one
as before, then its evident, what was be-
plane, will now be the reſiſtance of the plane
fecond, v feet, and S
fore the impreffion on the
a,
or cylinder, that is (by art. 321)
avv = 2 ha the reſiſtance, were the
2 S
particles of the fluid all driven directly forward, but fince (by art. 315)
they move round in all directions, it may be proved (fee art. 325)
I
AND MECHANIC,
199
that this refitance is double too much, and therefore
the true refiftance, that is, its refiftance is
vv a of the fluid.
45
V va
ha=R,
48
weight of the cylinder
324. The denfer the fluid, the greater is the reſiſtance, for the more
particles it contains, the more it mult refilt, and the denfer the body,
the lefs the refiftance for the more particks it contains the more po-
wer it has to overcome the refiftance, fo if D
and d that of the body, we'll have
the density of the fluid
vva D
R,
40 i
if v = 2 √
or v v
D
4 h Sd
D
ha D
>
and
d
cafe, the refiftance is
we'll have R= ha, in which
the weight of an equal cylinder of the fluid.
I
325. If a globe 'fig. 162 move uniformly forward in direction C A,
draw G BD || C A, draw the tangent D H, and let fall GHL DH,
let GD be the force of a particle of the fluid againft the bafe B, in
direction G D, then G H will be the force acting againſt D, in direction
DC, and this force is to the force in direction GD, as DC to D B,
whence the force again B is to that againſt D in direction GD, in a ratio
compound of CD to DH, and DC to D B, that is as
DB, fo the force of all the particles of the fluid against the
their force againſt the convex furface as the fum of all the
to the fum of all the DB's, which fums are as 2 to 1, that is,
ㅁ
​reſiſtance of a ſphere's furface, is but half the reſiſtance of the baſe,
or of a cylinder of the fame diameter.
V
VV
DC to
bafe, is to
DC s is
the
326. Whence (fee art. 323) becomes that is, if Athe
4 S
88'
diameter of a globe, its refiftance is the weight of a cylinder of
the fluid of the fame diameter A, and its length and if v v=
16
3
V V
85'
SA, or v = 4√
v=
S
3.
V V 16 S A
then
8 S
24 S
=A, viz. its refiftance
is weight of an equal globe of the fluid.
327. Let D
denſity of the fluid and d = -denſity of the globe,
then, fince a globe whofe axis is A, is
A, the weight of the globe will be
Ad
whofe length is 2 Aḍ
3 D
and (by art.
a cylinder, whofe height is
weight of a cylinder of the fluid
316) the weight of the globe in
207 THE UNIVERSAL MEASURER
=
the fluid is
d D
X :
D
weight of a cylinder of the fluid, whofe length is A
: but (by art. 321) the refiſtance of the globe moving
4
with the velocity acquired by falling in vacuo thro' the height Ax
d D
X:
D
:
is
d — D,
D
weight of a cylinder of the fluid whofe length is A
therefore, the weight of the globe in the fluid is — re-
fiſtance, and confequently cannot accelerate the globe; whence, the
greateſt velocity a globe can obtain by defcending in a fluid, is that
which it would acquire by falling in vacuo, thro' a space S, that is to
A, as d D is to D.
328. If v = 4√
globe in the fluid.
d-D
3 D
SA the reſiſtance is = weight of the
329. If a cylinder move in a fluid incloſed in a veſſel, inſtead of the
abfolute velocity, the relative velocity in the fluid muſt be taken in or-
der to find the reſiſtance, and the narrower the veffel the greater will
be the refiftance, for then the more particles of the fluid are driven
directly before the moving cylinder, and if the veffel be fo narrow as
none of theſe particles can diverge in all directions, but be all driven
before the cylinder, then the refiftance is the greateft poffible, and is
equal to the reſiſtance of a plane (art. 321) prefs'd by the fluid, whereas
if the particles of the fluid have liberty to deverge in all directions, the
force of a cylinder of water againſt a plane, is double the reſiſtance an
equal cylinder meets with, when moving in water with the fame velo-
city, as is plain from art. 321, and 323.
330. If A B (fig. 163) be a cylinder to the horizon kept con.
ſtantly full of water, with a hole at the fide or bottom B, conftantly
running out, let Sfpace defcended by gravity in one fecond of time
t = time in feconds of the waters running out, h AB the depth of
the veffel to the center of the hole, a area of the hole, now (by
theo. 167) if the cylinder of water ha, fall thro' half its heighth,
by its own weight, it will by that fall acquire fuch a motion as to paſs
thro' h uniformly in the fame time, and the quantity of preffure at
any given depth upon a given furface, being always the fame, the refore
the water in the hole B, is preffed with the cylinder of water h a,
whence, the preffure at B, will generate the fame motion in the
Spouting water, as was generated by the weight of the cylinder h a,
AND MECHANIC.
201
so in the time of falling thro'h, a cylinder of water will ſpout out,
whofe length or the fpace paffed uniformly over ish, and in the fame
time repeated another cylinder will flow out, and in a third part of
time, a third &c. therefore, the length of the whole cylinder run out
will be proportional to the time, and fo the velocity of the water at B
is uniform and therefore in the time of falling thro' half h, a quantity
of the fluid runs out the cylinder ha, with a uniform velocity at B,
= that acquired by a heavy body falling thro' h, and becauſe the
velocities of falling bodies are as the fquare roots of the heights, there-
fore, the velocities of the fluid fpouting out at different depths, will be
as the Iquare roots of the depths, and theſe velocities will be the fame
in any direction whether ſpouting upwards, downwards, fideways &c.
and if it ſpout upwards it will afcend nearly to the upper furface of the
fluid. From what is faid in this article, we have ta/2hSquantity of
water in feet, that runs out in the time t, (a, h, and S, being each in
feet) = 6,128 ta2hS ale gallons. Theo. 196.
331. But if there be a rectangular hole in the fide of a veffel con-
ſtantly kept full of water then becauſe the velocities are as the fquare
roots of the heights let e any height of this hole or flit, and y= its
breadth) which ratio's form the equation of the common parabola,
whoſe area ise ya in art. 330. whence the quantity of water
difcharged thro' the flite Bae (fig, 163) is the quantity difchar-
ged out of an equal hole plac'd at the whole depth ea, or at the baſe
Ba in the fame time, fo taking hea, we'll from the laſt theo, have
X 6,128 ta √ 2 hS, for the ale gallons run out in the time t,
332. If the yeffel is not kept full, viz. no water taken inat top, then
ha will conftantly decreafe, and confequently the velocity at B,
whence, in this cafe the velocity at B, is but half of what it is when the
veffel grows no emptier,
333. A fluid ſpouting thro' a hole, endeavours at a ſmall diſtance
from the hole to contract itfelf into a kind of fpire, by which its velo-
-city is fomewhat increaſed, and the thinner the hole is, the greater is
this increaſe, and if the hole be in a thin plate of metal it will acquire
a velocity near that acquired by a heavy body falling thro' the whole
height h; alfo, from the refiftance of the air, all bodies projected up-
wards fall ſhort of theſe projected in vacuo, by ſpaces which are as the
fquares of the heights, as in fmall heights of fluids fpouting, is proved
by experiments. There is great difference in fpouting veſſels on ac-
count of their forms, bignefs of the hole, &c. to help which, let A =
area at ▲ A the water's furface, S, a, t, h as in art. 3.30, then as 2 Aª
C C
402 THE UNIVERSAL MEASURER
a': A²::h:
hA 2
2 A²-a²
H, the height of a falling body to acquire
the velocity of the water at the hole B, and if the area of the hole be
but ſmall in refpect of A, then H=
h A²
2A²—0
h, as before in art.
330, but if A = a, or the veffel have no bottom, then H=
hA²
2A2-A³
=h, (ſee art. 441, and 442) whence it appears that the velocity of
the fpouting fluid can never be lefs than that acquired by falling thro'
h, the veſſel being conftantly kept full, but if no water runs in whilst
it ſpouts out, then it muſt be as 3 A² — a* : A²
fo in both cafes it may be
A2 h
n A
: A²::h:
A² h
=H,
За
3 A² a
=H, the height fallen thro' to
A² h
gain the velocity at B, fo (by theo. 167) as S: 2S:::-
NA -a²
2 S
to
✓
A h
¤ A ² a
A² hs
:= 2√:
DA
:the velocity, or
તુ
fpace uniformly paffed thro' by the water at the hole B, in one fecond
which multiplied by t and a the area of the hole, &c. gives 2 ta
: feet for the water run out in one fecond. Theo. 197.
A² h S
nA² -2
2
334. Since water will run thro' a hole in the fide of any veffel filled
therewith, it is evident that the fides as well as the bottom is every
where preſſed with the fluid, and that with a force proportional to the
weight of the fluid above the place preffed; let therefore, a area
of the cylinder's baſe (fig. 163), and fuppofe its height A B, divided
into an infinite number of equal parts, beginning at top A, 0, 1, 2, 3 &c.
p, then the fides at each of the parts will be preffed with the weights
O, a, 2a, 3a, &c. na, foo+a+2a+3 a + &c.:+na≈‡nnà,
the whole preffure againſt the fidesha, becauſe nna, is as the
whole weight, but the bottom a, bears the whole weight ha, of the
fluid, fo the preffure against the bottom is to that againſt the fides
as ha ha, or as 2 to 1.
:
335. If ABC Df (fig. 164) be any veffel containing a fluid, and BL,
ED, GFC, and HFOK, be Ls to the horizon, GB, FL, CD, ||s
thereunto, then if the fluid be poured in at the top A B, till the veffel is
filled to n L, it will rife in the other part to Fa, the fame level, whence
it follows, that if A and a be the baſes of two veffels of any kind or any
ſhape, H and h their heights perpendicular the horizon, the preffures
AND MECHANIC.
203
will be as ha, this confirmed by experiments, for if a veffel fQK, be
made fo as to open like a pair of bellows, and a tube F G H, of any
diameter G H, fixed to it by pouring water in this tube, it will raiſe
the upper baſe F a, loaded with a weight f, nearly that of a column
of water of the fame bafe Fa, and h HK, that of the veffel and
tube.
336. The force or preffure is every where directedly against the
inner furface of the veffel (by theo. 157) fo at f, it is directed upwards,
at L fideways, and at K downwards, and as the fame heights are
equal, viz. the preffure at L and F is equal, becaufe B L GF, alfo,
the preffures at D, O, C, Q, are equal becaufe EDHOG C,
&c. and at K the preffure is as HK.
337. Hence (and by art. 334) the preffure or ftrefs on any pipe or
tube full of water is as the diameter of the pipe and the height of
the water above that place, fo, as the internal preffure on any length
of the pipe, is to the ſtreſs it fuffers as to fplitting, fo is 2 X 3.1416 to 1.
338. Let A and B, be the bulks, or magnitudes of two bodies,
and d, their denfities,G and g their weights or gravities, now the
denfer bodies are, the m more matter they contain, and the greater they
are the more matter they alſo contain, and the weights being as the
quantities of matter, it will be, as AD: G:: Bd: g and if AB,
then as D: G::d: g, or if D-d, then as A: G: B: g, but if
Gg, then as AD: 1 :: Bd: 1, or as A :d : : B : D.
339. Since the ſpecific gravity of bodies, are as the weights of equal
bulks, they are therefore as their denfities, fo whatſoever is faid in
reſpect to the denfities of bodies, is true in regard to their ſpecific
gravities.
340. That is, if the weights be equal, the magnitudes are inverfe-
ly as the fpecific gravities, viz. as A: d: : B : D.
341. Let a the fpecific gravity of a body A, B its weight in
water c the fpecific gravity of water, then (by art. 316) the weight
loft by the body in the fluid is
of water, ſo (by the laſt art.) as A
AB
=
the weight of an bulk
с А
B: A: C:
A - B
= a, that
is, the fpecific gravities of bodies will be as their weights (A) in the
air directly, and as (AB) in the fame fluid inverfely, becaufe c is
conftant.
342. If the fame body is weighed in feveral fluids.and be fpecifically hea-
vier, then from a =
CA.
we have c-
ax: A-B:
A
B
A
204 THE UNIVERSAL MEASURER
AB, hence the fpecific gravities of thefe fluids are as the weight of
the body loft in the fluids.
=
343. Let a and b the ſpecific gravity of a mixture of two bodies
A and B, c the weight of the mixture, d its ſpecific gravity, then'
if z the weight of the body A, c z will be
2
art. 341) and
C
a
that of the mixture, then if_Z_
that of B, and (by
C
Z
will be the
weights loft in the fluid, and
b
C
Z
+
a
a c
b- d
have z
X
or d =
>
ba
letters, the reſt being given.
d
a b c
be taken =
b z — azt a c
345. If we take mz and nc —
&
we'll
&c. for any of the
oto
11
Z
C - Z
-Z,
then
+.
a
b
will be +공
​m
a
bd m
C
whence, a =
, or as
: m
bc-d n
a
::I: a.
346. But if one of the bodies (A) be ſpecifically lighter than the
fluid, it will be as m :
C
n
:: Ia and thus may a table of
d
fpecific gravities be calculated, and an irregular folid &c. meaſured
thereby, thus, let w the weight of a cubic inch of water, or any
other Auid that the folid is to be immerfed in, d the fpecific gravity
of the folid, then as I wd:: folid content in inches: weight in oun-
ces, &c.
347. A fluid is a body, whofe parts yield to any force impreſs'd
and are eaſily moved amongſt themſelves, there are two kinds of fluids,
as bodies, viz. elaſtic and non-elaftic, a fluid is elaftic when it can be
reduced into a leſſer ſpace by compreffion, fuch as air, which differs
from other fluids in theſe particulars, (1) it may be preffed into a lefs'
fpace, and fo can no other fluid, (2) it cannot be congealed or any how
fixed, and all other fluids can (3)its denfity every where decreaſes from
the earth's furface upwards; but other fluids are of an uniform den-
fity throughout, &c.
Non-elaftic fluids are fuch as cannot be reduced to a lefs bulk, fuch
as water, &c.
348. If the particles of air have all the fame elaſtic force, the com-
prefling force must be as the number of particles prefs'd, but if there
be two equal bulks of air and the denſity of the one ben times the
AND MFCHANIC.
205
denſity of the other, the latter contains n times more particles than the
former. Hence, if all the particles of air have the fame elaſtic force,
the force of compreffion, (by art. 338) is as the denfity.
349. The elasticity of the air is the foundation of the air pump;
for when the pifton is forc'd down to the bottom of the barrel and rais'd
up again the air in the receiver will expand itſelf and part of it will
enter into the barrel, fo that the air in the receiver and that in the
barrel will have the fame denfity, which will be to the firſt denſity as
the capacity of the receiver, is to the capacity of the barrel and re-
ceiver together, and by repeating the motion of the pifton a ſecond
time, the density of the air will again be leſſened in the fame ratio, and
fo on; by which means the air in the receiver may be reduc'd to the
Icaſt denſity, but can never be entirely exhaufted, for the air which is
exhauſted is only puſhed out by the ſpring of that which is left behind,
if therefore every particle were fuppofed to be exhauſted, the laſt
particle would be pushed out without a caufe, contrary to art. 233.
Hence, if the capacity of the receiver, that of the barrel, d
the denfity of the air in the receiver before the pump begin to work,
then
As, n+m:n::d:
dn
I
n+m
the denfity
+m:n::
dn
n+m n+mf
dnn
of the air
| in the re-
2
ftroke of
the piſton
dnn
dnno
ceiver after
or turn.
3
n+m:n::
the
Theo. 198
n+m² n+m³
3
and univerfally
n+m/s
350. Hence, fince the fpring or elaſticity of the air is the force it
exerts againſt the force of compreffion, it follows that the air does the
fame thing by its fpring, as a non-elaſtic fluid does by its weight.
351. Since the lefs fpace a body of air is confined in, the greater
muſt be the force of compreffion that confines it, it follows that the
compreffing forces are inverfely as the ſpaces which contain the fame
quantity of air.
351. Let there be two unequal cylinders A and B, whoſe diameters
are d and r, heights — a and e, filled with two different fluids of den-
fities, n, m, and fuppofe the fum of thefe columns in equilib. with a
nother column C, whofe diameter is r, height = b, filled with a fluid
of m, denſity, then (by art. 338) paddn, perrm, and pbrrm, are
the weights of the columns, fo paddn+perrm = pbrrm, or
206 THE UNIVERSAL MEASURER
addn+errm=brrm, and if r1, then addn+embm,
or dd
=
:b-e:m
bm-na
>
or if dr, then e e=
m
na
thefe quantities.
&c. for any of
352. But if one of thefe columns (A) be air, and their diameters
all the fame and the fluids B and C of the fame denſity, fuppofe a po-
wer at B (fig. 165) to ſuſtain the column BD, of the fluid B, or C, but
if the column of air E F, be added the fame power B can only fuítain
the column BC of the faid fluid B, then its evident the column C D,
of the fluid B (being the difference between BD and B C is
= the
elaftic force of the air in EC (fee art. 350) and becaufe cylinders of
equal bafes are as their altitudes, it will be as the weight of the fluid
in D B, is to that in DC, ::DB:DC= and (by art. 351)
DC
DB,
as the force of the air when confined in E C : its force when confined
EC
in EF EF: CE=
EC
E F
EF
fo,
is as the force in E F,
EC
E F
DC
therefore
whence, as BD:CD:: EC: EF, or
CE
D B
as BD (b): B D - BC (be) :: BE-BC (h-e): EF (c)
ergo, bcbheb-ehee, which by compleating the fquare
&c. we'll find e√:bc+ h-u b + h
+- &c. if you'll
have other of thefe quantities.
2
2
353. If A IK C, be a compound barometer (fig. 166) cloſe ſtopp'd
at A, and open at C, enipty from A to D, filled with mercury from
D to B, and with water from B to E, the diameter of the leſſer tube
FK C, to that of the greater tube FIA, as I to d, the denſity of
mercury to that of water as m: 1, take the points H. G, in the fame
horizontal line with B, then becauſe the air preffes equally in all direc-
tions, the fluid in G K B, will be in equilib. with itſelf, as will alfo the
fluid in HIB, whence all the compound between G and H, is in cqui-
librio with itfelt; confequently, the column of mercury D H, is in
equilibrio with the column of water G E and a column of air of the
fame baſe conjointly, and fo will vary with the fum of the variations
of each of thefe, let v the variation of the air's weight, which is
meatured by the fpace the mercury moves in the common barometer
in a given time, e = the fpace which the water moves thro' at E, in
the fame time, then (becaufe cylinderic fpaces are inverfely as the
AND MECHANIC.
207
t
e
=
ſquares of their diameters), it will be as d d :e :: 1: the ſpace
dd
moved thro'or variation at B, therefore G E, the difference of the legs
EK and KB, will vary in its weight by e +
d
dde te
ad
Alfo,
face the fpace moved thro' by the mercury at B, is that moved
e
H
thro' at D=—, the difference D I will vary its weight
d d
this variation of weight is both the former, and fince
altitude of water, it will be as m:1::
dd
ddele:
e
by 2, but
is an
a d
dde +edde
L
e
:
equal the
d j
md d
2 e
= √ +
dd
height of the mercury of the fame weight, confequently
ddee, whence, e =
I.
vdd m
2m-dd-
dd — 1
or as ev ddm: 2m-dd
354. The body of a water pump (fig. 167) nm HC, is called the
pumps barrel. AG the cistern into which the water comes and runs
cut at a fpout A, the part EF CH, which goes down from the barrel
into the water at B, is called the pipe, or fucking pipe, the piſton ID
moves in the barrel by means of a lever or handle PG, &c. when the
pilton descends to HC, the air contained between it and the valve v,
(in the pipe) being reduced into a leffer fpace will be condenfed, and
by its elastic force preffes down the valve v, and fo forces open the
valve D, (of the bucket in the pillon) and fo rifes above the piſton,
then by railing the piston, the weight of the atmosphere preffes down
the valve D, fo that no air paffes thro' the bucket, by which the air
that remained between v and D, will be rarified, and as that in the
pipe is denfer than that above v, it acquires an elaſtic force fit to raiſe
the valve v, and fo enters into the barrel till the air in both is of the
fame denfity, then the atmoſphere preffing on the furface of the water
E F, in the well, caufes it to rife into the pipe B v, and in moving the
piſton in this manner the water will rife into the barrel, and from thence
into the ciftern, provided the length of the pipe E H, between H and
the furface of the water in the well (fee queſtion 178) do not exceed
35 feet,but becauſe part of the air which enters into the barrel is forc'd
down againinto the fucking pipe, by the fhutting of the valve, the air
cannot be (fee art 349) intirely taken out, and fo this height is fettled
at 34 feet which anfwers experiments. Let a DH the diſtance
208 THE UNIVERSAL MEASURER
+
moved by a ſtroke of the piſton, b = EH the height of the pipe, c➡
34 feet, e the height to which the water rifes in the pipe (at the
Erit ftrokes of the pilton and ſuppoſe the diameter of the barrel equal
that of the pipe, otherwiſe they muſt be reduced to the fame diameter
by dividing the capacity of the barrel from HC to the end of the ftroke,
by the fection of the pipe, thus if the di meter of the pipe be d, and
that of the bariel D, and the length of a ſtroke I, then 1D D the
capacity, fo IDD dd, lo, if the pifton moves I inches in the barrel,
it anſwers to IDD÷dd inches in the pipe, the piſton in every ſtroke
fhould defcend clofe to the valve v, at the pipe's head, otherwife the
rifing of this valve will be partly flop'd by the waters weight above it,
which ſo much ſtops the working of the pump. Then, when the piſ-
ton is railed the height a, the air in the pipe would be rarified into the
fpace a + b, did the water not rife in the pipe, but the water rifing to
the height e, this air is reduced into the ſpace a+be, fo (by art.
351) its denfity before the ftroke, will be to that after the ftroke in-
verſely as a + b —- e : b, but becauſe the air in the pipe was in equilib.
with the atmoſphere before the ftroke, it will be of the fame denfity,
and fo its elastic force will be as c, whence as a + b
e: b: C:
the denſity or elaſtic force of the air after the ftroke, which
bc
ab-e
together with the weight of the water e, are in equilib.
moſphere immediately after the ftroke, viz. e +
bc
a be
1.
with the at-
=c, fo, e a
or putting
eb-ee-ca-ce, or: a+b+c:xe:_ee = ac,
?z=a+b+c, then 2 ze -eea c, which by compleating the
fquare &c. gives e — z — V:zz + aç:.
355. (By theo. 168) we have c for the velocity with which it en-
ters the pipe at F, and (by theo. 169) c — √/b, for the velocity with
which it enters into the barrel at H, whence the pifton moving uni-
formly in the barrel. The water in the pipe follows it with a vari-
able velocity, the greateft velocity being to the leaft as √c: N/C -
b.
356. But it's evident, 2 pump diſcharges moft water with the great-
eft eaſe, when the water in the pump rifes clofe after the bottom of
the pilton for th n there is no vacancy between the water and bottom
of the pifton, and fo no part of the ſtroke is loft, nor any of the water
hindered, as is the cafe when the water moves fafter than the pifton,
now becauſe the pifton, and confequently the water in the barrel
moves uniformly it is evident that the uniform velocity of the water at
the bottom of the pifton, will be the velocity of the pifton, when thę
AND MECHANIC.
209
pump is in its beſt perfection. Let S the ſpace defcended by a heavy
body in the firit fecond of time, then (by theo. 167) as /S : 2S ::
25√c 2 Sh
Nc-b:
نه
= 2 √/Sc — 2 √Sbv, feet per ſe-
cond, the uniform velocity of the piſton.
357. Suppofe the higheft elevation of the pifton to be Dv, and let.
Ev + Dvb put v the velocity of the pifton, D the diameter
of the b₁re, d = that of the pipe, V
that of the pipe. V the leaft velocity of water
that rifes in the pipe, which (by the laft art.) is found V = 2/Sc
2Sb, now if (DD) the bore of the barrel be greater than (dd)
that of the pipe, its manifeft that the water will run out of the pipe
into the barrel with a greater velocity, in proportion as the bore of the
barrel is greater, fo it will be as DDV::dd: v, whence DD v
2√ Sc― 2 Sb. Theo 199.
=dd V, or V=
DD v
dd
muen
Any one of thefe quantities may be found if all the reſt are known,
by which a water pump may be fitted to the beſt advantage, but here
obferve that the water as it rifes in the pipe meets with ſome reſiſtance
from the fides of the pipe fo the velocity as before found may be a
fmall matter too, and that the water may meetwith the lealt re-
fiftance, let the diameter of the bucket's cavity be that of the pipe,
and the infides of the barrel and pipe truly cylinderical; the difcharge
of water is the fame, whether the valve v be placed at the bottom of
the barrel, or bottom of the pipe at E or near E, as is eafily gathered
from the foregoing demonftrations, when the valve v is placed at H,
it will be eaſier come at when it wants repairing, which it will oftner
do than if it were placed near E where it is always kept wet.
38. Since the rifing of water in a pump depends intirely on the air
being taken out above it, the parts thereof fhould be fo tight as to let
in no air, and then the water (fee art. 354) will rife to the height of
34 feet below the loweſt deſcent of the pifton, but by reaſon of the
water's weight &c. (in this fort of pump called the common, or fucking
pump it cannot be carried much above the pifton. fo in heights above
34 feet there is ufe made of two other kinds of pumps working in
frames and forcing the water above them, &c. called forcing, and lirt-
ing pumps.
Dd
3
210 THE UNIVERSAL MEASURER
PROBLEM CC.
To find the path of a projectile in a non-refifting medium.
359. Suppofe a body projected from the point A (fig. 168) in direc-
tion A C with a velocity v, fit to carry it over a given diſtance d, in
Let b the distance fallen by a heavy body from a ſtate
fine LCAB, which Am E B the path of the
the horizon AB, cits co-fine, r = radius,
=
the time t.
of reit in that time, S
projectile, makes with
then (by theo. 48) T
re
C
= A G and
e S
===
C
G H (e being = A H) now
becauſe the ball, or projectile in motion, is by the force of gravity con-
tinually compelled to leave its right line or direction A C, it muſt
therefore defcribe fome curve, as AmE B, where AC is a tangent to
it at the point A, but as gravity acts always || G H, it does not alter
the velocity of the body in that direction, ſo if we ſuppoſe E H to move
parallel to its felf along with the ball, its plain the motion of the ball
in this direction may be looked upon as uniform, and then as dt:;
time of the bails defcribing A m, and (by theo. 166) as
r.e
ret
A
C
dc
2
rreeb
tt: b
ret
:
Gm
dc
d dc c
the diftance fallen thro' in that time
lo, G H - Gm=mH=es÷c:
rreeb
d dc c
ddesc-rreeb
ddcc
and when m H is a maximum we'll have dd Sc—2rrebo, whence
e=
s c d d
2 rrb
ssd d
AD and then H m becomes ED=
again
4 brr
if e be taken AB, then if mo=
ddesc-Treeb
ddcc
whence
C
Scd d
rrb
- AB, which is double to the value of A D laſt found, and
fo proves the ball to be higheſt when half its flight is performed, when
e=
Scd d
2 rib
=AD, then
eS
C
GH; becomes
S Sd d
ärrb
=FD, which
S s d d
45rb
whence the path Am EB, is a common
is double to ED =
parabola, for in that curve, the ſub-tangent F D is double ED.
Theo. 203.
360. From this theorem deduced the principles of gunnery, where
note,
AND MECHANIC.
211
361. The angle B A C, which the gun's axis makes with the plane
of the horizon A B, is called the elevation.
362. ED the greateſt height to which the ball rifes in its flight, is
called the height or altitude of the projection.
363. If the ball were thrown directly upwards with the fame velo-
city, the height A P, to which it would rife, is called the impetus.
364. The diftance A B, between the gun A, and B, where the ball
firſt ſtrikes the horizon, is called amplitude, random, or horizontal
range, which is greateſt when the elevation is 450, becauſe the fine and
S S d a
co-fine of 45° are equal viz. c➡S, and then A B
for SXS,
is greater than any S x c.
rrb
365. Becauſe
s s d d
ED, and S
4 rrb
s c d d
rrb
A B, therefore as, E D
s s d d
:AB::
scdd
4
4 rrb
rrb
sc, hence as c: s:: 1B: 4ED,
fo 4 EDA B, when sc.
366. When the ball is projected directly upwards then oo, and
ssd d
rs, and then
S,
4 rrb
45° then rr―sses, or ssrr, then ED =
dd
=ED, becomes =AP, but if S = fine of
4b
rrdd
dd
8 rr d
85
whence, as A P:ED::
dd dd
b8b
:
:: 2 : I.
2: But
per laſt art. 4 E D
≈ AB, fo as, AP: AB :: 1 : 2, i. e. the greateſt amplitude is always.
double to the impetus.
1:
367. Its manifeft, while the ball performs its whole fight, a heavy
body from reſt would fall from G to B, fo let tangent LCA B
a = A B, t = time of the balls flight, r = radius, then (by theo. 47)
n
as rn::a: =BC, and (by theo. 166) as b: 1 fecond ::
na
r
na
Γ
na =tt, but if n = tangent 45°, then n = r, and fo=~,
r b
in fecond, and a in feet.
368. Becauſe the motion of the ball, or the
A C and A B is uniform, therefore as a (AB)
n
: c✔ = time defcribing A m.
ar b
:
GH, along the line
na
✔nà (t) ::e (A HI)
ΓΟ
212 THE UNIVERSAL MEASURER
369. Let a amplitude, Sfine, c = co-fine of the elevation,
and y = fine of its double. Alſo, let A, C, S and Y, denote the fame.
things in fome other projection of an equal impetus, then from
sc dá
rrb
AB, it will be, as A : a
CSd d
cs d d
rub
:: SC: sc,
but (bỳ
rY
I
rrb
theo. 40) SC = ry, and scry, whence, as A: a ::
ry :: Y: y, and if Sfine 45° then Yr, and as r: A:: y: a.
370. When Sfine 45° then A —
fine 45° then A - greateſt amplitude — twice im-
petus, íuppoſe = 2 m, (lec art. 364 and 366) therefore as r: 2m:
y:
2 m y
r
=a.
371. Let vthe velocity with which the ball fets out, which being
thrown directly upwards would riſe the height A Pm, fo (by theo.
169) the ball múlt fall thro' A P, to acquire fuch a velocity, which in
direction A C being uniform, it will (by theo. 167) be as, b: 4 bb::
m: 4 bm=vv, fo m
V V
4 b
372. Let h = E D, the height of the projection, then (by art 365)
as csa: 4 h,
h, ergo, h
get as follow, viz.
373. tt=
s a
:
с
4C
a, from thefé equations we
na
r b
2 myn
rr b
4 s cmn
rrb
4 sc mrs
rrb c
4 s s m
rb
(ſee art
176) =
SS V V
rb b
374. a =
2 m y
V V V
r
2rb
SCVY
r b
4 scm
4ch
,=
S
r
S
375. h =
SS V V
S sm
a =
=ttb.
4 C
4 rb
r
376. (By theo. 47) as (A D) ½ a : (D F) 2 h, : : r ; n =
4r h
ર
(tangent LCA B) and (by art. 167) tt =
na
=2a, fott=
4 rha
rba
D
r b
and becauſe is conftant, therefore t t∞ h, or t∞ wh,
b
be the elevations and ranges what they will.
AND MECHANIC.
213
377. From art. 373 and 376, we have tt=
rh, alfo by art. 273 tt=
SS V V
rb b
v vrh
m
laſt equation, rb btt =
4 s sm
rb
=
4h, fo s sm
orrbbit = ssv v, fo per
whence vbt, or (by art.
h
375) ▼ =
24/rbh
&c.
S
i
a=
378. If S the fine of elevation, cits co-fine, a amplitude,
impetus, h height of the projection, t = time of flight, and n
= tangent elevation of a gun g, and S, C, A, I, T, H, N, the like
things of another gun G, r = radius, then (by art. 374)
4 rs c
r
4 sch
აა
, or rh—iss, as alfo, r HISS, now ifh H, then iss
ISS, fo asi:
379. If i I, then
✔H::s: S.
I :: S: s.
r H
rh
or SShss H, hence, ash:
r
ა ა
11
$ S
rh
380. If s = S, then "H=h, or HihI, hence, as I: i ::
H: h.
381. If a A, then ics
ICS, hence, as I:i:: sc: SC.
382. Alſo, when a A, then Schs CH, fo as, H: h:: Sc:
sC, but (by art, 176) rs = cn, fo—=—, whence 4th
C
S
Γ
П
n
4rH
N
or Nhn H, therefore, as H: h:: N:n:: Scs C::I:i, (by art.
381 and 382) :: TT:tt (by art. 376).
383. Ife the horizontal diſtance of any object and q. the time of
flight thereunto, of the gun g, and E, Q, be the like things of the
gun G, then (by art 368) as A: T :: E :
TE
te
Q., and alfo
a
q, now if q =Q, then TEate A, and if Hh, or T=t, then
Aea E, and then as A: E::a: e.
384. If v and V, be the velocities of the balls from the gun g and
G, t and T, the times of flight over a and A, then (by theo. 153),
TV A, and tv ∞ a, hence, as TV: A:: tv: a :: E: e, and if
Tt, then as V : v :: A ; a ¦ ¦ E: e (or if V = v) as T : t : : A :
:a:: E : c.
214
THE UNIVERSAL MEASURER
SS V V
385, (Bý art. 373)
r bb
SS V V
tt, and
rb b
TT; and if t-
T, then ssv v
SS V V, or s v➡S V, hence, as V
vs: S, or if
T
t
s =S, then
V
=
or Tvt V, fo as V : T :: v: t, and when
V
neither are equal, it is as V : v :: : whence velocity, time and
T t
S S
fine of elevation here, are the fame as velocity, time, and ſpace in uni-
form motion.
386. But the velocity of a projectile in any point of the path, is ag
the fecant of the L of its direction above the horizon, for A H, the
horizontal velocity is the fame at all points of the curve, and the ve-
locity A G at A in the path Am E B, is the fecant of the L ofcle-
vation G A H, hence, the velocities at any two points in the curve
Am E B, equally diftant from E its middle, are equal.
387. If the body or ball be projected on an inclined plane A m, let
✔ the velocity of the ball in A, or the fpace it moves thro' in time 1,
b the fpace fallen thro' by a heavy body from reft in that time, s —
LCA m, c fine L G A P, or A Gm, z= fine L P A m, or Am H
fine L Am G, and a Am, then (by theo. 48) as c:a :: z:
Gm, but (by theo. 152) as v :
az
= AG, and as c:a::s:
sa
C
C
a z az
I
1 :: (AG) 22: 27=t, time of defcribing A G, alſo (by theo. 166)
as b:
с CV
sa
1 :: (Gm) sa : tt == time, falling from reft thro
с
c b
Gm, and that tt is manifeft, becauſe A G and G m are both de-
fcribed in the fame time, therefore tttt, viz.
Gm
a a zz
sa azz
or
c b
CVV
CC V V
whence, a Am=
S C v v
>
or a ∞
SC V V
zzb
ZZ
conſtant.)
(becauſe b is
388. If A dm d, then e is the vertex of the path A e Em, in re-
ſpect of the plane A m, and then (by art. 366) de = 4 Gm=
(becauſe a=
SCV V
zzb
)
SS v v
4zzb
of the oblique projection.
sa
4¢
which call h, viz. deh, the height
AND MECHANIC.
215
a z
389. Let t = time of flight, from A to m, then (by art. 387) t =
Sv
27=2, (becauſe a = SCVV).
C V
Z D
390. Hence, ash:t::
zzb
SS V V
SV
SS V V S V
:
:
4zzb
zb
2
Ꮓ Ꮓ
391. Becauſe a-
S C V V
zz b
therefore, z and b, continuing the fame,
a will be as Sc, which will be greateſt when Sc,
LGAP.
392. Let m the impetus, then (by art. 371) m=
or Lm AG
LMAG=
V V
lete=AH
4b
n = tangent ↳ H Am, Um H, or L m, radius = 1, then (by theo.
L
47) nemu, and y a➡u (y fine L HA m and zits co-fine Am H)
prza≈ e.
393. Whence e = za =
4 S cm
SC V V
SC V V
scm
or e ∞
or
zb
Z
Z
Z
394. Thefe things are true whether m be above, or below, the
horizon A B, and from thefe equations, we get thefe following, viz.
395. a =
SC VV
45 cm
e
u
b z z
Z
22
y
396. h
S V
2 S
397. t =
b z
2
b
S S V V
4 b z z
Ꮮ Ꮓ .
Ssm
sa
es
us
11
11
4C 4C2
4yc
398. Theſe things may alſo be proved from the problem itſelf, thus,
Let radius=r, and c, s, d, b, h, a, e, be the fame as in art 359 and
es Freeb
369, then becauſe,
с ddcc
Hmh, (and by trigonometry)
rraa÷cc=dd, and ab, (taking b = BC, d = A C, a = A B)
C
5
_=h, ==X:ae-ee: but (by art. 176) —
we'll have
es
ees
с
a c
са
t (t = tangent LC A B) ſo h = -x: ae-ce: from whence
ra
r
raha
we get A Hea+aa-
t
216 THE UNIVERSAL MEASURER
399. Ife and h, with a the greateſt
elevation, then (per art. 369)
Sca
2rr
amplitude be given to find the
prefent amplitude, whence h =
C.
e s ees
ca
es
becomes h
rree
C
2 acc
(becauſe by art. 176, rs = A, and
cc
-rree-ttee, which folved gives t
-
te
r
rr+t
rr
: rr + tt: Xce
2 arr
=2arrh=2aret
r/:aa-2 ah-e e:+ ra
e
e
(but when e AL)= ±
r√ : aa + 2 ah — ee:+ar
(but if m =
tt I: xee
ra-
+
fet
th
the impetus be required, we'll have m=÷a=
dius r being taken 1, and 4 et 4h, when e A H, but 4et +
4h, when te AL, if the L of elevation (t) be required, fo as m the
impetus may be the leaft poffible, then by making t variable in m =
rr+ ttxee we'll have 2 tte + 2 trh rre ett0, = 2
4ret4rth
rh
minimum, whence t ==¹h+√:
e
rrhh
r:
+rr but (by theo. 47)
ee
rhen (n= tangent Lm AH), fot=Fn+v:nn + rr:, but
(by art. 176) n'n+rrss (s = fecant L m A H) ſo t =
foto + S,
whence Lm A GLGAP (as by art. 391) fo the required leaft im-
Hm+ mA serh
petus will be =
2
en
r
Alfo, becauſe ==h, we have (by art. 399) ‡ a =
a, or as r:tLGAB::e:a,
21
errtte
416-450
(n =
rd
(s = fe-
S
tangent m A H), the object's elevation or depreffion.
400. From the laft art. we'll have e=
23r :t
rr. tt
ī
cant Lm A H, the inclination of the plane and d = A m) whence d➡
2 sax:tn:
rr+tt
vation.
2 sa
cc
x:tn: (cfecant 4 m A H, the ele-
401. The greateſt random upon a horizontal plane being (by art.
364) when the elevation is 450 but by reafon of the air's refiftance,
this elevation muſt be fomewhat less than 45°, and the ranges above
45° will fearce go fo far, or thoſe fo much below 450. Alfo, the
lighter matter bullets are made, the greater will be the refiftance,
AND MECHANIC.
217
for take two equal quantities of any matter fuppofe lead, and of the
one make a globe, but of the other a fohere of a double furface, then
its plain the refiftance of the fphere will be double to that of the globe,
(every thing elfe being the fame) for a double area or furface mult
ſtrike a double number of particles of the fluid. Hence, with the
fame charge of powder a ball will go much farther than a charge of
fmall thot of equal weight. But in the practice of gunnery the refif-
tance of the air is not regarded, becauſe when it is, the work is very
tedious, and indeed need not to be obſerved much, becauſe the greateſt
random of every piece is had by trial made with a proper ball aud
charge of powder, by which any other randoni &c. is found; in this
its plain this refiitance is in fome meaſure conſidered, the theorems for
this purpoſe where the refiftance of the air is confidered are as follow.
Let t tangent of elevation, sits fecant, m = height, e — the
height fallen through to acquire the greateft velocity in the medium,
cco-fine of the elevation that gives the greateſt random, z = fine
of any elevation to an amplitude a, and A = fine of that elevation in
vacuo; then, if the refiftance of the medium be as the fquare of the
ball's velocity, we'll have (fig. 168).
1
1. h=ED=
tt m 4 tttmm
2t5 mm
+
$ S
3 esss
+ &c.
3e85
2. a A B =
3. AD=
4tm
S S
2 ttm m
3 est
1614 mm
ni
+
3ess
+ &c.
2tm
5. S
4 tt mm
083
+
4t4 mm
+ &c.
3 es +
4.cc=
96e
235 m
nearly.
5. 2A+
a a
a
4
+ &c. nearly.
192e-512m
+
óem 192 em
For the demonſtration of theſe, fee art. 432 and 436.
Note. When the refiftance is
the terms divided by e' are = 0,
refifting medium as before.
eis
o, then e is infinite, and then all
then thefe theorems hold in a non-
But if the body be projected directly upwards with a given velocity.
g, then q being put the uniform velocity acquired by falling from
reft in the fluid in the first fecond of time it will be 6, 4/2qe = the
greateſt velocity acquired by falling in the fluid 7, mex 2,302555
times log. of ± £8±29, fee queſt. 233, and thoſe following it,
age + vr
E e
218 THE UNIVERSAL MEASURER
PROBLEM
CCI,
Being as a SUPPLEMENT to the foregoing problems,
fbewing the nature of fluxions, and flucnts, with ſeveral uſe-
ful articles in mechanics, thercan depending.
402. Fluxion, according to Sir I. NEWTON, is the fame with ve
Jocity, is by all allowed to be but a fimple idea.
403. Whence, the fluxion of any quantity is but an idea of fome-
thing inconceivably fmall, by whoſe continual motion the quantities or
Дuents are generated.
404. The very fmall parts of the fluent which are generated in very
fmall parts of time, are called moments or increments, or if the fluent
decreaſes, the parts continually deftroyed are called decrements.
405. Moments being the effects of fluxions, fluxions are as moments
generated in the fame time, for the effects of like caufes are propor-
tional.
406. The velocity, variation, or quickneſs of increaſe (or decreaſe)
of any fluxion is called the fecond fluxion, and the fame thing of the
fecond fuxion, is called the third fluxion, &c. for as the parts of a
fluent may be generated, falter or flower, fo theſe parts will cauſe a
variation in the fluxion &c.
407. If z be a flowing or variable quantity its fluxion is written thus
z, the fluxion of that fluxion, or fecond fluxion of z thus Z, the fluxion
of that fluxion, or third fluxion of z thus z, &c. but becauſe theſe
points above the letters are often wrong printed, I fhall ufe different let-
ters for fluxions.
408. In the following articles, e, y and R, are taken for the abfciffa
ordinate (or femi-ordinate) and curve refpectively, and their refpective
fluxions are denoted by u, v, and z,
409. A conftant quantity has neither increaſe nor decreaſe, and fo
can have no fluxion, fo bu, is the fluxion of e b, as alfo of ebb
f, jfb and f, be both conftant, or ſtanding quantities, and on the con-
trary, e b is the only fluent that will be found to u.b, unleſs by the
nature of the queſtion, as fometimes it happens, there must be fome-
thing put to this fluent, and then it is called the correct fluent, or fluent
corrected.
410. Let A R B (fig. 210) be a curve, draw F n parallel to, and very
near EB; alfo, draw CB parallel to Q_A, and D n, and tangent to
the curve, in the point let e A E, y EB, R arch A B, then if
=
AND MECHANIC.
219
· and
the moment of y be r n v, that of e will be Br EFu,
that of R = n B = z that is v, u, and z (becauſe the moments are
as the fluxions) are the fluxions of the ordinate, abfciffa and curve re-
Ipectively.
411. If by the parallel motion of y the area or ſpace E A B be de-
fcribed, then uy, viz. the very fmall rectangle FE Br, will be as the
fluxion of the faid fpace, which if the equation of the curve be cen—
y, will be u cenuy, but (prob. 185) the area, viz. the fluent of
this fluxion u y is =
u,
се
n
whence, the fluxion of any fluent ceª
is nucer¹ and on the contrary, the fluent of the fluxion n ce"-1
is =ce", that is, the fluent of any expreflion confitting of conſtant
quantities, with only one variable quantity and its powers and fluxion,
is found by the rule in prob. 185, and the reverſe of that rule will
find the fluxion of fuch a fluent.
412. From the laft general fluxion may be had the fluxions of furds
or of fractions, thus the fluxion of: 2ae-ce: is
2 u
e u
V: Jac
ee:
for here n, fon
I
*
†, and the fluxion of 2 a e
ee is =
2 au
u, is: 2 au
2 eu, and ne¹ is X 242-00
플
​2eu: XX2ae—cɩ
therefore, n ce
I
2
as before. Alſo,
a
- I
=axe
we have n =-
I,
fo n
I -
2, therefore,
C
I
ane¹¹u—— IXaeu, for the fluxion of, &c.
for any fuch like.
au
ee
413. If the fluxion of e y, the product of two variable quantities be
required, let each be added to its moment, and then multiplied to-
gether i. e.eu:x:y+v=ey + ev uy uv, wherein
u v, being very fall in refpect of the other terms may be rejected,
and then it will be eyev + yu, from which taking e y the given
product there leaves ev + y u, for the moment of ey, which is as the
fluxion required, in like manner the fluxion of the product of Rey,
three variable quantities will be found Rev Ruy+zey, &c.
for any other number of variable quantities. Hence, if e = R= }
we'll have zeu for the fluxion of c e, alfo 3 e eu for the fluxion of
e e e, and in general nen-
u for the fluxion of en' the fame as in art,
411, alſo, if R≈c+
a
y
we'll by what goes before, have z — 1 —
#
220 THE UNIVERSAL MEASURER
аг
,
Jy
2uv
a e
again if R = 2, or R =,2 e y—¹, then z =
-ave
y
if Raa-ee: then z
a u
ave
y
YY
, if R
ce:
e u -cu
=V: 2ae-ee: then z =
&c. &c.
√:2ae—ee:
y y
-
:aa
e u
414. In exponential expreflions, or thofe whofe indexes are variable,
as y put Rhyperbolic log. of y, then ye becomes e R, whoſe
e
fuxion is =ez+uR = ~+u R, becauſe (art. 416) z = ~
y
У
415. By thefe articles any expreffion may be put into fluxions, and
tho' the reverſe methods give the fuent, yet if the fluxion is altered
as it often happens, it mult have a different fluent, which when all
other methods fail may be had by infinite feries, as directed in prob.
185, therefore to find the fluent of any fluxion, try to find fuch an ex-
proffion, as being put into fluxions may be exactly the fluxion given,
which if it be, you have hit on the fluent required, otherwife you muſt
ufe infinite feries.
416. It is proved in the conſtruction of
I
Ite
feries &c. gives the hyperbolic log, of 1+e, whence
fusion of the hyperbolic log. of 1 +e, but
U
being put into a
U
is the
1 + e
is fluxion of
1 + e, ÷ : 1 ÷e: i. c. the fluxion of any expreffion, divided by that
expreffion is the hyperbolic log, of that expreiñon, whence the fluent
of fuch a fluxion will be the hyperbolic log. of the faid expreffion,
thus the flucnt of
ש
C
=
is the hyperbolic log. of e, or 2,3058 tabular
log. e, alfo the fluent of u =
au
✔: yy — a a
ya; again, the fluent of
a+e+:2aeee: fluent of
+ Veeaa, &c. for any fuch like.
a u
is e 2,3058 a x log. y
V:2 ae+ce:
u
V:eetaa:
is 2,3058 axlog.
= 2,3058 log. e
AND MECHANIC.
221
417. Sometimes one fluent may be found by another fluent given,
called the compariſon of fluents. Thus, given A
U
fluent of
/:aa‡es:
ee u
eeu
to find B = fluent of
here
x € =
: aa fee:
Viaa+cc:
e u
√: alee + ceee:,
then from a notion of fluxions it appears that
the laſt expreſſion wantsa a eu in the numerator to make it a com-
Jaa cú Lee eu, whofe fluent will be vatc´+<*
eu
√: dae e + eeee:
plete fuxion, viz.
=
tevaatee: from which take the fluent of
the part added, which fluent as that of
is given A, will be a a A, whence
e u
√: a²e² + c*:
±
aa eu
:
v:aatee:
eva a tee:
=B, the fluent ſought; again, given A=fluent of
Π
7
aaA
to
7:2
:22 ee:
-
to find B = fluent of
eeu
Here as before, the fluxion of
√ a 'e
therefore, from e√: aa
-eeee:
√:aa -ee:
evaa-ee: is =
a a ve
ceea
2
a
ee: take
Я а
the fluent of
-aa e u
1 an
-
and there leaves :aa-ec:
/:d C
1
+ Aaa
Note. A
Vida — ce
B, the fluent fought.
2,3058 log. e + №: a+e, in the first of thefe ex-
u
amples, viz.
=
fluent of
by the laſt art, and Aarch
viaa +ee:
e
of a circle, radius unity and fine
= fluent of
u
for per
d
√:aa—cc:
fimilar As (fig. 210) as EB (v: aa-ee:): QB (1): :r B (u): nB
z, there are other methods more general, for the compariton of Auers
but they are very prolix, and fo are here neglected.
418. In fluxional equations, where only one of the two variable
quantities (e and y) enters, you may fubftitute for the ratio of the two
fluxions`(u and v) and fo find the required quantity; thus, ifa u v◄ ▼
=y× uu+vv)², let u—n v and then an ▼ * = y X DAVV + V
1
222
THE UNIVERSAL MEASURER
2
whence yan÷nn÷1
which in fluxions (taking m =
fluxion
3
whence unyam n
of n) is vam— 3 annm:÷nni
-3 annnm:÷nn
-
J
now let s fluxion of w, and put ww-
nn+1, then w w 1=nn, in fluxions wsnm, which fuítituted
in the value of u, gives u =
aws 3 awsx:w
-I :
6
W
45 W
S
4aw
21
4
2
- 3 s w — 3, and therefore e (the fluent of u)
за
a
W +
2 W W
3annta
2Xn+1
as required. Again, ifa y uv v v
+
≈uu+vv|, put nyu, then aynv nav v
1
ſo y =
I
:X
confe-
nn
i
m
nn-1, which in fluxions is v =:3nn+2
an
quently nv":303+2n- 1: X, therefore e ==:
n
a
annnnnn – 2,3025 log. n, as required.
X:
d
419. Fluxional equations, may alſo be folved by an affumed feries,
as directed at the latter end of algebra, only here you muſt have e-
quations for both the required quantity and its fluxion or fluxions,
— v, or u — v — evo, to find e in the terms of y,
thus in
u
I fe
affume e Ay
=
fluxion of e, viz u
5
Byy+ Cy3+Dy4+ Eys+&c. and then the
Av+2 Byv+ 3 Cy v+ 4 Dy 3 v +5 Ey¹v
+&c. which values of e and u, fubftituted in u v — evo, gives
Av + 2 Byv +3 Cyyv+4 Dyyyy+ &c.
-v-Ayv-Byyv-Cyyyv-
Whence A- 1 = 0, or A
&c.
1 = 0, or A = 1, 2 B — A = 0, or B, 3 G
I
cc.}
O.
==
2
I
2,3,4
yyy
+
2,3
— B = o, or C = ; B = 1, 4D — C = o, or D = C=
— 1
2,3
confequently e = Ay+Byy+Cyyy+&c.=y+LY +
2
yyyy
y s
5
+
+ &c.
2, 3, 4
2,3,4,5
AND MECHANIC.
223
yu
Again, to find y in terms of e, in ce euyu av, or av
yu ceeu≈ 0. Here affuming y= Ae+Bee+ Ceee+
Det + &c. as before we'll get
aAu+2aBeu + 3aCeeu + 4a Deeeu +5aEetu
O
A eu . Bee u Ce e eu - Det u
ceeu
&c. for a v
&c. for y u
=
C
Whence A = 0, 2 a B = A = 0, 5 a CB+c, or C = a
D = C=
D=C=C
C_, or D =
C
, 5E=D=
с
or E-
3, 4 a a
3, 4 a a
ce3
+
3 4
3 a
and confequently y Ae+Bee+Ceee+&c. =
w
*
3
C
4a
3,4,5aaa
ce+
3,4aa
ces
+
3,4,523
+
себ
+ &c.
3, 4, 5, 6a+
n
---
v
a
Again, to find e in terms of zin à u u + ee v v a avv = 0, (v
being put for the fluxion of z), firft, for the indexes of the affumed
feries, work as directed, at the latter end of algebra. Thus, write
z" for e, and n z n¹ y for u, and then the indexes of z in the equa-
tion, will be 2n 2, 2n, and o, the two leaſt of which (2n-2ando)
being put equal each other, we get n=1, whence the indexes (20 –
2, 2n and o) are 0, 2, 0, and the differences o, 2, to which by con-
tinually adding 2, thefe turn out 0, 2, 4, 6, 8 &c. each of which
added to the value of n. gives 1, 3, 5, 7, 9, &c. for the indices of
the affumed feries, therefore, put e Az + B z ³ + C z ³ + Dz?
+&c. and to facilitate the operation, let v = 1, then u — A + 3 Bz³
+ 5 G z + + 7 D z + &c. which two values of e and u, fquared and
fubftituted in the given equation for ee and uu, it becomes
6
3
6
5
aa AA 6a a ABzz+10aa AC2+14aa AD 2° + &c.
+ 9 aa BBz30 aa BC z + &c.
6
6
a a
A A z z + 2 A B z +
+ 2 AC z + &c.
+ BB z 6
0.
+ &c.
Whence A Aaaaa, fo A = 1, 6 aa B —— A A, ſo B =
6ad
I
2, 3, 4, 5 a
I
14.36a4
I
2,3 a a
A A
6a2
10 aa AC + 9a a BB + 2 A B = 0, ſo C =
14a a AD 30a a BC+2 AC+ BB = 0, foD-
confequently e
Ζ Ζ Ζ
+
2,3 a 2
5
2, 3, 4, 5 at
224 THE UNIVERSAL MEASURER
Z
2, 3, 4, 5, 6, 7 a
6
↓
required y, in a aez — 2 aauva euu+eeey
- I,
will be n
we get n = 2,
o (z being the fecond fluxion of y, and u'and v, the fluxions of e
and y as before) then putting en for y, the indexes
D 1. I and n --- I, where making n
whence, the differences being o 2, we must put y
+Ce+ De +&c. from which making u1, we get
8
3
↳
I,
2
Ae² + Be
V = 2 Ae -
22 A 12 Be² + 30Cet +
4 Be³ + 5Ces 8 De' + &c.
2
Thefe values of y and z fubſtituted in the
56 De°+ &c.
given equation, &c. as
&c. this feries, is
4
before, we'll find y =
ес
e
6
8
e
e
2 a
2
6 as
8 a'
the fluent of
2. e u
aa tee
=y, alfo
aa
"
au tee
=2,30581 ax log.
Required y and e, from the two equations y zru, and rýrz
ez.
Let e
ARBR² + CR³ + DR+ + ER5 + &c. (R= fluent of z)
and yaRbR² + c R³ + d ReR5 &c. then by fubftitution
and tranfpofition, our two equations will become.
3
a Rz÷b R R z÷ c R ³ z + d R¹z+ fR5z +&c.
raz+arbRz+31cR³z + 4rdR³z+5₁f R¹z,
BR2z+ CR³ z + DR¹z + &c.
-rAz
ar BRz
3rCR¹z— 4rDR³z 5 г FR¹ z,
C.
and
}
-rz
AR z
= 0.
Whence, A , a
2 r B, b b =
3 r C, c = 4 r D, d=
5rE,
by first equation and a
A
B
C
1, b
b:
d
&c
2 r
3 r
4 r
by fecond equation, therefore, 2 r B=1, 3г C
C
A
4r
4 ID=
2 r
I
C=0, D =
B
5 r E =
a
Γ
1
4 &c. confequently, B
2 r
B
I
D
E=0, F
2,3,413
I
3,4 rr
2, 3, 4, 5, 6г5
&c. Alfo b3r Co, c=4rD=
&c. &c. Whence ya R+bR²+&c.
5,6 rr
I
2,315
R3
R
+
2, 3 rr
R$
2,3,4,51
R7
2,3,4,5,6,7 *°•
&c. and eAR+ BR² + &c. =
AND MECHANIC,
225
RR
R4
+
2г
3
2, 3, 41
Note. Here y
(fig. 210) — A B,
Ro
2, 3, 4, 5, 6гs
fine, and e
&c.
fine, and everfed fine of any circular arch R
radius Q_B —QA =r, for per fimilar As as r
: (BE) y :: (En) z: (Br) u, ſo y z— ru, alfo, as r: (QE)r—e
(Bn) z : (nr) v, whence ry
ry=rz ze, the fecond given equa.
tion.
420. To folve problems or queſtions in fluxions, you muſt always
by the nature of the problem get the moment of the thing wanted,
whofe fluent is the anſwer required, theſe moments are fo fmall, as in
all cafes, to be looked upon as generated uniformly, or made up of
ftreight lines; moments increaſing or increments are poſitive, but if
decreaſing, called decrements, are noted with the fign
about moments holds in fluxions art.
what is faid
421. When any quantity or expreffion, is required to be the great-
eft, or the leaſt poſſible, its evident at that time its variable parts can
neither increaſe, nor decreafe, whence any expreflion is a maximum,
or a minimum, when its fluxion is made o, in fuch cafes where
there is but one variable quantity, the fluxion of that quantity being
in every term, divides off, fo in fuch queſtions I leave it out, whereby
it fuits either this article, or the prob. for that purpofe, in thefe artic-
les (fee art. 410) I take y, e, and R, for variable quantities, and v, u,
and z, for their fluxions, when any thing elfe happens, it is mentioned.
Thefe few foregoing articles, if well confidered, will teach the prin-
ciples of fluxions; what follow, are fome very curious articles by way
of illustration.
422. It has been proved that the refiftance of a body moving in a
perfect fluid is as the fquare of the fine of the incident angle, it there-
fore AR B (fig. 210) be fuppofed to move in fuch a fluid in direction
CB parallel to its axis QA, then the curve in the point B, (viz. the
line n BD, being a tangent to the curve in that point) makes with the
faid direction CB, the incident angle r Bn (art. 410) therefore, the
fines of the Ls being as their oppofite fides, it will be as z z: v v (viz.
Bn: rn) :: force of a particle of the fluid against the bafe B Q
VV now the quan-
: force of a particle againſt the fide at B, i. e. as 1 :
tity of the fluid which ſtrikes againſt the curve in
i. e. the reſiſtance of Bn, is
whence, the laſt ratio 1 to
*
Z z
B to refill its motion,
rn (v) becauſe its motion is LnF,
V V
ZZ
VVV
becomes v to
22
?
or becauſe z z =
Ff
226 THE UNIVERSAL MEASURER
F
uu+vv) v to
V V V
uu+vv
the moment of refiftance of a plane, out of
y y v v
>
ΓΓ yy
that is, v to
which by the equation of the curve exterminate u u, and the fluent of
what leaves, is the anfwer, thus if the figure be a circle, radius QB
=QRr, then per limilar As QBE and rn B, as QE; EB
In rB, whence uu
v v v x. rr y y:
uu - vv
V V V
fo v to
becomes v to
ΓΓΕ
y y vv + v v
YY, whofe fluent is y
y v
:rr
•
-yy:
rr
Το
rry
ΓΓ
TVV
3rr
or 3 rr to 3гr-yy, and when y = r it is 3 to 2, fo
is refiftance againſt baſe, to refiftance against circle or cylinderic fur-
face
Again, If e=cy", then (art. 410) uncy
n
V V V
I
v,
fo vv =
conny 2-² vv, which fubftituted in v to
for uu, it be-
uu + vv
V
comes v to
if the figure be a plane triangle moving
+ n
cenny
ее
in direction of e its perpendicular, bafey, then cc =
and n
and n=1,
y y
i
V
then the laſt general ratio is v to
>
or (art. 410) y to
У
Cc+I
cc - I
that is, eeyy to yy, fo is reſiſtance of the bafe to refiftance of the
fide, if the triangle move in direction parallel to y its bafe, then cc=
YY ſo inſtead of e e+yy to yy, we'll have y y +ee to ee; re-
yy
ce
fiftance baſe: refiftance fide.
*Note. What is here meant by a plane figure, is à priſmatic folid of
any given depth, whofe bafe is that (plane figure) floating on and paral-
lel to the furface of the fluid, and its depth, or thickneſs perpendicular
to the faid furface.
423. The quantity of thefluid ftriking againſt z (B n) the furface of
the fmall folid, formed by A D B, turning about AQ, will be as y v₂
fo our ratio 1 to becomes
V V
ZZ
y v to
y v v v
Ꮓ Ꮓ
y v v v
uu+vv
then as in
the laſt article with the nature of the generating curve, expunge u u
the fluent of what leaves will be the refiftance. Thus, if the quadrant
A BR, be turned about AQ, it will form a hemi-fphere, and as before
AND MECHANIC.
227
!
YYY, fo the laft ratio becomes y v to
yy vu
rr-yy
rry-yvyy
E there-
rr
fore (art. 410) y y toyy-YYYY, and when ry, it will be
2 to I,
A
4 rr
:: refiftance bafe to refiftance folid. Again, if (as before)
суп e, our general ratio, will become y v to
y v
Z ń
ccnny
t
and if the folid be a cone, formed as before R B A, being fuppofed a
ftreight line, A E≈e, radius E B of the baſe =r, then n = 1, and
c, when the cone moves in direction E A of its axis, fo the laft
e e
r r
ratio becomes y v to
I to
сс
f
CC - I
:
y v
CC + I
whence (art. 410) 1 y y to
I
Z
yy
2004 2
or
or eerr torr, or ss to rr (s being: - fide of the
cone) refiftance bafe to refiftance fide, or convex furface of the
cone, &c. for any fotid.
CQ__||
424. Let (fig. 211 ) A E = e E B = y, A Q = c; Bn=&rn=7;
EFu, arch CBR, and fuppofe A QL, and C Q|| horizon,
and that a heavy body is falling freely in the curve R BA, from R,
at reſt; it has been proved that the velocities of falling bodies (in
Vacuo) are as the fquare roots of the heights failen thro', fo the velo-
city at any point B is as CB = √ Q E = ✔: c—e:, and the
heights fallen thro' are as the times and velocities conjointly, whence
z=ty:c-e i fot
e: =
moment of the time of falling
in any curve, by whofe
of what leaves will be
==
Z
Vic-c:
equation you may exterminate z, the fuent
T, the time of defcent, thus if a e RR (a
any conftant quantity) then (art. 410) ae R, fo "√ a
whence t =
ua
2:ce-ee:
-
2 Ve
Zị
=the fluxion (art. 219) of a circular
arch, verfed fine e and radius c, and therefore thefe two mo-
ments or fluxions t and z, are always as a to c, whence T muſt be
always to that arch as a to c, but when e=c the faid arch becomes
a femicircle whofe diameter is p c, and then as c: pc::a: pa
=T, which being no ways affected with c, will still be the fame, let
the diſtance defcended be what it will, that is all bodies defcending, ot
all bobs of pendulums vibrating from any different points R, B, in fuch
1
१
228 THE UNIVERSAL MEASURER
a curve (called a cycloid) are performed in the fame time. Again, if
RBA (fig. 211) be a circular arch, arO A its radius, then
per fimilar As, or the property of the
whence t
ur
V:c_e: XV: ae ée:
circle, z=
ur.
V: ae ee:
which by an infinite feries
(prob. 185) will give the value of T, or by the compariſon of fluents
(art. 417) when ce, T=3,1416 √²x: 1+
3,3 cc
2, 2, 4, 4 d d
+
the arch R A.
3, 3, 5; 5 cc
2, 2, 4, 4, 6, 6, a a a
a
X:
4
C
2, 2 a
+
+&c. time of deſcribing
425. If PAR (fig. 211) be a veffel filled with water, ſtanding upon
A, its vertex, axis AQL horizon, a hole at A to let the water run
out, its plain the furface of the defcending water defcribes the folidity
of the veffel's cavity, therefore, let c AQ, e any height A E,
yy, upen = uyyy, is =mo-
and becauſe the velocity of
y
FL
=
BD, &c. as before, then if pe
ment of folidity viz. as the fluxion of it,
water, or any other heavy body is (e) the fquare root of the height
fallen thro', there Pe
pe" u
√e
= pe■ - u = t (ſee the laſt art.) — mo-
ment of time of evacuation, whofe fluent (art. 410) is
pe
n + ½
Σ
= T,
fo in a cylinder, or prifm ftanding upright, no, p=1, whence 2 /e
-T, a cone ſtanding upon its vertex, n = 2, fo√ecece = T, a
common parabolic concid, ftanding on its vertex, n = 1, ſo veee
=T, if R AP be the fegment of a fphere ftar.ding upon A its vertex,
2 € /1/
then Puya e
ve
(wnce) px:
ee:
t in fluents, T=px:
10 ac² — 6c
$4
15
2 ae
3
उ
3
if the fegment ſtand on its baſe
RP, then e muſt be = QE, put r radius of the fphere, then p u x
rr
ce
:=pux:rre
Ve
e t, fo in fluents px: 2rre
P
:=
T = (if c = e and a = 2 r) p X :
5a a √ c - 4 c
} e
Let
10
time in which this veſſel would be emptied with the firſt
or greateſt velocity, for inftance, fuppofe it to hold 69 gallons, and to
run out 1 gallon in the firſt ſecond of time, then at that rate it would
}
T
AND MECHANIC.
pc
be emptied in 69 feconds
- cc: it will be as
5
5 aa c
- 4 C
10
fo on for any veffel.
VC
229
Q), then its folidity be pcxa a
Xaa-cc: is to Q, fo is px
C
to T, the truc time of emptying in feconds, and
426. If R B D P (fig. 211) be the fruftum of a cone ſtanding on the
leſſer baſe B D, let e = EQ, its height, c A E, the height of the
remaining part of the cone, then px: c+c=Px: cc + 2 ce
tee:=yy (QP) fo per laft art. x: cc + 2ce + eé:
ve
pu
ee:=T,
t, fo (art. 410) Px: vex:2cc++ce + ee: T, if the
fruftum fland on the greater baſe RP, put e=QE its height, c =
QA the height of the whole cone, then px c-el = px: cc.
2ċe+ee:=yy (ED) ſo
pu
ve
X:cc_2ce+ee
2ce +eet, in fluents
pvex:2cc- ce+ee=T, if theſe two times are equal, the
quantity run out in the firſt caſe, will be to the quantity run out in
the fecond cafe, as 2 cc + ce+ eeto 2 cc — — ce + 3 ec,
where, if co, in the firft, and ce, in the fecond, then e AQ,
4
or the fruftum becomes a cone, and then the laft ratio, will become,
중
​16
as 3:15, or as 3:8 :: quantity run out when the cone ftands upon
its vertex, to that when it ftands upon its bafe, at an equal orifice,
and in the fame time, &c. for a fruftum of any other ſolid.
427. Suppoſe the curve R A P (fig. 211) to be a heavy flexible line
hung over two pins R, P, parallel to the horizon, draw Cr and QA,
each L RP, alſo BD and r F each || RP, and Bn a tangent to the
curve in B, the part B A is fuftained by 3 forces, firft its gravity in
direction Br, which is as the weight (fuppofe n) below the point B of
fufpenfion. Secondly, it is drawn at A, in direction || RP, or BD for
if the curve were cut aſunder at A, it would hang as two Ls, on the
two pins. Let a this force, which having nothing to do with the
chains weight, is conſtant, or ſtill the fame. Thirdly, the force in di-
rection Bn, by which it is ſuſtained; let A Eẹ, EB=y, Bn=z,
rov, Bru, then as v: a (force at a || BD): : u :
a u
V
= (force
at B || QA) n, by which the ratio of weights hung to a flexible line,
fo as to make it form any propofed curve may be found, as follows.
230 THE UNIVERSAL MEASURER
428. If it muſt be a femicircle, radius A Q = QR = Q P = a;
Q_E=e, EB =y, then y ya a — ee, in fluxions yv=- euj
-ay, which in fluxions (patting m
ау
>
a l
whence n =
becomes n =
e
fluxion of n, and a as before conftant) is m =
ayu
ee
W
a e v
now if
ayu
a e v
e e
a u
the force preffing the curve in the point B upon the fmall ſpace z;
then m = W2=
QB::r B:n B, whence,
but fimilar As (fig 210) as EB:
=z, fo wz=
wau
ayu - ey
y
ee
У
=
(becauſe y u = —
eu au vy
a u e e
eu, or v
= (becauſe
y
yee
aa u
yy + eea a)
whence w =
y ee
yyee
2. Again, if aey, be the
се
equation of the curve required, then in fluxions raer
a a
V V
fquaring each fide, u u =
2 I
rraae
—z inſtead ofu u, we get z✔:
T
uv, and
which put in :uu+vv:
V V
+vv in which
2
rraae
Veerryy:
ry
putting y y for its equal a ae **, we get z =
or ſubſtituting rraa e ²r
* u u
uu-
ee+rryy then from ra er
uurryy for vv we get z =
Lee
:frer:Xe
11
e
a u
I
uv, and n =
we get n=
u
I
น
rru
ru
in fluxions m=
WZ=
2 r
2
rre
rre
rer
}
wu
re-e
&c. for any
e
re:eeTrry y
Veerryy fo w =
curve.
429. If the ratio of the weights be given, and the nature of the curve
required. Let p fluxion of n and that of v, alio m
ofn as before, then n being, its fluxion m =
a v q
VV
fluxion
aup and
inuu+vv-zzo taking z conftant, we'll have (art. 410) uqt
AND MECHANIC.
231
vp
vp = o, or
=q, which two equations cleared of q, we get m
น
av vp
a up
арх u u
V V apzz Hence if the
•
U v V
u v v
V V
U V V
curve R A P (fig. 211) be preffed in each point by a force which is
as any power r, of its diſtance from RP, we ſhall have m — ze¹ —
whofe fluent (z being as before, conftant, and writing
a pzz
a
U V V
II
r = I
e
for
a,
to make the equation of like terms) is
ΙΤΙ
z a
V
the fluent of which will be the equation of the curve re-
quired, and may be found when r is known. Thus,
430. If the equation of the curve forming the ſtronge't arch poffible
be fought, here it is plain all the particles of the arch must be of the
fame weight, each weight tending towards the horizon perpendicularly
whence in this cafe ro, fo our equation becomes e vaz, which
fquared is e ev va azzaa uu aavv, fo vvx: e e a a
aa:au, but in this cafe, (art. 428)
aa uu, and y : ee
R = n
a u
vv:ee
敢
​=2, fo: eeaa:=R, for the length of any arch
V
aa:=0,
AB (fig. 211), and when A B=R=0, then :ee
or ae, therefore, if the abfciffa e, begin at the vertex A, we muſt
take ae inftead of e, and then Ree-a a: will become R
=√2aetee: the equation of the catenary (AE= e, E B : Y,
AB=R (fig_211) a = any conſtant quantity,) this equation in flux-
ions is
:aa+2a e tee: Xuu
auteu
૧ તે
= z ſquared is
Vizaetee:
2 ae + ee
<=z z = ➡uu+vy, take u u from each ſide then a au u
a u
/: aaetee:
2aeee
= vv, fo
v, then (art. 416) 2,3025 ax log. a + c +
√:2ae ee: but when eo, y=2,3025 a log. a, therefore, the
fluent corrected (art. 409) is 2,3025 a ×log, a+e+v:2aetee:
a te+ 1:22 +ee:
- 2,3025 a log. a=2,3025 a × log.
a
=Yx
232
THE UNIVERSAL MEASURER
in the fame manner, you'll find y=2,3025 a× log
R+: aa + RR.
a
by any of which 3 equation the catenary may be conſtructed. Laſtly,
if the point B be fo taken as L n BELn Br=45°, then rn=r Bj
(vu) and then R
a u
V
2.
Note. its plain that any flexible line (of little, or no weight) being
preffed by the wind or any other fluid uniformly in direction perpen-
dicular to R P, will put itſelf into this curve.
431 If the preffure be perpendicular to the curve, then draw O B
perpendicular to Bn, viz. to the curve in any point B, now if O B be
the force in direction O B, O E, will be the power of that force in
direction Q A, or C B, but by the ſimilar As rn B and E O B, as O B
:rn::m:
v m
Z
:
v v m
Z Z
V m
Z
power in direction B C and (art. 427) as z : v :
(becauſe as before m=
a
apzz: 2p, the fluxion
u
u v v
the preffure against the curve perpendicularly, in any point B, now the
quantity of the curve preffed being z, if as before this preffing force
be as any power r, of the diftance from R P, we'll have ze
=z,
a F, or =
pa
rti
—, fo zuer
u
=pa " + 1
I
u
>
Σ
where zas before being con-
ftant, we'll have the fluent
I
ze
I
va
where the index r
r+ I
being known, the fluent, or required equation of the curve will be
found, thus, if it be required to find the nature of a curve that a flexi-
ble line will form when filled with a fluid of any kind. Here by the
nature of fluids, the preffure of any point B, is as the height C B,
whencer 1, ſo our laſt equation becomes zeev a a, which fquared
is v va¹ = z ze¹=e¹×: uu+v v :, whence e eu ≈ v√: a *
4 — et :
whofe fluent is the equation required, but if the fluid be air, the al-
titudes C B, QA, &c. may be taken as equal (becauſe in fuch ſmall
heights of the atmosphere the preffure differs but little) and then ra
and ſo e z — a v, the equation of the circle (art. 421).
4
4
432. To find the velocity of a body defcending, or afcending in a
right line RC Q, (fig. 210) in a perfect fluid, where the refiftance is
as the fquare of the velocity. Let c velocity (uniformly generated,
in the time r) with which the body begin its motion at R or Q,, y=
that in the point C, T= time of deſcribing RC or QC, e = R C, or
==
AND
233
MECHANIC.
QC, put u, v, t, for the moments of e, y, T, and fuppofe the firſt ve-
locity c would be deſtroyed in the time n, by moving uniformly over a
diſtance d, then as n: d::r: c (for the ſpaces c, d, deſcribed with
the fame velocity, are as the times r, n, of defcription), whence,
r
n
cr
CC
again, as n c::r: Cr, or C (becauſe
n
c
d
=
=) that part of
now the
motion which would be uniformly deſtroyed in the time r;
reſiſtance at R, being to that at C as cc to y y, its plain the velocity
deſtroyed in the time r, by a force reſiſtance at C, will be CCX
=
yy meaſure of reſiſtance. Let b
YY=YY
yy
сс d
=
сс
व
velocity, generated in the
medium, or fluid, by the force of gravity in the time r, then b added
to, or taken from, according as the body aſcends, or deſcends,
d
gives bf, the whole force affecting the motion at C, but by
u
the laws of motion, =r, and fr = ±v, fo f = ± = (in this
У
vu
y v
u
caſe, becauſe y decreaſes while e increaſes) —, and therefore y
-
У
ru
+b=f=-, whence u =
dyv, and t=
yy+bd
У
d
- rdv
yy + ba
the fluent of the firft is (art. 416) e —— 2,3025 × 1 dxlog. yy +
bd: but when eo, then cy, and fo the correct fluent is e
2,3025 xdx log. y y bd: +2,3025 xdxlog.: cc
ra
d
2
bd:
cc + bd
yy+bd
:
is xD, where D =
and I, (radius = 1) fo the
2,3025 × d x log. :
yy+bd
cc+bd
:=2,3025 xlog.:
-rdr
if bda a, then the fluent of t =
>
yy-bd
C
difference of the two circular arches,
a
ล
time of the whole afcentis TX
ra
e
a
Note.
when t=
and
y
are the tangents of the arches to be uſed, but
2
rd v
yy+bd
a
or the body defcends, then T, the time of defcent
Gg
234
THE UNIVERSAL MEASURER
I a
is = 2x difference of the hyperbolic log. of
a+y and
at c
a-y
a
C
ra
26
20
fo, if c, or the body fall from reft, T
B M
Note. b = 321 X
B
X 2,3025 log. a+y
a y
where B to M, is as the fpecific gravity
of the body to that of the fluid, and 324 being the velocity acquired
by gravity in the firft fecond of time in vacuo.
433. If the body move directly forward, then bo, whence u
dy
V fo e=
I
2,3025 x d × log. y, which corrected (by
o ×
taking e = 。 and y = c) gives e = 2,3025 d x log., alfo t — —
y
2
-
r d y — ² v, ſo T =+
rd
y
= (when corrected) *_-d
rd
n C
I =
y
У
(becaufe ncrd).
434. Since any expreffion is got a maximum, or a minimum, when
its moment is made o, if therefore, the fluxion of any moment be
made o, fuch a moment (allowing the expreffion) will be had a
maximum or a minimum, if therefore, things be as in art. 410, 423,
423, 424, then u, uy, uyy, y✔uu+vv, √: 0 0 + vv:
.
y v v v Z are as the moments of the abfciffa, area, folidity, fur-
uu -f v v
Ve
>
face, length of the curve, refiftance, defcent, refpectively, fo if u +
b √ : u u + v v+ cuy+duyy+fy: uu+vv: +
uu Avv
hy v v v
uu + vv
:=m, a maximum, or a minimum, (u alone vari-
e
ub
able) we'll have 1 +
+cy+dyy+
fyu
✅: uu + vv :
✔: uu + vv:
2 hu y v v v
+
wherein the co-efficients a,
gu
✔: uue + v ve:
u u + v v
b, c, d, &c. may be any conftant quantities, pofitive or negative, as the
nature of the queſt. requires, any two or more of thefe terms madeo,
the equation of the curve &c. may be found, when the terms fo taken
are a maximum, or minimum; thus, if the length of the curve be a
ub
minimum or the area a maximum, then cy +
√:uu+vv:
b
= 0, fo,
yv: uut vv :
b
u = (putting r=-
— _—_-_) r u, fquaring cach
C
fide, yyx: uu+vv: rruu, forr
the equation of the circle, (art. 422).
C
yy: Xuu yyvv,
AND MECHANIC.
235
435. If the equation of the generating curve be required, when the
furface is a minimum or the folidity a maximum then dyy+
=0, whence
+
d
11 (putting
fy u
4/:uu+vv:
uf=dy:uu + v v :, or y ✔: uu+vv : =
f
d
-) ru, and fquaring each fide, y y x: uu+
vv=rruu, the fame as before, viz. the folid is a globe, if the
Jength and greatest diameter of a folid be given, to find the equation
of the generating curve, when the convex furface is a minimum, or the
folidity a maximum.
Note. The ordinate is always fuppofed given, becaufe when the
equation of the curve is determined, the abfciffa may be found to it,
fo here we have 1 +
=0, or putting aquaring
fy u
y: u u÷vV;
each file &c. we get uux: yy
a v
yyyaa:
vyv
-aaaav v, whence, u =
whofe fluent, (art. 415) is e equal 2,3025 a x lug.
ad
y+ YV (after correction) the equation of the catenary (art.
a
430) wherein y muſt always be greater than a, otherwife it will not
hold. If the curve of fwifteft deicent be required, the abfciffa is given,
gu
=0, or gu—— z√e, fo gguu≈zze, or
Z ve
we'll have 1 +
uua
(taking agg)
zze, whence z =
4
uva
24/2
the fame as in
art. 424, the equation of the cycloid (eQE, fig. 211) if the points
R, A (g. 211) or the abfcifià A Q, ordinate RQ, and length R BA,
gu
of the curve be given, then 1 +
+
: uue
vve:
my o
u-b
√:uu+vv:
હૈ
o, or (becaufe zz-uu+vv) we'll have 2e+gu+ube=o,
fo z=-=
u
u b = (writing a = g, and b—b)
a u
-+uu,
whofe fluent fhews in this cafe, the nature of the curve of fwifteft de-
fcent. Further if the velocity at any point B,
the diſtance C B (e), the laſt equation will be
be as any power n of
a u
z
= =
+ub=ae-
I
u+ub, which in fluents is R =
સદ
+eb, (bo and n)
I
n
2a√e, crea, the equation of the cycloid as before.
If the length and greatelt diameter of a folid be given, required the
equation of the generating curve, when the folid is that of lealt refif-
tance, here 1 +
- 2 by uvv v
Z Z Z Z
o, whence, (putting a 2 h)
236 THE UNIVERSAL MEASURER
ayurvv=z'; again, if inſtead of the greateft diameter, or length,
the folidity be given, then dyy +
2 hyu vvv
Z z z z
=0, whence y zª
auv³. Alfo, if the length, greateſt diameter, and folidity be given,
2 h y uv v v
then 1+dyy +
-
Z Z Z Z
2
given, then fra hy uv v r
fyu
+
Z
Z Z Z Z
putting a = 2h, it will be a vvv
=o, if the length and furface be
=0, whence fz³ = 2 h v v ▼ and
zzz, which being no way affected
with any of the variable quantities (e, y, R) will ſtill be the fame, and
fo the figure will be a conoid or the fruftum of one, and it will be the
fame when the length, or ordinate and area of the generating plane is
2 hyurry
given, for then cy+
=0, fo aurr F = Z Z Z Z.
