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UNIVERSITY OF MICHIGAN
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CB12
SCIENTIA
ARTES
VERITAS
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Faei Defensor etc.
Obunum.
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bior
Lux Stereometria:
OR THE
A R T
Of Meaſuring
Surfaces and Solids,
Theoretically and Pra&tically
DEMONSTRATED;
And Apply'd
To the moſt Difficult CASES
in G AUGING, after a New
and more Exact Method, than any yet
Practis d.
To which is Prefixd
A ſhort View of Decimal Arithmetick,
and the Extraction of the Square Root :
Together with a Deſcription of the ſeveral
ſorts of Superficial and Solid Bodies: and a
TABLE for finding the Solid Content of
Round Timber, c.
d!
By FAMES LIGATBODT, Philomath.
.
LONDON:
printed for FRANCIS HUBB ART in Duck-
lane, and HVGH NEW MAN at the Graſ-
hopper in the Poultrey, 1701.
-వరం . . . . . .
- - -
ఆ వస్తు సంక, నా
అవును
--
To His Ingenious
A N D
wa
WORTHY FRIEND,
Mr. J. Brightland,
.
T T HIS
Treatiſe
Is Humbly
DEDICATED
W
By His Obliged
Humble Servant
2
wacht
The Authoz
aminsgrike it
រ
ST
1 ។
I go
(
០៦
-
:.
Hilfsmance
315
S.
fue on-le-e
The Preface.
Ince-my former perfor-
mances of this kind
bave met with ſo wel-
come a reception from the
orld, I ibought my ſelf
Obliged by the Meaſures
of Gratitude, to think of
Jomething more wortby. of
their applauſe. Hom far I
bave ſucceeded in this di-
tempt, the Publick is 10
judge: I can only ſay, that
look upon this as the beſt of
.:
A 4 my
The PREFACE.
my performances, and flatter
my ſelf with the hopes that
the Opinion of the World
will juſtifie my thoughts, by
giving it the preferencea I
bave often obſerved with re-
gret, that the pratica)
part of Geoinetry is com-
monly fo fever'd from the
Theory, that the Profeffors
of the Former know little or
nothing of the Ladier, to the
infmite Scandal of their
profeffion. By this méarar,
they are uncapable of doing
any thing but by Roat, and
only follow the Cuſtomary
rules,
The PREFACE.
rules, without ſeeing the rea-
fons they are built upon, or
knowing how to Meaſure
any uncommon Figure that
lies out of the beaten Road.
Whereas if they apply'dibem-
ſelves to tbe principles of
Geometry, in wbich the
Various forms of ſuperficial
and ſolid bodies are examind
and mutually collated, tbey
would be able to give a
rational Account of their o
peration, and extend their
Practice beyond the verge of
a fet rule. Upon this ac-
connt, V bave bere infifted
upon
...
The PREFACE.
3
upon that merbod of blend-
ing tbe Tbeory and Practice
together, and Uſhering in
the rules of Menfuration by
the Speculative Qualities of
the bodies that are to be
Meaſured. - My chief deſ
Sign was to give a plain and
rational Scheme of the Art
of Gauginga mbich is the
moſt difficult part of Practical
Geometry: And I am hopeful
that thoſe who are concerned
on bis Majeſty's revenue of
the Exciſe, will be ſenſible
ibat I bave in ſome Meaſure
compaſsid my end. There are
other
The PREFACE.
***
os
ture
other incident propoſitions,
relating to the Menfuration
of Board Glaſs Wainſcot, &c.
W bicb, tho they are not any
direkt part of Stereometry or
Gauging, yet they are uſe.
ful for Iunſtrating the Do-
Trine of Solids ; for the Na-
that of Surfaces, as well as
that of Surfaces døes upon
Lines and indeed the beſt
way of tracing the Knowledg
of any thing, is reducing it 10
its firſt and ſimpleſt principles
.
It was neceſary to pre-
mile a ſhort view of Decimal
The PREFACE.
Arithmetick and the Ex-
traction of the Square Root:
for the proportion between
Surfaces and Lines is beſt
Illuftrated by the nature of
Square Powers: and the Ope-
rations in the way of Gaug
ing, are not to be perform d
witbout Decimal Arithmetic,
and extracting Square roots.
If ny buſineſs bad not oca
cafrond d neceſſary abſence
from Town, Ibad taken care
to have made tbis reariſe
more Correct and Methodi.
cal: However, fuch ar it is,
I hope it will nor diſpleaſe tbe
Reader.
THE
THE
CONTENTS
PART 1
0
P. 8.
p. 11.
Chap 1 F Decimal Arithme
tick, Page r.
Notation of Decimals.
ibid.
Addition of Decimals.
P. 4-
Subfraction of Decimals. P. 7.
Multiplication of Decimals.
Diviſion of Decimals.
Extraction of the Square Roof. p. 14.
Chap. 2. Of Magnitude or bigneſs, p.27.
Geometrical Problems : ſuch as are moje
aleful in the art of meaſuring, P.30.
Menfuration of plain ſuperficies, ſuch as
Board, Glas, Wainſcof, Painting,
Paving, &c.
P. 35
Chap
The CONTENTS.
Chap. 3. Menfuration of Solid! Food
and Stone,
47.
Chap. 4. Of Round T'imber, p. 61.
ATable for finding the ſolid content of
Round Tumber by theCircumference p.73
PART II.
Chap. 1. The Art of Meaſuring
Sarfaces and Solids Theoretically and
Practically
demonſtrated,
p. 81.
Chap.2 Mexfuration of Regular and
Irregular Solids çalfo of imboſſed Solids
P. 103
Chap. 3. of Gauging. p. 125.
Chap.4. How to Gauge Brewers Tums
of varions Forms and Situations, p. 137.
Of CaskGauging, p. 139.
To find the Vacuity of Spheroia al
Cask, pofited with its Axis parallel to
ebe Horizon,
p. 145.
To find the Vacuity of ſtanding cask
pohted
The CONTENTS.
1:
pofated with its Axis Perpendicular to
the Horizon,
p. 147.
Some uſeful Remarks in Gauge
ing; whereby a Man may not only informa
bis Judgment, but, at ſome time or e-
ther, facilitate his works, p. 149.