Z Z Z Z
436. If a body, inſtead of afcending directly upwards (art. 432) &c.
be projected obliquely, it will defcribe a curve as ABC (fig. 209)
whofe equation may be thus found, draw BQ, perpendicular to the
horizon, or parallel to AH the axis of the curve, and P BLAH,
let A P = e, P B = y, arch A BR, then their fluxions, BM-Nь
— u, M N — Bb = v, BN = z. Alfo, let r = fecond fluxion of e
and s➡ third fluxion thereof, now if Q = velocity of the body at B
in direction PB, the decreaſe of velocity in the faid direction is caufed
by the refiftance alone (art. 95) and must therefore be — —
་
fluxion of, and therefore, as v : z::-q: -
motion in direction B N and as ▼ : u ::
tion B M, alfo fo is to Qu
qu
V
9, the
qz = decreaſe of
V
Z
= that in direc◄
velocity in direction B M, whofe
fluxion+q muſt be as the whole alteration of motion in that
Y
direction, both gravity and reſiſtance confidered, from which taking
qu, the part owing to the refiftance, there leaves for the effect of
gravity, whence as
V
Y
42:: force of gravity:
Q
force of reſiſtance, in direction B N, but in falling bodies, the velocities
are as the times, therefore the velocity generated by gravity in the time
(——, ſpace ÷ velocity) of defcribing Bb, with velocity Q, will be as
受
​AND MECHANIC,
137
रं
for another expreffion of the effect of gravity, ſo
Q==, but to clear this equation of, let 2Q q = −
으로
​VVS
ΓΙ
the fluxion thereof be divided by QQ=, the fluent, and then e
====
Z
qz becomes
whence
ZS
2rr
the weight, of the body, or gravity=1.
q
for the true force of refiftarce
437. If the refiftance be as the product of the fquare of the velocity
and (d) the density of the medium, then being abfolute velo-
¥
Z
=
=reſiſtance at that point, and therefore
eity at B, we'll have QQ z z
ofde-
TY
Z z
FF
= velocity deſtroyed thereby in the time (—)-
fcribing B N, which put — —
q z
V
found before, and you'll have d=
the effect of the fame refillance
S
S
2=227, by writing = for
Z
I
its equal, whence, 5=2rzd, or s s = 4 r r z zd d = 4 dd ×: um
+▼▼: Xrг, now (art. 419) we'll find e =
aly +
2ady 3
+
2
9
addy+
4 ad³ys + a³ dys
+
6
60
I 20
+
5 a³ d³ y° + 4 ad*y* &c.
when the body is in the defcending part of the curve, and if the figns
are made negative in every term, where the odd powers of dare found
the feries will then be for the afcending part of the curve. If there
be no refiſtance, or d=o, then eayy, foy y, the equation
2
of the common parabola, as in art. 359, perameter =—, but if it be
as pr: the denfity of the body: denfity of the medium, and if the
body be a globe, the diameter =D, then the above denſity d =
3
2 PD
.
(art. 327) and P X × =—=h, the perameter of the peras
P-I
2
bola, that would be defcribed in vacuo, by a body having the fame ve-
locity at the vertex A, whence a —
2p- 2
ph
is found, when y is any given ordinate.
= nearly, whence
?
238 THE UNIVERSAL MEASURER
438. In the two laſt, as alfo in fome foregoing articles (as is com-
mon in the folution of fuch problems) I have made fome one of the firle
Auctions, of fome one variable quantity (e, y, z.) conftant which both
avoids trouble, and ferves as a 'ftandard, to refer the variable fluctions
of the other quantities too.
439. Fluctions of all orders, are contemporanecus, or generated
with their reſpective velocities in one and the fame infinitely fmall part
of time (art. 402) alfo (art. 409) any variable quantity can have but
one fluxion, of the fame order but any flexion may have many fluents
out of which one mult be taken, that fuits the conditions of the prob-
lem.
440. Since,
ZS
2rr
is to 1, as the refitance is to the gravity (art. 437)
we may from the given equation of any curve, find the ratio of the
relittance to the gravity, when the body defcribes that curve; thus, if
the curve be a common parabola, and a cy y, in fluxions a u — 2 y v,
in fecond fluxions ar 2 v v, in third fluxions sao, whence
Z S =0, and therefore to defcribe fuch a curve, the body must meet with
2rr
no reſiſtance, but move in vacuo. Again, in a cubic parabola, where
ae— yyy, we have a u≈ 3 y y v, and ar➡6 y vv, allò, sa — 6 v v V,
which fubftituted in
for the refiftance
when the gravity is 1, again if RBA (fig.
a z
ZS
'
for r and s, gives
210
12 y v
210) be a quadrant of 2
circle, you'll find
2 S
2rr
2 Z
3 CB
= (per circle)
2 Q B
that is, as 3 CB
: 2 QB :: reſiſtance : gravity, and v the velocity, will be crery where
as № 3 E.
e-
441. In an upright veffel A B A B (fig. 163) kept conftantly full of
water &c. to find the quantity thereof diſcharged in any given time
thro' an orifice at the bottom B. Let the defcent of the water's
center of gravity, acquired by the fall of its furface from A to B,
AB, which will allo be as the defcent of the faid furface A A at B,
R equal the quantity of water in the veffcl. v, u and z, the refpe&ive
fluxions of V, e and R. Alfo, put A = arca of the water's fu face,
and a area of the orifice at B, then VR, will be e momentum of
the water in A B AB, whofe fluxion F = V z÷Rv, fuppofe at the
beginning of the evacuation the column of water A B A B, becomes
g Bmn Bg. Now the time of running out muit be inverfely as the
orifice (for the greater the one, the leis the other) and the defcents
of heavy bodies being as the fquares of the times of defcent, it will be
as AA :c::aa:
the defcent of the furface AA to qg, and
a ae
AA
&
AND MECHANIC.
239
therefore, e
fluxion u
a ae
AA
a au
A A
the difference of theſe two defcents, whofe
, multiplied by A, gives A X: u
a au
f, for
A A
which (it is
AB, there-
the fluxion of the momentum of the water in g Bmn Bq which
plain) muit bez, and becauſe A e =
z, and becauſe A e the water in AB A B,
fore Ae R, which values of z and R. fubftituted in FV z + R v,
- F, again, fince A e = water in
we get A V u +Aev
aa Vu
A
A B A B, its fluxion A u, multiplied by e, the defcent of the furface
F, the fluxion of the fame momentum, and there-
A, A, gives A ue
fore A Vu 4- A ev
¤ A Va + A ev
I
aa Vu
A
Aue or (putting n = 1
S
á a
}
A A
Aeu, or nVu+eveu, multiply each fide by
u+e¹v=e¹u, whofe fluent (art. 413)
en- and then, n Ven-
n+ [
€
e
a a
is Ve" =
fo V =
e
I
:, whec e
e
nt I
ni
A A
аа
a a
A A e
(2-
I
V
A
1
A A
2 A A
a a
= 1 +n becauſe`n — 1
*
442. If the water in the veſſel A B A B, defcends without being fup-
plied. or its furface A, A. defcends by its own gravity, it is plain
that the velocity acquired in the defcent thro' A q. is accelerated,
whill that in the defcent thro' Bm is for a ſmall time uniform, and
therefore deſcribes a double ſpace (theo. 167) in the fame time to what
it would defcribe if accclerated, fo that in this cafe, to find the differ-
ence of defccnts; mentioned in the lalt art. we mult take the defcent
thro Aq, from 2e, twice the defcent thro' Bm, in order to have
AA
2 4
3 201
the fluxion of the faid difference, which wrought as in
the laſt art, we'll find V =
A A e
3 AA
a a
443. If a body B (fig. 211) or a pendulum be ofcillating in a cy-
cloid R AP, and be retifted by an uniform force, which is to the force
of gravity as m to 1, and alſo by the fluid or mediu u in which it moves
where the refiftance is as the fquare of the body's velocity, 10 find the
țime of ofcillation Stc, let das in art. 432, or to the final arch Bm,
as the whole motion of the body at B to that loft in moving thro' Bm,
by the refiitance of the fluid, put e B = R, Bm = 2. A e = a, AR or
the pendulum's lengthb, hE=e≈ the diſtance the body mut
freely fall in vacuo by a force its ſpecific gravity in the fluid, to ac-
quire the velocity it has at B,then becauſe (art. 424) the deſcents from
240 THE UNIVERSAL MEASURER
any points R, e, B, &c. in a cycloid to A, the bottom is made in the
fame time, it will be as b is to 1 the force of gravity fo is a
a
R
b
R to
that part of gravity which accelerates the motion of the
body at B, and becauſe the uniform velocity acquired by falling thro'
any diſtance e is (theorem 167)=2e, therefore, as d: 1 :: 2e:
e
2 the reſiſtance cauſed by the fluid, which added to m, the other
d
part of reſiſtance, and that taken from
a - R
a R
"
gives
m
b
b
2, for the whole accelerative force of the body at B, now (theo.151)
time of defcribing B m, and (theo. 152) vft,
t ====, fo
S
Z
V
V2e
a R
that is:
2 e
Z
m
X
velocity generated in that time,
b
e
u
which, its plain, muſt be =√2e, the fluxion or increaſe of ✔✅ 2 e, the
velocity at B, hence, we have
a — R
ce
ប
m
or putting
b
7
amb, it will be — — +
R
2 e
++믈​=
=0, which (art.
419) folved gives e =
CR
a - 2 c
2 R R R
:X: RR-
V
+
2 b d
3 d
4 R4
8 R 5
3,4dd
3,4,5d3
&c. which feries fubftituted in
of e, will give the fluxion of the time, whofe fluent is found, T=
Z
=t, inſtead
√2e
3,1416 Vox: 1+
CC
2 CCC
&c.
the time of one ofcilla-
6dd
9 d d d
tion. When the body comes to I, the height of its afcent, then e=,
and from the above value of e we'll find R 2 C — 4 CC
16 c c c
+
3d
3 dd
=arch B A I, whence arch ID=22=2c+40-
16cc
&c.
3 d
9dd
(becauſe camb) 2 mb+
4CC
16cc c
800.
3d
9 dd
The End of the Second Part.
茶茶
​THE UNIVERSAL
MEASURER
**
AND
MECHANIC.
PART THIRD.
CONTAINING
1. Decimal arithmetic made easy, with new constructions; and the
extraction of roots.
2. Deſcription, conſtruction, and uſe of Coggleſbal's fliding-rule.
3. Multiplication of feet and inches, generally called cross multi-
plication.
4. Superficial meaſure of planes and of folids; with the methods of
taking dimenſions with, or without inches.
5. The feveral artificers works.
6. Menfuration of folids.
7. Surveying, plotting, and dividing of lands.
8. Gauging in all its parts. With the reduction of measures, def
cription of inftruments, new improvements on the fliding-rule,
gauging without inches, &c.
9. Questions and folutions; exercifing menſurations, furveying,
gauging and mechanics: as the mechanic powers, pendulums,
muſical ſtrings, barometers, pumps, mills, engines, gunnery, &c.
THE UNIVERSAL MEASURER
SECTION 1. Notation of Decimals.
EXPLANATIONS,
1.TF the denominator of any fraction be 1 and cyphers, fuch a frac
tion is called a decimal fraction, as for,, &c. are decimal
fractions. Now if for the unit I in the denominator we write a comma
or full point. and fuppofe every figure in the numerator to poſſeſs the
place of a cypher in the denominator, then the above fractions will
ftand thus 5,05 &c. whence decimal fractions may be expreffed
without their denominators, and fo wrote in one line like whole numbers,
being always careful of the comma and cyphers on the left hand of the
figures; for the comma is in the place of units, the next to it the place
oftens, the next in the place of hundreds, &c. i. e. =,5.or=0,5,
which is 5 tenths or no whole numbers 5 tenths, and x35,35 = 3
tenths 5 hundred parts, or 35 hundred parts of an unit, and 35
2035
no tenths, 3 hundreds, 5 thouſands, or 35 thoufand parts of
an unit, &c. Cyphers on the right hand fide of decimals are of no value
and may be rejected, except for illuftrations fake, 100=10000
i.
S
75
5
1. e. ‚5=‚50=,5000, &c. each of theſe being half of an unit; for if
an unit be divided into 10 parts 5 is half of them, if 100 parts
50 is alfo half of them.
=
2. If an integer or unit be divided into 10, 100, 1000, &c. = parts,
the figures in the decimal fhew how many of fuch parts are to be taken;
fo in ,25 the integer, whether a foot, yard, pound, &c. is divided into
100 parts, and,25 denotes 25 of them; alfo,2345 denotes,2345
parts of 10000. Now becauſe 10×10≈ 100, and 100 X 10 = 1000,
it follows that the 10's are 10 times greater than the 100's, the 100's
ten times greater than the 1000's &c. if the figures were all
≈; fo that
if ever fo many figures follow the, it is greater than them all, even if
they were fo large as to be all 8's or all 9's; i. e. unity is greater than
›99999, π is greater than,c99999, &c.
TỔ
:
3. Therefore,2345 may be taken for,2345897, or ,0028 for
,00287648 &c. for all the figures after 5, viz. 897 do not make
, nor all after 8, viz. 7648 &c. do not amount to,c001. But if
two or more decimals are to be wrought with, and you would be more
cxaćl, you may take the fum of all the figures after the breaking off ti-
gures, and add the number of points to any one of the breaking off fi-
gures; thus if,23456,007892, and ,913479602 were the decimals,
and you would work with three places in each, viz.,234,907, and, 013,
here the fuum of two of the following figures in each is 56+89+ 47
AND MECHANIC
}
3
192, and becauſe it is nearer 200 than 100, add 2 to fome of the
Breaking off figures 4, 7, 3, fuppofe to 3, and then the 3 contracted
decimals will be,234,007 and ,015 which in working will anſwer
nearly the fame end as if you work with the three first taken ones. For
the fame reaſons 3,1416 may be taken for 3,1415927&c. But if the
figure following the breaking off figure be under this increaſe of 1 may
be too much, as 3.78912 you may take 3,79 in a conimon way, but to
be more exact 3,789 may be taken, &c. for others as you'll find made
ufe of in the following pages, where 3 or 4 places of decimals are thought
fufficiently exact in common cafes, which method fpares a great deal
of labour.
1
In addition and multiplication thefe methods may anfwer, but'in fub-
traction and divifion it is plain; if one of the places be increaſed by uż
hity, &c. and not the other, the error muſt confequently be greater.
Reduction of Decimals.
To reduce a vulgar fraction to a decimal one of the fame value.
Rule 1. Annex or bring down cyphers to the numerator, and divide
it by the denominator till nothing remain, (if poffible) and the quotient
is the decimal required.
Ex. 4. 5. and 6. Reduce,, and, each to decimals.
2)1,00(,5=
ΙΟ
20) 1,00(,05=15
100
200) 1,000(,005=100
1000
7. If the numerator and denominator of any fraction be each multi-
plied by one and the fame number, the fraction will ftill retain its first
value, fo in , if the numerator r and denominator 200 be cach mul
베이어
​-
tiplied by 5, we fhall have to roos,005.
8. Hence to determine tlie value of any decimal, fay how often the
Whole denominator 200 in the numerator 1, it goes ro times, fo make¨
a comma in the quotient place; then fay how often the whole denomi-
nator 200 in the whole number r and one cypher, it goes no times again,
fo write o after the comma; again, how often 200 in 1 and two cy-
phers, it yet goes no times, fo write a fecond cypher in the quotientz
again try with 200 in 1 and three cyphers, and find it goes jult 5 times
fo fetting 5 in the quotient after the fecond o you find thedecimal,oog
I
To the given fraction. Obferve the fame method in all examples
of this kind, and you'll eaſily find the value of the decimal, which is
the chief thing in decimals; for by writing 5 for o, 5, or 0,5 for ,005,
the error in the end may be much greater than if you-uſed no decimals
or parts at all, fince decimals are only for working fuch parts as are lef
4
THE UNIVERSAL MEASURER
i
|
than a unit of the integer of the whole part. i. e. having taken the
decimal of theſe parts, you join it to the right hand fide of the whole
numbers, and ſo work with that mixt number inſtead of the given one,
as 7,5 for 7,5 being the decimal of 4.
Ex. 9. Reduce to a decimal.
TT
13)5,000(,384615,384615, &c.
5 the remainder.
Here I annex 3 cyphers and bring down 3 more, and then find that
5 remains, which 5 is to the firſt dividend, and therefore the figures
in the quotient will return again or circulate in the fame order as before;
fo therefore add as many figures to the quotient as you pleaſe with
continuing the divifion the more the nearer the truth you'll come, as
by ex. 2 and 3. tho' it never will be quite exact. Theſe are called re-
peating, circulating, or infinite decimals.
Ex. 10, 11, 12. Reduce and
3)1,00(,3333 +=}
9
1 &c.
and 34 each to decimals.
I I
T
3)2,00 5,6666+2
18 2,6667-
3)11,00 53.666 +2
23,667-
9
20
18
2 &c.
2 &c.
{ + } = +
See example 3.
}=}}
The fign + annexed to any number ſhews it to be too little, but
denotes it ſomething too much.
Ex. 13, and 14. Reduce
Reduce 5 inches to the decimal parts of a foot, and
alfo of a yard. Here 12 inches are one foot, and 36 one yard, ſo we
only have and to reduce.
T
5
12)5,000(,4167- feet.
8 the remainder
36)5,000(,1389- yards.
32 the remainder
Ex.*15, 16, 17. Reduce 2 or 2,5 inches to the decimal of a foot;
alfo 17s 6d, and 84d, each to the decimal of a pound ſterling.
24)5,000
40)35,00
480) 17,000
24 in. =,2083 f.
17s 6d=,875 £•
8d 35416£.
Otherwife.
12)2,5000
2in.,2083f.=
=12
AND MECHANIC.
5
the
1
Firſt. Becauſe 2 inches is 5 half inches, and 24 half inches — the
integer I foot, therefore divide 5 and cyphers by 24 or 2,5 and cyphers.
by 12. Secondly. 178 6d is = 35 fixpences and 40 fixpences,
integer I pound. Alfo 8d 17 half pence and 480 halfpence
pound. And thus may coin, weight, time meaſure, &c. be reduced. So
if the integer be a foot, the decimal of 14 inches will be,125 = 14
=11 = 4, of 9 inches ;75 = r = 4 of an inch
==,
75≈≈
of 3 inches =,25
=,0416 &c. &c.
T
Addition and Subtraction of Decimals.
Rule 2. Theſe are performed in all refpects like whole numbers,
regard being had to place units under units, (viz comma's under com-
ma's) tens under tens &c. both in the integers and decimals.
Ex. 18.
30057
.0002ܪ
,06
,083
,1489
Examples in Addition,
Ex. 19.
785623
,02
Ex. 20.
25,687
1032,02
Ex. 21.
7.5
27,07
,5
4,5
123,009
,079
2432,0689
6,0008
1,384623
3494,2759
163,5798
Examples in Subtraction.
Ex. 22.
Ex. 23.
Ex. 24.
Ex. 25.
From 3123
Take ,0102
25,02
270
I
3.98
,3021
21,04
,028.
269,972
98769
,01231
Multiplication of Decimals.
Rule 2. Having multiplied the factors together as if they were whole
numbers, prick off as many decimal places in the product towards the
right hand as there are in both factors, viz. in both the multiplier and
multiplicand. For if 10 and a 100, or 100 and 1000, &c. be the de-
nominators of two decimals, their products will be 10x100=1000,
and 100 × 1000=190000, i. e. there will be as many cyphers in the
product as there are in both factors.
Note. If there be not a fufficient number of figures in the product to
prick off, you muſt fupply that want by prefixing cyphers.
THE UNIVERSAL MEASURER
1
Ex. 29:
Ex, 26.
Ex. 27.
Ex. 29%
Mul
3,024
32,12
0347
,000627*
by 2,23
24.3
,0236,
1',0002
9072
9036
2082
1254
6048
12848
1041
627000
6048
6424
694
6,74352
780,516
,00081892
,006271254
30. When the factors together contain more places of decimals tham
you chuſe to have in the product, they may be contracted in the working
to any number of places, and the product true to that number as if the
factors had been multiplied at large. Thus, place the units place of the
multiplier under the number of decimals in the multiplicand that you
would have in the product, which done fet all the other figures in the
multiplier in an inverfe order; then begin to multiply each figure in the
multiplicand by the figure in the multiplier which ftands under it, fetting
it down under the units place in the multiplier, and the reſt as in com-
mon multiplication, the fum of all theſe products is that required. The
greateſt difficulty is to get the firft figure, begin therefore 2, 3, or 4 fi-
gures before that which you are to fet down, ſo you'll have the tens to
carry to it, which increaſe muſt be carefully obferved.
Ex. 31. 32. 33. Let it be required to multiply 3,141592 by
52,7438, and to have 4 places of decimals in the product. Alſo 257,356
by 76,48, and to have no decimals in the product. Alſo, 9,248264 by
0,725234 and to have 5 places of decimals in the product.
Ex, 33%
,248264
Ex. 31.
Ex. 32.
3,141562
257.356
8347,25
84,67
432527,
Ï570796
18015
17378
62832
1544
497
21991
103
134
1257
20
943
25
19682
,18004
165,6995
Since (by· Ex. 3′)` 3 or 4 places of decimals are fafficient, and this
way of multiplying anſwers that end with ſo much eafe, you'll find it
often ufed in the following pages, and to make it yet more plain. See
*-theſe examples-here at large.
AND MECHANIC.
Ex. 31.
Ex. 32.
Ex. 33.
3,141592
257,356
52.7438
76,48
,248264
725234
25132736
2059848
993056
9424776
10219424
1744792
12576368
1544136
496528
21991144
1801492
1241320
5283284
19682,58688
4961528
570796/0
17378148
65,6995|001296
,18004 9493776
34. To find a vulgar fraction nearly to a given decimal one. (See
prob. 173) But if the decimal expreffion be a circulate it may be done
thus, for every cypher in the firft circulate write a 9, ſo you have a
vulgar fraction the given decimal, which vulgar fraction by reduction
will turn out the fame decimal, fo in,189189189 &c. here the cir-
21 = 17
culate is 189 ſo 189189189 becomes 18x in lower terms
in the loweſt terms, alfo,666 &c. ==†, and, 123123 &c. ÷ ÷3
=-13, alſo,15666 &c. =,156 =,15 and ſo on, obſerving to om-
mit the one in the denominator when you write 9's for o's. A fraction
is brought into lower terms by dividing its denominator and numera-
for by any number that meaſures them both without a remainder.
TIT
23
35. To find the value of any decimal, is the converfe of finding the
decimal of fuch a value, therefore, in ,587 and in ,587333 &c. how
many pence and farthings fterling.
Ex. 36.
9587
12
Ex. 37.
2587333 &c. =
12
,587
I 2
₫ 7,044
add
₫ 7,044
₫ 7,047996
4
4
4=000 of 12
qrs 0,176
qrs. 0,191984
d 7,048
4
qr. 0,192
Anf. 7d. o qrs. 176587h.
Anſwer,587333 &c=7do qr.,191984, or nearer 7d 0 qr.‚192.
Ex, and 39. In,5234 of a flat foot how many fquare inches, and in
1251212 &c. of a yard in length how many feet and inches?
}
8
THE UNIVERSAL MEASURER:
1
5234
144
,1251212
༢
20936
3753036
20936
12
5234
4,5043032
,125 +4
3
2
,375 13
12
4.500 144
fq. in. 75,3696 An. o feet but 4.50436 in. nearer of 4,50413 in.
In thefe and many other examples of the fame kind, you muſt prick
off as many decimals in every product as you have in the multiplicand,
and if any figures remain towards the left hand they are whole numb-
ers of the fame name with the multiplier. In circulating decimals if
you put the vulgar fraction for its equal you'll have its juft value. (See
ex. 37 and 39,) wrought both ways.
Divifion of Decimals.
Rule 4. Perform divifion as in whole numbers; then prick off as ma-
ny decimals in the quotient: as that their number added to that of the
divifor may be equal to thofe in the dividend: fo if the decimal places
in the divifor and dividend be made equal by annexing cyphers, the
quotient will be whole numbers till you bring down more cyphers and
then you get decimals; ſo after divifion is ended if any thing remain,
you may bring down cyphers and continue the quotient at pleaſure. See
ex. 3.
Ex. 40.
24,3)780,516(32,12
Fox. 41.
,0236),00081892(0347
When the quotient figures are too few the want must be fupplied
by prefixing cyphers to the quotient. Divifion and multiplication are
the converfe of one another.
Ex. 42, 43, 44, 45. Divide unity by ,7854, by 282, by 231, and
by 2150,42.
,7854)1,000(1,2732+
5752
remainders
282)1,000(,003546+
130
2150,42)1,000000(,000,465+
rémainders
231)1,000(,004329+
67
5470
46. Any whole number is multiplied by a vulgar fraction when it is
multiplied by its numerator and divided by its denominator. So to
multiply any whole number by is but to divide it by 282. But
(by ex. 43) =,003546+, therefore multiplying by,003546 and
dividing by 282 are the fame, or multiplying by 282 and dividing by
003546 give the fame anfwer. And thus the 4 laftexamples being di
*H
AND MECHANIC.
A
viſors are turned to multipliers, or being multipliers are turned to di-
viſors, and ſo may any other by making 1 the dividend and the given,
factor the divifor. And becaufe multiplication is eaſier than divifion,
you may thus turn any conftant divifor into a conſtant multiplier or
factor.
18
47. Since, by the laſt rule, the number of places in the quotient muſt
be to the difference between that number in the divifor and dividend,
you may therefore know by obſerving the divifor and dividend what
places there muſt be in the quotient, and fo contract your divifion, which
will greatly leffen the work when there are many places in the divifor
and dividend:
48. Divide 70,23 by 7,9863..
Contracted.
7,9863)70.2300(8,7938.
628901
Ujjyo
55904
7492
7187
305
239
66.
63
3
The fame at large.
7,9863)70,2300(87,938
638904 |
1
63396
55904
५०
7491 i 90
7187 167
304 | 230.
239 | 589.
64 1,6414
6318904
017516.
For every figure you write in the quotient leave out one in the divi-
for towards the right hand; having due regard to the increaſe that
would arise from that figure fo left out, and then divide on as ufual.
+
:
}
3
49. The fquare root of any number is fuch a number, as being mul-
tiplied by its felf may produce the firft taken number, which is called-
the fquare of that root; fo 2 is the fquare root of 4, 3. of 9, 5 of 25,
and
4 is the fquare of 2, 9 of 3, 25 of 5, &c. In the fame manner the.
cube of any number is known by multiplying the number twice into its.
felf; ſo 8 is the cube of 2, for 2 ×2×28, alfo 64 is the cube of 4.
8000 is the cube of 20, &c. and 2 is the cube root of 8, 4 of 64, 20 of:
8000, &c. by which method the following table is calculated,
827 64 125 216 343 512 729 1000
Cubes
Squares
Roots
4
9|16||25| 36| 49
21 31 41 51 61
51 61 71
71
64
81
100
8
9
ra
B
1
1
19 THE UNIVERSAL MEASURER
be-
50. When any number as 2578,3692 is given to be extracted, you
muft firſt divide it into periods by points made over every fecond figure
for the fquare root, and over every third figure for the cube root,
ginning to point at the units place whether whole numbers or decimals,
fo the above number pointed for the fquare root ſtands thus 2578,3692,
and for the cube root thus, 2578,3692, this ſhews what the firſt figure
in the root will be, by comparing the firſt period with the above table,
and alfo how many figures will be in the root, there being a figure for
every full point except one, fo the fquare root will confift of 4 places,
and the root of the firſt period 25 is 5; but the cube root will have only´
3 places, and the nearest root in the table to the first period 2 is 1, for
the fquare of 10 being 100, the fquare of 100 being 10000, the
cube of 10 being = 1000, the cube of 100 being = 1000000, &c. it
follows that two figures cannot have above one figure in their ſquare
root, nor three figures more than one figure in the cube root.
=
51. Hence, alfo, if you arc to take the fquare root of any decimal
it must be made to confift of an even number of places, otherwife you
cannot have the true root of the denominator. Alfo for the cube root
of a decimal, it muſt be made to confift of 3, 6, 9, &c. places, by ang
pexing a cypher or two; for if you were to take the cube root of a,8
and call it 0,2 (becauſe by the table 2 is the cube root of 8) then
P,2x0,2X,02 is (by rule 3d),008, which (by explanation 49) fhould
be =
0,8, fo 0,2 is not the cube root of 0,8, but the cube root of
0,800 will be the cube root of 0,8.
To extract the Square root.
Having pointed the given number as before directed into periods of
two figures each, take the neareft lefs fquare root of the first period to-
wards the left hand (which is foon had by the table) and fet it in the
quotient, and take its fquare from the faid firft period for a dividend,
double the quotient for a divifor, then fee how often it may be had in
that dividend, fo as the figure thence arifing being annexed to the divifor
and multiplied by the faid figure, the product may be, or the nearest
Jefs to the dividend that can be; fubtract this product from the divi-
dend, and to the remainder bring down the next period for a new divi-
dend, double the whole quotient for a divifor and proceed as before.
Thus go on until all the periods are brought down, if nothing remain
you have the true root, but if there be a remainder bring down cyphers
two ata time, and fo carry the root into decimals at pleasure.
AND MECHANIC
tt
Ex. 52. What is the fquare root of 321489?
321489(567 the root.
25
1
106) 714
636
1127) 7889
7889
714, and
Here the neareſt leſs root to 32 the first period is 5 its ſquare 25 fs
there remains 7, to which bring down 14 and the dividend is
5 doubled is 10, fay how often 10 in 71 (leaving out the 4 becauſe the
figure arifing is to be annexed to 10) in goes 6 times, fo fet 6 after 10
and it will be 106 the first divifior, which multiplied by 6 gives 636 and
taken from 714 leaves 78, to which bring the next period 89, ſo 7889
will be a fecond dividend, the whole quotient 56 doubled is 1 1 2, which
feems to go times in 788, fo 1127 is the fecond divifor, which multi-
plied by 7 gives 7889 to the dividend; fo 567 is the exact fquare root
of 321489, as will appear by multiplying 567 by its felf, which is the
way to prove the fquare root, and when there is a remainder it muſt be
added.
Ex. 53, 54, 55. Required the fquare root of 32,1489, of 321,489.
and of ,0321489.
32,1489(5,67
321,4990(17,930i
03214890(0,1793
25.
root
I
root
I
root
106) 714
636
27) 221
189
27) 221
189
1127) 7889
7889
349) 3248
3141
349) 3248
3141
3583) 13740
3583) 10790
10749
J0749
358601) 41000o
358601
remains
remains 051399
1
12
THE UNIVERSAL MEASURER
2
root
161) 206
161
Ex. 56, 57, What is the fquare root of and of 11 ?
÷ =,666666(,8164
64
1= 1,50(1,2247
I
22) 50
root
44
1626)
10566
9756
242) 600
484
16324) 81066
65296
2444) 11600
9776
remains 15770
24487) 102400
171409
remains 10991
Extraction of the cube root.
58. Having pointed the given number into periods, as by ex. 50.
take the neareſt leſs root to the first period and write it in the quotient,
take its firſt cube from the faid first period, and to the remainder bring
down the firſt figure of the next period for a dividend, ſquare the quo-
tient or firſt root and treble it for a divifor, perform diviſion and ſet this
quotient figure after the laft. Again, cube this whole quotient and take
it from the two firſt periods, bring down to the remainder the firſt fi-
gure of the third period for a ſecond dividend, and multiply the ſquare
of the quotient by 3 for a divifor, perform divifion, and place this quo-
tint figure after the laſt; thus go on until all the periods have been
ſubtract d from, and if at laſt any thing remain, bring down cyphers
and carry the root into decimals as far as you pleaſe. It is to be ob-
ſerved, that after you have got 2 or 3 figures in the quotient, you may
bring down a whole period at a time, and ſo get three places of figures
more by divifion, as in ex. 60, &c.
Ex. 59. Required the cube root of 1728.
1728(12 root
I
3) 7
7
6
t
Here the neareſt leſs root of the first period 1 is 1 whofe cube is
1, which taken from the firſt period I leaves ó, to which bring down
7 the firſt figure in the next period and you'll have 7 for the dividend,
AND MECHANIC.
13
}
and 3 times the ſquare of 1 = 3 for the divifor which goes twice in 7,
fo fet 2 after I in the quotient and it makes 12 whofe cube is 1728,
which taken from 1728 the two firſt periods (in this cafe all the given
number) leaves o; fo 1728 is the exact cube of 12.
Ex. 60. What is the cube root of ,067507824239 ?
,067507824239(,407178 the root required
,064 the cube of 4
3 times of 448)
35
,067507
two first periods
,064000
cube of,40
3 timesof,40,4800) 35078
3
times
33600
1478
,0675078243 first periods
of,407,067419143 cube of,407
,496947),000886812
496947
3898653
3478629
4200249 &c.
Ex. 61, 62. Required the cube roots of a gallon of ale and one of
wine; viz. 282 and of 231.
282)6,558 = root
216 = cube of 6
6x6x3=108) 00,0
540
282,000 two firſt periods
274,625 = cube of 6,5
6,5X6,5X3=126,75) 73750
63375
103750
101400
14
THE UNIVERSAL MEASURER
231(6,136 root
=
216
=cube of 6
6x6x3=108) 150
108
231,000
two first periods
226,981
=
cube of 6,1
6,1 × 6,1 × 3 =111,63) 40190
33489
67010
66978
In Ex. 60. becauſe the cube of,407 is ſo near
the given number
I take in 4 of the laſt figures in the root by divifion, in each of the laſt
examples, two are taken in by divifion, but if this method in any caſe
ſeem not to hold true, you may proceed according to the rule, viz.
by cubing the quote for every new figure annexed &c. this method of
extracting the cube root is from the converging feries (prob. 171. art
84.) by which method any root may be taken, but other roots being
of little uſe, and are much eaſier done by (prob. 171) are here omitted.
Note. If at any time your cubed quotient be too much, as it would
be in ex. 61. if you took 9,6 (viz. the number arifing by dividing 6,60
by 108) inſtead of 0,5 you muſt take a unit leſs in the laſt figure &c.
as is done there.
cube
Take theſe examples for exercife.
its root
fquare
its root
2150,42
12,9081] [
71
2,6457513 }
14 = 1,333+ i
1,1006
,000762 1
0,0275+1
7121,102
19,238
4,00006712
2,000016+
7612,812
19,67+
300
17,3205+
192
13,8564 +
T
900
SECTION II. Deſcription, conſtruction, and uſe of Cogglefbal's,
or the common fliding-rule,
63. On this rule are two lines, one on each ſide of the ſlider, one of
them is exactly the fame with that on the flider, numbered 1,2, 3, 4,
5, 6, 7, 8, 9, to the middle where they again begin at 1 and ends at 9;
thefe are commonly called Gunter's lines being the fame with the line
AND MECHANIC.
15
of numbers on Gunter's fcale. The other line on rules fometimes
marked with D at one end, and girt line being fet at 12 being the gauge
point for timber. (Alſo, on ſome of theſe rules are A. G. the gauge point
for ale gallons, and W. G, for wine ;) this line is of a double radius to
the other two, the diſtance here between 5 and 6 being double that be-
tween 5 and 6 on the other two lines, &c. in any other like numbers;
this line begins commonly about 4 and ends about 40, numbered 4, 5,
6, 7, 8, 0, 10, 20, 30, 40. D.
Ι
64. Thefe lines are numbered from the left hand towards the right
according to the order of figures; thus, if the firft 1 (beginning at the
left hand) be 1, they read 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40,
50, 60, 70, 80, 90, if the firft i denote 10, they read 10, 20, 30,
40, &c. to 900, if the first be 100, then they read 100, 200, &c. to
9000, if the firſt I be 0,1, they read 0,1. 0,2,0,3.0, 4. 0,5. 0,6.0.7.
0,8. 0,9, 1, 2, 3, 4, 5, 6, 7, 8, 9, &c. thefe you are carefully to
ob ferve, or you cannot know the value of the anſwers.
65. The ſpace between every figure is divided into 10 parts called
the large divifions, and each of thefe into 10 other parts called fmall di-
viſions, which are hundred parts the other being tenths; to find any
number on theſe lines fuppofe 346, for the 300 take the figure 3 and
for 40 take 4 off the large divifions, and for 6 take 6 off the ſmall ones,
ſo you'll have the point where 346 ftands, and at the fame point will
alfo ftand,346, or 3,46, or 34,6, or 346000, or,00346 by taking the
figure 3 to be 2,3, or 3, or 30, or 300, or 300000, or,003, &c. Alfo,
if you would find 4567891 on the lines, for 4 take the figure 4 (always
take the figure on the flider the fame with the first figure in the given num-
ber) for 5 take 5 off the large divifions, for 6 take 6 off the fmall divifions,
then becauſe there are no more divifions, you muſt compute for the o-
ther figures as near as you can, fo 7891 being to compute, take ſome-
what more than three fourths of the next fmall diviſion (becauſe 75 viz.
0,75 is = but 0,78 is a little more) that is the required point where
4567891 muſt be is at 63+ fmall divifions paſt 5, large ones paſt the
figure 4, which is alfo the place of 4,567891,004567891 or 45,67891,
according as you take figures 4 to be 4 or ,004 or 40, &c. But when
there are not fo many ſmall diviſions as are mentioned, which is the eaſe
in all common fliding-rules, you muſt take what diviſions there are and
compute for the reft as above directed: fo if between any two figures 8
and 9 on theſe rules there are but 5 large divifions, and you would find
875, for 8 take the figure 8, then for 7 you muſt ſuppoſe them fubdi-
vided fo 7 will be 31 of them and 3 will be but parts of one of
16 THE UNIVERSAL MEASURER
thefe divifions, which is fo fmall that you can only take good meaſure
of 31 divisions pait the figure 8 for 875, or even for 87 and any number
of figures to follow it. This method of numbering the rule called no-
tation ſhould be well unde itood, for it is the greatest difficulty you'lk
meet with in the uſe of this or any other rule. This method of nota-
tion will ferve for any rule or ſcale.
=
66. The conftruction of this ſcale is fuch that if any 4 numbers be
directly proportional, the diſtance on Gunter's lines between the firſt
and fecond terms is to the distance between the third and fourth terms,
i. e. the distance between the firt and fecond terms being doubled or
added to itfelf gives the distance between the firit and fourth terms
which plainly thews this rule to be a logarithmetic ſcale of the numbers
2, 3, 4, 5, 6, 7, 8, 9. and is thus made. From a nice diagonal fcale
take the numbers 301, 471, 602, &c. (riz.) the logs. of 2, 3, 4, &c.
and fet them in a right tine one after another, fo you have the true
distances on the rule between the figures 1. 2, 3. 4. &c. for the large
divifions take the logs. of 1.1 1,2 1,3 &c. and 2,1 2,2 2,3 &c. from
the fame diagonal fcale and fet them after the lines 1, 2, 3. &c. on the
rule, &c. for the ſmall divifions, by which means Gunter's lines are foon
conftrufted, the line D or gauge point line is made the fame way only
its divifions are double to the reſpective divifions on the other lines.
Note, For diftinction I call that line which begins 1 on the left hand
of the rule the former line, and the following line beginning at near
the mid rule I call the latter line, both of which are conftructed alike.
67. The uſe of the ſliding-rule muft be eaſy from the conſtruction
thereof; becauſe in any 4 proportionals, the diſtance taken with a pair
of compaffles between the firſt and ſecond terms, will the fame way ex-
tend from the third to the 4th term. Now in a fliding-rule the compaf-
fes are fupplied by the flider; whence this
Rule 5. Set the firſt term on the flider to the fecond term on the
rule, then against the third term on the flider ſtands the fourth term on
the rule; or, fet the firft on the rule to the fecond on the flider, then
againſt the third on the rule ftands the fourth on the flider; ſo that the
firſt and third terms must fill be found on one line and the ſecond and
fourth terms on the other: or you may ufe the ſecond term for the
third and the third for the fecond. This rule anfwers for multiplication
and divifion. For, as I is to one of the factors (in multiplication) fo is
the other factor to the product. Alſo in divifion, as the divifor is to the
dividend fo is I to the quotient.
+
AND MECHANIC.
17
I
68. By ex. 64, if the first 1 be 1 the laſt 9 will be 90, ifit be 10 the
Laſt 9 will be goo, i. e. the last 9 is always 10 times the firit g`or ço¨
times the firſt 1. Hence in 4 proportionals when the ſecond and third
terms are each more than 80 or 90 times the firft term: or contrary,
fuch a proportion will fall off the rule and flider; in fuch cafes it is
belt to fuppofe the fmall term to be multiplied by 10 or 100 or 1000,
&c. or the great one to be divided by 10 or 100, &c. till they will come
on the rule and lider, and then multiply your answer by the fame
number that the first term was multiplied by, or that either the fecond
or third terms were divided by; but if the first term be divided, or ei-
ther of the other two multiplied. then the anfwer or fourth term muit
be divided by the fame number to give the true anſwer. Now to mul-
tiply by 10, 100, &c. is but to annex a cypher or two to a whole num-
ber of in a mixt number to remove the decimal point 1 or 2 places
ncarer the right hand, and divifion is just the reverfe of this, fo that this
may easily be performed in your mind. And here it is to be obſerved
that the line Dis of a double radius, the numbers on the flider will be
fquares to the numbers on it: ſo that if you divide by ro on the line D
it will answer to 100 on the flder, &c. for by ex. 65, the first and fe-
cond terms being taken. the value of the other figures on the lines are
thereby known, and by obferving the faid example it must be easy to
bring any numbers on the lines by removing the decimal point &c.
M
69. If I coft 7 s and 6d, what will cot at that rate? Here as 1 on
the flider: 7,5 (for 5d is 0,5 of a thilling) on the rule :: 3 on the flider
: 22,5s on the rule, the answer. Bat to have the ufe of the fliding
rule, fhorter, cr in fewer words. let A and B denote Gunter's lines on
the rule and ſlider, and D the girt, or gauge point, then the last propor-
tion will ſtand thus, as I on A: 7,5 on B:: 3 on A : 22,5 on B the
andwer.
70. If I coft 7.5 what will 300 col at that rate? As I on A : 300
on B:: 7.5 on A : 2250 on B the antwer.
or,
71. If I cuft 300 what will 750 coft Or, multiply 750 by 300.
(By ex. 68) as I on A: 750 on B:: 3,00 on A to 2250 cn B; 07, 23
I on A: 300 on B::7,50 on A to 2250 on B; or, as 100 on A: 300
on B:: 750 on Å tɔ 2250 0ũ B. which multiplied by 100 (becauſe the
fecond or third term was divided by 100, or the firft term multiplied
by 100) gives 225000 for the anſwer.
72. When the proportion is increaſing (as in the 3 laſt examples)
find the two first terms on the former lines; but if it is decreating, find
them on the latter lines, as in examples 73 and 74.
C
18
THE UNIVERSAL MEASURER
I
73. If 221 coſt 3, what will coft? That is, divide 22,5 by 3. As
3 on A: 22,5 on B:: 1 on A to 7,5 on B the answer.
74. Divide 2250 by 300. As 300 on A :: ron B':: 2250 on A
7.5 on B. the anſwer.
ર
75. Divide 225000 by 750. By example 68, it will be as 7, 50 or
A: 225000 on B :: 1 on A: 30000 on B, or as 75,0on· A` :· 225000
on B:: ro on A: 30000 on B; or, as 750 on A: 215000 on B::
160 on A to 30000 on B, which divided by roo (becauſe the first term
was fo divided, or the firft term divided by ro, and the fecond or third
multiplied by ro, 10 times to being 100, or the fecond or third
term multiphed by roo) gives 300 for the answer.
76. If the fquare of 3 require 7, what will the fquare of 60 require?
As 3 on D (viz, the figure 30 muſt be taken becauſe 3 is not on that
line, for it begins with 4) 7 on the flider :: 6 on D (this 6 by thë
queftion fhould be 60, but it is only 9,6 becauſe 30 is taken for
fa
this line D in this cafe is divided by roo) to 0,28 on the filder, which
multiplied by the fquare of 100, viz. by roooo (fee ex. 68) gives
2800 for the anſwer.
3.
As I on the flider: 10
1,23 on D, which multi-
77. What is the fquare root of 15129?
D:: 1,5129 ( fee ex. 68) on the flider :
plied by 100 (becaufe 15129 on A was divided by the fquare of 100)
gives 123 for the fquare root of 15129.
78. Between any two given numbers fuppofe 4 and 9, to find a mean
geometrical proportional. If you multiply the two numbers together,
the fquare root of that product will be the mean proportional fought.
Or, by the ſliding-rule, as the lefter number on A: the faid leffer
number 4 on D: : the greater number 9 on A:he mean proportional
C; or, as 9 on A : 9 on D :: 4 oh A to 6 on D as before
on
79. To find the first of two raean proportionals between any two
given numbers; you must take the cube root of their product which
will be the proportional fought: this cannot be done on this fliding-
rule for want of the line E, which fee in fection 8.
I
88. On the edge of this fliding-rule is a line divided into 100 equal
parts ending at 12 inches which is a foot decimally divided, and ferves
to find the decimals of inches when the integer is 1 foot; thus, againſt
6, 9, 1½ &c. on the inches ftands,50,,75,12 &c. on the faid deci-
mal line, i. e. 0,5 for the decimal of 6 inches, 0,75 for that of 9 inches,
0,121 or 0,125 for that of 14 inchies, &c.
81. On this fliding-rule, is a table marked at top D. L. S; D. viz,
pence, pounds, fhillings, pence; this table fhews the value of 50 feet
timber called a load, at all rates per foot from 6 to 24 pence, ſo
AND MECHANIC.
you would know what 50 feet comes to at 10d, or 10ld, red, or road,
per foot, here under the firſt D, and againft 10, and the firft or fecond
or third line below 10 you'll find 2l is 8d the price at rod a foot, or
21 28-8d the price at 10d a foot, or al 35 9d the price at 1od per
foot, &c. See Ex. 91.
4
82. Laftly, It is plain from what goes before that if the fecond or
third terms fall on the line D the fquare root of the firft term muit
alſo be on D. Alſo, if any of the terms be a product of two numbers,
a mean proportional between ſuch two numbers malt be found, and
tifed on D, inſtead of ſuch a product.
SECTION III. Multiplication of feet and inches.
There are feveral ways to work multiplication of feet and inches,
but the beſt is as follows: where the 12th part of a foot, whether lin-
eal, fuperficial, or falid, is called an inch, the 12th part of an inch is
called one part, and fo on, as in this table.
12 inches
12 parts
2 feconds
2 thirds
#2 fourths
make one
f foot
inch
part
fecond
third
confequently
(f. xf. give feet
f. xin. give inches
in.in. give parts
in. x parts give fec.
parts parts, thirds J
divided by
12.gives
r
feet
inches
parts
Lecond
That is, feet into feet give feet, feet into inches give inches, and di
vided by 12 give feet and inches remain; inches into inches give parts
which divided by 12 gives inches and parts remain, &c.
84.
85.
F. 1. P. F. I. P.
4 : 6
2: 8:3
83.
F. I. P.
8:9
3:6
24: 10
3
8:
9.
26:3
198 :
୪
09:0
4
4: 6
18:
7:6
:
36
0: 0:10
30:7: 6 217
10.
aote.
an inch parts
an inch parts
J.
2: 3:0:0:0
9 : 0:10 : 1 : 6 the anfwer
In Ex. 84. I firft multiply 10 inches by 8 feet, faying 8 times rois
80 inches, which divided by 12 quotes 6 feet and 8 inches remains,
fet 8 under inches, and carry 6 to the feet, faying 8 times 4 is 32 and
Ở is 38, fet 8 down and carry 3, faying 8 times 2 is 16 and 3 is 19,
which is fet down under feet, again for the fecond row, 9 inches into
To inches gives 90 parts or 7 inches and 6 parts, fet down 6 under parts
then multiply the 24 by 9, and to the product 216 inches add the t
and the fum is 223 inches or 18 feet 7 inches as you fee fet down in
the work, add thefe two rows together pricking at 12's under inches,
20
THE UNIVERSAL MEASURER
parts, &c. and you'll have 217 feet 3 inches and 6 parts for the Anf.
Ex. 83. is wrought the fame way. In fx. 85, begin with 6 inches
by two feet and fo get the firit row, then begin with 6 parts and 8
inches and ſo get the fecond row, laftly, with the 3 parts get the
third row, always pricking at 12's except in feet, which prick at 10's.
N
F. I. P. S. T.
5
6
2
8 4 3
༣ ༣ 7 8
25
010 4
2 I 0 10
O
XW N a
4
4.10
5
6
O
6 10 3
2 6
8 0
I 4
8 6 11 0
But if there are more figures in
the firſt than one, it will be eafier
to work by aliquot parts, thus,
multiply all the feet, inches, &e.
in the multiplicand, by the feet
in the multiplier as before, and
take fuch aliquot parts of the
27 7 3 5 1 8 8 11 o multiplicand as fut the inches,
parts, &c. in the multiplier, which added together gives the anſwer.
Here foliows a table of aliquot parts. Take ex. 83 and 84 by this me-
thod.
11 inch
1
one twelfth
one fixth
21
31
41
5
6 inches {
is
71
8 1
91
ΙΟ
II]
I
3
9
one fourth
one third
one third and one twelfth
one half
one third and one fourth
one third and one third
one fecond and one fourth
one half and one third
one third 1 third and 1 fourth
I
of a foot.
24
ΙΟ
6
8
9
20.
UJ
3
198
8
4
4
6 of 8 ft. 9 in.
12
5 = 1 of 24
6 2 6
feet to inches.
uf 24 ft. 10 in.
217
3 6 the product
30 7 6 product
In ex. 83, I first multiply 8 feet 9 inches by 3 feet and fo get the
firft row 26 3, then 6 inches being half a foot I take half of 8 ft. 9 n.
which is 4 f. 4 1.6 p. In ex. 84 multiply 24f. 1si. by 8 f. and then
91. being and take thoſe parts of 24 f. 10 i. viz.
f. 10 i. viz. 1 and 2. But if
the feet in the maluplier are above 12, (as in ex. 87 they are 17 ft.)
it will be eaſier first to multiply the feet together not regarding the inch-
es, then take fuch aliquot parts of the feet in the muiuplier as fuits the
inches, parts, &c. in the multiplier as before.
AND MECHANIC.
21
Ex. 87.
Ex. 88.
75 9
97
8
17 7
8
9
525
781
75
8 61 of 17 ft.
4 3
4 of 17 ft.
25 3
of 75 ft. 9 in.
4 = 57 ft. 8 in. x 8 ft.
of 97 ft. 8 in.
48 10
24 54 of 97 ft. 8 in.
854 7 product
18 II
3:
3 = 1 7 3 h. c
1331 11
3 me product
Ex. 89.
Ex. 90.
40 10 3
30 6
4
jo f.
to
159 5 8=37f.71.5px. 4f. 1200 = 4u -
37 7 5
4 8 8 6
12 6 5
12 6 5 8
1 6 9 8 6=ditto
of ditto
65.
8337f. 7 i. 5 P.
25 = 30 ft. x 10 in. = 300 ie.
0 7 6=30ft, × 3 p. = 9º p.
I 6 = 1 of 40 10 3
20 5
177 1 5 0 6 prvüuct
I 8
5
1
6 = do × 6 parts
1247 9 0 7
6 product
Thus multiplication of feet and inches, like practice, may be
wrought feveral ways.
91. There is yet another article belonging to the fliding rule, viz.
when one or more of the terms is a fum or difference: thus, if 10
10 give
20, what will the fum of 30, 40, and 50 give? or, (which is the fame)
what will 120 give? Set 10 on A to 20 on B, then againft 30, 40,
and 50 on A ftands 60, 80, and 100 refpectively, which added toge-
ther gives 240 for the answer. Alſo, if 9 gives 20, what will the fum
of the ſquares of 30 and 40 give? Here, becauſe the ſquare of the fe-
cond or third term is to be uſed it must be found on D, ſo by rule 5,
the first term 9 (viz. 3 its fquare root by ex. 68,) muit alſo be on D,
i. e. the 30 on D mult be 3, fo 40 will be 4, and then as 3 on D: 20 00
A :: 3, 4 on D: 20 and 35.5 +on A reſpectively, whofe fum is 55,5+
which multiplied by 100 (becaufe D is divided by the ſquare root of
100) gives 5550+ for the anfwer. Alio, if the difference of theſe
fquares were to be uſed it would be the fame on the rule, only taking
20 from 35,5,+leaves 15.5+, which multiplied by 100 gives 1550 +
for the anfwer. This is evident, if you confider, that the fum of all
the parts is equal to the whole.
22
THE UNIVERSAL MEASURER
XXXXXXXXXXXXXXXXXXXXXXXXX
SECTION IV. Superficial Meaſure.
92. The area and dimenfions are commonly kept in one name; that
is, if the area be required
in
inches
feet
yards
the dimentions are taken in
inches
feet and inches
yards feet and in.
then thoſe odd parts being reduced to decimal parts of the integer, the
area is had by working as the nature of the figure requires, or without
decimals, the area may be had by multiplication of feet and inches,
but eaſieſt of all by taking the dimenfions by decimally divided rules,
and then the odd parts are ready reduced to hand, and is alfo truer
than to take the dimenfions in feet and inches &c. and to reduce the
inches to decimal parts of a foot, for by ſuch reduction often fome-
thing remains, befides the inches are divided into more parts decimally,
than among themſelves, which is only into 8th's or half-quarters, fo
that a foot contains 12 times 8 or 96 parts, but decimally divided it
contains 100 parts, hence, when the area is required in feet, take the
dimenfions in feet and for the odds look not at the inches, but at the
line on the edge of the fliding-rule decimally divided into a 100 parts
in the length of a foot or 12 inches, fo may you have your dimenfions
in feet and decimal parts of feet true, there are alſo yards decimally di-
vided for the fame end, and for the like reaſon, Gunter's chain is divid-
ed into 100 links, &c. in moft cafes the content is required in inches,
feet, or yards, and being in any one of thefe it may be reduced to any
other of them, by theſe three tables.
Superficial Meaſure.
Long Meaſure.
1. F. Y.
I. F. Y.
12 = 1
144 1 મેં
1296 = 9 = I
36 = 3 = I
Solid Meaſure.
J. F. Y.
1728
45656 # 27 — I
To reduce meafures from one denomination to another, obferve this
general
Rule 6. Lefs numbers are brought into greater by multiplication,.
and greater into lefs by diviſion.
În
Ex. 93, 94. In 54,5 yards in length how many feet and inches? And
54,5 ſquare or ſuperficial yards how many feet and inches.
54,5
3
163,5 feet in length
12
་
1962,0 inches in length
·54,5
9
490,5 iquare feet
144
70632,0 ¡quare inches
AND MECHANIC.
23
Ex. 95, 96. In 0,52 parts of a folid foot how many folid inches?
Allo, in 1730,63 fquare inches how many fquare feet?
$728
0,52
144)1730,63) 12,014+, fquare feet
47 the remainder
898,56 loud inches
To meaſure a ſquare.
Rule 7. Multiply any fide of the fquare by itfelf, and the product
it its area. By theorem 22.
Ex. 97. If each ſide A B B D of a fquare (fig. 17) be 7 feet 6
inches what its area.
By feet and inches.
7 6
76
56 3 the anſwer
Decimally.
7:5
7,5
56,25 the anfwer
By the fliding-rule. As I on A: 7,5 on B:: 7,5 on A: 564 on B,
To measure a rectangled parallelogram.
Rule 8. Multiply the length by the breadth for the area. Theo-
Xem 22.
By feet and inches.
IQ
2
Decimally.
9 2
}
{ breadth
length
}
{
1 0,75
2
21,50
21.6.
anfwer
Sliding-Rule. As I on A: 10,75 on B:: 2 on A: 21,5 on B.
To measure an oblique angled parallelogram, as a rhombus or rhom
boides.
Rule 9. Multiply the perpendicular let fall from one of the obtafe,
angles upon the oppofite fide, by that fide, for the area. Theorem 23.
Ex. 99. Given the fide A B D C 2 feet 10 inches, the perpen-
dicular D P C E 9 inches to find the area. Fig. 21.
By feet and inches..
:
210 6
→ fide
9
-perpendicular
Decimally.
2,875
75
4
210 6. aniwer
Sliding-rule. As I on A: 2,875 on B:
To meaſure any plane
2,15625
0,75 on A: 2,15+on B.
triangle.
Rule ro. Multiply the bafe and perpendicular one by half the other
or multiply them together and take half the product; by either method
you may find the area. Theorem 24.
24
THE UNIVERSAL MEASURER
Ex. 100. In a right angled plane triangle C BP. given the bafe BP
7 feet 10 inches, and perpendicular CP 2 feet 3 inches, required
the area. Fig. 11, 12, 13.
Feet and inches.
7 10
Decimally.
7.8333
2 3
15 8
I II 6
17 7 6
=
bafe
perpendicular
=
2,25
2,17,025
8,8125
28 f. 9 in. 9 p.
Sliding-rule. As 2 on A: 7,833 on B:: 2,25 on A : 8,81 on B';
the fame as if you take CB and a perpendicular to it from P.
=
Ex. 101. In an oblique angled plane A A B C, given the bafe A B
10 feet 3 inches, the
6 feet 10 inches, to find the area.
Fig. 11, 12, 13.
Feet and inches.
CP
:
A
ΙΟ 3
=AB
3 5
= CP
30 9
4 3 3
Decimally.
3.416
CP
10,25 = A B
35,02083
Sliding-rule As 2 on A: 10.25 on R:: 6.83
on A: 35,02 on B; or, as I on : 3.416 on
B:: 10,25 on A: 35,02 on B, the anſwer.
35 0 3
102. The plane A is the moſt uſeful figure in menfurations, &c.
for if a figure be ever fo irregular it may be reduced into As, by draw-
ing or meaſuring diagonals, and Ls from thole diagonals to their op-.
pofite angles, by which the area of every A may be found; and then
it is evident, the fum of all the areas of thefe As will be the area of the
irregular polygon, if there be bended fides, you may take offsets as
taught in furveying, or by rule 24 and 25.
To meaſure any plane A by having its 3 fides given.
1
Ex. 103. In the oblique ▲ A B C are given the 3 fides A B≈ 56,
BC= 41, and C A = 29, required its area.
Rule 11. From half the fum of the 3 fides fubtract each ſide ſeve-
rally, multiply the faid half fum and 3 remainders together accord-
ing to continual multiplication, and the fquare root of the last product
is the area. Theorem 73.
AND MECHANIC.
25
7 firſt difference
29
A C
63 63 63
63
41
BC
56 41 29
56 = A B
7 22 34
441
2) 126
lum fides
63 = fum
In the following examples, you muſt
fuppofe the dimenfions to be taken as
directed in ex. 92; ſo that let them be
what they will the area will be of the
fame name.
22 fecond deffer.
9702
34 third differ.
329868(574,34 = area
25
107) 798
1144) 4968
11483) 39200
114864) 475100
↓
Sliding-rule. As I on A: 22 on B:: 7 on A: 154 on B; and, as
I on A: 154 on B:: 34 on A: 5236 on B. Then as 63 on A: 63 on
D:: 52,36 on A (for 5236 is off the rule) to 57,4 on D, which (by
ex. 78 and 91) multiplied by 10, becauſe 5236 was on A, divided by
the fquare of 10, gives 574 area.
To meaſure a trapezia.
Rule 12. Take the fum of the two Is and diagonal on which they
fall, half the product is the area; or, the one multiplied by the other
gives it. Theorem 25.
Ex. 104. In the trapezia A H G I given the diagonal A G≈ 28,2
the Le H 10,5 and the Ia8, to find the area. Fig. 14.
10.5
8
18,5
14,
260,85
I =
Le H
a I
fum Ls
2
diagonal
arga
Sliding-rule. As 2 on A: 28,2 on
B:: 18,5 on A: 260,85 on B the
anfer.
To meaſure any irregular polygon..
Ex. 105. In the irregular pentagon a bade (fig. 15) is given the
diagonals a c 24, and a d = 20,72, the Ls b n = 10, dq= 124.
and du 10,2, required the area. Here two Ls b n and dq fall upon
one diagonal a c, fo the figure (fee ex. 102) is divided into a trapezia
a b c d and a Aade.
10 = b n
12.5=dq
22,5 = fum is
12 diag. a c
trap. abcd
270,0
20,72 = da
5, I
du
A dae
105,672
270
=
trapezia a b c d
375,672 = area required
D
26.
THE UNIVERSAL MEASURER
Sliding-rule. As I on A: 22,5 on B: 12 on A: 270 on B the
area of the trapezia; as 2 on A: 20,72 on B:: 10,2 on A: 105,6 on
B, &c.
To meaſure any regular polygon.
Rule 13. Multiply the let fall from the center of the polygon
to the middle of one of the fides, and one of the fides, and the number
of fides into one another, half the laſt product is the area, or the pro-
duct of any two of thefe three into half the third gives the area. Theq
rem 27.
Ex. 106. If each fide C D of a regular pentagon be 500, and the
LQP 360, what is its area? Fig. 25.
a fide
500
5 = number of hides
2500
180 - 2
450000
atea
Sliding-rule. When three numbers are
to be multiplied together, you may
turn fome one of them into a divifor, or
find a mean proportional between them.
1. Thus, as 5 on A on B:: I on
: 0,2 on B, then as 0,2 on A: 500 on B:: 1,8 on A (for 180 is off
the rule) to 4500 on B, fo (by Ex. 68.) 450000-area. 2. Or (by Ex.
78) as 5 on A: 5 on D:: ƒ on A (for 500 is off the rule) to 5 on
D. So (by ex. 68) 50 is a mean proportional between 5 and so. Then,
as I (viz. the 10) on D: 180 on A:: 9.5 on D (for so is off the rule)
45 on A; fo (by ex. 68) 450000 is the anfwer.
If the QP be given the fide CD (by theorem 48) may be found,
and the contrary. Thus, as fine half the (C QD): half the oppofite
fide (CD): co-fine of that L (viz. fine LQCD or Q DC): the L
QP; now if we call the fide CDI and fo by this find the area as
before, we'll have a factor for all regular pentagons, (fee theorem 36)
and in this manner is this table calculated to the regular polygons men-
tioned in it.
Names of the
polygons
number half the s tareas, the theſe multipli-
fides being fers turned into
Trigon
Square
Pentagon
Hexagon
Heptagon
Octagon
Nona goa
Decagon
of
at the
fides
center
unity.
divifors
3
600,00′
,433013
2,3094
450,00'
1,000.00
1,0000
5
300,00′ 1,72047
,5812
6 78
300,00'
2,59807
3849
250.42
3,6340
,2752
220,30′
4,8284
,2071
9:
200,00′
6,1818
,1617
TO
180.00'
7.6042
•1299
AND MECHANIC.
'29
toy. To ufe the table. Multiply the fquare of the polygon's fide by
the tabular number belonging to that polygon, or divide it by the divi-
for, the product or quotient will be the area. So if the fide of a regular
hexagon be 10; here 10 x to 100, then under areas, &c. and a-
gainſt hexagon ftands 2,59807, fo 100 × 2,59807 259,807 its a
rea, or 100÷,3849 259,807 the fame.
Sliding-rule. As I on D: 2,590n A:: io (fee ex. 63) on D: 259,8
on A; or, as ;3849 on A: 10 on B:: 10 on A: 259,8 on B, the a
rea. &c. for others.
To meaſure a circle.
Rule 14. Multiply half the periphery by half the diameter, or one
fourth of their whole product is the area.
Ex. 108. Given the circumference A B CD314.16, and diame
ter AC = B.D=100, to find the area. Theorem 28. Fig. 43.
157,08 = A B C D
50
7854,00
diameter
area
liding rule.
As 4on A: 314,16 on B:: 102
on A: 7854 on B, the anſwer.
The diameter of a circle gives, to find its arca.
Rule 15, Multiply the fquare of the diameter by 0,7354, or divide
it by 1,2732 (fee ex. 42) fo you'll have the area. Theorem 82.
fo
{
100} = diameter
}=
Sliding-rule.
As 1,2732 on A: 100 on B:: IQƠ
10000 — diameter on A: 7854 on B; or, as fon D:
0.7854
7854,0000 = area
0,7854 on A :: 100 (fee ex. 68) on
D:7854 on A the anſwer.
Note. 1 to 9,784 (by prob. 173) will be found as i 4 to 11; hence,
as 14: 11: the of any circle's diameter to its area.
The circumference of a circle given to find its area.
Rule 16. Multiply the of the circumference by ,07958, or divide
the thereof by 12,566, and you have the area. Theorem $3.
F
fo {314,16}= circumference
98696.5056 circum. 12,566)98696,5056(7854,20area
85970,=.07958 inverted
7854,26 — area
D:,079
Sliding-rule. As I on D: ,079 on A: : 3,1416 on D (for 314,16
is off the rule) :,7854 on A, fo (by ex. 68) 7854 area; or, as 12, §66
on A: 314,16 on B:: 314,16 on A : 7854 on B, as before.
=
28'
THE UNIVERSAL MEASURER
1
•
Having the diameter of a circle, to find the periphery. Or, the
periphery, to find the diameter.
Rule 17. Multiply the diameter by 3,1416 and the product is the pe-
riphery. (See theorem 82. But, by problem 173) as I 314167
: 22; or nearcr, : 113: 355 :: the diameter to the circumference.
But 7 to 22 will do in moſt cafes.
Ex. 109. If the diameter be 32, what is the periphery?
conftant factor = 3,1416 or as 7:22:32 or as 113: 355 :: 32
given diameter-
32
reqd. per. 100,5312
} area ==
22
32
7)704 or 113) 11360
area 100.57 or
100,5309
The reverſe of any of thefe ways will find the diameter when the pe-
riphery is given; i. e. as 3,1416: 1, or as I: 0,3184, or as 22: 7,
or (to be moſt exact) as 355: 113:: the periphery of any circle: its
diameter, and on the fliding-rule may thefe proportions be wrought on
A and B.
To meaſure any fector, femi-circle, or quadrant of a circle.
Rule 18. Half the product of the radius into the arch, or half the
one into the other gives the area. Theorem 29.
Ex. 110. If the radius A B
218,56, what is the area? Fig.
109,28=arch BC
radius A B
}
50
5464,00
=
arca
A C be 50, and the arch B C =
9.
Sliding-rule.
As 2 on A: 218,56 on B:: 50 on
A: 5464 on B, the anſwer.
To measure any fegment of a circle.
It is evident that if from the area of the ſector (fig. 153) SQA B
(found as before) you take the area of the AS A B there will leave the
arca of the fegment A QBG; therefore,
Ex. 111. Let the arch B QA 28,656, radius SAS B≈ 16,
'he verſed fine or height of the fegment =6=QG, and chord A B≈
24,98, to find the fegment's area.
chord A B
area AS A B
28,656 = arch A Q_B
24,98
8
radius SQ
5
SG
2
229,248
area ſector SA QB
124,90
124,9
area AS AB
104,348
area fegment A QB
"
Note. The area of any hyperbola, circular, and eliptical fegments
may be found to what exactneſs you pleaſe by the feries in problem
189.
3
AND MECHANIC.
29
Às 2 on A: 28,656 on B:: 16 on A: 229,248 on B area fector,
and as 2 on A: 24,98 on B:: 10 on A : 124,9 on B≈ area ▲, their
difference is 104,348 the anſwer.
But if the ſegment be greater than a femi-circle, you muſt find the
areas of the ſector WEDC (fig. 112) and A W CE the fame way,
then its plain per fig. that their fum is
area of the fegment W ED.
112. To conſtruct the table of ſegments; fuppofe the radius of the
circle (being unity) to be divided into 1000 parts, and thro' every
part chords be drawn, then the femicircle will be cut into 500 fegments
the verfed fine of the firit fegment being,001. Now, the half chord
of any fegment being a mean proportional between the verfed fine of
that fegment and remaining part of the diameter, the faid half chord
and arch (ſee prob. 56, 57, and 58.)may be found, and ſo the area of
this fegment by the above method will be found,0000421, in like
manner, the height of the ſecond ſegment being =,002 its area will be
found=,00011919 and fo on for others. From fuch a table the arca
of any circular fegment may be had by this,
Rule 19. To the given verſed fine annex 3 cyphers, and divide it
by the diameter of its circle, feek the quotient under V. S. or verfed
fine, and take out the number againſt it under feg. or area of ſegments,
this number multiplyed by the □ of the diameter gives the area of the
fegment, for (by theorem 9.) As the diameter of any circle : any ver-
fed fine thereof :: the diameter of any other circle (viz. 1,000 that
of the table) to its like verfed fine, which proves the first part of the
rule, and (by theorem 36.) As I the tabular diameter or radius
the area of any ſegment thereof :: the fquare of any other circle's ra-
dius to the area of its like fegment, which proves the fecond part of the
laſt rule.
Ex. 113. Let the verſed fine be 6 and radius 16, as in ex. 111.
?
32) 6,000 (,1875
againſt which ſtands
1019
32
fquare diameter 32=
1024
280
4076
2038
101.90
256
24 &c.
Sliding-rule, As 32 on A: 1 on B::
which in the table ftands,1019, then as
on D (fee Ex. 68) : 104,3 on A.
area
104,3456
6 on A: 18 on B againſt
I on D:,1019 on A:: 32.
36
THE UNIVERSAL MEASURER
Becauſe the radius of the table is 500, viz. 3 places of figures un-
der V S. and this quote, is 1875 viz. 4 places, I therefore take the
fegment for 187 and ſay by the rule of three, as 187: its ſegment:
187,5 1019 its fegment, and thus you may find a proportional part
or fegment where great exactneſs is required, otherwiſe the ſegment for
187 or 188 will do; if the fegment be greater than half a circle you.
muſt find the leſſer ſegment as above, and take it from the area of the
whole circle. The area of a circular fegment may be had nearly by
theorem 84, and the area of an eliptical fegment by theorem 85, but
to go through theſe and other curve-lin'd fegments or ſpaces would be
almoſt endlefs, and when done would be to little purpoſe in practice,
for fuch figures are feldom met with in meaſuring, and if they be, how
muft their forms be known, which must be had before their proper
rules can be applyed, it is then certainly beſt to make uſe of a general
rule which will come near any plane whatever, let it be what it will,
See rule 24, and 25.
How to meaſure a parabola.
Rule, 20. The area is had by multiplying the axis and ordinate, the
one by of the other, theorem 77.
Ex. 114. If the axis eu (fig. 27) be ro,5 and the greateſt ordinate
A B 15,6, what is the area of the common parabola A u B?
15,6 A B
10.5 = u e
163,80
2
3) 327,6
109,2 area
Sliding-rule.
As 1,5 (viz. 3 ÷ 2) on A : 15,6 on
B:: 10,5 on A
anfwer.
To meaſure an oval or ellipfis.
109,2 on B, the
Rule 21. For the area, multiply the product of its two diameters by
,7854.
Ex. 115. If the tranfverfe diameter TS (fig. 26) be 21, and the
conjugate I G = 15,6, what is the area?
15,6 = GI
=TS
21
327,6
prod.
327,6 prod.
7854 factor
257,29705 area
See theorem 80.
Sliding-rule. As 1,27032
(viz.,7g54) on A is to 21 on
B fo is 15,6 on A to 257,29 B.
A mean proportional between the two diameters of any ellipfis is —
the diameter of a circle whoſe area is that of the ellipfis, fo that what
rule ever gives the area of a circle by its diameter only, the fame rule
will give the area of an ellipfis by having its two diameters.
AND MECHANIC.
31
Having given the two diameters of an ellipfis to find its periphery.
Rule 22. To twice the fquare root of the fum of the fquares of the
two diameters, add of the ſhorter diameter, and the fum will be the
periphery nearly, fo if the tranfverfe diameter be 40 and the conjugate
30
then 40 1600 and 30 900, their fum is 2500 whofe
foot is 50 doubled is 100, to which adding of 3= 10 gives 110
for the required periphery nearly.
To meaſure an hyperbola,
Rule 23. To 80 times the tranfverfe axis add 39 times the abfciffa,
multiply that fum by the fquare root of the product of the faid abfciffa
and axis, and that product again by four times the product of the abſciſſa
and conjugate axis, this laft product is a dividend, then to 16 times the
tranfverle axis add 3 times the abfciffa, this fum multiplyed by i5 times
the tranfverfe axis gives the divifor, which divifion being made the quote
is the area, this is but changing the figns in theorem 80, which in the
fame words (by reading difference for fum) will give the area of an
elliptic fegment, but to be very exact you muſt uſe the ſeries in prob. 189.
Ex. 116. Let the tranfverfe Tu (fig. 28) be 30, abſciſſa or height
C10, and conjugate axis= 18, to aproximate the area.
i
30= Tu
80
30 = Tu
16
240080X Ta
add 390 = 39 × ù G
480
add
30 = 3 × C
4
2790
510
mult. 17.32 = □root 10 X 30
48322,3 product
720 4 times 18 x 10
34792416 produ&t — dividend
mult. 450 = 15 times 36
229500 divifor
2295|00)247924|16(151,60] = area nearly.
2160 &c. the remainder
This area, by the feries problem 89, taking in 6027 terms would be
151 687; whence it appears that the above rule may ſerve in common
practice, only differing in this example by (151,687 → 151,601 =)
,086.
Ex. 117. If 30 and 18 be the tranfverfe and conjugate diameters
of an ellipfis and s a 10 the height of a fegment s ba b (fig. 136.)
then by the last rule,
1
32
THE UNIVERSAL MEASURER
from
2400
take
390
80 times f T
39 times fa
from
-
480 16 times fT
take
30 3 times fa
2010
mult. 17,32 =□ root of 10 X 30 mult.
34813,2
mult. 7204 times 18× 10
450
450 = 15 times 30.
202500 divifor
25065504 divid. 202500)25065504(123,78 area nearly
Ex. 118. If the diameter of a circle be 30, and the verfed fine or
height of a fegment thereof be 10; then, by the laſt rule, all is the
fame as in the laſt example, only multiply by 4 times 10 inſtead of 4
times 18 x 10, and by 15 inſtead of 15 times 30.
450
fo 34813,2
mult. 40 4 times 10. 15
1392528
6750)1392528(206,3 A nearly
dividend 6750 = divifor
How to meaſure any fegment or curve-lined ſpace.
Rule 24. Meaſure in a right line from one corner of the figure to a-
nother, and at right angles to this line take as many breadths as you
chufe, but eſpecially the greateſt and leaft breadths, add all thefe
breadths together and divide the fum by the number of them, which
gives the mean breadth, this multiplied by the length gives the area, or
find the area of every part by itſelf, thus multiply half the fum of two
breadths nerreft together by their diſtance (fee theorem 26) and the
product is the area of that part, do in the fame manner with every
part, and the fum of all theſe areas will be the area fought.
119. In meaſuring, the dimenfions must be taken
hence the area is the leaft and moft exact, becaufe
fhorteft.
to each other,
lines are the
Ex. 120. Let it be required to find the area of the ſpace D Em B,
fuppofing D B 12,245, ED = 6, m H 5,49, z y = 3,83, and
yby HHD 4,081. Fig. 168,
br. at B
breadths S, ED
By the fecond part of the rule.
2)4.081 y B
I
2,0405 =
mul. 3,88 y z
y B
7,917140 area ABzy
add
5,49mH
3,88 zy
numb. 4) 15,37 fum
=
3,84
mean breadth
19,119485
multiply 12,245
=
length
23.445345
47,0208
area
50,481970
mzy H
mEDH
area fought
AND MECHANIC
32
3.88
y z
уч
5 49 = m H
9,37
fum
4,685
half ſum
multiply
4.081
y H
ун
19,119405 — area y z m H
5.745
of E D and m H
multiply
4,081
HD
23,445345 = mEDH
But to be more exact, the following rule, taken from theorem 138,
is to be uſed
Rule 25. Take 4 breadths equally diftant from one another, (the
two outermost of which must be at the broadest and narrowest parts of
the figure; and if the figure be very irregular, it muſt be taken at fe-
veral times, that the fides or curves between the two outermoft breadths
may have a gradual curvature; this muft alfo be obferved in folids ta-
ken this way,) and to 3 times the fum of the two innerinoft breadths,
add the fum of the 2 outermoſt ones, the laſt fum multiplied by part
of the length gives the area.
Ex. 121. Let things be the fame as in the laſt example.
5,49 m H
3.88 zy
6=ED
o breadth at B
6 = ſum
9,37
3
28,11=
triple fum of inner. br.
28,113 times that fum
34,II
multiply
1,53 = ¦ D B
DB
52,1883 area
Sliding-rule. As 8 on A: 34,11 on B:: 12,24 on A : 52,19–
on B, the area required.
The dimenſions uſed in this and the former example, are the fame
with thoſe in ex. III, where by the true method the half fegment turns
out 52,174, which compared with thefe fhews, that the first method
by rule 24 is fomething wider than the fecond method by that rule;
but the error in rule 25 appears inconfiderable, and as it comes fo near
a circular fegment, it may be depended on as near in any fegment
whatever; confequently this rule. is fufficient for the whole of fuper
fcial meaſure.
E
THE UNIVERSAL MEASURER
Superficial meaſure of folids.
To find the fuperficial content of any cube, prifm, cylinder, &c.
Rule 26. Multiply the girt or periphery of the folid by its length,
and the product is the fuperficial content of all but the bafes, to which
you add the area of each bafe, viz. the double area of one baſe, you'll
have the area or fuperficial content of the whole folid. Theo. 132.
Ex. 122. If each ſfide of a cubebe 10,5 what is the whole fuperficial
ontent?
10,5
multiply
4
a fide
fuch fideš
42,0 = periphery
10,5 = length
441.0
curve ſuper.
220,5 area both bafes
661,5 = whole furface
10,5 = a fide
10,5
110,25 = area 1 bafe
2
220,50 — area both baſes
Sliding-rule. Becauſe a cube has fix = fides, therefore the area of
pne fide is a fixth of the whole fuperficial content. So as ở viz. 0,166
+ on A: 10,5 on B:: 10,5 on A: 661,5 on B the anſwer.
Ex. 123. If each fide of the baſe of a triangular priſm be 10 and
its length 25,5, what is its fuperficial content.
The area of any equilateral A or trigon is had by rule, or ex. 107.
thus,
0,433013 ib. no for a trigon
100
43,3013
2
of a fide at bala
area of either baſe
85,6026 = content both bafes
25,5
length
30 = 10×3= periphery
765 curve fup.
2dd 86,6026 bafes
851,6026 anfwer
Sliding-rule. As ↑ on A: 30 on B:: 25,5 on A : 765 0a Ę. And
as I on D: 0.433 on A:: 1,0 cn D: 43,3 on A (See ex, 68).
Ex. 124. If the length of a cylinder be 255, and periphery of its
bafe 10, required its fuperficial content ?
,07958 fée rule 16
100
periphery
7,958 = area bafe
2
15,916 — area at both bafo
2559 = curve fuperficies
2565,916 = the whole fuperficie
255 = length
10 = periphery
2550 ≈ curve fapes.
AND MECHANIC.
Sliding-rule. As 12,56 on A: ro on B:: 20 on A : 15,91 on B
the double area. And as I on A: 255 on B:: 1000 A: 2550 on B、
whoſe fum 2565,9 is the anſwer.
To find the fuperficies of any cone or pyramid.
Rule 27. The product of the flant height and periphery of the bafe
divided by 2, or half the one multiplied by the other gives the curve
fuperficies, to which adding the area of the baſe, the fum is the whole
fuperficial content, theorem 131.
Ex. 125. If each fide A BBC-CD=DA (fig. 119) of the
bafe of a pyramid be 10,5 and the flant height v A➡vẹ 30, what
is the fuperficial content ?
10,5 = A B
4
fuch fides
42,0 periphery bafe
=
15 = 플 ​A
630 = curve fuperficies
110.25
740,25
area baſe
whole furface
10,57
= A B
10,5
525
1050
110,25 = baſe
Sliding-rule. As 2 on A : 42 on B : : go on A: 630 on B. And
as i on A : 10,5 on B :: ro,5 on A: 110,25 on B, whoſe fum 740,25
is the anſwer.
Ex. 126. If the periphery of the bafe of fame cone or pyramid be
45, and its flant height 380, what's the convex furface ?
190
45
8550
£
=
flant height
periphery at bafe
convex fuperficies
Sliding-rule. As 2 on A is to
350 on B fo is 45 on A to 8550
on B, the antwer.
To find the fuperficial content of any tapering ſtreight ſided folid.
Rule 28. Take the periphery of the two bafes or ends in one fum,
and multiply that fum and the folid's length one by half the other, the
product is the curve fuperficies, to which adding the areas of the two
baſes you get the whole fuperficial content. Theorem 130.
Ex. 127. There is a conical fruftum A BEF (fig. 127) the peri-
phery at A B the greater bafe is 52,5, at the leffer bafe E F is 30,7§,
and length viz. flant length A FEB 40, required the fuperficial
content of all except the greateſt baſe.
36
THE UNIVERSAL MEASURER
3
52,5
30,75
30.75
30.75
83,25 um
20 플 ​AF
85970
945,5625 fquare of 30.75
inverted factor
1665,00 curve fuperficies
66189
752,47 ALB
8510
2417,47 anfwer
472
76
572,47 area leffer baſe
Sliding-rule. As 2 on A : 83,25 on B:: 40 on A: 1665 on B,
and as I on D: ,07958 on A :: 3,075 on D (See ex. 68.): 7,52 —
nearly on A, fo 752% is the area of the leffer bafe, hence the antwer
is=2417.5.
Ex. 128. If there be a cylinderoid, prifmoid, fruftum, &c. whoſe
peripheries at the two ends is 502 and 35, and ſlant length 600, what
is the convex iurface.