How to Inch a ſwelling Cask, Juch as
the Middle of a Fruftum of a ſpheroid,
Jo as you may know the Content of every
Inch or Inches, from Head to Head,
being erected with its Axis Perpendick-
fer to the Horizon,
p. 151,
To inch a round or ſquare Tun, as the
Fruftum of a Cone, cr Square Pyramid
P. 157.
csato & flestess
1:
.
pie
Lux
بها الآن
.
Lux Stereometrie :
OR, THE
A RT
::
OF
Meaſuring Surfaces and Solids:
CH A P. I.
of Decimal Arithmetick.
Notation of Decimals.
A
Decimal is that by which is diſtin-
guiſhed the Parts of a Unite, and
is decreaſed from Unity to ſo many
Tenth Parts of a Unite: for Uniry is di-
vided into 10 Parts, and every roth Pare
se
Lux Stereometria.
4.
is called a Prime, every hundreth Part
is a Second ; and every thouſand Part a
Third,
ds. So as whole Numbers increaſe
by Tens, from the Units place towards
the left hand, fo Decimals decreaſe by
Tens, from the Unites place towards the
right hand; as may appear by the follow,
ing Example.
A
IN
3 4 5 6
mong Unity, or an Integer,
Primes, or Tenth Parts
Seconds, or Hundred Parts.
Thirds, or Thouſand Parts.
+ Fourths, or 10 Thouſand Parts,
Fifths, or roo Thouſand Parts.
Sixths, or 100000 Parts.
For 1o Primes is one Unite; and 10
Seconds is one Prime ; and 10 Thirds
is one Second; and to Fourths is one
Third, &c. So that every place towards
the left, is 1b Times leſs then the pre-
ceeding
Lux Stereometrie
..........
cceding Figure : For preponing Cyphers,
leffen the following Figure in a Tenfold
manner; for (.2) is two Primes, or two
Tenths of an Integer or Unite ; but (.02)
is but . Seconds, or 2 hundred Parts of
an Unite. Cyphers after a Decimal, nei-
ther augment nor diminiſh the value of
the Decimal.
A Decimal is always diſtinguiſhed from
a whole Number, by a Prick or Period,
2$ in the Numbers following
15,3 365-3,262.15 1625.32 273.2
The Numerators are only fer dower
the Denominators being knower by the
number of Places in the Numeraror; for
if the Numerator confift baç of one place,
as I, it is if of two, as 22, it is 22
if of three, as 235, it is 235
dc.
I hall inklt no more upon notarion of
Decimals, the foregoing being fufficient,
but (hall proceed to Addition,
100)
1030)
a
B
4
Lux Stereometria.
op
Addition of Decimals.
ADdition is the adding
of Summs toge-
ther, and making one intire Summ
of two, or more; you muſt take particu-
lar care in placing whole Number under
whole Number; and Decimal under De-
cimal, and Unites under Unites, and
Tens under Tens., he adde
Suppoſe 325.7 to be added to 463.72,
I place them one under another, in man-
ner following ;
To--463.72
Add —-325.7
Summ
789.42
ito
Obferve always to prick off as many
Decimals in the toral, as there is in the
biggeſt Decimal given to be added.
Example
Lüx Stereometria.
5
:
4
Example.
To :
365.23
Add} 23.20
2.03
7
.
*
Summ
390.40
:
An Example of Addition, as to Time
ber or Board Meafure.
3
>
Note, That all'intire Quantitics, as
Feet, Yards, Ells, Ounces, Pounds, and
hundred Weights, are divided into a
hundred Parts; therefore the one Fourth
of any of theſe is .25, the half is 59, and
the three fourths is .75.
There is three Boards of theſe following
Dimenſions, What is the Summ of Feet
and Parts ?
Feet Parts
The firſt
Board
246.25
The ſecond
43 50
The third
16.26
11
306.01
The Summ
The total Summ is 306 Foot, and one
hundred part of a Foot.
B 3
.
6
Lux Stereometria
Addition of Money.
A Pound Sterling is likewiſe divided
into 100 parts; fo .o5 is the Decimal for
one Shilling, and to for two Shillings
and .15 for three Shillings, cc. Suppoſe
I were to add
3 Shillings and 4 pence to
2 Shillings and 3 pence,
The Decimal for 3 Shill, is 5.15 and
The Decimal for 2 Shill. is jo
The Decimal for 4 pence is .0166667
The Decimal for 3 pence is .0125
bona 2b, no!
-2791667
pbouit
Ic is very hard for the Learner to know
the value of this totál, being he is not
come the length of Multiplication, which
is the only Rule uſeful in this caſe, the
general Rule is, every prime in the total
Summ is 2 s. value; and every 5 Seconds
is 1 s. and the Superplus of the Secords
above 5, is ſo many ten Farthings, and
the number of Digits in the Thirds Place
is ſo many Farthings; all the reſt of the
Decimals to the right being of no value
;
the .2 primes
in the total is 4 s. and the
3 5 ſeconds of the 7 is i s, and the two re-
3 perc is .013;
omi ti Summ
maifiing
}
Lux Stereometria. 7 7
maining is 20 Farthings; the thirds ad-
ded thereto, makes 29 Farthings, which
is 7 d. farthing; but by reaſon the Num-
ber is above 25, the Farthing must be cut
off; the Summ is ss. d.
So much for Addition of Decimals
one
:
Subſtra&tion of Decimals...
Subſtraction is the taking a ſmall Summ
from or out of a greater; as if you wou'd
take yl. 5 s. 6d, out of job 56.64
the Remainder muſt be 34 or if from
265.2, you would take 153.1, deci la M
Example
From 105.25 foot of fuperficial Mea-
füre, ſubftra&t 97.39 foot.
105.259
C10352
97.255 Or,
973-5,
from 2
8.00
61.7
From .876
Take 324
From 725.2
Take 322,6
Remains 552 Remains 402,6
B 4
8
Lux Stereometria.
Multiplication of Decimals.
IN Multiplication of Decimals, the Me-
thod is the fame as in whole Numbers,
only you are to prick off ſo many Deci-
mals towards the right hand in the Pro.
duct, as there is Decimals both in the
Multiplicator and Multiplicand : As, fup-
poſe I were to multiply 26.75 by 35, there
muſt be two Decimals prick'd off in the
Product, by reaſon there is two in the
Multiplicand. 719 LI
Example 1.
S
Of mix'd Numbers.
Multiplicand—26.75
Multiplicator
.35
13375
8025
ORTA
Predua
936.25
Oo
Exa nple 2.
Lur Stereometria
9
2.
Example 2.
246.32
24.61
24632
147792
98528
49264
6061.9352
:
Example 3
Example 4.
32.6
6.32753
32.64
T: 6.52
2531012
3796518
1265506
1898252
206.5305792
4
Here you ſee in the Fourth Example,
there is in the Multiplicator and Multi-
plicand, ſeven Decimals; therefore I prick
off as many in the Product as there is in
both.
Notes
Lux Stereometria
Nore, That as whole Numbers multi-
ply'd by whole Numbers, increaſe their
value; fo Decimals multiply'd by Deci-
mals, decreaſe their value, by reaſon the
Product is removed farther from Unity
than either of the Decimals given to be
multiply'd, as ſhall appear by the follow
ing Example.
8:20
.03 •75 0:32
-01
2
006
0300
128
x
It will ſometimes fall out, that there
are not ſo many Figures in the Produđ,
as there are Decimals in the Multiplicand
and Multiplicator; in ſuch caſes you muſt
place Cyphers before the Figures till they
be equal; as you ſhall ſee in the following
Work.
boy
6.5
Example.
QT70870s
004 .42
.22
O02 07 ODO 2.147
ino
བསམ་པས་ལན་པ་ནམ་མ་
Prod: 1.000008 - .0297 09130" 3088
buon rri yung
om
Divi
Lux Stereometria
oor
ܕ ܐ ܕ ܐ -
o the right hand, in the
bre
Diviſion of Decimals.
WE fall in the
next place proceed to
Diviſion of Decimals, which is the
moſt difficult of all the reſt; all the diffi-
culty being to find the true value of the
Quotient.
The general Rule is, when you have
finiſhed your Diviſion, to prick off fo
Qnocient, as will make the Decimals in
the Diviſor equal in Number to theſe in
the Dividend, and the Figures to the left
are whole Numbers. plast one want
Sorel Toro
Example 1.
Let 64.326 be divided by 32,4" or
32.4) 64.326 (19 LON
3 192
2766
You ſee in the foregoing Work, that
there is three Decimals in the Dividend,
and
12
Lux Stereometriæ.
and one in the Diviſor, therefore I make
that r in the Diviſor, and 2 in the Quo-
tient equal to the 3 in the Dividend.
Example 2.
-325) 53.62321 (161.19 w
2012
296 23. dois de
out to al 2 982 osoby
571
240
smiling ni bere Example 3.
Let 28 be divided by 32.6 In this Ex-
ample the Diviſor 32.6 is greater then 28,
the Dividend; in this, and all other ſuch
Caſes, you muſt place a competent number
of Cyphers behind the Dividend ; and if
it be a whole Number you are to divide,
you muſt prick off the Cyphers from the
whole Numbers, and then proceed in
your Diviſion, as you were to divide
whole Numbers,
32.6) 28.00000 (.8585
I 920
2900
1920
290
Ex.
Lux Stereometrie. 13
4.05) 900 (18.0
Example 4.
To divide a Decimal Fraction by a
Decimal Fraction,
Ler 900 be divided by os
♡
10
40
Let .gooo be divided by .0005
.0005).9000 (1800
40
000
:
Would not one think it very ſtrange,
à Decimal Fraction divided by a Decimal
Fraction, ſhould bring forth a whole
Nomber in the Quotient; I ſhall make
it very plain that it muſt be ſo, and no
otherways, by reaſonable Demonſtration.
By the firft Example, you ſee I divide
.900 by .os, and the Quotient I find to
be 18.0, becauſe there is three Decimals
in the Dividend, and one in the Diviſor;
I make up the number of thoſe in the
Dividend, by taking one Decimal from
the Quotient, and adding to the 2 in the
Diviſor.
Now
Lux Stereometria.
Now the Nature of the Queſtion is
this ; I deſire to know how many times
S
Seconds in nine Primes? The Anſwer
is 18.
For there is 18 times 5 Seconds in
9 Primes.
If it fall out at any time, that there is
not ſo many Figures in the Quotient, as
will make theſe in the Diviſor equal in
Number to thoſe in the Dividend then
you muſt prefix Cyphers before the Quo-
tient to the left hand, aa in the follow
ing
Example
4) 13779 (03444
I Shall give you an Example as to the
Uſe and Application of Diviſion of Deci:
mal Fractions, to the end that thoſe that
intend to uſe Decimak, may the better
underſtand what they are going about.
craSuppoſe l-were to divide to Shillings
amongſt 20 Men, the Decimal
for 10
Shillings is .50; therefore I divide my
Decimal .so by the number of Men 20,
and the Quoricot will be .025 ; multiply
by 12, and double the Prodaćt, it will be
Six Pence.
Vid booty
I could give ſeveral Examples of this
kind, but the various Examples that will
hap-
Lux Stereoinetrix: 15
happen in the following Work, will, in
its proper place, give you a clearer De.
monſtration than can be here expected
fo I ſhall refer the Learner to the Pras
Etice in General, and proceedeto my
intended Worki
kung
The Extraction of the Square
Root.
origine in
2
Firſt, A Square Number is any Digit,
or any other Number ; which
being multiplied by or into it ſelf, pro--
duceth a Square Number; as 9 being
multiplied into it felf, ſaying 9 times 9
produceth the Square Number, 81; the
Root of which is g.
Secondly, Square Numbers are eithes
ſingle, or Compound.
Thirdly, A lingle Square Number is
that, which is produced by the Mul-
tiplication of one ſingle Figure by it
Telf, and is always leſs than 100; fo 36
is a fügle Square Number, produced by
6; likewile , multiplied by it ſelf, pro-
dueeth the Square Number 49
Fouribly,
.
16
Lux Stereometriæ.
Fourthly, all the ſingle Square Num-
bers, together with their reſpective
Roots, are fet down in the following
Table.