502
35
537
300 half the height
101100 the anſwer
Sliding-rule. As 2 on 1: 537 on B:: 60,0 on A: 16110 on B
(See ex. 68) fo 161105 is the convex fuperficies required.
To find the fuperficies of any globe or ſphere.
Rule 29. (By theo. 133.) the ſurface of a globe is to 4 times the
area of its greatelt circle, fo any of the factors for the area of a circle
(being multiplied by 4) will be factors for the furface of a globe when
the like things are given, that is 3,1416 (≈ 4 times 0,7854) when
the axis is given, or,31832 (= 4 ×,07958) when the periphery is
given, &c.
Ex. 129. If the axis of a globe or fphere be 10,5, what's the fuper-
ficial content ?
10,5
10.
31
}
axis
110,25 fquare of the axis
22 fee rule 16
7) 2425.50
376,5 furface
Sliding-rule.
As I on D is to 3,1416 on
A fo is 1,05 (for 10.5 is of the
rule) on D to 3,465 on A fo (by
ex. 68.) 346,5 is the anſwer.
AND MECHANIC.
37
Having given the length or height of any part of a globe or ſphere
to find its fuperficial content, theo. 135.
Rule go. Multiply the height of the given part, the axis of the globe
and 3.1416 into one another, the laft product is the curve furface, to
which add the area of the end or ends for the whole furface.
Ex. 130. Given the axis B D (fig. 128.) 10,5 of the globe Bv DL
: 1,75 the height of a fegment Hvd, to find the curve fur-
and Ev=
face.
10,5 axis
1,75 height fegment
18,375 product
6141,3 inverted factor
55125
1837
735
18
II
57,720 the anſwer.
Sliding-rule.
As 0.318 on A is to 10,5 on B
fo is 1,75 on A to 57,7+ on B th e
curve fuperficies of the fegment H vd
and is alfo = the curve fuperficies of
any part or zone mHdG whofe
length m Hd G is = 1,75
To find the fuperficial content of any of the five regular bodies.
See problem 157.
1. The tetraedron's furface being 4 and equilateral As; therefore,
4 times the area of one of thefe As will be the furface of the tetrae-
dron, or being a pyramid, its ſurface may be had by rule 27. Now if
the fide of an equilateral ▲ be 1, its area (by ex. 107) will be 0, 433013
which multiplied by 4 gives 1,732051 for the furface of a tetraedron.
whefe fide is I.
2. The furface of the octaedron being 8
and equilateral As, it
is therefore double to that of the tetraedron, fo 3,464102 is the furface
of an octaedron, when each fide thereof is unity.
3. The hexaedron is a cube, which fee in ex. 122.
4. The furface of the icofaedron is contained under 20 and equi
lateral As, and is therefore 20+ 433013 = 8,66026, when its ſide iş
unity.
5. The furface of the dodecaedron is contained under 12 = penta-
gons, ſo 1,720477, being the area of a pentagon whoſe fide is unity, 12
times 1,72047720,645724, will be the furface of a dodecaedron,
whofe fide is unity, and thus you get this table, in which for the foli-
dity's See Ex. 181.
38
THE UNIVERSAL MEASURER
[Tetraedron 1
1,732051)
If the
Octaedron
3,464102
(0,11785)
0,47140 |
fide of
Hexaedron
6,
I,
the
Icofaedron
8,660260
| 2,18.169
Dodecaedron J (20,645724)
17,66312
Rule 31. Multiply the fquare of the fide of the propoſed body by
the tabular fuperficies belonging to that body, and that product is its
furface.
Ex. 131. If the fide of a tetraedron be 10, what is its fuperficies?
1,732051
100
Sliding-rule.
As t on D is to 1,732 on A fo is 1,0 on
173,2051 anſwer D to 1,732 on A, ſo (ſee ex. 68) 173,2. anf.
Ex. 132. If the fide of a dodecaedron be 13, what is its fuperficial
content.
20,645724
169 of 13
3489,127356 anſwer
Sliding-rule. As I on D is to 20,64 on
A ſo is 1,3 on D to 34,89 on A, (fee
ex. 68) fo 3489 is the answer.
In rule 30. Is the method of finding the furface of any fegment or
of any fruitum of a globe, but to find the furface of any or polygon
on the furface of a globe, this following rule is general.
Rule
32. Take the angles in degrees and decimal parts of a degree,
to their fum add 360, from this fum take the product of 180 into the
number of angles, multiply the remainder and the ſquare of the fphere's
axis and,0044 (3.1416÷720) into one another, the last product is
the area or fuperficies of the polygon. Theo. 76.
Ex. 133. and 134. If the axis of a globe be 16, what is the area
of a ſpherical A on that globe, whofe 3 Ls are 360 8′ and 46° 18′ and
104°, viz. 36,1333 and 46,3 and 104. Alfo, what is the area of a
ſpherical pentagon, whoſe 5 Ls are 70, 801, 130, 150, and 160, and
axis of the globe 20.
AND MECHANIC.
39
*. For the A, (degrees)
A. For the pentagon.
36,1333
46,3
70 degrees
80,5
130
add
104,
186,4333
360,
150
160
add
360
546,4333
950,5
fubtract 540 = 180 × 3
180 X 5
900 fubtract
6,4333
50,5
,0044
3,1416÷720,0044
,02830652
25616 the axis
7,24646912 area A
,22220
axis 20
400
area pent. 88,88
Ex. 135. But to find the furface of any folid which properly is not
known.
Rule 33. Take 3 equi-diffant girths or periphery's each at right Ls
to the curve length, or flant length and to 4 times the middle girth add
the other two girths, this fum multiplyed by a fixth of the curve
length, gives the furface. Theo. 13.
Note. The two extreme girths muſt be taken the one at the greateſt
bulge, and the other at the leaft, in the figure, and the mean girth in
the middle between them, and if the figure be very irregular, or there
be feveral fuch bulges it muſt be taken at feveral times as if it were fo
many feveral figures, and the fum of all the contents will (nearly) be
that required, or if you think 3 girths not fufficient, you may take 4,
and work by rule 25.
Required the fuperficies of the folid QVQv (fig. r40) the perip-
hery at V v the greateſt bulge, being 40, at Q the leaftoar M m
tho middle, between Qand V35, and the curve length Q V=Q-v
≈30,
Girths.
o at Q
40 at N
140 = 4 X 35
180 their fum
5 to Qr
900
T
Sliding-rule. As 6 on A is to 180 on B⋅ fo is
30 on A to 900 on B the furface of Q_N▾,
and if the other part QNle this part al-
ready meaſured, then twice 900 1800 the
furface of the whole folid; or, as & on A is
1800-B fo is 60 on A to 1800 on B.
40
THE UNIVERSAL MEASURER
}
XX
(XXX XXXXXXX
SECTION V. Artificers work.
*XXX
Artificers have different ways of meaſuring according to the cuſtom
of each craft, fome give the content in feet, fome in yards, fome in
fquares, fome in rods &c. they alfo differ according to the custom of
the place, fome give the content in one denomination, and fome in a-
nother. Now as a general rule for all the various methods amongſt
thefe workmen, it is beft to take the dimenfions in feet and decimal
parts of a foot (ex. 92.) and fo find the content in feet which (by rule
6.) may be reduced to what name you pleaſe. for it is feldom that ar-
tificers ufe ought but fuperficial or flat meafure, and becaufe walls,
doors, floors, windows, roofs, &c. are rectangles the areas of fuch are
always found by rule 8, but if any other form come in practice it can-
not but come under fome of the foregoing rules, 2 or 3 examples in
every cafe will clear up this fection.
Of Bricklayers work.
The principals are walling, tileing, partitioning and chimney-work.
Bricklayers commonly meafure their work by the rod fquare of 16
feet and an half, ſo that one rod in length and one in breadth contains
(16,5 × 16,5) 272,25 fquare feet; but in fome places they allow 18
feet to the rod i. e. (18 x 18) 324 fquare feet, and in others 21 feet
is a rod with 3 feet height i, e. 63 fquare feet, and here they do not
'regard the wall's thickuefs, only moderates the price accordingly, fo
when you are to meaſure fuch work enquire which of theſe ways mult
be uſed, and then divide the product of the length and breadth in feet
by the proper divifor, fo will you have the content. But where brick
walls are meaſured by the rod, and reduced to the ftandard thickneſs
of 14 brick thick, this is the
Rule 34. Multiply the fuperficial content of the wall in feet by the
number of half bricks it is in thickneſs, one third part of that product
is its content in feet, at the ſtandard thickneſs.
Ex. 136. If a wall be 724 feet long, 194 feet high, and 54 bricks
thick, how many ſtandard rods doth it contain?
19,25 feet
multiply 72,5
1395,625
br. thick=
II
3.) 15351,875
272) 5117,291
rods
= 18,8
Note. In reducing feet to rods,
it is uſual to reject the odd,25
and divide only by 272. but the
divifors 3 and 272.25 may be re-
duced to one factor or divifor, and
fo be a conftant gauge point.
Thus, 272,25 X3816,75
1
AND MECHANIC.
4I
which divided by 11 quotes 74.19, which is a conſtant diviſor for 51
bricks thick and in this manner is this table conftructed, for the thick-
neſs between 1 and 4, and may be continued at pleaſure.
I
multipliers divifors
bricks
thick
,00245 | 405,27
1 1/
I
,00367 |
272,25
,00489 |
204,19 |
2호
​,00662
163.35
3
,00734
136.12
3 1/
,00857 1
I 6,68
4
.00884!
102. Į
12.2 mm
Sliding-rule.
As 74.19 (the gauge point
for the given thickneſs 5) on
A: 19,25 on B:: 725 on A
: 18,8 rods on B. the anſwer.
The uſual way to take dimenfions of a building is to meafure halfround
it (at the middle) on the out-ſide, and half round it in the in fide with
a cord, this gives the true compafs of the building in which the thickneſs
of the wall is confidered and if it be all of a height, meaſure its height
at any place from the bottom of the foundation, but if the height be
unequal, you muſt take feveral heights, and their fum divided by the
number of heights gives the mean height, the pykes on the end walls
being As are meaſured as fuch, or two ones being put together makes
a rectangle of the fame height and breadth of the building.
Ex. 137. If each fide wall of a building be 45 feet long on the out-
fide, each end wall 15 feet broad on the in fide, the height of the build-
ing 20 feet, and the pyke at each end wall 6 feet high, the whole, 2
bricks thick, how many ſtandard rods, as alfo how many fquare yards
not minding the thickneſs, as if it were a ftone building &c.
add
Feet.
90. fum fide walls
30 that of the ends
120 compaſs
20 height
2400 area without pikes
15×6= 90 pikes add
2490
feet in the whole
mult. 00489 fact. 2 bricks thick
12,17610 rods, anfwer.
For the fquare yards.
9)2490 arcam feet
276. yards
3
Sliding rule, for the rods. By
the table, the divifor for two
bricks thick is 204,19, ſo as
204, 19 on A: 30on B:: 90
on A: the rods excluſive of the
pikes, and as 204,19 on A :
15 on B:: 6 on A: the rods contained in the pikes.
Sliding-rule for the yards. As 9 on A is to 120 on B ſo is 20 on A
to 266 on B the yard's area pikes excluded; and as 9 on A is to 15
F
42
THE UNIVERSAL MEASURER
on B fo is 6 on A to 10 on B the pikes; fo 2663 added to 10 gives
2763 the area in yards as required.
Note. Stone buildings are fometimes reckoned amongſt maſon's
works.
Of Chimneys.
}
In meafuring a chimney, the ufual way is, if it ftand alone without
leaning againſt a wall, &c. girt it about below the mantle, and take that
for the length, and the height of the room, or chimney fo far as it
keeps the fame girt, the product of theſe two is the content, but if it
ftand againſt a wall you muſt meaſure it round to the wall for the girt,
and take the height as beföre; for arches, fome take the breadth and
half breadth of the arch, and multiplies that fum by the length of the
arch, for the content; fome make deduction for windows, doors, and
the vacancy in chimneys between the hearth and mantle, &c. and fome
make none.
Of Tileing, and Slating.
Tileing is meaſured by the fquare of 100 feet, viz. 10 feet long and
10 feet broad. Slating fometimes by the yard fquare, and fometimes
by the rod of 49 fquare yards, &c. To take the dimenfions, meaſure
with a cord the length of the ridge, then having a ſmall weight at one
end of the cord, put it over the ridge and let it go down to the eave,
and take the other end to the other eave, fo you have the breadth of
the roof, which multiplied by the length of the ridge gives the content,
Note. Double meaſure is commonly allowed for hips, vallies,
ters, &c. And in tileing it is common to allow double meaſure at the
faves fo much as the projecture is over the plate, which is generally
about 18 or 20 inches.
gut-
Ex. 138. If the breadth of a roof with the ufual allowance at the
caves be 24,5 feet and the length of the ridge 45 feet, how many yards
fquares, and rods are in that roof?
24,5 feet breadth
45 length
9) 1101,5
49) 1225
feet
yards
2,5 roods
100) 1102,5
feet
11,025
9:
Sliding-rule. As 9 on A: 45 on B:: 24,5 on A: 122 on B the
yards area, and by fetting ICO and 441 (viz. 9 times 49) on A inſtead
of 9 you'll get the anſwer in fquares and in roods.
Note. Some make an allowance in roofs for the fpaces taken up by
chimucys,
&c. but others make nonę.
AND MECHANIC.
43
Of Carpenters work.
Carpenters work, as flooring, partitioning, and roofing, are meafured
By the fquare of 100 feet like tileing, and fome places by the fquare
yard.
Ex. 139. If a floor be $7,25 feet long, and 28,5 feet broad, how
many fquares?
feet 57,25 length
28,5 breadth
L
100)1631,625 feet
16,31625 fquares
Sliding-rule.
As 100 on A: 57,25 on B:
28,5 on A: 16; nearly, anfwer
Ex. 140. If a partition between two rooms be in length 80,5 feet,
and in height 124, how many fquares are contained therein ?
12,25 feet
80,5
100)906,125 feet
9,86125 iquares
Sliding-rule. As 100 on A:
12,25 on B::80,5 on A : 90,86
on B, the anſwer.
Ex. 141. If a houſe within the walls be 40,5 feet long and 20,§
feet broad, how many fquares?
40,5 feet
20,5 breadth
1.00) 830,25 feet in the flat
8,3025 fquares in the flat
add 4,1512 half the flat
12,4537 fquares in the roof
Sliding-rule.
1
As 66,66 (viz. of 100) on
A: 40,5 on B:: 20,5 on A:
12,45 on B, the anſwer.
Note. If the roof of a building be truly pitched, the area and half
area of the flat or floor will be the area of the roof, or as 2 is to 3 fo is
the area of the floor to that of the roof, on which footing the last ex.
is wrought.
Of Plaiſterers work:
Plaiſtering, rendering, and cieling, are all meafüred by the fquare
yard, in fuch works it is beſt to take dimenfions with a decimally divid
ed yard, when a room is plaiſtered or rendered, take its compafs in the
infide girting in chimneys, &c. where ever the work comes, this mul
tiplied by the height of the room gives its content.
Ex. 142. If the compafs of a room be 47,2 yards, and its height
4,3 yards, what is the content?
yards 5,41 height
45.8 circuit
247,778 content
Sliding-rule. As I on A: 47,2 on B:
4.3 203 nearly.
44
THE UNIVERSAL MEASURER
Note. Whiting and colouring are both meafured by the yard like
plaiftering. Deductions or allowances are to be made according to the
different cuſtoms of places.
Of Joiners work.
This work is alſo given in fquare yards, in taking dimenfions, they
meaſure round the room, or round the floor upon which it ſtands for
the breadth of their work, and for the height, they meaſure with a cord
the height of the room denting in the line where the plane comes, and
girting over cornifhes, pannels, mouldings, &c.
Ex. 143. If a room of wainſcot (being girt over the mouldings &c.)
be 5,41 yards, and the compafs thereof 137,4 feet, what is the content?
yards 5,41 height
45,8 circuit
Sliding-rule.
As I on A: 45,8 on B:: 5,41
on A: 247,7 on B, the anſwer.
2:7,778 content
Note. Doors, window-fhutters, and and all fuch like as are wrought
on both fides are called work and half work; in fuch cafes find the
content as before, take half thereof and add to it,
content; or, as 2:3:: the content before found
work and half work.
which gives the
the content at
Ex. 144. If the window fhutters about a room be 70 feet broad
(viz, all their breadths in one fum 70 feet) and the height of each
fhutter 6.5 feet, what's the content at work and half?
6,5 feet
70
2)455.0 work
227.5 half work
91682.5
75,8 yds. work and
Sliding-rule, As 6 (viz. 9 times 2 divided
by 2) on A: 70 on B:: 6,5 on A : 75,8
on B, the content at work and half.
In fuch examples as this, if all the
heights be you may take all the breadthis
in one fun, if all the breadths be you
may take all the heights in one fum, which
is caſier than to work them by 1 at a time, and equally true.
Of Painters work.
Painters work is the fame with that of joiners, both in taking dimen-
fions and meaſure.
Ex. 145. If a globe be painted whofe greateſt circumference is 10
feet, how many yards of painting is thereon?
10 periphery
periphery
100
,31832 fee rule 19.
9)31,832 r feet
Sliding-rule.
As 9 on A: 100 on B: : 0,318 o
A: 3,53+ on B, the answer,
3,537 yards
AND MECHANIC.
45
Of Maſons work.
Maſons work is fometimes meaſured by the foot fquare, and fome-
times by the running foot, (viz. only fo many feet in length) where
obferve that for every foot in length, running mcafure, there ought to
be a foot in breadth, or girt if it be a folid wrought about, and this
makes running meaſure agree with fuperficial meafure, as it always
ought to do, if a ſtone be hewed into any folid form you may by fome
of the rules in the laft fection find its area.
always meaſure ſo far as their tools come.
times reckoned amongſt maſons work.
with mafons work.
Mafons like joiners, &c.
Stone buildings are fome-
Paving is alfo reckoned in
Ex. 146. If a wall be 112,25 feet long, and 16,5 feet high, how
many rods at 63 fquare feet to a rod.
112,25 feet
16.5
61852.125
29.4- rods
Sliding-rule. As 63 on A: 112,25
on B: 16 on A: 29,4- on B.
A perſon that cannot even multi-
ply may eaſily find the feet. &e. in
any rectangle; thus, this wall being 16 feet high, if you take its length
1124 feet and meaſure it on your rule 16 times and a half,
find 1852, feet for the area of the wall.
Of Glaziers work.
you will
Glaziers meaſure by the fquare foot, and commonly take dimenfions
nearer than any of the foregoing, for they'll go to the 8th, or 10th of
an inch.
Ex. 147. If there be 4 panes of glafs each 4,23 feet high, and 1,25
feet broad, how many fquare feet?
4,23
5 = 4 times 1,25
21,15 feet required
Sliding-rule.
As I on A: 4,23 on B:: 5 on A :
25,38 on B, the anſwer.
Éx. 148. If the diameter of a round window be 2,23, what is the
content?
2,23
2,23
4,9729 feet
U
sliding-rule.
As I on A: 2,23 on B:: 2,23 on A : 4,97
on B the anſwer.
}
Note. Round and oval windows are meaſured as if they were
that length, &c. becauſe glafs waftes in cutting for fuch forms.
full
口
​46
THE UNIVERSAL MEASURER
·To measure Board and Plank.
pro-
Board and plank are fometimes fold by foot running meafure, which
is had by meaſuring the planks length along its middle, and by the flat
foot, which is had by multiplying the length, taken as before in feet
and decimal parts of a foot, by the breadth in the middle taken in inch-
es, that product divided by 12, gives the fuperficial or flat feet, and
fometimes by the folid foot, which is moſt reaſonable, becauſe it con-
fiders length, breadth and thickneſs, viz. one multiplyed into the
duct of the other two, and if they all three be inches, the product di-
vided by 1728 (the folid inches in a cubical, or folid foot) gives the
folid feet, but if any one of thefe 3 dimenſions be in feet as generally
the length is, then divide the laft product by 144, or if two of theſe
3 be in feet, then divide by 12. But it is eafielt, and will require no
dividing, to take all the 3 dimenſions in feet and decimal parts thereof,
fee the following examples.
Ex. 149. If a board be 16 inches broad and 13 feet or 156 inches
long, how many fquare or flat feet is therein?
vulgarly
inches 156 length
common method
13 feet
16 breadth
16
144)2496
inches
12) 208
17
feet
17 1
decimally
13 feet
19333
17,329
1. As 144 on A : 16 on B:: 156 on A : 174 on B,
2. As 12 on A: 16 on B:: 13 on A: 17 on B,
3. As I on A: 13 on B:: 13 on A : 17½ on B,
the anſwer,
in
fquare feet.
Any of theſe 3 ways (as alſo by feet and inches) may be taken, buť
the laſt being molt expeditious and exact too, if the dimenfions are ta-
ken decimally, it is moſt in uſe.
Ex. 150 If the length of a plank be 15.2 feet, its breadth 0,32
parts of a foot, and thickness 0,25 of a foot, what is its content fu-
perficial and folid.
15,2 length
multiply,32 breadth
4,864 fuperficial content
4,864 fuperficial content
,26 thickneſs
1,216 folid content
Sliding-rule. As 1 on A: 15,2 on B::,32 on A: 4,86 on B the
fat feet, for the ſolid feet find a mean proportional between any two-
of the 3 dimenfions (the breadth and depth are commonly taken for
this purpofe). Then it will be as 1 (or the fquare root of any other
divifor) on D: the length on A :: the mean proportional on D: the
content on A, or turn fom one of the 3 dimenfions into a divifor (fee
AND MECHANIC.
47
ex. 106) fo here, as,25 on A:,25 on D :: 32 on A:,28 on D the
mean proportional between,25 and 32; then as 8 on D: 15,2 on A
::,28 on D:,28 on D: 1,22 feet on A the folid content.
Ex. 151.
1. If a board be 5 inches broad, how much in length
make a flat foot?
will
Rule 35. Since the product of the length and breadth gives the area,
'tis evident that if the given area be divided by one of them, the
quotient will be the other of them.
5)144 inches in a flat foot
anfwer 28,8 inches in length
Sliding-rule. As 5 on A ,144 on B:: I on A: 28,8 on B.
Ex. 152. If a board be,82 feet broad, how much in length will
make 2 flat feet?
anf.
,82)2,50 flat feet
3,05 feet length
Sliding-Rule. As 0,82 on A: 25
on B:: I on A : 3,05 on B.
Theſe two examples confiders the board to be an = breadth through-
out, but if it be broader at one end than at the other, you may take a
breadth as near as you can gueſs in the middle of the part to be cut
off, and with that breadth find the length as in the laſt ex. and then
find the content by ex. 150, which if it be too much, or too little, you
may try a leſs, or a greater breadth, &c. fo by a few trials may come
near the length required, but to do it exactly and at once, ſee Queſt. 19.
Of Square Timber.
Square timber meaſure is the fame with folid plank, therefore the
length of the tree, the breadth and depth, or thickneſs, taken in the
middle thereof, and multiplied into one another gives the true folid
content if the bafes or ends are equal, (for then it is a prifm) but in
tapering timber it gives the content fomewhat too little as appears by
ex. 172.
Ex. 153. If a piece of ſquare timber be 13,2 feet long, and 1,52
feet fquare, (viz. 1,52 broad and 1,52 deep) how many feet of timber 2
A
1,52 breadth
1.52 depth
2,3104
2,3104 area in the middle
:
13,2 length
30,49728 anfwer
Sliding-rule. As I on D: 13,2 on A :: 1,52 on D: 30,5 on A
the folid feet required.
48
THE UNIVERSAL MEASURER
Ex. 154. If a piece of fquare tapering timber be 15 foot long I foot
broad, and 0,7 of a foot deep at the leffer end, 2 feet broad and 1,4
feet deep at the greater end, how many folid feet is therein?
Note. If the tree (whether fquare or round timber) be not contain-
ed under ſtreight planes from baſe to baſe, ſo that you judge its breaath
and depth taken in the middle not exact, you may take thefe dimen-
fions at each bafe or end, and half their fum will give that in the mid-
dle, fo in this ex. half the fum of the two breadths 1
half the fum of the two depths 0,7 and 1,4 is 1,05.
1,5 breadth
1,05 depth
1,575
15 length
23,625 folid feet.
}
and 2 is 1,5 and
Then,
taken in the middle of the tree.
Sliding-rule, fee ex. 106. and 150.
As 1,05 on A: 1,5 on D:: 1,5 on
A: 1,25 on D a mean proportional
between 1,05 and 1,5. Then as I on D: 15 on A:: 1,25 on D :
23,521 on A. the anſwer.
Otherwife. As 15 on A: I on B:: 1 on A:,066 on B, which is
15 turned to a divifor. Then as ,066 on A : 1,5 on B :: 1,05 on A
: 23,621 on B the anſwer.
►
Ex. 155. If a piece of ſquare tapering timber be 13 foot long, 6
inches deep and 1 foot 6 inches broad in the middle, how many folid
feet?
1,5 breadth
5 depth
,75
18 length
13,50 felid feet.
Sliding-rule, here o, 5 turned to,
a divifor is 2. Then as 2 on A
18 on B:: 1.5 on A: 13,5 on B.
the anſwer.
Note. Some meafurers of timber take half the fum of the breadth
and depth taken in the middle of the tree, for the fide of a mean fquare
which fquared and multiplied by the length they fay gives the content.
But it gives it too much, and if the difference between the breadth and
depth be great, this error is fo too, as in the laft example, the depth
15 feet added to the breadth 1,5 feet gives 24 feet whereof is 1 foot,
which multiplied by 18 feet the length, gives 18 feet for the content,
too much by 4.5 feet.
Ex. 156. If a piece of bafed fquare timber be 6 inches fquare
at each end, how much in length makes 11 folid feet.
AND MECHANIC.
49
Rule 36. Becauſe the product of the length, breadth, and depth gives
the content; therefore, if the given content be divided by the product
of any two of them, the quotient will be the third.
feet {5} = g. fide
,25)1,5 g. content
6 feet anſwer
Sliding-rule. As 0,5 on D: 1,5 on A ::1
on D: 6 on A the anſwer.
Note. If the timber be tapering, fee
queftion 20, fection 9.
Of round Timber.
The ufual way is to girt the tree in the middle with a fmall cord,
then one fourth part of that girt ſquared and multiplied by the length
gives the content, but if the tree be uneven in the middle fo that a true
girt cannot be taken there, then you muſt girt it at each end, and take
half the fum of theſe two girts, which anfwers to a girt taken in the
middle, whereof muſt be uſed as before, when you've taken the girt
with a ſmall ſtring, double it and meaſure that double on the inches on
a fliding rule, &c. half this is a fourth of the girt in inches, which
fquared and multiplied by the length in feet and divided by 144 gives
the content in feet, or as a general rule upon the fliding rule As 12 on
D: the length in feet on A :: a fourth of the girt in inches on D: the
content in feet on A, but if you meaſure the girt upon a foot rule de-
cimally divided, you need not divide by 144, and on the fliding-rule
it will be as 1 on D: the length in feet on A:: a fourth of the girt in
feet and decimal parts of a foot on D: the content in feet on A. This
is the method generally practifed becauſe of its eafe and expedition,
but it always gives the content too little; as is thus proved, if the cir-
cumference of a circle be 1 then of 1 is 0,25, which fquared gives
I 44 I
,0625 for the area of the girth but (by rule 16) the true area of fach
a circle is ,07958, fo that if the timbeṛ be baſed, (viz. a cylinder)
the true content will be to that found this way, as ,07958 :,0625
which is nearly as 22 to 18, fo thofe that buy round timber by this
meafure, have nearly one fifth part allowed for chips, and more if the
timber be tapering as you'll fee in ex. 174. But if the diameter of a
circle be 12 inches its area (by rule 15) is 113,0976, whofe fquare
root is 10,635, ſo that if inſtead of 12 on the line D you take 10,635
on it, or by the pen divide by 113,0976, inſtead of 144 you'll have the
true content nearly. But cuftom being fo much in favour with the
erroneous gauge point 12 that it is needleſs to offer any other method.
Ex. 157. If a piece of round timber be 20 foot 3 inches long, and 6
Enches fquare in the middle, how many feet of timber is therein ?
G
50
THE UNIVERSAL MEASURER
Note. That by faying round timber is fo much fquare means that
of the girt is that much.
common way
6
inches {} girt
{{
65 4
4
36 girt
20,25 length
144)729,00 area girt
5,06 feet content
Sliding-rule. 1. As 12 on D: 20
folid content in feet. 2. As I on D:
A as before.
+
{{
decimally
feet {5} 4 girt
,25 area girt
20,25 length
5,0625 feet, content
on A :: 6 on D: 5,66 on A the
20,4 on A :: ,5 on D: 5,06 on
Ex. 158. In a piece of round timber 18 feet long, 0,92 feet girt at
the leffer end, 1,8 feet girt at the greater end, how many ſolid feet?
Here the fum of 0,92 and 1,8 is 2,72 (ſee ex. 154) half of which
1,36 for the girt in the middle.
feet 342 of 1,36
34 girt in the middle
mivt.
,11564 girt
18 feet, length
2,0808 anſwer
Sliding-rule.
As I on D: 18 on A:: 34 on D
: 2,08 on A the anſwer.
Ex. 159. If the length of a walking-ftick be 3,5 feet and an inch
quare, what's the content?
feet 3,5-length
,254 girt in inches
144),875
,0061 feet content
1728 in. in a folid foot
10,5408 folid inches anſwer
Length in feet.
20, 25
18
Sliding-rule. As 12 on D: 3,5
on A : 5 (for 0,5 is off the rule)
on D: 0,61 on A; fo (by ex. 68)
,0061 part of a foot is the anſwer,
After thefe methods the follow-
ing table is calculated.
4 girt in inches.
4.08
8
Content in feet.
5,00
2,08
6,8
TIETE
6,5
32
15
18
за
12
24
14,2
4,4
67,5
72
64,3
AND MECHANIC.
Note. To meafure a growing tree is the fame thing when the di
menſions are taken, which may be had by a ladder and a long ſtaff,
with a cord to take the girt, or you may take the height by problemi
130 &c to 135, and find the girt by problem 162, 163, &c.
To find the folidity of any cube. Fig. 123.
Rule 26. Multiply the fide by itfelf, and that product multiplied by
the fame fide gives the folidity. Theorem 86.
Ex. 160. If each ſide of a cube be 62,06, what is its folid content?
60,2} given fide
3624,04 its fquare
60,02 given fide
218167,208 folidity
Sliding.rule.
As 1 on D: 60,2 on A ::0,602 (for
60,2 is off the rule) on D 21,8 + on
A, ſo (by ex. 68) 218000÷ is the an
anfwer.
To find the folidity of any parallelopipedon, prifm, or cylinder.
Rule 37. Multiply the area of the bafe by the length of the folid,
and the product is the folid content. Theorem 86.
Ex. 111. If the length of a
baſe 7,2, what is the folidity?
7,2} a fide of the bafe
51,84 area of the baſe
If length
777,60 folidity
Ex. 162. If a picce of
prifm be 15, and each fide of its
'
Sliding-rule.
As i on D: 1
on D: 15 on A :: 0,72 (for
7.2 is off the rule) on D: 7.776 on A
fo (by ex. 68) 777,6 is the anſwer.
bafed hewn timber or flone, &c. be 25 feet
long, 9 inches deep, and 25 inches broad, how many folid feet?
common way
inches 25 breadth
9 depth
decimally
2,0833 feet
75
25
by feet and inches
2
I
9
225 area bafe
25 ft. height
1,562475 area bafe = 1 6
144)5625 (39,061875 content
multiply 25
25 0
о
Sliding-rule. 1. As 9 ch D: 9 on A::25
12 6
F O
0 6 3
on A: 15 on D, a mean proportional between
9 and 25. 2. As 12 on D (becauſe 12 is the 39 Ο 9 anfi.
fquare root of 144), 15 on A :: 15 on D: 39,00 folid feet on A the
anſwer. Otherwiſe. See ex. 106. 1. As 0,75 on A : 1 on B:: I on
A: 1,33 on B, 0,75 in a divifor. 2. As 1,33 on A: 25 on B:: 2,083
on A: 39,06 on B the answer.
52
THE UNIVERSAL MEASURER
}
Ex. 163. If there be a triangular prifm, the baſe of its triangulaš
baſe being 131inches, and thereof 12 inches, and the folid's length
zo feet, how many folid feet is contained therein?
common way
13,5 inches
I
6 = 1 of 12
feet and inches
decimally
1,125 feet
II
6
I
I
81,0 area baſe
20 feet height
1,125
I
I 6
I
ΙΟ
10
2
height
11,25 anfwer
II 3 0
144 1620
11,25 anſwer
Sliding-rule. r. As I om A: 1,06 on D:: a mean proportional
between 1 and 1,125. 2. As I on D: 10 (of 20) on A:: 1,06 on
D:11 on A the anſwer. Otherwiſe. 1. As 6 on A: 6 onD:: 13,5
on A: 9 on D, a mean proportional between 9 and 131. 2. As 12 on
D: 20 on A:: 9 on D: 11,25 on A the anſwer.
Ex. 164. If the length of a cylinder be 8 and the diameter of each
bafe 2, 1, what's the folid content ?
་
2, I
2.I
4,41 diameter
,7854
3,463614 arca bafe
3,463614
8 height
27,708912 folidity
Sliding-rule.
,7854 turned to a divifor is 1,2732,
whoſe fquare root is 1,128; therefores
(ſee ex. 68) as 1,128 on D: 8 on A ::
2,1 on D: 27,7+ on A the anſwer.
To find the folidity of any cone or pyramid.
Rule 38. Find the area of the baſe according to its form, by fome of
the rules in fection 4, which area multiplied by a third part of the fo-
lid's axis gives the folid content. Theorem 89.
Ex. 165. If A v CD (fig. 119) be a fquare pyramid, whofe axis
vP is 15, and each fide A BBC = C D D A of its fquare bafe
A B C D 10, 1, what's its folid content ?
10, I
IO, I
102,01 area bafe
5 axis
+
510,05 folidity
Sliding-rule.
As I on D: 5 on A: : 1,01 (for 10,1
is off the rule) on D: 5,1 on A; fo (by
ex. 68) 510 is the anfwer.
AND MECHANIC.
53
1
Ex. 166 If there be a pyramid whofe bafe is a regular heptagon
ach fide whereof is 10 and the axis of the pyramid 68,1, what is its
folidity.
IO
10
100
}
}a fide of bafe
that fide
3 6339 fee ex. 107
363,39 area bafe
22,7= axis
3
8249,08 fondity
Sliding-rule.
Set the fquare root of ,2751 (viz.
3,6339 or 3,6340 in a divifor) on D:
22,7 on A then against 10 on D ſtands
8249 on A the anſwer.
Ex. 167. If the axis v D of a cone v A B (fig. 127) be 68, 1, and
A B the diameter of its bafe 10, what is the folid content?
10
diam. bafe
0}
100 its
7854
78,54 area baſe
v D
22,7
1782,858 folidity
!
Sliding-rule.
See ex. 164. As 1,128 on D: 22,7
on A: 1 (for 10 is off the rule) on
D: 17,83-on A, fo (by examp. 68)
1783- is the anſwer.
Ex. 168. If the cone be elliptical, (viz. of an elliptical bafe) the
greater diameter of the ellipfis 15,2, the leffer 10, and cone's axis 22,
what's the folidity?
15,2
IO
152
22 axis
3344 product
3344
multiply 7854
3)2626,3776
2 =
875,4592 folidity
Sliding-rule. As 1,128 on D: 7,34 on A: 12,38 (a mean
proportional between 15,2 and 10) on D: 875,4 on A the anſwer.
To find the folidity of a fruſtum of any pyramid or cone.
Rule 39. For the fruftum of a pyramid, whoſe baſes are fimilar re-
gular polygons, to a fide of the greater bafe add a fide of the leffer bafe,
from the fquare of that fum take the product of the two fides, multiply
that difference by the tabular number (ex. 107) which reſpects the
form of its baſes, this laſt product multiplied by a third part of the fruf
tum's height gives its folidity, Theorem 107.
Rule 40. To the product of two fides of the two baſes, add part
of the of their difference, this fum multiplied by the tabular num-
ber (as before) and then by the truftum's height or length gives the
folidity. Theorem 106.
54
THE UNIVERSAL MEASURER
Rule 41. Multiply the difference between two fides of the two bafe
by 0,52 that product added to a fide of the leffer baſe, this fum fquare
and multiplied by the tabular number and fruftum's length as before,
gives its folidity nearly. Theorem 124.
Note. Theſe 3 rules hold true in conical fruftums by uſing the di-
ameters of the baſes as fides, and ,7854 as the tabular number, or if
you take the peripheries inftead of the diameters you muſt take,07854
inſtead of ,7854, and if the bafes of the fruftums be not regular poly-
gons, as rectangles cllipfis &c, the two following rules are general.
Rule 42. Multiply the area of the greater baſe by the area of the
leffer baſe, to that product add the faid two areas, this produces the
folidity after being multiplied by a third part of the fruftums height,
let the baſes be in any form whatſoever provided they be fimilar and
parallel. Theorem 108.
Rule 43. To four times the area taken in the middle of the folid
parallel to its two bafes, add the area of each baſe, this fum multiplied
by one fixth part of the folids length gives its folidity. Theorem 138.
or 140.
Note. This laſt rule holds true in all ftreight fided fruftums, whether
of cones or pyramids, as alfo in cylinderoids and prifmoids, and very
near in any folid whatever, as appears in prob. 190, whence this rule
(like rule 25) is in itſelf fufficient for the whole of folid meaſure, and
there is no difficulty in it, but taking the faid middle area, which in
practiſe may be done as eaſily as an area at each end, and in curve-
lined folids it cannot be harder to come at than the figures form which
is to be known before you can find its content by the rule adapted to
ſuch a form &c. But in ftreight lined folids (by theo. 9)half the fum
of
any two like fides at the bafes is equal to a fimilar fide in the middle
between thefe two bafes, fo in theſe folids it is had without meaſuring
In all cafes of folid meafure the length is to be taken perpendicularly to
the breadth or breadths, as is evident from the theory of menfurations.
Part fecond.
Ex. 169. If AbCD be the fruftum of a pyramid whofe axis p
is 24, AB, a fide of the greater bafe 13, and a b, a fide of the lelle
bafe 8, what is its folidity? Fig. 119
AND MECHANIC.
$5
fubtract
First, by rule 39.
13 = AB
add 8
+
21 fun
a b
441fum
104 = 13 X 8
337
8
I
PP
P P
2969 folidity
Secondly, by rule 40.
8
} = { 13
add
Thirdly, by rule 41.
5 difference of fides
0,52 multiply
104 product
84 of 25, differen.
1121/1
24
= p P
PP
2696 folidity
Fourthly, by rule 43.
21 fum of A Band a b
4 times ar. mid.
AB
☐ ab
441
2,60
add
169
8
=ab
64
10.60 mean fide
674 fum
112,36 its ☐
4p P
24 P
= P
2696,64 folidity
2696 foliditÿ
The beſt way to work thefe examples by the fliding-rule is dedu-
ced from rule 43; thus, as I on D: 4 (3 of 24) on A:: 21 (the fum
of 8 and 13) and 13 and 8 on D: 1764 and 676 and 256 on A ref-
pectively, the fum of thefe 3 numbers gives 2696 for the anfwer.
Fach of theſe methods gives the fame content except that by rule
41, which differs from truth by oo00,64, the reaſon is becauſe 0,52 iš
not a true factor nor can there be any one multiplier fixed for this pur-
pofe, as is proved in prob. 187, yet the faid factor never can exceed
an error of ,08, therefore, if you chufe to work by it, it may ferve
with the word nearly.
Ex. 170. Suppoſe the laſt mentioned fruftum to be one of a trigonal
pyramid (viz. a trigon) whofe axis or length p P is 24, â fide of the
greater baſe 13, and one of the leffer 8 as before, to find its folid
content.
3,72 &
Sliding-rule. If you divide 6 by 0,433013 (the tabular number for a
trigon, fee ex. 107) the fquare root of the quotient will be
conftant gauge-point for fuch fruftums; then, as 3,72 on D: 24 the
whole axis on A: : 21 (twice a fide in the middle) and 13 and 8 (a fide
at each end) on D : 3 fuch numbers on A as being added together gives
1167,368 for the anſwer. It is to be noted, that 0,433 may ferve as
the factor in common uſe inſtead of 0,433013, as is done in this ex-
ample.
56
THE UNIVERSAL MEASURER
By rule 23.
13
3∞
8
21 fum
44! its
fubtract 104 product
337
433 tab. number
145.921
8 = axis
+
1167,368 folidity
Ex. 171. If each ſide of the greater baſe of the fruftum of a hexa
genal pyramid be 13, each fide of the leffer bafe 8, and length 241
what's the folidity ?
By rule 41.
0,52 multiplier
5 difference
112,36
mean fide
2.59 tabular number
add
8
2,60 product
a fide leffer bafe
10,6 mean fide
112,36 its
391,91128 product
24 axis
7005,86972 folidity nearly
Sliding rule. By rule 41. As 0,62 (ſquare root of 0.3849, fee ex.
107) on D: 24 on A :: 10,6 (ſee ex. 68) on D: 7005+on A, the
anfwer nearly.
Otherwiſe, as in the laſt ex. If you divide 6 by 2.5984, the tabu-
lar number for an hexagon, the quote will be 2,309, whofe fquare
root is 1,518, a conftant gauge point for fuch fruitums. Then, as
1,518 on D: 24 on A :: 21 and 13 and 8 on D : three fuch numbers
on A as being added together gives 7004,208, (ſee ex. 63)the folidity.
Note. Any of theſe 3 examples foregoing, may be wrought by the
pen, by any, or all of the 5 foregoing rules laſt laid down.
Ex. 172. If there be a fruftum of a rectangular pyramid, whofe
length is 15, breadth at the greater end 2, depth there 1,4. breadth
at the leffer end 1, and depth there 0,7, to find its folid content?
Note. This ex. is the fame with ex. 154, where the content is found
23,625. But here the true content is found 24,5; hence in ex.
154, the breadth and depth (in meaſuring hewn tapering timber
the ufual way) ſhould not be taken in the middle, but fomething nearer
the greater bafe, if you would come near the true content.
AND MECHANIC
57
By rule 42.
I breadth
,7 depth
,7
2.8
2,8 area at greater baſe
2
1,4
1,4 fquare root
areas
2,8
1,96 (1,4 fquare root
I
24) 196
96
2,8
,7
By rule 43,
7} area {greater bafe
leffer bafe
6.34 times area middle
9,& ium
خولها
2,5 length
=
24,50 10lidity
,7. area at lefs baſe
4,9 fum
5 = length
I
3
24,5 folidity
Sliding-rule. If you take 4 times.
the area in the middle and the area
at each end in one fum, the fet
on the rule will be eafy. Thus,
as 6 on A : 15 on B:: 9,8 on
A: 24,5 on B the anſwer.
I
Otherwife. Find 3 mean proportionals, one between 1 and 0,7, one
between 2 and 14, and one between 2,1 and 3, (the fum of the two
breadths 1 and 2, and the two depths 0,7 and 1
and the two depths 0,7 and 1,4 which anſwers to
twice a ſide of a fquare in the middle) then it will be as I on D : 2,5 a
fixth of the length on A: : theſe 3 mean proportionals on D: 3 fuch
numbers on A as taken in one fum gives the content.
Ex. 173. If there be a fruftum A BEF (fig. 127) of a cone whofe
length Dy is 24, diameter A B≈ 13 and E F 8, at the two bafes,
to find the folidity.
First, by rule 40.
13 = A B = 13
8 EF=
8
Secondly, by rule 41.
,52
5
104
5 their difference
2,60
add 8,33
3)25 its fquare
add 8,
EF
112,33
8,33
diff.
.7854 mul,
88,223982
24 length
2117,375508 the content
10,0} mean diam.
112,36 its fquare
mult. 7854
88,247544
24
2117,941054 content
Sliding-rule, by rule 43. Here 6,78547,639 whofe fquare
root is 2,76, a conftant gauge point, conical fruftum, fo as 2,76 on D
: 24 on A:: 8 and 13 and 21 on D: 201,26 and 532,9 and 1383,≈
on A whofe fum is the anſwer viz. 2117,36.
Sliding-rule, by rule 41. As 1,128 on D: 24 on A :: 10,6 on D.
the mean diameter: 2117,9 on A the content nearly.
H
$8
THE UNIVERSAL MEASURER
Ex. 174. Let there be a fruftum of a cone whofe length is 18, and
the periphery at the {greater }
by rule 43.
1,8
leffer
92 § peripheries
}
2,72 per. mid. twice
7,3904 its ſquare
3,24 fq. 1,8
,8464 fq. 92
11,4848 fum
mult. .07958
913960384
3 = length
2,741881152 folidity
baſe {1,92}
what's the folidity?
Sliding-rule. by rule 43.
:
If you divide 6 by 07958 the qua
tient is 75,89, whofe fq. root is near
8,7, a conftant gauge point for coni-
cal fruftums, when the peripheries
of the bafes are uſed; fo, as 8,7 on
D: 18 on A: 2,72 and 1,8 and
,92 on D to 3 fuch numbers on A,
as being added together gives 2,74
the anſwer. This ex. is the fame
with ex. 158, where the content
the falſe way is but 2,08.
To find the folidity of any prifmoid, or of any cylinderoid.
Rule 44. To twice the length of the greater bafe, add the length of
the leffer bafe, multiply that fum by the breadth of the greater baſe;
alfo, to twice the length of the leffer baſe, add the length of the great-
er bafe, multiply that fum by the breadth of the leffer bafe, now the
fum of theſe two products multiplied by 0,785 4 and then by a fixth of
the length gives the folid content. Theorem 139.
Note. Length and breadth of the bafes here means theſe two things
which multiplied together and then by any factor, will give the areas
of the bafes. Hence, if either bafe be a fquare, ora rectangle you need
not multiply by 7854.
* Ex. 175. If there be a cylinderoid whofe length is 20, with two pa”
rallel, and elliptical unlike bafes, the tranfverfe diameter of the greater
bafe 13, conjugate 8, tranfverfe diameter of the leffer baſe 10, and its
conjugate 5,2, what is its folidity?
,7854 factor
mult. 459,6 fum
13
12
2
2
26
20
300,96934
10
1.3
20 = long
36
33
6)7219,39680
5,2
-8
288 1ttprod. 171,6
171,6 2d prod.
499,6 Tum
1203.2328 folidity
Sliding-rule. The fum of the 2 afore-
faid products 459,6 being had the
fet is eafy. Thus, as 1,2743 on A
1 459,6 on B : : ₺ of 20 3,33 on A : 1293,2 on B. Sęc ex. 821
AND MECHANIC.
59
Ex. 176. Let AdCD (fig. 169) be a prifmoid whofe height c P
is 20, the greater baſe A B C D a fq. each fide whereof is 13, the lef
fer baſe abcd, a rectangle, whofe length is 12 and breadth 5, what's
the folidity?
Note. When the bafes are floping as in this fig. the L length of the
folid will fall from a corner, or fide of one bafe upon the plane of the
other bafe produced, and if the bafes are not parallel, it muſt be meaſur-
ed at twice, by meaſuring a hoof &c. from one of the baſes, to make
them parallel, fo that the figure may be a prifmoid &c.
26 twice AD
679 fum product
24
- twice da
12 ad
13=DC
20 CP
38 fum
37 fum
13 = DC.
5=dc
6)13580
494 first product
185 ſecond prod,
2263folidity
494
first product
679 fum
Sliding-rule. As 6 on A: 679 on B:: 20 on A: 2263,3 on B the
anfwer. See the laft.
To find the folidity of a hoof, of a cone or pyramid.
·
Rule 45. For the greater of two elliptic hoofs, from the ſquare of
the diameter of the hoofs baſe, take the fquare root of the product of
the two diameters, multiply the remainder, by the leſs diam. of the hoofs
bafe and its height, and 0,2618 into one another, the laft product di-
vided by the difference between the two diameters, gives the folid
content. But for the leffer hoof, multiply the ſquare root of the pro-
duct of the two diameters, by the greater diameter, from that product
take the fquare of the diameter of the hoofs bafe, multiply the remain-
der, the diameter of the hoof's bafe, and its height, and 0,2618 intớ
one another, the laſt product divided by the difference between the
two diameters, gives the folidity. Theorem 116.
Ex. 177, and 178. Required the folidities of the two elliptical hoofs
TSn (fig. 136) and TS m, the heights being SR24, diameters
Tn 13, and 5 m = 8.
ift, For the greater hoof TS n. 2d, For the leffer hoof T Sn.
13 = T n
8 = Sm
104 product
10,198 fquare prod. 8 and 13
13 = Tn
132,574 product
60
THE UNIVERSAL MEASURER
10,198 fquare of 104
8 Sm
81,584
132,574 product
64. =fquare S in
68,574
fubtract 169 fquare of 13
multiply
8-Sm
87,416.
548,592
multiply 13 Tn
multiply 24 SR
1136,403
13166,208
multiply 24 SR
multiply ,2618
5)3446,75+
689,35+
27273,792
multiply 2618
5)7140,1+
1428,02 + folidity gr. hoof
The ſliding-rule requires ſo many mean proportionals (fee ex. 82.)
that it is eaſier by the pen.
Note. The dimenſions here are alſo thofe in ex. 173, where the
fruftum is found 2117,37+, and here
the
greater
leffer
} hoof is
{ 1428,02 +
689,35 +
proof, fum 2117,37 †
Ex. 179, and 180. If Tmsn (fig. 136) be the fruftum of a ſquare
pyramid whofe height SR is 24, a fide T n of the greater baſe 13, and
a fide s m of the leffer baſe 8, the folidities of the two hoofs TS n and
TS m, may be found (by theorem 115) thus, to 26 (twice 13) add
8, multiply the fum 34 by 13, and the product 442, multiplied by 4
(a 6th of 24) gives 1768 the folidity of the greater hoof. Alfo, to 16
(twice 8) add 13, multiply 29 that fum by 8 and 232 that product
multiplied by 4 (a 6th of 24) gives 928 for the folidity of the leffer
hoof, by comparing this with ex. 169 you'll find theſe two hoofs to
make juſt 2696, the whole fruſtum.
Sliding-rule. 1. As 24 the length on A: 6 on B :: 1 on A :0,25
on B. (See ex. 106) Then, As 25 on A: 13 on B:: 26 (twice 13)
and 8 on A: 13,53 and 4,15 on B whoſe fum is 17,68. And, as 25
on A: 8 on B: 16 (twice 8) and 13 on A: 5,12 and 4,16, whoſe
fum is 9,28, each of thefe multiplied by 100, becauſe we uſed the di-
vifor 25 instead of 0,25 (See ex, 68) gives 1768 and 928, for the two
hoofs as before. The parabolic and hyperbolic hoofs may be had by
theorems 117 and 118, or any hoof nearly by rule 43. But as hoofs
are of little ufe in practice except the elliptic ones in finding the drip of
a tun &c. I think theſe fufficient.
To find the folidity of any of the five regular bodies.
AND MECHANIC.
GI
Thefe bodies being made up of pyramids whofe vertexes all meet
in the center of the body, it is but finding the folidity of one of their.
pyramids (they being in each body equal) and multiplying it by the
number of pyramids which compofe the body, fo have you its folidity,
and thus are the folidities got in the table. (ex. 131) whoſe uſe is
Rule 46. (by theorem 37) Multiply the cube of the fide and the ta-
bular folidity belonging the body, the one into the other fo you have
the folidity.
Ex. 181. If each fide of a dodecaedron be 20, what's its folidity?
7,66312 tab. number
8000 cube fide
61304,96 1olidity
Sliding-rule. Ift, 7,66+ turned to a di-
vifor is 0, 1305, whofe fquare root is
0,36; then, as 36 on D: 20 on A ::
20 on D: 6,13 on A, fo (by ex. 68) 61300+ is the anſwer, &c. for
any other of theſe bodies.
To find the folid content of any globe or ſphere.
Rule 47. As I: 0,5236 :: 21: 11 (nearly) fo is the cube of any
globe's axis to the globe's folidity, theorem 93, or multiply the circum-
ference and fquare of the axis, one by the other, a 6th part of that
product is the ſolidity. Theorem 134.
Ex. 182. If A P the axis of a globe (fig. 170) be 20, what is the
folidity?
,5236 conſtant factor
8000 cube of axis
4188,8 folidity
Sliding-rule. As 1,382 (the fquare
root of,) on D: 20 on A:: 20 on
D (fee ex. 68): 4188,8 on A the anfwer.
To find the folidity of any fegment of a globe.
Rule 48. From three times the axis of the whole globe, take twice
the fegments height, multiply the remainder, the fquare of the fegments
height, and 0,5236 into one another, the laſt product is the folidity,
this is when the fegments height and axis of the globe are given, but
when the fegments height and diameter of its bafe are given; then, to
three times the fquare of the half diameter of the fegments baſe, add
the fquare of its height, multiply that fum, and the ſaid height and
0,5236 into one another, the laſt product is the folidity. Theo. 93.
Ex. 183. Let K PL (fig. 170) be the fegment of a globe, whoſe
height P D is 4, and K L the diameter of its baſe 16, or A P the
globe's axis 20, to find the folidity.
62 THE UNIVERSAL MEASURER
ift, When K L and DP are given. 2dly, when AP and PD are given.
8
KD-LD
64 the fquare of it
20
axis P A
· 3
3
60
192 3 times that ſquare
fub. 8
82 DP
add
16 fquare of PD
52
208
16 =
fquare DP
4 PD
832
832
05236 factor
435.6352 folidity
,5236 factor
435,6352 folidity
4 on D
Sliding-rule. As 1,382 on D: 4 on A :: 8 and
134,04
and 33,51 on A, fo 3 times 13404 = 402,12 added to 33,51 gives
435,63+ the anſwer, (fee ex. 91) by wnich it will be eafy to fet the
fecond operation alſo.
To find the folidity of a fruftum, or middle zone of a globe.
Rule 49. To twice the fquare of the axis or greateſt diameter, add
the fquare of the leaſt diameter, multiply that fum, and the length of
the zone, and 0,2618 into one another, the laſt product is the folidity.
Theorem 91.
Ex. 184. Required the folidity of the half zone HK K (fig. 170)
whofe length C D is 6, diameters HI 20, and K L 16.
20=HI
400 its ſquare
2
800 twice fquare HI
256 fquare K L
add
1056
,2618 factor
276,4608
6=CD
1658,7648 folidity
Sliding-rule. Firft, 0,2618 turned
to a diviſor, and then the ſquare
root taken is 1,954; fo, as 1,954
on D: 6 on A:: 28,2 (√2 × 20)
and 16 on D: two fuch numbers
on A, as being added together
gives 1658,76 the anſwer.
Note. The fame rule ferves for
the fruftum or middle zone of a
fpheroid.
If the folid be leſs than a half zone as the part K L n m. (fig. 170)
Then,
Rule to. To half the fum of the fquares of the two diameters, add
to the fquare of the height, that fum multiplied by the height, and
by 0,7854 gives the folidity. (by theo. 141) But if it be the like part
of a spheroid, you must multiply the faid of the fquare of the height
by the fquare of the fpheroids greateft diameter, and divide that by
the fquare of the spheroids axis, and then add the quotient and work
as before. Theorem 141.
AND MECHANIC.
63
is 30,
Ex. 185. Suppofe fig. 170 to be a ſpheroid whofe axis A P is
greateſt diameter H I 20, K L the greateſt diameter of the part K L nm
to be meaſured 16, leaft diameter n m 1c, and height D E 4, to find
the folidity, of K L n m.
add
256 fquare K L
100 fquare n m
2)356
178 half fum
4.74
182,74
4 = DE
730,96 product
mult.
16 fquare DC
32
2
400 fquare 20
9|00)128|00 900=fquare 30
3) 14222
3
4,74 = fquare 4 multi-
ply fquare 20, and divide fquare 30.
730,96 product
>7854 factor
574,095984 folidity
The pen is eaſier for this example than the fliding-rule.
Ex. 186. If fig. 170, be a globe, and the folidity of the part K Ln m,
be required, K L, nm and D E, being as m the laſt example. Then,
178
add
1066
10 66
188,66
fum fquares K and nm
754,64 product
:7854 factor
=iquare D E
522,69+ folidity
Sliding-rule. As
1,128 (fee ex. 164)
}
4 = =DE
754,64 produc
on D: 4 on A :: 16 and 10 and 4 on
D: 3 fuch numbers on A, that if to the fum of the two firſt you
the laft, that fum will be 592,69 the anfwer. See ex. 91.
22
add
To find the folidity of any ſpheroid, right or oblate.
Rule 51. Multiply the axis, the fquare of the greateſt diameter,
and 0,5 236 into one another, the laft product is the folidity. Theo 92.
Ex. 187, and 188. If the axis of a right ſpheroid be 30, and diam-
eter of its greateſt circle 20, or the axis of an oblate fpheroid 20, and
diameter of its greateſt circle 30, what's the folidity of each.
Ift, For the right spheroid.
400 diameter
=
30 = axis
13000
5236 factor
6283,2000 folidity
Sliding-rule. (See ex. 182), 1. As
2d, For the oblate fpheroid,
900
diameter
20 = axis
18000 B
,5236 factor
9424,8000 folidity.
1,382 on D: 30 on A :: 20 on
D: 6283,2 on B the right fpheroid. 2. As 1,382 on D: 20 on A
:: 30 on D: 9424,8 on A the folid content,
64
THE UNIVERSAL MEASURER
To find the folidity of any fpheroidical ſegment.
Rule 52. With the axis of the fpheroid, and height of the fegment,
find the folidity exactly by rule 43, then it will be as the fquare of the
faid axis is to the fegnent thus found, fo is the fquare of the fpheroids
greateſt diameter to the folidity required. Theo. 90.
Ex. 189. Suppoſe K L P (fig. 170) to be the ſegment of a spheroid,
(either right or oblate) wherein P D = 4, D L = 16, A P=20, as in
ex. 183 and let H I the greatest diameter of the ſpheroid be 15, what
is the folidity. 1. By Ex. 183 the folidity of the fegment, if it were
the ſegment of a globe, is found 435,6; then as 400 (□ 20) is to
435,6 ſo is 225 (15) to 245, the folidity fought. By the fliding-
rule. As the axis 20 on D: 435,5 on A:: 15 the greateſt diameter
on D: 245 on A the anſwer. The middle zone of a ſpheroid is had
exactly by rule 49.
—
Ex. 190. Let the fame things be given as in ex. 184, then by the
fliding-rule. As 1,954 on D: the height 6 on A :: the two diameters
29 and 16 on D: two fuch numbers on A; that if to twice the first
you add the fecond, the fum will be 1658,75 the anſwer, and is a
better way then that, by the fliding-rule to ex. 184, for a part of this
zone. See Ex. 185.
To find the folidity of a parabolic conoid.
Rule 53. Multiply the area of the bafe by half the axis. Theo 87.
Ex. 191. If Eu F (fig. 139) be a parabolic conoid, whoſe axis u P
is 60, and E F the diameter of its baſe 40, what's the folidity?
40 EF
1600E F
EF
30 UP
48000
7854 factor
37699,2 folidity
Sliding-rule.
As 1,59 (the fq. root of one
divided by half,7854) on D: 60
on A :: 4,0 on D (see ex. 68)
37699 on A, fo 37699 is the
anſwer.
To find the folidity of a parabolic conoid's fruftum.
Rule 54. To the fquare of the greateſt diameter, add the ſquare of
the leaſt, multiply that fum and the conoid's fruftum's height, and
0,3927 into one another. Theorem 102.
Ex. 192. If E FTS (fig. 139) be the fruftum, whofe height IP is
38,4 diameters EF 40, and S T 24, what's the folidity?
AND MECHANIC.
65
1600 EF
add 576 =OST
2176
mult. 38,4 = IP
83558,4
›3927 factor
32813,38368 folidity
Sliding-rule.
As 1,595 (the fq. root of,
on D: 38,4 on A :: (See ex 68)
40 and 24 on D: 24127 and 8686
on A whoſe fum is 32813 the anf.
To find the folidity of a parabolic fpindle.
8
Is
Rule 55. Every parabolic fpindle being of its circumfcribing
cylinder, multiply 0.41888 (viz. Ts of 0,7854) and the axis, and the
fquare of the greateſt diameter, into one another and the last product
is the folid content. Theorem 101.
Ex, 193. IfQVQv (fig. 140) be a parabolic fpindle, whofe length
QQ is 60, and V v the diameter of the greateſt circle 40, what's the
folidity?
1600 fquare Vv
mult. 60 axis ee
96000
$41888 factor
40212,48, folidity
•
4,0
Sliding-rule. As 1,545 (the fquare root of
10,41888) on D: 60 on A:
on D (fee ex. 68.): 402,1248 on A,
fo 40212,48 is the answer.
To find the folidity of the fruftum of a parabolic fpindle.
Rule 56. To twice the fquare of the greateſt diameter, add the
of the leaſt diameter, from that fum take of the ſquare of their dif-
ference, multiply the remainder and the length, and 0,2618 into one
another, the laſt product is the folidity. Theo. 100.
Ex. 194. If M M m m (fig. 140) be the middle zone of a parabolic
fpindle, whofe length n n is 40, greateſt diameter V v 32, and leaſt
diameter M m = 24, what is its folidity?
32 V v
32 Vv
1024 fquare Vv
24 M m
fub.
2
8 difference
2624 fum
25,6 0,4fq. differ.
2598,4
40 na
2048 tw. that fq.
add 576 fquare M m
2624 fuma
64 fquare diff.
0,4
103936,0
25.6 fq. diff.
8162, factor inverted
207872
62361
1039
832
27210,4 folidity
I
66
THE UNIVERSAL MEASURER
Sliding-rule. As 1,954 on D: 40 on A;: 32 and 24 and 8 on D ;
10723,3 and 6031,9 and 670.2 on A refpectively, ſo twice 10723,3
21446,6 added to 6031,9 gives 27478,5 from which take of
670,2268,08, and there leaves 27210,42 the folidity.
Ex. 195. Let it be required to find the content of the laſt mentioned
fruftum by rule 43; in order to do this we muſt have a diameter in the
middle between V v and Mm, which by theo. 65, or by menſuration,
is found 30, becauſe rule 43 is general for all folids, from that rule
may be deduced this general one for meafuring any folid by the fliding-
rule, viz.
1024 = ſquare V v 32
3900 fquare twice 39
576 fquare M m 24
5200= fum
.1309 one fixth of 0,7554
6306,8
=
40 =nn
27227,2 folidity
27210,4 the true folidity
16,8 difference
Rule 57. As 2,764 the fquare root of 6,7854 is to the height or
length on A fo is a diameter taken at each end and one in the middle,
on D, to 3
fuch numbers on A, that their fum is the anſwer. Or if for
the diameters you take the peripheries, then the gauge point is 8,7 (See
ex. 174) to be uled with the peripheries as 2,764 is with the diam.
Ex 196. Let ex. 183 be wrought by rule 43, where a diameter
mn taken in the middle between D and P is 9, height D P 4, greateſt
diameter K L 12, lealt diameter at P = 0.
144 = fquare K L 1 2
324 fq. twice n m 9
=fq. diameter PO
468,0 fum
9031, inverted factor
4630
1404
000
42
61,26
4 = DP
245:04 fulidity
Sliding-rule. By rule 57. As 2,764
on D: 4 on A :: 12 and 18 (twice
9) on D: 75,04 and 170 on A whoſe
fùm is 245,04, the fame as turns out
by the particular rule ex. 183, which
rule 43, will alfo give the true folid-
ity of any of the folids in this fection
except ex. 177, 178, and 194; and
in thefe the error would be lefs than
what might arife from gueffing at the
form of the folid; for in the laft you
fee the error is but 16,8 in fo great a
number as 27227,2, Ijudge it need-
efs to give any rules &c. for the circular and elliptica! fpindles, the
AND MECHANIC.
fecond fegments, flices, &c. which may be eafily done from their re
fpective theorems, if curiofity require it, practice will not call for any
ſuch difficulties, if what is faid of rule 43 be obſerved.
SECTION VII. Of Surveying.
XXX
What is here meant by furveying, is the meafuring, plotting map-
ping or protracting, and dividing of ground, and if what goes before be
underſtood this will be ealy; for in prob. 129 the chain and other in-
ftruments for this purpoſe are defcribed and applied to practice. Alſo,
land being always computed by fuperficial meafure, you have that done
already in fest. 4; where befides the general method rule 25, you
have general rules for all the uſeful formas of figures particularly ex-
emplified, fo that let a field be in what form it will, its content may be
had by ſome of thefe rules. In this fection I fhall give fuch examples
wrought by the faid rules, as are molt ufeful in furveying, and be more
tedious in mapping dividing. &c.
A Table of Long Meaſure.
| Link | Foot
Inches
Yard
7.92
12 36
Links 1.515
4,56
Perch | Chain | Miles
198
792
63360
25
100
8000
Feet
3
|
16,5 | . 65.
5280
| Yards
5:5 1 22
1760
| Perches!
4
320
| Chains
801
That is, 36 inches, or 4,56 links, or 3 feet, make 1 yard; alo
*92 inches = 100 links = 66 feet : — 22 yards
22 yards=4 perches, each =
I chain, &c. Chains and links are fet down and wrought in chains and
decimal parts of a chain. See prob. 129, deff. 21.
Examples in Long Meafore.
Ex. 197. In 15 chains 25 links,.
chains 25 links, how many feet?
15,25
66 feet in one chain
1006,50 feet anſwer
Ex. 198. In 29 chains how many feet?
29
66
1914 feet, anfwer
68 THE UNIVERSAL MEASURER
A TABLE of Square Meaſure.
Ex. 199.
In
27,30
4
chains how many poles roods or perches
Ex. 200. In 2580 links how many chains?
109,2 roods &c. anſwer.
100)2580
Ex. 201. In 25,8 chains how many links?
25,80 = 25 chains 80 links
100
25,8
2580 links, anſwer
Note. Links are turned into chains by pricking off 2 decimals-
Inches
Inches 1
Links
Links
62,764
I
Feet
Feet
144
2,295
I
Yards
Yards 1296
Perches 39204
Chains 627346
Acres 6272460
Miles 4014480600
120,755
19
I
Perches
625
10000
272,25
130,25
I
Chains
4356
1484
16
I
Acres
100000
43560
4840
Їбо
10 I Miles
4000000 2787400
3097600 102400 63001940
I
This table reads t' us, I inch = 1
inch, 62,764
inches=1
link, 144
inches 2,295
links1
foot, 1296 inches 20,775 links
9 feet
= 1 yard, &c.
Examples in fquare meaſure.
Ex. 202. In
1671925 fquare links, how many fquare chains?
10000) 1671925
167,1925 = 167 chains 1925 links anf.
Ex. 203. In 1232,52 fquare chains, how many fquare links?
10000
12325200 ſquare links anfwer
Ex. 204. In 152 ſquare
links, how many ſquare chains ?
,0152 fquare chains anſwer.
AND MECHANIC.
69
£x. 205. În 152 ſquare links, how many fo parts of an acre?
,0152 acres, or rather parts of an acre anfwer.
Ex. 206. In 1278642 fquare links, are 12,78642 acres.
Note. Square links are turned into fquare chains, by pricking off 4
places towards the right hand for decimals, or fquare links, and the
reſt are ſquare chains; alſo ſquare chains are turned into ſquare links
by annexing four cyphers, and acres into fquare links by annexing 5
cyphers, or fquare links into acres by the contrary viz. pricking off 5
decimals, and fquare chains into acres by pricking off one decimal, or
removing the decimal point a place nearer to the left hand, &c. fo,
Ex. 207, In 2560276 fquare links, or 256,0276 fquare chains.
are 25,60276 acres &c.
Ex. 208. In 20 acres, how many ſquare yards?
4840 fquare yards in an acre
20
anfwer 96800 yards
Ex. 209. 99220 fquare yards, how many acres?
4840) 99220 ( 20,5 acres anſwer
Ex. 210. In 358,75 fquare chains, how many acres?
358,75 chains
acres 35,875
4roods in one acrè
roods*
3.500
40 poles in one rood
roods, poles or perches 20,000
Ex. 211. In 9,9 ſquare chains, or in 0,99 acres, how many roods
Ex. 212. In 172
yards to the pole.
fo is 121 (fq. 11,
cuſtomary.
As
and perches?
0,99
4
acres
8,96 roods
40
38,40 poles
ftatute acres, how many cuſtomary acres of 6
144 (fq. 12 the yards in 6 yards) is to 172
yards in a ſtatute pole) to 144,72 acres
the
I
76
93
THE UNIVERSAL MEASURER
fo 76
4 roods in one acre
144) 304 (2 roods
288
16
40 poles in any acrè
144) 640 ( 4 poles
576
64
Anſwer 144 acres 2 roods 4 +44 poles, at 18 foot or 6 yards to the
64
141
pole, are to 172 acres ftatute at 5
yards per pole.
Note. Some land meaſurers, work altogether by acres roods and
perches, but it is readieſt to work decimally, and to ſet down chains
and links as a mixed decimal number (vide def. 21, problem 129.) as
you'll find done in this fection.
The two foregoing tables and thefe examples, refpect the ftatute
pole, of 5 yards in length. But in feveral parts of England there are
poles of different lengths, called cuftomary, fome 18, fome 21, fomet
24, &c. yards in length; but in all places 4 roods make an acre,
-and 40 fquare poles a rood, fo that the différence is in the pole only;
now to reduce one of theſe meaſures to the other, obferve this general
Rule 58. As the ſquare of the feet in a cuſtomary acre is to any num-
ber of ſtatute acres fo is the fquare feet in a ſtatute pole to the cuſtom-
ary acres, as by ex. 212.
In Ireland 7 yards in length is their ftatute pole or perch, in fome
places the inhabitants know not the length of their pole, but ſay that fo
many gallons of oats will fuffice one of their acres for feed. Now in
many places it is reckoned that 60 gallons of oats will fow a ſtatute,
or plantation acre. Therefore, it will be as 60 gallons it to 121 (the
fquare of 11, the yards in a ſtatute pole) fo is the number of gallons
that ſow a cuſtomary acre, to the ſquare of the yards in that cuſto-
mary pole, as by the laſt rule.
Ex. 213, If 97 gallons of oats be fufficient for fowing fome cufto-
mary acre, how many yards are to the pole of length in that acre?
gall.yards gall.
If 60
121
97
6|0) 11737
97?
195,6166-whofe fquare root is 14
nearly, the half yards required.
AND MECHANIC.
梦​等
​Ex. 214. If 120 gallons fow an acre, how many yards to the pole
in length in that acre ?
If 60 121
120?
120
60)14520
242 (15,5 yards anfwer
I
25) 142
125
305) 1700
1525
175 remains
Ex. 215. What proportion doth an Engliſh ſtatute acre, bear to an
Iriſh ſtatute one? Firſt the half yards in an Engliſh pole are II whofe
fquare is 121, and in an Irith pole are 14 half yards, whofe fquare is
196; therefore, As 121 : 196 :: an Engiſh to an Iriſh acre, which
nearly as 11 to 17.
is
Ex, 216. In 52,04 cuſtomary acres at 6 yards to the pole, how
many ftatute acres?
Firſt,5,530,25 yards. Second, 636 yards.
Then, If 30,25
36
52.04
□
52,04
30,25) 1873.64 (61,9 + answer. by rule 58
2839 remainder
Theſe are the principle examples in reduction of land meaſure, if
any other cafe is wanted, it may alſo be done from the two foregoing
tables
Of meaſuring Land.
Ex. 217. If the fide of a fquare field, be 7 chains 60 links, what,
acres roods and perches are contained in that field?
Ex. 218. In a long fquare piece of ground 10 ch. 75 links long, and
2 chains broad, how many acres roods and perches?
72
THE UNIVERSAL MEASURER
(217)
(218)
7,62
7.65
by rule
10,75
2
75} by rule
8
57,76 fquare chains
21,50 fquare chairs
5,776 acres
4
3,104 roods
40
4,160 perches
anf. 5 acres 3 roods 4,16 perches
2,15 acres
4
,60 roods
40
24,00 perches
anf. 2 acres 24 perches.
Note. When chains and links are multiplied together, if you cut 4
figures off towards the right hand, they will be ſquare links or decimal
parts of a chain, but 5 figures pricked off are acres and decimal parts
of an acre.
Ex. 219. If a field in form of a rhomboides, length 2 chains I link,
perpendicular breadth 90 links, how many acres, roods, and perches?
F.x. 220. In a triangular field whoſe baſe is 7 ch. 10 links, perpen--
dicular 2 ch. 30 links, how many acres, roods and perches ?
(219)
2,1} by rule 9
1 by rule 10
(220)
3,55 = half 7,1
2.3
,8165 acres
1}
4
,189 acres
4
756 roods
40
30,240
Anſwer, o A. o R. 30,24 P.
3,266 roodɛ
40
10,54 perches
Anfwer, o A. 3 R. 10,64 P.
Ex. 221. In a trapezia, whofe diagonal is 28 ch. 20 links, perpen-
diculars 10 ch. 50 links, and 8 ch. how many acres, roods, &c, ?
10,5 by rule 12.
8
18,5 fum of the perpendiculars-
14,1 half the diagonal
26,085 acres
4
,340 roods
40
13,600 perches
Anſwer, 26 A. o R. 13,6 P.-
AND MECHANIC.
73
Ex. 222. In a trapezia with 2 parallel fides 3202 links and 40681,
and the perpendicular diftance between them 2500 links, and the other
two fides 3400 links and 2800 links, how many acres, &c.?
Firſt way.
40,08
30,02
}
two fides
2)72,10 fum, (by theo. 20)
Second way.
28 >
oth er two fides
34
36,05 half fum
25. perpendicular
88,125 true area
2:02 fum
31 half fum
mul. 36,05 half fum of the fides
111,755 falfe area
fubtract
88,125 true area
23,620 error
Some Surveyors, do no more with 4 fided fields but multiply by half
the fum of two oppofite fides, by half the fum of the other two oppoſite
fides, for the area, which is a grofs error, and always gives the
content too much, as appears by the laſt example wrought both ways,
and fhews that the error in 88 acres is 23 acres, the true way is to take
a diagonal and two 1s, or if any two of the 4 fides be parallel, it is
equally true, (by theo. 26) if you multiply half the fum of thefe two
parallel fides by their diſtance afunder, as per laft ex.
Ex. 223. If the diameter of a circle be 20,02 yards, required its
area in acres roods and perches. Here, you may find the content in
ſquare yards, and then reduce it to acres &c. or you may turn the
diameter from yards to chains and links, thus. As 22 yards is to I
chain or 100 links fo is any number of yards to the chains and links
anſwering, then find the area by rule 15. See both ways.
Firft.
20,02} yards diameter
400,8004 its fquare
7854 factor
314,78863416 area in yards
Here, becauſe 314,79 is nearly
I uſe 314,79 inftead thereof.
4840)314,79(0 acres
4
4840)1259,16(o roods
40
4840)50366,40(10,406 percheg
314,78863416 the area in yards,
K
74 THE UNIVERSAL MEASURER
Secondly. If 22 100 20,02
100
22)2002
91 diameter in links
-8281 its fquare
E
,7854 factor
है
6503,8974 area in links
,65038974 area in chains
,065038974 area in acres
4
,260155896 roods
40
10,406235840 perches
Ex. 224. To meaſure hills, vallies, &c. (fig: 104) meaſure round
the bottom z y of the hill, which ſuppoſe 15,88 chains, and round its
fummit or top TD 10,12 chains; then take the length of the longeſt
fide z D 14,1 chains, and alfo the length y F 10 chains of the ſhorteſt
fide, half the fum of theſe two fides is 12,15 chains for the mean
length; alſo half the fum of the two peripheries 15,88 and 10,12 is
13 chains for the mean periphery: then 12,15 × 13 = 157,95 fquare
chains 15,795 acres, the fuperficial content of the hill, which is true
if the peripheries were taken parallel to each other, and the hill not
too rugged; but if it be very uneven, you ſhould take more peripheries
and work by rule 24 or 25. This way will do for ſmall hills, but for
large ones it is better to meaſure them as planes, as taught in the fol-
lowing examples.
Note. What is faid of hills alfo holds true in vallies, they being but
as hills furveyed in the infide, but in juſtice the content of a hill or
valley, ſhould be no more than the plane on which it ſtands, for fince
grafs &c. always grows to the horizon, its evident that if you ſup-
poſe a hill to be raiſed upon any plane, that the fame quantity of grafs
on the plane, will by fprouting thro' the fides of the hill be no ways in-
creaſed, becauſe growing upright, it will be a flope to the fides of the
hill.
Ex. 225. Let ABCD (fig. 171) be ſome piece of bending ground
whoſe content is required. Firſt, meaſure along its middle E F, taking
breadths at right angles thereunto as you go on, and ſuppoſe the dimen-
fions to be as fet down in the figure: (ſee deff. 30, prob. 129.)
AND MECHANIC.
75
number of breadths
breadths
1,2
I
2
3.
4
divide by num. breadths
See rule 24
multiply
area in acres
2,2
3,08
4,02
4)10,50 fum
2,625 mid. breadth
16,71 = EF
4,386375
Note. In fuch figures as this, where the ends flope, &c. fuch ends
muſt not be taken for breadths, becauſe they are not fquare to the
lengths FF, in fuch cafes the breadths at ends muſt be taken within
the ends as you ſee from B and D, for the length E F, if it be taken in
the middle, it gives the true mean length, however the ends flope, pro-
vided they be ftreight. This method in fuch forms holds pretty true,
and is much eaſier than to divide fuch figures in trapezia's, As, &c.
See ex. 119.
Thefe examples may fuffice on this head, the next is to take the
plans &c. of fields.
To take the plan of any field by the plane table.
Ex. 226. Let it be required to take the true form of the field
ABDEFG, (fig. 172) upon paper.
1. Caufe marks or ſtaffs to be fet up at every L or corner of the
field, as at A, B, D, E, F, G, and alſo if there be any place H, in the
field that you would have in your map, caufe a flick to be fet up there
at H alfo.
2. Chufe any point C, in the field, from whence you can efpy all
the marks fo fet up, there place your inftrument parallel to the horizon,
with a ftreight ruler, or index moveable upon its center, under which
index, upon the table muſt be faſtened a ſheet of white paper, then
thro' the fights upon the index, or along its ftreight edge, efpy every
mark one by one, and meaſure the diſtance between the center at C, of
your table and each mark, and from a diagonal ſcale from C, along the
edge of the ruler pointing to the faid mark, lay this meaſured
diſtance, this do for every corner in the field, and where the diſtances
end on your paper as at A, B, D, E, F, G, join with lines, fo will
you have a true plan of the field before ever you go out of it, this me-
thod of taking a plan by the plain table (if the weather is dry fo as not
to wet the paper) is as eaſy and expeditious as any, for in a manner,
you have the map on paper, and the dimenfions all at once; your own
judgement will direct you, that fuch places as H, whether brook, tree,
1
76 THE UNIVERSAL MEASURER
houfe, &c. is not to be joined in with the marks which compofe the
out, or ring hedge A BDEFG, and if this hedge be much bended
you must take fo many more marks or take offsets between the prin-
cipal marks as in ex 236.
To take the plan of a field by the femi-circle, or theodilite, at one
itation.
Ex. 227. Required the true map of the field A BDEF, fig. 173.
1. Having fet up marks at every angle in the field, and at all the
objects you would have in your map, at any point C, where you can
fee all theſe marks, place your inftrument flat, and fcrew it faſt ſo that
along the diameter where the degrees begin you may elpy any one of
the marks, fuppofe that at A, then turn the index efpying every of the
other marks thro' its fights obferving what angles are made, or degrees
cut, by looking at thefe marks (the fame as in prob. 137) which let
be as fet down in the figure.
2. Next meaſure the diftance between C, the center of the inftru-
ment and each mark, which let be as fet down along the lines, CA,
L B, &c.
To lay down this plan.
1. With a chord of 60° fweep a circ le, on whofe periphery lay the
chords of 74°, 90°, 62°, &c. as per figure, and by these chords draw
the lines CA, CB, CD, &c. on which from a diagonal fcale lay 4 C,
3,8 C, 3,3 C, &c. and where theſe ends as at A, E, D, &c. draw the
black lines A B, BD, DE, &c. fo have you a true plan of your ob-
fervations.
Ex. 223. If you meaſure but one diſtance as C A, and meaſure all
the hedges A B, BD, DE, &c. it will be the fame for to lay down the
plan by thefe dimenfions; firft lay down all the angles at the center C,
as before, and the line C A, then from the fame diagonal fcale take the
line A B, and with one foot in A crofs the line C B in B, join A B, alſo
with the line BD in your compaffes and one foot in B, croſs the line
CD, join B D, and thus go on till all the hedges be laid down.
To take the plan of a field at one (lation by the chain only.
Ex. 229. This is the fame as before, only you take all the Ls at C,
with the chain inſtead of the theod. &c. thus, fet down a ſtaff at O, over
which put the ring at the end of the chain, where ſtand, and let your
afliftant go with the other end towards A, till you fee him and the mark
A, in the fame right line, there (tick down a ſtick which is atb (fig. 173)
then dirc&t him towards B, and there fet down another flick which is
AND MECHANIC.