Squares. | |
36
49
81
Fiftbly, When the Root of any Square
Number is required, it being leſs than
100, and yet is not exa&ly a fingle
Square expreſe in the Table above,
then
of that
Lux Stereometriæ.
fingle Square Number, expreſs’d in the
faid Table, which ſ being leſs) is near-
eſt to the given Square; as if it were re-
quired to find the root of 60, it would
be found to be 7; and I being given,
the Root that belongs to it is 3.
Sixthly, A compound Square Number
is that, which being produced by a Num-
ber (that conſiſts of more places than
one) multiplied by it felf, is never los
than 100 ; fo 729 is a compound Square
Number, produc'd by the Multiplication
of 27 multiplied by it ſelf.
sve
Seventhly, The Root of any Number
under 100, may be eaſily diſcovered by
the Table of bingle Squares; but to ex-
tract the Root of a compound Square
Number, conliſting of Integers, oblerve
theſe following Precepts.
Firſt, Point the given Number, viz
place a Point over the firſt Figure, 10-
wards the right hand, omicțiog every
other Figure place one over the 3d, sth.
and ſo on, according to the number of
Places,
The Figures thus diſtinguiſhed are call'd
Points (or Squares ) and as many Points
as are in the Number given for Extra-
&ion, of fo mapy Places (or Figures)
C
cono
18 Lux Stereometria.
an
eonſiſts the Root, if the Number be
exact compound Square.
Secondly, When you have thus pre-
pard your Number, then draw a crook-
ed Line on the right hand of your Num-
ber ; behind which you are to place the
Root, as you do for a Quotient in Divi-
Root, as you do
Thirdly, When your Nomber is pre-
pard, find out the Square Root, or the
firſt Squarc (or Point, by the Table of
ſingle Squares) and place that Root be.
hind the crooked Line and ſubſtra& the
ſaid Square from the firſt Point.
Fourtbly, To this Remainder bring
down the next Point (or Square ) and
place it on the right Hand thereof, which
call the Reſolvend.
Fifthly, Double the Root of the firſt
Point, and place it on the left hand of
the Reſolvend, diftinguiſhing it there-
from by’a crooked Line, thus;
)? which
call your Divifor.
Sixtbly, Seek how often this Diviſor
is contain'd in all the Figures of the Re-
ſolvend, except the laſt towards the right
liand, and place the Anſwer in the Quo
tient, and alſo on the right hand of the
Diviſor.
Seventhly,
Lux Stereometrie.
19
Seventhly, Multiply the Diviſor by the
Figure laſt placed in the Quotient, and
Subſtract that Product from the Reſol..
vend, ſubſcribing the Remainder.
Laſtly, To this Remainder bring down
the next Point for a new Reſolvend, and
proceed as with the firſt Reſolvend, re-
peating the work of the 5th, 6th, and
7th Præept aforegoing, and concinue lo
to do, till the Extraction be finiſhed. A
few Examples will make it plain.
q
Example i.
I deſire to know the Root of 729.
Firſt The Number being prepar'd, by
Pointing it (according to the firſt Pre-
*
cept ) ſtands thus ; 729 which conſiſts of
two Poincs (or Squares ) and therefore
the Root fought is to conſiſt of two F:-
gures or Places.
Secondly, Draw a crooked Line 6 as is
before taught ) and the Work will ſtand
chuss
of 272961 na
Thirdly, The greateſt Square in the firſt
Point, 7, is 4, whoſe Root is 2, where-
C2
wich
w
Lux Stereometria.
20
with proceed, according to the 3d. Pre-
cept, and the Work will ſtand thus ;
52962 grue de base
die
4
Rem. 3
Fourthly, To the Remainder 3, bring
down your next Square 29, and work
according as you are directed in the sth,
6th, and 7th Precepts aforegoing, and
the Work will ſtand thus ;
Root
729)27
4
0
47( 329 Reſolv.
Düne Diviſ. 329 Product
11 Boro Podano 100,7
Here you are to note, that che Points
Being all brought down, and o remain-
ing, ſhews the Work to be finiſhed ; and
alſo that 729 is an exa& Square Number,
whoſe Roor is found to be 27, as will
appear; for if you ſquare 27 (which is
the Proof of the Extra&tion of the Square
Root)
!
z
S.
.
$
Lux Stereometria.
Root) you will find the Product to be
729, which was the Number propos'd.
Note likewiſe, That if at any time
when you have multiplied the Number,
Itanding in the place of the Diviſor, bý
the Figure laſt placed in the Quotient or
Root) the Product be greater than the
Reſolvend, she work is erroneous; to
correct which, you are to put a leſs Fi-
gure in the Quotient; but if the Remain-
der be greater than the next Diviſor,
put a greater Figure in your Quotient,
and proceed according to the 7th. Pre-
cept, bc.
It is alſo worth obſerving, that the
reaſon why every other Figure is Pointed,
is becauſe the Square of any one single
Figure never exceeds 2 Places
.
The working of two or three Queſtions
more (at large) will ( I preſume) make
the Extraction of the Square Roor (of
an exad Square Nnumber ) wtelligible
enough to an ordinary Capacity.
ci
C3
More
Lux Stereometriæ.
More Examples.
ch
What is the Square of 15129. gada
Firſt, Prepare the Number according
to the firſt and ſecond Precepts, and your
Work will ſtand thus :
151296
151296
Rem. o
Then the Operation of the third Pre.
cept being performed, the Work ſtand-
eth thus :
***
15129( I
DOI
Then proceed according to the 4th.
Precept, and the Work ftandeth thus : 1
SI Reſov.
15129(1
I
The
Lur Stereometrią. 23
The Operation of the sth. Precept ob-
ſerved, the Work ſtands thus :
Diviſ. 2 51
15129/12
I
-
The Operation of the 7th. Precept
obſerved, the Work will ſtand thus :
iſt. Div. 22) 5. Reſolv.
Prod.
44
Rem. 07
Laſtly, The Operation of the 5th, 6th,
and 7th, Precepts being duly performed,
the Work will ſtand as you ſee:
15129(123
I
1ſt: Div. 22) 51 Refolvend.
44 Product
2 Div. 243) 729 Refov.
which ſhews 15129 is an exact Square
Number, found by fquaring 1 23.