77
at a, then meaſure the diſtance a b, which let be 1,2 C, fo is the angle
A CB meaſured by the chain, and thus you may meaſure all the angles,
at the ftation C.
To lay down this plan.
1. From the diagonal ſcale take one chain, with which in your com-
paffes and one foot in C, fweep a circle on whofe periphery from the
fame diagonal ſcale lay down all the Ls as you fee them in your rough
draught (def. 30. prob. 129) and thro' theſe points draw the line CA,
C3, CD, &c. (meafured in the field as per ex. 227) at pleaſure, on
which from a diagonal feale lay their lengths &c. and join A B, B D,
DE, &c. ío have you a true map of your obfervations, or you may
measure out one line between your ftation C, and any of the marks
A, B, D, xc. and meafare all the hedges, as directed in the laſt ex.
To take the plot of a field at two ſtations, by only meaſuring the
diffance between the two ftations, and obferving the Ls.
Ex. 230. Let A B C D E (fig, 174) be a field to be mapped.
I.
Chufe two convenient points F and S for your two ſtations, then
work as in problem 139 which is the very fame with this example, only
here for variety, I take the two (tations within the field, which ſhould
be pretty far afunder, that fo the lines of obfervation F A, FB, &c.
and SA, SA, &c. may not meet in very fmall angles A, B, &c. for if
they do, it is not fo plain to fee where the angular points are.
2. Place your inſtrument as at F, letting the diameter point to S your
ſecond ſtation, there having it flat, fcrew it faft, and thro' the fights
upon the index efpy all the marks or corners of the field E, A, B, C, D,
marking down the angles of obfervations which let be S FE 10º EFA
84º, A F B 73ɔ, B F C 1453, C F D 46°; then meaſure the ſtatio-
nary diftance F S, which fuppofe 80,2 C, there at S place your inftru-
ment and make obfervations to F, and every mark as before, and fup-
pofe the Ls ASF 422, ASB 442, BSC 80°, CS D 840, DSE 540,
and ES A 56°.
X
To lay down this field on paper.
1. On a large ſheet of paper draw a line S F, on which from the
diagɔnal ſcale, ley the ſtationary diſtance 80,2, then on F and S, with
the morda 6ɔ' freep two circles, make LSFE 10° drawing F E
at pleafure, ɔn 3 make the L (ES A+A SF) FS E — 96°, drawing
$5 to mex FE in 3, fo is E one of the marks or corners in the field,
-
agio make ↳ E FA
L
A, another mark &c.
=
84°, and L ESA 56°, meeting in A, fo is
and thus go on till you have all the corners or
marks, which join with lines E A, A B, &c. and its done. This ex.
78 THE UNIVERSAL MEASURER
requires that when you are at either ftation, you must fee the other
ſtation and all the marks, but when that cannot be done, you may fet
lines from your ſtation to the unfecn marks &c. leaving ftaffs fet up in
thofe lines to go by. Otherwife, make obfervations to all the marks
in fight, and to new ones, towards thoſe unſeen, which is all a one as
to take ſeveral fields adjoining to one another, &c. as is plain to un-
derſtand.
To take the plot of a wood, morafs, &c. by going about the out-
fide of it.
Ex. 231. Let A B CDEFG, be fuch a wood &c. (fig. #75)
1. Set up marks at every corner of the wood &c. and at every ob-
ject you would have in your map.
2. Place your inftrument (viz. any thing that wall take, or meaſure
an angle) at any of theſe marks, fuppofe at A, there obferve the two
neareſt marks B and G, and fuppofe the index between looking at B
and G, moves over 1000, which is the meaſure of the L BAG; then
remove your inftrument to another mark B, meaſuring the diſtance,
or hedge A B 21 ch. 8 links, there look out for the next two neareſt
marks A and C, and mind what degrees are included, fuppoſe 760
LA B C, remove to C, meaſuring the hedge B C 15 chains 20 links,
there as before, take the L B C D &c. by going thus round, take all
the Ls and meaſure all the hedges. Alſo, if there be any place as H,
out of the hedges that you would have on your plan, make two obfer-
vations, to it as one from. A and another from B.
See prob. 137.
To lay down theſe obſervations.
1. Draw a line A B making it by the diagonal ſcale = 21,08 chains,
upon-A make an L of 1000, and upon B one of 76°, make A G by
the diagonal fcale = 25,06 chains, and BC= 15,2 chains, and thus
lay down all the fides and Ls as you find them (def. 30. prob. 129)
in your paper, ſo you'll have a true map.
Ex. 232. Note. When the two laſt lines in a map are ſet off, being
produced they will meet without laying down their lengths, or the L
which their meeting makes, fo if this L and theſe two lines be meafur-
ed and compared with the like L and two hedges taken in the field, it
will prove, if the plan is truly made.
Ex. 233. To know if the angles in the field are truly taken.
2
Take a from the number of the inward angles, and multiply the re-
mainder by 180, this product, if the work is right, will be equal to the
fum of all the angles in the figure, if there be outward angles, you add
their complements, fo as in the lalt figure, the number of Ls (exclude.
}
AND MECHANIC.
79
ing LC) are 6, then 180 X 4 (6 — 2) is ± 720 ± 69525, the
fum 695 of all the inward Ls added to 25, the comp. of the cutward
L C, to 180°. This is eafily proved, fince the fum of the thrce Ls
of any plain A is known to be 180°, as alfo that the fum of the exter-
nal Ls of any right lined figure is --- 3600,
Ex. 234. The laſt ex. may be done by the chain only. Thus (in
fig. 176) fet up marks at every corner A BCD, &c. in the field, then
begin at any of theſe marks as at A, and meaſure the hedge A B 50 ch.
8 links, then hold you one end of the chain at B, and let your affiſt-
ant go with the other end untill he fee himſelf in a line with the marks
B and A, there flick down a flick which is at c, then let him move the
end of the chain, until you fee him and the mark Cin one line, there
fet down another ſtick at d, and meafure the diftance dc 1,21 chain,
again meaſure the hedge B C 75,29 chains and holding one end of the
chain at C, or putting the great ring over a ſtick there, move the other
end to e, in a line with BC, and then to f, in a line with CD, then
meaſure the ſtreight diflance e f, 1,7 ch. next meaſure the hedge C D
70,5 chains, and upon D take the Lg Dh 1,4ch. as before, and thus
may all the fides and Ls be taken by the chain only.
To plan this figure.
1. Draw a line A B at pleaſure, laying 50,08 chains from A to B,
then with the radius one chain on B, fweep the arch c d, on which lay
1,21 chain from c to d, thro' d draw B C at pleaſure, and lay 75,29
ch. from B to C, upon C with the radius 1 chain, fweep the arch e f,
lay 1,7 ch. from e to f, thro' f draw C D &c. as is plain by the figure,
all theſe numbers are taken from the diagonal fcale. where,
I
Note. That one fcale muſt be uſed to all the fides, but you may ufe
a larger ſcale to the Ls if you pleaſe, for great care and exactneſs
fhould be uſed in the Ls both in taking, and laying them down, for a
fmall matter in an L where the hedges are long, will caufe a great er-
ror. Here as in ex. 232, the laſt line laid down D A, will meet B A
in A and be 62,11 ch. if the work be right. Alfo, in the next ex.
are an L and two fides, to prove the plan.
Ex. 235. To take the plot of a field A B C D E (fig. 175) by go-
ing the inſide of it, with the chain only.
This is performed nearly as in the laft ex. by meaſuring the hedget
and at every corner, or ſet up mark, for inftance, at A put the ring as
the chains end over a flick, and ſtretch out the other end to a, in aline.
with A E, and alfo to b, in a line with A B meaſure the diſtance freight
from a to b, 1 ch. 29 links, and thus do at every corner A, B, C, D,
&c. as you meafure the lengths of the hedges, and fuppofe the hedges
and angles to be as fet down in the figure.
80 THE UNIVERSAL MEASURER
R
To lay down this figure on paper.
This is fo eafy, from the numbers on the figure, and what is ſaid in
the laſt ex. that I think it needs no other directions.
Ex. 236. To take the plan of a field A B C D E (fig. 177) by dia-
gonal lines.
To do this you muit meaſure ſo many diagonal hires as will divide
the field into As, and then it will be eafily laid down (by prcb. 14, or
16) in all the foregoing examples, the hedges are ftreight, but in this
ex. for variety, I have taken them bended, and what is here done in
crooked hedges is to be obferved in any other method of plotting &c.
Having taken a view, and a rough draught of the field, meaſure in
a line from A to B taking offsets as you go on, audfet them (fee def.
30 prob. 129) at the end or along the offset line which muſt always
be to the ranging line A B. Alſo, along the faid ftreight line å B, ſet
pown the diſtance between every offset, and always take care to off-
fet to every turn, or bend in the hedge, thus go round the field, or
rather the trapezia which you are about to take; and the greater trape-
zias you take the better, for the fewer the parcels the truer the con-
tent in this reſpect, alſo the ſhorter the offsets the better; becauſe
without much labour it is difficult to get them fquare to the ranging
line, &c. when you have thus gone round the trapezia, take a diagonal
A C, fo will you be able (by prob. 16) to lay down the trapezia
A B CD, for by your aforesaid draught the fide A B is = 3 + 1 +
1,5 + 2 =7,5, the fide B C = 2 + 5 +8 = 15, the fide C D = 3
+5
+ 2 + 2 + 2 + 1 = 10, and the fide DA= 2 + 2 + 3 + 15 + 1
8,5 the diagonal, AC 12.94, you may alſo meaſure a diagonal
from B to D in the field, which applied to your plan will be proof
thereunto, for the offsets, they are ſet off as directed in prob. 36,
from the fame diagonal fcale, that you take the other lines A B, BC,
A C, &c. from. Thus, fet 3 chains from A towards B upon A B,
there raiſe a 1, making it = 1 chain, at one chain further raiſe a 1,
and make it = 1,3 ch. and ſo on. Then over the tops of thefe s
draw the curve, or bended line, which will form the thing required.
Ex, 237. Any four ſided figure (right lined may be readily laid
down by having its four fides and one L, (fuppofe L A) given; thus
make the LA (fig. 177) = what you took it in the field, drawing A B
and A D till they be — what you took them, then upon B, with BC
in your compafies firike an arch, and upon D with D˚C, croſs it in C,
join B C and D C, and its done, on which you may ſet the offsets, if
any, as before.
AND MECHANIC.
ફ
#
Ex. 238. There are ſeveral other methods for taking the plot of
à field, &c. but they all end in the taking of fides and Ls &c. thofe
here laid down are the moſt eaſy, plain and uſeful, and if underſtood,
it will be eafy not only to underſtand, but even invent other methods;
that in ex. 236, as is true as any to depend on where it can be
practifed, becaufe it meafores diagonal lines, which cannot eafily fail
in taking the field truly, if it be uneven, becaufe in fome degree they
must go over thefe uneven places, &c. If you have a pocket compafs
you may fet it to any line you are meafuring, and fo have the compafs
or bearing of the places in your map. Alfo by the help of a compafs,
you may take the Ls in a field, by ſetting it at one mark, and cbferva
ing the bearing of the next two nearest marks as is easy to underſtand,
for if one bear north, and the other weft, then its plain the L at the
mark where you ftand is 90°. &c. &c.
Ex. 239. How to take a perpendicular in the field. (fig. 14)
This may be done with any inftrument that will take an L, thus,
as you are meaſuring along any diagonal A G, try new and then with
your inftrument, by looking at the marks A and I, till you find the L
A a I=90°, fo will a I be the which meaſure, this may alſo be
done cafily by a mafon's fquare; thus, as you meaſure along AG,
try now and then with the fquare, till along one fide ofit you ſee the
marks A or G, and the fame time along the other fide of it, fee the
mark I, and that is the point where the will rife, which in this fig
ure the L of the fquare will be at a, there flop and meaſure the Lal,
then come back to a, and as you meaſure along a G, take in like man-
ner the Le H, now having fet down in your rough draught the dif
tances A a and ae or e G, and the Ls a I and e H, you may from
then lay down the figure. Thus, draw a line AG, on which lay Aa
on a, raiſe the La I, on it lay a I, lay a e frem a to e, on e raile the
Le H on it lay e H, then join the points A, I, G, H, which will be a
true plan of the right lined trapezia A IGH. If there be more tfa-
pezia's than one you may take, and lay them down (by one at a time
joining to one another as they lie in the field, taking offsets if the out
hedges are bended, as in ex. 236) either by this method, if Ls canbe
eafily had, or by ex. 237.
Ex. 240. How to meaſure ſtreight.
Either let him that goes with the fore end of the chain, flick dow
his ſtick, in a line with him that goes behind, and the mark behind the
faid hindmoſt man. Otherwife, he that goes behind may dire& him that
goes before, by fome mark in the line, called a foremark, the other
being called a back mark.
L
82 THE UNIVERSAL MEASURER
Ex. 241. To fet out a line, when both ends cannot be ſeen at once.
Set up a long pole as at A, (fig. 14) and one at G, and let there be
fo many men between A and G, as may fee to direct one another into the
fame right line A G, then where they thus ſtand, ſet up marks within
fight of one another, you may run a line by back marks at a venture,
and by it find the true line you want. Thus, fuppofe you want a line
from A to C, and by reaſon of fome obftacle you cannot ſee between
A and G. (fig. 14) Firſt, meaſure as near the line as you can gueſs,
ſuppoſe from A to I, where you can ſee the mark &, there take the
LA IG, which if it be a right L, the fquare root of the ſum of the
fquares of A I and I G, will be A G, or if it be any other L, you
may by trigonometry find the fide A G, having firſt meaſured IG,
for then you'll have two fides and their contained 4, when a field is
to be divided into ſhares, then its content ſhould be had very near,
then it is beſt to uſe ſuch methods for plotting, as will take diagonal
lines &c. in or through the field. (See ex. 238) But to plot a town-
fhip, county &c. it will be fooneft done if you go round it taking the
Ls &c. as before directed, and as you go on meaſure the fides with a
parambulator, or furveying wheel, which is done by driving a wheel
before you, made for that purpoſe, with which you may furvey as faſt
as you can walk, for the pointers' on this wheel fhew what diftance
at any time it has moved.
Ex. 242. To find the content of any plot when laid down.
If a field is furveyed by ex. 236. its content may be caft up by rule
11, or if by ex. 239, its area may be had by rule 10, fo that in theſe
two cafes, if you want only the area the field need not be laid down
on paper, in all cafes, the contents of the offsets are had by rule 24.
Let it be required to find how many acres is contained in fig. 177.
플
​of 3 X I.
Offsets between
1 + 1, 3 × of 1.
X
1,3,8 × 1 of 1,5.
of 2.
,8 +,04 ×
X
1,5
1,15
A and B,
1,575
,84
2,00
2 X of 2.
Offsets from
8,25
B to C,
6,00
1,50
1,5 X
3 X 1 of 2.
3,5 X
1 of 8.
of 5.
Offsets from
3,00
2,50
C to D,
2,00
し​,25
39,25
45,29
}={
Subtract
114,105 Total
X
I + 1, 5 × 1 of 2.
1,5 ÷ 5 ×
,5 X 1 of 1.
A ABC,
EDA,
10,5 Area of D E A.
10,3605 Acres, the anfwer
}
I + 2 ×
1
of 2.
of 2.
}found by rule 11.
AND MECHANIC.
83
[ 1,5
Offsets from
3,5
D to A,
4,5 >
25
5 J
Sum
( 1,5 × of 2
X
2.
1,5 + 2 × 1 of 2.
2 + 1x
IX
I + IX
LIX 1 of 1.
10,5 = the area of D E A.
of 3.
of,5.
3
Thefe offset areas are all found by rule 24; and as the content of
any plot is thus found, it ſeems unneceffary to give any more examples.
Becaufe by the figure, the hedge D E A bends inward from the rang-
ing line D A, it is plain the content of the part DE A muſt be taken
from that of the field. In any of the other methods, when the field is
laid down on paper, be careful to meaſure diagonals and perpendiculars
on the fame diagonal fcale, by which you'll eafily find the content.
Ex. 243. How to divide a plot when laid down.
When any plot is truly laid down by a good diagonal ſcale upon
a large fheet of paper, i. e. your paper ſhould be fo large as that every
line may in your map, have its true length in chains and links (or
links be ſeldom or never regarded) fee prob. 23. This done, you are
taught in the first part of this book, to divide or cut away any plot by
art every way poffible. But as there is yet another eafy, general and
practical method, I fhall here give an example thereof. It is thus,
when the plot is laid down, and its content found, (in doing this you
need have no lines on your plot but the black ones or out hedges, for
the pricked lines in the foregoing plots, being only for illuftration,
may in your plot be drawn dry, and not pricked, for needleſs lines ferve
only to blind a figure or plot) and you know what way the dividing..
hedges muft range, and the quantity of ground every new inclofure.
is to contain, then meaſure off fuch parts in your plan, and ſee what
diſtance is thereon between every new hedge, which mark down, then
go into the field and meaſure out the fame diftances upon the fame
lines, as on your plan, there cut out little holes or ſtrike in marks in
the ground, and meaſure every parcel thus laid down in the field, and
if the areas of theſe new inclofures do not fuit thofe in your plan, or
are not the areas they ſhould be, you muſt vary the lines, till they do
agree.
Note. Since moſt ground is uneven, it meaſures to more in parcels,
than it does altogether.
Let A B C D, be a field of 120,55 acres, to be divided between
two men, z to have 20,55 acres, andy to have the reſt, (viz.) 100
acres, to be hedged off parallel to the hedge D C. Fig. 178.
84 THE UNIVERSAL MEASURER
1. Draw a line de parallel to D C, as near good as you can gueſs,
then meaſure the trapezia de CD, which fuppofe 90,21 acres, this
is too little, by 9.79 acres, fo you must move the line d e nearer to
A E, until you find it cut you off a content of 100 acres, which fup-
pofe it does when at E F, then meaſure the diſtance D F and CE by
your diagonal fcale, then go and meaſure out the fame diſtances with
your chain in the field, from D towards A, and from C towards B,
and meaſure the trapezia D C E F, and alfo the figure E F A B, and
if the are 100 and 20,55 acres, refpectively, its done. Otherwiſe,
you muſt remove the line E F in the field till you find by meaſuring that
tanfwers. To find nearly, how far the line d e muſt ſhift, fuppofe it
1 chains, because it cuts off an area too little by 9,79 acres or 97,9
quare chains, and content divided by length gives breadth, therefore
97,9 divided by 10, gives 9 chains 79 links, and fo much may you
remove d e towards A B, on your plan, and then meaſure and fee
what the content is, which if it differ a little, you may divide and try
again &c.
Ex. 244. Note. If you are to cut off a fhare next to any bended
hedge A zB (fig. 178) it is beſt to add the offsets in A z Bto the fhare
and then lay it off as before from the ftreight line A B, but if the
hedge bend out-wards, the offsets muſt be taken from the faid ſhare
as is plain by the figure. Alfo if the hedges which the dividing line
cuts, be bended, it is beſt to work with thoſe cut offsets by themſelves,
and then add or fubtract from the right line part, as the fig. directs.
Ex. 245. Let A B CDE, (fig. 179) be a field containing 27 acres,
to be divided equally amongſt 3 men, and all the hedges to meet at
a pond, or watering place g in the field.
I
1. Since no point in the out hedges is given to begin the new hed-
ges from, draw a line A g, from the angle A to the well at g, which
let be one of the new hedges. Alſo, join C g, and meaſure the tra-
pezia A go CB 6,65 acres, now if this 6,65 had been 9 acres viz. § of
27 acres the fields area, then C g would have been another dividing
hedge, but 6,65 wants 2,35 of 9, ſo if you add a ▲ of 2,35 to A g
CB 6,65 you'll have the fecond dividing hedge, therefore, take the
neareſt diſtance 8 chains 62 links from g to the ftreight line CD, then
becaufe the area of a ▲ divided by half its baſe gives the L &c. there-
fore, 2,35 divided by 4,31 (half of 8,62) gives 5 ch, 54 links, which
fet from C to H, and join g H the ſecond dividing hedges, in like man-
ner by joining g D, you'll find the area of the A g DH will want the
area of the Ag D I, to make it 9 acres, fo I g will be the third and
alt dividing hedge, now meaſure the diſtances EI and C H on your
AND MECHANIC.
85
plan by the diagonal ſcale, that you may know what chains and links,
to meaſure from the like corners in the field, where make marks at I
and H, and meaſure the 3 parts AgH CB, HDI g, and IE A g,
which if the plan and work be true will be very near good, but if there
be a ſmall difference you may rectify it, by regulating the lines, as
before. In this manner you may divide land at pleaſure. where
Note. That in uneven ground, you muſt take dimenſions, ſo, if you
can, as that they may go over theſe uneven places, fo will your plan be
the truer, and confequently lefs varying the divifional hedges in the
field.
Rectangular, and triangular pieces of ground, are eaſily divided
by rule 35, and prob. 111, &c. as in the two following examples.
Ex. 246. If a rectangle be 20 ch. long, how many chains in breadth
will make 5 acres, or 50 fq. chains, alfo how many yards in breadth
will make 2 acres.
chains
20) 50 given area
anfwer 2,5 chains
yards
4840 yards in one acre
2
4410) 9680 given area
anfwer 22 yards.
Becauſe 22 yards is 1 chain, 22 X 20 = 440, the yards in 20 chains.
I
Ex. 247. Let there be a triangular piece of ground A B C (fig. 89)
whoſe area is 300 chains, and baſe B C 40 chains to be divided into
3fhares by lines drawn from the vertical A, to the baſe B C.
1. By rule 10, the L d A muſt be 15 chains, becauſe 40 × 15 and
÷2300 the given area, a third of which is 100, this ÷ 15 the L
gives 6 chains 66 links, for the baſe of 100 chains the third ſhare, or
by prob. 111, you need only divide the baſe B C into 3 = parts, and
draw lines between each of theſe parts and the LA, and its done.
Ex. 248. If a field be 80 graffes, and its content 192,57 acres, how
many acres are there to a graſs.
As 80 graffes is to 192,57 acres fo is 1 graſs to 2,407 + acres anf.
fo that if any number of graffes be multiplied by 2,407, the product
will fhew how many acres are in that number of graffes. As for
Ex. 249. How many acres muſt be to 5 and 1½ and 61 grafies re-
fpectively.
2,407
5
2,407
1,5
2,407
6,75
acres 12,635 5 G. ac, 3,61051 ac. 16,24725= 61 graffes
86 THE UNIVERSAL MEASURER
I
Ex. 250. If one half of the laſt mentioned field, muſt abate 1 rood
per acre, and the other half thereof have added 1 rood per acre, how
much to a grafs in each.
I
Becauſe a rood is the fourth of an acre, and every acre is to abate a
rood, therefore, every whole is to abate of itſelf, fo the 4 of 2,407
is 0,60175, which taken from, and added to 2,407 gives 1,80525 per
grafs for the abating part, and 3,00875 per grafs for the increaſed part.
Note. In cafes of abating &c. you muſt take care that the ſum of the
abatements be the fum to be added, for, if you abate more than
you add, you'll have land to ſpare, and the contrary if you abate leſs
then is to be added, as you may eafily perceive.
XXXXXXXXXXXXXXXXXXX
XXXX
SECTION VIII. The practice of Gauging.
Ex. 251. Gauging, is to know what quantity of liquor any veſſel
will hold, or how much liquor is in it when it is quite full or partly fo.
All liquid meaſures are computed by folid inches, whence gauging is
no more but folid meaſure and therefore may be all done by fection 7
only, there the dimenfions are taken to the outſide of the folid, but
here they muſt be taken within, to get the folidity of the cavity, or
part holding the liquor, as is evident.
Ex. 252. Gauging (for eaſe fake) is almoſt all performed by the
ſliding-rule, and may be done by Coggleſhal's fliding-rule as by fect. 7.
But fince there is one on purpofe for gauging, it may not be amiſs to
defcribe it here.
Ex. 253. This rule is in form of a parallelopipedon, three faces of
which have each a ſlider, tho' there are other forts, but this is what
I uſe here; on one face, ftands A, at the end of the rule, and B at the
end of the flide, thefe are the lines mentioned in ex. 69, here is alſo
a line marked MD which is the fame with the lines A and B, but reads
contrary way, and begins at 21,5042, and ending at 2,15042; now
if from the logarithm of 10, or 100 you take the log. of 2,15042 or
of 21,5042 the remainder fhews the diſtance of unity or 1 from the
end of the line M, D, and is exactly the fame with the diſtance of the
points M B, (on the line A) from the end of the line A, the line mark
N, on the flider is the fame with the line B, above it; on this face are
feveral other marks as A at 282, the folid inches in a gallon of ale, W
at 231, theſe in a gallon of wine, Cat 3,14 the periphery of a circle
whoſe diameter is unity, SI at 0,707 the fide of a fquare infcribed in
that circle, diameter unity s e at 0,886 the fide of a fquare area of a
circle, diameter unity. At each of0,785 area of a circle diameter unity
and,0795 periphery unity, ftands a black point.
AND MECHANIC.
87
Ex 254. On a fecond fide of this rule ftands two lines, the one mark-
cd D, being the fame with that line on the common ſliding-rule (the
other marked D on the ſlider is alfo the fame) the other marked E is
a line of a triple radius to the line E or its radius is
of the lines A,
B, or C, being all three the fame, fo that if the line A be made, all
the other lines B, C, D, E, &c. M D may be made from it, on the line
18,95,
D are feveral marks as W G, A G, M S, MR, at 17,15, I
46,36, and 52,32, being the gauge points for wine and ale gallons,
and for malt bufhels, in fquare, and round veſſels, T P, the gauge
point for a pound of tallow nett weight and ſtands at 6,32.
Ex. 255. On a third fide of this rule is a line on the flider marked
N, and is the fame with the lines A, B, C, and on the rule is two lines,
the one marked S L. or fegment, 1 y, or ullaging a lying cafk, the other
marked S S, or fegment, s t for ullaging a ſtanding caſk. Theſe lines
may be conſtructed thus, firſt for the line S L, take a caſk which a-
grees neareſt with the forms of the common ones in practice, and fup-
poſe its bung diameter to be divided into 100 parts, then planes paſs-
ing thro' each of thoſe parts parallel to the axis of the caſk, will dis
vide it into ſo many unequal parts or flices. Now its evident if you find
the contents of thefe flices in gallons, and to the content of the firſt, add
that of the ſecond, you'll have a number for ,01 and,02,
and adding
the content of the third flice to the laſt fum you'll have a number for
03, &c. by continually adding a ſlice more to the laft fum you'll have
a table of meaſures, by which the line S L may be laid down. In like
manner, if for a ſtanding caſk, you imagine the axis to be divided into
100 parts and planes to paſs thro' each of them parallel to the caſks
bung diameter, it by this means will be divided into 100 unequal parts,
whofe contents being feverally found, and the laft being always added
to ſome of the preceding ones, you'll have a table of meaſures for the
lines. Becauſe ale gallons are moft common in practice, thefe lines are
fitted to that meaſure.
Ex. 256. The line A L, (a new line for ullaging) is made thus, let
8any fegment, under fegment in the table of fegments, d-diame-
ter of any cylinder, and 1 its length,.
ddls
282
=m, the meaſure of a
flice thereof parallel to the axis, in ale gallons, whofe end is fimilar to
the ſegments, cr
the content of the fame fegment in wine
ddls
231
gallons, But if ddls
=m, then
282 ddl
>
whence ariſes this pro
232
S
m
88 THE UNIVERSAL MEASURER
portion on the ſliding-rule viz. as₁/282, on D: Ion C::d, on D:
$
m on C, ſo that if 282 be divided by every number under fegment in
a table of fegments made to a radius of a 1000, or 1000 verfed fines
(but the table in this book taken from Mr. Shirtcliffe's book of gaug
ing is only made to 500 fegments or half the circle) and half the log,
of each quotient, taken from the fame fcale of = parts, that the line
N was taken from, and thoſe diſtances laid on the line A L, you'll
have the line required.
Ex. 257. On the fourth and laft fide of this rule, is a line of inches
divided into tenths, under it ftands 3 other lines marked fpheroid.
2d variety, and 3d variety, for finding the mean diameters of cafks
when the difference between the head and bung diameters is under 8,
inches and are made by theorems 121, 122, and 123. See ex. 337.
Ex. 258. On the infide of the fliders B and C, are placed two lines,
the one numbered from 13 to 36 inches, and the other from 0,42 to
3,60 gallons, which is only as a table to fhew the content of any cylin-
der at one inch deep in ale gallons, whofe diameter is between 13 and
36 inches; on the inſide of the little flider are two lines for reducing
the 4th variety of cafks to a mean diameter, as per ex. 257, conſtruct-
ed by theo. 124. What is faid in the foregoing fections of the lines
A, B, D, on the common fliding-rule, are here to be underſtood of
the lines A, B, C, D, and N, they being the fame lines on both rules,
only, the line D on this rule begins at 1 and ends at 10, yet their uſes
is the fame, both in meaſuring and in gauging. The ufe of all theſe
in gauging follow.
Ex. 259. Extraction of the cube root by the fliding-rule.
The cube root is extracted by the lines D and E, as the fquare root
is by D and A (ſee ex. 77.) therefore, as the firſt 1 on E is to the firſt
I on D fo is any number on E to its cube root on D, fo if the cube root
of 2197, were required, it will be as I on E: I on D::2,197 or E
1,3 on D, fo 13 is the cube root of 2197.
Note. As in taking the fquare root, you divide the number given,
by the fquare of 10, fquare of 100, &c. fo in the cube root, when the
number given is too large you muſt divide it by the cube of 10, or of 100,
&c. for what number ever you divide by, the anſwer muſt be multipli-
ed by the root of that number; fo here I divide 2197 by the cube of
10 viz. by 1000. If I were to divide it by 100, or by 16, thefe have
no perfect cube roots to multiply 1,3 the cube root of 2,197, by, (See
ex. 68.
AND MECHANIC.
89
1
Ex. 260. Having given any 3 numbers (2, 8 and 9) to find a fourth
which may be to the third, as the cube of the fecond is to the cube of
the firſt. By what is faid in the laſt ex. it is plain, that as 2 on D is to
9 on E fo is 8 on D to 576 on E the anſwer. Again, if the three given
numbers were 2, 80, and 9. Here when 2 on D is fet to E, 80 is off
the line D, but againſt 8 (viz. 8,0) on D flands 576 on D which mul
tiplied by (1000) the cube of 10 (becauſe 80 was divided by 10) gives
576000 the term required.
The lines of numbers A, B, C, N, and D, are uſed in gauging vef-
fels by the directions in fect. 2, as you'll find in the following exam-
ples. For the uſe of the line MD, and the gauge points M R, and MS,
fee malt gauging, SL, and S S, fee ullaging, &c.
Ex. 261. Defcription and uſe of the gauging rule, or rod.
This rule is commonly 4 feet long, and for convenience, is made.
to fold in four joints, and hath 4 fides or faces, on the firſt face are 2
lines, the one marked A G, for ale gallons, the other W G, for wine
gallons, called the diagonals. Becauſe when the end which is cut a
flope, is put in at the bung of the caſk, and to the bottom of the head
(try to both the heads to know if the bung be in the middle) the num-
ber cut at the bung, fhews how many ale or wine gallons this caſk will
hold, and is near good in the French wine hoghead, or London beer
barrel, the lines being made for fuch caſks. Thus, as the content of
any caſk is to the cube of its diagonal, fo is the content of any fimilar
caſk to the cube of its diagonal, whofe cube root will be a number on
the diagonal line for that meaſure, whence veffels gauged this way
ought to be fimilar and of the fame variety. Otherwiſe, its not true.
On a fecond face is a line of inches from 1 to 48, decimally divided
and alſo, upon the fame fide ftands Oughthred's gauge line, being a
line of one third of areas of circles in wine gallons, by which you may
gauge a caſk. Thus, let the bung diameter be 0,71 wine gallons, then
its double is 1,42, to this add the head diameter, fuppofe 0,58, the
fum 2 gallons multiplied by 30 inches the caſks length, gives 60 wine
gallons for its content, this holds true in a ſpheroidical catk.
On a third fide of this rule is a line of parts from 1 to 96, which
with Oughthred's gauge line, makes a table of circular areas in ale
gallons. As for example, if the diameter of a circle be
area of that circle upon the other edge of Oughthred's
above 1 gallon, the diameter being found on the line of
fourth fide is a line of numbers, the fame with the lines A, B, C, and
N, on the fliding-rule, on which are the fame gauge points, AG,WG,
M
30 inches, the
line is a little
parts; on the
до
THE UNIVERSAL MEASURER
or
&c. which works with a pair of compaffes inftead of a flider. Thus if
the length of any caſk or veſſel be 30 inches and its mean diameter
26 inches, then ſet one foot of your compaffes in the gauge point
AG, or W G, and extend the other foot to 26 the mean diameter,
with that extent and one foot in 30 the length, turn the compaffes
twice over and you'll find the last moved foot to fall upon 57 ale,
go wine gallons the content. With this line of numbers, there is alſo
a line of fegments (SL) for ullaging a lying cafk. Thus, extend from
the bung diameter on the line of numbers to 100, or radius on the line
of fegments, that extent the fame way reaches from the dry inches on
the line of numbers, to a referved number on the line of fegments, or,
which is better, on the line of numbers extend from the bụng diameter
to the dry inches, then on the line of fegments that extent reaches
from 100 to the reſerved number. Then extend on the line of num-
pers from 1 to the caſks content, that extent the ſame way reaches
from the referved number to what the cafk wants of being full.
But if you use the wet, instead of the dry inches, you'll have the ale
gallons in the cafk. There are other four feet gauging rods made for
London caflis, &c. Thus, fet the veffel level and pour in a gallon of
water, and putting in your rod down right make a mark for 1 gallon
where the furface of the water cuts the rod; pour in one gallon more,
ſo you'll have a point on your rod to mark for two gallons, and fo by
putting in the red for every gallon poured in you'll have marks for
any number of gallons in the veffel whether ftanding or lying. And fo
when you come to gauge fuch a veffel, you have nothing to do but
put in your rod down right, and the wet part will fhew what gallons
are in that veffel. The black dots on theſe rods made for quarts, are
made by pouring in a quart at a time, and at the end of every quare
mark a dot on your rod, and as you did a figure for a gallon. Theſe
rods are made up for the following veffels, viz. 1 a line for a butt
ſtanding, 2 a line for a butt lying, each of which contains 108 gallons
beer meafure, 3 a line for a hogshead of 54 gallops, 4 a line for the
þarrel of 36 gallons, 5 one for the kilderkin of 18 gallons, 5 one for
firkin of 9 gallons, all beer meaſure; fo that you may eaſily know
which of thefe veffels you are gauging by putting in your rod, for if
the top of the veffel cuts y gallons it is a barrel, if 54 gallons the veſſel
is a hogshead, &c. The London coopers are obliged to make theſe
veffels and alike. For the fame purpoſe there are other lines put on
fach gauge rods for wine meafure, as for a tun, butt, puncheon, hog
head, tierce, barrel, rundlet, and anchor, of 252, 126, 84, 63, 43,
31, 18, and 9, gallons.
སཱཝ
AND MECHANIC
91
Ex. 262. The Engliſh wine gallon by act of parliament, is to cons
tàin 231 cubic or folid incles. By which is meaſured all wines, bran-
dy's, ſpirits, ſtrong waters, mead, perry, cyder, vinegar, oil, and ho-
ney. And as I pound troy weight is to 1 pound averdupoize weight,
fo is the cubic inches in a gallon of wine to thofe in a gallon of ale. Now
(by Ward's mathematics, page 35) 1 pound averdupoize is 14,02
12 p, w. troy, therefore, as 12 oz is 14 oz 12 p.w. fois 231 to 281
car = 282 folid inches as the ale or beer gallon is now fetled. The
ſtandard, or Wincheſter corn bufhel, to contain 8 gallons of wheat,
every gallon to weigh 8 pounds troy, each pound to weigh 12 ounces
Bach ounce 20 penny-weights, and each penny-weight to weigh 32
grains of wheat taken out of the middle of the ear. Now fuch a bufhel
is found to contain 2145,6 folid inches, whofe fill of common ſpring
water is found to weigh 1131 ounces 14 penny-weights troy. But a
cylinderical veffel 8 inches deep and 18,5 inches diameter, is 2150,43
folid inches content, which tho' it exceed the ſtandard buſhel of 2145,6
by 4,82 fölid inches, a cylinder of theſe dimenſions is now fixed on for
the ſtandard buffel, there being no other dimenſions ſo convenient
without running further into decimals. Now the eighth part of 2150,43
is 268,8025; the folid inches in a corn or malt gallon, but the decimal
,0025 being but ſmall is left out and 268,8 taken for the ſtandard corn
gallon, therefore, let the fhape of any veffel be what it will, if its folid
content be 2150,42, or 268,8, or 282, or 231, folid inches it will
Hold a ftatute bufhel of malt, or a gallon of malt, or å gallon of ale, or
Beer, or a gallon of wine.
Note. The Irih gallon of wine, oil, or ale, contains 217,6 folid
inches.
A TABLE of Ale Meaſure in London.
Kil. Bar. Hd
3
4
2
Söl. Ín. Pints Qts, Gall. Fir.
13536 | 384 | 192 | 48
9024256128 32
4512 128 64 16
2256
64 32
282
2
Hogfh ead
Barrel
Kilde rkin
Firkin
8
I
Gallon
4
I
Quart
701
2
Σ
Pint
35
I
92 THE UNIVERSAL MEASURER
A TABLE of Wine Meaſure.
Sol. In. Pints. Qrts.
Gall. Rund. Bar. T. Hd.
Pnn. But. Ton
гon
58212 2016 1008 252
14
8
But
29106 1008
504
126
7.
Puncheon 19404 672 336
84
4
4,662,662
42
632
2
I
מ -
1,33
H
Hogfhead 14553
504
252
63
3/1/1/00
2
ખેત
I
Tierce
9702 | 336
168
42
2,33
1,33 I
Barrel
7276252 126
31
1,75
I
Rundlet
4158 144
72
18
I
Gallon
231
8
I
Quart
573
2
Q.
Pint
2831
I
O
Sol. In.
A TABLE of Beer Meafure in London.
Pints. Quarts. Gall. Fir. Kil. Barrel. Hd. Butt. Ton
Ton
60912
Butt
30456
1728 864
864432
216
24
108 12
12 6
6
Hogshead
Barrel
Kilderkin
15228
432 216 54
10152
288 144
36
5076
144 72 18
Firkin
2538
72
3:6
9
I
642 H
工
​42 +
I
2
#
I
2
I
Gallon
Quart
Pint
273]
282
8
4
I
701
I
354
I
A TABLE of Ale and Beer Meaſure in the Country.
Barrel
rioginead
14382
Sol. In.
Pints Qrts
Gall.
Fir.
Kil.
408204
SI
6
3
Bar. Hd.
11
I'
9588
272136
34
4
2
Kilderkin
4794
136
68
17
I
Fickia
2397
68
34
I
Gallon
288
Quart
70플
​2
Pint
354
I
AND MECHANIC
93
tallow grofs
30,28
Note.
That one
pound of
tallow nett
31,4
hard foap
green foft foap
is equal to
27,14
folid inches.
25,67
white foft foap
raw ſtarch
25,56
34,8
It is common in gauging to take the dimenfions in inches (in the in-
fide of the veſſel) and give the content in gallons, which done may be
reduced to other meafures as you pleaſe by theſe tables, thus (fee rule 6.)
263. In 5768 gallons of beer, how many barrels country meaſure?
34) 5768 (169 barrels
17) 22 ( 1 kilderkin
5 gallons
Anſwer 16 bar. 1 kil. 5 gal.
Ex. 264. In 68502 gallons of ale, how many hogfheads country
meaſure ?
51 )68502 ( 1343 hog. 9 gall. Anfwer
9 remains
Ex. 265. In 2520 gallons of wine, how many tons?
252) 2520 ( 10 tons Anſwer
Ex. 266. In 0,5768 parts of a gallon of ale, how many quarts,
pints, gills and noggins?
›5768
2,3072
4 quarts in one gallon
2 pints in one quart
,6144
1,2288
94576
2 gills in one pint
2 noggins in one gill
Anfwer 2 quarts o pints 1 gill 0,4576 nog.
Ex. 267. In 67854,2 folid inches, how many gallons of ale, and
rundlets of wine?
"
1
1
94 THE UNIVERSAL MEASURER
282 ) 67854,2
4158) 67854,2
gallons 240,61773
rundlets 16,318949
4
18
quarts 2,47092
gallons 5,741082
2
pints
0,94184
quarts
2,964328
2
2
gills
1,88368
pints
2
1,928656
Anfwer 16 R. 5 G. 2 Q
noggins
1,76736
1,9 Pint in wine.
Anſwer 240 Gall. 2.Q.0 P. 1 G. 1,76 Nog. in ale.
Ex. 268. In 1 gallon of wine, how many gallons corn?
Inches Gall. Inch. Gall.
Aš 268,8 : i :: 231: 0,8594
Í
I
268,8 ) 231,000
0,8594
Ex. 269. In 1 ale gallon, how many wine gallons?
As 231: I :: 282: 1,22
That is, a wine gallon is to a corn gallon, as 1 to ,8594, which is
nearly as i to,86, ſo if any number of wine gallons be multiplied by
0,8594 the product will be corn gallons, and for the like reafon (ex.
269) if any number of ale gallons be multiplied by 1,22, the product
will be wine gallons, &c. for others.
Ex. 270. Having in the foregoing tables fet down the divifors, viz.
how many folid inches makes any of the meafures therein mentioned,
if you fuppofe the dimenfions in fection 6 to be taken in inches, then
any of the folidities in that fection, divided by 282, or by 231, &c.
will give the content of the folid in ale gallons, or in wine gallons, or
in malt bufhels if you divide by 2150,42, &c. but if you reduce the
factors in that fection and thefe divifors 282; 231, &c. into one factor
or divifor, the work will be much fhortened, and yet performed by the
fame rules as you'll find in the following examples, where obſerve, that
when you gauge a fuperficies it is fuppofed to have one inch of depth
for a plane furface can hold no liquor. Alfo, all fuch figures as are
meaſured by uning the circular factor 0,7854, &c. are called circular
figures, and all other figures are called right lined figures
Ex. 271. All, or any of the aforefaid divifors may (by ex. 46.) be
turned into multipliers. Thus,
AND MECHANIC.
95
282) 1,0000 (,003546 1
ale
231) 1,0000 (,004329
268,8) 1,0000 (,003720
2150,42 )1,0000 (,0004650)
is a multi-
wine
gallons in right
lined figures.
plier for
corn
bufhels.
malt
See exp. 289. Or, if you make ,7854, or more exactly ,785398
your dividend inſtead of 1,000, you'H have multipliers for circular a-
areas; or, making ,785398 your divifor to the dividends 282, 231₂
&c. you'll have divifors for circular figures; thus,
282
231
268,8
2150,42
And
}
{1785.
,002785
785398003399
,002921
,0003652
282
359,05
,785398
ale gall.
Swine gall.in circular
corn gall.
corn buſh.
figures.
231 294,12 are divifors for the fame.
268,8 342,25
2150,40 (2738,
Now to gauge by the fliding-rule. Take the fquare root of any of theſe
divifors, and ſet it upon the line D, (fee ex. 82,) and it will be a con-
Lant gauge point forthat meaſure.
So the
fquare
282
231
16,79
15,19
2150,42
is
46,37
root of
359,05
18,94
294,12
17,14
2738
52,32
M S. malt bufhels, right lined figures
and is marked
A G. ale gallons,
on D with
circular figures becaufe moft
WG. wine gallons, S veffels are fuch.
MR. malt bufhels, circular figures.
272. In this manner you may find multipliers, gauge points &c. for
any other meaſure, and mark fuch gauge points on the line D, as you
pleaſe.
How to gauge fuperficial figures.
Ex. 273. If each fide of a fquare be 7,6 inches, what is its content
in ale and wine gallons, at one inch deep.
7,6 by rule 7.
7,65
57,76 area in inches
,003546 multiplier for ale
20482396 ale gallons.
Or thus, by divifion
282) 57.76
,2048 ale gallons
1,22 See ex. 269
249856 wine gallons
96 THE UNIVERSAL MEASURER
Sliding-rule. As {16,79 or
on D: 1 on C:: 76 (for 7,6 is aff) on
S
D: {20,48} on C, (which divided by ſquare of 10, See ex. 68.)
S,2048 ale
2,2498 wine S
is
gallons.
Ex. 274. Required the content of a rectangle in ale gallons, whofe
length is 100,5 inches and breadth 20 inches.
20
2010,0 area in In. by rule 8
,003546 multiplier for A. G.
7,127460 ale gallons, Anf.
Sliding-rule. As 282 on A: 20
on B: 100,5 on A: 7,13 nearly
on B the anſwer. Or as 16,79 on D
: 44,83 (a mean propor. between
100,5 and 20) on D:: I on C;
7,13 on C the ale gall. required.
Ex. 275. If the baſe of a plane triangle be 260 inches and its per-
pendicular 1 10 inches, what is its area in corn gallons, and pounds of
tallow nett.
130 fee rule 10
130} 10
14300 area in inches
,00372 mult. for C. I.
53,196 corn in. Anf.
31,4) 14300 (455,4 pounds tall. nett
130 remains.
Sliding-rule. As 268,8 on B: 110 on
A:: 130 on B: 53,2 fere on A, the
C. G. And as 31,4 on B: 130 on A ::
110 on B: 455,4 on A the tall. pounds.
the diagonal 2820 inches, the perpendicu-
lars 105 and 80 inches, whats the content in corn gallons, and tallow
pounds, grofs.
Ex. 276. In a trapezia,
805
5} fee
105 fee rule 12
185
1410
diagonal
260859 area in inches
,00372 mult. for corn gall.
970,36200 corn gallons
30,28) 260850 (8621,2 tall. lb. grofs
6120 remains
Sliding-rule. As {269,8} on B
is to 1410 on A fo is 185 on B to
5970,362
28621,25
{
%}
on A the anſwer.
When you
take dimenſions of a trapezia &c. ftretch a diagonal line
the longeſt way corner-wife, with a cord, and there let it lie faft, then
with another cord, take the neareſt diſtance between the two oppofite
angles and the faid line, ſo you'll get the perpendiculars.
I think it needleſs to add any more examples in right-lined furfaces,
for few fuch forms are to be met with in the practice of gauging, and if
they were, the rules in fection 4 are fufficient; for if the dimenfions there
be in inches, then any area there divided, or multiplied, as directed in
ex. 270, will give the content at one inch deep, &c.
1
AND MECHANIC.
97
Ex. 277. If the diameter of a circle be 40 inches, what is its area
1
40
in ale and wine gallons.
40 fee rule 15
1600
1600 [cir. fig.
,002785 mul.for A. G. ,003399 mul. for w.
4,456000 ale gallons
. g.
5,438400 wine gallons.
}
on D: I on G :: 40 on
1600 fq. diam.
Sliding-rule. As {19,94 or ale gallons
4,45
or wine
+44 on C, the anfwer.
}on
D: SAAA
5,44
Otherwife. As 359,05 or 294,2 on B: 40 on A :: 46 on B : 4,45
or to 5,44 nearly on A, the anfwer.
Ex. 278. If the diameters of an oval be 72 and 50 inches, what is
its area in ale gallons.
{fee
72 { fee rule 21
50
3600 prod. diam.
36c0
,002785
10,126000 ale gallons
Sliding-rule. As 359,05 on-A : 72 on B:: 50 on A: 10,12ổ đà-
D. the anſwer. Otherwiſe, as the gauge point A G, on D : 1 on C::
60 (a mean proportional between 72 and 50) on D : 10,126 on G anf.
Ex. 178. If there be a fegment of a circle, whofe verfed fine is 10
inches, and diameter of the circle 50 inches, what is its area in ale
gallons.
This may be done by ex. 118, but that by ex. 113, being ſhorter
is commonly uſed.
Thus 50) 16,000 ( 200 againſt which under V. S. in the table of feg.
ftands,111823
mult.
2500 fq. diam.
279,5575 area ſeg.
2063546 mul. for A
,9913+ anfwer
Sliding-rule. As 50 on A: I on
B: 10,000 on A: 200 on B, a-
gainſt which under V. S. ftands
G.,1118 +, then as 282 on A:,1118
on B: 2500 on A: 0,9913 on B
the anſwer.
Ex. 279. If the area of any figure whatever be 521883, as in ex.
121, what is the area in malt buſhels, at an inch deep, the area 521883
inches, being every where the fame, or the veffel of the fame breadth
in that inch of depth.
2150,4) 521883 (242,2 malt bufhels, Anfwer
4862 &c.
Sliding-rule. As 2150,4 on B: 1 on A:: 521883 on B : 242,2,
(fee ex. 68) on A the anſwer.
N
98 THE UNIVERSAL MEASURER
Thus you fee, it is but dividing the area in inches by the divifor for
that meaſure you want, let the figure be what it will, or multiply the
faid area by the multiplier for the meafure wanted, ſo you'll have the
anfwer.
Ex. 280. By thefe examples (and fect. 4) you may gauge any plane
at one inch deep, and confequently any fimilar and bafed folid, for
the content at one inch deep, multiplied by any number of inches,
muſt give the content of that number of inches in fuch a folid. As for
example, if a cylinder lie parallel to the horizon, whofe length is 100
inches, diameter of each bafe 50 inches, and fo much liquor in this
veffel, as that putting a ſtick into it to the horizon the wet inches
may be 10, by ex. 278, the content of this fegment is found to be 0,9913
which multiplied by the length 100 inches, gives 99,13 ale gallons of
liquor in this cylinder.
Ex. 281. Again, if there be a floor of malt whofe content is 242,2
bufhels, at one inch deep, and the floor be 5 inches deep, then 242,2
multiplied by 5 gives 1211, for the malt bufhels in this floor, &c. for
any other.
Ex. 282. Becaufe 16,79 multiplied by 16,79 gives 282 the inches
in an ale gallon right lined figures, its evident, that if you divide a
ftreight rule into parts 16,79 inches to each part, and each of theſe
parts into 10 or 100 parts, fuch a rule will take the dimenfions in
gallons, and decimal parts of an ale gallon, by which you may have the
content without uſing the factor or divifor, as in the foregoing exam-
ples is ufed. This plain, if we confider, a board &c. 12 inches fquare
to be
to one a foot fquare, tor the area of both is the fame, viz.
144 fquare inches 1 fquare foot &c; therefore, the area is 1 foot
whether you take dimenſions in inches or feet. Alſo, 18,94 inches,
being the gauge point for circular figures A, G. If you put 18,94 in-
ches to a divifion or gallon, on your rule, and divide it decimally as
before, you'll have a rule to take the dimenfions of circular figures in
ale gallons, by which the area of a circle in ale gallons is = to the fq.
of its diameter, that of an oval the product of its two diameters, and
ay fuch rules be made for any other meaſure, or fuch meaſures
be marked at their proper places upon an inch rule, and the con-
ach fooner had by them than by the inches as is evident.
283. If each fide of a regular hexagon be 10 wine gallons (viz.
mes 15,19 inches) what is its content at 1 inch deep. By ex.
107 the required content is 259,807 wine gallons.
AND MECHANIC.
99
3
Ex. 284. If the diameter of a round ciſtern be 100 W. G. (viz. 100
times 15,19 inches) what is its content at 3 inches deep. By rule 15,
the content at one inch deep is 7854 wine gallons, which multiplied by
the depth, gives 23562 W. G. anſwer. But if you take the diameter
in circular wine gallons, it will be 88.6+ (viz. 88,6 times 17,14 the
circular wine gauge point), and its fquare will be (88,6 × 88,6 =)
7854 as before nearly. Hence, if you fuppofe the dimenfions in fect.
4, to be taken in theſe meaſures the areas there, will be the contents
in ſuch meaſures, at 1 inch deep.
How to gauge folids, or rather veffels.
If any of the folidities in fection 6 be in inches, and be divided by
any of the foregoing divifors, or multiplied by the multipliers anſwering
thereunto, the quote, or product thence arifing will give the mcafure
in gallons &c. (as by ex. 270) I fhall here, therefore, only inftance in
the moſt common veffels.
Fx. 285. If each fide of a cube be 30 inches, how many gallons of
wine and pounds of hard foap will it hold?
30 fee rule 36
305
900
30
231) 27000 folidity
116,88 wine gallons
Sliding-rule. As {15,19}
=gallons of wine
pounds of hard foap
27,14 ) 27000,00
994,43 pounds of foap
on D:30 on C:: 30 on D: {116,88
?
on C.
S
Ex. 286. If the length of a rectangular prifm, be 81 inches, breadth
25 inches, and depth 26 inches, what is the content in ale, wine,
malt, &c.
I
81} fee rule 37
25'5
2025
26
282)52650(186,7 ale gallons
231)52650(227,92 wine gallons
2150,4)52650(24,48 malt bufhels
31,4)52650(1676,75 pounds tallow ne
30,28)52650(1738,77 pounds tallow grofs
$
100 THE UNIVERSAL MEASURER
27,14)52659(1939,94 pounds hard foap
25,67)52650(2051,03 pounds green foft foap
25,56)52650(2059,85 pounds white ſoft ſoap
34,8)52650(1529,31 pounds raw ftarch
sliding rule. First, a mean proportional between fome two of the
3 dimentions, as 81 and 25 is found to be 45. (See ex. 82.) Then,
16,79
186,7 A. G.
15,19
46,37
the fq. root of the divifor,
5,6
227,9 W.G.
24,48 &c.
on D; 26 on C:: 45 on D1667,7
As
5,5
the mean propor. between
1738,9
5,2
81 and 45: the anfwer on
1939,9
5,06
C. (viz.)
2051,0
5,05
2059,8
5,9
L1529,3
Ex, 287. If the diameter of a cylinder be 564 inches, and its length
96 inches, what is the content in ale, wine, malt, tallew, &c.
dian. 56,5 fee rule 37
As
5635
3192,25
len. 96
359,05)306456(853,52 ale gallons
294,118)306456(1041,94 wine gallons
2738)306456(111,92 malt bufhels
32,68)306456(9238,29 green foft foap
38,55)306456(7949,57 pounds tallow grofs
39,98)306456(7665,23 pounds tallow nett
34,56)305456(8867,36 pounds hard foap
32,54)306456(9417,82 pounds white ſoft ſoap
44,32)306456(6914,62 pounds raw ftarch.
18,94
17,14
52,32
5,7
172 3
6,2
6,3
5.9
5,7
L. 6,66
Sliding-rule.
the gauge point, or fq. root
of the circular divifor on D
: 96 the length on C:: the
diam. 56.5 on D: the anf.
on C. (viz.)
( 853,52
1041,94
II1,92
| 9377,47
7949,57
7665,23
8876.36
9417,82
L 6914,62
Theſe two examples (from Leadbetters gauging) are fet down to
all the moſt uſeful meaſures and weights, on purpoſe that you may have
ready the gauge points &c. if wanted. (See ex. 270, and 271.)
AND MECHANIC
ΤΟΥ
Ex. 286. If there be a fruftum of a fquare pyramid, each ſide of the
greater and leffer baſes 13 and 8 inches, and depth 24 inches, what is
the content in ale and wine gallons. This is the fame with ex. 169.
which I fhall work by rule 43.
13
169 fq. 13
add 8
64 fq. 8
9,560016
21 twice a mid, fide
441 fq. 21
1,22
fum 674
of,003546 is =,000591
›39833.4
24 height
9,560016 ale gall.
tums of fquare pyramids in ale gallons.
11,66321952 W. G.
Sliding-rule, for ale. If
you multiply 282 by 6 the
fquare root of that product
will be 41,13 a conftant
gauge point for the fruf-
Then, as 41,13 on D: 24
on C:: 13, 8 and 21 on D : three numbers on C whoſe fum is 9,56
ale gallons anfwer.
Or if you fet the common right lined gauge point 16,79-on. D. to a
fixth of the length on C, the rest of the work will be the fame.
Ex. 289. The reafon why I take the fquare root of 6 times 282, is
becauſe 6 and 282 are both diviſors to the folidity, viz. 282 for the gal、
lons and 6 for the of the length. And here obferve, that in all fuch
cafes, any number of diviſors multiplied into one another, the laſt pro-
duct is one divifor for them all. Alſo, if you multiply any number by
the product of any number of multipliers, the answer will be the fame
as if you multiply that number by theſe multipliers one after another,
hence, a multiplier divided by a divifor, gives one multiplier for both,
and a divifor divided by a multiplier gives one divifor for both, &c.
1
Ex. 290. If feveral numbers are to be multiplied together continual-
ly, it is no matter which you take firſt, or which laft, the anfwer will
be the fame, fo in ex. 288, I have 674 and ,000591 and 24 to multi-
ply into one another, you'll find the product 9,56+, multiply them
in what order you pleaſe. Alſo, if a part of fome one of theſe num-
bers is to be taken, you may take that part of any one of them, that
fuits beft. Thus, in ex. 288, Ifhould by the rule 43, take of the
length 24, but Itake of the factor ,003546, and fo multiplies by the
whole length, and the product is 9,56+, take of which number
10
674 or 24 or ,003546, you pleafe, this is all evident from algebra.
!
Ex. 291. There is a fruftum of a rectangular pyramid, the length
of the greater bafe 27 inches, its breadth 12 inches, length of the lef
fer baſe 18 inches, its breadth 8 inches, and the fruftum's length 100
inches, how many ale gallons will it hold? First, the fum of 27 and
18 is 45 twice the length in the middle, and the fum of 8 and 12 is
20, twice the breadth in the middle, ſo per rule 43.
!
102 THE UNIVERSAL MEASURER
900 = 45 × 20 area mid. 4 times
324 = 27 X 12 area greater baſe
144 18 x8 area leffer bafe
=
1368 fum
100 axis
136800 fix times folidity, in inches
,000591 of,003546
80,8488 ale gallons anfwer
Sliding-rule. Find 18 a mean
proportional between 27 and
12, the length and breadth
of the greater baſe, and alſo
12, a mean proportional be-
tween 18 and 8, the length
and breadth of the leſs baſe,
which mean proportionals
18 and 12 in one fum gives
30, twice a mean proportional in the middle. Then as 41,13 on D:
100 on C, or as 16,79 on D: 10% on C:: 18 and 12, and 30 on D:
19,15 and 8,51 and 53,19 on C, whoſe fum is 80,85 the ale gallons
required.
Ex. 292. There is a fruftum of a cone whofe length, or depth is
100 inches, diameter at the greater baſe 18 inches, and diameter at
the leffer baſe 12 inches, how many ale or beer gallons will it hold?
Firſt the fum of 18 and 12 is 30 twice a diameter in the middle,
then by rule 43.
144 fq. 12 lefs diameter
324 fq. 18 greater diam.
900 fq. 30 twice mid. diam.
1368 fum
100 length
136800
,000464
of ,002785
63,475200 ale gallons anfwer
Sliding-rule. If you multiply
359,05 by 6 the fquare root of
the product 2154,3 is 46,41 a
conftant gauge point for all coni-
cal fruftums ale gallons, fo as
46,41 on D: 100 on C:: the
two diameters 18 and 12, and
their fum 30 on D : 14,9 and
6,7 and 41,9 on C whofe fum is
63,5 ale gallons for the answer.
Ex. 293. There is a fruftum of an elliptical cone, length 100, the
tranfverfe and conjugate diameters of the greater baſe 27 and 12,
12, and
of the leffer bafe 18 and all in inches, whats the content in ale gallons
See ex. 291.
=
144 18 X 8
32427 X 12
900 = 45 X 20
1368 fum
100 length
136800
,000464 factor
63.4752 ale gall. anſwer
Sliding-rule. Firſt find 18 a mean pro-
portional between 27 and 12, as alfo 12,
one between 18 and 8, then, as 46,41 on
D: 100 on C:: the two mean proportionals
18 and 12 and 30 their fum, on D: 14.9
on C, whoſe fum is 63,5 the anſwer.
AND MECHANIC.
103
།
Ex. 294. If the axis of a ſphere be 20 inches, how many ale gallons
will it hold?
1. The greateſt diameter in a ſphere is that thro' its center, which
is equal to its axis (viz. HI=AP fig. 170.) and the leaſt diameter
(as at A or at P)o, and G F or KL a diameter in the middle be-
tween theſe two, now K L doubled and ſquared, or 4 times fq. K L
or G F is (by the property of a circle or by meaſuring) = 3 times the
fq. of the greateſt diameter H I or axis A P. So by rule 43.
o= fq. leaft diam.
fq. great diam.
30000= 4 times fq. mid.diam.
10000=
40000 fum
,000464 factor
18,560090
100 length or axis
1856 ale gallons anfwer
Sliding-rule. By the opera-
tion you fee that the fum is -
4 times the fq. of the axis, i. e.
twice the axis ſquared, ſo it
will be, As 46,41 on D: 100
the axis or length on C:: 200
twice the diameter or axis on
D: 1856 on C. anfwer.
Ex. 295. If K LP (fig. 170, or fig. 180) be a bowl, bottom of a
kettle, &c. whofe depth D P is 15 inches, diameter K L 60 inches,
diameter at Po, and middle diameter m n 40 inches, what ale gal-
lons will it hold? By rule 43.
6400 fq. 2 mn, or 4 times fq. mn
3600= fq. K L
• = fq. at P
10000 fum
15=DP
150000
,000464 factor
69,600000 A G. anfwer
Sliding-rule.
As 46,41 on D: 15 on C::
60 and 80 (twice 40) on D: 25,1
and 44,5 on C, whofe fum is 69,6
ale gallons for the anſwer.
296. I have wrought each of thefe examples by rule 43, which has
turned out the very fame anſwers as if you had ufſed their particular
rules in fection 6. This faid rule 43 is as extenſive in gauging as in
meaſuring. See fection 6.
297. You may find the wine and malt, or any other gauge points,
as the ale one is found in ex. 288 and 292, which if marked on the line
D, inſtead of the common gauge points A G, W G, MR, &c. would
be of general ufe in gauging, for any round veffel what ever may be
done by this general
104 THE UNIVERSAL MEASURËR
Rule 59. As the gauge point on D is to the depth on C fo is a diam
at each end and twice one in the middle on D, to 3 fuch numbers on C,
as being added together, gives the required content, or you may take
the diameters in gallons (fee ex. 282) and the depth in inches, and then
the gauge point on the line D will be unity, and fo will be eafier. In
fhort, take dimenſions either this way or all in inches. This noble rule
is fufficient for the whole of gauging, for the error arifing by gueffing
at the veſſels form, may be much greater than can happen by this rule,
in the moſt common forts of veffels it gives the content exactly, and in
any form very near as is proved in prob. 190.
How to gauge malt.
The Duty upon malt is charged at fixpence the Wincheſter buſhel
of 2150,42 folid inches, fo that what form foever the furface of the malt
lie in, whether floor, cistern, couch, &c. find the content of that form
at one inch deep in malt bufhels, which multiplied by the depth in in-
ches gives the content in fuch buſhels. And here obſerve, that malt
being a dry fubftance, will not like liquor have an even furface, alfo the
bottom of the floor &c. may be uneven. Now to remedy all this you
muft take the depth or thickneſs of the malt in 6, 8, or 10 places or
more if needful (particularly, obferve to take it where you judge it to
be thickeſt, and alſo where thinneſt, ſee rule 24,) then the fum of theſe
depths divided by their number, will give the mean depth.
Ex. 298. There is a rectangular floor of malt, length 72 inches,
breadth 48 inches, and the
72 length
48 breadth
3456
5 mean depth
17280 folid inches
ך I
4277
5,4 inches, what is
3 depth is 5,6 the cont. in malt
41
5
419 | bufhels?
(4,4)
number of depths=5)25,0 fum depths
5 mean depth
17280 content in inches
,000465 mul. for M B.
8,035200 malt bufhels anfwer
By the fliding-rule. Firſt, as 48 on D: 48 on C:: 72 on C: 58,7
on D a mean propor. between 48 and 72, then as the gauge point MS
46,37 on D: 5 the mean depth on G::58,7 on D: 8,04— on C, the
anfwer. Otherwife, Set the length 72 on B to the breadth 48 on M D
then againſt the mean depth 5 on A ftands the content 8,04— buſhels
AND MECHANIC.
105
on B. This line M D, with the lines B A, are for gauging rectangular
floors of malt, for malt lies moftly on fuch floors, its uſe (viz. the ufe
of the line M D) in fuch cafes turn one of the dimenfions into a divifor
malt meaſure. (See ex. 106) Now when the figure 1 on MD denotes
I, then the brass p in M B over againſt it on A denotes 2150,42 the
cubic inches in a malt buſhel, by which means the leaſt number that can
be on A is 100 (viz. the firſt 1 on A is 100) but if this firſt one be 1,
then the line A muſt be divided by 100, and when A is divided by 100
then B and M D muſt be divided by two fuch numbers as being multi-
plied together may produce 100, i. e. the divifor for A, and the product
of the divifors for B and M D being equal, the fourth number found
on B will be the content in malt buſhels as is plain by ex. 253. Then
this is the
Rule. As the length on B is to the breadth on M D fo is the depth
on A to the malt buſhels on B, the dimenfions being all in inches.
Ex. 299. If the breadth of a couch, floor &c. be 56,2 inches, length
270, and mean depth 5,2 inches, how many malt bufhels is in that
prifm.
Sliding-rule. Here that 5,2 the depth, may be found on A, the firſt
I on A may be either 1 or 0,1,fuppofe it 1, then the fecond M B on
A is divided by 100 viz. 2150,42 becomes 21,5042, fo the length 270
and breadth 56,2 being each divided by 10 (becaufe 10 times 10 is 100
the divifor of A) will be 27 and 5,62. Therefore, as 27 on B: 5,62
on M D :: 5,2 on A : 36,7 on B anfwer.
Ex. 300. If the breadth be 72, length 140, and depth 18,2. Here
if the firft on A denotes 1, then the fecond M B 2150,42, on A will
be 21,5042, fo the divifor of A being 100, the product of the divi-
fors of the length 140 on B and breadth 72 on M D muſt alſo be 100,
fo let each of 140 and 72 be divided by 10, and they'll be 14 and 7,2
then as 14 on B:: 7,2 on MD:: 18,2 on A : 85,3 on B the bushels
required.
Ex. 301. Let the length be 1250, breadth 360 and depth 9 all in
inches.
I
Let the firſt i on A be 0,1, then the fecond M B 2150,42 on A will
be 2,15042, ſo A is divided by 1000. Therefore, let 1250 be divi-
ded by 10 and 360 by 100 (becauſe 10 x by 100-1000) and they'll
become 125 and 3,6. Then as 125 on B: 3,6 on MD:: 9 on A :
1884 bushels on B the anſwer. Theſe three examples are fufficient to
thew the ufe of the line M D, they or any fuch like may also be wrought
as ex. 298, by the right lined malt gauge point M S.
Q
106 THE UNIVERSAL MEASURER
Ex. 302. There is an elliptical ciftern of malt, whoſe longeſt dia-
meter is 72 inches, fhorteft diam. 48 inches, and mean depth 5 inches,
how many malt buſhels.
72
48
3456
,000365 factor
Sliding-rule. As MR 52,32 on D : 5 01
C:: 58,7 (a mean propor. between 72 and
48) on D: 6,3 bushels on C, anſwer.
1,26044
5 depth
anf. 6,3022 M B.
303. When barley is wet, or ſteeped in the cistern, it fwells out a-
bout a fifth part (or more if its very good corn) it is then called a couch
and bears that name till it hath been 30 hours out of the cistern, but if
it has been out above 30 hours, then on account of its fprouting &c.
it is double to what it was before being wet, it is then called a floor,
and fo continues till it come to the kiln to be dried. Now from the
ciſtern to the couch there is an allowance of} = 0,2, which taken from
leaves 0,8. Hence if any number of couch bufhels be multiplied by
o, it will give the nett bufhels. But from the couch to the floor there
is allowed. Therefore any number of floor bufhels multiplied by
0,5 gives the nett bufhels.
ફ્
The law having thus given the allowance of 4 bufḥels in 20, on the
ciftern and couch gauges, and 10 in 20 on the floor; if therefore an
officer do not advance in his gauges according to thefe proportions he
may expect his fupervifor will diary him for fuch a fault, for this you
mult write a reaſon for, damaged barley &c. for fuch barley will not
hold out to fuch allowances, tho' good barley will exceed them.
Ex. 304. If a ciftern's gauge of dry barley (or however very foon
after it is wet) be 13,8 bufhels, and the beft couch gauge you can get
be 15 bushels, whether does it apfwer the allowance or not.
Here the allowance of 13,8 is 2,76, and 13,8 from 15 leaves
1,2 which ſhould be 2,76, therefore it is too little by 1,56, i. c. 2,76
added to 13,8 gives 16,56 for the couch gauge, inſtead of 15 as it
was gauged to.
Ex. 305. If a floor gauge be 100,8, and the beſt couch gauge be
63,6 which will afford the moſt duty? Here (by ex. 303,) a floor
gauge is multiplied by ,5, and a couch gauge is multiplied by,8, now
to reduce theſe two factors to one factor, (fee ex. 290) it will be 0,5
0,80,625, by which multiply a floor gauge and it will be equal
the fame in a couch gauge, or,8,5 1,6, by which multiply a couch
gauge, and it will be equal to the fame in a floor gauge. So 100,8 X
AND MECHANIC.
107
1625 63 buſhels the couch gauge to the floor 100,8; but this couch
was gauged to 63,6 bufhels, fo the duty mult be upon the couch.
From theſe things, you may learn that malt is to be gauged feveral
times while it is in making.
How to gauge any tun, tub, back, cooler, &c.
Theſe veffels are of various forms, as prifms, fruftums of cones and
pyramids, cylinders, cylinderoids, &c. any of which may be gauged by
the general rule 43, as applied to the foregoing examples, fo in this
place I fhall only fhew the inching of fuch veffels viz. to find how much
liquor they will hold at every, or any inch of depth, which being known
and fet down in a table, you may by putting a ſtick into the veſſel and
ſeeing the wet inches, know by your table what liquor is in the veffel
in gallons, &c.
306. When the bottom of any ſuch veffel is parallel to the horizon,
or exactly level, its called an upright veffel, but otherwiſe an inclined
one,
307. What liquor is required to cover every part of the bottom of
an inclined tun, is called the drip, or fall of the tun, ſo G A P is the
drip of the tun, A GB C. Fig. 181.
308. That horizontal line D G that touches the higheſt part G of
the bottom A G, is called the horizontal bafe, and another line C g
parallel to D G, touching the loweſt point C of the top C B is called
the horizontal end.
309. If from C any point in the horizontal end a perpendicular CD
be let fall upon the horizontal baſe G D it is called the height of the
tun.
310. The ends of tuns are faid to be parallely-pofited, when the
length of the one end is parallel to either the length or breadth of the
other end, and the breadth of one end parallel to either length or
breadth of the other end, and ifit be an elliptical tun we may for bre-
vity's fake call the tranfverfe diameter the greatest length, and the
conjugate the greateſt breadth of the bafes or ends.
Ex. 311. Let there be an elliptical upright parallel-pofited tun,
tub, &c. length at top 65, breadth there 60, length at bottom 110,
breadth there 100, and the whole depth 12 all in inches, to find the
content at every inch deep.
Firſt, to find the diameters at every inch deep, you may meaſure.
them in the veſſel itſelf, or find them by this rule (from the nature of
the difference
the cone viz.) as the height of the tun 12 is to {45
}
1
1
"
}
f
108 THE UNIVERSAL MEASURER
between the two
1
to {3
{{3,75}
33
Slengths 110 and 65 fo is any propofed depth
breadths roo and 60
} the difference between the
S
}
lengths
breadths at 12 and 11 in-
ches height, now thefe differences being found for 1 inch, you may by
continually adding 3,75 the difference of the lengths to 65 the leffer
length, have all the lengths, and by continually adding 3 to 60, have
all the breadths as in the ſecond and third column's in the following
table, for the above proportion holds for any depth, and thus having
fet down all the dimenfions you may by rule 44 find all the contents as
fet down in the 4th column which reſpectively taken from 238,9595
the content of the whole tun, leaves the contents in the 5th column,
but having by the faid rule 44, found the 3 firft contents 11,4884 and
24,2766 and 38,4340, you may with more eafe find all the reſt, than
by the faid rule, thus (by theorem 127) take any 3 contiguous contents
and divide each by its dry inches, take the fecond quotient from the
third, add 3 times the remainder to the firft quotient, multiply the
fum by the dry inches belonging to the required content, the product
will be that content. Now the dry inches of the 3 above contents are
I and 2 and 3, and the dry inches of thefe following will be 4, 5, 6,
&c. fo 11,4884+ 1 (its dry inches) = 11,4884 and 24,27662=
12,1383, and 38,4340 ÷ 3 = 12,8113 from which take 12,1383
leaves ,6730 which tripled is 2,019, this added to 11,4884 gives,
13,5074, this multiplied by 4 gives 54,0305 fere, for the 4th content.
Again 13504 being the 4th content÷4, from it take 12,8113, the
3d content its dry inches 3 leaves ,6927 whofe triple is 2,8781 this
added to 12,1383 (the ad content divided its dry inches 2) gives
14.2164, this multiplied 5 its dry inches gives 71,OS20 or 71,1355 by
carrying
the decimal on further, and thus taking the 3 laft contents
38,4304 and 54,0305 and 71,1355, dividing each by its dry inches
3, 4, 5, and ſo on as per rule, &c. &c. You may foon compleat the
table, or by taking any 3 contents together in fuch a table you may
foon prove the 4th content, whether right or not, the 3 firft contents
are by rule 44 found,
×
Thus: 65 X 2 + 68 x 60 +681 X 2 + 65 × 631: X,000464=
11,4884 A. G.
ad.: 65×2+721 × 60 + 721 x 2 + 65 x 663: X,000464
2,2766 A. G.
3d.
ཀ་ཙ
X +
$2.65 25 24 76 x 60 + 764 x 2 -65 X 70: X,000464
38,4340 A. G.
$
13
ދ
!
4
}
AND MECHANIC
109
In this manner you may find all the contents, but the above method
is much easier. The uſe of this table is eafy, thus fuppofe you come
to the above tun and find 7 inches wet, then looking in the table under
wet inches and againſt 7 I find under content from the bottom 167,8240
ale gallons in the tun, or under content from the top, I find 71,1355
ale gallons which it wants of being full.
A TABLE fhewing the content in ale gallons to every inch
of depth in the upright tun Eyn C. Fig. 181.
Dry
Inches.
Content
Lengths Breadths
| Content from
Wet
from top Cn bottom E y
Inches
о
65
60
,0
238,9595
12
I
68,75
63
6 33 3/31
I
11,4884
227,4711
II
2
72,5
663-3
24,2766
214,6829
10
3
76,25
70
38,4340
200,5255
4
80
73
3
54,0305
184,9290
83,75
7623/3
71,1355
167,8240
87,5
83
89,8187
149,1408
100 76
8
91,25
831
I 10,1497
128,8098
5
95
862/
3
132,1982
106,7613
4
9
98,75
90
156,0338
82,9257
10
102,5
9335
181,7262
57›2333
II
106,25
963-33
209,3446
29,6147
32 1
I
12
IIO
100
238,9595
312. But all fuch tuns are fet to lean a little, for the benefit of clean-
ing them &c. and alfo have a conftant point the eaſieſt to come at,
fixt for taking the wet or dry inches, called the dipping place. Now
to fhew how fuch inclined tuns may be inched, let us fuppofe that we
have juft now inched the upright tun CEyn (fig. 181) whofe height
CE=ny 12 inches, and the feveral contents are as in the above
table, and let it be required to inch the inclining tun A CBG or rather
q C AG, becauſe the plane C q being parallel to the horizon the
veffel can hold no liquor above the point q, and fo the hoof q C B
need not be regarded, fuppofe the dipping place to be at n, and let zy
the diſtance between G D and A F, the horizontal bafe and bottom
of the tun be 2 inches, and let the hoof A P G, or the quantity of li-
quor juſt covering all the bottom of the veffel at G, be 15 ale gallons,
fo when the wet inches are 2, there is in the tun 15 gallons. This
quantity of liquor is commonly meaſured into the tun with a gallon &c.
juſt to cover the bottom and every gallon marked as you pour it in on
your rule (as by ex 261,) then the dimenfions taken parallel to the
furface of the liquor at top C q, and clofe by it at bottom G P, and
养
​tro THE UNIVERSAL MEASURER
the depth n z perpendicular thereunto, fo that if a circular veſſel lear
to one fide, the furface of the liquor will be an oval and the contrary,
now if we ſuppoſe the diameters at C q and G l', and the depth n
to be the fame as before, we fhall by the above method have this ta-
ble, whoſe uſe is the fame as before. Thns
Ex. 313. If the wet inches be 7 what liquor is then in the tun? Here
under content from ale gallons and againſt 7, under wet inches (tands
143,8098 A G. the anſwer. Alfo, if the dry inches be 7, then the tun
wants 89,8187 A G. of being full, &c. for any other
A TABLE, of contents in A G. to every inch of depth in
the inclin'd tun Á Gq C.
Fig. 181.
Dry
Inches.
Lengths Breadths
Content
from C q
Content
from A G.
Wet
Inches
I
65
60
,0
253,9595
14
2
684/1
63
11,4884
242,4711
13
3
72/31/20
663-3-3
24,2766
229.6829
12
4
764
70
38,4340
215,5255
II
5
80
734
54,0305
199,9290
ΙΟ
6
833
763-
71,1355
182,8240
9
871
80
89,8187
164,1408
8
9111
83-3-3-
II0, 1497
9
95
864
132,1982
143,8098
121,7613
8 76
10
983
90
156,0338
97.9215
5
11
1021
93/1/2
181,7262
72,2333
4
12
1061
96/3/2
209,3448
44,6147
13
IIO
100
238,9595
15
32
314. Backs or coolers, are often very broad and not deep, that ſo
the worts, &c. in them may fooner cool, they generally are of a prif-
matic form, and often have uneven bottoms, in fuch cafes you muſt
take the depths or dips in feveral places (as directed in ex. 298) and
fo find a mean depth; then fix a point in the cooler for a dipping place,
as that the depth there taken, may be equal to this mean depth : buż
if this cannot be done you muſt make a juſt allowance for the differ-
ence; fuch large bottomed veſſels are commonly taken at every tenth
part of an inch, and if they are prifmatic veffels, it is very eaſy to do:
for (by the foregoing examples) you need but find the content at one
inch deep, and one tenth of this content will be that of one tenth of
an inch deep, &c. for any number of tenths: in ſuch caſes you need
no table of contents &c. for the area at the furface being known (and
every where equal between top and bottom,) this area multiplied by
the depth in wet inches and parts of an inch, will give the content
fuch a depth.
AND MECHANIC.
III
Ex. 315. Let there be a prifmatic back, length 140 inches, breadth
120, depths at 10 feveral places in inches as here follow.
required the content at
NO
I
2
depth
5.1
4,8
3
5,0
4
5,2
5
4.6
6
4,9
7
4.5
8
4,7
9
5,2
10
4,9
No dep 10)48,9 fum
0,7 inch deep
4,89 mean depth
140 length
120 breadth
282) 16800 area baſe
I
59,57 A G. at 1 inch deep
0,7 propoſed depth
་
41,999 A G. at 0,7 inch deep,
41,699
And thus you may find the content at
any depth you pleaſe
Ex. 316. Again, fuppofe you come fometime to this cooler and find
the depth 1,2 wet inches, what ale gallons are then in it?
59,57 cooler at one inch deep
I,
1,2
71,484 ale gallons anſwer
317. In fuch veffels, there is always a point (the eaſieſt to be come at)
in the edge of the veffel marked for the conftant dipping place, and if
in the laft example you find the depth at the dipping place to be 5 inches
then from this 5 you must take the mean depth 4,89 and there leaves
0,11 inches fo that you muſt always take this 0,11 from the wet inch-
es taken at this dipping place, and the aforefaid content 59,57 multi-
plied by the remainder gives the true content of liquor in the cooler;
but if the mean depth exceed that taken at the dipping place, then the
difference muſt be added to the wet inches taken there, as is eaſy to
underſtand.
318. This will do for backs coolers &c. whofe bafes are equal, and
their fides ftreight, but if they be otherwife you muft (by ex. 311 &c.)
find the content to every inch deep, if you would be very exact, but
the common way is to take or find the dimenſions in the middle of every
inch (tho' fome take them in the middle of every 2, 3, 4, &c. inches)
and fo get a mean area, for one inch deep, and fo proceeding for every
inch of depth you'll have a table as in ex. 311.
*
319. Thefe veffels are very readily inch'd by the lines on the infide
of the two fliders (fee ex. 258) If they are circular and their diameters
between 12 and 36 inches. Thus, put the fliding-rule in the middle
of every inch of depth to take the greateſt breadth or diameter, putting
:
112 THE UNIVERSAL MEASURER
out the flider, or both ſliders, to the oppoſite fides of the veffel, then
by looking the underſides of theſe ſliders you have the diameter there
both in inches and ale gallons, at one inch deep.
320. If theſe veſſels are not quite circular but very near fo, you may
take the greateſt and leaſt diameters (each thro' the middle of the vef-
ſel) and half their fum may be taken as a mean content. If the diam-
eter is but 12 inches it is had by the fliding-rule its felf becauſe it is
12 inches in length, if the diameter be between 12 and 24 inches, it is
had by drawing one flider, but if above 24 inches you muſt draw out
both ſliders, the gallons on the infides of thefe fliders, are found by ex.
277. Thus if the diameter of a circle be 13,4 its content will be 0,5
A G. ſo againſt 13,4 inches you fet 0,5 on the line of gallons &c. for
any other.
321. It is found by tryal that every 10 gallons of hot wort will be
but 9 gallons when cold; therefore it will always be as I is to 0,9 fo
is any gauge of hot wort to the nett wort, fo if a hot gauge be 90,3,
then, As I : 90,3 :: 0,9 : 81,3 the nett content &c. for any other.
To gauge curve-lined tuns, as coppers, ſtills, &c.
322 The common way to gauge theſe veſſels, is thus, take croſs
diameters or dimenfions by which you can have the area or content in
the middle of every 4, 6, or 8 inches of depth, and this area multiplied
by that height gives the content of that part nearly, and thus having
found the content to every 6 &c. inches deep, the fum of all thefe con-
tents muſt be that of the copper, all except the crown whofe content
may be had by ex. 295, or by meaſuring it with water by an ale quart.
323. Otherwiſe by rule 43, take crofs diameters &c. at the top of
the copper, top of the crown, and middle between them, if the cop-
per have but one bulge viz. be as one fruftum, but if it be as feveral
fruftums take fuch diameters for every fruftum, and height thereof;
and fo find the content by the faid univerfal
Rule 43. Then its plain, the fum of thefe contents will be that
required.
Ex. 324. Let d d HH (fig. 129) be a copper with a riſing crown
de H, which is covered up to the diameter d q with 92,8 gallons,
ſuppoſe the top diameter H L D (to be the greateſt) = 115 inches, the
bottom diameter viz. that over the crown to be degrees, the leaft =
96 inches, the depth e L 30 inches, and diameters in the middle of
every 6 inches depth to be as in the following table.
AND MECHANIC.
113
To find the content the common way, the copper being circular.
Diameter in the mid-
Content of every 6 inches
depth in ale gallons.
Inches depth
dle of every 6 inches
6
113,1
6
6
6
109,2
105,28
101,36
6
97,33
213.72
199,26
185,16
171,66
158,22
Sum of contents = 928,02
Gallons to cover the crown
Coppers content
92.8
1020,82
Theſe contents in the 3d column are very readily had by the fliding-
rule. Thus, As the gauge point A G on D is to 6 the common depth
on C fo is each diameter in the fecond column on D to the content on
Cas per third column, which third column in one fum is 928,02 A G.
the content of the copper above the top of its crown, to which add
92,8 A Ġ. the liquor that covers the crown and you have 1020,82 ale
gallons the content of the whole copper. To do this by rule 43 there
wants a diameter in the middle between qd and Hd, which by the
table you fee is 105,28
DB.
13225 = fq. Hd
44335,41 = fq. 2 DB
9216 = fq. q d
66776,41 fum
30 = depth e L
2003292,3
Then
2003292.3
,000464 of,002785
.929,5276272 content of d Hd g
crown add
92.8
1022,327 + A. G. anſwer
325 A diameter taken in the middle of every 6 &c. inches cannot be
a true mean diameter even if the veffel be conical (fee rule 41.) much
leſs ſo if the veſſel is curve-lined as a copper, &c Hence the content
thus found muſt be fomewhat too little, and therefore that found by
the general rule 43 must be truer, i. e. 1022,327 + is a truer content
than 1020,82 and is much eaſier had.
Ex. 326. Let d d H H (fig. 129) be an elliptical bas'd copper, whofe
greateſt bulge or wideneſs is at D B out of the middle of the copper
and the crown d EHC to bend outward. Such veffels as theſe whoſe
greateſt bulge is not in the middle, nor at top or bottom, muſt be
meaſured at twice like two different fruftums, thus, take two croſs diam-
eters 89 and 89,5 at D B the greateſt bulge, two at L 80,5 and 80,$
P.
**
114 THE UNIVERSAL MEASURER
#
the top two at a 85,5 and 86 in the middle between Land a, by which
and the length L b 15, (by rule 43.) find the content of the fruftum
DB Hd. Again, find two crofs diameters 83 and 83,5 at E where
the crown begins; alfo two 86,5 and 87 at g in the middle between
E and b, which the length 21 Eb (by rule 43) find the content of the
fruitum B D d EH, which two fruftums added to the content of the
Crown d C H, gives the content required.
Suppofe that the crown dC H hold 32 ale gallons, Then
diameter at L 80,5 +80,8
6504.4
diameter at b 89,5 + 89
7965,5
4 times diam. at a 85,5 +86 + 4 =
29412,0
fum = 43881,9
15=16
658228,5
inverted factor,000464 464000,0
26329
3949
263
content of B D d LH = 305,41
Again
product of the bottom, diam. 83 + 85,5
6930,5
product of diam. at g, 4 times 86,5 + 87 + 4 30102,0
diameter at b 89,5 +89 7965,5
7965.5
fum = 44998,0
21 = Eb
944958,00
inverted factor
464000,0
content of DBE- 438,460
content of B D L = 305,41
content of the crown d CH
content of the whole copper
32,00
775,87 fum
327.
If you meet with a ftill or any veffel that hath two bulges, you
mult confider it as 3 frultums &c. The contents of fills are given in
wine gallons, and therefore in fuch cafe you muft multiply by the fac-
tor for wine gallons inftead of that for ale i, e. by,on05666 = & of
1003399, inſtead of,000464 = 3 0f,002785, all elfe is the very fame.
The globical part of a ftill &c. may be foon had by rule 50. More ex-
amples might be added, but I think thefe fufficient, to any one that is
acquainted with rule 43, and what is faid about it in this work.
AND MECHANIC,
115
:
Of caſk gauging.
328. Every cafk is fuppofed to be made up of two equal fruftums
(or one middle zone) join'd together at their greater baſes whoſe
common diameter B D (fig. 182,) is called the bung diameter, the
diameters d Hd H, at the leffer bafes are called head diameters
and E L their joint lengths (viz. C L=CE) is called the caſks length,
ſo when you come to gauge a cafk, take the bung diameter exactly in
the middle at C and fee if the two head ones be equal, which if they
be, it may be gauged as one fruftum, by uſing L E to both fruftums
inftead of C E or C L to one of them. But if the head diameters are
unequal you muſt gauge the caſk at twice viz. each ſide of B D as one
fingle fruftum.
329. Several cafks may have the fame length, bung and head diam←
eters, and yet have different contents on account of their different
curvatures as is plain by fig. 182, for the caſk G G, will hold more
than the cafk g g, tho' they both have the fame dimenſions B D and
Hd and L E, for this reaſon caſks are divided into 6 forms or varieties
as you'll fee in what follows, fome have but the 4 varieties mentioned
in prob. 187, amongft which in cafk gauging the 4th form may be left
out, for no caſk can have its ſtaves ftreight from head to bung, never-
theleſs the rules are applicable to conical fruftums.
I
[
2
Spheroid
circular fpindle
variety
3
two equal
fruftums
of a
parabolic spindle
equilateral hyp. fpindle
parabolic conoid
cone
Theſe forms are fet down according as they curve, the firſt variety
curving moſt, the ſecond next moſt, and ſo on to the laſt which hath its
ftaves ftreight from bung to head.
330 Now to find the content of a caſk in each form or variety, and
ſee which holds moſt, let us ſuppoſe each to have the fame dimenſions
viz. length LE (fig. 182) = 40 inches, bung diameter BD 32 in-
ches, head diameter H d = 26 inches.
Firſt, for variety firft, being a spheroidical cafk, this is done by
rule 49.
116 THE UNIVERSAL MEASURER
Thus 32×32×2 20.8
twice fq. bung diam.
26 × 26 = 676
fq. head diameter
Again, for wine
108960
fum 2724
length 40
108960
inverted factor=84290c0.0 Or, = 3331100,0
ale galions
101,15 W G. 123,49
Here 359,5 × 3 = 1077,15 a divifor for ale gallons. Alfo,
294, 12 × 3 = 882,36 that for wine, whofe fquare roots are 32,82
and 29,7. ſo by the
Sliding-rule for ale gallons. As 32.82 on D: 40 on C:: 32 and 26
on D: 38,03 and 25,1 on C, fo twice 38,03 76,06 added to 25,I
gives 101,16 A. G.
If you would have wine gallons, ufe the gauge point 29,7, the fac-
tors ,0009284 for A G. and ,0011333 for W G. are had by ex. 271
or by multiplying 359,05 by 0,2618 and 294,12 by 0,2618. See ex.
184 and 290.
Ex. 331. For the fccond variety, or that of the circular ſpindle.
Rule 60. From the fum of the fquares of the length and head diam-
eter, take the ſquare of the bung diameter, the remainder multiplied
by 14 times the fquare of the difference of the diameters is a dividend,
and 35 times the fquare of the length, added to 5 times the fquare of
the difference of the diameters, is a divifor, get the quotient and
multiply the difference between it, and the fum of twice the fquare
of the bung diameter and fquare of the head diameter, by the cafk's
,0009284
or divide by
,0011333
length, and that product multiply by {
{1077,15 for {aline gallons. or by the
}
S
}
32,82 for AG. on D is to the caſks length on
Sliding-rule. As {32,82 for AG.
29,7 for W G.
}
C fo is the bung and head diameters, and their difference on D to 3
fuch numbers on C, that if to the fecond you add twice the firft and from
that fum take 0,3 of the third, there will leave the content nearly.
All from theorem 95 &c. the diameters being 32 and 26, their differ-
ence will be 6. So per laft rule,
AND MECHANIC.
117
1600 fq. 40 Ed
676=fq. 32 B d
fum 2276
fub. 1024fq. 32 BD
1252
504
fq. 6 X 14
631008
the dividend
Again for W. G.
1085 10,720
108511
10851
5600035 times fq. 40
add 1805 times fq. 6
56180 divifor
=
fo 56180)631008
take 11,232 quotient
from 2724 fq. 26 added to 2 fq. 32
2712,768
40 EL
33331100,0,00113+invert. 108510.72
3255
326
482900,0,0009284 invert.
97660
2170
868
43
100,741 A. G.
WG.
33
3
122,979
Sliding-rule, for wine gallons. As 29,7 on D: 40 on C :: 32 and
26 and 6 on D: 46,42 and 30,64 and 1,63 on C, ſo 92,84 (twice
46,42) added to 30,64 gives 123,48. from which take 0,49 (0,3
of 1,63) leaves 122,99 wine gallons &c. for ale gallons.
Ex. 332. For variety third viz. that of a parabolic ſpindle, done
by rule 56.
1024 fq. 32
676 fq. 26
1024}
Again, for W G.
108384,000
inv. factor 33331100,0
2724 fum
fub. 14,40,4 × 36
2709,6
40= length
108384,0
4829000,0 inverted factor
97546
2168
867
43
100,624 A. G.
108384
10838
3252
325
33
3
W. G. 122,835
Sliding-rule, for wine gallons. As 29,7
on D: 40 on C:: 32 and 26 and 6 (their
difference) on D: 46,42 and 30,64 and 1,63
on C, fo twice 46,42 added to 30,64 and
the fum leffen'd by 0,4 of 1,63, gives at laſt
122,83 W. G. &c. for A. G.
118 THE UNIVERSAL MEASURER
Ex. 333. For the fourth variety, or that of an equilateral hyper-
bolic fpindle.
Rule 61. To the fq of the length add the difference of the fquares
of the bung and head diameters; this multiplied by 14 times the fquare
of the difference of the diameters is a dividend, from 35 times the
ſquare of the length take 5 times the ſquare of the difference of the
diameters, the remainder is a diviſor, take the difference between the
quotient of this divifion and the ſum of twice the fquare of the bung
diameter added to the fquare of the head diameter, and multiply it by
the fame factors for the fame meaſure as in the laſt example, or divide
by the fame divifors 1077,15, or 882,35, for wine. (Theo. 97)
1600= fq. 40
1024fq. 32
fq.32
2624 fum
fub. 676fq. 32
1948
mul. 504 14 x fq.6
981792 dividend
Sliding-rule, for ale. As 32,82
on D: 40 on C:: 32 and 26 and;
6 on D: 38,03 and 25.1 and 1,34
on C, fo twice 38,03 added to 25, 1
and leffened by of 1,34 gives (at
laſt 100,49 A. G.
56000= 35 times fq. 40
take 1805 times fq. 6
55820 divifor
fo 55820)981792
take 17,59 quotient
2724 twice fq. 32
from
2706,41
(added iq.26
40 = length
108256,4
inv. fact. 33331100,0 for W. G.
108256
10826
3248
325
32
3
122,690 W. G.
!
&c. for A. G. 100,505
Ex. 334. For the 5th variety, viz. two fruftums of a parabolic
conoid; here by
ale
Rule 54. the divifor for wine } must be
§ 395.05
294,12
} {
muſt be {718,1
{588,24} viz. twice
fo the factors will be 10013925 and,0017, and gauge
points 26,8 and 24,25.
AND MECHANIC.
119
1024 fq.32
Sliding-rule.
676 fq. 26
As
fum 1700
40
{
$26,8
on D: 40 on C:: 32
24.25
and 26 on D:
[
57,1 and 37.6
69,63 and 45,97
on
C, whofe fum is
{?
94,7 A. G.
115,6 W. G.
718,1)65000(94,69 AG.
1,22
115,6 W G.
Ex. 335. For the 6th variety, viz. two equal fruftums of a cone.
By rule 39.
58=32+26
53
3364
fub. 8321
32 × 26
2532
As
A G. 94,029
1.22
114,71538 W G,
Sliding-rule.
40 length
101 230
4829000.0 inv. factor
9.152
2026
810
[33
32,82 for A G.
29,7 for W G.
:)
on D: the caſks
length 40 on C:: 58 the fum diameters
and 6 their difference on D to two fuch
numbers on C, that if to
the first you
add the fecond, you'll have the con-
tent.
41
94,29 A G.
336. Theſe are the rules fet to the feveral forms of cafks, except
rules 60 and 61, which are approximated as directed after theorem 97,
yet very near true, each form is alfo wrought by the fliding-rule ac-
cording to the rule by the pen. (except variety 2d and 4th, and thoſe
as near as can be done eaſily) a method much fhorter and alfo truer
than that practifed by molt authors on gauging, viz. by the mean dia-
meters
but that nothing may be wanting in this fection, take that
method here.
To gauge a cafk by the mean diameter.
337. First to find the mean diameters by the fliding-rule (fee ex. 257)
on the line of inches, find the difference between the bung and head
diameters, and against it on the line belonging to its variety, you have
a number which added to the head diameter gives the mean diameter,
fo againſt 6 on the inches, you have 4,17 againſt ſpheroiḍ, 3,78 againft
2d variety and 3.39 againſt 3d variety which refpectively added to 26
the head diameter gives 30,17 and 29,78 and 29,39, for the mean
diameters of the fpheroid, parabolic ſpindle and parabolic conoid.
Otherwife, without the rule,
120 THE UNIVERSAL MEASURER
338. Multiply the difference between the head and bung diameters
(,69
1 2
I
}
,68
675
by
for the
3
variety.
,67
4
,525
5
,5 I
This product added to the head diameter gives the mean diameter,
i.e. reduces the caſk to a cylinder of the fame length, hence when
you have got this mean diameter work with it and the caſks length as
if you were gauging a cylinder, and you get the aniwer.
So in the 6 preceeding examples, the difference between the head
and bung diameters is 6.
[69]
1,68
which mul,675
tiplied by,67
,525
L₂5.1
gives
[4.14] this added (30,14)
| 4,08 | to 26, the | 30,
4.05 head diam. 30,05 of
4,02 gives the 30,02 the
| 3,15 mean dia-
| mean dia- | 29,15
[3,06 meter, L 29,06
(1)
2
YAW N
A
variety.
How the multipliers are found, fee theorems 121, 122, 123. 124,
&c. Now any of thefe mean diameters fquared, and multiplied by
the caſks length, and that product divided by 359,05 gives its content
in A G. &c. for W G.
As the gauge A
G. for ale or W
G. for wine fup-
pofe A G. on D
: 40 on C::
Or by the Sliding-rule.
(30,14)
30,08 |
Ham tu
(101,2)
{1}
100,7
12
on D:
100,6
on Cnearly for
3
100,5
the AG. in the
| 29,15
94,7
15
[29,06 J
94,0.
30,05
30,02
variety.
By comparing theſe with thoſe 6 foregoing examples done by the
true rules it appears, thefe contents found by the mean diameters are
pretty near good, whence the mean diameters in this caſe, found by
the conftant factors 0,69 and 0,675, &c. are truer than thoſe found as
above by the fliding-rule..
338. Thus you have the art of gauging a full cafk, if its form be
known, but to do that will require more work than the gauging of it,
as is plain by theorem 64, 65, and 66. Alſo tho' 6 varieties be fixt
on, yet there may be more as appears by fig. 182. But the uſual way
to determine the cafks form is by infpection, or as near as you think
good by looking at it, and indeed is a better rule than the Cooper re-
AND MFCHANIC.
121
gards in making cafks to be in fuch forms; hence a cafk may come
near ſome of thefe varieties by chance, for there is no other rule for
it that I know of.
339. Theſe things confidered, its evident that this method of cafk
gauging is at beft but guefs work, and therefore to have a more ge-
neral method, it will be beft, both in regard of accuracy and readineſs
in practice, to work by rule 43 which regards all forms alike, and to
take the 4th dimenfion, or diameter, in the middle between head and
bung. (See prob. 163.)
How to gauge any cafk without regarding its variety
Ex. 340 Let HBHdDd (fig 128 be fome cafk whofe length
LE is 40. bung diameter B D 32, head diameter Hd 26, and a dia-
meter m G, taken in the middle between the head and bung be 30,4,
all in inches, to fad its content in ale and wine gallons. By the gen-
eral rule 43.
1024
fq. of B D
676 -- fq. H d
3696,64 fq. twice M G
5396,64 tum
40 LE
Again for wine gall.
215865.60
7665000,
107933
12951
1295
151
215865,60
2464000, fee ex. 295
86346
12952
863
43
A G. 100,204
WG. = 122.330
46.4 for ale
Sliding-rule, Rule 49. As 42 for wine
{46.4 for ale }
on D: 40 on C:: 32 and 26 and 60 8,
19 and 12,8 and68,4
{22 and 26,32 and 334
25 and 16,3 and 83
100,2 ale gallon.
122.3 wine gallons.
(twice 30,4) on D:
S
on C
on C
8}
whofe fum is
This is the only method to be followed by any cafk gauger, whe-
ther he take it for eafe, accuracy, or expedition, I have fo much every
where praiſed this rule 43 in this fection that I think it needleſs to ſay
any more about it.
To find the ullage of any ſtanding or lying cafk, &c.
341. If a caſk ſtand upright upon one of its heads, it is called a
ſtanding cafk, but if it lye with its axis parallel to the horizon, it is
called a lying caík, now what I fhall do in this place, will be to find
what quantity of liquor is in a cafk part full, in each cafe.
122 THE UNIVERSAL MEASURER
Ex. 342. Let HHdd (fig. 183) be a cafk of the 3d variety of the
fame dimenfions with thoſe aforefaid, ſtanding upright and filled with
liquor to q z, wet inches OE 28, or dry inches LO=12, diameter
q z at the liquors furface 31,05, diameter m G in the middle between
O and L 29,09, to find the ullage.
This is plain by the figure, that it is but to gauge a fiuftum dtu HE
of the caſk if it be not half full, viz. only full to tu, or two unequal
fruftums B D d H and RD qz. ifit be above half full, viz. full to 0,
whence, if the caſk is above half full it is eafielt to find the content of
the empty part qd LHm z, and take it from the cafks content fo
you'll have the content of the full part d q Oz H, but if the caſk is
not half full, it is fooneft done by taking the content of that part d t u
HE. Therefore, working (as in ex. 340) with q z, Hd, m G and
LO, as bung head, middle diameters and length you'll find the con-
tent of the empty part qd L Hz to be 27,949 A G. Which taken from,
100,6 the cafk's content leaves 72,651 A G. in the cafk, very near let
the variety be what it will.
To work this example the old way by the lines N and SS on the
fliding-rule, firft find the content 103,6 of the full caſk, by its variety
and mean diameter, then as 40 the cafk's length on N is to the radius
of fegments 100 on S S, ſo is 28 the wet inches on N to a referved
number 71,6 on SS. Again, As 100 on B is to the whole content
100,6 on A fo is the reſerved number 71,6 on B to 72 A G. on A,
the liquor in the cafk, nearly of a gallon too little as appears by
that had by the pen before.
Note. If you uſe dry inches you'll have the empty content.
Ex. 343. Let Hdd H (fig. 184) repreſent a lying caík full to 0,
bung diameter BD 32, head Hd 26, Length L E 40 as before; dry
inches B O= 12 m G a diameter in the middle between the head and
Lung 30,5, all given in inches to find the ullage or meaſure of the
empty part B LOQ
Rule 43. To the fum of the areas of the fegments at the bung and
head, add 4 times the area of the middle diameters fegment, multiply
that fum by the length L E-LQ and divide by 1692 (viz 6 times 282).
or multiply by ,00591 for A G. the fame with ex. 291. It is plain
by fig. 128 or by fig. 184, that if from the dry inches B O 12, you
take HC 3 half the difference between the bung and head diameters
there will leave QH9 the dry inches at the head, in like manner you
get min 11,25 and QH 9 being had, annex 3 cyphers to each and
divide it by its refpective diameter, fo you will have 3 verfed-fines, with
which enter the table under V S. and write out the 3 numbers againf
AND MECHANIC.
123
them under fegments, (fee rule 19) and each of thefe 3 numbers fo
found, multiply'd by the fquare of its diameter, you'll have the three
aforefaid areas.
Thus
32
30,5
26
12,000
375
LI,250
3685/55
againſt which
in the table
269013
263052
9,000
346/3/
ftands
241327
Then this mult. by
the fq. of its diam.
(viz.) by
1024 = fq. 32
3721 = fq. 30,5
676=19.26
gives the
275.469
area of the 978,816
feg. =
163,137
fum 1417,422
len.
40
1692)56696,880
quotient 33,5078 A G.
Now this 33,5078 the meaſure of the empty part LB Q, taken from
100,624 the full cafks content leaves 67,116 A G. remaining in the
caſk, if the fame things were given and the wet inches 12, the work
would be the very fame as before, and you would have 33,5078 À G.
for the content of the full part.
344. This method of ullaging a ſtanding, or a lying cafk by the
middle diameter may always be uſed where great exactneſs is required,
but in common practice, as the line SS, on the fliding-rule ferved to
ullage a ſtanding cafk, So the line S L on that rule will nearly ullage
a lying one. Thus, as the bung diameter 32 on N is to 100, on S L
67,2
the referved number on SL
232,85
fo is
{
wet inches 20
dry inches 12
}
on N to
Then as Ico on B is to roo,6 the cafks content on A fo is the reſerv-
ed number
67,2 on B, to the ul-
lage on A viz.
(67,6 A G. in the caſk
32,8
erring fomething above
33,0 A G. drawn out of it
a gallon as per laſt method.
To find the ullage of a lying cafk by its mean diameter.
}
345. By ex. 338, find the mean diameter fuitable to the cafks form
then to twice the wet or dry inches, viz. the leaft of the two, or to
either of them when equal (viz. when the caſk is juſt half full ) add
the mean diameter, and from that fum take the bung diameter, half
the remainder is the height of a mean fegment, (fee rule 19) which
divide by the mean diameter and look for the quotient under V.S.
write out the number againſt it under feg. in the table of fegments,
which multiply by the fquare of the mean diameter, and that product
by the caſks length, and then by (003546] or di- [282 for ale
,004329 vide by ( 231 for wine
fo will you have the ullage required, that is, what is drawn ou if you
wrought with the dry inches (they being leſs) or what remains in the
124
THE UNIVERSAL MEASURER
cafk if the wet inches be ufed (they being leaft); for if a mean diam-
eter reduces a caſk to a cylinder this method alfo reduces any ſegment
thereof to the fegment of a cylinder. &c.
Ex. 346. Let the dimenſions be the fame as in ex. 333, and the caſk-
of the third variety; then by ex 338, its mean diameter is 30.05 which
added to 24 (twice 12 the dry inches) gives 54,05 from which taking
32 leaves 22,05 half of which is 1 1,025, this divided by 30.05 the mean
diameter quotes,367 nearly, againſt,367 under V S. itands,261284,
which multiplyed by 903 the 1q. of 30,05 the mean diameter, gives
235,9894 and this again by the length 40 gives 9439,5760, this di-
vided by 232 quëtes 33,47 A G. drawn out, or 33.47 A G. in, if the
wet inches be 12 Hence, by the true method, viz. by the middle
area, the ullages is 33,5079. By the line SL on the fliding-rule it
is 33; and by the mean diameter 33,47 which is nearer good than
that by the faid line SL the common way of ullaging, and is alſo eaſier
for it does not require the cafks content.
347. This lalt method of allaging may alfo be done by the new line
A L (fee ex. 256; on the fliding-rule. Thus, As the mean diameter
30,05 on N is to 18,95 un A L fo is 11,025 the mean dry inches on
N to 33,9 on A L, the referved number, Then, as that referved
number 33,9
1,9 on D is to 40 the cafks length on C fo is 30,05 the mean
diameter on D to 33 47 fere, on C the ullage required.
3-9. Let things be the fame as in ex. 3,6, and the caſk of variety
6, the mean diameter of the cafk is by ex. 338 29.06, which added
to 24 (twice 12 is 53 06, from which take 32 the bung diameter and
there leaves 21,06 half of which is 10 53, the mean wet, or dry inches,
Then, as 29.06 on N : 18.9, on A L:: 10,53 on N: 33.14 on A L.
Again, As 33.1 on D: 40 on C:: 29,06 on D: 30,7 A G on C,
the liquor in the cafk if 12 was the wet inches, or what is drawn out
if 12 was the dry inches.
་
I
350. If you fuppofe the dimenfions in fect. 6 to be taken by a rule
divided into (6,558 inches to a part, viz. the cube root of 282) equal
parts, then the answer to every figure or example in that fection will
be in ale galons. Thus. if the length of a fquare prifm be 15 times.
6,558 inches, and each fide of its fquare bafe 7,2 times 6,558 inches,
this veffel (by ex. 151) will hold 777.6 ale gallons. Alfo, if the
length of a cylinder be 8 times 6,558 inches, and its diameter 2. 1 times
6.558 inches, this veffel (by ex. 164) holds 27,708912 A G. and fo
for any other fond or veifel in that fection. Likewife, if you take the
cube roots of 231 and of 2150 42, they will be dimenfions for wine
ga
allons and for malt bufhels, it is an cafy method to gauge by ſuch a
rule, which may be divided into gallons and decimal parts of a gallon,
as I judge is very easy to underitand, by which you'll in effect, have
gauging in this book twice over.
AND MECHANIC.
125
A TABLE of the areas of fegments.
V.S. Seg. Area. V. S. Seg. Area. V.S Seg. Area. V. S. Seg. Area.
COI C55042
3000219
4000337
5000+70
6000618
7 000779
8 000951
91001135
010 001329
001553
12 001746
49 014247
50014681
51015119
52 015561
53016007
54016457
55 016911
89 034441
133062026
134 062707
135 063389
136 | 064074
137 064760
2 000119
045012554
46 012971
90 035011
47 013392
91 035585
48 013818
92 036162
93 036741
94 037323
138 065449
95 038909
139 066140
i 96 038446
140 066833
97 039087
141 067528
¡ 98 039680
142 068225
I l
99 040276
143 068924
56 017369
100 040875
144 069625
13001968
14002199
57 017831 ΙΟΙ 041476
58018296
145 070328
102 042080
146 071033
59 018766 103 042687
147 071741
60 019239
104 043296
148 072450
105 043908 149 073161
106 044522
150 073874
107, 045139
151 074589
108 · 045759
152 075306
·
109 046381 153 076026
110.047005
154 076749
III 047632 155 077469
112 048262 156 078194
61 019716
62 020196
63020689
64 021168
65021659
66022154
67 022652
68023154
69023659
70024168
71 024680
72025195
73 025714
74 026236 118 052092
119 052736
15002438
16 002685
17002940
18003202
19003471
20 003748
21004031
22:004322
2300,618
24 004921
25 005230
26 005546
27 005007
28 c06194
29 006527
30 006865
31 007209 75 026761
32 007558
33007913
34005273
35 008638
36 00,008
37 009383
38004763
39 1010148
40 010537
41 010931
42 011330
43 011734
44 1012142
13 048894| 157 | 078921
114049528|| 158 | 079649
115 050165 159080380
116050804160081112
117 051446 161 081846
162 082524
163 083320
164 084059
76 027289 120 053385
77 027821 121 054036 165 08,801
122 054689 166 C85544
78 028356
79028894
800-9435
81 029979
82030526
123055345
167 086289
124 056003
168 087036
~ ~ ~
125056663
136 057326
83031076 127 057991
84031629 128 058658
850:2186 129 059327
80032745
87033307
88 033872
130059999
169 087705
17088535
171 089287
172 090041
173 090797
174 091554
131060672 175 092313
132061348176 093074
126 THE UNIVERSAL MEASURER
V. S. Seg. Area. V. S. Seg. Area.
V. S. Seg. Area. V.S. Seg. Area.
177093836
223130605
269 | 170202
315 212011
178094601
224 131438
270 171089
316 212945
179 095366
225 132272
271 171978
317 213871
180096134
226 133108
272 172867 318 214802
181 096903
182097674
183|098447
184099221
185 099990
186 100774
187 101553
183 102334
189 103118
227 133945
228 134784
229 135624
230 136465
231 137307
292 138150
233 138995
273 173758
274 174649
275 175542
276176435
277 177330
278 178225
319 215733
320 216666
321
217599
322
218533
323 219468
324 220404
279 179122
325 221401
234, 139841
280 180019
326 222277
235 140688
281 180918
327 223215
190 103900
1236 141537 282 181817
191 104685
283 182718
329
328 224154
225093
284 183619
285 184521
286 185425
330 226033
287186329
3317220974
332 227915
333 228858
288 18723+
334 229801
289 188140
335 230644
290 189047
336 231389
291 189955 367 232034
237 142387
192 105472
238 143238
193 IC6261 239 144099
194 107051 240 144944
195 107842 241 145799
196 108636 242 146655
197 109430 243 147512
198|| 110226
244 148371
199 111024 245 149230
200 111823 246 150091
201 112624 247 150953
202 113426 248 151816
203 114230
249 152680
204 115035
250 153546
205 115842 251 154412
200 110050 252 155280
207 117460 253 156149
208 118271 254 15719
205 119000 255 157890
210 119897 250 158762
120712 257 159036
121529 258 160510
213 122347 259 161386
214 123167 260 162263
215 123988 261 163140
216 124810 262 164019
217 125634 263 164899
218 126459 204 165780
219 127285 265 166663
220 128113 266 167546
221 128042 267 168430
222129773
211
212
292 | 190864 338 232579
293 | 191775
339 233026
294 192684 340 235473
295 | 193596 341 236421
296 | 194509 342 237369
297 195422 343 238318
298 | 196337 344 239268
299 | 197252 345 240218
300 198168 346 241169
301 199085 347 242121
302 200921 348 243074
303201232 349 244026
304 201841 350 244980
305 202761
306 | 203681
307 204605
308205527
309 206451
310 | 207376
311 208301 357 251673
312 209227
351 245934
352246889
353247845
354 248801
355249757
356250715
358252631
313 210154
359253590
268, 169315
3601254550
314 211082|| 3601 254550
AND MECHANIC.
127
404297292
370264178
405 298273
373 267078 408 301220
374263045
375 269013
376 264982
372 266111 407 300238
406299255
V. S. Seg. Area.
361255510
362 256471
363 257433
364258395
365 | 259357
366260320
367 261284
368 | 262248
369 263213
371 265144
V. S. Seg. Area. V.S. Seg. Area.
396 289453431 | 323918 |
397 | 290432
398 | 291411
399 292390
400293369
401| 294349
402295330
403296311
43.2 324909
433325900
434 326892
435327882
436 328874
V. S. Seg Area.
466 358725
467 359723
463360721
469361719
470 362717
471 363715
437329866
472 364713
438330858
473 365712
439331850
474 366710
440332843
475 367709
441333836
476 368708
442 334829
477 369707
443 335822
478 | 370706
409 302203
410303187
444 336316
+793717-5
445337310
480 372794
411 304171
446 338804
431 373703
377 270951 412 305155
447 339798
482 374702
378 271920
413 306138
4+8340793
483 37 5702
379 272800
414307125
449 341787
484 376702
380 273861 415 308110
450342782
485377701
381 274832
416 309095
451 343777
486 | 378701
417 310081
382275803
383276775
418 311068
38. 277748 419 312054
385 278721 420313041
386 | 279694 421 314029
387 280668 422 315016
383 281642 423 316004
389 282617 424316992
390283592 425 317981
391 284568 426 318970
392285544
393286521
395 | 288476
427 319959
428
320948
394 287498 429 321938
439322928
495387699
196|383699
462 354795 497 389699
463 355732 498 390699
464356730 499 391699
465 357727500 3926991
452344772
453345768
454 346764
487 379700
488 380700
489 381699
455347759
456 343755
45734975 2
490 382699
491 383699
492 384699
458350748
4:9 351745
460 352751
461353768
493 385699
+94 386699
128 THE UNIVERSAL MEASURER
*
****
***
SECTION IX.
Practical Questions.
It is to be obſerved, that the 40 following queſtions belong to
menfurations, gauging, &c.
Question 1. How many hewn ftones of a rectangular form, each
foot long and 2 feet broad,, will pave a walk 40 yards long and
3 yards broad?
Divide 1080 feet the area of the walk by 7,5 feet the area of one
ftone, and the quotient 144 is the number of Itones required.
Queſtion 2. How many panes of glaſs each 7 inches fquare will fuf-
fice 4 windows, each 5 foot high, and 3 feet 7 inches broad?
The area 10320 inches of all the windows, divided by 49 inches,
the area of one pane, quotes 21030 panes anſwer.
IC
Question 3. How many rafts each 2 inches broad and 1 inches
thick, can be fawn out of a piece of equal fquared timber, the length
of each end being 174 inches and breadth 10 inches?
The area of each baſe of the timber is 17,5 x 10 = 175 inches,
which divided by 2,5 × 1.5 = 3.75 inches the area of one end of a
raft, gives 46,666 rafts for the anſwer.
Question 4. There is a room whofe circuit is 20 yards, and height
4 yards to be hung about with tapeftry 2 foot wide, all except a door
cafe whofe height is 8 foot and breadth 4 feet, what tapestry will do it?
From 60 X 12 720 feet the area of the room, take 8 x 4 = 32,
the area of the door cafe, the remainder 688 feet divided by 2 fee
the breadth of the tapestry quotes 344 or 114 yards, for the anſwer.
Question 5 How many bricks each 9 inches long 4 inches broad
and 3 inches thick, muſt be taken to build a wall 100 feet long 20 feet
high, and one foot thick?
Here 100 × 20X I = 2000 feet the walls folidity, which multiply.
ed by 1728 inches in a folid foot, gives 3456000 the walls folidity in
inches, which divided by 9 x 3 X 4.5 121,5 inches, the folidity of
one brick quotes 2844, 444 bricks anſwer.
Question 6. If a piece of round timber be 20 feet folid, how many
folid feet will it be when hewn to fquare timber?
If the diameter of a circle be 1 its area is 0,7854 and c,5 half the
diameter fquared, and then doubled, is the area of the greatest infcrib'd
fquare, therefore, As,7854: ‚5 :: 20: 12,732 folid feet anſwer.
AND MECHANIC.
129
Question 7. If a cellar be 18,75 feet wide, 10 feet long, and 5,2
feet deep, how many floors of earth are therein at 324 folid feet to
one floor.
Here 18,75 X 10 X 5,2975 folid feet in the floor, which di-
vided by 324 the folid feet in a floor of earth, or digging, quotes
3 floors 3 feet, the anſwer.
Question 8. If the folidity of a cylinder be 2150,42, and its height
10, what is the diameter of its baſe?
The folidity divided by the height quotes 215,042, which again,
divided by 0,7854 quotes 273,67, whofe fquare root is 16,5 the dia-
meter required.
Question 9. If the length of a ſhip's keel be 44 feet, depth of the
hold 9 feet, and mid-ſhip beam 20 feet, what muſt theſe dimenſions
be in another ſhip of the fame mould to carry a double burthen.
By theorem 31. If you cube any of theſe dinienfions, the cube root
of the double thereof, will be the like dimenfion of a ſhip of a double
fize, ſo 44 cubed is 85184, and doubled is 170368 whoſe cube root
55,44
,44 feet for the keel fought. Thus having found any one of the
dimenſions, the reſt may be had by the rule of three.
is
55,44::
20: 25,22 mid-ſhip beam.
9: 11,34 depth of the holds.
}
Thus, as 44:
By the Sliding-rule.
Becauſe the burthens are as I to 2, it will be as I on E is to
44
20
粥
​on D fo is 2 on E to
55,44
25,22
11,34
on D, the fame as before.
Question 10. If the length of a fhip's keel be 80 feet, mid-ſhip 32
feet, and depth of the hold 14,1 feet, what is her tunnage or burthen
in tuns. See Queſtion 169.
The uſual way to guage a fhip is to divide the product of theſe 3
dimenſions in feet, by 95, or by 100, if there be allowance made for
guns &c. Therefore, by the fliding-rule. As 9,78 (the ſquare root
of 95) on D is to 14,1 on C fo is 50,05 (a mean proportional between
80 and 32) on D to 380 fere, on C the anſwer. But if the divifor
be 100, or guage point 10, the burthen will be 360,96 tuns.
Question 11. If the axis of a globe be 4 inches and it's weight 4 lb,
what will be the weight of another globe of the fame metal whofe
axis is 8 inches.
As 64 the cube of the axis is to 4 its weight, fo is 512 cube of 8,
the cube of the like part of any other like folid to 32 lb. its weight,
of the fame metal.
R
130 THE UNIVERSAL MEASURER
Question 12. Whether is half a foot fquare, or half a fquare foot
greater? Half a fquare foot is halfot 144 fquare inches=72, but half
a foot fquare is but 36 inches, the fquare of 6.
Queft on 13. What is the difference between half a foot folid and
half a folid foot ?
Half a folid foot is half of 1728 folid inches-864, but halfa foot
folid is 216 being the cube of 6 inches or half a foot.
Question 14. If one fathoms of a cable rope weigh 17 lb. when it is
10 inches about, what would it weigh if it were 18 inches circum-
ference?
As 100 inches the fquare of the periphery of any cylinder's bafe is
to 17 lb. its weight or folidity, fo is 324 the fquare of any other cylin-
der's bafe's periphery of the matter or height, to 55,08 lb its weight
or folidity. See theorem 37.
Question 15. If an oak cheſt be 8 folid feet when meaſured on the
outfide and but 7 folid feet when meafured in the infide, what is its
weight A folid inch of oak weighing 0,537 parts of an ounce.
Since folidity is as weight; therefore, As one inch is to its weight
,537 oz. fuis 1728 inches (the difference between 8 and 7 feet) to.
928 oz. fere, the anſwer.
Question 16. What length of a rope will be fit to tye to a cow's
tail, the other end being fixt in the earth, that fhe may grafs juſt an
acre or 4840 ſquare yards, allowing the cow's length 4 yards.
The length of the rope and cow mull be the radius of a circle whoſe
area must be 4840. Therefore 4840 divided by 0.7854 is 6161,09
the fquare of the whole diameter, whofe fquare root is 78,49 half
whereof is 39,245 yards the anſwer. See rule 15.
Question 17. What mufl be the dimenfions of a cubical box, to
hold 200 oranges of a globular form, each 24 inches diameter ?
If the oranges be laid in rows upon one another, each will take as
much room as a cube would do, whofe fide is 24 inches; therefore,
the cube of 2,5 = 15.625 multiplied by 200 gives 31250 for the fol-
idity of the 2000 cubes, or that of the box. So thecube root of 31250.
is=15,5 fère, a fide of the box.
Queſtion 18. If a plank 14 feet long, I thick, and half a foot
broad, can be fold for 8d a foot running meafure, 7d a foot fuperficial
meafure, and 10d a foot fold meafure; which of thefe ways mult it
be fold to make the moſt money?
Its area is 14 X 1,5 21 feet, its felidity is 14 × 1,5 × 0,5 =10,5
feet, its length 14 feet; fo, 14 times 8d is 112d, its price at running
.meaſure, and 10,5 times rod is = 105d its price at folid meafure;
alſo 21 times 7d is 147 d value at flat meaſure which is the beſt way
to fell it.
AND MECHANIC.
131
Queſtion 19. If a board be to feet long, 8 inches broad at the great-
er end, and 6 inches broad at the leſſer end, how much in length at
the leffer end will make one foot.
Becauſe the ends are given in inches, and length in feet, we'll call
the foot to be cut off 12 inches (fee ex. 83) then to (48) twice the
product of 12 the part to be cut off and 2 the difference of the breadths
8 and 6, add (360) the product of the length 10 feet and fquare of
the leffer breadth 6 inches, multiply the fum (408) by the length 10
feet, from the fquare root of (4080) that product take the pr duct of
the leffer breadth 6 inches, and the length 10 feet, (60) and the re-
mainder 3.8 divided by 2 the difference of the two breadths gives 1,9
for the anſwer. See theorem 125
Question 20. If a piece of fquare tapering timber be to feet long,
9 inches fquare at greater baſe, and 6 inches ſquare at the leffer baſe,
how much in length from the leffer end will make a folid foot. See
theorem 126.
Firſt, 9 inches,75 F. and 6 inches,5 F. and their difference
3 inches
25. Then to (1,25 F.) the product of the height 10 F.
and the cube of a fide (0,5 F.) of the leffer baſe, add (,75) 3 times
the product of 1 F. the part to be cut off and ,25 F. the difference
between a fide of each bafe multiply (2,00) the fum by 100 F. the
íquare of the height, and from (5,849) the cube root of that product
(200) take the product (5) of the leffer breadth,5 F. and height 10 F.
the remainder 0,849 divid d by,25 F. the difference of the two bafes
quotes 3,4 fere for the feet in length require 1.
I
Note. If it be the fruftum of a cone you may multiply the part to
be cut off by of 0,7854, and then work with that product inſtead of
the faid part, and the diameters as with the fides before: Alfo, if any
part was to be cut off from the greater baſe, take the content of that
part from the content of the fruftum, and work to cut the remainder
from the leffer baſe as before.
Question 21. If a board be 24 inches broad at the greater end, 8
inches broad at the leffer end, and 20 feet long, where mult it be cut
fo that a foot in length may contain 156 inches area?
The content 156 divided by 12 the length quotes 13 inches for the
mean, or middle breadth of the piece to be cut out, (fee theorem 26)
then (by theorem 9) As 16 (24-8) is to 20 the length, fo is 5 (13
—8) to 6. 24 feet the diſtance between the middle of the piece to be
cut out and the leffer end.
Question 22. A hath a piece of fquare tapering timber 24 inches
fquare at the greater end, 6 inches fquare at the leffer end, and 60 feet
132 THE UNIVERSAL MEASURER
long; B bids him 12d a foot running meaſure, C offers him 18d a foot
folid; how muſt he cut it between B and C to make the moſt of it to
himself?
of one folid, it is evident, that
As the value of a foot in length is
a fide of the dividing fection must be the fide of a fquare prifm,
wherefore a foot in length will be to of a foot folid : but 3 of 1728
the inches in a folid foot is 1152, which divided by 12 inches or a foot
in length, quotes 96 inches the area of the dividing ſection, whoſe ſq.
root is near = 9,8, then, As 18 inches (24-6) is to 60 feet length,
fo is 3,8 inches (9,8 — 6) to 123 feet from the leffer end where the
tree is to be cut, and B to have the ſmaller end, C the thicker.
Queſtion 23. Things being the fame as in the laft queftion, fuppofe
the folid foot at 12 d and the foot in length at 18 d, where muft it be
cut to make moſt ?
3
Here, its plain that if ſuch a ſolid foot be cut out of the tree (by
queftion 21) as that its length be of a foot or 8 inches, the value of
this foot folid will be the fame in both meaſures, confequently, if the
tree be cut thro' the middle of this piece it will anfwer the queſtion.
So 1728÷8216, whoſe ſquare root is 14,7 fere a fide of the
dividing fection; then, As 18 (24—6) is to 60 feet fo is 8,7 inches,
(14,7—6) to 29 feet length running meaſure from the leffer end to
go at 18d a foot.
Question 24. Whether will fmall, or thick round timber waſte more
in fquaring.
By queft. 6. Any piece of round timber is to the fame when ſquar-
ed, as ,7854 to,5, fo there is no difference, i, e. the wafte is as the
thickneſs.
Question 25. If a tree girt 22 inches with a rope of one inch dia-
meter, what is its true girt?
In girding timber &c. the upper fide of the cord is made to meet,
then being ftretched on a rule, its plain the girt of the tree is taken to
be what this upper fide meaſures too, now if this upper fide be 22 in-
ches periphery it's diameter will (by rule 17) be 7 inches, but the rope
being 1 inch thick, the diameter of the tree or underſide of the rope
will be but 5) 7--2) inches; therefore, as 7: 22 5: 15 inches,
the true girth of the tree. Hence appears the neceffity of girding with
a fmall cord.
Question 26. If D L (fig. 185) be the length of an egg 4 inches,
d D1 inch, the periphery of the greatest circle G H 10 inches, a
periphery taken at e in the middle betweend and D=7 inches, and one
taken at a or m m, in the middle between n and L 6 inches, what
is the folidity? (See rule 43.)
AND MECHANIC.
133
100= fq. H G
196 = 4nn
100
fq. H G
58,1940
144 = 2 m m
23,5320
296 fum
244
I
length d D
6)81,7260
3=d L
296
732
13,621 in, folid.
,0795 factor
,0795 factor
23,5320
58,1940
Question 27. Required the folid and fuperficial content of an ellipti-
cal ring, whole diameters taken in the infide, are 28 and 38 inches, and
thickneſs of metal in the ring 2 inches diameter, the faid ring being cy-
lindrical.
Here 38 and 28 each added to 2, the thickness of metal in the ring,
gives 40 and 30 for the two diameters of the oval paffing thro' the
ring's middle, whofe periphery, (by rule 22) is 110 for the mean length
of the ring; then 2, viz. 4 X 110 X 0,7854 172,788 the folidity
in inches, and 3,1416×2× 1100 = 691,152 inches, the fuperficial
content.
Queſtion 28. If the walls of a building be 20 yards about on the
outſide, 16 yards about in the middle, 5 yards high, and 0,5 yards
thick, what is the folid content of theſe walls? (See theorem 26.)
Here, half the fum of 20 and 16 is 18 y. the circuit of the wall if
taken in the middle; fo 18 × 5 × 0,545 folid yards, the anſwer.
Question 29. Required the axis of the greatest cylinder, that can be
made of a given diameter 20 and diagonal 30; or, of the greateſt
cone under a given flant length. (See theorem 146.)
The required axis is 17,31, the given diameter or flant fide 30,
multiplied by 0,577 the fquare root of
=
Question 30. Required the axis of the greateſt cone that can be cut
out of a globe or fpheroid, whoſe axis is 30. (See theorem 145.)
This, as in the laſt queſtion, is found 17,31. But the axis of the
greatet infcribed cylinder will be = 20, =
fide of the globe or ſpheroid's center.
of 30, viz. 10 on each
Question 31. If a rectangular piece of ground is to be £2 for every
chain in length, and £3 for every chain in breadth, Quere, the length
and breadth, to as moſt land poffible may be had for £40.
To give a general folution to all questions of this kind, let p=£2,
9=£3, s = £40, e one of the dimenſions, viz. either length or
breadth, and a the other of them; then a eA the area and pe+
qas the money to be paid for that area, and by tranſpoſition, &c.
es
=a; whence, Aae—
Leep =m a max ; fo, (by
s-pe
૧
q
134 THE UNIVERSAL MEASURER
S
art. 221, prob. 191)
s-2ep
o, hence, e=
40
= 10,
9
2 P 4
a p
20
pe 40
20
=6, then 6 × 10 = 6,6 3 acres
9
3
3
and a =
S
the anſwer.
Quest. 32. Required the folidity of the greateſt cylinder that can be
cut out of a fquare pyramid, whofe axis is 60, and a fide of its baſe 30.
Firſt, (by prob. 191) the pyramid muſt be cut at of its axis, fo
the cylinder's height will be of 60 = 20; then, as 60: 30 :: 40 viz.
(3 of 60): 20, a ſide of the baſe of the pyramid cut off, which is alfo =
the diameter of the cylinder's baſe. So 20 X 20 X 0,7854 X 20 =
6283,2 anſwer.
Question 33. A weaver's beam 24 inches in circumference, on which
is 95 rounds of cloth of an inch thick, what is the length of the
web?
95
22
7050
2
✰
TZT
Firſt, As 22:7:: 24: 4 the diameter of the beam and xd
twice 95 is
=, twice the thickneſs of the ring of cloth, fo 84 - 25
= 'the diameter of the beamnandweb together. Then from 32041
the fquare (17) this diameter, take the fquare of (4) the beams
diameter, and the remainder 24985 multiplied by 0,7554 gives the
area of the end of the ring of cloth, which area divided by (viz.
xd by 22) gives the length of the web in inches, which divided by 36,
See rule 35.
quotes 99,02 yards the length of the web.
Question 34.
What length of wire will come out,
One fourth of an inch about,
From brafs in meafure jult a foot;
Pray, fir, try if you can do't.
ΣΤ
1
That is, out of a folid foot of brafs, what length of wire may be
drawn that is 0,25 inches circumference? (See rule 36.)
Here, 0,25 × 0,25 × 07958 =,004973, the area of that circle
whoſe periphery is of an inch. Then,004973)1728;000000(347468
nches, length, which divided by 36, and that quotient by 1760 will
fhew it in miles.
Question 35. If the perimeter of a circle, trigon, and fquare be each
equal to unity, which of them is greateſt ?
IX,07958,07958
area circle
Scircle
of
XX,43303,04811
trigon
1X1=1,0625,0625 the fquare
4
perimeter, the circle is the greatft.
by which it appears that
of all fuperficial figures
contain'd under the fame
Question 36. A globe, à cube a cylinder,
All three in furface, equal are,
In folidity, what do they diff'r.
AND MECHANIC.
135'
Suppoſe the ſurface of each to be 3,1416, then 0,5236 the
folidity, and 3,14164,7124, and the fquare root of that quotient
taken gives 0,821, the diameter and height of the cylinder which
cubed and multiplied by 0.7854 gives 0,42985, the cylinders folidity.
Alſo, 3.1416 – 6=0,5236 the fquare of a fide of the cube, whoſe
fquare root 0,723 being cubed is 0,3786, the cube's folidity. Hence,
of all folids under the fame fuperficies, the globe is the greateſt.
Question 37. If I take an angle of 50°, with a femicircle, which
does not ftand level, but makes an angle with the horizon of 33° 45′
what is the true angle of obſervation.
Let BC (fig. 98) be the edge of the inftrument, making an angle
ABC= 33° 45′ with the horizon A B, then its plain, while the point
of the index moves a flope from 3 to C, it would move horizontally
from B to A, fo if B C fubtend the obferved angle of 50°, B A will
fubtend the true or required angle, or AC if it be an angle of altitude;
fo (by axiom 2) As radius :-50°:: co-tine inclination 500 15: 410
the truc L of obfervation, or :: fine L inclination 33° 45′ : 27,6° the
true 4, if the obfervation was for an altitude Hence appears the ne-
ceflity of having your inftrument level if you are to take L's for diftan-
ces, or if you are to take altitudes.
Question 38. A whin, a thorn, a fheaf of corn,
Each to be meaſured are;
Come, let us mind, if we can find,
A rule fuch things to clear.
The ufual way to me.fure in regular folids, fuch as craggy ſtone,
lumps of metal, bulhes of thrubs, &c is to fill a veffel with water, and
then put in the fold, appote 5 ale gallons of water to run over the top;
of the veffel, then the content of this immerfed folid is 5 times 282=
1410 folid inches; or, if you take out the folid and meaſure the empty
part of the veill, it will give the fame content.
Question 58. If a tub ten inches deep and no more,
Hold in ale gallons juft half a fcore,
What mult the two diameters be,
To be in the ratio five to three?
This veffel being fuppofed a conical fruftum, find (128.284 inches)
the content of a conical fiulum whofe depth (10 inches) and propor-
tion of diameters (3 and 5) are the fame with thofe given in the quef-
tion, then fay, as this folidity (128,284) is to the given folidity (2820
inches) fo is (9) the fquare of (3) the leffer ratio (3) to (190,05) the
fquare of the leffer diameter, or fo is (25) the fquare of the greater
diameter, whofe fquare root is the diameter fought, to the fquare root
of 190.05 is 13.9 inches the leffer diameter, and as 3: 5 :: 13,9 :
23,1 the greater diameter.
136 THE UNIVERSAL MEASURER
Question 40. A cylinder its faid, there is to be made,
To take the leaft wood that can be;
Pray let us fee, what dimenſions muſt be,
To hold a gallon of brandy.
Divide (462) twice the given folidity by ,7854 and (8,38 inches)
the cube root of that quotient is the diameter, (fee theorem 147) by
which and queſtion 8 the height is found. (4,2 inches)
The 10 following queftions fhew how the contents of bodies may
be had by knowing their centers of gravity, and the contrary, fee theo-
rem 190.
Definition. If in any fide of any plane or furface, we find its center
of gravity, and fuppofe the plane to turn round, fo as this ſide deſcribe
a circle, as being the radius thereof; then a line perpendicular to the
radius on the point on which it turned, is called the axis of rotation,
and the diſtance between the faid point and the center of gravity, is
called the radius of gravity. Then to find the content of fuch bodies,
this is the
Rule 62. Multiply the area of the generating plane, the radius of
gravity, and 6,2832 (viz twice 3,1416) into one another, the laft pro-
duct is the content. Or the faid area multiplied by the periphery of
the circle deſcribed by the radius of gravity gives the fame content.
Question 41. If a circular fector A QBS (fig. 153) whofe radius.
SQ= 12 chord A B 6, arch A QB 7 nearly, be turned about the
center S, with the radius S Qperpendicular to the axis of rotation,
what is the content of the folid thus form'd
A Q B
6=180
1
By Art. 275
12 S Q
7
12 X 2 A B
42
7
area fector or generating plane
S C radius of gravity
3X7=21)144
SC 7 fere for rad. of gra. 294
=
6,2832 conftant factor
1847,2608 content anſwer
Question 42. If the fector (laft queftion) be turn'd about the point
Q, fo as Q S be perpendicular to the axis of rotation, what is the foli-
dity of the folid, form'd by this revolution?
6,2832 conftant factor
42 area fect or gen. plane
263,8944 product
263,8944 product
5 = QC rad. of gravity
1319,4720 folidity required
Question 43. What is the content of a folid, formed by the rotation
of the femicircle A B C (fig.
lar to the diameter C A,
186) about the tangent A T perpendicu-
14 the femi-periphery A B C 22.
AND MECHANIC.
137
11 = half of A B C
7=A0
77 = area femi-circle or generating plane
6,2832 conſtant factor
483,8064
7 = radius of gravity A O
3386,6448 folidity required
Question 44. But if the femi-circle ABC (fig. 186) revolve round
the tangent D T, parallel to the diameter A C, what is the content
of the parallel fo form'd?
By Art, 275
49=AO
4
22X3=66)196
nearly 3 =
Note This queſt. is uſeful
in finding the foliditics of
vaults &c.
7=0B
3=0G
4 BG radius of gravity
6,2832 conſtant factor
25,1328
77 area circle
1935,2256 folidity required.
Note. If the axis of rotation do not touch the generating plane, but
be at fome diſtance from it, the rule is ſtill the fame; alfo, if you turn
the factor 6,2832 into a divifor 0,157, then any of this fort of conciſe
folutions (for fo they are when compared with the common methods)
may readily be done by a fliding-rule. Thus, As 0,157 on A: the
area of the generating plane on B:: the radius of gravity on A,
the content on B.
سة
to
Question 45. If the rectangle EDCH (fig. 187) be turned about
the axis of rotation B A, at the diftance A E-8 from it, what will
be the content of the folid viz. of a hollow cylinder, formed by this
revolution, DC-EH being 10 and the breadth HC-ED=6?
The center of gravity of the generating plane EDCH will (by
Art. 274) be at G in the middle of E D, fo AE8+EG3=AGII
the radius of gravity.
Then 6,2832 constant factor
60=10×6= area EDCH
376,9920
IIA G radius of gravity
4146,9120 folidity fought.
Sliding-rule. As, 157
on A: 60 (10 × 6) on B
:: II on A: 4147 on B
the content of matter
in the hollow cylinder.
fought.
Question 46. Of half the rectangle (laſt queſt.) viz. H ED be ta
ken inſtead of the whole one H EDC, the hollow of the figure gene
S
138 THE UNIVERSAL MEASURER
=
=
rated thereby will be a cylinder, height H E, diameter A E, its
thickneſs at one end Ho, and at the other end D ED—6,
what is the folidity of this figure?
L
Firſt, the radius of gravity will be (A E 8 + ‡ E D 2) = 10, then
25 0,157 on A : 30 (area ▲ HE D) on B:: 10 (radius of gravity)
on A: 1885 on B, the content of matter in this hollow cylinder.
Question 47. Given the perpendicular B P 60, (fig. 188.) and the
fegments of the bafe A P 120, CP-90 of a plane A ABC, what
will be the folidity of a folid, formed by the rotation of this about
the axis CD?
Here, becauſe the A's A BP and C B P, have each their centers of
gravity G and g, we may reduce theſe two centers to one, (by theorem
185) or find the content at twice, thus, of C P 90 = ( g 60, the ra-
dius of gravity of the AC BP, and C G = (C P 90+ PG 40, viz.
+AP 120) 130; fo, as 0,157 : area's A B P and CBP 3600
and 2700: C G and C g 130 and 60: 2953537,6 and 1017878,4,
which added together gives 4071416 the anfwer. In like manner may
folidities be computed, when their generating planes confift of feveral
forts of figures.
If the contents of any fuch folids as thefe 7 laſt queſtions be known,
the center of gravity of the generating plane may, by the laſt rule, be
found; but to find the center of gravity of any folid (whoſe greateſt a-
rea is at one end, and leaſt area at the other end) without its equati-
on, &c. this is a general
Rule 63. To the greateſt area, viz. the area at the greater baſe, add
twice an area taken in the middle between the two bafes, multiply this
fum by the fquare of the folid's axis, and divide the product by 6 times
the folidity, for the diſtance of the center of gravity from the leffer
bafe. The fame method anfwers in a plane, by ufing breadths inftead
of areas; if you divide the faid product by 6 times the diſtance of the
center of gravity from the leffer baſe, the quotient is the content. See
theorem 190.
Queſtion 48. If B F the
of a plane ▲ A B C, (fig. 188) be 60,
and the baſe A C 40, what is its center of gravity?
First, by the property of the A, drawn thro' the middle of B P paral-
lel to A C, will be 20, fo by rule 63, twice 20=40+ AC 40 gives
80, and 80 X B P 60 ( 3600) is 288000 the product, which di-
vided by 7200 fix times the area or content (40 X of 60×6) of the
figures quotes 40 for the diſtance Bd of the center of gravity.
Question 49. If the length of a conical fruftum be 40 feet, or 480
inches, and diameters at ends 8 inches and 6 inches, where is its cen-
ter of gravity?
AND MECHANIC.
139
Firſt, by rule 43, the content of this fruftum is found 111589,632
when multiplied by 6, viz. 6 times the folidity in inches, then 98
twice fquare of 7, a diameter in the middle, added to 64, the ſquare
of 8, that at the greater baſe, gives 162, fo 162 ×,7854127,2348,
which multiplied 230400 the fquare of 480 inches, the axis gives
29314897.92 this by 111589,632 quotes 262.7 inches, for the
diſtance of the center of gravity from the leffer end.
Note. In fruftums of cones or others of circular bafes, you may uſe
the fquares of the diameters inſtead of the areas, and in thefe of pyra-
mids the fquares of like fides. Or, from rules 43 and 63, we may draw
a fhorter rule for fuch fruftums, the fame as by theorem 189,
Rule 64. To half the fquare of the ſum of a fide at each baſe, add
the ſquare of a fide at the greater bafe, this fum multiplied by the
axis is a dividend, then to the fum of the fquares of a fide at each
bafe, add the fquare of their 'fum, for a divifor, the quote of this di-
vifion, fhews the dillance of the center of gravity from the leffer end.
Note. Obferve to take fimilar fides at the bafes, and if conical fruf-
tums uſe diameters or peripheries inſtead of fides, or the ratio's of any
fimilar fides &c. at each bafe will do.
Question 50. If the greatest and leaſt diameters of a veffel, or =
headed cafk be 37 and 30,834, what muſt the length be, when the
content is the greatest or convex fuperficies the leaft poffible. fig. 207.
1. Let c Id Bd Ic, be fuch a caſk, 2 a= A B
30,834, the leaft
2y=dc37 the greatest diameter e I m half the cafk's length,
then (art. 435) e = 2,3025 a × log. Y+✔:yy
a
aa:
=18,5 fere,
which doubled is 37 Im I, the cafk's length required, and the
curvature is that of the catenary.
The 34 following queſtions concern more ufes of the centers of
gravity, the forces of the mechanic powers &c.
Question 51. Two men A and B are to have 5 s for carrying a piece
of fquare timber, length 40 feet, a fide at the greater bafe 8 inches,
and a fide at the leffer baſe 6 inches, A bears a foot within the greater
end, and B a foot within the leffer end, what money muft each man
have in proportion to the weight he bears?
The weights (by theorem 184) born are inverfely as their diftances
from the center of gravity; now by the laft queftion the center of
gravity from the leffer end is 2134 feet, which taken from 40 feet the
whole length leaves 184 feet its diftance from the greater end, and
becauſe each man bears one foot within the end, take 1 from each of
theſe and there leaves 2014 and 174, ſo as 2034 : 1774 oz. out of
fractions, as 773: 633:: the weight born by A, to that born by B,
confequently as (773+633) 1406:55:: 773: 25=9d nearly A's fh.
633:25=3d Beth,
s
I
I
$
140 THE UNIVERSAL MEASURER
1
ZN
Question 52. A piece of tapering timber 24 feet long, being laid ·
over a prop, is found to ballance itſelf when the prop is 13 feet from
the leffer end, fo that point is it's center of gravity, but removing the
prop a foot nearer to the faid leffer end, it takes a man's weight of 15
itone, ſtanding on the leſſer end to hold it in equilibrium, what is the
tree's weight?
By theorem 185. As I foot (the diſtance between the prop and
center of gravity) is to 13 feet, fo is 15 ſtone to 195 ſtone the anſwer.
By this, if a man know his own weight, he may know the weight of
any beam &c.
Question 53. An irregular folid 60 inches long, laid over a propſet
to its middle, and a weight of 25 ftone, fet 10 inches from the prop
towards the leffer end, holds the body in equilibrium, but removing
the prop 8 inches nearer to the leffer end, the weight 25 ftone is re-
moved 15 inches nearer thereunto before the body will be in equilib.
upon the prop, what's the bodies weight and folidity, of a ſtone of
fuch matter being a folid foot? See theorem 187.
Firit, 8 inches x 10 inches=80 and (10 — 2+ 15) 17 — 10 —7,
then 80÷711 inches, the bodies center of gravity diſtant from
the middle, (fo by the fame theorem) 10 inches x 25 ſtone = 250,
which by 11 is 21,875 ſtone, the bodies weight, then, as 1 ftone
is to foot fo is 21,875 ſtone to 5,468 folid feet anſwer.
Question 54. If A B (fig. 189) be a walking ſtick 40 inches long,
ſuſpended by a ſtring S D, faſtened to its middle, now if a body be
hung at e, 6 inches diftant from D, and a weight of 2 lb. hung at the
fmaller end A, the flick will be in equilibrium, but removing the body
to a, one inch nearer to D, the 2 lb weight on the other fide D is
moved to d, within 8 inches of D, before the ſtick will reſt in equilibrio
what's the body's weight?
By theorem 185. Multiply (12) the difference between (20 and 8)
the diftances of the 2 lb weight from the point D offuſpenſion, by (2):
the faid weight, that product (24) ÷ by (1) the difference between
(6 and 5) the diſtance of the body from the faid point D, quotes 24 lb
the weight of the body.
Note., This is a very eafy and fimple way to weigh any matter, it
requires nothing but any known weight and any walking ſtick &c, di-
vided into inches, or any other parts.
Question 55. If a two pound weight and a body be in equilibrio,
on a ruler, beam, ſtaff, &c. fufpended in the middle, or in any other
place, and if by moving the body 1, the weight is found to move 12,
before they again be in equilibrio, what is the weight of the body?
See theorem 183.
AND MECHANIĆ.
141
This queſtion is the very fame with the laft, but the folution made
more eafy. Thus, as I the diſtance moved by the body, is to 12, that
moved by the weight, fo is 2 lb the weight, to 24 lb the body's weight.
Question 56. If in a pair of fcales, a body weigh 90lb in one ſcale,
but being put into the other ſcale it only weighs 40 lb, what is its true
weight?
By theorem 186, 90 × 40 = 3600, whoſe ſquare root is 60, the
anſwer. By which you may know if the ſcales be true.
Question 57. If An and Cn be two cords (fig. 190) having each an
end fastened to the nail n, with a ſtaff or ruler A C, faftened between
them, as alſo a plumb line n P, all hanging at liberty on the nail n,
now if a weight of 2 lb be hung to the end C of the ſtaff, and a body
B to the other thereof, the ſtaff is cut be the plumb in e, fo, as I is to
20 fo is e A to e C, what is the weight of the body B?
By Art. 281, e is the center of gravity of B and W. therefore
By theorem 183 or 184, As A e 1 :: e C 20:: W 2 lb: B 40 lb
the anſwer, or (by the ſaid art. 281) it is all the fame whether the ſtaff
A C and weights B and W, be ſuſpended by the two cords n B and
n C, or at e by the one cord n e, or on a prop fet to the point e, the
equilibrio in any one cafe, holds in all.▾
Question 58. If two bodies W and b (fig. 191) be in equilib. (W
110lb and b = 2 lb) on a prifmatic beam AB, fufpended on D, what
is the diſtance D b, if D B — 2 and D A = 14?
Firſt, Suppoſe the weight of the beam A B = 16, then becauſe it is
of thicknefs the length of any part will be as the weight of that part
and alfo the center of gravity of any part DB or D A will be in the
middle thereof, fo (by art. 272) A DX÷AD+Dbxb=DBX÷
BD+DB× W, that is 14 X 7 + 2D b = 2 × 1 + 2 × 110, or 98
222-98
+2DB=222 ſo Db =
= 62 anſwer. If you take the
2
weight W = 1 lb, 2 lb, 3 lb &c. you may thus find the diſtances D 1,
D2, D3 &c. which is the conſtruction of the Roman fteel yard, for
weighing bodies at the end B, by moving a known weight on the other
arm D A.
Question 59. Two men A and C (fig. 190) bearing a weight of 30
ſtone upon, or hung to a leaver A C at the point e, A bears A e 2 feet
from the center of the weight and C bears Ce 6 feet therefrom, what
weight does each man bear?
This queſtion is the fame in effect with queſtion 51, i. e. the weights
born are inverſely as their distances from the center of gravity e, there-
}
!
142 THE UNIVERSAL MEASURER
fore, As 8 (2+6): 30 ftone::
{
S2: 7,5 ftone
: 22,5 ſtone
S
the weight born
by {}
C
A
&
the anſwer.
Question 60. What is the leaſt number of weights, and the weight
of each, that will weigh any number of pounds between 1 lb and 500
tons?
Ι
I
If you take the weights 1 lb and 3 lb, they'll weigh 1 lb, 2 lb, 3 lb,
and 4 lb, for I weighs 1, 3 weighs 3, 3 and I weighs 4, 1 from 3 weighs
2; alfo, if you take the weights 1 lb 3 lb and 9 lb, you'll find that by
them any number of pounds may be weighed between 1 lb and 13 lb,
&c. &c. Whence it appears that a ſeries of terms in geometrical pro-
portion, whoſe firſt term is 1 and common ratio 3, will weigh at one
draught any number of weights between I and the fum of all the terms
in the ſeries, therefore the required number of weights muſt be 14,
and their weights in pounds 1, 3, 9, 27, 81, 243, 729, 2187, 6561,
19683, and fo on to 14 terms.
Question 61. How muſt an axle tree be fixt in the fides of a conical
fruftum or bucket, whoſe flant length is 30, diameters 10 and 20, fo
that it may hang with the mouth downwards when empty, but upwards
when full ?
By comparing the equations in art. 278, it appears that the center
of gravity of a folid fruftum is nearer the greater end than that of a
hollow fruftum. Therefore the bottom of the bucket muſt be in the
greater end of the fruftum, and the required point in the common cen-
ter of gravity of the bottom and hollow fruftum; fo by the faid art.
278, divide 1500 (twice the greater diameter, 20+ to the leffer, 10
and × by the length 30) by 90 (3 times the fum of the diameters) and
the quote 163 is the hollow fruftum's center of gravity from the leffer
end, or (30—163) 13 from the greater end, now fuppofe the mid-
dle of the breadth of the bottom, to be within the greater end, and
its weight or folidity to that of the hollow fruftum, as 2 to 15, then
(by theorem 185 ) 13 × 15 ÷ 15 +2 = £4=114, the diſtance of
the axle-tree from the middle of the bottom's breadth.
I I
Queſtion 62. If A B C (fig.192) be a rowler (over a mine pit, well
&c.) of a A 6 radius with a crook B D E at one end A A, of B D 18
radius, what power muſt be applied to the handle D E to make the
rowler A C raiſe a weight W of 200 lb?
It is evident that the radiuſes a A and a D may be taken as two
leavers acting against each other at the center a, therefore, (by theo”
xem 184) as 18:6:: 200: 663 lb anſwer.
AND MECHANIC.
143
Question 63 If the radius of a wheel be 40, and a power of 500 lb
applied to its periphery does move a weight at the axle-tree of 40000lb
what is the radius of the axle-tree?
The property of the wheel my be eaſily underſtood by the laſt
mentioned figure, viz. a D being taken as the radius of the wheel and
a A as that of the axle-tree; therefore, As 40000 lb: 500lb :: 40:
for the anſwer.
Queſtion 64. If there are two blocks or pullies, the one fixt and the
other moveable, fuch as are in ſhips &c. for hoifting, and to the move-
able block there be a weight of 300 lb', what weight pulling at the
rope end will raife it?
The axis of pullies may be taken as leavers, whofe fulcrums, are
in their middles, or centers of the pullies, whence (by theorem 184)
you can by one pulley only raiſe a weight equal to the force that pulls
at the other end of the rope; but in any number of pullies any how
ordered, or in any machine whatever, (by theorem 183) the diſtances
moved over in the fame time, by the power and weight are equal, fo
in an equilib. (fee theorem 184) the product of the power and it's dif
tance moved, is to that of the weight and it's diſtance moved in the
fame time, fo in the above queltion, of 300 150 lb is the anſwer,
for you'll (in that cafe) fee the power defcend 2, while the weight
afcends one.
=
Question 65. How muft three pullies A, C, E, (fig. 193) be combined
fo as to raiſe the moſt weight W, with the leaft power or force P?
Let the uppermoft pulley A be fixt to a beam &c. as per fig. then
with a little confideration it will appear, that while the weight W riſes
I inch, the power P fettles 7, that is, (by theo. 183) one pound at P
will raife 7 lb at W, which will answer the queftion. For by the laſt
queſtion, a man with two pullies, can raife twice his weight, with 3
pullies he'll raife thrice his own weight, with 4 pullies, 4 times his
weight &c. half the number of pullies being fix'd, and the other half
number moveable, and all the parts of the rope parallel to each other,
but if each pulley have a fixed rope it muſt be confidered as a leaver
of the fecond kind, and fo will double the power of the foregoing laft
pullies, i. e. 4 pullies will increaſe the weight 16 times.
Queſtion 66. If by prefling with a weight of 30 lb at the end B (fig.
194) of a leaver BA 77 inches long, the end A fix'd at right angles to
the head of a fcrew, which by this preffure, preffes or raiſes a weight
I inch, what is this weight or force of the fcrew?
Here, in turning the ſcrew once about, the weight W moves but
I inch; but the power at B moves thro' 484 inches the periphery of
a circle whofe radius is A B 77 inches, therefore (by theo. 183) As I
nch 484 inches :: 30lb: 14520 lb the answer.
144 THE UNIVERSAL MEASURER
Note. While the ſcrew turns once round two of its threads will ap-
pear, from whence we have this rule, as the diſtance between two con-
tigious threads is to the diſtance moved by the applied power, ſo is
the force of the power to the force of the ſcrew in one revolution;
hence we fee the vaſt force of a fcrew, and that the nearer together
the threads are, the greater will the force be.
Question 67. If a weight P of 150lb, can be drawn up the fide B A
(fig. 195) of an upright wall 20 height, what weight W with the fame
caſe, will be drawn up a plank C A 30 long laid a flope from the top
of the wall to the bottom, the plank and wallbeing of equ al fmooth-
nefs?
Here its plain that while the weight W rifes to the
height B A
20, the power P that draws it (in direction parallel to CA) moves over
the diſtance C A 30, therefore (by theo. 183) As A B 20: A C 30:
P 150lb WV 225 lb the antwer, and fo is B C 22,36: 167,7 lb the
preffure againſt A C. See queftion 69.
:
Question 68. Let things be the fame as in the last queftion (fig. 196)
but instead of the weight W, being drawn in direction parallel to the
hypothenuſe CA, ſuppoſe it to be drawn up the fame Ĉ A, in direc-
tion parallel to the horizon CB ?
Here its plain, while the power P moves the weight W from C to A,
it moves itſelf from C to B, and the faid weight W afcends from B to
C, therefore, (by theo. 183) as A B 20:
SCB 22,36
AC 30,00
W 167,7 anfwer, and 275 the preffure againſt A C.
}:: P
:: P 150:
Question 69. Things being the fame as before, fuppofe the weight
v to be drawn, or futained in equilibrio (fig. 195 and 196) by a power
q 150lb in an oblique direction q v to the inclined plane C A, fo as
Bd=15, being perpendicular to the line of direction of the power,
q v, then in fig. 195, fuppofe d C = 18,6 ?
Firſt, If B e d is to the traction qv, then it's plain, Bd is L to
qv the direction of the power C B to A 3 the direction of gravity, and
Cd to A C the direction of the preffure againſt the plane; therefore,
(by theo. 159) As Bd 15: q 150 lb:: C B 22,36: v 223,6 lb, and
fo is Cd 18,6: 186 lb the preffure upon the plane A C, or if we take
the fines of the angles oppofite theſe fides, then the L dev being 90°
the LC dB comp. evd, and the ſides being (by theo. 48) as the
fines of their oppofite angles, it will hold, the power, weight, and pref-
fure againſt the plane are refpectively as the fine of B C d, the plane's
elevation, co-fine of the L of traction e v d (viz. Cd B) and C B d co-
fine of the power's direction above the horizon, and when v 9 infides
with A C, you have the fame proportion as in quellion 67, or with BC
AND MECHANIC.
145
as in queſtion 68; from this laſt queſtion it is plain, that the preffure
againſt the plain is greater when the time of the power's direction v q
is below the plane, than when it is above it (all elſe the fame if the Ls
of traction be =) for in fig. 195, d C is greater than it is in fig. 196.
Question 70. What weight will a wedge A B C, (fig. 197) raiſe by
applying a force or power of 150 lb to it's head at A B, the flant length
CA-CB and thickneſs A B = 4.
3
Here its plain that while the wood &c. into which the wedge driv-
en, opens its felf from H to B, or to A, the flant fides of the wedge
moves from B to C or from A to C; therefore, (by theorem 183) as
HAHB 2: B C = AC =3::150lb: 225 lb the anſwer, or
if the wedge be oblique, As HB+AH=4: BC+AC
=6::
150lb : 225 lb the anſwer, here it appears, the ſharper the wedge the
lefs force will drive it.
Question 71. If two bodies W and P be in equilibrio upon two in-
clined planes A B 40 length, and A C 30 length, their common L
AP 20, what is the ratio of the weights of thefe two bodies? Fig. 198.
By queſtion 67, As A P 20 AC 30 :: 1:20 P = 1,5
P1,5 and as
20: A B 40:: 1 40 = W = 2, i. e. (becaufe the bodies are in
equilibrio) W = 2 and P = 1,5, fo, as W: P:: 2 : 1,5 :: 4 : 3
:
ཏཱ
viz. the weight are as the lengths of the planes.
=
Question 72. If two barrels the one holding 10 gallons and the other
15 gallons, are to be carried hung over a horfe's back, how high mutt
the heavier barrel be raiſed to ballance the lighter?
By the laft queſtion, fuppofe the lighter barrel to hang down the L
AP (fig. 198) and the heavier down A C, then as A P: A C:: 10
15, fo (by theo. 13) the fquare root of the difference of the fquares
of 15 and 10 is 11,3 PC; therefore, as 10: 11,3, or as 1 : 1,13,
and ſo mult the fides of the horſe be raiſed, that the carriage may bai-
lance, and one fide be not heavier nor lighter than the other. But if
you fuppofe the barrels to be fufpended at A and B (fig. 149) the ends
of a leaver A B, and make it as A G:GB:: 15: 10, this point G
will (by theo. 185) be the common center of gravity of the two bar-
rels, whofe weights or contents are 15 and 10, and is the point in the
leaver to be laid over the prop &c. that the weights may be in equil.
Queſtion 73. If A C be an inclined plane, on which a cylinder open
at top mn, and perpendicular to the faid plane is drawn uniformly in
direction parallel to A C, required the angle of inclination A C B of
the plane with the horizon fuch that the moft water poffible may be
drawn out of a well at C, by the faid cylinder min bd, in any given
time? Fig. 195.
T
146 THE UNIVERSAL MEASURER
1. It's plain, the greater the angle A C B is, the fhorter will A C,
the way of the water be, and if n r be the furface of the water in the
cylinder, then the greater the veffel n r bd, or only it's height r b, is,
the more water will be drawn, whence the product of the fine of the
AC B, and height rb of the veffel muſt be a maximum, therefore
let e fine LA CB, then : 1 ée: it's co-fine (radius = 1)
and m bnd1, then the As A B C and n m r being fimilar it will
be by trigonometry as (✔: 1 —ee: fine LCA Borm rn): a (mn)
therefore m b—mr=1-
::e (fine mnr): mr=
a e
: I-ee:
√ : I
ae
√: 1 —ee'
I
rb, forbxe=e
22 e
√:1 ee:
and by (art. 421) 1—
aee
I
√:1-ee:
eee a
I ee
=m,a maximum,
o, fo we have
$ — c e | } = 2 ae — a e e e, and if we fuppofe a 1, it will be I
=
2 c — eee, wherein will be found € 0,4123 the natural fine of
24° 21′ the anſwer. When the depth and diameter of the cylinder are
equal, but if d the depth and a
diameter, the angle will vary, and
our equation become i eed=2ae
be found, whatever a and d be.
a eee, in which e may
Question 74 If a wall of equal thickneſs a B C D (fig. 199) be built
perpendicular to the fide of a hill, making an angle B a E with the
horizon a E of 200, what weight of the wall will be on each fide B.
and a, and whether will it overfet or not, if its height a DBC be,
10 feet, and it's thickneſs DC a B 4 feet?
Becauſe the wall is all of the fame matter and thickness, it's center
of gravity will be at G in the middle thereof G n (being
a B) will
be 5 half the height and n an B2, half the thickness, let Ge
be the horizon li e, then will Le Gn-L Ba E 20, fo (by theo.
48) As fine Lne G is to fine Le Gn 20°, fo is n G 5 to e n 1,82,
therefore, the wall will not overfet, be Ge horizon, and paffing
thro' the center of gravity G, falls within the breadth a B of the wall,
now for the weight, e in this caſe is the center of gravity, ſo (by theo.
185) as e B 3,82 is to e a 0,18, fo is the weight at a to that at B.
Question 75. A cart load of hay &c. 10 feet high and 4 feet broad
is to be drawn on the fide of a hill elevated 200, what muſt be the
height of the greateſt wheels poffible, to carry it without overſetting?
Let a BCD (fig. 199) reprefent the load, whofe center of gravity
is at G per laſt queſtion, produce Da and Ge till they meet in z,
then (by art. 23) a z, will be the diameter of the wheel, fuch that
the load will reft in equilibrio in z, the lower edge thereof, the axle
AND MECHANIC.
147
tree being at a, fo if the wheel be any higher the load will overturn
upon z, becauſe then G z the horizon, will fall without the fide Da,
then per fimilar As za è and G ne, asen 1,82 : e G 5,
5,5: ca: 0,18
: 0,54 feet ➡a z the radius of the required wheel.
Question 76. If Ba (fig. 156) be a beam, wall &c. whofe weight is
500, it's center of gravity at C, (C B being = 22) and leaning, fo as
to make an LA B a with the horizon of 60°, and if a prop A a, be fet
at A, 40 diſtant from B the foot of the object, making an L B A a, or
QA a with the horizon B A of 55°, what weight doth the prop A a
bear?