729 Product
C4
w What
24
Lux Stereometria. I
::
-ว
to $2990121.527
1991 $7 do not go 117
What is the Square Root of 10 bv 107
[fought
12299049(3507 the Root
9
iſt. Diviſ. 65)329 Reſolvend qo oT
329 Product
Varoldo
1 (non
2d. Diviſ. 700) 490 Reſolvend
ooo Product
ro og
3d. Divi17007)49649 Reſolvend vir
49049 Product 19 TDNE
Swo Warit
COLOC
Here you are to Note, That when the
Diviſor cannot be had in the Reſolvend
(according to the oth Přecept ) then, in
that caſe, you are to place a Cypher in
the Quotient, and alſo on the right hand
of the Diviſor ; and then bring down the
Reſolvend a ſtep lower ; and bring down
the next Square to it (as in the laſt Ex-
ample ) for a new Reſolvend; or let both
the Refolvend and Divifor Remain, and
bring duw, the next Point to it; as the
laſt
Lux Stereometria.s I 256
laſt Example again repeated, will fhew
you more plainly.
12299049(3507 the Root
.
9
Diviſor 65) 329 Reſolvend
325 Producto de
100
Diviſ. 7007) 49049 Reſolvend
1
49049 Product
What is the Square Root of
9042049(3007
9
6007) 042049 Reſolvend
42049 Product
Do
The
lo
.000 Product
26 : Lux Stereometria.
* The Operation after the other Method,
will be as followeth:
irilla orom
9042049(3007
9
60) 04 Reſolvend
600) 0420 Reſolvend
CO
6007) 42049 Reſolvend
42049 Produa
lo
But if there be given an exa&t Square
Number (to be extracted ) conſiſting
either of whole Numbers with Decimals
annexed, or of Decimals alone, then
point every other Figure in the Decimals,
from the place of Units towards the
right hand, as you did in whole Num-
bers towards the left hand , and as many
Poipts as there are in the Decimals, ſo
many Decimal Places there will be in
the Root ſought.
1. As Er
Lux Stereometria.
27
1. An Example of whole Numbers
with Decimals.
What is the Square Root of
141000, 25(372,5 the Roop
9
Divif. 67510 Refolvend
469 Prodnet
Diviſ. 745)4100 Reſolvend
3725 Product
..
Diviſ. 7505).37525 Reſolvend
37525 Product
.
::
Of Magnitude or Bigueſs.
1.MAgnitude is a continual quantity :
2. That is ſaid to be continual, whoſe
parts are held together by ſome common
Bounds, being either ſtreight, or crooked
Lines.
3. A
28
Lux Stereometria.
of 2 Magnitude, which may be called
6. Magoose or agreeable Magnitudes
3. A Bound is the utmoſt Part or Li-
mit
the Surface of ic.
4. A Magnicude is made continu'd, or
divided by thoſe things wherewith it is
bounded.
A Point is an indiviſible fign in
a Magnitude, incommenſurable.
6. Magnitudes commenſurable may be
mcafüred by one and the fame meaſure.
are equal
ivia
vonairs? (rors
Of a Line.
1. Magnitude is either a Line, or a
Lineate.
2. A Line is a Magnitude only of
Length 3 but a Lincate is the bounds of
of a Magnitude.
3. The bound of a Line is a Point
4 A Right Line is that which lieth
equally between its own bounds, and is
the nigheſt diſtance between its bounds:
b
2S d.
2
5. A
Lux Stereometria. 29
i
5. A crooked Line lieth contrarywiſe,
and may be the greateſt diſtance between
its bounds, as the Line c. d.
臺
​d
$
:
6. A crooked Line is either a Periphe.
ry or Helix.
7. A Periphery is a Circumference of
equal diſtance from the Center.
8. A Helix is a crooked Line equally
diſtant from the midſt of the ſpace in-
cloſed.
9. A Perpendicular Line is a Line
which riſes from a Point upon a Line
given, and doth encline to neither ſide;
ſo that the Angles at the point are
equal.
10. Parallel Lines are ſuch as are e-
qually diltant
ſo that if they were
drawn out infinitely, they would never
incloſe ſpace, as a, b.
SHIRA Superficies is a Geometrical Fi-
gure, incompaffed with a Line or Lines ;
and is either Round or Angular.
12. A
30
Lux Stereometria.
51. 12. A Round Figure is that which is
contain'd by one round Line, and either
a Circle or Elipſis.
Geometrical Problems :
Such as are moſt uſeful in the Art
of Meaſuring
1. Upon a Right Line given, te
Vlianpo erect 4 Perpendicular.
TO
onds *
oni
OTO
D
Ivide the Line given into two equal
Paris, then ſet the foot of the
Compaſſes in a, and deſcribe the Arch c;
likewiſe fer the Compaſſes in b, and de-
fcribe the Arch d , then raiſe the Line
from e, to the laterfection of the two
Arches.
103 prin
2. How
Lux Stereometria.
31
2. How to raiſe a Perpendicular upon
the end of a Line giver:
:
de
*
8
1
B
a
Set the foot of your Compaſſes in the
Angle a, and deſcribe the Arch bc; and
the Compaſſes at the ſame diſtance, divide
the Arch into two equal Parts at f and ;
then ſet the foot of the Compaſſes in f,
and make the ſmall Arch e, ſo that it may
be as oppoſite as can be conceived to the
end of the Line at a ; then place the Com-
paſſes at 8, and make the Arch d; and
draw a Line from a, to the Interſection
of the two ſmall Arches, d, and e.
3. T.
::
2
132
Lux Stereometria.
**3. To let fall a Perpendicular from
a Point giver, upon a Line.
..
bo
Let the Point given be a, and the Line
whereon it is to fall, be bc;
ſet the foot
of the Compalles in the Point at a, and
make upon the Line bean Arch; then
divide that pare of the Line which lies
between the Extremities of the Arch, in-
to equal Parts at 0, then lay a Ruler from
a to the Point ato, and draw a Line, and
-you have your delire.
bas tan
4. TO
Lux Stereometria.
33
dicular, en
right
To find the Perpendicular of a
Triangle,having the Baſe and Hypothenu.
fal,
From the Square of the Hypothenuſal)
(which is the longeſt ſide of a Triangle)
fubftract the Square of the Baſe, and the
difference is the Square of the Perpendi-
cular, the Roor whereof is the perpen-
oot wh
Example.