By theo. 191. Let radius=unity, then the natural co-fine of 600
22 × 500 X0.5,
is 0,5, and the natural fine of 550 is =0,8191, fo
· 5500
32.764
40 X0,8191
་
167,8, the weight born by the prop A a, which taken
from 500 the whole weight leaves 332,2, for the weight preſs'd at B,
and (by art 28 :) is in direction of a line drawn from B to the meet-
ing of A a (produced) and a line drawn thro G, perpendicular to the
horizon.
Question 77. Things being as in the last queftion, and the prop A a
at the fame place, what weight doth it bear, when the object B a pref-
ſes with the greateſt force, and the prop fupports with moſt eaſe?
By theo. 192. When the object makes an L with the horizon of
51° 50' then it preffes the prop A a with the greateſt force poſſibles
which prop at the fame time mult make equal Ls with the object and
horizon, therefore the comp. of 51° 50′ is 380 10' whofe natural fine
is 0,6180, and 180 51° 50′ = 128° 10 half whereof is 64°5′
the angle of the prop's inclination, whofe natural fine is 0,8995, fo
22 X 500 XO,618
188,9 the anſwer.
40 × 0,8995
Question 78. A beam B C (fig. 200.) whoſe center of gravity is at
G, or a leaver B C bearing a weight G parallel to the horizon AD, upon
two looſe props C D and B A, making on angle B AD with the horizon
of 50°, what angle muſt the prop C D make therewith, to fupport the
beam in equilibrio, BC being 8 and C G = 12?
CG
1. Thro' G draw FG the horizon, and produce AB to meet in
F, thro' F and B draw FB A fo is B A (by article 281) the other prop,
then, (by theorem 47) as BG8: G C 12:: tangent L BFC 400
(comp. LFB G 500): tangent L GFC 51,32 whofe compliment is
38,28 LFCG or ADC the anſwer.
148 THE UNIVERSAL MEASURER
Question 79. If the weight G (laſt queſtion) be 500, how much of
it doth each prop bear; or, if it be hung by the two ropes F B and
FC, what is the tenfion of theſe ropes, or the weights they bear?
Draw I H parallel to F B, then, (by article 281.) as fine L FIH
91° 32′ (LA 50° +LD 33° 28′ and taken from 180º): 500 (the
weight denoted by F H):: fine LHFI 51° 32′ : 391,7 fere, the weight
upon A B, and fo is fine LF HI 40: 321,5 the weight upon C D,
that is, the weight borne by A B is to that borne by C D as 391,7:
321,5 or as 39:32; therefore, as (39 + 32) 7 is to 500
225,3
theſe
{32:
39: 274.8} the juft weight borne by {CD or FC, in the fe
A
B or FB,
fo is
directions. But, if you would know the preffure in any other direction,
as that to the horizon: then, (by theorem 157) as radius (L BE A)
: 274,6 (force in direction B A) :: fine LEAB 50° to 210,3 the pref-
ſure in direction BEL to the horizon, and :: fine LE B A 40°:
176,5 the puſh outward in direction EB parallel to the horizon. The
line may be done for the prop C D.
Question 80. Fig 157. If DCE be an upright wall height CD 20,
built at the foot of a flope wall A BC, of the fame height A B, and
flope fide B C 30, and the vacant triangle B C D, between the two
walls be filled with earth, fand, &c. with what force doth it prefs each
wall?
By art. 288, As go: 20: 1:, the weight fuſtained by the wall
CB, fo (1) that fuftained by CD E, in direction D C, and
{
as BC 30: AC 22,36::
41.72
90
4959, the preffure in direc-
tion A C perpendicular to the wall D C E, and acts at L in direction
GL (CL being of 2013.)
Question 81. If a folid inch of the wall DDE (laft queftion) be to
a folid inch of the ABCD as 4 to 5, what muſt be the thickneſs of the
faid wall, that it may not be overfet by the preffure of the A ?
1 BD 22,36 × 1 DC 10223,6 the content of the ABCD, then
(by theo. 193) Eex: y+P:=
ACXDCX W
BC
>
alfo, DCX EC=
the wall's (CDE) content. Likewiſe, ife be the center of gravity of
the two weights P (preffing upon C) add y (upon R) then, by theorem
1 C F. X Y__ (R being in the middle of C E, becauſe CD E is
185)
P+ y
I
equal thickness) fo Ee=Cc+CE.
CEXY
++CE; there-
P+y
AND MECHANIC.
149
ACXDCXW
fore, Eex:y+P:=:y+P:XCE=
or=
ОВС
A CX □ DCX w
OB C
becauſe the force in direction A Cacts atL and
LC= DC. Then, (by art. 288) P = w
DCXW
BC
which put
in the laſt equation for P, we get: y+w
ACX
DCXW
BC
: XC E =
CE
DCXW, but by the queftion, y = DCXC Exn and
DB C
W = ACX DC × m (n being taken = 4 and m = 5) which ſub-
ſtituted in the laſt equation for y and w we'll have, after reduction com-
pleating the fquare &c. CE = √
mx□ACX O D C
+: 1
BC
DC
X
mXAC
BC
2 n
է
DC
:1+ : X
B C
mx AC, but to have the
20
DC
equation fhorter,let 1—
zzmmxAC
=z, then E C =№:
mx□ACXDC
BC
3
BCX n
+
16 n n
zmx AC
40.
in numbers it is, CE =
5 X 500 X 400
4X 25 X 500
+
2X5 X 22,36
9 X 10 X 10 1
= 6,1 anf.
3 X 4 X 4
3 × 900 X 4
Question 82. What must be the thickneſs of the ſtone wall C D E
(fig. 157) to reſiſt the preffure of the A of earth B CD, the weight
of ſtone being to that of earth as 3 to 2, and allowing a heavy body
to looſe of its weight in fliding down an inclined plane B C, on the
account of friction ?
Here n = 3 and m = 2, but m muſt abate of it's weight becauſe
of friction as is found to anfwer to experience in planes middling ſmooth
ſo m = (2 — of 2, or 4 of 2), ſo to have m and n out of frictions
it will be as 3
:: 9:4, that is, n = 9 and m— 4,
9 and m4, which put in
the last general
4
equation, gives
4, 500, 400
+
3, 900, 9
4, 16, 500
9, 16, 81
2, 4, 22, 36 — CE√
800000 32000
+
178,88
2700,9 11664
3, 4, 9
5,3 the anſwer.
But if the preffure of the ABCD in direction DC downwards be
106
150 THE UNIVERSAL MEASURER
taken away, or which is all the fame, the two laft terms in the equa-
tion being but finall may be neglected, and then the folution will be
very eaſy; thus, CE=mx□ & C × □ D €
= √
100*2000
10800
CE = √ 800000
24300
30XOBC
=√5 × 500 × 400
3 X 4 X 900
=√ 92,6 = 9,6, for the anſwer to queſtion 81, and
= 432,9 = 5,7, which is but a ſmall matter more
than 5.3 the laſt anſwer; and therefore is to be uſed in fuch cafes be-
fore 5,3, becauſe this 5,3 is the breadth on the equilib. ſo 5,7 muſt
do more than ballance.
Question 83. If the wall C D E (fig. 157) have no breadth at top
D and a double breadth 2 C E 11,4 at bottom, how much more
CD next the preffure being up-
preffure will it then refift, the fide
right and the other fide D E flant?
Firſt, half of 5,7 is 2,85
= 2,85 = CR, R being the center of gravity
when the wall is of thickneſs, but 4 of 11,4 = 7,46´+, when no
breadth at top; fo, as 2,857,46: the force of the former: that
of the latter. That is, the wall C D E, will endure a preffure a-
gainſt the fide D C before it turn over upon E, which is as 2,85, and
if it become the wall A B C it muſt be preſſed againſt B A with a force
as (of 11,4) 3,8 before it overturn upon C, but a force to overſet
upon A by preffing at B C is as 7,46 +, each of theſe preffures be-
ing in the fame direction and at the fame height above the bottom,
and thus appears the reafon why tapering walls &c. are strongest.
it
The walls here mentioned are fuppofed to be equally ſtrong, or ftick
well together. (See queſtion 108 and 109.)
Queſtion 84. How many folid inches are in a pair of bellows A B C
when the fides AB, AC, are each a circle of 10 inches diameter, and
makes an angle B A C with each other at the pipe A of 30°, and what
is the greateſt quantity of air they can poffibly hold? (Fig. 201)
1. Its plain by the figure, that if another pair of bellows A CD,
be laid upon A C, with their pipe over C, theſe two pair of bellows
fo joined will form an oblique cylinder A B C D, half of whofe con-
tent will be the anſwer; fo by trigonometry, as radius: A C 10 ::
fine LBAC 30°: LCP5 height of the cylinder A B C D there-
fore 0,7854 ABX CP 0,7854 X 100 X 2.5 196,35 folid
inches, and the content must be the greateft poffible, when the fides
CA and C B are L to each other, and then CPCA AB=10,
whence 100 X 0,7854 × 5 = 392,7 folid inches, for the greateft
content poffible.
AND MECHANIC,
151
Question 85. This and the 29 following queftions are about wheel
carriages, the ftrength and ſtreſs of timber, walls, &c. A carriage of
8 hundred weight is to be drawn up the fide of a hill making an L
with the horizon of 300, what force will be fufficient to draw it when
the radius of each wheel is 24 inches or 2 feet?
(By prob. 196, art. 265). The carriage goes with moſt eafe when
the traces &c. are parallel to the hills fide &c. Therefore, (fee fig.
147) as fine radius: CE 2 feet :: fine LCE H 600 (comp. hill's
elevation): C H 1.75 which ſquared and taken from 4 the fquare
of C E leaves 0,9375 whoſe ſquare root is 96. Then, as C E 2: 8
hundred weight :: 0,96 : 3,84 hundred weight, the force fit to hold
it on the hill's fide.
Question 86. Things being as in the last queftion, but fuppoſe the
traces, or direction, to make an L with the hill's fide of 200, what
force will the carriage require in that direction to hold it on the ſide
of the hill?
By theo. 180. As fine 700: fine 900 :: 3,84 laſt weight found :
4,1 hundred weight, the anfwer.
Question 87. Things being the fame as in queſtion 85, what force
will be fufficient to hold the carriage on the top of the hill or obſtacle,
whofe height is 0,25 feet (2 feet 1,75 feet) in a direction parallel
to the horizon ? See theorem 182.
1. Multiply 3,75 the difference between 4 the wheel's diameter and
0,25 the object's height by 0,25 the faid height, the fquare root of
0,9375 that product is 0,96. Then, as 1,75 feet (radius 2 feet
object's height 0,25 feet) is to 8 hundred weight, fo is 0,96 to 4,3
hundred weight the anfwer; thus you fee that in any pofition of the
traces, that parallel to the object is the beft.
Question 88. If a wheel carriage is to be drawn on rough uneven
ground, how much eafier will it be drawn on wheels of 3 feet radius
than on thofe of 2 feet radius? See theorem 181.
As † : 4 :: 2 : 3, fo is the force required to draw the greater
wheels to that required to draw the leffer ones.
Queftion 89. There are two wheels, the radius of one is 2, and of
the other 5, which will have the moſt advantage to drive an obſtacle
before it ? See art. 267.
I
As 1/2-1: 15-:: 1:2, fo is the force required by the lef
fer wheel to that required by the greater one. Hence, in this cafe which
feldom happens, fmall wheels may have the advantage; but in the
foregoing queſtions great ones have it.
Question 90. Whether is the friction on great or finall axle-trees,
the moſt?
152 THE UNIVERSAL MEASURER
1. The friction or rubbing parts may be taken as a force acting a-
gainſt the force to draw the carriage, and therefore (theo. 183) the
greater the velocity of thefe parts, the greater will the friction be;
fo if the wheels bear the fame weight, and turn about in the fame time,
their velocities (viz.) that of their axle-trees, will be as the diameters
of the faid axle-trees. Hence, fmall axle-trees have lefs friction than
great ones, and in any machine, the fmaller the parts that rub againſt
one another, the lefs is the friction.
Queſtion 91. If a board be 2 inches thick and 10 inches broad, how
much more weight will it bear edge way, than broad way? See art. 301.
1. In all the like ſections, the ſquare of the depth, or fide of the
fection the horizon, multiplied by the breadth, or fide parallel there-
unto, is as the ſtrength. Therefore, 2 × 2 × 10 40, and 10 × 10
X 2 = 200, ſo, as 40: 200, or as 1:5:: the ſtrength broadway :
the ftrength edge way.
Queſtion 92. If any beam &c. making an angle with the horizon
of 600 can bear 200 Cweight, what weight will it bear in the fame
place, when parallel thereunto. (By art. 300) This 200 C. weight
multiplied by 0,5 the natural co-fine of 60° the inclination, gives 100
C, weight the anfwer,
Question 93. An iron bar bearing 500lb bends thro' 1,8 feet, (b)
then is juſt a breaking, a ſteel bar bears the fame weight (c) but de-
flects only 0,1 foot (B) then is juſt a breaking, with what velocity (v)
muſt a ball 5 lb weight (w) be thrown againſt cach to break it, ſee the
latter part of art. 311 ?
}
==
1. dbc÷÷2w = 1,8 × 500 ÷ 10=90, then, (theo. 166) as
√ 16 : 32 : : vd (√ 90) : 76 nearly feet per ſecond v, for the
iron bar; and taking 0,1 for 1,8 we get v 9 nearly, for the fteel.
Hence, force to break the iron: force to break the ſteel :: 76 × 500
:9 × 500 :: 76: 9 the anſwer.
I
TO
Question 94. How much weight will an oak beam 10 feet long and
1 foot ſquare bear, before it break, when a like piece of oak a foot
long, and of a foot fquare beares 320 lb to break it? See art. 308.
Here e = 1,
d=0,1,g=0,5, v=320lb, w=0,44 lb, the weight
of a peice of oak 1 foot long, and 0,1 of a foot ſquare, a =
d d =‚01, E = 10, D = 1, A = D D = 1.
DDDeex: w+v:-dEED Dw_318,02
Whence y=
འ +
31802 anf.
ddd Ee
,Q}
Question 95. How long muft the laft mentioned beam be, to break
with its own weight? See art. 308.
AND MECHANIC.
153
1. From dEEw Deew+2 Deev, we get E≈
Deew+2Deev
d w
Tr
640,44 120,6 feet, the anſwer.
044
Question 96. If a ſquare priſm of oak 1 foot long and o,1 of a foot
fquare, bear 320 to break it, what are the dimenfions of a fimilar oak
priſm, that breaks with its own weight? See art. 308.
1. E=:ew+2 ev:÷w=640,44 ÷ 0,44 = 1455,54 feet the
length, and as I : 0,1 :: 1455,54: 145,554 feet, a fide of it's bafe.
Question 97. If a fquare prifm of oak fupport 320 lb weight, before
it break, required the dimenfions of the ftrongeſt ſquare beam of oak
poffible, fimilar to the faid prifm, being a foot long, and 0,1 of a foot
fquare? See art. 308.
Here e = 1, v = 320lb, w = 0,44 lb the weight of the ſaid priſm
2 ex: w+2 v: 2X:0,44 +640 :
then E=
3 w
1,32
= 970,37 feet the
length, and as 1 :: 0,1 :: 970,37: 97,037 feet a fide of it's bafe,
:W + 2: XE Ee EEEw=
and the weight it will bear is y
2 eee
132300000 lb, nearly.
I
Question 98. If a fquare oak beam 10 feet long and 1 foot fquare,
31802 lb, what will each prifm bear when it is flit into two triangulaṛ
ones, fupported the fame way as the beam was, with a fide parallel to
the horizon?
1. Here (ſee theo. 195) we have P≈4, p=†, D=d, E≈e, G=g
= 1 E = 1 e, c = 1 C = 1, w = 440 lb the weight of 10 folid feet
of oak, the ſquare beam, v=31802 lb, which duly fubftituted in the
faid theorem, we'll get y = :w+6v : ÷ 8: 23907 lb, the weight
borne by each prifm., and is much above half of 31802 lb, the weight
borne by the whole beam.
Queſtion 99. Whether is a fquare beam ftronger, when it lies corn-
er uppermott, or a fide uppermoft? See art. 302.
7
78
Let a area of the fquare, dit's fide, Dits diagonal, then
da is the ftrength, when the weight preffes one of its fides, and
82 (for DD-2 dd) da nearly, the ftrength in
direction of its diagonal, i. e. as : 13 :: 6 : 7 : : the weight it will
bear upon a fide to that it will bear upon a corner, for in both cafes,
the length and weight, of the beam is the fame.
7
U
154
THE UNIVERSAL MEASURER
Queſtion 100. Whether are round, or ſquare beams ſtronger? See
art. 302.
÷
1. Suppoſe two beams, the one round and the other ſquare, of the
fame matter, length, content, and confequently weight, then if d=
the diameter of the round beam or cylinder, it's ftrength will be
,2451 ddd, and D the fide of a fquare the circle will be ✔,7854
dd=,887 d=D, but D3,236 ddd, is the ſtrength of the
fquare in direction of it's fides, and by the laſt queſtion of,236ddd
,275 ddd, the ſtrength of the fquare, in direction of it's diagonals.
Therefore, theſe three ftrengths are as the numbers 245 for the round
236 for the ſquare on one of it's fides and 275 if on one of its corners.
I
Question 101. If a round beam 1 foot diameter, 10 foot long 320 lb
weight, bear 30000lb, what weight will it bear if hollowed, diameter
of the hollow o, foot? See art. 302, and 308.
उठ
Let D = 1, the diameter of the cylinder d = that of its bore, a =
the area of the ring e 10, the length w=320, v = 30000, then
G=g=3e=4E, in the plane of a circle we have d, in the p☛
riphery thereofd, in each caſed nearly, foD a is nearly as
the ſtrength of the hollow cylinder, and 0,245 D D D, as the ſtrength
of the round beam, but a D D – dd: X,7854 =,75 × 7854
,589, alfo as ,7854 DD X 10: 320 ::,589 × 10 : 230 = z,
then as, 245 DDD: ew+ev:: Da (,196 D):
whence y =
,196 ×: w+2v:
DD
,245 X 2
ez +cy,
,245 Z
24913 lb anfwer.
5
TO
Note. If the area a, be very fmall, 3 D, ſhould be uſed, but if a
be pretty great, and d but fmall, it is better to take D. This laſt
anfwer will be truer if we take D, half the fum of theſe two, tho'
The difference is very little.
I I
Queſtion 102. If a fquare beam 1 foot fquare (D) 10 feet long (e)
weight 440 lb (w) bear 31802 lb (v), what weight (y) will it bear
when cut into a fquare pyramid of the fame length and baſe?
theo. 195.
See
of it's circumfcribing prifm, therefore
{e, D=d, P = p, C=c, by which
w+ v = 1 z+y, whence yv+r
31985,33 lb. Whence, the part will bear
1. Becauſe a pyramid is
z=w, alſo G = 4 c, g
the faid theorem, becomes
w31802183,33
more than the whole.
Note. Theſe beams fince queft. 90, may either be fixt at one end
or fuported at both ends, either parallel to the horizon or make any
angle therewith provided, any two beams in the fame queftion have
the fame poſition. In this queſtion the beam is fixt at one end, if it
be ſupported at both ends it muſt be two = pyramids.
AND MECHANIC.
155
Queflion 103. If a cylinder of any kind of metal 1 foot diameter
(d) 10 feet (e) length, weight (w) 900 lb, fixt at one end, and at the
ether bear a weight of 5000 lb before it break, (v) what weight, or
force (y) will be fit to twine it about? See art. 302, and 304.
To
1. 245 ddd, or 5 of,7854 d d d, is as the lateral ſtrength, and
( b d d) b d d = 3 of 3,1416 ddd (becauſe 2 d x 3,1416 —b)
is as the twisting ftrength, but as of,7854 of 3,1416:: 5 : 8
5 ÷
:: lateral ſtrength : twifting ftrength. Therefore, as 5: ew+ev
:: 8:
4 we + 8 ve
5
==87200 lb anſwer.
•
5:
Note. Becauſe of 3,1416, is always the fame, let d be what it
will; therefore, the ftrength of cylindrical pieces, or of any fimilar
pieces of timber, being twifted, will be as d d d, the cubes of their
diameters. Therefore, if a force of 5 lb weight twine a hazel rod of
half an inch diameter, a force of 40 lb, will twine one of 1 inch dia→
meter, for as, 125 (cube of 1): 1 :: 5 lb : 40 lb.
Question 104. A cylinder one foot diameter (d), 10 feet long (e)
900 lb weight (w) fixt at one end parallel to the horizon, (as in the
laſt queſtion) and at the other end bears (v) 5000 lb at it's outmoſt
ftrength, if it be laid on a plane parallel to the horizon, what weight
or force (y) will pull it afunder? See art. 303.
1. Here Ge, gd, s=d, fo: ew+ev: xd=gsy,
dex:w+v:
IO X: 450 + 5000: × 16
g s
whence, y =
174400 lb the anſwer.
,
5
X
Question 105. If the laſt mentioned cylinder, be fufpended by one
end, what weight at the other end will pull it aſunder? Seeart. 303.
1. Becaufe in this cafe its own weight 900 lb, is alfo employed in
pulling it afunder, therefore (fee the laſt queſtion) 174400
173500 lb to part it at top.
900=
Question 106. A beam lengthe, weight w, fixt at one end, bearing
a weight v, at the other end, what weight will it bear in the middle,
when fupported at each end, and alſo, when fixt, or nailed down at
each end? See art. 311.
1. When fixt at both ends, then
I
ex+w+ex { v, is as the
ſtreſs, if only one end be fixt, but the other fixt end bears weight,
therefore, half of ew+ev =
nailed down at both ends; but
÷
TO
w+ev, is as the ſtreſs when
ew+ev, α ſtreſs when fixt at
one end, and ew+e v, when fupported at both ends, hence thefe
ſtreſſes are as 1,8 and 2 fo the weights they'll bear, are as 8,1 and 4
156 THE UNIVERSAL MEASURER
1
Question 107. Two fimilar rectangular beams, or plates, of the fame
depth d, lengths e and E, breadths b and B,
borne v and y, whether is the ſtronger?
The plates &c. being of the fame matter.
weights w and z, weights
See art. 299, and 307.
See art.
Ez + Ey: and taking
1. Bdd xew+ev:= bddx:
dd out of each fide, and dividing by e and E, it will be e B x : w
+v:= Ebx:z+y: but becauſe the folids in refpect of length
and breadth are fimilar, e B = Eb, for as B: E:: b:e, therefore,
W+2v=Z+2 y. Hence, if w and z, the weights of the folids, be
neglected, as they may in fmall ones, we'll have v=y, that is, they
will bear equal weights, as per art. 299.
From theſe queſtions appear the impoffibility of things being infinite-
ly great, for the ftrength being as the cube of the depth, and the ftrefs
as the product of the matter and length, it's plain the ftrength increaf-
es in a lefs ratio than the ſtreſs, and therefore, any wall, tree, beam,
mechanic engine, and even man or beaſt, may be taken fo large, as to
break, or be cruſh'd by it's own weight, whence it is, that little ani-
mals, machines &c. (fimilar and of the ſame matter) are more active,
will carry more weight, leap further &c. for their fize, than great ones,
becauſe the ſtrength increaſes in a lefs proportion than the fize, or
weight, and fo cannot reſiſt any violence in the fame ratio of ſize.
Queſtion 108. A wall is to be built with a plane rectangular face to-
wards the wind, what muſt be the form of its backſide, that it may be
equally ftrong to refift the wind? See art 305.
1. If the parts of the wall ſtick well together, the back-fide there-
of muſt be a ſtreight flope, or the thickneſs of the wall a plane triangle
whofe hypothenufe is the faid flope. But if the wall is built of loofe
materials, the faid flope fhould be a common parabolic curve.
Queſtion 109. In what form muſt a tower &c. be built to be equally
ftrong throughout, in refifting the wind? See art. 305.
As in the laft queftion, when the weight preffes uniformly, or
the parts
ftick well together, then a cone ſtanding on its baſe is beſt.
Otherwife, parabolic conoids are equally ftrong throughout; for a cone
or pyramid fixt at bafe, or a wedge fo fixt at it's thicker end, and its
two parallel fides, parallel to the horizon, a weight, or power pulling
at the other end, will no fooner break it in one place than in another.
Alfo, if a beam &c. has each fide cut in the form of a parabola, fixt
at bafe, pull'd at any where, it is equally ftrong between the bafe, and
place fo pull'd at, and therefore equally fit throughout to fupport it-
felf. Alfo, if a fpring be made by the first part of the ſaid art. 305,
the ftrefs will in every part be proportional to the ftrength.
AND MECHANIC.
157
Queſtion 110. A veffel, or wall, to hold a bank of earth, or any
fluid body, in what form muft its out-fide be, that it may be equally
ftrong throughout, when the in-fide is ftreight, and upright? See art.
305. and the laſt queſtion.
A wall to anſwer this end, if the parts flick well together, fhould
be concave in form of a femi-cubical parabola. But if of looſe materials
then a right line or floping plane ought to be it's figure.
Question III. A beam, or foot-bridge, horizontally fupported at
each end, into what form muſt its under-fide be cut that it may be e-
qually ſtrong throughout to ſupport a variable weight? (Fig. 202)
See art. 306.
1. The form of the under-fide muſt be A QB, a half ellipfis, but if
the beam fhould be equally ftrong throughout to fupport its own weight
then QA and QB fhould be two common parabola's, in which cafe
of the beam may be cut away without any lofs of ftrength.
Question 112. A bow is to be made out of a cylinder 10 feet long,
and I foot diameter, into what form muſt it be cut to be equally ſtrong
throughout in bending, and what is its folidity when fo cut (fig. 203)
See art. 305.
1. This is the fame thing as to make a ſpring equally ſtrong &c. If
the bow is to be made tapering from the middle then its form A DEB,
muſt be two conoids A DE and BDE of two equal cubic parabola's,
joining in the middle on one common baſe at E, diameter there equal
I foot, and therefore, it's folidity will be to that of the whole cylinder
asto 1, or as 3 to 5, and therefore of fuch a cylinder may be cut
away without lofs of ftrength, but if the bow is to be made of half a
cylinder, and the infide AD B, to be every where of the fame breadth,
then the out-fide A E B, must be two common parabolic curves, A E
B E, then it's folidity will be to that of the half cylinder as to 1,
or as 2 to 3, fo of the half cylinder may be cut away.
$
ड
2. Otherwife, If we fuppofe A B perpendicular to the horizon, and
a weight w laid upon A to bend the bow, then fince this weight w acts
in direction A B, it may be fuppofed to act at C, in any part of A B, as
at the end of a lever CD, whence (by theo. 190) the ſtreſs at D, will
be as CD, or as A C, or as A D, becauſe, CD, A C, and AD, are
all nearly right lines. Therefore, if de E, the depth of the body
A E B e D be — it's breadth, then if the ftrength which is as bdd, be
every where as the ftrefs, that is, A C be as b dd, it will be the very
fame with the first part of art. 305, and fo the anſwer as before. In
like manner, if E A e be a prop fet oblique to the horizon, to ſupport
a weight w, it's evident, this prop to be equally ſtrong throughout,
must be cut into the form of half the bow, with it's greater bafe fet
on the horizon Fe E. Alſo, if a piece of timber &c. be bended in
158 THE UNIVERSAL MEASURER
feveral places, the force to break it in any point D, will be as a per-
pendicular DC, let fall from that point upon A B, the line of direction
of the force, let the bending be what it will, and in any direction.
Question 113. If a beam A B (fig. 161) 10 feet long, fupported at
both ends, bend thro' a diftance CD 6 inches from it's middle, what
will this deflection be if the faid beam be 20 feet long? See art. 310,
and 311.
=
1. Let e 10, E20, a CD-6 inches -0,5 feet, then b and
a EEEE
d being in both, it will be as e eee:(0,5)a:: EJE E E
eeee
8 feet, the anſwer; for as e: w:: E: it's weight, the weights be-
ing as the lengths, but if the beam be a breaking with 6 inches de-
flection, then as ee:a::EE: E Ea÷ee=2 feet, the deflection
of the longer beam, when it breaks.
If two props E and F (fig. 204) are to be fet to the flender cylin-
der De He D, ſo that the bending may be the leaft poffible, then
De De muſt be = of DHD, the whole length of the cylinder?
Theſe queſtions are of great uſe in the conftruction of machines,
teaching how to make their parts equally strong, and of the leaft
weight &c.
Alfo, If the ſtrength of any fmall piece of any fort of timber &c. be
known, the ftrength, or weight, that any beam of the fame matter will
bear may by theſe queſtions foon be found. Thus, if a ſquare prifm
of oak, a foot long, and an inch fquare, be horizontally ſupported at
each end, it will bear in the middle 320 lb weight before it break, and
fo may any other body be tried; and thus, the proportion of ftrength
of the following bodies is found, viz. oak, box, yew, plumb-tree 11,
elm, afh, 8, walnut, thorn 71, red fir, hollin, elder, plane, crab-tree,
apple-tree 7, beech, cherry-tree, hazel, 6, alder, afp, birch, white-
fir, willow 6. Iron 107, brafs 50, bone 22, lead 61, fine free ſtone 1.
A good hempen rope of an inch circumference drawn in length with
1000 lb will break. See Emerfon's mechanics, for thefe experiments.
The 26 following queſtions are about pendulums, mufical cords,
founds, bells &c.
Queſtion 114. A pendulum length 11, ➡ CSB (fig. 158) fufpended
at S, at 1 within the end, what weight B muſt be at the lower end,
that the pendulum may fwing the fafteft poffible, when a weight C, of
100 lb is fixt at top.
1. Let S CC=1, SB = b = 10, and the points O and G, the
Bbb+Ccc
centers of ofcillation and gravity, then (art. 294)
bBcC
AND MECHANIC.
159
2 B b - 2 C c
SO=2SG=
B+ C
b b
which multiplied croſs-wife and re-
duced, is B B : 4 Cbc-Cbb-Ccc
-- 22 =
B-
CCcc
:xB=-
put
b b
CC cc
.com-
b b
C Ccc
4 C b c + Cbb + Ccc, then B B-2 z=-
b b
plete the ſquare &c. then you'll find B≈z+√: zz —
b b
70,5 +√ : 4970,25 — 100 :
100 : = 70,5 + 69,6 = 140, 1 lb anſwer.
Question 115. A pendulum 11 long, fufpended 1 within the upper
end, what weight B muſt be at the lower end, that the pendulum in
vibrating may have the greatest momentum poffible, with the moſt eaſe
when a weight C is fixt at it's upper end, of 100 lb? See queſt. 253.
1. Let c=SC=1 (fig. 158) b SB 10, then (art. 292.)
b=SB =
B b b + Ccc
is as the time tof vibrating, which time is as v the
Bb C c
velocity, and v B, is as the momentum, therefore B:
√:
Bbb+acc
Bb+Cc
α v B∞ the forces of the body, B, or C, therefore by making this
BB Bbb
expreffion, or:
Bb
BBC c c
C c
: it's equal a maximum, we'll
(by art. 223, or 421,) Balone variable
b √ : B B B b b + B B C c c :
Bb Cc
C b c c - 3 bb Cc
2 b b b
3 BBbb + 2 B C c c
√:B³b²+BBücc: X✔:Bb-Cc:
=0, which reduced is B B+:
3
: XB
CC ccc
bbb
Cbcc-3bbCc
, put 2 z =
2 b b b
CC ccc
bbb
then it will be B B+22=
&c. gives Bz+√:zz+
Σ
1000 15,16 lb the anſwer.
*CC ccc
bb b
which by completing the fquare,
:=7,25 +√: 52,5625 +
Hence, any machine working beams, as levers, &c. with an acceler-
ated motion, is in its beft perfection, when if the power Bbe in equilibrio
with the weight C, and there be taken 1,516 B inſtead of B, or nearly
1,5 B, viz. the faid power B increaſed one half; for it is plain, if a
pendulum ſwing flower it loſes time, and if fafter it will be too much
fhaken or ſtreſs'd.
160 THE UNIVERSAL MEASURER
Note. In the following queſtions about pendulums, any body will
anſwer the fame end, whofe length between the point of fufpenfion and
center of ofcillation is the length of the pendulum. (Queft. 116)
If a cylinder, or priſm, of equal matter every where, and 58,8 in-
ches length, be fufpended cloſe by one end, it will be a ſecond pen-
dulum, for of 58,839,2 &c.
123456
T
Queſtion 116. If in any latitude, a heavy body by the force of it's
own gravity fall thro' a diſtance of 1394 inches in the firſt ſecond of
time, what muſt be the length of a pendulum in that latitude to vibrate
feconds?
By theo. 175. As the fquare of 3,14159 &c. is to 1 fo is 139 in-
ches to 19,6 inches, which doubled gives 39.2 inches the length of a
ſecond pendulum viz. of one that vibrates 60 times in a minute.
Question 117. If the length of a fecond pendulum be 39,2 inches,
what is the length of a half ſecond pendulum?
See theo. 174.
As fquare 2 feconds (4) is to fq. 1 fecond (1) fo is 39,2 inches to
9,8 inches the anſwer.
Question 118. Pray what length muſt a pendulum be,
To vibrate once in feconds three;
Where fecond pendulums have their lengths,
Thirty-nine inches and two tenths.
By queft. 117. It will be as fq. 1 (1) is to fq. 3 (9) ſo is 39,2 to
352,8 inches anſwer. So that if a rope or cord have one end faſtened
to the top of a houſe &c. and at the other end there be a weight, and
if it vibrate once in 3 feconds, the length of the cord or height of the
houſe is 352,8 inches.
Question 119. If a pendulum meaſures mean or equal time when it
vibrates in an arch of 5°, what feconds will it lofe per day when it
is made to vibrate in an arch of 70? See theo. 177.
From 49 = fq. 7, take 25 = fq. 5, there leaves 24, and 3 of 24 is
103, the required number of feconds loft per day.
Question 120. If a pendulum beats feconds when the height of the
barometer is 30,25 inches, what time will it beat when the ſaid height
is 36 inches? (Art. 443.)
are each very fmall, fo 2m b +
1. Here m and
d
&c, ID, will be 2 mb +
4 a a
3 d
4CC
16 cc
9dd
3 d
ID nearly, the difference
between the arch's of defcent and afcent B A and A I. Now if m be
negative orm, or fuppofed to accelerate the motion of the pendu
Jum, fo as juſt to overcome the reſiſtance of the medium, the pendu-
AND MECHANIC.
161
lum will then vibrate ftill in the fame arch, and then I Do2mb
4aa
+++ whence 3 m b d = 2 aa; therefore as m, the force acting uni-
3 d
formly on the body to counterballance the reſiſtance of the medium, is
to 1, the weight of the pendulum, fo is 2 a a to 3 b d. Hence, if the
force m, which keeps the pendulum in motion, be conſtant, a, the arch
deſcribed will be as bd, or if m be variable and b d conſtant, then
a is as m.
2. Again, when mo, c will be a, and then T
a a
+
2 a a a
3,1416√bx:1
&c. which ſuppoſe = 1 ſecond, the time of def-
6 d d 9 d d d
cribing the arch a, when the height of the barometer is 30,25K,
let h36 the other height, and c arch defcribed at that height in
the time t; then becauſe /b da, or (becauſe b. is conſtant)
H
da, and (it's plain) the denſity of the medium or height of the
barometer is inverfely as d, it will be as h: H::a: a h
Hd D, whence, if 3,1416 √.b× : 1 =
c and as h: H::d:
a a
odd
&c.=T=1,then 3,1416√b×:1+
a a
cc
as T:t:: 1+
:I+
6 d d
6 D D
cc
6 DD
&c.;t, therefore
where if a and d be known,
the true anſwer may be had; but the following way may give a gueſs
and is much eaſier, thus, fince h∞ d and H ∞ D, let a be ſuppoſed
equal 6, then as T:t: : 1+
a a
сс
äа
:1+
:: 1+
6hh
6 HH
6 h h
a a
I
I
I +
::I+ : I +
216
:: I fecond: 1,0008 feconds
181,5
6 H h
the anſwer, which (tho' fo great an alteration as 5 inches in the
barometer can rarely happen) is fo fmall as fcarce to be regarded, the
alteration which happens by heat lengthening the rod of the pendulumt
is much more, tho' in thefe parts it is not fo much as in hotter climates,
but becauſe heat raiſes mercury and expands metals, a cylinderic tube
filled with mercury, might be fo taken for the rod of a pendulum, as
nearly would correct this error.
3. If h and H be inverfely as the denfities or any other two perfect
fluids, the above work holds true, as well in the fmall arch of a circlę
as in any arch of a cycloid for which this question is intended,
X
162 THE UNIVERSAL MEASURER
4. In the above value of T, when d is infinite,, or the refiftance of
the medium = 0, then T = 3,1416b, the fame with art. 424.
Allo, T = 3,1
3,1416b the fame thing whether m beo, or — any
finite number; hence, if the refiitance m be uniform, the vibrations
will be ifochronal, and performed in the fame time as if there were
no refillance, if the bobs of pendulums be pretty denfe, as of lead,
iron, &c. the refiftance they'll meet with in air, is fo little as not to
be obferved.
Queſtion 121. If there be 12 threads in an inch length on the ſcrew
of a pendulum, and the clock gain 3 minutes per day, or 24 hours,
how many threads mult the bob belet down to meaſure mean time?
See theorem 177.
Here of 3 X 12 is 2 threads nealy, for the anſwer; but if
the clock lofe thice mi.utes per day, the bob mutt be raiſed near two
threads. By thefe two laft queitions any pendulum clock may be re-
gulated, the lali method being eaficit is molt practifed.
Question 122. Required the length of a fecond pendulums rod, it's
bob being a globe (as commonly they are) of 2 inches radius. See
art. 292.
Let 229,2 in. the length ofa 2d pendulum, r3 the radius of the
þɔb, to find d, the distance between the centers of the bob and pin on
which the pendulum hangs, then 21=d+
2 tr
5 d
>
O. IT
d+ whence
ӣ
dd4r2d1, and by compleating the fquare, dd2d1+1 ==
11 -,4rr which folved gives d=1+v:11
4rr:= 32-16 in-
ches, aufwer. Or (39,16 — 2) 37.16 inches, the length of the rod
between the pin, and periphery of thẹ bob.
Question 123. It's a (fig. 206) be 30 inches the length of a pendu-
Jum's rod between the pin s, on which it hangs, and the bob a, «ndits
weight to that of the bob as 1 to 2, what is it's true length, the ra-
dius d'a of the bob being 1 inches?
The rods of pendulums are commonly light in refpect of the bobs,
but if their weights be confiderable, then the common center of of-
cillation of the rod and bob, points out the true length of the pendu-
lum, lét O be this center, B the center of ofcillation of the bob and A,
that of the rod, then comparing this question with theo. 194, we'll have a
SA of Sa=20, A1, Sd = (30+ 1,5) = 31,5, B = 2,
2 da
and b (by art. 292) = 31,5+
5 da S
Aaa Bbb
SB whence
= 31,5+
0,9
31,5
= 31,53
SO 29 inches fere, anfwer, then
Aa+Bb
AND MECHANIC.
163
by queft. 117) As 29: 39,21.fecond (1): 1,35 whofe fquare
root is 1,19 fecends the vibrations this compound pendulum will make
in i fecond.
Question 124. If a pendulum 39,2 inches long, vibrate feconds at
London (latitude 51° 32′) what time will it gain or lofe, and how
much per day, when removed to Whitehaven, (latitude 54° 56′ ).
See question 271.
Since the earth turns round its axis in 24 hours it's evident that every
particle of matter on it's furface will in that time defcribe a circle the
greateſt of which will be that defcribed by a particle under the equi
noctial, viz. on the middle of the earth and confequently will have the
greateít velocity, which velocity manifeftly endeavours to throw away
every thing from the center, fo that where this velocity is greatest, the
force of gravity (which impels) towards the center muſt be lealt, from
which confiderations, Sir Ifaac Newton has proved, that the gravity
at the poles is to the gravity at the equator, as 692 to 689, therefore
the decreaſe of gravity at the equator is parts of the whole, buể
this decreaſe at the equator is to that in any other latidude, as the fq
of radius is to the fquare of the fine of that latitude, therefore, as I
(radius 1) is to ,6131 ( fine 51° 32') fois 3 (decreafe of gravity.
at the equator) to 1,8393 (decreafe of gravity at London) which adds.
ed to 689 the gravity at the equator gives 690,8393 the gravity at
London, in like manner you'll find 691,009 for the gravity at White-
haven. Then as 690,8393 is to 691,009, fo is 39.2 to 39,21 inches
the length of a fecond pend. at Whiteh. by queſt. 117, the clock will be
found to gain 4,176 feconds per day. From this folution it appears that
689 lb at the equator will weigh 691,009 1 at Whitehaven, i. e, if a
perfon can fuftain 691.009 lb at the equator, he will be equally ftrains
ed with 689 lb at Whitehaven, which is the reafon that the clock
gains time there, for where gravity is quickelt, the vibrations of pen
dulums mult be fo too, gravity being the caufe of that motion.
queftion 16.
ཀཉྩི ༡ – ཨ
Question 125. If a muficat ftring weighing 8,64 grains or oorb
length 2 feet or 30 inches, be (tretched with a weight at one end of
10 lb, how many times will it vibrate in one fecond, allowing a heavy
body to defcend 16 feet in the firft fecond of time from a ſtate of
reft? See theorem 179.
Here ✓ : 2 × 16 × 10
2,5 X.,0015
times anſwer.
=v
320
0037
№
85333 = 292,₤
-Queſtion 126. If a ſeconds pendulum, be made with a thread and
leaden bullet, and put to fwinging in an arch of 12 inches, and the bat
$64
THE UNIVERSAL MEASURER
4
Jet be obſerved to fall of an inch fhort of 12 inches the first vibration
how many feconds will it fwing before it be at reft? (Queſtion 120,
art. 443.)
1. Since I Dmb+
&c. if d be infinite, or the refift-
4 сс
3 d
ance of the medium o, then, I D 2mb, i. e when the reſiſtance
m is uniform, ID the difference between the arches of defcent and
afcent, B A and AI is ſtill the fame, let the arch defcribed be what it
will, and (ftep 4th queſtion 120) all fuch arches are defcribed in the
fame time, and the refiftance this pendulum meets with from the air,
being fo fmall as to be neglected, it will fwing fo many times before it
be at reft, as there are quarters of an inch in 12 inches, viz. 48 times
or feconds.
Question 127. With what weight muſt a cord 6 feet long, be
ftretched parallel to the horizon, that it may vibrate half feconds al-
lowing 16 feet for the defcent of gravity in the firſt ſecond?
theorem 179.
wl
See
Here, ✔ =t, viz. v = 0,5, or 0,25 X 32 f = 6 w i. e.
2 sa
61
32f
8f=6w, or fw, hence, the tenfion muſt be 3 fourths of the
cord's weight.
Question 128. There are two mufical cords of equal lengths and
diameters, with what force or tenfion, muft each be ftretched, to
found a fifth? See theorem 178.
Here, the vibrations being as 3 to 2, the tenſions must be inverfely
as the fquares of thefe numbers viz. as 9 to 4.
Definition, If two mufical ftrings vibrate in the fame time the con-
cord is moft perfect, and more agreeable to the ear than any other,
and is called unifon, and fo on as here fet down.
If the times of the vibration of two muſical ſtrings be,
As
0-01
it is called
unifon, which is moft pleafing to the ear,
diapafon or octave, lefs pleating,
diapente or fifth, lefs pleaſing,
diatefaron or fourth &c. of which this laſt
and the next that follows it in order are not ſo pleaſing to the ear, and
are therefore called imperfect concords, nor are there above feven notes
(befides the half notes called flats and fharps, by which the natural
notes made half a note higher, or lower, as the nature of the mufic
requires) in all the infinite variety of tones fit to merit a place in mu-
fical compofition.
Question 129. If two mufical ftrings of the fame kind A and B have
their tenſions equal, and their diameters as 5 to 6, what muſt be their
lengths to found a diapente or 5th? See theo. 178.
AND MECHANIC.
165
In this cafe the vibrations being as 2 to 3, or as 3 to 2 it will be, as
3:2::51:6 L, whence, 10l 18 L, or 11,8 L, fo as 1 : 1,8
:: L: 1, that is the length muſt be as 1 : 1,8.
Question 130. There are two mufical ftrings of equal length whoſe
weights are as 5 to 6, what muſt be their tentions to found an octave?
(By theorem 178.) As 1 : 2 :: √
6
F
ergo 2 √ =
5
f
or 20 F6f, whence, as 20: 6, or as 10: 3 :: f: F, i. e.
the tenfion of the lighter ftring must be 10, to make it vibrate twice
while the heavier ſtring whoſe tenſion is 3, vibrates once.
Question 131. There are two mufical ftrings of equal tenfions whofe
lengths are as 3 to 1, and their vibrations are as 5 to 6, in the fame
time, viz. in fefquiditonus, what is the ratio of their diameters? See
theorem 178.
As 5:6:: 3 d: D, or as 5 : 18::d: D, that is, the diameter of
the longer ftring is to that of the fhorter as 5 to 18, and vibrations as
5 to 6. From theſe queſtions may muſical inſtruments be ftrung to the
beſt advantage.
Question 132.
If in a minutes time you eſpy,
The moving wings of a fmall fly,
Nine thoufand times to fhift;
What number then muſt be the ſwings,
In the fame time of a bee's wings,
To found a perfect fifth?
As the fly is lefs than the bee, let her fwings in the fame time be
more than thoſe of the bee's, then to make a fifth, the vibrations in
the fame time being as 3 to 2, we'll have as 3:2:: 9000: 6000 anſ.
Question 133. If a flute 10 inches long be unifon with a muſical
ſtring, what muſt be the length of another flute of the fame bore and
force of wind, to found an octave with the ſaid ſtring?
Definition, In wind inftruments of mufic, the found being made by
the vibration of a column of elaftic air contain'd in the tube, the time
of vibration or tone of the inftrument, nuft vary with the length and
diameter of the faid column of air which compreffes it, in the fame
manner as the tones of muſical ſtrings vary with their lengths, diam-
eters and tenfions, from whence (by queſtion 128) the length of the
required tube will be 5 inches.
Question 134. If the axis of a ſpeaking trumpet be 10 feet, and the
area of the orifice at the out end 0,5 foot, how many times will it
magnify the voice of the fpeaker? See queſtion 137.
If we ſuppoſe, a perfon plac'd in the center of a fphere, whoſe ra-
dius is equal the length of the tube 10 feet, and there fpeak without
166 THE UNIVERSAL MEASURER
}
C
ST
the tube, it's then plain his voice will be equally diffuſed thro' the whole
furface of the fphere, but ſpeaking thro' the tube with the fame ftrength
of voice it will be only diffuſed thro' that part of the ſphere's fun face
which fubtends the out end of the tube, and becauſe this part of the
faid furface is but fmall, it may be taken for the area of the faid end
of the tube, and then it will be, as 0,5 the area of the faid orifice is tớ
1256.64 the furface of a ſphere whofe radius is 10, fo is unity (1) to
2513.28 times anſwer.
Question 135. If the axis of a fpeaking trumpet (otherwife called
the itentorophonic tube) be 10 feet, the diameter at the leffer end I
inch, and the diameter at the greater end 8 inches, what muſt be the
diameter at every foot in length when the trumpet is made to the beft
advantage?
}
Since the force of the voice is propagated thro' a feries of elaſtic
bodies of air in the tube, it is evident from art. 258: that if thefe
portions of air be in geometrical progreffion, the voice will receive the
greateſt augmentation poffible. Whence, if the required diameters
divide the axis into 10 equal parts, thefe parts being fall may be
taken as cylinderical bodies of air in geometrical.progreffion,, but cylin-
ders of equal altitudes are as their bafes, and thefe cylinders being by
fuppofition indefinitely fhort, will be nearly as their diameters, whence
the diameters of the tube, muſt be in geometrical progreffion, and
therefore if they be 11 in number, the leaft being 1 (Q) and the
greatest 8 (Qe n-¡fee art 258, it is required to find e the common
ratio of the diameters n being equal 11, fon-110, therefore, Q
being equal 1, we have, e =8, or e= io √81,232, fo the leaſt
diameter viz. being multiplied 1,232 gives 1,232, the diameter at
a foot length from the leffer end, and IX 1,232 gives 1,515 inches
the diameter at two feet from it, and fo on. From this it appears that
a fpeaking trumpet formed by the revolution of the logarithmetic curve
about its axis, will augment the found, more than one of equal length
of any other ſhape.
10
Queſtion 136. What must be the weight of a bell to found the lower
octave, with a ſmaller bell of the fame kind that weighs 25 pounds?
The found of a bell confills of a vibratory motion of its parts much
like that of a muſical (tring, now by the foregoing queſtions, the length
of a ſtring gives the lower, or graver octave to a like ftring half that
length, or half the length gives the acute or fharp cave to the whole
length, fo as 1 (the cube of 1) is to 8 the cube of 2) fo is 25 lb to
200lb anſwer.
It is proved,by experiment that if a bell be rung in the exhaufted
receiver of an air pump, it will give no found, whence it follows that
1
AND MECHANIC.
167
1
I
found is nothing but the tremors and vibrations of the particles of a
fonerous body impreffed on thoſe of the air, and as a pool of ſtagnat-
ed water, being any how ftruck on its furface, will by that ftroke be
put into waves, which are greateſt where the ſtroke is made, and de
cline gradually in a ſpherical form around the ſaid ftroke till at laſt
they have no motion, fo the air being (truck, the waves, or pulfes of
it give the greatest found nearest the itroke and as thefe pulfes weaken
the found is leffened, till at last it cannot be heard Hence, founds`
are ftrong or weak, according as we are nearer to, or farther from
the founding body, and fince (by theo. 178) the vibrations of a found-
ing body are each made in the fame time and thefe vibrations being
the cauſe of found, it's evident that the intervals of aerial pulfes will
be fo too, and that all founds whether loud or low, which are excited
by the vibrations of the fame body, are of one tone; alſo, thoſe bo-
dies which vibrate flowell have the graveſt or deepeſt tone, and thofe
bodies which vibrate quickelt, have the fharpeft or fhrilleft tone, founds
are loudest when the preffure of the atmoſphere is greateſt viz. in dry
weather for then the particles of moiſture in the air are feweſt, ſo it's
elaiticy is greatett, sit Ifaac Newton, (by comparing the vibrations of
theſe aerial waves, or pulfes with thofe of a pendulum, whoſe length
is equal to the height of a homogenial atmoſphere every where of the
fame deality with the air at the earth's furface, the bulk of a particle
of air equal to one of water or falt, the air to the vapours in it as ro
to 1 &c.) proves the motion of found to be uniform and at the rate of
1142 feet per ſecond, but in every 10 miles we must allow about half
a mile when the wind blows (trongly against the found, and deduct the
fame when it blows with it, but when it blows crofs the found it makes
little alteration in its velocity, all which agrees with experiments of
firing guns &c.
Question 137. If a body A, be placed 3 feet diftant from a lighted
candle and a body B, 10 feet therefrom, what proportion of light-hath
each ?
If the fame quantiy of light be placed in the centers of two different
ſpheres its evident the furface of each fphere will be illuminated by
this equal quantity of light inverfely as the faid furfaces, but the fur-
faces of fpheres are as the fquares of their diameters or radiufes there
fore, as9 (fq. 3) is to reo (fq. 10) fo is the light received by B to
that received by A, the fame is to be obferved in heat, cold, found, &c.,
Question 138. If it be ro feconds of time between dropping a ftone.
6lb weight into a pit, and hearing it ftrike the bottom, how deep is
the pit and with what weight doth the ftone fall at bottom, allowing
found at the rate of 1142 feet per fecond, and the defcent of heavy bo-
dies 16 feer in the firſt fecond of time?
168
THE UNIVERSAL MEASURER
I. Let d the pit's depth, then as 16 feet: 1 fecond::d://
d time of the ftonc's defcent, and as 1142 fect : 1 fecond ::
d.
1142`
d:
= time of the aſcent of found, now by the queſtion the ſum of
theſe two times is 10 ſeconds, that is, i √d↓
4
d
=10=t,
1142
fod +285,5 ✔d=1142 t, and by compleating the fquare, d + 285,5
285.5, whence/d=√: 1142 t
vd+O
28515
=1142t+O
2
+fquare of 285.5
2
2
285,5 =35,6 nearly, which ſquared is d
2
1267,36 feet the depth fought.
2. Now the body falling 16 feet-S, the firſt ſecond of time, it then has
16 feet velocity, which in the fame time will carry it uniformly over
2 S, twice that ſpace, therofore, as S: 2S::d: 2 vd S, the
uniform velocity acquired by falling thro' d fo 6 lb x 2 √ ds = 6 x
2 X 35,6 × 4 = 1708,8 lb, the required force with which the ſtone
ftrikes the bottom.
I
3. If d = 1, then dS= 0,5 and 6 × 2 √ ds = 6 lb, or the
force of the ftroke body's own weight, but if do, then this force
iso, now if the velocity be the fame, this force or momentum is the
fame, let the body move in what direction it will, but if it fall directly
downwards, it's evident it preffes with it's own weight more than the
momentum, whence 1708,8 +6=1714,8 lb (in this queſtion) is the
force wherewith the body preffes at the bottom of the pit the moment
it comes there. This is called preſſure, and ought in all fuch cafes to
be conſidered from momentum, if the body move down, an inclined
plane, it's weight thereon may be found by what goes before.
Question 139. If a man ftanding at the fide of a river oppofite to a
bank or fome houfes &c. on the other fide, hear his voice reflected
back in 3 feconds of time (called echo) what is the river's breadth.
When found ftrikes any large obftacle (ſee art. 259) it is reflected
back with the fame velocity, fo in this cafe, the found moves twice
over the river in 3 feconds; therefore, half of 1142×31713 feet
anfwer.
Note. To increaſe the hearing with a tube, it ſhould be made as
directed in queſtion 135, the longer the better, and fet the leſſer end
to the ear, and the wider end gathering in more pulfes of air will
mightily increafe the hearing, &c.
Here follow 26 queſtions concerning the maxima of bodies moving
in fluids, how to conftruct mill's engines &c. to the best perfection the
force of moving bodies &c.
AND MECHANIC.
169
Question 140. If the length of a ſolid be 50, and the radius of the
greater bale 20, required the radius of the leffer 'baſe, and form of
the fold, that moving in a perfect fluid in direction of its axis with the
leffer baſe foremost, it may be the leait refilted poffible, or leſs, than
any other folid of the fame length, bafe and furface. Fig 210.
1. In art. 435, we have a y u v-v vzzzz for the equation of the
curve generating this folid, which (art. 418) is found e- X:
nnnnnn-2,3025 log. n: and y =
a
nnn + 2n
I
+
a
na
na
I
I
which made a minimum (n variable) we get
X:3 nn
+2:
a
nn
Dn
I
o, which folved gives n =
fo y:
3.08
=
=
fere for
√ 3
nd
d
the leaft femi-ordinate, alſo, if in the value of e, we write
I
for Dy
№ 3
T
its equal, we'll have e=
5
2.2025 log
3, which taken from
12a
a
the aforefaid value of e=
a
x: ¿nnnn+nn- 2,3025 log. n:
leaves e= 1x: Innnn +na- — 2,3025 log. n √ 3 : by
✔
a
which values of e and y (e, by the queſtion being 50 and y = 20)
and taking e to begin at the middle of the leaſt ordinate, we get 50
I
n
x: n³ +20+ 20 ×:ñ¹ +nn — — 2,3025 lag. a √3:
from which n is found near =3,5, then
2,5, whence y=
3.08
a
nn +1
ny=20=
we find a =
na
1,2 the required radius of the leffer baſe;
now if the folid be form'd by the rotation of the ſpace EBRQ about
the axis É Q, then R Q = 20, E Q = 50, E B ='1,2, and to find
EQ - EB
as many femi-ordinates Fzy, with their reſpective abfciffes E. Fe,
as you pleaſe, you may taken any numbers between
3,5 fuppofe=3, 2, 1, &c. and you'll have y =
3 and
=13:33, 5a
na
Y
170 THE UNIVERSAL MEASURER
1,6, &c. and e=Lx:1m* + nm
â 4
5
12
2.3025 log. √ 3 :
easily conſtructed. L
E 30, 5,8, 0,34, &c. by which the figure is
appears that the leaft ordinate E B cannot be = 0. yet if E Q and Q R
be unlimited, the conftant quantity a may be taken as great as you
pleafe, and then y =
3.08
=EB, will be very ſmall.
a
Question 141. If AB (fig 208) be the breadth of a river or canal,
required the pofition of the food-gates D A=D B, refifting the water
with the greateſt eafe?
Firit, Upon BD produc'd let fall the perpendicular A E, now the
longer either gate is, fuppofe the gate B D, the greater will be the
preffure of the water againſt it. Alſo, the longer timber of the fame
diameter is, the weaker it is; hence, in each of thefe cafes the refift-
ance is inverſely as the length of the gate, and therefore, in both caſes
together it is as
but per fimilar As, as
I
OBD
"
BD:BC::
BABE, or becaufe B CA C, and is conftant, it will be
which is as the refiftance of
25 PD: I :: IBEX
OBD
the gate BD; again, the force with which the gate A D refifts the
preffure of the gate BD at D, is the two forces ED parallel to B D
and fo avails oothing, and A E perpendicular, which fo, is the only
active force (the fum of thefe two forces being the one force A D,)
bence BEXA E a maximum, i. e. pattinge A Eanda BE,
then: a a
ee:xe=aae-eee, fo (by art. 221) aazee,
shen e = a √÷=1577; the line of 35° 16′ = Ls D-B´A and DA B
as required.
Queſtion 142. If the length of a conical fruftum be so, and the ra-
dius of its greater bafe 20, required the radius of the leffer baſe, that
moving in a perfed fluid, in direction of its axis with the leffer baſe.
foremost, may meet with lefs refiftance than any other ſuch fraftum of
the fame axis and bafe? Fig. 216.
Bet Pa F, be a right angled triangle by whofe rotation about F P
a cone is form'd in which E B a F, forms the required fruftum, mov-
ing in direction F E, putra e, r B=FE=b=¿o, nF=2=29,
then F Faa-a-c2ae-ee, and n B =
a ÷
p
IT
bb+ce, alfo [ar=eg this compared with : (art.423) gives
AND MECHANIC.
171.
aa
2ae-ee or ſuppoſing the refiftance of the baſe (a a) to be
65400
1, we have
eex: 2 ae—ee, for the reſiſtance of the furface des
bo tee
ſcribed by a B, to which a
>
and we'll have
that of the circle deſcribed by BE
aabb-2aebb +eebb
2aebbeebb Lazee
for the whole re-
bb tee
fiftance of the fruftum defcribed by E Ba F, which by the quellion
muit be a minimum, and therefore making e variable we'll (art. 421)
hare
2abbee+2ebbbb-żabbb
bb fee
b b
=0, whence ee+c=bb,
b
a
and by compleating the fquare &c. e == √:4aa+bb: -
2 a
BE 12, anſwer,
BE
ba
ભ
8 nearly, fon F 20 — or 8 = r F
— or 8 = r F
Queſtion 143. If the axis of a folid be 50, and its folidity 3612264
(S) required irs greatest diameter, fo that moving in a perfect fluid in
direction of its axis with the leffer baſe foremoſt may meet with lefa
retiitance than any other folid of the fame length and ſalidity ? Fig,
210.
1. The equation of this folid (art. 435) is a uvv ▼ = Y Ż Ż Z Z j
which being folved gives (art. 418) e 3 ann +
=
2X:00+
and y =
=
a n where if a = o, then yo, ſo in this cafe the curve will
meet the axis, or the leaft ordinateo, but when a =0, e={a,
therefore, that the abfciffa's and ordinates may begin together, or be
o, at the fame time this a, muſt be taken from the above value of
e, and then e=
Заво + a
2:00+1*
a
2
200-20
2X:00 +1
where it 29
pears that a muſt be leſs than unity, otherwiſe the numerator will be
n
negative, then from y=
y =
ann-annnn
an
and e=
we get
00 + 11
2X:00 T
2 € which made a folid ( C = 3,1416) we get cayya
non
4ce wc÷:0-non, whofe finent (theo. 78) is
3SX: 0
S, fo
4C
4 ce e e
3X0-000
=ece=12503o, therefore, a - 800-
172 THE UNIVERSAL MEASURER
3 S
✔:1250004 4 C
:
nearly, whence n = 1, and therefore, a
>=240, which doubled is
fo
a n
2ex: no 11 = 8331, 1o y = √1 + 1
nnnnnn
S
no
480 the required diameter, i. e. if this folid be formed by the rotation
of the curve A B R, about A Q e=50, and QR = y = 240,
the content of this fold 3612264, and by taking n = many
833; always) you may have as many
abfciffas and their refpective femi-ordinates as you pleaſe, and ſo draw
the curve.
1
fractions between 0 and (a
+
Question 144. If a perfect fluid in direction C B, ſtriking againſt a
plane ǹ B P, make an LC B n therewith of 30°, Required the ratio
of the forces of this fluid to push the plane forward, and to turn it
about. Fig. 210.
=
co-fine
1. Draw Cn Bn and nr CB, let e fine Ln C B
LCB n, radius = 1, then I
-ee fine ↳ CBn; then if CB,
which fuppofe = 1, exprefs the whole force of the fluid, Cn will ex-
prefs that part of it which tends to move the plane forward in direction
CB and nr being that direction, mult therefore expreſs the part of
that force which tends to turn the plane about; but the whole force to
move the plane forward being as fine incident L C B n, viz. as I
(radius): 1 (CB) : : 1 ee (fine LC Bn): 1 ee force
Cn, and by trigonometry, as radius (1) : C n(1 —ee)::e: e
e e e force n r; whence, as force Cn to push the plane forward :
force n r to turn it :: I ee: e eee :: 1:e :: (in this queſt.)
1:40,75, i. e. as radius: the co-fine of the L of incidence, anſwer.
2. If e
eeem a maximum, then I 3 e e = 0, ſo é =√1}{
0,57733 the natural fine of 35°16′ whofe comp.is 54′44″; hence
the water has the greateſt force against the rudder of a fhip, &c. to
turn the ſhip, when it makes an L with the keel of 54°44′. Alfo the
wind blowing parallel to the axis of a wind mill, gives the fails the
greateſt force to turn, when the plane of the fail makes an L with the
Laid axis of 54°44′•
fo
Queſtion 145. If the velocity of the wind be at the rate of 3 feet
per
fecond, with what force will it trike a plane of 10 feet area when
it blows perpendicularly againft it? See queftion 178, and art. 323.
Wind being a ſtream of air, any thing ftruck thereby may be look-
ed upon as in that fluid, and a cylinder of air, a foot bafe, and a foot
height (viz, a foot folid of air) weighing ,07268 lb, we'll have a =
,07268 lb, v 3 the given velocity S=16, fo (from art. 323) R =
,07268 vv
=,00113 VV,07268 h, and taking A =
v v a
45
67
AND MECHANIC.
173
the area of any plane in feet, and Fforce in pounds weight, we have
this general expreffion ,00113 vv A,07268 h AF,,10170 lb,
in this cafe, becauſe A 10 and v = 3, or v v ÷ 9.
The velocity of the wind may be very eaſily computed thus, take
a feather or fome light body, and letting it go in an open plane where
the wind is not moleſted, obferve the time of its flight by a half fecond
watch or pendulum, then meaſure the diſtance it has flown, fo you'll
have it's velocity, by this method the Rev. Dr. Derham found that
the velocity of a very great wind was at the rate of 66 feet per fecond
and that at a medium its velocity is at the rate of 17,6 or 22 feet
per fecond.
Question 146. If the area of a wind-mill's fail be 50 feet, and the
velocity of the wind 20 feet per fecond, with what force will it ſtrike
a fail, when the fails are fet to turn to the beſt advantage?
By question 144, the force of the fluid in direction A B, is to its
force in direction C D, as a a:
aaaaaa7
eee
a aeeee
a
oras a a:
3
a
(becauſe ea✔) which by reduction and taking a = 1, will be as
2
I: 4, or as I :- or as 5,19: 2, which is nearly as 13:5 ::
5519
the whole force of the wind, to that part of its force which blows a-
gainſt the fails when they are pofited in the beſt manner for turning,
therefore, per laft quethon,x,00113 v v A,000435 vv A=F
the required force, and because v≈ 20, and A = 50, we'll have
,000435 v v A = 8,7 lb. the force conftantly prefs'd againſt the given
fail, which multiply'd by 4 the number of fails (if they are all equal in
area) gives 34 8 lb, the force againſt them all.
Queſtion 147. Things being as in the laſt queſtion, fuppofe the 4 fails
to be equal and alike, and the diſtance between the center of gravity
of each fail and the center of the axle-tree to be 5 feet, required the
velocity of the fails, when the mill is charged with ſuch a weight as to
perform the greateft effect?
By this queſtion, if the mill be charged with a weight of 174 lb, it
will be in equilib. with the force of the wind, fo (by queſtion 150) of
174 lb is 77 lb with which the mill being charged it will perform
the greateft effect, and (by queftion 150) of 20 the winds velocity is
6 feet per fecond for the fails velocity, alfo, 5 X 34,8
T
3
174 lb,
the force with which the axle-tree is prefs'd to turn, and if d = the
diltance between the center of gravity of any fail and that of its axle-
tree, then,00113 vv Ad F, the force to turn the machine.
174 THE UNIVERSAL MEASURER
Question 148. If the four rectangular fails of a wind-mill be each 8
feet long, 2 feet between the lower end of each fail and center of the
axle-tree, and velocity of the wind 30 feet per fecond, what must be
(b) the breadth of each fail, when fet in the beſt manner for turning,'
that the axle-tree may be in equilib. with 174lb weight?
=
Becauſe the force of the preffure acts at the center of gravity of each
fail, we may take A = 8 xbx ( 2 + 1 of 8) 6 = 48 b
Then (by
queſtion 146),000435 v × 48 b = F174 lb, or 18,792 b 174,
174 9,3 the breadth of all the 4 fails, therefore 9,3
18,792
fo b =
42,4 feet, breadth of each fail,
if charged with 77 lb weight =
and the mill will then work beſt
of 174lb.
Queſtion 149. If the wind blow at the rate of 30 feet per fecond,
perpendicularly against a wall 200 feet long, and a feet high, with
what force doth it endeavour to overfet the wall ?
The wall being a rectangle of equal thickness, its center of gravity
will be in its mi idle, viz. at 10 feet height, then A = 200 X 20 X 10
=40000, fo (by quetion 146),00113 v v A = 4,52 v v = 4068 lb,
anfwer.
Queſtion 150. If a ftream of water eonftantly preffing or ftriking the
pallets of a wheel or engine, mill &c. with (w) 810 weight of water,
be juſt fit to give the machine motion with what weight (p) of water
muit the pallets be ſtruck when the effect of the machine is the great-
eſt poſſible, allowing the whole weight on the machine to looſe by.
friction when the velocities of the pallets and ftream are equal?
1. Let v = the velocity of the ftream, e the required velocity
of the pallets, then ve the difference of theſe velocities, or that
with which the pallets are ftruck, now the force (art. 319) being as
the fquare of the velocity, and pand w the weights that balances the
force of the ftream when its velocity is v and v e, we have as v v :
व
p :: ve² : PX
A
nifeſt PX ve
jigl
V V
2
V
V Y
e
w, if there were no friction, but its mas
muſt ballance both w and the friction, now let a
t
the
=0,1 (≈ 1 of 1) then a wap the friction when the velocity of
caw + eap-
the pallets is v, therefore as v ; a w†ap::e:
friction when the velocity of the pallets is e, but to be more univerfal,
let this friction be expreſs'd by
enw+emp, then Ex
V
V
AND MECHANIC
175
en w+emp, or £x v—c":
V
W=
ej
v teu
v - ej e
X
vten
___x v—c)² = wv+caw+émp, whence
Y
emp, which multiplied by e, gives ew=
v + en
emp the momentum, which by the queſtion
vten
must be a maximum, fo making e variable (prob. 201) and taking the
Auxion we'll get vvv-
o
4vve + 3 vee
2 myye — mnyeezo, but if we divide this equation by v — e,
2 veen + 2 eeen
it will be abridged to v v― 3 e v— 2nee
2 m v ve
V
mavve
=0,
now taking m and n each 0, 1, we'll find e 0,3 v nearly, whence
2
w = 1 x
: V -
el
ef :-emp
=
V
vten
39 P__, whence p =
P=
103 W
103
39
to 10, alſo, e w =
39p × 0.3 v
103
2139,2, or the velocity of the pallets
117 pv
to that of the ftream e to v, as 3
the momentum.
`general equation) whence e
1030
2. If we fuppofe the friction to arife only from the motion of the
weight w, then m = 0, and ſo v v—3ev — 2nee- o (from the laſt
2 V
6,13
2 V
= (ifn=0,1)
3+ √9+80:
=0,32 0, fo w = 1×
V
0,67 v
X
X
0,4489 44,89
P
vten
1,03 V
1,03
103-W
As fo
44,89
103
P = 1858,5 or the velocity of the ftream to that of the
pallets v to e, as 100 to 32, allo e w➡
143.648 D V
пози
рех
V
_44,89pvo,32☀
V
ven
103
the momentum.
3. If the machine have friction arifing from the weight of its parts,
it may be confidered in the weight w, thus, fuppofe it were known
that ro weight, applied to the pallets (the engine having no charge)
would give them a velocity v, then w being put 820 inſtead of 810,
the power p will be found fo much greater as to anſwer this weight.
If there were no friction at all, then m and, are each≈0, and
our general equation becomes v v — 3 ve — 0, foev, whence
176 THE UNIVERSAL MEASURER
W=
Px v− c 2
PX V
V V
the momentum.
4
= p, fo p =w= 1822,5, and ew=
17 P
27
Queſtion 151. A rectangular fluice conftantly 25 inches (a) deep of
water, by drawing up this fluice-gate 9 inches, the water ftriking the
pallets of a mill's wheel &c. is just fit to put the machine in motion,
how far muſt it be drawn up, when the engine performs the greateſt
effect poffible, allowing no friction?
1. If an engine be in equilib. with a weight w, and that weight be
madew, the engine (queftion 150) will perform the greateft effect,
and if the orifice be rectangular, (art. 331) the weight of water dif-
charged will be as the area of a parabola, of the fame bafe and height
with the water in the fluice, fo let b≈ 25 — 9 = 16, e—a— require
ed height, then 22abbow, and aa-÷bb:
2 ×
aabb ava
= 3
-ee, whence ava-bb-eve, fuppofez, then
—
3
∞w, avaee, that is,
Z
=
cec=zz, foe = z³ = 12,116, and 25 — 12,884 inches anſwer.
See queltion 155.
Question 152. A water mill is to be built where there is a fall of
water of 24 feet, whether will a wheel of 18 feet radius with 6 feet
fall, or one of 16 feet radius with 8 feet fall, grind moſt corn with
leaſt water.
ww
The velocities of falling bodies being as the fquare roots of the
heights fallen thro' we'll have (per queſtion 150) 18 × √ 6 = mo→
mentum of the greater wheel, and 16× √8 that of the leffer wheel
therefore, as 18√ 6 : 16 √ 8, or as № 243: √256:: the quan-
of corn grinded by the greater wheel to that by the leffer wheel in the
ſame time, and with the fame quantity of water, fo the leffer wheel
will perform better,
Question 153. If a current of water have 8 feet perpendicular deſ-
cent, what must be the diameter of a wheel to receive the greateſt
force poffible from the faid current ſtriking A the middle of the wheel.
Let a 8 feet — BE (fig. 212) the height of the fall, and e-GE
AD the radius of the wheel, then per laſt queſtion. ✔ a- .e.is as
the velocity of the water at G, and (by queftion 150). GA (e) X
√:a_e:=e√: a e: is as the force at G, which by the queſtion muſt
be a maximum, viz. e \/:a—e: =m, or which is the fame, aeeeee
m, fo (by art. 221) ea 5 feet, the radius fought, The
= 1
AND MECHANIC.
177
ftream fhould always ftrike the wheel directly, for if it ftrike it ob-
liquely it preffes the wheel againſt the other ſide, and fo increaſes the
friction, beſides, the advantage is lefs, for if q m be the direction of
the fluid, it ſtrikes the wheel at G, with a velocity: BG: draw Am
Lqm, then Am x BG the momentum in this oblique ftroke,
but if B G, be the direction of the fluid, then A GXBG the
momentum, &c.
I
3
3
Queſtion 154. If the ftream (laft queftion) have the fame fall B E,
but run in direction q m making an angle B q
m making an angle B q G with the horizon of 300
what then muſt be A m the diameter of the moſt advantageous wheel?
Having found A Ga=5 by the laſt queſtion, then by trigo-
nometry, as radius 1: G A 5 ::s LAGm 30: Am 2 feet, anf-
wer. If it be required to find what angle B q G, the ftream must make
with the horizon, ſo as to ſtrike the wheel with the greateſt force
poffible, lets its fine, then : 1.
I ss: its co-fine, and by trig-
onometry, sa Am, a : 1 ss := Gm, and 3sa √: 1 —ss:
Gn; then, Ga+GB Bna+sa√: I—ss, and by
=
the laſt queſtion, ✔: Bn: xsa =
a maximum by the queſtion, or,
i. e. aaax: ss + 2 sss: I
Z
theo. 149) 2 s +6 ss √ : I ~ss: —
S S :
=
S:
1
sa √ ÷ a + ÷ savi —ss
s a | × 3a+sa√: 1 SS:
ss: —m a maximum; fɔ (by
2 S
4
VISS
o, (s being variable)
which by reduction is: Iss: = 4 s
3 3 s, and by´involution
and tranfpofition 1 165-24 s4+1os s, which equation folved
gives s0,7214 the fine of 46°30′ Thus, if the water run down a
trunk, &c. making an angle with the horizon of 46°30' and the radi-
us of the wheel be found to anſwer this angle, &c. and the water ſtrike
its pallets perpendicularly, then the advantage is greateſt.
Question 155. If a rectangular flood-gate before a mill, engine,
&c. be 2 feet broad, and 6 feet deep of water, and it being drawn up
half a foot, the water running against the pallets be fit to give the
wheel motion, with what force doth it ſtrike the pallets, and how much
farther muſt it be drawn, that the machine may perform the greateſk
effect poffible, fuppofing the ftream to ftrike the pallets perpendicular-
ly, at the bottom of the orifice, the friction being nothing.
1. Here 2 × 0,5 = 1 foota, the area of the orifice, and h=5,75
feet the depth of water above the middle of the orifice, fo ha= 5,75.
folid feet of water, but a cubic foot of water is 6,25 lb, and 1121
Z
198 THE UNIVERSAL MEASURER
62,5 ha
ICweight, fo
I 12
3,208 C. or more accurately, (queftion
(6 — 0,5) to 3.6666,
4 of 5,5 × 1,914 =
151) thus, as 6 is to 4 (fquare of 2) fo is 5,5
whofe ſquare root is 1,914, then of 6 × 2
5,982 feet a, and h=6 feet, the depth of the fluice, ſo ha = 5,892
folid feet of water, which by the queftion ballances the engine, there-
fore (queſtion 150) ½ of 5,982—13,257 feet of water, with which
the pallets being prefs'd, the engine will perform the greateft effect,
fo 13,257 ÷ 6 (h) = 2,209, the area of the lower part of a parabola,
whofe height (a) is required, the axis of the whole parabola being 6,
and greateſt ordinate 2, ſo as 6:4::6
a: 4 — a, whofe fq.
√:
—‡
root is : 4a: then of 6 x 2 of 6-ax√4=
2,209, that is, 126: 4-a: +a√: 4-a:=3,3135, or
8,4865, which by involution and multipli-
aaa= fq. 8,4865 = 72 almoſt,
18 aa + 108 a 108, this cubic equa-
1.25 feet nearly, anſwer.
6-ax√: 4—a:
cation is 144 72 a
which reduced gives a a a
tion, folved gives a
12 a a
ах
ક
But becauſe this method is tedious in practice, the first method may
ferve for a guefs, thus of 1, the firſt mentioned area is 2,25, which
divided by 2, the breadth of the fluice, gives 1,125 feet (less then 1,25
by 0,125, the true height) for the height, the fluice-gate is to be drawn
from the bottom, when the engine works to the beſt advantage.
up
Note. the pallets or paddles of a wheel, fhould be juft fo many, and
fo fet as that the water may ſtrike them, perpendicularly one after
another; and if the water is little, it is better to have boxes than
paddles, becauſe the preffure is increafed by the weight of the water
in the boxes &c.
1
Question 156. If the breadth of a ſtream of water be 20 feet, depth
feet and velocity 5 feet per fecond, with what force doth it prefs a
plane fet perpendicularly to it? See art. 321.
Here v5,
15, and a 20 X 480, the area of the plane, but be-
caufe the plane is prefs'd to be over-fet, or driven directly forward,
by the force acting against its center of gravity, we muſt take a =
=20X
4× 2 (half of 4) 160, thien s being 16, we'll have ha=
X
5 × 5
32%
anſwer.
vva
2 S
X 160 = 125 folid feet of water = 7812,5 lb (62,5 × 125)
The velocity of any ftream may be found by dropping in fome light
matter that will fwim, and meaſuring the diftance it moves in one fe
J
AND MECHANIC.
179
tond, by this means you'll find that the velocity of a ſtream is much
greater near the middle, than at the fides, owing to the friction &c.
againſt the fides and bottom, where the water is fhallow, now if you
take half the fum of theſe two velocities, it may ſerve for the mean
velocity of the water.
Question 157. If the depth of a fluice be to feet, and there be a hole
in it's bottom 2 feet area, what quantity of water will run out in t
minute, or 60 feconds, and with what velocity, the fluice being
conftantly kept full? See theo. 196.
Heret 60, a 2, h= 10, then 6,128 ta 2 hs 6,128 ×
ta√ =
120 × V: 320: = 13089,4 ale gallons, that will run out in a minute,
and with a uniform velocity v = √2 s h = √ 320 — 17,8 feet per
fecond.
Queſtion 158. If there be a rectangular flit 2 inches wide and &
inches deep, at the top of a fluice kept always full of water, what
quantity of water will be difcharged thro' this orifice in a minute's
time? See ait. 330.
Here h 0,5 feet a 2 inches x 6 inches = 1 foot, t 60 fe
conds, and s always16, ſo of 6,128 ta / 2 hs = 245,12 № 16
=980,48 ale gallons anfwer.
Question 159. Two equal conical fruftums depth e 30, diama
eters of the greater and leffer bafes 20 and 10, each filled with water
&c. each ftanding upright, the one upon the greater and the other up-
on the leſſer bafes, required the ratio of the time in which each will
be emptied by an equal orifice in the bottom of each veffel? See art.
426. Here C = 66 and c→ 30.
C
1. As 10 (2010): 30 :: 20: 60 the axis of the whole cone,
and fo is 10 to 30, the axis of the cone cut off. Then as 2cc +
ce+ee is to 2 C C — Ce+ee::3360:5160 :: 28: 43::
velocity at the leffer baſe: velocity at the greater bafe, and in this
cafe time being inverfely as velocity, it will be, as 43: 28: time of
running out at the leffer baſe: time of running at the greater baſe.
Queſtion 160. If a column of water a B (fig. 213) 20 feet high,
be kept conftantly full of water, what angle muſt a ſpout a, at it's bot
tom make with the horizon a D, to throw the water to D, five feet
diſtance from the faid fpout a? See art. 366 and 333
If a hole at a, be fmall in refpect of the cylinder's &c. bafe, then
the height fallen thro' to acquire the velocity of the fpouting water at
a, muſt be 19, half of the columns height, then this height douỡ.
led gives 20 the amphitude of the fpouting water under an angle of 45
1
180 THE UNIVERSAL MEASURER
and for this angle to any other amphitude it will be, as this greateſt
amphitude 20 is to the radius 1, fo is any other amphitude, 5 to 0,25
whofe fine belonging, is 14° 45', half of which is 7° 22′ for the re-
quired angle, and then the path of the water is the parabolic curve
a e D, and taken from 90° leaves 82° 38′, for the direction of the
ſpout, and then the path is a u D. And thus, any fuch caſes may be
folv'd as in the queftions of gunnery.
Note. If the area of the ſpout be equal to that of the column's baſe,
then (by art. 333) the greateſt amphitude is double to what is here
taken.
Question 161. If each ſide of a cube full of water be 2 feet, what
preffure doth the whole cube fuftain ?
See art. 334.
lb, and each ſide of the cube 2
A cubic foot of water being 62
feet it will be 2 X2 X2 X 62,5
500 lb the preffure againſt the
bottom, half whereof is 250 lb the preffure againſt each fide,
and becauſe there are 4 fides, therefore, 250 X 4 = 1000 lb the
preffure againſt all the fides, fo 500 + 1000 1500 lb, the whole
preſſure ſuſtain'd by the cube, i. e. it is prefs'd with three times the
weight of the water that fills it.
Queſtion 162. If the breadth of a fluice bé 20 feet, and its depth
4 feet, with what force is the bank or wall prefs'd that holds it in.
Becauſe the depth is 4, the center of gravity will be 2 from the bot-
tom fo 20 X 4 X 2×62,5 = 10000 lb anſwer.
the
This compared with queſtion 156 may ſeem ſtrange, that the force
of ſtanding water fhould be greater than that of running water,
reafon is;
If the ſtream were at reft then the plane in it would
have no preſſure for then it is equally ſupported on either fide by the
ftagnant water, but in this queſtion there is no water on one fide of
the plane and fo the other fide is prefs'd by the water againſt it, &c.