The Square of the Hypothenuſal
but is not in part 420.25
The
Square of the Baſe is
The Difference is 376.69
The Root or Perpendicu
lar is
om
19.4
oslu i sili bus 2
5. Having any two fides of
angled Triangle to find the third.
The Square of the longeſt fidé, or Hy-
pothénuſal, is equal to the Sum of the
Square of the Bafe, and Perpendicular.
Example.
Sappoſe the Bafe and Perpendicular of
Triangle be 48 and 36, The Square of
D
48 is
43.56
34
Lux Stereometria
48 is 2304, and the Square of 36 is 1296;
the Sum of Both is 3600, which is equal
to the Square of the Hpothenuſe, the
Root whereof is the Hypothenuſe.
(BOUNT 900035
WS
IT
oda lani u od tu silun 903 Daud
N the following Chapter I
Thall deſcribe the Meaſuring
of ſuch Magnitudes as are moſt
commonly uſed among Me-
chanicks, fuch as Carpenters,
Joyners, Glafiers, Painters, and
Maſons; being done in a more
exact and ſhort Method than
has been yet Practiſed, both
by Decimals and the Rule of
Practice, which was never ap-
plyed to Solids before. smed
ora 10 92 oris oi trupra londo
ciclibraqnos bnse sind sie osiç
asipang l CHAP
to istrabroqyol breskars stories?
to twop2r; : bris od start
zie
Lux Stereometria.
35
CHAP II.
:
Menſuration of Plain Super-
ficies, ſuch as Board Glaſs,
Wainſcoat,
Painting Pav-
ing, &c.
..
IN Superficial Meaſure the Square of 12
Inches is a Foot, being 144 Inches.
DOCI
Note, That the Squaring of a Number is
Multiplying it in it Self.
di voler Slug lorong
If you would know how many Foot
Square is in a Yard Square, you muſt mul-
tiply the Feet in a Yard, Viz. 3 by it felf,
and the Product is 9, the Square Feet
in a Yard. u 01 loquerem
nanish Absord 29.nl o buc2001
Since1 CDI 92 am von
obivia bus 70110 113 yo o doyo
D 2
Example
36
Lux Stereometrix.
II. Example
How many Inches Square are tbere ir a
Tard Square.. 19
artera M
1 zulo base 23.00 askarte
36
36
216
si lo sreupe od: 108
001
1296 Square
Square Inches
are
The General Rule is to multiply the
Length by the Breadth (let it be Inches,
Feet or Yards) and the Product is the
Content LTY 1901 srij yigit
Bubor? gnt brus
I ſhall Suppoſe a Plank to be 20 loches
Long, and 16 Inches Broad, I demand
how many ſuperficial Feet are in it ? Mul-
tiply the one by the other, and Divide
1) and
by 144.
Length
Lux Stereometria
37
:
Ax", Heatheter :
Length 20
Breadth 16
2
120
144) 320 (2. 22 Feet
320
320
3
I ſhall work the ſame Queſtion by the
Rule of Practice, for a Proof of the other.
Perhaps the Learner may be a ſtranger
to this Rule; therefore I ſhall explain it
as far as Room will permit.
75
Firſ, You are to multipy the Length
by the Breadth, being Stated in Feet and
Inches, faying, once 8 is 8, and once n is
1, then take the one third of 1 Foot 33
Inches,which is 6Inches and an half, beilig
added together as follows, the Product is
2 Feet 2 Inches and an half, the Content
of the Plank. The reaſon you take the
one third is. becauſe 4 Inches are the one
third of a Foor
D 3
Foot
38
Lux Stereometrix.
*
Foot. Inches ]
or 61
Length. I
8
Breadth: 1:
04
I
8
6.
odio id not go patrow lindi
abges1 s 12.30-23 Equalis
Secondly, There are 3 Planks, one 3
Foot long and 2 Foot Broad ; another 2
Foot and an half long, and 2 Foot and a
Quarter Broad, the third is 4 Foot and a
Quarter long, and 3 Foot and 3 Quarters
Broad how many Square Feet are there
in all theſe Boards or Blanks.diverton
Baboro dolor 90er De
Note, That the Decimal for one fourth of
Å Foot is 125,
for one half 50 and for
three Quarters 75.1 A 10
Length.
:::::..
Lux Stereometria.
39
2.
Length 2 2.50
Breadth. 3 2.25
5.25
3.75
Foot
6. 1250
2125
500 2975
500 1275
:
5. 6250 15.9375
5.6250
Number of Feet 27.5625
By the Rule of Practice.
Long 2:6
Broad 2
4 : 3 3
3:9 %
3
5:0 12:96
7 1 3 : 2 }
bi si bros: 77 15: 11 $ (6)
15.IT
06:00
27:6 Equal to the former:
Thirdly, There is a Plank 10 Inches
Broad at one end, and 7 at the other, and
64.5 in Length, how many Square Feet
D4
are
40 Lux Stereometria.
are in it.
Add both ends together,
and take half the Summ for the mean
Breadth.
IO
7
scramidi
17
8.5
5. the half
Length 64
Breadth
5
8.5
3225
5166
144) 548. 25 (3. 807
1162
Io50
420
Fourthly, There is a Fiece of Painting
16 Foot and an half in length, and 12 and
a Quarter Broad; bow many Foot is there-
in.
1911moi gilt alaptors
earloul ei meio e di asset
Feet
brie 19dro orisa bans. 39 500
1991 Staup, meme von eine
Lux Sterremetrie.
41
Feet
16,
Inches Decimal
6
16:50
3
12.25
I 2
5.82.50
32
106
4
O
? 3300
3300
I 650
2
202
202.12510
In the foregoing Queſtion, in Pradice
it is harder than if the Number of Feet
were under 10. For you muſt first mol-
tiply the Number of Feet by it ſelf, then
tay 13 Times 6 Inches is o Foot, which
beeing added to it makes 198 Foot; then
the Quarter of 16 Foot and an half is 4
Foot i Inch and an half, which Amounts
to in all 202 Foot and one Inch and an
half. To bring it into Yards, Divide by
o for there is o Foot in a Superficial Yard,
as you may fee by the following Exam-
ple,
Example
ਦਾ ਕਰਨ
42
Lux Stereometrie.