Question 163. If a heavy body falling from a height of 15 feet into
clay, fnow, foft earth &c. do make an hole 5 inches deep, how deep
would the hole be if it fell from a height of 20 feet.
If a body fell upon foft or yielding matter, its evident it muſt take -
up fome time in making the dent, or cavity, and therefore, in that
time may be faid to move in a reſiſting matter or medium, but by art.
317) the refiftance is as the fquare of the velocity, and (by theo. 166)
the diſtances fallen thro' are as the fquares of the velocities, therefore,
it will be as 15: 5 :: 20: 6 inches deep anſwer. But if the body
fall upon a hard unyeilding ſubſtance, its then plain there will be no
time spent in deſtroying the generated velocity by the fall, and fo by
(theo. 155) the effect will be as the velocity, and thus the effect of a
AND MECHANIC.
181
ſtroke may be as the ſquare of the velocity, or as the velocity, or any
way between two, according as the matter ſtruck, yields, or not yields
to the ftroke, &c.
Queſtion 164. In what time (T) will a conical fruftum ſtanding up-
right on its leffer baſe, be emptied thro' a hole in it's bottom of one
inch diameter, the depth of the fruftum being e = 30 inches and dia-
meters 20 and 10 inches?
fquare 20
1. As 10 (20 — 10): 30 :: 10 : 30c, the axis of the cone eut
off, now becauſe the diameters and orifice are circles, we may leave
out 0,7854, the circular factor, and take A= ſquare 20 400, and
a ſquare 1 = 1 (theo. 197) then becauſe the veifel is ſuppoſed to
be full of liquor and none to run in while it empties, n will be 3,
alfo s 16 feet 192 inches, we'll from 2 Ta✅:
=
A es
n A A— a a
૧ for the quantity run out in T feeonds, of a cylinder, heighte
and area baſe =A= upper bafe of the fruftum, get T =
9
2 a
√:
nA A a a :=144 feconds (q being 30 × fq. 20) that is, an up-
e A As
right cylinder full of water depth
30, and diameter 20 inches
will be emptied by a hole in its bottom of one inch diameter in 144
feconds of time, then (art. 426) as 2 √e: p√ex: 2cc++ c
+: ee, fo is 144: 168,6 feconds the anſwer.
Note. here ✔p, the perameter of the cone is =
11, fo
Р 48
P
diam. 20
axis
60
11
2. If the fruftum ſtand on the greater bafe, with the fame orifice
there, then (queſtion 159) as 43: 28:: 186,6: 121,5 feconds, the
time of evacuation.
182 THE UNIVERSAL MEASURER
Here follow 26 queftions concerning the ſpecific gravities of bodies,
the properties of the air, the conftruction of weather glaffes,
pumps, &c.
A TABLE of fpecific gravities of Bodies.
coal
caft brafs
fteel
copper half-pence
fine brafs
9,000 logwood
8,915 ice
8,35
beech
8,100 afh
Fine gold
19,640
ftandard gold
18,888
brazil wood
lead
11,340
box wood
fine filver
11,092
bee's wax
ftandard filver
copper
10,536 oak
1,520
1,03 I
1,030
955
,920
,913
,908
,854
,820
7,850 yew
›760
iron
pewter
7,644 elm
,750
7,471
crab-tree
,700
tin
7,320
cedar
,613
caft iron
7,000
fir
,580
lead oar
6,200
cork
238
pebble ftone
2,700
glafs
FLUIDS.
2,600
flint
2,570
quickfilver
14,000
common ftone
2,500 || urine
7 1,032
brick
2,000 milk
earth
1,984 fea-water
1
chalk
1,793 ale
day
1,712
common water
fand
1,520
common air
1,031
1,030
1,028
1,000
.,001,2
in refpect
Thefe are mean fpecific gravities of thefe folids and fluids,
of their goodness, fineness, drynefs, texture, &c. alfo heat and cold
will make fome difference; this table not only fhews the ratio's of the
Specific gravities, but alſo the weight of a cubic foot of each in aver-
dupoife ounces, for a cubic foot of common water is found to weigh
very nicely 1000 averd. ounces, ſo a cubic foot of lead, will be 11340
oz. one of copper 9000 oz. one of niercury or quickfilver 14000 oz.
one of air 1,2 oz. &c. for any other, and may, if needful, be reduced
(by ex. 262, fect. 8) to troy weight.
Theſe ſpecific gravities are found (by art. 341, 342, &c.) thus let
the fpecific gravity of water be 1000 c, then if the body be
heavier than water firſt weigh it in air, and fuppofe its weight there
= 9000 oz. —A, then having a cord faftened to it let it go into
water and obferve nicely what weight will hold it in equilib, when its
*
AND MECHANIC,
183
covered with water fuppofe 8000oz.B, then
€ A
A
B
a=90000Z
the fpecific gravity of the body fought which 9000 looked for in the
table is found to be copper, and fo as 1000: 9000 :: 1: 9:: the
fpecific gravity of water to that of copper; alfo, if the ſpecific gra-
vity of the body be 9 you'll have c =
I for that of the
a A — a B
A
the weight of the
fluid, but if the body is lighter then water, then tie a piece of metal to
it to make it ſink in water, and then taking e
compound in water, and a, A, B, C, as before, we have
CA
Ate-f
та
the ſpecific gravity of the light body. Thus, if I take a piece of
metal whoſe weight in water is 31 oz. and tie it to a bag of beans and
the weight of the whole in air be 101 oz. and in water 7 oz. then,
CA
A + e - f
1000 X JOI
ICI +31-7
•
101000
125
808 oz. that is the
Specific gravity of water is to that of beans as I to o,808, hence a
cubic foot of beans weight 808, oz. or 50 lb averdupoife, and thus
may you find the ſpecific gravity of any thing what ever, but for frag-
ments, dut, powders, &c. it will be beft to weigh them in clofe boxes,
or metal buckets, obferving to ballance the weights of fuch boxes &c.
both in air and water.
Question 166. If a crown &c. made of gold and filver together,
weight 128, fpecific gravitiy 16, how much of each of thefe metals is
in the mafs?
See art. 344.
a b c
b
From d
we'll have z =
-d: xaç
bz
dz +ac
:ba
a: xd
4224
33 the weight of gold in the
19-11: × 16
128
:19 16: XII X 128
the mixture, fo 128
3395, the weight of filver therein.
Note. I have here taken the fpecific gravity of gold to that of ſilver
as 19 to 11, becaufe that of the crown is 16 to fuch parts.
Question 167. If the fpecific gravity of a man's body, water and
cork be as 10, 9, 2,25, how much cork will make him fwim?
If he be made with cork tied to him of the fame fpecific gravity
with water, (9 d) its plain if a little more cork were added he
gould not fink, fo from (art. 344) d =
a b c
:b-a:x:ztac
we'll have
184 THE UNIVERSAL MEASURER
(fuppofing the man's weight 150lb = z) c =
zd x : b = a:
ax: b-d:
150 X9X:: 10+ 2 1:
150 X 9 X 7.75
155lb, the
10 x 0,75
10 X : 24 — 9:
weight of the man and cork together, fo 155-150
the answer.
5lb of cork,
Question 168. If a folid inch of fome matter weigh 8 ounces, avoird.
what is its ſpecific gravity, weight of water being 0,578 oz. per inch?
By art. 346, it will be as the folidity in inches I : the weight in oun-
ces: I to weight, or
8
I X 0,578
13,8408 (0,578 being the oun-
ces weight of a folid inch of water) i. e. as 1 : 13,0408 :: the ſpecific
gravity of water that of the matter.
Question 169. If in water, a cube of fir, fink 3 inches downright,
and if each fide be 12 wide, what ounces is its weight.
Since a folid inch of any fluid bears a folid inch of any matter ofthe
fame ſpecific gravity, it will be, as I folid inch of water is to 0,578697
ounces its weight, fo is (12 X 12 X3) 432 folid inches the wet part
of the cube to 249,997 104 oz its weight anſwer.
By this method it will be eaſy to find the burden and weight of any
'fhip &c. for, if you find the folidity of the wet part of a fhip when ſhe
ſwims empty (in inches) and multiply it by the weight of a cubic inch
of fuch water as ſhe ſwims in, the product is the ſhip's weight in oun-
ces. And if in like manner, you find her weight when loaden, its
plain the difference of theſe two weights will fhew the weight of her
burden, which may be reduc'd into tuns; thus, a cubic foot of fea-
water weighing 10300z. or 64,375 lb. and 2240 lb being = 1 tun, fo
as 64,375 lb 1 foot :: 2240lb: 34,78 feet, the weight of a tun of
fea-water, ſo if you find the folidity of the fpace in the inſide of the
fhip in feet, between the light, and laden marks, and divide that ſo-
lidity by 34,78, you'll have the true weight of the burden in tuns,
which is a much truer method than the common one in queſt. 10.
→
Question 170. What muſt be the thickness of metal in a hollow
cube, made of 4 folid inches of copper, that it may fwim in one inch
deep of water?
Let s =
4 the folidity of metal, a a fide of the required cube,
and e1 the inches, D the fpecific gravity or denſity of copper,
and d — that of water, which are as 9 to 1, now a a e➡ the bulk of
the wet part of the cube, and s bulk of the metal, fo (by art 338 )
AND MECHANIC,
185
daaeDS whence a D6 inches, therefore 6×6×6=
de
216 inches the folidity of the required cube, from which taking 4 the
quantity of metal leaves 212, the folidity of the cavity whofe cube root
is 5.963 inches, the length of the infide of the hollow cube, fo6—5,963
0,037, half of which is ,0185 inches, the required thickneſs of the
hollow cube.
Question 171. If the axis of a copper fphere be 10 inches, what
muſt be the thickneſs of the ſhell, that it may fwim in the air?
=
To give a general folution to queſtions of this nature, Let D= 10
inches the axis of the ſphere, d the axis of the cavity or fhell, e =
the inches of depth that the ſphere is to be immerfed in the fluid, p =
0,7854, and m to n as the denſity of the fluid to the denſity of the
metal, then pDDD the ſphere's folidity, and pddd the
cavity's folidity, their difference is 3 px: DDD-ddd the content
of metal; alfo, p Deepeee the folidity of a fegment of the
fluid in which the fphere is to fwim, or be in equilib. with, fo (by art.
338) mpx:DDD-ddd:=npx: Dee
eee: fo,
ddd=DDD : 3 Dee+cee:, but in this queſtion the
—
2 111
fphere is to be wholly immerfed in the fluid, fo e D, and then, this
laſt general theorem becomes d d d = D D D — " D D D = D D D
n
X: I
: fod=DVI
m
n
m
n
-, now by the foregoing table the
denfity of copper is to that of air as 90000 (n) to 12 (m), or lower,
3
ก
as 7500 to 1, whence d=³√✓ I— — XD='√ I-
3
7499
7500
m
D-d
X 10 = 0,99978, whence
inch, the thickness of metal required, but
2
I
X 10.=
7500
=,00011 parts of an
Note. There muſt be no air in this ſphere, for if it be filled with air,
it will ballance the fame bulk of air without, and fo cannot be in
equilib. therewith.
Question 172. If two folid feet of fome light matter as wool, fea-
thers &c. weigh 4 lb, how much will it weigh when preſs'd into half
a foot bulk?
A a
186 THE UNIVERSAL MEASURER
(By art. 316,) the weights loft by theſe two bulk are, as to 2, or
as I to 4, now a folid foot of air, being 1,2 oz. its plain the mat-
ter will lofe 1 times more in 2 feet bulk than in a foot bulk, fo 1
X 1,2 = 1,8 oz. hence 4 lb 1,8 oz. is the anſwer; hence, the clofer
light fubftances are tied together, the more they will weigh.
Question 173. The denſity of gold being to that of braſs as about
2 to 1, if 2 lb of gold be in equilib, with 2lb of brafs when the air
is in a mean gravity, or the mercury in the barometer ftands at the
height of 28 inches, what will they differ in weight when the mercury
tands at 31 inches ?
I. A cubic inch of gold weighs 10,36 ounces troy and a cubic inch
of air of a grain, when the air is in a mean ſtate, ſo as 10,360z,: †
grains :: 12 oz. :
grains, which 12 oz or 1 lb of gold wil!
24
72,52
lofe when the air is in a mean ſtate, (2) as 28 : : : 30:
196 49
=
15. gr.
60
the weght of the air (1 inch) when the mercury ftands at 30 inches
height, then as 10,36:
15
49
180
:: 12:
507,64
by a pound of gold in that ftate of the air, now
45
grain more loft
126,91
24
=,37 and
45
126,91
72,5
,3546 and,37 ,3546,0154 grains, which doubled is,0308
gr. more lofs in 2 lb when the air is heavieft, now if the bulks of brafs
and gold were ➡, their loffes would be — and ſo the equilibrium ftill
kept, but the bulk of brafs being double to that of gold it muſt loſe
double,0308=,0616 gr. and therefore,0616 —,0308 =,0308 gr.
that muſt be put to the braſs to keep the equilib. the anſwer.
Quefiion 174. If the folidity of the cavity of a pair of fmith's bel-
lows filled with air be 495 inches, and the bellows be preffed together
in a ſecond of time, required the velocity of the air thro' the nöfe of
0,6 inches area?
Here 495 folid inches, the quantity of air voided in a fecond, being
divided by 0,6 inches the area of the orifice thro' which it paffes gives
B25 inches = 68,75 feet for the length of a column of air that would
be generated in one fecond of time, and is therefore the required ve
locity of the air per fecond thro' the nofe of the bellows pipe.
Queſtion 175. If the capacity of the receiver of an air pump be to
that of the barrel as 8 to 1, and the denſity of the air in the receiver
1, what will be the denfity of the air there after 3 ftrokes, or turne
of the pilton?
AND MECHANIC.
187
Here, d = 1
=1,n=1, m = 8 and s = 3, ſo (by theo. 198)
198), of
729
as 729 :: the density before the firſt ſtroke to that after the third
ftroke, or becauſe the rarity is inverfely as the denfity it will be as I g
729 the rarity before the firſt ſtroke to that after the third ſtroke.
=
Question 176. The denfity being I as in the laft queftion, and
the capacities of the barrel and receiver equal, fuppofe each = 1;
how many turns must be made to rarify the air in the receiver 100
times?
From theo. 198, we have 2
log. 2 log. 100, whence s =
= 100, fo by the logarithms s. X
log. 100
log. 2
6,6 the turns required.
Question 177. If any quantity of air is reduc'd into a fourth part of
it's firſt bulk, what more force is required to keep it there? See art.
351.
4
If the fame body of air is confin'd in of the ſpace, it muſt require
times more force, for it there acts with 4 times more force.
Note, Gunpowder fired is nothing but an elaſtic fluid, and froni
hence it appears that the greater quantity of this fluid is contained in
the fame ſpace, the more violent will the exploſion be.
To
Question 178, It is proved by experiments that mercury in the ba
rometer fettles of an inch, when the barometer is removed 85 feet
directly upwards, now if the mercury at that time on the earth's fur
face, ſtand in the barometer at the height of 30 inches, and the den-
fity of the air be every where the fame, what muſt be the height of
this body of air called the atmoſphere ?
If a tube be filled with mercury and hung perpendicular to the hori
zon, the upper end being clofe ftopped and the lower end open, the
mercury in that tube will ftand at the height of 30 inches, the cauſe
of which can be nothing but the preffure of the air acting on the bot
tom of the tube, for if ever fo little air get in at top, the fluid falls to
the ground, whence a column of mercury 30 inches height is in equis
librium with a column of air, of the fame bafe, or, one of mercury
of an inch height in equilib. with one of air of the fame baſe and 85
feet height, fo it muſt be, as 0,1 inch 85 feet: 30 inches : 25500
feet
5 miles nearly for the anſwer.
=
2 Hence 25500 X 12 = 306000 inches, the height of a column of
air in equilib, with one of mercury of the fame bafe and 30 inches in-
ches height, therefore, as 30: 306000, 1: 1 10200:: the density
of air, to that of mercury.
188 THE UNIVERSAL MEASURER
3. The particles of air near the earth's furface muſt bear the preffure
of thofe above them, whence, the air cannot be every where of the
fame denfity, but muſt be lighter the higher we go, and fo its height
will be indefinite, but if we fix the boundary of the atmoſphere, at
that height where it has power to reflect a ray of light, which is the ut-
moſt limit of twilight, then this height by trigonometry will be found
44 miles.
1
2
4. If the baſe of the aforefaid column of mercury be one foot, then
1 foot x 30 inches (= 2,5 feet) it's height gives 2,5 feet folidity whofe
weight is about 2187,5 lb, which muſt alſo be the weight of a co-
lumn of air of the fame bafe, whence it appears that every foot area
on any furface, as man's body &c. is preffed with 2187,5 lb weight,
which vaft preffure he could not fuftain were it not that the air acted
equally in all directions and fuffers him to feel this weight no where.
5. But this preffure acting on the ſurface of any fluid, caufes it to
rife into any pipe or tube where the air is by any means taken out of
fuch a tube, and becauſe mercury is the heaviest of all fluids, it riſes
to the least height of any, hence we ſee the reaſon of fucking, and that
it is but taking away the air out of the pipe &c. and the fluid follows,
to the height of 30 inches if it be mercury, but if water, to the height
of 35 feet, (the denſity of mercury being to that of water as 14 to 1)
and no higher.
6. The mercury is by obſervations feen at all heights between 28
and 31 inches, which fhews the ſtate of the air to be variable this 3
inches difference is called the ſcale of variation, being that graduated
part at the top of the barometer between the greateſt and leaft heights
of the mercury; there has been feveral inventions to enlarge this
ſcale, fome by bending the top of the barometer, becauſe fluids rifing
to the fame height muſt run further in an oblique direction than in a
Lone, others have had the tube conical becauſe as the bafe decreaſes
the length muſt increaſe to contain the fame ſpace, but on the account
of friction &c, the upright barometer is ſtill the beſt: you may make
a weather glafs with any fluid; thus, take a tube, or long vial bottle,
and let it be ſomewhat more than half fill'd with the fluid, then ſtop
the mouth of it wirh your finger &c. and turn it down into a veſſel of
the fame fluid, then take away your finger &c. and ſet the veſſel and
tube in it, againſt a wall or in a frame for that purpofe (out of the fun-
fhine for heat as well as preffure affects fluids) then as the preffure of
the air increaſes or decreafes, upon the furface of the fluid in the veffel
it will cauſe that in the tube to rife or fettle, for the preffure of air
above the fluid in the tube cannot change becauſe the top of the tube
AND MECHANIC.
189
is cloſe ftopped, it is beft to make theſe kind of barometers when the
air is in a mean ftate viz. when the mercury in the common barome-
ter ſtands at the height 29,5 inches, this being the mean between the
two extreams 28 and 31.
7. Any fluid may be exhauſted by heat, ſo if a tube is heated by
the fire till the air is gone out of it, and then its open end (the top
end hermetically fealed up) be immerfed into a veffel of mercury, the
mercury will rife in the tube to the common height, and if the tube
be not above 28 &c. inches long the fluid will run over at top, if it be
open there. See art 349.
Queſtion 179. If mercury in the barometer fland at the height of
30 inches, how high would it rife in the fame tube if there were 20
inches height of water in it. See art. 351.
Its plain that the water and mercury together muſt be in equilib,
with a column of mercury (b) 30 inches height, therefore the denfity
of mercury to that of water being as (m to n) 14 to 1, it will be e
na
b- = 30-32 = 28 4, for the height of mercury
mb — na
m
m
4
न
284,
in the tube fo 20 + 28 = 48 4 inches, the height of the mercury
and water together anſwer.
Question 180. When mercury in the common barometer ſtands at
the height of 30 inches, what muſt be the ratio of the diameters of a
compound barometer of mercury and water that each of theſe fluids
may have 20 inches height in the compound tube. See art. 351, and
the laſt queſtion?
dd =
:b-e:xm
: 30
=
na
20: X 14
20
= 140=7, fod=
20
7, that is, as I: 7:: the diameter of the mercury to that of
the water.
Question 181. If a tube 35 (h) inches long 30 (b) inches of it filled
with mercury, and then the orifice ſtopp'd with the finger and turn'd
down into a veffel of the fame fluid, the finger then taken away, it is
plain there will be 5 (c) inches of air in the tube, quere (e) the height
of the mercury in this tube, that in the common one at this time being
(b, 30 inches? See art. 352.
h - b
b+ h
Here e =+√: bc+
:+
45 if be taken,
2
2
but 20 if— be taken before the furd fign, but 45 is more than 35
the whole length of the tube, ſo 20 inches is the answer.
4
190 THE UNIVERSAL MEASURER
Again, fuppofe the tube to be but 23 inches long, all elſe the fame
then =+√ : cb +
: +
30+23
2
2
h
b+h
:+
2
2
+√:150-130—23
2
=39,24 inches for the greater root, and 13,76 inches
for the leffer root, which is the anſwer.
Queſtion 182. How much air muſt be in a tube 10 inches long, that
the mean height of the mercury in it may be 8 inches? See art. 352.
Here is given b=29, the mean height in the common barometer
= 10 the propoſed tube's length and e8 the mean height in it,
to find c, ſo it will be, c=h_c_hetee 10-8 —
80+64
=
b
h
29,5
16
=2
=I
13,5 inches of air, anſwer.
29,5
29,5
Question 183. What muſt be the diameters of the two tubes (fig.
166) that the variation in this compound barometers leffer tube may
be to that in the common barometer as 10 to 1? See art. 353.
Here it is, as ev:: ddm: 2m dd. 1, or fo is 14 d d : 28
←dd 1, but by the queſtion, as e: v :: 10: I, therefore, as 10
I : 14 d d : 28 — d d — 1, whence d= 3,3541 viz. the diameters
muſt be as 3,3541 to 1, (the denſity of mercury being to that of wa-
ter as 14 to 1) in like manner, you may find a ſcale of variation as
you pleaſe, (and uſe any other fluid with mercury as well as water)
for if 2 m- d d I be taken = o, or d d 2 m I, then d
√: 2 ni 1: = √ 27 = 5,2, that is, if the diameter of AIF be to
that of CKF as 5,2 to unity, the ſcale of variation here will be in-
finite in refpect of that in the common barometer, but for all this, the
common barometer is better, for this compound one is difficult, both
to make and keep in order.
=
Question 184. If F B (fig. 165) be a barometer whofe ſcale of va-
riation is DG, how much muſt this ſcale be bent out that it may be
3 times as long?
Let G L be the tube F B, and with three times D G in your
compaffes and one foot in D, interfect G L in L, fo is G L the requir
ed fcale, for its plain while the fluid would rife perpendiculary from
D to G, it would fill the ſpace G. F. See queſtion 178.
G.F.
Question 185. If mercury in the common barometer at the bottom
of a hill ſtand at the height of 30 inches, and at the top thereof, at the
height of 29 inches, what is the height of that lill? See art. 349
AND MECHANIC.
191
The height of mercury in the barometer being as the density of the
air and this denſity decreafing in a geometrical ratio as the altitudes
increaſe in arith, progreffion, its evident, the latter will be inverſely as
the logarithm of the former, that is (fee queft. 178) as log. of
I
29,4
I
29,9
: 86,45 feet, that is, if the barometer be
10
: 85 feet :: log, of
carried fo high till the mercury fall an inch, it muſt be raiſed 86,43
feet higher to make it ſettle of an inch more, and if in this manner
the heights be found for every of an inch fall of the mercury its
plain the fum of all theſe heights will be the height required. Thus
L: 85,58, and
As the log. of: 85,29, and :: log. of
29,9
I
I
I
29,7
fo is log. of : 85,86 and :: log. of : 86,16, then 85+
29,6
85,29 +85,58 +85,86 +86,16
29,5
427,89 feet the anſwer.
Again, fuppofe that by carrying the barometer to the bottom of a
pit &c. the mercury in it rife 5 tenths of an inch viz. there ftand at
30,5, then it will be as log, of :85: the logs. of L,
I
I
I
30,2 30,3 30,4
I
29,9
I
30
I
30, I
I
and to 84,72, 84,44, 84, 16, 83,89, and
30,5
83,61 relpectively, the height which caufes the mercury to rife
of
an inch as you are a going down the pit; then, 85 +84,72 +84,44
+84,16 +83,89 + 83,61420,82 feet the pit's depth. In this
manner you may conſtruct a tube to fhew the heights either afcended
or defcended, or both, and fo by moving with a barometer either up-
wards or downwards, you'll fee the diftance to every tenth of an inch
&c. fall or rife of the mercury, but if the height is but fmall, this way
may ſerve viz. as 0,1 inch : 85 feet: : 0,5 inch : 425 feet anſwer,
which 2,89 feet in the firft cafe and (425-420,82) 4,18 feet in the
fecond cafe from truth.
Queſtion 186. If A B (fig. 214) be a tube open at the end B, with
a ring of lead &c. laid about that end to make it fink that end fore-
molt in deep water, when fufpended by the other end A, which is
cloſe ſtopp'd, how far in the tube will water rife, when it is 300 feet
under water, the faid tube being 10 inches long?
By art. 351. As AD: AB:: the force of the air in the whole
tube A B, to its force when preffed by the water &c. into the ſpace
A D, now (by queſtion. 178) the preffure of the air in a mean ſtate
192
THE UNIVERSAL MEASURER
40
9 to I,
fuftains a column of water (14 X 29.5 inches height of mercury) 413
inches high, but mercury is not quite 14 times heavier than river wa-
ter, ſo we'll take this height =400 inches, fo if the tube be funk 400
inches it will there be half full, becauſe then the tube full of air, will
by the water at that depth, be prefſſed into half its firſt ſpace, and for
any other depth it will be (by the above proportion) 3600 inches =
300 feet is to 400 inches the ſtandard height as (608
that is, the tube at 300 feet under water will be 9 parts full and I part
empty, be its capacity more or lefs, whence if the tube were much wider
at bottom B, than at top A, the water would not riſe ſo high in it,
becauſe the greater the diameter, the leſs the height to contain the
fame ſpace, and thus we have the nature of the diving bell, with which
you may be let down into the water of a good depth &c. hence, alſo
is the nature of an inftrument called Sea-gauge, which being funk to
the bottom of deep water, there ſtrikes, and loffes a weight which
carried it down, and then rifes to the top again and fhews how far the
air in it has been compreffed, by which the depth is known as above,
See queítion 190.
Note. No other fluid like air will endure compreffion, for by ex-
periments they either preſs thro' the pores of the veſſel or break it.
The thermometer, a tube commonly fet with the barometer, to
fhew the degrees of heat and cold, is made of fpirits of wine ting'd
with cochineal.
Question 187. If the fucking pipe of a water pump be 28 feet long,
(b) and the pifton can be moved 8 feet every ſtroke (a) how many
ftrokes must be made before the water come to the top of the pipe?
See art. 354.
Suppofing a column of water 31 feet high (c) at that time to be in
equilib. with the atmoſphere, and the diameter of the barrel that of
of the pipe.
Here, 2 z = a+b+c=8+ 28 +31
Z --
67, fo z=33,5 whence
√:22- ac: 4 feet nearly, for the height to which the
water riſes the firſt ſtroke, now this 4 feet taken from the pipe's height
28 feet leaves 24 for a fecond (b) the rest of the letters being as be-
fore, we'll have e≈ z
ac: 4,2 feet, the height the
water rifes the fecond ſtroke, and this taken from the laſt b 24 leaves
19,8 for a third b, and ſo on as before, you'll find 4,6, 5,1,8,7 feet,
for the heights of water raiſed by the 3d, 4th, 5th, ftrokes, whoſe
total is 28 feet, all but 1,4 feet, hence the water at the 6th ftroke
has only 1,4 feet to rife in the pipe, after which it must run into the
barrel.
AND MECHANIC.
193
Question 188. If a pipe be (b) 20 feet long, and a column of water
35 feet (c) in equilib, with the atmofphere, what length (a) muſt a
ftroke be made with the piſton to raiſe the water in the pipe (e) 10 feet?
See laſt queſtion.
Here a
be+ce
c-e
e e
be
c_e
+e=2000° +10=18 feet anf
Note. In theſe queſtions the pifton is fuppofed to defcend every
ftroke, cloſe to the head of the pipe and then afcend to the higheſt
elevation making every ſtroke equal, for if it do not defcend from the
higheſt to the loweſt elevation, there may no water rife in the pump,
as is thus proved from the laſt queſtion, its evident when the pilton is
put cloſe down to the pipe, and then raiſed 8 feet above it, the air in
the 28 feet of pipe would be expanded into (28 +8) 36 feet, did not
the water rife (fee queſtion 187) 4 feet, whence the faid 28 feet of
air is in this cafe expanded into (36 — 4) 32 feet; then as the pifton
defcends, the water by its tendency downwards fhuts the valve v (fig.
167) fo that neither air nor water can pafs into the well, now its plain
when the pifton has defcended ſo low as to drive this 32 feet of ex-
panded air into its firft fpace of 28 feet, it will then be in equilib. with
the external air, fo if the pifton fall any lower, the air below it will
become denfer than that above it fo forces open the valve D to make
its eſcape, therefore 32 — 28 4 feet, the ſpace that the piston muſt
fall from the higheſt elevation to thus comprefs the internal air, or
make it of equal denfity with the external air, but if no air efcape
thro' the valve D, no water can rife in the pump, whence, if the firſt.
ftroke be 8 feet the fecond muſt be more (in this cafe) than 4 feet, or
no water can rife, and fo on for
any other.
Question 189. If the diameter of a pump's barrel be 6 inches, the
diameter of the pipe 2,1 inches, its height above the water 16 feet,
the height of a column of water in equilib. with the atmoſphere 31 feet
each ſtroke of the piſton 2 feet, and the defcent of gravity 16,1 feet,
what muſt be the velocity of the piſton per fecond that the water may
follow it, or the pump work to the beft advantage? See theo. 199.
Here, D6, d=2,1, b=16+ 2 =18, c = 31, and s = 16,1
2 d d
whence, v=
8,82
X:√:SC√ sb:=
X: √ 499,1
D D
36
8,82
36
№ 289,8 := X: 22,34 - 17,02 :=
3,82
36
-X5,321,3034
feet per fecond, or (1,3034 X 60) 78,204 feet per minute, fo78,204
Bb
194 THE UNIVERSAL MEASURER
4 (becauſe in one ftroke are 4 feet, viz. 2 up and 2 down) nearly
20 ftrokes in a minute, anfwer.
I
3' 4' 59
34 feet
Question 190. If by any means water can be made to aſcend in a
tube filled with air, and none of the air go out, its manifeit that the
air in the tube will by the rifing water be driven into a lefs fpace, and
fo have a greater fpring or elaftic force; as for example, if the air be
comprefs'd into half the first ſpace, or the veffel fill'd with water, then
it's fpring will be double, or have the force of twice the atmofphere.
Now becauſe one atmoſphere can raife water to the height of
in vacuo, it follows that if there be a fpout made a little below the
furface of this rais'd water in the tube, that it will fpout to the height
34 feet directly upwards provided the tube be conſtantly kept full of
water to the fame place, for the force of the air in the tube being
double to that of the external air, one half thereof will be ſpent in
overcoming the force of the faid external preffure, fo the other half
muſt raiſe water to the fame height that the preffure doth in vacuo, in
Jike manner if the air be compreffed into 1, &c. of its firſt ſpace
the water will fpout to the heights of 68, 102, 136 &c. feet, from
whence we have the conftruction of the engine for quenching fire in
cities &c. and may be done by a common pump, by faſtening leather &c.
about the piſton at the head of the pump ſo that no air can paſs, and
then having made a few ftrokes to raiſe the water above the fpout, lec
the ſpout be opened and the water will flow in a continual ftream in
any direction fo long as the workman pleaſes to keep the water at the
fame height. But if to the barrel of the pump (fig. 167) you fix an
air veffel or hollow cylinder d q, all clofe ſtopped except at the bot-
tom d, there be a hole to communicate with the pump. Now its plain
when the water in the pump rifes to d, it will run into the tube d q
and drive the air into a fpace T q, whence, the water in the pump will
be prefs'd with both this air T q, and that above it at A, in the cif.
tern or barrel, and then turning a top-cock at A, it will run out in a
continual ſtream, the aforefaid engine confifts of one large air veffel, and
two pumps.
Quest.191. The engines for raifing water by fire, called fire-engines.
are thus, A B C (fig. 215) is a copper veffel partly filled with water to
DE, which being fet over a fire and made to boil will fill the upper-
part D BE, with a vaftly elaſtic vapour, whofe ftrength when fit to
work the engine, is known by its forcing open a valve at e, charged
with a fufficient weight, this elaftic fteam is then let into the barrel
abcd, by turning a cock at F, where by its elaftic force it raiſes the
pifton G, which drives the air before it thro' a proper cock at top,
AND MECHANIC.
195
after this, that the piſton may by its weight defcend, a little cold water
from a fountain fgn is let in at the bottom by turning a cock at k,
which condenfes the hot fteam in the barrel into 3000 times lefs fpace
than it was in before the jet of cold water, this makes a fufficient va-
cuum for the piſton to defcend in, the piſton G, and lever H I being
thus put
in motion, do accordingly raife and deprefs the pifton K in
the barrel of the foregoing pump &c. L M on the other fide, which
by the pipe MW, draws the water from the depth W.
There are now feveral forms of fire engines with late improvements,
but they all work on thefe principles, though their forms are much
more compleat.
X
By an experiment made on a fire engine, that made 16 ſtrokes in a
minute, content of the fteam barrel 113 gallons, the boiler requir
ed to be fupplied with water at the rate of 5 pints per minute, fo 113
× 16 ×8 = 14464 pints of fleam per minute, therefore, as 5: 14464
:: I: 2893: the rarity of water to that of team, that is, I of wa→
ter will be 2893 of ſteam whofe force is equal the preffure of the at
mofphere, but this number is fettled at 3000, as aforefaid, i. e. the
3000 parts of fteam is reduced into 1 part of water by the jet of cold
water; hence, we may compute the force of fteam, as is done of air
in the laſt queſtion; thus, when 3000 is confined in 1, 1, 1. &c. of
3000 it will be as ſtrong as 2, 3, 4, &c. atmoſphere's, from thefe quef-
tions alfo, we have the rarities of water, air and fteam, as 1,860 and
3000.
Note. The forementioned cocks F and k, are made to open and
fhut as need requires, by the working of the engine.
Question 192. As there are 3 forts of water pumps in ufe, the firſt
(fig. 167) called the fucking pump, is fufficiently defcrib'd in the theory.
&c, it cannot be amifs here to explain the other two forts?
Firſt, a forcing pump of the common fort, is thus conftructed (fig.
216) A B is the barrel ſtanding in the water at W, G C is the piſton,
and C a folid piece without any hole or valve, becauſe no water is to
pafs thro' it, this piece fhould be leathered very fit ſo that in working
neither air nor water may pafs between it and the barrel at a diſtance
as D, below is fixed a valve in the barrel, between this and the low-
eſt deſcent of the pifton, as at H, there goes off a pipe, in which as at
E is fixed a valve and diaphragma like that at D, now the piſton be-
ing drawn up from C towards A, rarifies the air above D, and ſo the
water rifes into the ſpace C D, then when the pifton is forced down,
the water finding no paffage thro' C or D, is forced into the pipe at
196 THE UNIVERSAL MEASURER
H, and thro' its valve E into the cistern at F, (which may be fet at
any diftance above the pump) and there run off by a ſpout.
Secondly, a lifting pump, is a forcing pump of another form, (fig.
217) F B is a barrel fixt in a frame ILM K which is alfo fixed im-
moveable with the lower end in the water, EQHO is a frame with
two ſtrong iron rods moveable thro' holes 1, K, L, M at the upper
and lower parts of the pump in QH, to the bottom of this frame is fixed
an inverted pilton B D, with its bucket and valve at top D, upon F
the top of the barrel there goes off a pipe F H, either fixed to the bar-
rel, or moveable by a ball and focket, in either cafe it muſt be fitted
tight that no air get into the barrel, in this pipe as at C is fixed a va-
lve. Now when the pitton frame is thruft down into the water, the
piſton D defcends, and the water below will rush up thro' the valve
D, and get above the piſton, where upon the frames being lifted up
the pifton forces the water thro' the valve C up into the cistern H,
there to run out.
Note. This fort of a pump is fet fo far into the water as that the
piſton may play below the furface of it.
The 37 following que lions contain the full practice of Gunnery.
Question 193. If the elevation of a gun, (viz. the angle which its
axis makes with the horizon) be 30° and the amplitude 2000 feet,
to what height will the ball rife in the air, alſo, what muſt be the ele-
vation when the amplitude is the greateſt poffible? See article 364,
and 365.
As fine 60° (co-eleration) is to fine 300 (the elevation) fo is 500
(the fourth of 2000) to 287,5 feet, the height of the ball's path, and
the required elevation is 45°. To this elevation guns are fet when
fired on trial, and the amplitude meaſured in feet, and marked on the
gun, which is a ſtandard to compute other ranges by, and becauſe theſe
ranges vary with the product of the fine and co-fine of the elevations,
it follows that any two elevations equally diftant from 45°, will have
the fame amplitude; that is 30°, and 60°, alſo 40°, and 50°, &c.
will give the fame amplitude.
Note. In all theſe queſtions about gunnery the force or charge of
powder is fuppofed to be the fame, except it be faid to the contrary.
Question 194. When the impetus is 4000, what muſt be the two
elevations, CA and D A (fig. 218) to ſtrike an object B, at 6969 feet
diſtance on the plane of the horizon? See art. 369.
As twice the impetus 8000, or greateſt amplitude is to radius (fine
90°) fo is and other amplitude 6969 to 60° 32′ the double of its
AND MECHANIC.
197
elevation, fo half of 60° 32′ = 30° 16′ = L DA B, of the path
A F B, the leffer elevation, whofe comp. is 59° 44′ = LC A B or the
path AEB, to the greater elevation.
Question 195. If the elevation of a piece be 30º 16', what muſt be
the impetus, to hit an object on the plane of the horizon at 6969 feet
diſtance (by the lalt queftion) ass 60° 32′ (twice 30° 16′:s 90° ::
6969: 8000 half whereot is 4000 feet the required impetus.
Question 196. If a ball continues 12 feconds in the air when pro-
jected under an angle of 30°, what will be the time of its flight when
the elevation is 60° ?
(By art. 373) As s 30° any elevation is to 12 feconds the time of
the balls flight fo is s 60° any other elevation to 20,8 fecond the time
of flight in that elevation.
Question 197. If a ball continue 12 feconds in the air when the ele
vation is 450, what will the impetus be?
See art 373.
Here 144 the fquare of 12, multiply'd by 16 the defcent of gravity,
gives 2304 feet for the greateſt amplitude, half of which is 1152 feet,
anfwer.
Question 198. What is the time of flight, when the elevation is
320, and amplitude 5280 feet? See art. 373-
As radius is to tangent 320, the elevation, fo is 5280 feet the am-
plitude to 3299, whoſe ſquare root is 57,44,4 times the required
time in feconds, fo 4 of 57,44 = 14.36 feconds anfwer. Theſe two
queſtions are of great ufe at fea, and in adjufting the fufee &c.
Question 199. If the elevation of a piece be 45°, and the impetus
1800 feet, and it ſtrike object whoſe horizontal diſtance is 3000 feet,
what is the time of the flight? See art. 368.
1. As (57600) 16 times 3600 the double impetus, or greateſt am-
plitude, is to 1 the natural tangent of the given elevation 45° fo is
9000000 the fquare 3000 the horizontal diſtance to 156,25, whoſe
fquare root is 12,5 feconds anfwer.
Question 200. With what velocity doth the ball leave the gun, when
the impetus is 6400? See art. 374.
1. The fquare root of 32 times the greateſt amplitude 12800 is
640 feet anfwer, that is, if the ball move uniformly, it will paſs over
640 feet in a ſecond of time.
Question 201. If Ey (fig. 218) the height of the projection be 287
feet the amplitude A B 2000 feet, with what force will the ball from
A ſtrike an object m, whofe horizontal diſtance is A H 800 feet and
198 THE UNIVERSAL MEASURER
altitude m H (above the horizon) 276 feet, the balls weight being 20
pounds?
1. Draw m n parallel to the horizon AB, then its evident the ball
would defcribe the path m En, with that velocity that it ftrikes m with,
therefore the velocity fit to defcribe m En will be the anſwer, fo, as
ma (Hy) 200 is to radius fo is twice Fa 23 (2 Ey - 2 m H) to
tangent 6° 30′ the elevation. Then (by art. 369) as fine 130 (twice
this elevation) is to its amplitude mn 400 to is radius to 1780, the
1780 x 32238,6
20lb 4772 lb the force
× =
greateſt amplitude, fo (by the laft queft.)
feet, the required velocity, then 238.6
with which the ball ſtrikes the cbject, and leaves the gun with the fame
force. See queftion 138.
Question 202. If the elevation (angle BAC fig. 218) be 300 and
the amplitude. A B 4000 feet, where muft the piece be placed to hit
an objecì m, whofe perpendicular height Hm above A B the level of
the piece 400 feet? See art. 398.
1. As tangent 30° elevation, is to 400, objects height, fo is 40co,
amplitude to 2807017 this taken from 4000000, half (4000) the
amplitude fquared, leaves 1192983, whofe fquare root 1092 added
to half the amplitude 2000 gives 3092 feet — A v. If the ball were
to ſtrike the object n, after it has performed half its flight, but if it be
to hit an object m, before half its path is pafs'd over, then 3092 taken
from A B 4000, leaves A H≈ 908 feet for the anſwer.
Question 203. Let things be as in the laſt queſtion, but fuppofe the
object to be m 400 feet below A B the level of the piece? Fig. 168.
1. Having got 2807017, as in the laſt queſt. add it to 400000 the
fquare of 2000 (half A B), and the ſquare root of that fum added to
2000 the faid half amplitude gives 4609 A L, that m is below L
being the object.
Question 204. If the impetus be 4000 feet, what muſt be the ele-
vation C AB (fig. 218) to hit an object n, whofe horizontal diftance
A v is 5600 feet and height vn 812 feet?
(By art. 399) From 8000 twice the impetus take 1624 twice 812
the objects height, multiply 6376 the remainder by 8000 twice the
impetus and from 51008000 that product take 31360000 the fquare
of Av 5600, and there leaves 19642000 whofe fquare root 4432, add-
ed to, and taken from 8000 twice the impetus, gives 12432 and 3568,
then as 5600 ≈ A v, is to rad. fo is 12432 to tangent 65° 45', the
greater elevation and fo is 3568 to tangent 320 30' the leffer elevation
required.
AND MECHANIC.
199
Question 205. Let things ftand as in the laft queftion, but let m
(fig. 168) be 812 feet, viz. the object m to be 812 feet below the
level of the piece, and A L its horizontal diftance 5600 feet &c. as
before?
1. To 1624 twice Lm 812 add 8000 twice the impetus, multiply
9624 that fum by 8000 twice the impetus, and from 76992000 the
product take 31860000 the fquare of AL 5600, then 6755 the fq.
root of 45632000 the remainder added to and taken from 8000 the
greatest amplitude gives 14755 and 1245; laftly, as 5500 radius
:: 14755 : tangent 69° 12′ the greater, and fo is 1245 to tangent
12° 37 the leffer elevations required.
Question 206. If A L(fig. 168) the horizontal diſtance of an ob-
ject m, be 5600 feet, and its angle of depreffion LA m 80 15' and the
elevation (angle LAC) of the piece 12° 37', what must be the
impetus fo as to hit the object? See art. 399.
As 1,476 four times the fum of the natural tang. 0,224 and 0,145,
of 120 37½ L LAC and 5° 15′ L L Am, (but if the object were ele-
vated you muſt take 4 times the difference of theſe natural tangents)
is to 1.0.49, the fquare of 0,224 added to unity fo is 5600 to 4000
feet anfwer.
Question 207. If a gun be planted on the top of a tower, what
muſt be its elevation with the leaft impetus poffible, to ſtrike an ob-
ject whofe horizontal diftance is 5600 feet, and angle of depreffion
8° 15'? See art. 391. and 399.
1. From 90° take the object's depreffion 8° 15', half of the remain-
der 81° 45′ is 40° 52′ the answer, but if the object were elevated
8° 15', then 90° 8° 15′ = 98° 15', half of which is 490 7
would be the anfwer.
Question 208. Things being as in the laſt queſtion, what is the
impetus ?
1. As radius is to tangent, elevation 40° 52′ fo is 5600 to 4850,
half whereof is 2425 the required lealt impetus, art. 399.
Question 209. If the elevation (B A C fig 163 be 320 30' and its
impetus 4000 feet, and the inclination of a plane (angle B A m)
8° 15', required A m, the amplitude of this projection? See art. 400.
As the fquare of the fecant of 32° 30' the elevation, is to the fe-
cant of 8° 15′, the inclination, fo is 4 times the impetus 16000 to a
fourth number, which multiplied by the fum and difference of the tan-
gents of 320 30 and 80 15' the elevation and inclination, gives 8992
feet Am, if the plane is depreffed, but 5658 feet Am when mis
above the horizon.
200 THE UNIVERSAL MEASURER
Question 210. If the inclination of a plane be 80 15', what muſt be
the elevation of a piece with a given impetus, to throw the ball the
fartheft poffible on the ſaid plane?
This is the fame with queſtion 207, and therefore half the fum and
difference of 900 and 8° 15′, is 49° 7 ′, for the elevation of the piece
when the plane is elevated, and 40° 52′ for the faid elevation when
the plane is depreſſed.
Queſtion 211. If by 16 lb of gun-powder the impetus be 4000 feet
what powder muſt be taken that the impetus may be 5000 feet?
As 4000 feet is to 16 lb fo is 5000 feet to 20 lb anſwer. But in
great charges of powder there is a confiderable part of the powder
blown out unfired, ſo that the velocity of the ball in ſuch caſes cannot
be exactly in the fubduplicate ratio of the quantity of powder, there-
fore this method ferves only as a guefs. See queſtion 177.
NS
Question 212. Given the diſtances CS8900 N S 80000 each
in feet, (fig. 219) and the angle NSC 30° on the plane of the ho-
rizon, at s a gun is planted, whofe elevation is 45° and impetus 40000
fo that it might throw its ball to N, at C another gun is planted whofe
impetus is 900co feet, what muſt be the elevation of this gun C, fo as
both guns be fired at the fame inftant of time, the balls may hit each
other in the air, and the ball from s, be put out of its path by that
from C?
Becauſe the two guns are fired at the fame time, and the balls are
to meet, therefore the lines of flight must be equal, fo by art. 382 the
heights F G and D E of the two projections will be equal; whence,
(by art 378) as 300 the ſquare root of 90000 C's impetus is to 200
the fquare root 40000 S's impetus fo is fine 45° S's elevation to fine
28° 12′ C's elevation required.
Queſtion 213. If s repreſent the ſouth point of the horizon and N
the north point thereof (laſt queſtion) to what point of the compafs
muſt the gun C be directed?
Becauſe D E the height of the path of the gun's, is (by art. 365)
Ns the greateſt amplitude = 20000, which is alfo (per laft queſt.)
FG the height of C's path, therefore it will be as tangent 28° 12′
C's elevation is to radius fo is 20000 the height of its path, to 37500
which doubled gives 75000 = C Q = CG or GQ, for C's half
amplitude. Now as the paths interfect each other in A, where the
balls meet, fo the amplitudes crofs each other in B, the horizontal
point directly under A, and becauſe the times of flight are equal, we
have (per art. 383) as S B: CBSN: CQ, fo by plane trigono-
metry, as C B, or as CQ 150000 : fine L NSC 30º :: S B or S N
1
AND MECHANIC.
201
80000 : fine LB CS, 15° 30′ which added to L NSC 30, gives L
QBS=NBC= 45° 30′ or fouth 45° 30' eafterly for the direction
of the gun C.
Queſtion 214. Where will the two balls mentioned in queſtion 212
fall, fuppofing them to be of the fame metal, non-elaftic, the weight
of the ball from C to that of the ball from s as 5 to 3 (C to B) and
that when the ball B croffes the path of the ball C, Cftrikes it with its
whole force, the centers of the two balls being then in one right line
parallel to the horizon? Fig. 220.
Becauſe when the balls meet in the air their centers are equally dif-
tant from the horizon, it follows that what alteration is made by the
ſtroke will only be in their velocities and directions. Whence, in this
cafe the balls may be looked upon as moving in their horizontal ranges
SN and CQ, and becauſe the times of flight between meeting in B
and falling at N and Q, are equal, therefore, B N and BQ will be
as their velocities, and becauſe theſe velocities fhew N and Q, the
point of fall if the balls did not interrupt each other by their meeting,
fo the velocities after the ftroke will point out the required places of
fall Now becauſe C ftrikes B perpendicularly, therefore, B Qx C
C's momentum againſt B, but B ſtrikes C obliquely in the direction
SN, fo from N upon QC, let fall the LN A, then (by theo. 157)
BA will exprefs B's velocity againft C, fo BA x B B's moment
againſt C, then by trigonometry, as fine LCBS 134° 30′ is to CS
8900 fo is fine L BCS 30° to B C 6275, and fo is fine L BCS 150
30' to BS 3450, then NS 8000
NB 4550, and C Q
150000 CB 6275 = BQ 143725, again, as radius: NB 4550
:: fine L A NB (co-fine L N BC 45° 30′) 44° 30′ : A B 3200, then
becauſe the balls tend to meet, we'll have (by theorem 160)
PQ XC - BNX B 143725 × 5 — 3200 X 388628, the com-
5 + 3
-
C + B
BS 3450
mon velocity after the ftroke, which is the diſtance moved by C, in
the direction BQ ſuppoſe to R, viz. BR = 88628 feet) before it fall
which would alfo be the place of B's fall, were it not that B is affected
by the aforefaid oblique motion, draw Nu parallel to CQ making
NPA B, and Pu= BR, now B N, B's force being (by theo. 157)
refolved into the two forces BAPN and B PAN, whereof A N
is not altered by the force B R of C, but the other force NP is there·
by made Nu, whence B after the ftroke will move in the line Bu
and come to u in the fame time that C comes to R, (fee art. 259)
therefore, the ball from C will fall at R, which is fouth 45° 30′ eaſt-
C c
**
202 THE UNIVERSAL MEASURER
k
1
erly and 95903 (CB+BR) feet diftant from the gun E, and the
ball from s will at the fame time fall at u which is alfo fouth 45° 30°
caſterly and 91828 feet (N P+P u) diſtant from N.
Queſtion 215. Suppofe the weight of the ball B (laſt queſtion) to
be 500, and that of the ball C 1, where will the balls fall? By reafon-
ing as before we'll have 1437251-3200X500_143725—1600c0a
-1456275
I +500
501
2906,5 feet, for the common velocity or diſtance
501
moved after the ftroke, which fhews that C is reflected back, and
would be followed by B, were it not for B's oblique ftroke, whence
the directions will be the fame as before, viz. fouth 45° 30′ eaſterly,
and 6275 (CB) — 2906,5 Cd3369,5 feet, d being the place where
the ball from C falls, as alfo P N 3200 — 2906,5 — Ne 293,5 feet,
e the place of fall of the ball from the gun s.
Queſtion 216. Let the weights and velocities of the balls B and C,
be as in the laſt queſtion, but fuppoſe they ſtrike each obliquely viz.
when the line y z parallel to the horizon joins their centers which line
y z, with P E the line or plane which the balls touch when they meet
on which from N and Q_let fall the perpendiculars NP and Q_D, which
Ls are as the velocities with which (fee art. 259) the balls directly
meet each other, then by trigonometry, as radius: NB 4550 :: fine
LNBP 45° N P 3220, and as radius : BQ 143725:: fine L DBQ
DQXC-NP X B
89° 30′ QD 1437 10 then (by theo. 160)
C X B
142710 YI:- 3220 X 500
I + 500
=> 2926 or =+ 2926 (if B be pofi-
tive and C taken negative) for the direct velocity of both balls after the
ftroke, and becauſe the balls move from B towards Q and N, and the
velocities B D and BP being parallel are not altered by the ſtroke,
thefe velocities muſt be on the fide P of y z, but if the balls had mov-
ed from Q and N towards B the velocities BD and BP must have
been fet on the other fide of y z upon B E, fo produce QD till RD=
2926, and make Pu 2926, join BR and B u, fo after the ſtroke
the ball B moves in B u and falls at u 299 feet (P N - Pu) fouth
eaft of N, in the fame time that Cmoving in B R falls at R 146636
feet (QD+QR) north weft from Q
Queſtion 217. Suppofe the balls B and C (laft queſtion) to be per-
fectly elaſtic, where then would the bodies fall?
AND MECHANIC
203
Let Q the weight 1 V
143710 the velocity of C, and q➡500
the weight, v=3220 the velocity B, before the ſtroke, then (by art.
258) V-
29 V - 2qv Q V + q V -
Q+9
29 V
Q+q
29
q
QV-94-29
149563 the velocity of C after the ſtroke
Q+9
and (by the fame article 258)
2QV +297
Q+q
2633 for
the velocity of B after the ftroke, and becauſe the balls being elaſtic
does not alter their directions, therefore make DR=149563 feet and
Pu=2633 feet, fo is R, and u the places of fall as before.
Queftion 218. If the diameter of a cannon's bore be 8 inches, what
muit be the diameter of it's ball?
That the bullets may not be fo big to burft the piece, nor too little
fo as part of the powder may fly out between the ball and gun, it is
thought fit to make the balls diameter about parts of the guns dia
meter, therefore, as 20: 19:: 8: 7,6 inches anfwer.
19
Question 219. If the weight of a cannon's ball be 48 lb, and the
weight of the cannon 8000 ib, what must be the weight of another
cannon whoſe ball is 36 lb ?
As 48 lb: 8000lb :: 36lb : 6000lb anſwer.
Thefe pieces are made vaftly heavy in order to be fufficiently strong
fo as not to burit by diſcharging &c. The weight of a ball being 33
lb. the thickness of metal at the breech of the cannon fhould be 6
inches and at the mouth 2 inches &c. for others, the ordinary charge
of cannon is to have the weight of the powder equal to half the weight
of the ball, their lengths fhould be fuch, as that the whole charge of
powder may be on fire juft as the ball quits the piece, for if it be made
too long, the weight of air to be driven oat before the ball hinders its
force, if too short, the powder goes out more of it unfired, the great*
eſt random of a piece is about 10 times as far as the ball will go point
blank, viz. when the piece is laid parallel to the horizon. Cannons are
diſtinguiſhed by the diameters of the balls they carry, and are in length
from
4 to 12 feet.
Question 220. If a cannon weighing 6450 lb, give a ball of 24 lb
weight an uniform velocity of 640 feet per fecond at the breech of the
piece, how much will the cannon recoil in a fecond, if free to move!
The momentum of the cannon and ball are equal becauſe the pow
der acts equally in all directions (fee queft. 177) fo (by theo. 154)
640 x 24 = 2,4 feet per fecond the velocity with which the cannon
6400
begins to recoil,
204 THE UNIVERSAL MEASURER
Queſtion 221. Let things be the fame as in the laſt queſtion but fup
poſe the cannon to be fixed and its length 12 feet, required a preffing
weight equal to the force of the powder? See art. 252, 253 and
254.
Firſt, as 640 the uniform velocity of the ball is to 1 fecond, fo is
I
24 (twice 12 the cannon's length) to 24= parts of a ſecond the
640 26/
time of the balls paffing thro' the cannon at that rate, now fince
the powder conſtantly acts on the ball whilft in the gun it will there-
fore drive it along with an accelerated velocity, which velocity in the
fame time of paffing thro' 12 feet, produces an uniform one of 24 feet
as above. But in accelerated velocities the ſpaces paffed over, are as
I
: i
ſquares of the times, therefore, as 263 feconds fecond or as
ช
I
3
1:26 (7115) fo is 12 fect to 8533 feet, which the ball would
be carried thro' in one fecond by the accelerating force of the powder;
now, the weights of bodies being as the accelerated forces and theſe
as the ſpaces paſs'd thro' in the fame time, therefore as 16 feet the
defcent of gravity in the firſt ſecond, is to 24 lb, the weight it gives
the ball, fo is 8533 feet, the ſpace paffed by the powder in one ſe-
cond, to 12800 lb the whole force of the powder; but if the gun be
free to recoil, then this force 12800 lb, is part of it ſpent in giving the
gun a velocity of 2,4 feet (fee the laſt queſtion) per fecond; but the
whole force produces a velocity of 640 feet per fecond, therefore, as
640: 12800::2,4: 48 lb, the weight ſpent in giving velocity to the
gun, fo (12800-48) 12752 lb is fpent in preffing on the gun and ball
giving it a velocity of (640 - 2,4) 637,6 feet per fecond.
Queſtion 222. If a ball or globe, of caft iron falling from reſt in
vacuo, (viz. in a non-refiſting medium) do in the firft fecond of time
acquire an uniform velocity of 32,2 feet per fecond, what will this ve-
locity be if the fall be in open air? See the next queftion and theo.
As 7000 the ball's abfolute gravity is to (7000 1,2) 6998,8 its
ſpecific gravity in the fluid viz. air, fo is 32,2 to 32,19 feet anſwer.
Question 223. What is the greateſt velocity a ball of cast iron 6 in-
ches, or half a foot diameter (A) can poffibly acquire by falling in the
air?
1. The denfity of cast iron being to that of air as 70000 to 12, which
is nearly as 5833 to 1 (d to D) we'll have (by article 327) 4 Ax:
5823 I
d-D
of a foot X
=× 5832=3888 feet, the
D
AND MECHANIC.
205
height fallen in vacuo to acquire the greateſt velocity, that is,
4
A
3DX:
d-De, but (by art. 321) e
V V
V V
4 s
therefore,
4 A
4× 16, I
3 D
d-D
4 S
I
D
x:d—D:=e=
x: d_D:=e==, whence, as (by art. 328) v = 4√ :
SA:
5833
3
- 1x16,1x=500 feet, that is, the great-
eft velocity this ball can ever acquire by falling in the air is that which
would carry it over a space of 500 feet nearly, in every fecond of time,
otherwife if v = 2qe; then per laft queft. as d: d-D:: 2s:
=qfov v = 2qe=4sex: d-D: (and by writ-
2 s d
d
ing
3 D
2 s D
d
x: d-D for its equal e) then v v = 2 qe
D; but if we take e A
d
4Aqsx:d
then vv2qe==
se
3 D
-4x:d-D:
= 'fas x:
d-D
D
'16 As
3 D
X: d-D as before, hence, if the e
mentioned in the theorems to art. 401, be taken =÷Ax:
d-D
D
then q muſt be 2 s, twice 16,1, but if e =
4 d A
3 D
then a muſt be
-
- 2 sd — 2 s D, and in theſe theorems where q is not engaged, e
d
must be taken =
4 d A
3 D
(in this cafe) 3888,6.
Question 224. If a ball of caſt iron half a foot diameter, be thrown
directly upwards into the air, with a velocity (g) fit to carry it over
640 feet in the firft fecond of time, how high will it go, and what will
be the time of its afcent
Here per laſt queſt. e = 3888,6, g = 640, and q = 32,19, fo (by
theo. 7, art. 401) m = 2,3025 e × log. of 1 +
g g
29e
:= 2,3025 X
then for the time,
eX,4208 = 3811 feet for the required height,
Xarch of a circle whoſe radius is✔ 2 qe, and tangent =g,
=
q
thus, as /2qe√: 2X 32,19 X 3888,6 498,8, is to 640 (g)
206 THE UNIVERSAL MEASURER
fois rad. 1, to 1,283, which anſwers to the natural tangent of 52,066
dègrees, then becauſe 57,30 is the radius of a circle whofe periphery
is 360°, it will be as 57,3 is to 52,066 fo is 498,8 to 443,3 the faid
arch which divided by q
32,19 quotes 13,7 feconds, for the time
of the ball's afcending.
Queſtion 225. If a ball of cast iron half a foot diameter be let fall
in the air from a height of 3811 feet, what time will it be in falling,
and what velocity per fecond will it acquire by that fall ?
Here m =
:3811, and by queſt. 222 and 223, q= 32,19 and e→
as a log. and put n = 1, divided by its
3888,6, then take
m
2,3025 e
natural number, and we'll have v = √:aqex : I
n: and t
2,3025 log.
I + √: I n:
I √:1 n:
:X√, thus,
29
m
2,3025 e
14214 which looked for in the table of logs. its natural number is
I
found 2,64, then
=,378=n, ſo I n,622, there-
2,64
fore, v = √: 2 × 32,19 × 3889,6 × ,622 : = 360 feet, for the dif
tance that would be difcribed in one fecond of time with the velocity
e
acquired by falling, alfo, t=√ :X2,3025 log.:
29
= 15,5 feconds, for the time of failing.
I + √,622
I - √,622
Question 226. From what height in the air muſt a ball of cast iron
half a foot diameter fall, to acquire a velocity of 400 feet per fecond?
Here v = 400, and by the foregoing queſtions, e = 3888,6 and
q=32,16, then, from the laſt queſtion, m 2,3025 e log. 2qe
2,3025 × 3888,6×,4418 = 4047 feet, anſwer.
2qe-vv
Ify the greateſt velocity, or vv 2 qe, then the divifor 2 qe-
V=0, fo2 qe
2qe, in which cafe m, is infinite, that is,
2q e V V
0
the ball by defcending in the fluid can never acquire its greateſt velo-
city, tho' the higher or longer time it defcends the nearer it comes to
it, hence, (and by theo. 166) a falling body can never acquire an uni-
form velocity neither in a refifting, nor in a non-refifting medium, tho'
in either cafe it will in time come fo near it, as that it may be taken
as ſuch, for in vacuo, the ſpaces fallen thro' in each fecond of time
feparately, are as the odd numbers 1, 3, 5, 7, 9, 11, 13, 15, 17, &c.
AND MECHANIC.
207
where its plain that the farther thefe numbers are continued the lefs
they'll differ, or come nearer to equal ſpaces in equal times, &c.
4000
Question 227. With what velocity muſt a ball of oak wood 1227
feet diameter, be thrown into a pool of water 2 feet deep, that it may
go to the bottom, and there have 3 feet velocity per fecond?
The denſity of oak being to that of water as 920 to 920, which is
as 1 to 1, (dto D) we'll (by queſt. 223) have e —
I 2 7 7
4108
X 4000 = 1000=0,409, and by the queſt. m
if we take na natural number whofe log, is
4 A d
3 D
= + x +
2 and v = 3, now
m
2,3025 X2e
we'll have
g=nv=36,57 feet per fecond for the required velocity; alfo, the
2 e
ball's falling through the fluid is t = —
35,57 0,27 feconds.
0,818
X:g-v:
V:=
X
مذ
g
109,71
From theſe two equations g nv and t =
2 e
V g
X:g-v: we have
m = 2,3025 X 2 ex log. 1 +
g
2 €
X:n — I
I:.
2 e
g
4000
Queſtion 228. If a globe of oak wood 1323 feet diameter be pro-
jected into ftagnant water, with a velocity of 300 feet per fecond how
far will it have moved in 5 feconds of time?
Here t=5, g = 300, and by the laſt queſt. e= 0,409, whence m
= 2 × 2,3025: log 1+,818×2,3025 × 3,2635 =6,217
feet anfwer.
2 €
In theſe two queſtions, for varieties fake, I have fuppofed the den-
fities of oak and water to be equal, in which cafe q is o, and fo the
ball may
be confidered as moving by its innate force only; for it can-
not be affected in this cafe, by gravity, for (by quelt. 222) q being as
the velocity generated by gravity, it follows that in m=2,3025 e log.
8829: (fee theo. 7 art. 401) q muſt be negative if the ball
vv± 2qe
defcend, and pofitive if it afcend, but qo if it do neither, or then
the denfities of the ball and fluid are equal, in the other two cafes the
denſity of the ball is meant to exceed that of the fluid, and thus taking
q— or +, and having given g, e and m, you may find v the velocity
per fecond, of the ball at any height m, either defcending or afcend-
ing, and taking vo, we have the theorem in queſtion 224, &c. you
208 THE UNIVERSAL MEASURER
may find any one of theſe letters. if all the reſt are known, but if the
denfity of the fluid exceed that of the ball, you may take m=2,3025 e
× log. 88+29e
: when the ball is made to defcend in the fluid, and
v v + 2qe
m = 2,3025 e log.
Simpſon's Effays.
£ g - 2 qe
: when it afcends therein. See
29€
1000
Question 229. With what velocity muſt a ball of oak wood 1227
parts of a foot diameter, be thrown into a pool of water 2 feet deep,
that it may go to the bottom, and there have 3 feet velocity per ſe-
cond? (v)
The denſity of oak being to that of water as 920 to 1000, which
is as 23 to 25 (d to D) we'll (by queſtion 223) have e—
3000
4 A d
=
3 D
× 2 × 1121 = 0,37628, and (by queſtion 222) as 23: 32,2 ::
(2523) 22,8q, fo by the question, (m being
log of 88+29e.
vv+2qe
m
2,3025 e
2) we'll have
=2,310689, whofe natural number
by the log. tables is 204,45-88 +29_88 +2,087 there-
v v + 2qe
g g
11,087
fore, g: 204,45 X 11,087,-2,087:47,6 feet per fecond
anfwer.
If the required velocity were fuch, as juft to make the ball go to
the bottom of the water, then vo, and fo 204,45 = 1 + g g
I
2qe
therefore g=√: 204,45 1: × 2,087 : = 20,5 feet per fecond,
anfwer.
Queſtion 230. If a ball of caſt iron 6 inches diameter be projected
in the air with a velocity of 64 feet per fecond, what muſt be the an-
gle of elevation, to give it the greateſt horizontal amplitude? See
art. 401.
Note. In the 5 first theorems in art. 401. m denotes the height
fallen in vacuo to acquire the projectile velocity, or impetus =m,
fo (by art. 374) here m =
=64 feet, and (by queſtion 223)
☐
64
64
e= 3888, whence c=√ :
96e 235 m
=7072, the natural
192e512 m
fine of 45° 01′, whofe complement to 90° is 44° 59′ the required
elevation. Now putting t = 0,9995, and s=1,414 the natural tan-
AND MECHANIC.
209
gent and natural fecant of 44° 59' the elevation, we'll have (m
as before) a =
4 t m
S S
64
32 tt
16t
m m
+
:X
= 126,8
+
S
3 e
the anſwer, this amplitude and elevation in a non-refifting medium is
128 feet, and 45°, fo the difference is very fmall. Here it may be
obferved that the elevation, varies with m and e, &c. Thefe 37 laſt
queſtions, are more than the practice of gunnery can ever require,
but they exercife other curious parts of mathematics, and fo are
thought worth a place here; in gunnery there are feveral other ob-
ſtacles to be minded, fuch as the make of the gun and ball, if they be
exactly round, the ball to be of the fame metal, &c. See the fore-
going queſtions.
In theſe queſtions the ball is fuppoſed to have its velocity from the
force of gun-powder, but other forces may produce this velocity,
As, firft, a fpring, or bow. 2. By water &c. See queſtion 160. 3. By
air fee question 190; for, if two tubes or gun barrels, AI and FI
(fig. 166) be joined together at I, and the air in A I be by a fyringe,
&c. thrust into the fpace HI, and then by opening a valve at I
where a ball is lodged, the faid confined air will push the ball out at
F with great force, this is the nature of the air or wind gun.
4. By
fteam (fee queſtion 191) whoſe vaſt force is there computed, it is prov-
ed by trial, that if a piftol barrel clofe ftopped with a few drops of
cold water in it, have the end where the water is, put into the fire,
and when it begins to fimmer out at the touch hole, the air in the bar-
rel is then exhauſted, then ſtop the touch hole cloſe up and in a little
time the water will be converted into fteam, and will blow out the
cork or ball at the other end with as great violence and noife as a
charge of gun-powder, &c.
The 30 following queftions
explain frictions fpouting fluids,
whirling bodies, waves, &c. &c.
Queſtion 231. Which of the mechanic powers has moſt friction?
1. The ſmoother bodies are (except they be poliſhed ſo fine as to
ftick together which can happen to few bodies) and the lefs fpace they
have to rub on, the fmaller is the friction, whence in the leaver, if the
prop is ſharp pointed, the friction is little or nothing.
2. In the wheel, the friction upon the axis is as the weight upon it,
the diameter of the axis and the velocity of the axis periphery, (ſee
queſt, 90) this fort of friction is but ſmall.
3. A rope of 1 inch diameter, whoſe tenfion is 5lb, going over a
pulley 3 inches diameter requires a force of 1 lb to bend it, and the
I
Dd
:
210 THE UNIVERSAL MEASURER
force of bending a rope, is as the fquare of it's diameter and tenfion
directly, and the diameter of the pulley inverfely, (fee queſt. 128) &c.
and 15 lb over a fingle pulley will draw up but 14lb weight, but there
is a great difference in ropes arifing from the temper of the weather,
the difference of their fiffness, &c.
4. In the wedge, the power to overcome, muſt be at leaſt, as twice
the bafe to the height. (See queftion 70)
5. In the fcrew there is molt friction, for it will fuftain the weight
in any pofition when the force that raiſes the weight is taken away,
whence, to raiſe the weight there muſt be at leaſt a double force ap-
plied to that intended in queſt. 66.
6. In a fluid running thro a tube, its plain, the parts rubbed against,
are the fides of the tube, whence the friction is as the diameter of the
tube and velocity of the fluid, or as the periphery and velocity. But
the friction is greater in reſpect to the quantity of the fluid in ſmall
tubes, than in large ones, and that inverfely as their diameters, but
the abfolute quantity of friction in tubes is very fmall, except the tubes
be very long, and the velocity very great; in rivers, the velocity is
greateſt where the water runs deepeſt, and leaſt at the fides where it
is fhallow, it is common to take half the fum of theſe two velocities for
the mean velocity of the ſtream. See queft 156.
7. If any tube, the area of whoſe ſection is a, be divided into feve-
ral ſmall equal ones whofe number is n, the area of any one of theſe
fmall ones will be
and the friction being as the diameter it will be
a
n
a
I
as v
or as
n
n
will be as
fion.
n
(by taking a 1) and fo the friction of them all
which is as ✔n, the increaſe of friction by fuch divi-
8. The refiftance of a plane moving thro' a fluid is (by art. 317) as
the fquare of the velocity, or (by art. 323) = the weight of a column
of the fluid whoſe baſe is the plane, and height but in a globe
=article 326.
s
TV
4 S
9. In all bodies viz. woods and metals, the friction is nearly the
fame if they be oiled or greaſed, in wood acting againſt wood, greafe
makes the motion eafier, metals of the fame fort acting againſt each
other have more friction than metals of a different kind, the fofter and
AND MECHANIC.
211
2
3
rougher bodies are the greater the friction, thefe experiments have
been made (fee Mr. Fmerfon's mechanics) to determine friction, viz.
a cubic piece of foft wood 8lb weight, moving upon a ſmooth plane
of foft wood at the rate of 3 feet per fecond, the friction is of its
weight, but if the wood be rough it is near the weight, other foft
wood upon foft wood very ſmooth, the friction is about 4 the weight
foft wood upon hard, or hard upon foftor, the weight hard wood.
upoa hard wood, or the weight, poliſhed ſteel on Iteci or pewter
the weight on brafs the weight, on copper or lead, the weight,
&c. from which we learn, that there can be no general rule laid down
to determine the quantity of friction in all cafes.
I
M4
$
10. The friction (all things elfe being the fame) increafes with the
weight nearly in the fame proportion, as alfo with the velocity in moſt
cafes ; a greater furface alfo caufes more friction with the fame weight
and velocity, in fome cafes, but if the body move on clay, foft earth
&c. the friction may be increaſed by the rubbing parts being too fmall
thefe things are all very plain if we confider friction as a power acting
againſt the motion of the engine &c. for the only caufe of friction is
the roughness of the rubbing parts which rough particles may be con-
fidered as fo many obitacles over which the body is to be drawn.
11. Heace, if by increafing the power to overcome the friction the
weight on the rubbing parts be alfo increas'd, then there must be fome-
thing more added to the faid power, &c.
12. Alfo, if by any means the velocity of the rubbing parts can be
made lefs, the friction will be leffened nearly in the fame ratio, &c.
13. From thefe articles you may compute the friction of any engine
&c. by ellimating the friction of every part by its felf, and then the
fum of all thofe frictions will be that of the engine, &c.
Question 232. If in weighing 5lb, it be found that 1 grain weight
will give the fcales a caft, what weight will give them cait when they
weight 500lb? See art. 10, laft queftion.
As 5 lb is to 1 grain fo is 500lb to ICO grains anfwer, nearly.
Question 233. If a cylinder C A A (fig. 192) of A A forty inches
diameter require a weight w of 20 lb to make it turn in form of an
axle-tree round its diameters, what must be the diameters of two pins
or gudgeons, a, and C, that turning round on them (in the fame time)
the friction or force of turning may be only 3lb?. Sce art. 12, quelt,
231.
As 20lb is to 40 inches fo is 3 lb to 6 inches anſwer. And if you
lay the axle-tree of one wheel upon the periphery of another
wheel, the friction will ftill be leffened, fuch whcels are called friction
wheels.
212
THE UNIVERSAL MEASURER
Question 234. If the power P (fig. 193) being 2 lb weight, hold
the weight w in equilib. and it be known that the friction be one third
of the power, what weight muſt be added to the power to overcome
the friction, or give motion to the weight w?
1
Here its plain of (P) 2 lb muſt be added to the power P, but then
this additional weight caufes a greater preffure on the pullies, and
therefore (ſee art. 11, queſtion 231) requires of its weight to be
added, and that again for the like reaſon will require of its weight
and ſo on, ad infinitum, whence the weights to be added will be 4, 3,
,&c. the fum of which decreafing geometrical feries is (by ex.
47 in algebra) = 1 lb the anſwer.
I
3
Question 235. If in a jet d'eau ADF (fig. 222) the refervoir
A B be conſtantly kept full of water, whofe height L I, above d, the
Spout of the conduct pipe C D d, is 5 feet 1 inch, its found by expe-
riment that the water fpouts to the height d F of 5 feet, to what height
would it fpout if the water's furface L A in the refervoir were LI 10
feet?
1. The water would rife to the height G, fo that d G would be
LI, were it not for the air's refiſtance, and from thence fall down in
the ſtreams F H and F F., &c. now (by queft. 224) the deffect G Fis
as the fquare of LI ord G, the height of the water's furface in the
refervoir, that is, as 25 (fq. 5) is to 1 fo is 100 (fq. 10) to 4 inches,
therefore, 10 feet 4 inches 9 feet 8 inches the anſwer. But ſmall
jets fall fhort more than in this ratio, being more retarded by the air's
reſiſtance, and the greateſt jets never rife 300 feet high for the velo-
city of the water at d, is then fo great that it is diffipated into ſmall
drops like rain, by the air's refiſtance, and for the fame reafon, water
falling from great heights as from the clouds &c. defcends in fmaller
drops the higher it falls.
•
But to have a fet to rife to the greateſt height, thcfe things are to
be obferved. The adjutage (viz. the hole at d, thro' which the
water ſpouts) fhould be in a thin plate of metal, and fhould be larger
as the refervoir is higher and the larger it is the higher the water will
fpout (fee queft. 224) if there is plenty of water to fupply it, the con-
duct pipe ought to turn in a curve as D E d, and not with an elbow.
2. The diameter of the adjutage to be nearly as the fquare root of the
height of the refervoir, and if the velocity in the pipe of conduct is
to be the fame at all heights of the refervoir, that the friction may not
increaſe ſo much, then the fquare of the diameter of the pipe of con-
duct, muſt be as the cube of that of the adjutage. Otherwiſe, the
diameter of the conduct pipe fhould, at leaſt be 5 or 6 times as big
AND MECHANIC.
113
I
as the diameter of the adjutage, and if the height of the refervoir be
50 feet, the diameter of the conduct pipe 6 inches, that of the adju-
tage
ſhould be 1 inch, &c. Laftiy, the fpout fhould not be exactly
perpendicular in order to hinder the water falling on the adjutage a-
gain. When water is carried a great way through pipes, the friction of
the pipes will leffen the velocity of the water at d, and the jet will not
rife fo high.
Queſtion 236. To make an artificial fountain?
Fig. 223.
Let A B I be a veffel filled with water in at a pipe A B going near
the bottom B, now its plain, if this veſſel be turned bottom upwards,
there can no water come out at A, but will all run down a pipe A D
and there ſpout out in a ftream D H, and fall upon another like veffel
EFG, and fill it thro' another like pipe E F, which done the fountain
may be turn'd upon the end B, and the water will fpout out at H, and
fill the veffel A B I, by running in at the pipe A B as before, and thus
by turning the fountain, it may be kept conftantly playing.
Note. If with your finger you flop the orifice A of the pipe A B, the
fountain will ceaſe to play becauſe then no air paffes into the veffel
ABI and by taking off the finger it will ſpout against D) as before, &c.
An artificial fountain is very eafily made thus, fill a ſtrong bottle half
full of water, into which put a tube to reach near the bottom of the
bottle, then ſtop the top of the bottle cloſe, ſo that no air paſs between
it and the tube, this done, blow through the tube what air you can, fo
will the air in the bottle be condenſed, and the water fpout out at the
tube (ſee queſt. 190) any of theſe fountains playing in the fun-fhine,
and a black cloth placed behind them, the ftream will fhew all the co-
lours in the rain-bow.
Question 237. If the diameter of the rim of a ſpinning wheel be 24,
the diameter of the whorle that carries the feathers 4, and that of the
bobbin or twill 2, what is the difference between her twining and tak-
ing up?
Theſe things being carried about by the rubbing of the wheels band,
it follows, that while the rim makes i revolution, the feathers makes
(24 4) 6, and the twill (24÷ 2) 12 revolutions, hence, the
tity taken up is (12 6) 6.
quan
Note. The lefs a thread is twined (provided it have fo much twine
as to hinder the fibres from drawing out) the ſtronger it is, and will
bear more weight. From hence it appears, that the twill and feathers
together twine the yarn, and if their revolutions are equal the wheel
takes none up, if the feathers ftand ftill fhe twines none, and the grea-
ter the rim, the fafter fhe works.
214
THE UNIVERSAL MEASURER
I
Queſtion 238. If when the plane of the fail of a ſhip makes an L
with the keel of 1 point, or 11° 15′, the leeway is two points, what
would the leeway be if the faid angle were 3 points, or 33° 45′ ?
Fig. 224
+
Let S A be the fail to which draw S K to reprefent the force of
the wind againſt the fail, and to SD the pofition of the keel, draw
DK, then (by theo 157) the force S K is refolved into the two for-
ces, SD, the direct force, and DK, the leeway force, let DE and
DS be as the reſiſtances the ſhip has a head and a fide with equal ve-
locities, and fuppofe S C the way of the hip. Then (by art. 317) As
reſiſtance a head with velocity SD is to refiftance a head with velocity
DE fo is SD to DC, and, as refiftance a head with velocity
DE is to refiftance a fide with velocity DE fo is DE to S D, and as
refillance a ſide with velocity D E is to refiftance a fide with velocity
DC fo is DE to DC, from thefe three proportions we get, as
reſiſtance a head with velocity SD is to refiftance a fide with velocity
DC, fo is SDX □ DEX DE to □ DE X □ D C X S D fo is
SD X DE to DC, but the refiftances are as the forces producing
them, therefore, as SD: DK::SD×DE: □ DC, DE× DK
hence DC is a mean proportional between DE and D K, let radius
— 1, tangent L DSK co-tangent LDS At, then (by theo. 47)
as 1 : t:: SD: tx SD = DK, and as SD: DC:: 1: tangent
DC
LDSG=
but DC DEX DK, and DK = tx
S D
>
√
SD, therefore tangent L DSC-✔✅tXDE hence as refiſtance a
SD
Alfo, in the
fide is to reſiſtance a head with the fame velocity, fo is radius × co-
tangent LDS A to tang. fquare of L DSC the leeway.
fame fhip, DE and SD are conftant, fo it will be as
tangent of
78° 45' is to tangent of 22° 30' fo is co-tangent 33° 45′ 56° 15′
to tangent of 12° 37′ the leeway required.
Thus by the logs. the tangent of 78° 45′
10,7013 fotangent 78° 45′ 10,70132
(comp. 11° 15′)
5,3506
=
take
and tangent 2 points, or 22° 30′ = 9,6172 2 from
and tangent 56° 15′ = 10,1751 ÷ 2 =
✓
5,0875 S
anſwer tangent of 12° 37′ = 9,3541 leaves.
Note. Here the hull and riggin are not confidered, but the error from
that is but fmall, and on point of exceſs.
Question 239. If the wind in direction WS (fig. 225) fall on SA
the ſail of a ſhip, making no leeway, what are the ratio's of the
AND MECHANIC.
215
:
forces, with which the fhip is compelled to turn (the fail being
at her head) and in direction of her keel S 1), as alſo to windward.