:
Ioais Example.
9) 2021250 (22.4583
22
41
52
75
30
3
In 202 Foot one Inch and half there
are 22 Yards and above a Quarter of a
Yard.com
1 There is a peice of Pavement 6 Yards
and an half long and Five Yards and 2
Quarters broad, how many Square Yards
are therein.
Najno isech datar to THEO
2100 By Decimal, sibat 2009
os doch rant apie 6.50
100
5.25
kel.
3250 (gad xperia
1300
3250
34.1250
Ву
Lux Stereometria.
43
By Practice
6.
32.
34. og
Note, In the Practice way, you muſt
multiply the half Yards by 5, which is 2
Yards and an balf; then 5 Times 6 Yards
is 30,and 2 and an halt is 32 and an balf;
and the Quarter of 6 Yords and an balf
is 18. In all it is 34 Yards, and the Eighth
part of a Yard. OD
yleister
If a Board be s Inches Broad and 19
Long, what is it over or under a Square Fuor?
Multiply the Length by the Breadth,
and the Product ſubduct from 144,
and the Difference or Remainder is what
it wants.
If your Product had been
above 144, then you was to ſubduct 144
from it, and the Remainder was what it
was above.
1
Example
44
Lux Stereometria.
Example
R
19
...
95
95
Wants 49 Inches.
4 Plank be 9 Inches lo defire.co
årew what Longth will make a Square Foot?
Divide 144 by g, and the Quotient is the
Anſwer.
9) 144 (10
54
se o
a ñgard to un breadth 18 Inchar,
Longeb will de a Square Foot ? 4 10
1191 *p oomilica
18)-144 (8 11.
u boix. I ar 000
There is a Pane of Glaſs 4 Foot long.
tika ligot and an haif Broad, How many
perficial Feet are thenin?
Lux Stereometria.
Anſwer.
Length
Broad
4
O
89
1000, The Content in Feer
The fame Queſtion by the Rule of
Practice.
Example.
25 JJ Feet Inches.
Long. 4.
Broad 2. 6
8.
2.
.
AH 10.
*
There are 3 Panes of Glaſs, one is in
Length 6 Foot and an half, and in Breadth
2 Foot and a Quarter another 4 Foop
long and 3 Foot Broad : the third is 8
and a Quarter long and 6 Broad. I de
mand how many Foot in all? Anfrer
:
46
Lux Stereometria.
Anſwer.
.
Feet
Length 6.
Inches
6.4. 8. 3
Breadth 2.
2
-3. 3. 6.0
13
0 12 49.6
7.
Feet. po Inches
Inches sin side
The firſt Pane 14.2***
The Second
The Third 49.
ao
76. 1 $ TheContent of all
9001
CHAP
led na bar 10010
il1903.
br. 1001
i bas
O boog getrol 1500
Lux Stereometria.
47
CH A P III.
Men/uration of Solid Wood
and Stone. .
15
many Solid
bere is a Piece of 63 Inches Long, 32
T
feet are Feet therein ?
You are to Multiply the Length by the
Breadth, and the Product by the Thick-
neſs, and the laſt Product Divide by the
Inches in a Solid Foot, viz. 1728, which
is the Cube of 12, for 12 Multiplied in
it ſelf is 144, and that by 12 is 1728.
?
।
3
vagy
I
Example
2
8
Lur Stereometria.
Exameplé,
Decimally.
n9N
Length
Breath0132 SA
is
2018
ŠAM 911
Thick
B
10083's baie
bilo s ria
2016
ni boilustig
5:30
2
(918) 30246 (17.5
12960
* * 8640
C000
Pradically
Lux Stereometria.
49
.
Practically.
Feet Inches
Long S. 3
Broad 3
8
.
Io.
6
6
2.
14
Thick 1:
3
-
14
3
6
:
17
6
DO
.
There
50
Lux Stereometria .
There is a Piece of Timber 18 Inches
Long, and 15 Broad, and 6 Inches Thick at
one end, and 4 at the other; how wany Selid
Feet are therein ? You muſt find a mean
Thickneſs by taking the half of the Sum
of the ends and work as before.
:
Example:
ww
Decimally.
Length 18
Breadth 15
.::
90
18
270
Mean Thick 5
1723) 1350. Co (78
13040
1216
Pra&ically
Lux Stereometria
51
Practically
Foot Inches.
6
6
L.
:
I.
6
4
2
45
I
* IO
5
is 73
The
The
3
is
2
0.9
?
The foregoing Queſtion is very har!
in the Rule of Practice, being the wholo
Solidity is under a Foot, for after you
have multiply'd the Length by the
Breadth, you are to find the one thir:1
and the one fixth of the Product, which
is the Solid Content or three fourths of
a Foot and one Eighth of an Inch.
E 2
There
2
52 Lux Stereometria.
There is a Stone 3 Foot 6 Inches long, and
2. Foot 3 Inubes Broad and 2 Foot Ihick,
what's the Content.
Length 3 : 6
Breadth 2.
3
7
10
7.10
Thick 2 : 0
15 :
09.
3
Oli niongrot na
BATU 990 ads bacto TT
bo
sont so to satis en 30
Lux Stereometria.
53
of the Regular Polygons.
T Here is a Stone in the form of an Oet-
agon, having 8 equal fides, cach
ſide being 6 Inches Broad. I demand
the Content in Solid Feet. ic beeing 16
Inches long, you muſt Multiply a Line
drawn from the Center, co the Middle
of
any ſide, by half the Sum of the ſides
then Multiply that Product by the
Length, and you have the Content in
Inches, the which divided by 1728, gives
you the Contents in Feet.
1
The ſides 8
The Inches 6
1392
16
48
8352
1392
Half the Sum 2+ 22272 Solid Inches.
The Line
5.8
192
I 20
1321 the Sine fcial Inches.
5
54
Lux Stereometriæ.
1728) 2227.2 (1.2
4992
15360
1536
Here I find the Solid Content to be
Foot and above one Fourth of a
Foot.
The ſame Rule is uſed in all the Rex
gular Polygons, as is uſed in the forego
Of
Lux Stereometr...
55
. .
Of Triangle's Trapeziums,
and Rhomboides.