Draw SCLSA, and CD LSD, then (by art. 318) the force
acting upon the fail in direction SC, is as the fquare of the incident
LW SA, but (by theo. 158) the forces in direction SC and S D
are as S C: SD, or as radius 1: SLS CD, therefore, as radius I :
S, L W SA
force in direc-
SLWSA (force in direction SC):: SLSCDS, LASD:[]
S LWVSAXSLASD force in direction S D, and the force to
turn the fhip about, is that in direction D C L the keel S D, fo taking
its oppofite LCS D, inſtead of L S CD, we'll have
XS LDS C, or S, L W SAX co-fine L AS D,
tion D C, draw, SPL W S, and DG ISP, then (by art. 318) the
forces in directions SD and DG (that to windward) are as [SD :
GD, or as radius 1: S, L DSG,
:□
co-fine LWSD,
therefore, as radius 1: S, WSAXS, ASD (force in direction
SD) :: c S, WSD: S, WSAXO co-fine W S D x S.
ASD force in direction G D, whence, if A the area of the fail,
and v the velocity of the wind continue the fame, (fee art. 320) the
forces of the fhip forward, to turn her, and to windward are refpec-
tively as v va X□S, W SAX S, ASD and v v A XS, WS A
X co-fine ASD, and v v A XS, WSA X□ co-fine W SDS,
A S D, or as S, ASD and co-fine ASD, and co-fine W SDXS,
A S D, as required.
Question 240. Things being as in the last queft. let it be required
to find the velocities inftead of the forces? See art. 317.
The fquare of the velocity of the fhip,in any direction is as the force
of the wind upon the fail in that direction, or as (its) the refiftance.
the ſhip meets with in the water, confequently, the fquare root of any
of the expreffions for the force, will be as the velocity in that direc-
tion, thus, v XS, WSAX √: AXS, ASD: and v x S, WSAX
: A x co-fine ASD and v xS, WSAX co-fine W SD X
:AXS, ASD: are refpectively as the velocities, forward, to turn
and to windward.
:
Note. If the LWS A, be ftill the fame, and SDPC be a half
circle upon any line S C, then the force in any direction SD of the keel
will be to the velocity in that direction as the cord SD is to its fquare
root SD, for the LSD C is ftill = 90°.
Question 241. If the wind blow at right angles to the keel of a fhip,
what angle muſt the plane of her fails make with the keel, that ſhe
may fail the fwiftelt pollible? See quelt 239.
1
216 THE UNIVERSAL MEASURER
the
In this cafe WS (fig. 225) is LDS, fo the required L ASD, is
comp. of LWS A, let e fine L ASD, then S, WSAXS,
ASD—I—ee xe (radius = 1) = e — eee, which being a maxi-
mum, you'll find e = √,57735, the fine of 35° 16′ aniwer. In
this polition of the wind, L WSD 90°, its co-fine iso, fos,
o, the force to wiod.
cube S, WV SA, or cube
WSAX co-fine WSD (0) x S, ASD
ward, and S, W SA X co-fine ASD
co-fine ASD, which is greatelt when WSA 90º or ASD = 0,
that is when the wind falls perperdicularly on the fail, the fhip turns
fwifteft.
Queftion 242. How muft S A (fg. 226) the fail of a ſhip whofe cen.
ter of gravity is G, be inclined to the plane of the horizon HO, that
the perpendicular force CD of the wind againſt the fail S A may nei-
ther raife nor depreſs the head F of the ſhip, but keep her ſteady?
=
From C the center of gravity of the fail upon H O the horizon or
keel of the fhip, let fall the perpendicular C B then if D C be the force
of the wind againſt the fail, S A then D B, is the force that generates
the ship's progreflive motion and B C the force lifting the fhip upwards
now the force DB (= b C) acting at C, in direction D B, endeavours
to turn the ſhip, round an axis paffing thro' G, with a force as C B
X BD, viz. the abfolute torce B D x diftauce CB, and this is the force
depreffing her head; alfo the force raifing her head is BCX BG,
viz. = the force B C in direction B C endeavouring to turn the fhip
round an axis thro' G, the contrary way, x diſtance B G (fee art. 280)
hence the force to raife her head, is to the force to fink it as BCX B G
to CBX BD, that is B G to B D, therefore when the point D falls in
G, the fhip's head is neither raiſed nor depreffed by the wind, fo to
anfwer this queſtion, the fail S A mult be fo fet as that a line joining
the centers of gravity of the fail and ſhip, may be at right angles to
the plain of the faid fail, but this point G must be at the meeting of a
line HO parallel to the horizon or axis or keel of the fhip, and to
pafs thro' the center of preffure or refiftance which the ſhip has by the
water in her motion, and a line drawn perpendicular the horizon thro'
the center of gravity of the ſection of the ſhip and water.
Question 243. If SA (fig. 226) be an artificial kite, or a fail in that
form, with a line and ball e faftened to the end A, to keep up the
other end S to the wind, blowing in direction W C parallel to the ho-
rizon HO, now if this fail and ball be 10 ftone or 140 lb weight, and
its area 1000 feet, and the wind blow at the rate of 40 feet
per fe-
cond, and the angle of incidence WC be 300, what angle HD C,
with the horizon H O, will the cord CE D, faſtened one end to C the
fails center of gravity and the other end at D make ?
AND MECHANIC.
217
f. Draw C GLSA, and GELHO, fo is GC the direction of
the force of the wind againſt the fail, and the force of the wind to
keep it up is (by art. 318) as fquare incident LSC W, or C G E、
for becauſe W C is parallel HO, L W C G = LCG H, whence,
CGE comp. C G H = comp WCG-SC W, now becauſe E G
is L horizon C G L fail, and E C the direction of the cord, therefore
CG, GE and EC, are refpectively as the force of the wind, weight
of the fail and force pulling at the cord. Let v 40, velocity of the
wind, SS, incidence 30°, A area 1000, then (by quest. 145)
,00 113 S Sv v A,00113 X,25 X 1600 x 1000 = 452 lb
of the wind CG, and G E 140 lb the weight of the fail, by which
and the included L CGE 30° the Ls G CE, and GE C are found
= 15° and 135°, fo LHDC= 45, the answer. Here its plain if C G
=GE, the wind can only hold the fail in equilib. if C G be less than
GE it cannot more the fail, &c. and the greater CG is in refpect of
GE, the higher it will rife in the air, and the length of the cord CD
makes no alteration in the L HD E.
=
force
Question 244. If a plate of iron SA (fig. 226) be fastened with a
cord CD to C its center of gravity and D the bottom of a fream of
water running at the rate of 10 feet per fecond, with what force will
the cord be ſtretched, fuppofing the weight of the plate = 140 lb its
area 17,4 feet, and the angle of incidence S C W≈ 30°, and, the
plate put up in the ſtream as a kite in the air.
By queft. 145, we'll have 0,978 v v SSA,978 X 100 X,25 X
17.4 = 425 nearly, the force of the ſtream CG, and by this, and the
lait queſt. GE = 140, L C G E = 30°, fo the L of altitude HD E
45. And then as S, LGCE 150 is to G E 140 lb, fo is S, LC GE
30° to E C nearly 270 lb the anfwer, all found per laft queft. radius
being 1, and then S 0,5 the natural fine of 30°.
Queſtion 245. When the wind in direction WS (fig. 225) makes
an L WSD with the keel S D of a thip of 60°, what LASD muit
the plane of the fail S A make with the keel, that the fhip may make
the moſt way a head poffible ?
By queft. 241, we are to divide the L WSD into two fach parts.
or L's, WS A and A SD as that the fquare of the fine of WS A,
multiplied fine of ASD may be a maximum; let S fine WSD, a
fine WSA, e fine ASD, c = co-fine WSD, radius r, then,
to give a clear folution, let ANP (fig. 134) be the arch to be divided
whofe fine is PGS, and co-fine C G = c, into the two arches A N
and NP, whoſe fines are Nza, and PO=e, co-fines & 2 and CO,
E e
218 THE UNIVERSAL MEASURER
now becauſe L POI LIGCLs at B, and z, each = 90° there.
fore, comp. OIP comp. CIG=0PI=IC G, fo the As V OP
GIC, BOC, and z N C are fimilar, therefore as CN: Cz:: CO:
COX Cz
CN
- CB, and as CN: Nz::PO:VO-GB-
whence C B - GB =
NzxPO
CN
COXCZ NZX PO
Nz
CN
= CG,
That is,
I
: rr -aa xrr-e:
by theo. 44
2 fquared
2
aere, for CO
=√:rr-ee: and Cz = √:Ir—。a:
:rr-
erraaa√rr…eerS
2aae e
2 a e
3
aa tee
+
X √rr—ઠX
rr
I r
Vrr-ee
S S.
from 2 and
13
4
whence
5 Хаа
6
And, therefore
SS aa + 2 z ae +ee, taking z =
if radius r=1
e = √ SS-a a÷zza a
ing the fquare &c.
aae: a* SS— a°+zza:
mum by the queſtion.
4 ass
6 a 3
3
+6z¹a
2:a + S S — a ° + z za
6
tion folved will give the value of a required.
If in this laft equation we write I
z (its equal) we'll have 4 a SS — 6
4 a S S - 6 S S a
3
2: SS a+ — SS að;
which by tranfpofing
a+
2 √ : a* SS
3
of
or z=c,
-za, by compleat-
- za 3a maxi-
1:320, which equa-
SS for z z, and / 1 — SS for
a³ 4-6 a
6 S Sa³
— aº + aSSa:
a+a
− √ 9—955 — 4 5 — 6 Sa a
2₁/:aa-a+:
9-9SS
√9-95S = 0,
9-9 SS, fquaring each fide and multiplying
+
by 4 a a
4 a 4, gives, SSX: 16 — 48 a à † 36 a
= 36 X: SS
2 4
4
a +a
2 — S² a
whence, 36 a
4
-
- 36 a a
12 S Saaz
16 SS, and a
4
36-12SS
16 SS
aa=
or a 4
: I SS
36
36
Xaa-SS, which by compleating the fquare &c. and putting
I +SS=n, we'll have a = √÷
nn 4SS
:=0,62041
4
9
2
the natural fine of 38° 22′, ſo 60° — 38° 22′ = 21° 38′ anſwer.
AND MECHANIC.
219
Note. It appears that the fine of the difference of 38° 22′ (LWSA)
and 21° 58′ (LASD) is of the fine of 60° ¸ LWSD) their
fum, and will ſtill hold fo, let 2 WSD be what it will, and by theſe
fines you 'll find that tangent L WS A is always double tangent L ASD
by which fuch quetlions may readily be folved when S is given.
Queftion 246. Let things be the fame as in the laft queftion, but
fuppofe the velocity of the wind to be 9 (V) and that of the fail, or
fhip a head 3 (v) ?
Let S W (g. 205) be the velocity of the wind, and Sb that of
the fail, then its plain, while the fail with thefe two velocitics from S
to b, the angles WS A and WSD, becomes the Ls Wba and Wb D
(the LASH being fill equal angle Sb A, becauſe the fail moves
parallel to itself, draw LS LAS A, alfo L R and WQ, each 1:
SD) then by the lait queſtion, we'll have S Wbax SS baa
maximum, as will otherwife appear thus, draw We to the fail.
Aba produced, then fince WS is the direction and velocity of the
wind on the fail AS A, it is plain we (fce theo. 158) is that force of
the wind which drives the fail from b to S, which (by art. 318) is as
We, or as incident 4 W be, fo (fee queſtion 239) S Whe
XS L Sbe= (as above) force in direction b S, now V9, being =
SW and v = Sb, let Sline and z co-fine of L WSD (60°)
alfo put y Sa, eab, then Sa and L a being tangent of the Ls
Sba and L ba, to the equal radius b a, we'll (by the note to the laſt
2 y, fo LS3y,
=
alfo
43
e
=ttan-
y
queftion) have La2 Sa
gent La Sb, (if the radius be Sb = 1) then by fimilar As, as v : y
ЗУУ
3y:
V
SR, fo 3 y y F
Sb, alſo as ve:: 3y:
3 ev
F
V
bR SR-
V
LR, but by trigonometry QW =
SV, and b = z V
v, whence again, as S V: z V — v
3 ey
V
: (bR) 3xy — v v
V
, therefore, 3 yy - vv = :
Z P
:X3ey, (or
SV
becauſe y y +ee- vv) 2 y y ee-:
Z V
:3ey, which divi-
S V
ded by y y, &c. gives ee
z V
3e
+:
X
=2, which by com-
yy
S V
Y
220 THE UNIVERSAL MEASURER
pleating the fquare &c. gives
e
=√2+2x1
Z V
F
:
y
S V
Z V
S V
:=t=1,155, the natural tangent of 49° 07', whoſe com-
plement is 40° 53′ for the required L Sba LHSA, fo as that the
effe&t may be a maximum. If the fail be at reſt, or v = 0, then t
z V— v
z V
V
√:2+2x
- 1x
:=√:2+
SV
S V
9 z z V V
4 SS V V
32 V
:√:2+
3Z
:
13 323 = √ : 2+
9X0,25.
3X0.5
2 S
2 S V
,79!7
- L Sba
927
4 SS
4X0,75 2xv 0,75
the natural tangent of 38° 22′, whofe comp. is 51° 38′
LHSA, the fame as would be found by the laſt queſt.
for LW SD being 60° the LHS W, will be 120°, whofe na-
tural fine is .75 (for 1800 120º 60º) one third part of which
is + √,75 =,2887 the natural fine of 16° 44', then LHS W 1200
16° 44′ = 103° 16′, half of which is 51° 38′ for the LHS A, as
before.
उ
Question 246. If the wind or any fluid fall perpendicularly on the
plane of a fail, what angle muft the faid plane make with the direction
of its motion, that the effect may be the greateſt poſſible, ſuppoſing the
velocities (V and v) of the wind and fail to be equal? Fig. 205.
Here VWS, vSb, Let a fine ↳ WS A (= rad. 1) which
the direction of the ſtream makes with the plane, and e fine of the
required HS ALS ba, then becauſe A A || A a, the L W de=
LW SA, then (by the laſt queſt.) the force in the perpendicular di-
rection as is expreffed by w e, ſo the force fin direction bS, will be
☐ wex
Sa, but by plane trigonometry, ev
Sb'
S d, alfo, as 1 : V
Va-vex
e v
V
دم
Sae E and ev
a
e v
Sa
::a: Va-ev-ew, fo□ew X
a
Sb
- e VV aa-2 Vv a ee + v v e³ = f (=vv by queſt.
240) which by the queſtion muſt be a maximum, fo V V aa-4 V vae
+3ec v v = 0, therefore ce
4 V a
VV aa
>
whence e=
3 v
3 V V
2 Va
4
V Vaa
V Vaa
Va
V
3`Y
9 V V
3 v v
: -
= 0,3333 the natur-
3 V
AND MECHANIC.
221
al fine of 19° 29′ the anſwer, from this laſt equation we have, As
the velocity of the fluid (V) is to the velocity of the plane (v):: the
fine of the L made by the plane and its direction (e) to the fine of the
angle made by the plane and direction of the ſtream (a) when the ef-
fect of ſuch a machine is the greateſt poſſible, but this proportion is
a V
impoflible if e, be more than unity, for e the natural fine of any
3 V
L, cannot exceed radius (1) or unity, if a and e are each radius (1)
or the fail at right angles to the wind and goes directly before it, then
v = V. Hence, if the wind can produce a velocity in a fhip &c.
greater than of its own velocity, its plain the fhip may run fwifter
upon an oblique courfe than when the fails directly before the wind.
3
उ
Queſtion 246. What angle muſt the plane of a fail (of a wind-mill
&c.) make with the direction of its motion that it may turn or move
with the greatest freedom poffible, the velocity (v) of the fail being
of the wind's velocity (V) and the wind blowing perpendicularly on
the direction of the fail? See queſt. 144
Let things be the fame as in the laft queft. but one, then S being
radius =1, the co-fine z will beo, and ſo t = √ : 2 + 2 ×
z V — v
- X
S V
z V V
SV
Q V V
=√:2+
:+
4 V V
3 v
2V
= √2+8 2
+ 3 = √2,1406 +0,375 = 1,855, anfwering to the natural tangent
of 61° 40′ the anſwer. If vo then t√2, which anſwers exact-
with queft. 144, now the velocity of the fail of a wind-mill being as
its diſtance from the axis of motion, if we fuppofe v = the fails velo-
city at 1 foot from the faid-axis, then the velocity at 2, 3, 4, &c. feet
diſtance, will be 2 v, 3 v, 4v, &c. and that the effect may be the
greateſt, the fail muſt be ſo twiſted as every where to make an angle
with the direction of its motion, whofe tangent = √: 2+
9 V V
3 V
:+
2V
4 VV
Queſtion 246. If CSB, be a lever, S the prop, SC a = 6,SB =
b=30, the weight at C=w= 500, what weight or power P placed
at B, will give w the greatest momentum poffible? Fig. 158.
Note. Here are 4 queftions of this number 246, in a miſtake, but
the number is regulated after queſtion 294.
212 THE UNIVERSAL MEASURER
=
1. Let n the weight that will ballance w, then per lever b n
aw, or n= aw, let n+q=P, 2 s = velocity acquired by gravity
in the firſt fecond of time equal 32 feet v equal the velocity gain'd
or acquired by w in the fame time, then as a :v::b: veloci-
ty of P, therefore, wv+
vb P
a vw + vb p
a
a
b v
a
= the fum of the
motions of w and P, and muſt be 2 Sq, the motion of q, becaufe
by this motion they are both generated, i. e. a v w + vb P = 28 qa
P—n
(becaufe qPn) 2 Sax: P-n := (becauſe n =
w
a_W) 2 Sa
aw:
whence, w v =
X: P
a w
b
: whence v =
2 Sax: bp
2 Sax: b Pw aww:
bx:bP+aw:
bx: bPaw:
momentum of w, which by the queft.
muſt be a maximum, and therefore (art. 223, w variable) we get bb PP
2Pbaw+aa w w, which folved gives w =
bp
0,414 DP, fo P = 2,417 aw=
a
See queſtion 253.
I + 2
xbP=
a
2,417 X 6 X 500
30
=241,7 anfwer.
2. It is plain by the figure, that S B may be the radius of a wheel,
and S C the radius of its axle-tree, the procefs will be the fame, and
any 3 of the 4 letters a, b, P, w, being given the 4th may be had from
the laſt equation, and when a = b it is a pulley, and then P 2,417
× 500 = 1208,5.
2 + √2 b P in v =
2 Sax:bP aw: for
3. By writing
a
bx : D P + a W:
its equal w, we get v =
2 Sa
X: √2−1 = 13.248 = 2,6496
b
a
b
feet, the velocity acquired by w in the firft fecond of time, and the
velocity acquired by P in the fame time is = × 13,248
b
a
a
b
13,248 feet. And in cafe of a pulley, ab, then 13,248 is the ve-
locity acquired by each weight in the firft fecond of time, when the
momentum, or force of each weight is the greateſt.
bp
4.
In queft. 253, we have w
which written in the aforefaid
2 a
AND MECHANIC.
223
value of v, inftead of w, we'll get v=
2 Sa
bp - ibp
x:
b
b P÷÷b p
2S.a
3b
2 feet, the velocity acquired by w in the firſt ſecond of
b
time, and X
a.
2 Sa
3- b
2 S
3
= 10, the velocity acquired by P in the
I
fame time, when w is raiſed by. P in the leaſt time, and in cafe of á
pulley, or b a, 10 feet in the first fecond of time is the velocity
acquired by each weight,
32, the uniform velocity of
Here, If S
gravity then S, becomes S, the fame as ve, queſt. 150, &c.
Queſtion 247. If a machine is to move a weight of 5000 lb at the
rate of 2 feet per fecond and the power to do this can only move 10
feet per fecond, what is the weight of the power?
1. In cafe of an equilib. (fee theo. 183) the momentums of the po-
wer and weight muſt be equal, fo.to put the machine in motion, fome-
thing muſt be added to the power, or its. velocity increaſed, ſo 5000
X20XP, whence p- 10000 1000 lb. for the equilib. anf.
福
​=
Question. 258. If a veffel of water &c. receive a ſhake or ſtroke,
the water will undulate up and down the two oppoſite fides of the
velle, now if the breadth of this moving water be 18 inches, how
many times will it undulate in 1 ſecond?
I
#
1. If a beam &c. be horizontally fufpended by its middle and fet
a moving, it will vibrate up and down till it recover a pofition
parallel to the horizon and then reft, fo will water in a veffel-
after the reffel has leaned to one fide, and then fet upright a-.-.
gain, whence it follows, that the motion of waves, or afcent and def-
cent of water up and down the two oppofite fides of a veffel, is the
fame with the fwinging of a pendulum whoſe length is half the breadth.
of the wave, or undulating water, therefore (fee queftion 117) As 9-
inches (the of 18 the given breadth) is to fq, 1 fecond fo is 39.2
inches to 4.35+ whofe fquare root is 2,08 the number of times
per fecond anſwer.
་
Queſtion 249. If the breadth of a wave in the fea &c. be 18 inches,
what is the velocity of the water in that wave, that is, how many in-
ches will it move per fecond?·
By the laft queition, when the waves are 39,2 inches broad the was
ter will undulate in one fecond of time, then becauſe wayes like pen-
dulums are moved by gravity, it will be as 39,2 : fq
the fquare of the required velocity, that is, : 39,2,
inches per fecond anfwer.
39,2 io is 18 to
18 : = 26,56
:
224 THE UNIVERSAL MEASURER
Queſtion 250. If A B (fig. 227) be a circle made about any heavy
body whoſe center of gravity is at C in the center of this circle, and
if about this circle's periphery a ftring be wrap'd and one end S thereof
be faſtened to a pin, and the body let go, how far will it defcend in
the firſt ſecond of time, by unwinding itſelf about the ſtring, the radius
A C being 10 feet?
Let O be the center of ofcillation to A, the point of ſuſpenſion, A B
being parallel to the horizon, then (by queſtion 122) A Od+
0,4 rr
=r+0,4 r (becauſe in this cafe d=r) = 1,4 r = 14
d
r = 14 feet.
Now when a body ofcillates, its whole force or weight is in O, (art.
291) and ſo when it begins its motion, it begins to fall freely as if it
were placed in O, and at the fame time begins to whirl about the cen-
ter of gravity C, now becauſe A, C, O, are ſtill in the fame right line,
the body conftantly whirls in the fame manner, and therefore as A O
is to A C fo is the velocity of the body freely falling to its velocity
when whirling, ſo is 14 to 10, whence, as 14 is to 16 (the defcent of
gravity in the firſt ſecond of time) fo is 10 to 11 feet anſwer.
In this queſtion the body is fuppofed to be a globe with the ſtring
about its greateſt circle; in any other body the fame is true, by look·
ing upon the body as fufpended at A, at the unwinding of the ſtring
and finding the center of ofcillation O, to that point A.
Queſtion 251. Whether will a cylinder 36 inches long, 20 inches
diameter and 15 lb weight defcend faſter, by unwinding a cord from
about its middle (or rowl as it falls,) or by having the cord wrap'd
lengthways and fo turn end over end as it defcends, and in whether cafe
will it ftretch the cord moft? See art. 292.
By the laft queftion, as of 36 is to of 36, viz. as 24 is to 18,
or, as 4 is to 3, that is, is as its velocity turning endways and as
of 10 is to 10, viz. as 12,5 is to 10, or as 5 is to 4, that is, is as
its rowling velocity, therefore, as is to fo is 16 to 15, fo is the
rowling velocity to that turning end ways in the fame time. Secondly,
fince the weight &c. of any body lies in its center of gravity, the body
in this cafe may be confidered as fupported at the points A and O, and
then (by art. 272) CO and CA, will be as the forces in A and O,
and therefore, as AO: CO:: weight of the body: force at A, which
is equal the tenfion of the cord S A, and is (per laft queft.) always the
fame; whence, (24—18) § and (12,510) 2-5, that is, as 6 is to
15
I
15
15
2,5, or as 12 is to 5 fo is the tenfion of the ftring when the body turns
endways to its tenfion when the body or cylinder rowls.
AND MECHANIC.
225
Queſtion 252. If a pair of wheels were placed at the top of an in-
clined plane perfectly ſmooth, whether would they ſlide or whirl
down it?
If in the two laft queftions, inftead of fuppofing the motions to bę
1 we fuppofe them to be down an inclined plane, its manifeft the re-
fult will be the fame, and the place of the cord will be fupplied by the
friction of the wheels, if the plane be rough, or by teeth in them if it
be ſmooth, whence, as a body does not whirl in freely falling, its plain
the wheels will flide down this plane, and that the reaſon of carriage
&c. wheels whirling, is owing to the friction or roughnefs of the
ground, &c.
Queſtion 253. If two weights P and w (50 and 70) one at one end
of a rope, and the other at the other end thereof over a pulley be let
go at liberty, with what force is the
rope rais'd?
1. If w act with a force = F, and P with one f, then its plain that
F f is the force with which w is urged towards the horizon, whence
becauſe both bodies (by reafon of the rope) acquire the fame velocity
the motion generated in w alone, or in that part of the whole expreff-
fw
but if the rope were not, the mo-
Ï' qui W
tion of w, would only be F, therefore the lofs of motion by the action
ed by
W
P+w
will be
Fw.tw
of the rope is F
P + w W
FP + fw
P ÷ W
for the force ftretching
I
[
the rope, or that fufficient to caufe the faid lofs of motion, and if F =
w and f➡P, then 2 P w
P+W
=58 the faid ſtretching force, which dou-
JP-PR
bled (becaufe this force is the fame at each end of the rope) gives
4 Pw
P+w
116, the weight on the axis of the pulley while the bodies
are in motion.
1
2. If instead of the bodies moving in direction perpendicular to the
horizon, one of them (as w) moves along an inclined plane BO (fig.
211) by the perpendicular defcent of P, the rope over a pulley at the
angular point O, and the height B E of this plane be to its length B O
as a to b, then by the property of the inclined plane we'll in a cafe of
equilib have a w➡bP, and therefore
aw
force F, by which w tenda
Ff
226 THE UNIVERSAL MEASURER
to move down the plane OB, which written in
FP+ fw
>
P + W
for F,
F w
and P for f, it becomes
X:
P+W
a+b
b
: for the tenfion of the
rope in this cafe, where if the angle OBE of inclination be 30°, then
as b; a:
;a::
: 1, therefore
P W
a+b
X:
P+w
b
P w
x 2 =
P W
T
43, the tenfion of the rope, and doubled is 87, the weight born by
the pulley in the time of motion.
Sax:bw-a P:
3. Since (queſt. 246) v =
= 4 feet, in the
bx:bwa P:
firft fecond of time, that the body w would defcend in direction to the
horizon and P afcend the fame diftance directly upwards, or the con-
trary, therefore as 4 feet: 1 fecond :: any perpendicular height
;☐ time in feconds, of moving thro' that height, and in cafe the bo-
dies hang at liberty over a pulley, then a = b, and fo v=
2 feet, in the first fecond of time &c.
S w
SP
w+P
Note. S 16 feet the defcent of gravity in the first fecond.
4. If the weight w is to be drawn up the inclined plane BO (fig.
211) by the power P defcending perpendicularly at the other end of
the rope over a pulley at O, in the leaft time poffible, let b B O the
plane's length, and a EO its height, then fince as above, the force
aw bp - a w
tending to move w being
a w
as
we have P
=
Б
b
the force f by which the bodies are accelerated,
sb
✔ (theo. 151) we have t∞ (b being = 1 and
T
: b p
b
a w:
b
for
and fo from t of
b Paw
b
=f)
which being made a maximum (b variable) we get
b P2 aw, now if b, be as radius 1, then a will be as fine Linclina-
tion O B E, then P2 a w, or
P
2 W
=a=0,3571=fine of that angle
&c. If a b. then the weight w, and power P are at liberty over a
pulley, and then P
2 w, whence, the power muſt be equal to twice
the weight to raiſe it in the leaft time poffible but (queft.246) p=2,417
w, when the momentum is greateft. Hence, in any combination of
pulleys, if it be as velocity of w: velocity of P :: 0,414 P ; w (or ::
P: w) then will w have the greateſt momentum (or be raiſed in the
AND MECHANIC.
227
lealt time), alío by (queſt. 115) P = 1.5 w, when the machine per
forms the greateſt effect with the moſt eafe or freedom, but if the mo
tion of the machine be uniform then (queft. 150) P=2 w, when it
performs the greateft effect.
Question 255. If the breadth of a houſe be 24 feet, what muſt be
the height of the roof above the eaves, that water &c. may be the
leaft time poffible in running down it?
Let IH (fig. 146) = 12 feet, half the breadth of the room, and A Í
fine
the height of the pike, A H being the flope of the roof, put e
LIH, the inclination of the roof with the horizon, which is alſo
fine LIBH, becauſe L BHA 90°. Let a = 12 = IH, and d
= 16 feet, the defcent of gravity, then (by theo. 172) ✔A B will be
as the time of falling down A H, therefore by trigonometry,
=
a
: a :: radius 1 :
√: 1-ce:
a
:: radius 1 :
1:
:
a
CCN I
ev: 1- ee:
e e
=AH, and as e: (AH)
Ice:
2
W: i
ee:
A B, but as /c: I fecond ::/: A B :
: time of falling thro' A H, which by the queſtion
mull be a minimum, fo (by theo.
fine of 450 whence A I muit be
theo. 172) for fince the times of
we have e
=
natural
148)
IH, 12 this is alio evident (by
falling down A B and A H are equal
thefe times are leaft when AB is leaft in refpect of IH, and that will
be when I H is = A B.
LAB.
3
Queſtion 256. What must be the inclination of a plane with the ho
rizon, that a heavy body fliding, or rolling down it, may ftrike an ob-
ject with the greateſt force poffible, the object being perpendicular to
the horizon? Fig. 163.
Let e fine LC AB, which the plane C A makes with the horizon
A B, radius = 1, then : 1 ee fine L A G H let AP be the ob-
facle perpendicular to A B, now (by theo. 157) if GA express the
force of the heavy body, then H A, will be it's force against the ob-
ſtacle P A, but (by theo. 170) the velocity, or force G A, is as ✔✅G H,
or as e; therefore, by trigonometry, as radius 1 : √e (AG):
V:-ec: (fine LAGH): vex√:1—ee:= Vie
:e-eee:, =
the force in direction H A, now if this expreffion be a maximum, then
(by theo. 148) e- 3eee 0, or e=
natural fine of
35° 16' for the LGA H, as required.
Question 257. A beam B a (g. 156) is to be fupported in a given
pofition (La BP) by a prop C m, of a given length, infifting on a ho
rizontal beam B A fupported at its ends A and B, what muſt be the
228
THE UNIVERSAL MEASURER ·
;
pofition of the prop C m, fo that the beam A B on which it ftands may
be the leaſt ſtrain'd, the beam being of equal thickneſs ?
Suppofe B n perpendicular Cm, then (by theo. 191) the ftrefs or
and therefore the force tend-
preffure upon
the prop Cm is as
ing to break A B in m, will be as
I
В n
I
but (by art. 298) the
В m
I
(ftrength being inverſely as
ftrefs), whence as
I
is to
Bmx Am
Ifo
B m
ſtrength of B A in m, is as
Bmx Am
fo is 1 to A m, but (by
the queft) this ratio 1 to Am viz. A m is to be a minimum, confe-
quently B m mult be a maximum, but B m is to fine LB Cm as Cm
is to fine LCB m, whence Bm is greateſt when fine L B Cm is ſo,
and that is when it is = 900, or the prop Cm is to Ba.
Question 258. What is the fide (a) of a cube of wood, which
fwimming in freſh water is i inch (e) above the water's furface, but
in fea-water 2 inches (z) dry, or above the furface?
Here aa Xa- t=
the wet part in freſh water, and a ax a
the wet part in falt water, and (by art. 338) 100 aax a
Z
e =
1000=206
103aa× a—z (100 being to 103 as the denſity of freſh water to
that of falt water), which reduced gives 3 a = 103 2 —
100 = 106, ſo a = 106 ÷ 3 = 35
fo
inches anfwer.
Question 259. If an upright veffel conſtantly filled with water 9
inches deep, have a hole at the bottom which runs at the rate of 3
gallons per fecond, what length muſt a tube be, fixt to the hole and
of the fame bore therewith, that the water voided at the tube's bot-
tom may be 4 gallons per fecond. Becauſe the diameters of the hole
and tube are equal, the quantity of water roided will be as its velocity
at the hole and tube (viz. at the top and bottom of the tube, which velo-
cities being as the root of the heights fallen thro', it will be as 3 gall.
is to 3 fquare root of 9 inches the water's furface above the hole) fo
is 4 gall. to 4 (the fquare root of 16 inches the faid furface above the
tube's lower end) fo 16 — 9 = 7 inches the tubes length anſwer.
Hence it appears that a veffel will be fooner emptied thro' a pipe
at the bottom, than thro' a hole there &c.
Question 260. If a cannon ball in latitude 530 be projected direct-
ly upwards to the height of 2640 feet, where will it fall? Fig. 229.
AND MECHANIC.
229
t
1. The time of the balls afcent and deſcent being equal it will be as
16.1 feet fecond :: 2640 feet: 163,98, whofe fquare root
doubled, is 25,6 feconds the time ball is in motion, now in that time
the cannon will be carried thro' a ſpace of 23666 feet, by the rota-
tion of the earth about its axis, thus as radius is to co-fine of any la-
titude fo is the fpace moved by any point at the equator to the ſpace
moved by a point in that latitude in the fame time.
2. Let E T B, be an arch of the earth in latitude 530 q the earth's
center, then if the ball is projected at E, the point E in 25,6 ſeconds
will be carried to B, over the arch EB 23666 feet, now the ball
(carried by a motion compounded of the earth rotation and its pro-
jectile motion) will defcribe an ellipfis whofe focus is at q, and bodies
in motion defcribing equal areas in equal times, the elliptic area E A tq
circular area ETB q, or area t A Et area tq B, but becauſe
TA is very ſmall in refpect of Eq=tq, the area t AEt, may be
taken for that of a parabola, fo let b=Eq=21000000 feet, the
earth's radius, d= 23666 EB, h=TA= 2640 feet, 2tB
the required diſtance, then bahx:d-a:, fo 3ba4hd
4 hd
4 hd
4 ha, whence a =
nearly =
3b+4h
3 b
•
3,967 feet that
the ball will fall wefterly from the gun, becauſe the earth moves eaſt-
erly.
Σ
3. But if the ball were fufpended in the air directly above the gun
&c. at the height of 2640 feet, and there let fall, (ſuppoſing it to fall
16,1 feet in the firſt ſecond of time, and the earth to be a globe &c.
as before) it would trike the earth at of 23666 feet weft of the gun
&c. becauſe in this cafe the ball would not be affected by the earth's
motion as when projected from the earth for then the earth's motion
is communicated to it before it be projected, and fo confequently it
muſt move by the joint effect of theſe two motions or velocities.
The 32 following queſtions fhews the nature of circular motion.
Question 261. Whether will a fling be more ftrain'd, in whirling a
ſtone about at the rate of 2, or 5 feet per fecond?
v— 2
Let CA = CQ = a = the length of the fling (fig. 228)
CACQ
and u=5, and one end of the ſtring faſtened to the center C, with
with the ſtone at A the other end of it, then if the ftone (being at reft)
were ftruck by any power or force it would move in a right line as
A H, were it not for the ftring which pulls it out of the faid line or
tangent, a diftance HE Ad, and fo caufes it to move thro' the
*
230 THE UNIVERSAL MEASURER
arch A F, inſtead of the diſtance A H, and thus the ſtone conftantly
endeavours to fly off from the center which is called the centrifugal
force, with this force it acts on the center C, and this is called the
centripetal force, thefe two together are called the central forces of
the revolving body, alfo A H the space which the body A moves thro'
in a conſtant particle of time by an uniform velocity is called the pro-
jectile force, or velocity of the body, and if this AH be fuppofed very
imall hen the arch A E or its chord are each equal A H, join QE,
fo wil the QE A be a right one and by tinalar As, as à Q (a) :
AE::AE:
a
Ad, the central force, whence, if A £=
the projectile force = u = ·5, or = √ = 2, then we have as
U U VV
that is, as uu: vv, or as 25: 4 :: the firefs on the firing when the
ftone moves at 5 feet, to its thiets when the velocity of the one is at
the rate of 2 feet in the fame time, thus the greater the central force
or the more the ftring is ftrain'd, the falter the tone is whirl d about.
Qufion 262. What must be the velocity, and time of one revolu-
tion (called the periodical time) of a body muring round a circle when
its central force is double to its gravity, the circle's diameter being (a)
60 feet?
Let Pthe periodic time, v the velocity and f = the central
force, c=3,1416, then (per laft queft.) as AE: I::ca:
ca
A E
==
P, the time of defcribing the whole circle, the motion being uniform,
ca
ссаа
therefore P=
3,1416 a
A E
ca.
whence v = : Ca, fort =
but f= (laſt queſt.) whence f=
ссаа
P
$
PP
P Pa
сса
PP
or f is as
2
a
(becauſe c c is conftant) hence f is as and as
that is, the
PP
F F
a
pp
central force is directly as the fquare of the velocity and inverſely as
the fquare of the periodic time in the fame circle, and therefore PP vý
— a a, or P v = a, again, as ca: P::AE;
A EXP
v p
Et the -
са
fecond is to
са
time of defcribing A E, and (per falling bodies) as
16,2 feet ſo is (tt)
v v P P
ccaa
to the diſtance fallen in the time t, by the
AND MECHANIC.
231
16 2 v v PP :. .: 16,2 PP
ccaa
force of gravity, hence as a
: 1 :: 16.2 PP : cca:: PP
CC a
162 P
P
сса
PP: 0,615 a :: the central
force, to the force of gravity, now by the queftion, 2 PP =0,615X
60. fu? = √ 18,5 = 4,3 feconds (nearly) for the time of one re-
volution, and as 4.3 feconds : 60 x 3,1416
X : : 1 : F = 43,8 feet per
fecord aniwer.
Quezon 263. If a pendulum 3 feet in length be made to defcribe
a conical furface, what muſt be the diameter of the cone's bafe def-
c1b2d by the bab when the pendulum makes each revolution in one
fecond of time?
y
Let A B (fig. 127) be the length of the pendulum, A B≈a,
the diameter fought, v D= =h, the cone's height, c and P, as per laft
queit. now as the 900, or revolving body. endeavours to leave the cen-
ter D, by the central force in direction D A, it is compelled towards
it by the force of gravity in direction parallel to v D, but by the laſt
queftion, the central force is to that of gravity as
cca
cca
fore as
10 2PP
: 1, there-
16,2PP
•
ža (DA) : h (v D) ergo &a=
ccah
or
16,2PP
16,2 PP2cch, and 'per falling bodies) as 1 fec.: 16,2 feet ::
PP: 16,2 PP, the distance fallen thro' in the time of une revol, of the
pendulum, which call d then from the two laft equations we have d
16,2 PP
2cch, fo`h
d
2 CC
16.2
=0,82 feet =
19,72
D√: OPA - [] ;D: =DA =
Γ □ D:=DA= 2,9 fere, fo A Ba= 5,8
feet an wer.
Note. By the queft. the distance d muſt be 16,2 feet, being the
diſtance defcended in the given time one fecond
Queſtion 264. If a pendulum be made to defcribe a conical furface
whole axis is to its diameter as 9 to 2 (h to a) in what time (P) will
the pendulum mass oné revolution ?
From the laſt queſt, we have 16,2 P P=2cch, ſo P=cy:
h
1416 4 : BATH! := 3,1416 × 1,05 = 3.3 feconds nearly,
Let the ameter (a) of the core's baſe be what it will, P
will be equal 3, 3 feconds nearly, if the axis à be
3,3
by the lait quethon.
9 feet, as is plain
232 THE UNIVERSAL MEASURER
Queſtion 265. If a pendulum 3 feet long (a) be whirled about in
(P) 2 feconds of time in a horizontal direction, what is its gravity to
its centrifugal force?
1. As I fecond : 16,2 feet ::☐ 2 feconds: 64,8 feet, the force
of gravity generated in the time P of one vibration of the pendulum,
cca
and (by queſt. 262) is the central force, fo by this queſtion
as 64,8: (cca)
PP
PP
6 x
3,1416
4
14,78: 162: 37 nearly, :: gra-
vity central force, that is, if the pin on which the pendulum hangs,
be pulled with a force of 37 in direction of the pendulum, it will be
pulled by a force of 162 downwards in direction perpendicular to the
horizon, &c.
time
Question 266. If a pendulum, or cord 2 feet longa, with its
bob, or ball 3 ounces weight w, be whirled about in half a feconds
P, fo as to deſcribe a circle whofe radius is the faid cord, with
what weight or force f, will the cord be ſtretched? See queft. 262.
1. When gravity is 1, the central force is
T
a
сса
16,2PP
0,615 a
PP
— (becauſe a here — radius) 232, therefore, when gravity=w,
PP
the central force f ==
1,23 a w
PP
1,23 X 2
0,5 X 0,5
3
= 29,52 oun-
ces' anfwer.
Queſtion 267. If a cord 2 feet long (a) with the weight of its bob
3 ounces w,) be whirled about fo as to defcribe a conical furface in
half a ſecond of time (P), with what force (f) will the cord be ſtretch-
ed? See queſtion 263.
I Let h
the axis of the cone, r the radius of its baſe, then
gravity being as h and the central force as r, the tenſion mull be as
a, the fide of the cone, or length of the cord, now 16,2 P P
2cch,
I
we have 1,186 ✔h
P, or 2,376 √ h = 1, ſo h
5,665376
a w
6
0,1767, then as h: w:: a:
33,89, the ten-
h
0,1767.
fion required, in ounces.
2. √: aa- -hh:
1,93, whence in this cafe, the ftrefs on the
pin to which the cord is faſtened in theſe 3 directions, viz, of the cord,
perpendicular, and parallel to the horizon, are as the numbers (a) 2
(h) 0,1767 (r) 1,93.
AND MECHANIC.
233
3. If a 39,12 inches = 3,26 feet, the length of a fecond's pen-
dulum, and we put gravity 1 = 1,23 ª we'll find P =√: 4,0098:
PP
a
= 2 feconds, fere the time the pendulum m kes one revolution, or
two fwings, then 1,23d = f = 1 (when a = 3,26 and P = 2 )
PP
the force of gravity, and hence, if a pendulum is any how moved by
gravity alone, the ftrefs on the pin to which the pendulum is faſtened,
is only that cauſed by its own weight the fame as if it were at reſt.
Question 268. If the diameter of the equator or greateſt circle of
the earth, be 42000000 feet (a) in what time muſt the earth make one
revolution about its axis, that all bodies there may loſe their weights,
and be as liable to fall off as to abide thereon?
This muſt be when the forces with which bodies cleave to, and re-
cede from the periphery of that circle are equal, viz. when gravity
is = centrifugal force, fo (by quest. 262) PP 0,615 a, therefore
P√/0,615 a 5083" or 84′ 43″ anfwer.
=
=
=
Question 269. What is the centrifugal force at the equator to the
power of gravity there allowing the diameter of that circle to be
42000000 feet (a) and the earth to turn round its axis in 24 hours or
68400 feconds (P)?
1. (By queſt. 262) As PP: 0,615 a:; 289: 1 nearly, :: gravity
centrifugal force anfwer. But in other latitudes, fee the follow-
ing queſtions.
Question 270. If at the equator the gravity be to the centrifugal
force as 289 to I, what will it be in any other latitude fuppofe in the
latitude of 600 (fig. 228)? See queſtion 124.
Since the time of revolution of a body at the equator AQ, and
any parallel of latitude b B, is equal, it will be (by queſt. 262) as cen-
AQbB
trifugal force at AQ: that at b B::
:: AQ : b B,
PP
PP
AQ¸¢= 1b B, and f =
centrifugal force at
cf
centrifugal force at
let a =
A Q, then by this proportion we'll have
a
Bb, but in any latitude b B, becauſe the direction of gravity is always
towards C, the earth's center, therefore the centrifugal force is not
(as at the equator) oppofite to the whole gravity, for produce a B the
direction of the centrifugal force to D, and B C the direction of
Gg..
gras
234
THE UNIVERSAL MEASURER
vity to F, from D upon C F let fall the DF, then if BD be the
whole centrifugal (f) force, we have B F for that part of it which acts
directly against gravity, and by fimilar As, as B F: BD::AB: BC
=EC, therefore as a ; c :: cf. ccf F the centrifugal force at
a
aa
Bb oppofite to gravity, whence if a radius =1, then c 0,5 =
co-fine latitude 600, then as a a : cc:f: F, that is, as I: 0,25 ::
I
289×4
21:
:: centrifugal force A Q, to centrifugal force at B b,
which oppoſes gravity there. Hence, as (cc) the fquare of the co-
fine of any latitude is to 289 fo is the centrifugal force, to the gra.
vity there.
Queſtion 271. If the force of gravity at the equator be 1, what will
it be (G) at the poles, it being proved by obfervation that a ſecond
pendulum at the equator, is 39,1 inches long, and at London, 39,2
inches? Fig. 228.
:
Produce CF to G, and draw G D perpendicular Da, then if B D
be the whole centrifugal force at the equator, and B G the whole
decreaſe of gravity cauſed there by it, and fince (by the laſt queſtion)
in the latitude B, the centrifugal force which acts directly against gra-
vity is F B, the line F B will alfo exprefs the decreafe of gravity in
the latitude B, fo the difference G F will be the increaſe of gravity at
B, fo by ſimilar As as BG: FG :: BG: DG :: □ B C : Q
a C ::□radius: fine latitude at B, :: decreaſe at equator in
creafe at B, now (by queſtion 116,) the gravities at places are as the
lengths of pendulums vibrating feconds there, that is, as 392 391 ::
gravity at London : gravity at the equator, and by the laſt proportion
as I (radius 1): 6131 ( fine latitude 51° 32') :: GI (de-
creafe gravity at equator): 6131 G 0,6131 (decreaſe of gra-
vity at London) to which add the gravity at the equator I and we
have 0,6131 G +0,3869 the gravity at London, fo as 392 391::
0,6131 G +0,3869 : 1, whence G 1,0043 &c. for the gravity at
the poles, N and S, that is, as 229 230, or nearer, as 689: 692 ::
gravity at equator: gravity at poles :: leaft gravity on the earth's fur-
face to the greateſt thereon.
:
Queſtion 272. If any two equal bas'd cones of earth &c. with their
vertexes in the earth's center, and baſes in its periphery, be in equilib.
with each other, what is the equatoral diameter A Q(fig. 228) to
the axis NS, allowing the earth to revolve in 24 hours about NS?
The centrifugal force being greateft at A, or Q, and nothing at
H, or S, and this force acting at A, directly againſt gravity whoſe ſeat
AND MECHANIC
235
of action is at B, its plain, that a column of particles C A, must be fo
much heavier´ (viz. fo much longer if the matter is the fame) as to
make up that part of gravity loft at A, by the centrifugal force there,
which force does not alter the column of particles C N. Whence (and
by the laſt queſtion) as 689 : 692 :: the axis NS: diameter A Q
which proves the earth to be an oblate fpheroid flattened at the poles
N and S, and rais'd at the equator, or middle A Q, and fince A Q
is about 8000 miles, it will exceed N S by about 35 miles?
queſtion 274.
See
Question 273. If the earth were not to revolve about its axis, whe-
ther would pendulums gain of lofe time, and how much per day in
the latitude of 54° 30′, fuppofing the earth a globe?
If c the natural co-fine of the latitude 54° 3c', then (by queſt.
269) if the gravity at that latitude be 289, the centrifugal force there
will be cc0,33721249, ſo 289 -0,33721249 288,6627875,
that is, if 289 be the gravity when the earth is at reſt, 288,6627 &c.
will be the gravity when it revolves in 24 hours in latitude 54° 30′,
now (by queſtion 116) the lengths of pendulums being as the gravities
and the vibrations as the fquare roots of theſe lengths it will be, as
√288.6627875: 289: 86400 (feconds in 24 hours, earth being
in motion): 86451,83 (feconds when the earth is at reſt in 24 hours)
fo the difference 51,83 feconds is the time gain'd per day.
Question 274. Whether or no is it likely that the earth is of an uni
form denſity throughout ?
1. The earth at the creation being in a fluid ftate, it is reaſonable
to fuppofe that the heaviest matter fubfided firſt, according to the laws
of gravity, and therefore that the earth is more denfe and compact the
nearer we go to the center; whence it follows, that the centrifugal
force (queſt. 269) may be forewhat more than 289 viz, becauſe there
are more particles between the equator A, and center C, (fig. 228)
than there would be if the earth were of an uniform denfity as is there
fuppofed, therefore (queſtion 272) 689 : 692, is nearer the ratio of
NS to AQ, than that of 229 to 230, and alſo anfwers nearer by ex-
periments and this is the form of the earth, as fettled by Sir ISAAC
NEWTON.
Question 275. Whether or no doth bodies near the carth's furface
gravitate in directions, that pafs thro' the earth's center?
1. Its evident, that if a heavy body be let tall any where near the
earth's ſurface, it will ftrike the horizon perpendicularly, now the ho-
rizon is a plane or tangent line, touching the furface in the point where
the body ſtrikes it, and therefore the earth being a ff-heroid, this line
236 THE UNIVERSAL MEASURER
or direction of the body being at right angles to the tangent in the
point of contact with the curve, would not pafs thro' the center, ex-
cept at the equator and the poles, for at thofe places the faid tangent
is at right angles with the diameters of the generating ellipfis Hence
any other place, fince the earth's center is the center of attraction
to which all bodies from its furface would fall (ifnot interrupted, it
follows that the path is not a right line but a curve to which the line
of direction of the body is a tangent, to the point where it would firſt
begin its defcent.
ia
Queftien 276. If the direction of a falling body in latitude 450 were
continued thro' the earth, how far from the center would it cut the
axis, the equatoral diameter being about 8000 miles, and the axis
7955 miles?
45° the
1. Let E be the given place (fig. 229) in latitude 450, b E the di-
rection of the body at right angles to the horizon HO, which bE
produced will meet the axis NS in G, now LOHS is
latitude, fo L E G H will be co-latitude, whofe natural tangent
call t, to the radius 1, let c NS and a the equatoral dia-
meter Q_A, and e =NF, alfo, puta =zz, then, by the property
=
сс
of the ellipfis F E=z: 2ce-ee: and F G = zz X: c - e,
but by trigonometry, as Radius 1 : LEPF::FG: FE, fotz z
x:c-e:= 2:2ce-ee, or tz X:c_e: =√: 2 ce ce:
which fquared is ttzz: xcc ceee:=2ce-
2 ce- ee, by tran-
fpoûtion ttzz cc=2 ce+2 ttzzce-ee-ttzzee=2Sce
-bee, by writing Sc+ ttz z and — b — — I ttzz, which
equation by compleating the fquare &c, gives e =
ttzz C C
b
:
Sc
+v:
SSC C
b b
NF, but the folution will be eafier if we put e=
FC, and a, c, z, t, as before, for then we'll havettzzeecc-ee,
C
3982-5
fo e = √itt22+1 = √1,009+1:
3082.5
=2807=F C,
1,419
now FG-FC=zze e = 2843 — 2807 = 36 miles = CG anſ.
Queſtion 277. In what latitude will a right line drawn from the
point of iufpention of a plumb line and continued thro' the earth's cen-
ter, make the greateft angle poffible with the faid plumb line, fuppof-
ing the equator diameter to the axis as 230 to 229 ?
Let the point E (fig. 229) be the required place, fo will EG (being
horizon H O) reprefent the plumb line, join CE, fo is LCEG
AND MECHANIC.
237
that made by E G and a line E C paffing thro' the earth's center, let
e CF, and make it SCCA :: 1:1÷0 (riz. : : 01:0
=
115
114.5
) then (by the properties of the ellipfis) EF=÷1¬¤xil—ee:
and F Gene, therefore CEFE + □ CF :=
V: 1+n-nee: and GF=√1+0:x:1+aee: then by fimi-
ar As, as GE FE::CG: Cy (Cy being LEG)=
ne vi- ee
V: I nee
and by trigonometry as CE: Cy:: radius 1: fine LCEY =
✔
ne VI - ee:
I nee In-nee
+
་
I
which by the queft. must be a maxi-
mum, and then e will turn out = √.7071067 — CF, whence
FG=1+n:xe=713328. FE,7101241. CE-1.0021857,
Cv =,0043878, Ls FEG 45° 7′ 39″, FGF = 44° 52′ 21″ the
co-latitude and CEG 15' 3", the angle of the plumb line, in lati.
tude 45° 7′ 39″.
Queftion 278. In what time at the equator would a body by the
force of gravity, fall freely from the furface of the earth to its cen-
ter?
Becauſe (fee queſt. 268) when the earth revolves in 84′ 43″ time,
the power of gravity is thereby deſtroyed, it follows that while the
earth would at that rate make of a revolution, that a heavy body
would fall from its furface to its center, therefore 84′ 43″′÷4=21′
II" anfwer.
Quelt on 279. If the axis of the earth be 7940 miles, and the dia-
meter of the equator 7974 miles, in what time (in latitude 45° 14′)
would a heavy body by its own gravity freely defcend in a right line
from the earth's furface to its center? Fig. 229.
Let c CN, a=CA, CF-e, F = 21' 11", the time of defcent
at the equator and t = time of defcent from E the given latitude along
EC, which muſt alſo be the time of revolution in the circle whofe
radius is CE, (ſee the lait queft.) now the force being inverſely as
the diſtance from the center, it will be as CA: CE:: force with
which the body defcends from E, to that with which it defcends from A,
which forces are as the centripetal forces, and becaufe (by queft. 262)
pp=f, or P = √, it will be as
Fit
a
PP
a
CA
CE
::T
CE
CA
238 THE UNIVERSAL MEASURER
a a
: t, but by the property of the ellipfis, 2x: cc-ee EF, fo
c c
= □
DEF+OCF=CE, that is, aa-
aaee+ccee
C C
=OCE,
ccaa
put d
=
CC
then CE = √: aa - dee: whence the laſt pro-
✔a
portion becomes, as aandeel:
=t=20′ 11″ anſwer.
a a deel 4
T:aa-deel
::T:
a
Note. Í here take the ſurface of a ſpheroid, to be the fame with
that at the furface of a globe, for the difference is fmall.
very
Queſtion 280. It is found by obfervation that the moon revolves a
bout the earth in nearly 27,3 days, and that her mean diſtance from
the earth's center is to that from the earth's ſurface as 60 to 59, now
if for the revolving body or ſtone mentioned in queſt. 261, we place
the moon, for the center we place the earth, for the centripetal force
or fling, the earth's power of Attraction, and for the projectile force
the Almighty power of God at the creation, what will be the law of
this centripetal force.
PP
1. A body (fee queft. 268) revolves at the earth's furface in 5083
feconds, let P = 2362000 the ſeconds in 27,3 days the moon's perio-
dic time, a 60, half diameters of the earth her diſtance from the
earth's center, the faid body at the ſurface being 1 of theſe half dia-
meters from the earth's center, and put f the centripetal force = the
nth power of the diſtance a, viz. fa" = (fee queſt, 262), then
if n be taken ≈0, 1, 2, I, 2, &c. we'll have Pa, P = 1,
P=√, P=a, PP=a a a,&c. refpectively, this laſt equation is the law
of the whole planetary ſyſtem as conſtant obfervations have made out,
viz. the fquares of the periodic times are as the cubes of the mean
diſtances, of any of the planets revolving about the fun, or of any of
their moons about them, thus, in this queftion, as cube of 1: fquare
of 5083: cube of 60: fquare of 2362000, 27,8 days, hence f
=—, that is, the centripetal force is inverſely as the▷
2=a
1
I
aa
of the diffance from the center.
Queſtion 281. How high above the earths furface muſt a body 400 lb
weight, be raifed, to weigh only 300 lb, the earth's radius in that
place being 3980 miles (d)?
AND MECHANIC.
239
Let e the required height above the earth's furface, then becaufe
(by the laſt queſt.) the power of gravity is inverfely as the fquare of
the diſtance, it will be as 3c0: 400, or lower as 3:4::aa: a el
ſo a + c = a × √ ÷ = 3980 × √ ÷=4596 — 3980 = 616 miles
anſwer.
Question 282. If a ball 4lb weight at the earth's furface be carried
up in the air fo high that it weighs but 3 ib, with what velocity would
it begin to fall, and what would its velocity be at the earth's furface,
gravity at that place being 16,1 feet in the firſt ſecond.
Let a the earth's radius (3980 miles) in feet, c=24266880 feet,
in 496 miles the height of the ball (by the laft queftion) above the
earth's center,e any diſtance at first defcended from that height,
now the dillance defcended in the firſt ſecond of time being as the
gravity or weight at that place, it will be as 4: 16 ::3: 12yž
feet = d, the diſtance fallen in the firtt fecond, from the height 616
miles above the horizon, therefore 2d the uniform velocity at that
2 d c c
place, fo (by queft. 280) as ce|
el: 2d::cc:1=
2d cc,
C
the
velocity generated per fecond by falling thro' e, and therefore, the
velocities acquired by falling thro'e, 2 e. 3e, 4e, &c. will be 1, 2, 3, 4
>
2 dce
&c. to v, fo (by theo. 86)
and v =
2
C
e
2 dce
√ic-e:
now
dce
ife be taken =3252480 feet in 616 miles we'll have v=21:
=
13310 feet per fecond anſwer.
c-e
Question 283: Things being as in the laſt queſtion, in what time
would the ball be in falling to the ground, the ſaid ball, here as in the
laft queſtion, being fuppofed to meet with no refiftance, but to fall
freely in vacuo ?
Let v, c, d, and eas in the laſt queſtion, then as there we have
2 dcc...
C
for the expreffion generating the velocity v, and the velocity
dce
being inverſely as the time, therefore from v = 2:
: we'll
C
.e
have r÷2№:
ace/
C e
C
: = 2√:
for the
C
2 vacxve
expreffion generating the time t, which may be put into a feries &c,
dce
240 THE UNIVERSAL MEASURER
I
e
by theo. 86, or multiplying it by
it becomes
X
c - e
2 cd
C-e
whence, if radius
Vice-ee:
= I, S the fine, and a the
arch whoſe verfed fine is 2 €÷c, we'll have t = √x:S+a:
now=1418, and 20,26795, anfwering to an arch of
C
42° 56′ whofe length is 0,7494, and that of its fine 0,6811, hence,
0.73940,6811
× 1418 = 507 feconds = 8′ 27″ anfwer.
4
Question 284. If a folid foot of fome fort of metal weigh 1000 lb,
(b) at the earth's furface, and a cylinder of that metal 3252480 feet
high (e) and bafe 5 feet area (A) be fet upright on the ground, what
weight (w) will it weigh, the earth's femi-diameter at that place be-
ing 21014400 feet (a)?
1. The weights of bodies being inverfed as the fquares of the dif
tances from the earth's center, (fee queſt. 280) and b A being as the
weight at the ſurface we have as a + el
1
bA::aa:
b A aa
expreſſion generating the weight, now this expreffion b Ax
a + el
a a
the
a+el
compared with what is done in prob. 185, part 1ft. will give b'Ax
=w=8444000000 lb nearly answer,
a e
ate
Queſtion 285. Required the weight of a cylinder fet perpendicular
to the horizon, whofe bafe is 3 feet area (A) and height (e) infinite,
and a folid foot of metal of that cylinder 1000 lb (b) at the earth's
furface, the earth's femi-diameter at that place being 21014400
feet (a)?
1. From the laſt queſtion, we have w➡bA×
ae - bax
e
now if e be taken infinitely great, then a in the factor
e
ate
ate
may fafely be taken = 0, becauſe an infinitely great quantity, cannot
be made more by addition or lets by fubtracting a finite quantity from
it, and therefore
e
e
=
===
1,
whence w
bA=
a+e
ote
63043200000 lb anſwer.
AND MECHANIC.
241
I
Question 286. If a heavy body fall from an infinite height (freely
as in vacuo) to the earths furface, what velocity will it have there?
Let a 21014400 feet the earth's radius, d=16 feet, the force
of gravity there, viz(the distance freely fallen thro' in the first fecond
of time at the earthis furface) e the infinite diftance of the body
above the horizon, then (fee queft. 282) as a +ej: 2d :: aa:
which (fce the lait queft.) gives the velocity
2 da a
ate
V:
e
2:dax:
: 24'da = 36768 feet per fecond anfwer. Hence the ve-
a + e
locity of any body, acquired by falling to the earth's furface can ne-
ver exceed 35768 feet per ſecond, be the body ever ſo heavy, fall ever
fo far and with all the freedom pollible.
Queftion 237. Where mutt a tub of 20 inches diameter, be placed
to hold the most liquor pothible, and how many ale gallons will it hold
more at that place than at the earth's furface?
1. The figure of the earth, viz. land and water being nearly fpheri-
cal, and by reafon of the earth's largenefs, aveifel 20inches diameter,
viz, the area of fuch a diameter may at the earth's furface, belooked
upon as a plane, and fo to hold no liquor, but this plane being placed
at the earth's center, and liquor poured upon it, there the liquor will
rife to the form of a half globe before any of it run or, 10 (by ex. 294)
928 gallons, fuch a vefïel would hold more at the earth's center than
arits furface, anſwer.
Question 288. Required the ratio of the diameters, and bulks of the
fun, earth and moon? Fig. 230.
60,
1. Let E be the earth, m the moon or any other planet, then is the
LmEd, the moon's apparent femi-diameter, as feet from E, the
earth's center, and would be the fame if feen from its furface at D,
(the moon being in the horizon) becauſe DE | dm, this ▲ by obfer-
vation, at a mean diſtance of the moon from the earth, is 1876,5
half ſeconds, and (by question 280) the diftance E m is found =
fo by plane trigonometry as radius 1:tLdEm938,25":: Em60
: dm0,27523, parts of the earths femi-diameter becauſe E m is 60
of fuch parts, therefore as 1 : 0,27523 :: the earths diameter to the
moon's diameter, which is nearly as 109: 30. Again, if we fuppofe
m to be the fun, the LdEm is found by obfervation, = 16′ 26″ the
fun's apparent radius, and (by queit. 280) Em, the diftance between.
the fun and earth will be found = 20000 fere, femi-diameters of the
Hh
242 THE UNIVERSAL MEASURER
earth; therefore, as radius 1 : tangent d Em 16' 06" :: Em 20000
: dm 91,47, fo as 1 : 91,47, or nearly as 109: 10000 :: the earth's
diameter the funs, otherwife, let R 3864, half 11 and r=3753
half ſeconds the apparent diameters of the fun and moon, A = 10000,
and a 30, their real diameters D25000 and d 60, their diſtan-
ces from the earth, then becauſe any body appears larger the bigger
it is, and lefs the further it is diſtant, we'll have as DR: dr:: A : 2
&c. hence, their diameters are as 10000, 109 and 30, and their bulks
2s 10000]³, 109)³ and 30) anſwer.
3
Note. L Dm E is called the horizontal parallax of the planet, and
is the earth's apparent radius as feen from the planet at m, this L at
the fun is about 105" but at the moon at a mean is 60', and may be
thus found, as m E the diſtance between the earth and planet: DE,
the earth's radius:: radius 1 : tangent L Dm E.
Queſtion 289. Required (Q to q) the quantities of matter in the fun
and moon, their apparent diameters being 32′ 12″ and 31′ 165""
(R and r) and diſtances from the earth D 20000 and d= 60, and
that the heights of the tides on the ocean, at the new and quarter
oons are about as 6 to 4? Fig. 230.
m
**
1. Any of the planets by their attraction will diſturb the waters or
the earth (as alfo on one another if any water &c. be on them) but the
moon by its nearneſs and the fun by its largeneſs affect the waters on
the earth the moſt, that by the planets being very ſmall on this earth
where the force of attraction is moſt from the moon becauſe of its be-
ing
fo near to it, and is always feen to anfwer in the tides, in their
conſtantly obſerving the moons fouthing &c. Now at the new moon
its evident, the fun and moon act jointly together, and at 90º diſtant
against each other, fo in the former cafe the waters rife with both their
forces (F'+f=t,) and in the latter, with the difference of thefe forces
(fF 4) whence as F:f:: 1:5:: force of fun : force of moons
nowe = DE, ttangent LdEm half of 31′ 16″, and N➡
10000 the denſity of the fun to n, that of the moon, then as tLdEm
2
: radius 1 :: dm : Em, whence dt = dm, fo dmxn=nd ddttt
&q,
q, which (ſee queſt. 280) divided by del□ Sm, gives
nd ddttt
3 =
nd ddttt
dd-2detee
3
d+2e+ zee
d+ dd
4 eee
+&c. for the force with which the moon
at the point S, from which taking
attracts the earth's furface
nd³ fs
3
nd f³ the force with
dd
AND MECHANIC.
243
e
which ſhe affects the center E, there leaves ntttx: 2+ + 2e &c,
∞f which becauſe e is very ſmall in refpect of d, we may leave 3e
u
f, and by the
out and then ent 3 f, but t is as n, fo en rrr
fame way of reaſoning we'll have e R³ α F, whence as e NR 3
en r³, or as NR3: nr 3 :: F: f, and therefore, as N;n ::
nr3
}
R R R
I
5
..
3
3
:: 10000: 48911 :: denfity fun ; dẹn-
rrr
A
a
whence as R:r::
3864 3753
fity moon, now, by the laft queflion, we have as DR: dr:: A: a,
3
a
:
A 3 3
therefore as R³: r³::
D
I'
fo as
D³ d³
A 3 N
2 n
:
&: 4 (for by art. 335, as
D3
3
n:
30°
F:f::RN: r3n:
3
3
A ³ N : a ³n :: 10000 X 10000 : 30 × 48911 :: 10000
,0013 :: quantity of matter in the fun : quantity of matter in moou,
anfwer.
Note. From hence it appears, the forces with which the planets
diſturb the earth are as the cubes of their apparent diameters, mul-
tiplied by their denfities, or as the cubes of their true diameters mul-
tiplied by their denfities and divided by the cubes of their diflances,
or as QD D D.
Question 290. What are the denfities and quantities of matter in
the fun and earth.
Let N 10000 the denfity of the fun, to n, that of the earth, Q
10000 the quantity of matter in the fun to q, that in the earth, D
20000 and d = 60, the diſtances of the fun and moon from the earth,
P 365, 65 and p = 27,3 days their periodic times. if either the dif-
tances or periodic times be given, (ſee queſt. 280) the other may be
found, then becauſe the power of attraction is inverfely as the fquares
of the diſtances, (ſee queſt. 280) from the centers of forces, it will be
е
F:f:: force of fun on the earth: force of earth oạ
D D
वे
the moon, the earth being nearly at the center of the moon's orb, and
the fun the center of the earth's (and other planets) motion, but (ſee
D d
PP PP
queft. 262) as F: f:: Ppp
hence, as F:f::
D d
PP PP
therefore as Q: 9:
DDD ddd
P P
PP
e
DD
q
:
dd
:: 10000:0,1512
244 THE UNIVERSAL MEASURER
the anſwer for the quantities of matter, now let A = 10000 and a =
109, the diameters of the fun and earth, then (by art. 338) as
a
9
0
:: Nn:: 10000: 39539, the ratio of their denfities; in the
fame manner, the quantities of matter and denfities in the planets
Saturn and Jupiter are found, becauſe theſe two planets have moons
revolving about them.
Question 291. If a body weigh 10000 lb, at the furface of the fun,
what will it weigh at the furfaces of the earth and moon?
Becauſe (by queſt. 280) the force of attraction is inverfely as the
fqu re of the distance, therefore, the quantities of matter in theſe bo-
dies divided by the fquares of their diameters, the quotients will be as
the weights of bodies on their furfaces, fo 10000 ÷ 10000, and
,0512 ÷ 109 and ,0013 ÷ 30 are as 100co and 431 and 146,
that is, the body 10000 lb on the fun, would weigh 431 lb on the
earth, and 146 lb on the moon anſwer. From thefe 4 questions it
appears, that if the diameter of the fun be taken≈ 10000, that of the
earth will be 109, and of the moon 30, and their folidities are as
3
3
3
10000, 109 and 30, their quantities of matter as 10000,0512
and .0013, their denfities as 10000, 39539, and 48911.
Note. If two meridian altitudes of the fun, moon, or any ſtar be ta-
ken by two perfons at the fame inſtant of time about 69 miles afun-
der, thefe altitudes will differ about one degree; hence, the earth
being nearly spherical, and 69 miles make one degree on its furface,
it will be as 10: 69,5 :: 360° : 25020 miles the earth's periphery,
and as 22:7:: 25020: 8000 miles nearly, for the earth's diameter,
by which the diameters diſtances &c. of the fun, earth, and moon may
be had in miles from the foregoing ratio's.
Queſtion 292. Required the ratio of the forces (F to f) of the earth
and moon to produce tides at each others furface, their denſities being
as 39539 to 48911 which is nearly as 3 to 4, (N to n) and their dia-
meters feen at each others furface viz. the moons apparent diameter
and horizontal parallax as 16' to 60′ (r to R)?
1. In queſtion 289, we have enrrr f, where e is as the dia-
meter of the body acted on n the denfity, and r the apparent diameter
of the acting body, ſo if A = the earth's real diameter to a, that of
the moon's, then we'll have Anrrr f, and a NRRR of but
(queft. 288) we have R:r:: A:
r A
R
= a (becauſe D= d) whence
AND MECHANIC.
NIC
245
r
as F: f::a NR 3: A nr³
::
гA NR 3: Anr³3 :: NRK: arres
R
Anr³::NRK:nrr:
III nearly, i. e. the earth will raife the waters 11 times higher at
the moon, than the moon can raiſe them at the earth, but its likely
there are no waters in the moon becauſe no atmoſphere is feen about
her, for if any bright ſtar is feen near her it's light is not darkened
which could not be if ſhe had an atmoſphere like our earth.
Queſtion 293. How many miles from the earth's furface is the cen-
ter of gravity of the earth and moon, (ſee prob⋅ 197 and queſt. 291.
1. Their diſtance being 60 of the earth's femi-diameter, and quàn-
tities of matter as,512 to ,0013, or nearly as 40 to 1, (1) we'll have
as 43 +1: 1 :: 60: which multiply'd by 4000 the miles in one
femidiameter gives 5854 miles, and 5854-40001854=LS (fig.
230) E, being the earth and m the moon.
6
2. Now fince the earth and moon act on each other by attraction
they must be always in equilib. upon this point L, which point must
be the center of the moon's orb. (and not E the earth's center) and
fo is that point about which the earth and moon refolves to maintain
their equilib. therefore L is that point which defcribes the earth's
orb about the fun, whence the earth
e fun, whence the earth is fometimes in her orb a a, and
fometimes out of it, but ſtill near it becauſe LS is but finall in refpect
of Em, but the center of the moon mult defcribe a much more irre.
gular curve than that of the earth, becaufe it is at greater unequal dif
tances from the fun, for the fame reaſons (not the fun) but a point
about 0,8 of the fun's femi-diameter diftant from its furface is the cen-
ter of the folar fyftem about which the fun and all the planets revolve
for if the fun and planets in their orbs were all in one right line, this
point would be their center of gravity, now the fun being always very
near this point, and its diftance from it continually varying by the
different pofitions of the planets, its motion about it muſt be very ir-
regular whilſt that of the planets being far diftant from it, will be
nearly uniform and circular.
ཏྭཱ
FINI S.
I i
[246]
DEFINITIONS
AND
EXPLANATION 5,
of the Terms uſed in this WORK.
NOTE. P. fignifies problem, to be found in part 1ft and 2d. A.
article in part 2d. Ex. example in the first 8 fections of part 3d.
And Qu. queftion, in fection 9th of the 3d part.
ABSCISSA, A. 183.
Action, and Reaction, A. 234.
Air, and its properties, A. 347. Qu. 178.
Amplitude, A 364.
Angle, A. 120. of incidence and reflection, A. 260.
of Traction, the angle made by the direction of a power with
an inclined plane.
of inclination, the angle an inclined plane makes with the
horizon.
Acute, obtufe, oblique, P. 120.
Arches, P. 37, gothic, &c.
Axis, A. 240. In geometry the fame as perpendicular.
Balance, Qu. 56.
Barometer; Qu 178.
Bellows, Qu. 174
B.
Bofe of a figure, denotes the line or fide on which the perpend. falls.
Body, A. 235.
Bridge, and butments, P 37.
Bodies immerfed in fluids, P. 199.
Catenaria, P. 35.
Celerity, A. 230.
C.
Center, the point in the midft of a circle, fphere, &c.
of motion, maguitude, gravity, percullion, ofcillation, A. 240,
and 241.
Central &c. forces, Qu. 261.
Chain, P. 229.
Circle, and its parts P. 53, and 54, as alfo P. 179, the periphery of
every circle is fuppofed to be divided into 360 equal parts called
degrees, every degree into 60 equal parts called minutes, cach mi-
nute into 60 other equal parts called feconds, and lo on for thirde
fourths, &c.
EXPLANATION OF TERMS.
247
Chord and Cord, is any right line O TO, Fig. 43, drawn within
circle less than its diameter CA.
Cone, a folid formed by the revolution of the baſe and hypothenufe,
of a right angled triangle about its perpendicular, fuch as a round
fpire fteeple, P. 155.
Conic fections, P. 183.
Density, A. 236.
D.
Diameter is any right line C A drawn thro' the center of a circle,
dividing it into two equal parts. Fig. 43-
Diagonal is a right line drawn in any fig. connecting its oppofite an-
gles, dividing regular figures equally and irregular figures unequally.
fo A D and AC are diagonals in Fig. 15 &c.
E.
Equations, fimple, quadratic, adfected, their folutions, P. 171.
Equilibrium, A. 247,
F.
Force, any thing that acts on a body to put it in motion.
Friction, A. 242.
Fruftum, (A BEF, fig. 127) is the lower part of a cone or pyramid,
when the top part is cut off, or the middle part of a globe, &c.
when two equal and oppofite fegments are cut off, this fruftum is
called a middle zone.
G.
Gravity, the weight of bodies relative, fpecific, abfolute, A. 238.
Hoofs, of all kinds, A. 208.
Hydroſtatics, A. 312.
H.
Hydraulics, A. 313.
I.
Irregular figures, or polygons are fuch as confift of unequal fides and
angles.
Irregular folids, fuch as uneven timber, craggy ſtone, &c. whoſe di-
menfions cannot be taken.
Impetus, is the force wherewith one body ſtrikes another.
L.
Leaver, or lever, a bar or pole to raiſe weights.
Level, femicircle, theodolite, protractor, plane table circumferenter,
&c. The inftruments uſed in furveying meafuring heights, levelling,
&c. P. 129. alfo compt. fupplement Ar. Co. ule of Gunter's fcale
logarithms &c. definitions and inftruments uſed in trigon. P. 120.
Logarithms, conſtructed, P. 174.
Mechanics, A 224.
M.
Mechanic powers are 6 viz. lever, ballance, wheel, fcrew, pulley,
wedge.
Motion, momentum, &c. P. 192.
Moving force, any active force or power that moves a body.
248
EXPLANATION OF TERMS.
1
Ordinate, A. 183.
0,
Ofillation, fwinging of a pendulum or other pendulous body.
Oval or ellipfis, an imperfect circle, &c. or the plane of the ſection
ariling by cutting the cone oblique to its axis for it and its dia-
meters, as alfo the hyperbola, &c. fee P. 30, 32 and 180.
1
P.
;
Parallelogram, or long fquare, is a figure with two equal fides and
two equal ends, when the fides and ends are at right angles the
parallelogram is right angled, as doors, tables, &c. when the ends
and fides bevel, then the parallelogram is oblique.
Perpendicular, is the fhorteft right ine that can be drawn from any
point or angle to the oppofite fide, and cuts it fquare-wife or at
right angles,
Parallel lines are fuch as are equi-diſtant in all their parts and if in-
finitely extended would never either meet or converge.
Point, is that which hath no parts, or the beginning of all magnitude,
by the motion of which a line is generated or form'd.
a line is understood to have length, but neither breadth nor
thickneſs, the motion of a point in a conftant direction, traces out
a ftreight line,SA
In a variable direction a curve, crooked, or mixt line.
Periphery, or cucumference, implies the circuit or compafs of any
figure or thing,
Perimeter, the fum of the fides or outmoſt line of any body.
Parameter, or latus rectum P. 181.
Paraboli, A. 185, P. 31.
Property, équation or nature of any figure or curve, A. 186.
If a femi-parabola be turned about its axis the infcribed ſpace is a
parabolic conoid, and if the whole parabolic curve be turned about
its greatest ordinate a parabolic fpindle is generated.
The fame reasoning holds in refpect to hyperbolic curves. Alfo
ifa femi-ellipfis be turned about the tranfverfe axis an egg-like folid
is formed, called a light or prolate fpheroid, but if half the ellip-
tic curve be turned about its conjugate axis a turnip-like folid is
generated called an oblate fpheroid.
Primils are regular tapering folids, whofe bafes may be triangular
fquare or polygonfal, ending in a point perpendicular to the center
of the bale, as a fpire, fteeple, &c.
Prifius, are regular folids, clothed with equal parallelograms, (and
may be triangular, fquare or multangular,) the ends being equal
polygons. A round prifim is called a cylinder, fuch is a rolling ftone,
a fquare one is a cube comprehending, fix equal fquare plains as a
die.
I
One bounded by 4 equal parallelograms, and two equal ends
is called a parallelopipedon.
EXPLANATION OF TERM S.
249
Prismoid, the fruftum of a pyramid whofe bafes are parallel but
difproportional, if one baſe be an ellipfis and the other a circle,
it is called a cylinderoid.
Preffure, P. 199.
Pendulums, &c. P. 195.
Percuffion, the ftriking of one body against another.
Pifton, pump weather glaffes, &c. P. 199.
Prop, any article that bears up or fupports a heavy body, that under
a lever is called a fulcrum.
Q.
Quantities, proportional progreffive analogies, &c. P. 173.
Quadrant, a quarter or fourth part of a circle, as A E D. Fig. 43.
R.
Radius, that extent of the compaffes with which any circle's periphery
is deſcribed or half the circle's diameter.
Rhombus, or oblique angled fquare, a figure of 4 equal fides, making
oblique angles with one another. Fig. 23.
Refiftance, of fluids. P. 199.
Rhomboides, an oblique angled parallelogram. Fig. 21.
Regular polygons, in general are fuch figures as confift of an equal
number of fides regularly and equally diſpoſed about a center, re-
folveable into fo many equal triangles as the polygon hath fides.
As any number of equal fides and equal angles make a regular
polygon, fo fome number of regular polygons of the fame kind-
conftitute a regular folid body, of which there can be only 5, for
3 plane angles (at beft) must be taken (fee P. 154) to make a folid
angle, the fum of which must be less than 4 right angles, other-
wife their angular points being joined (P. 156.) together will all
lye in one plain reftitution or ſubſtitution, A 43.
Random, range, P. 200.
S.
Sector, a part cut from the center of a circle, oval, &c. and may be
greater or lefs than a femi-circle as SA QB (fig. 153) is a ſector
but the part B GA Q is a fegment.
Slice, fecond fegment, &c. P. 186.
Solidity, or folid content, thews how many equal cubes of a known
dimenfion any folid is compoſed of, as for example, if a folid body
be 15 feet in folidity, whatever its form be, fuch body may be cut
into 15 pieces, each 1 foot folid, or pieces that will make
I
a fide of each 1 foot.
Segment, a part cut off, &c.
Series, in order or fucceffion, &c.
infinite, fums &c. P. 110, 172.
15 cubes
Similar, alike or homogenial, are all fuch figures or bodies whofe
angles are reſpectively equal and fides proportional.
Sine, tangent, fecant, &c. P. 54.
Sphere, a hollow globe, formed by turning a femicircle about its
diameter.
250
EXPLANATION OF TERM S.
Springy, or elaftic and non-elaftic bodies, A. 257,258.
Superficial content, area, flat content or furface, fhews how many
fquares of a given dimenfion, or pieces that will make up fo many
fquares, a plane or curve furface may be cut into, or is compofed
of; convex furface fhews how many fuch fquares will clothe or
cover the outſide of any ſolid.
Strifs and ftrength, P. 198.
T
Triangles of all kinds P. 120, 184.
Trapezia, or quadrangle, any figure of 4 unequal fides or angles.
Trapezoid, is when two oppofite fides of a trapezia are parallel.
V.
Velocity, laws of motion, uniform accelerative, &c. P. 192.
Valve, fee pamp, P. 199.
Vibration, the fwinging of a pendulum back and forward.
Several more terms you'll find explained in the parts of Algebra,
Trigonometry, Geometry, Menfuration, Mechanicks, &C.
Books fold by G. ROBINSON and J. ROBERTS
in Pater-nofter-Row.
1. The Second Edition in Folio, of the Conftruc-
tion and Principal Ufes of Mathematical Inftruments,
tranflated from the French of M. Bion, chief In-
ftrument Maker to the French King: to which are
added, the Conftruction and Ufes of fuch Inftru-
ments as are omitted by M. Bion, particularly of
thoſe invented or improved by the Engliſh, by Ed-
mond Stone. The whole illuftrated with thirty Folio
Plates.
2. The Principles of Mechanics, explaining and
demonstrating the general Laws of Motion, the Laws
of Gravity, Motion of Defcending Bodies, Projectiles,
Mechanic Powers, Pendulums, Center of Gravity,
&c. Strength and Strefs of Timber, Hydroftatics,
and Conftruction of Machines. The Second Edi-
tion 4to.
3. The Doctrine of Fluctions; not only explaining
the Elements thereof, but alfo its Application and
Ufe in the feveral Parts of Mathematics and Natural
Philofophy. The Second Edition, with large Ad-
ditions, 8vo.
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