:
A Triangular Stone whoſe Baſe is 3 Foot
Perpendicular 2
8 Inches, what's the Content? You muſt
multiply the whole Baſe by half the
Perpendicular; and the Product is the
Superficial Content.
A Trapezium. You are to multiply
the common Baſe by half the Sum of
the Perpendiculars.
... **
A RBombus. Multiply the longeſt fide
by a Perpendicular let fall from the blunt
Angle to the Oppoſite ſide, and the Pro-
duét is the Content in Inches or Feet.
The Superficial Content of theſe be-
ing found as before directed, you are to
multiply it by the Depth, Thickneſs or
Length, and you have the Solid Content
either in Inches or Feet.
E+
1
56
Lux Stereometria.
The Triangle.
Baſe
Foot Inches
3
Halt thePers I
pend.
4 8 Superficial Content
DO
You are to multiply this by theLcogth,
and you have the Content in Soil Feet
and Inches.
The Trapezium.
The Bale
4 3
Half the Sum of S200R
the Perpend. Lisesti
8.6 Super Contest
Lux Steroemetric
57
$
The Rhombus.
fr
Side
6 4
Perpend
33
IO
7
20 7 Superficial Content.
Length
8 7
104 6
10 31
174 9 1 Solid Content:
$
berit
58
Lux Stereometria.
.
Of Pyramids.
.
A Square Pyramid. You muſt multi-
plythe Area of the Baſe by one third of
the Height and the Quotient is the Solid
Content,
Bot
Example
The Baſe fide is 34
3 4
o
I I
)
IA
One third of the
Height
4
44 5 the Solid Cont.
Of Round and Triangular Pyramids
You are to uſe the fame Rule.
Lur Stereometria.
59
A Globe or Sphere.
A Globe is two thirds of a Cylinder,
whoſe Diameter and Altitude are equal
to the Diameter of the Globe.
There is a Globe whore Diameter is
61.7. Idemand the Content in Solid
Feet.
Note, That two thirds the Area of
Unity is . $23598, which is the Con-
tent of a Sphere, whoſe Diameter is
Unity; the which divided by 1728
Quote 0003041 ; If the Cube of the
Diameter of any Globe be multiply'd by
this, the Product is the Content in
Square Feet
:
A Globe
to
Lux Stereometria.
A Globe Diameter
61.7
:
The Square is
3806.89
The Cube is 1234885093
NA
..
ja au w 234885.093
SİNO 90 g or
9 95403722
704655279
cia on
tub 70.5615567813
The Content of the Globe is 70 Foor
and above an balf.
CHAP
Lux Stereometria .
61
CHAP IV.
Of Round Timber
:
THE Common way uſed by Carpenters
for Meaſuring Round Timber, is to
girdle the Tree in the Middle, and mul-
tiply that Circumference by it ſelf, and
the Product multiply by 7, and chat Pro-
duct divide by 88, and multiply that
Quotient by the Length and the Product,
divide by the Square Inches in a Foot viz.
1728, and the laſt Quotient is the Square
Feet contain'd in that Tree or Piece of
Timber.
non
Example
62
Lux Stereometrią.
Example.
There is a Tree 72 Inches in Circumfe-
rence and in Length 94 Inches ; How many
Square Feet are therein?
: ما نه
72
1.70
72
144
CUM 10+
5184 IMI antolon
il para carrin e Vos
u vidunt Sabors 4.3
1.88) 3628838412.3 vih
bo 108 bogen 90
Bopitore
00208 9122lydobavi
3200 1649240869
37107 1909 190
conto
94
50
1728) 38756.2 (22.4 Foot
4196
7502
490
There
Lux Stereometria. 63
bu There is another way uſed by Carpen-
ters,to girdle the Tree in the Middle, and
take one fourth of that for the true
Square, which is very falſe, for they ſhall
looſe in a Tree which is not above so
Inches Circumference, one fifth of the
true Content, and the larger the Tree,the
greater the Loſs
They pretend that the Lofs in the
Meaſure is an allowance for the Chips or
Slabs, to bring it to a true Square: but
still this cannot prove it, neither accor-
ding to Art, nor truth; for you are to
find the true Content, and make your
Bargain accordingly.
os
I ſhall here lay down a true and exa&
Method, the which if you Practice, you
will find it both eaſy, juſt, and according
to Art.
Before I proced any further, I ſhall
ſhew you the way to find the Square of
the Gauge Point for Round Timber ; and
then how you ſhall ufe it.
You are to Multiply the Diviſor for
finding the Square Inches in a Circle, viz.
1. 2735 by the Square Inches in a Cubi-
64 Lur Stereometria
R.
cal or SolidFoot viz. 1728 and the Product
is the Square of the Gauge Point, the Roor
whereof is the Gauge Poist,or 46.9.
Over on Example. The story
podran
TOTO 1: 2735 napos
et di 1000
1728
:
1075 101880
03 97 uomo ob
Toto
25470
89145 outing
ob
12732
may pide would bedram
-
2200.6080 (46.00
olso
600
8460 No
Hairy wolf sanit
9980
Datumupa ibu
idup the dailograria d & To
Luz Stereometria.
.
65
2
To prove this Work, multiply the
Square of the Guce Point v Z 2200*6080.
y the Arca of Uni viz. 7853, and the
Prodia will be 1728 353 &c.
I fall in the next place thew you the
ple of the Square, of the G uge Point,
as alſo the D fforenci berween it and the
ulgar way, which is by the Rule of
Proportion, as 14 is to it, ſo is the
Square of the Diameter to the Area in
quare inches,
Example
The Diameter
of a Itse is
14.2
The Square of it is 584.64
58564
58564
14) 6342,04 (453.00
74
4.2
C004
f
.
66
Lux Stereometriæ.
This Quotient is the Square Inches, or
Superficial Area of it, which muſt be
divided by 1728, and Multiplyed by the
length, being 50, and the Product is the
Content in Square Feet.
:
anj bricot Examples
Example, on a adu
30
Doctoid W
1728) 453.00 100 2621
10740
50
3720
2640 13.1050
912 is to stop
28
Lux Stereometriæ. 67
I ſhall work the ſame Queſtion by the
Square of the Gauge Point in half the
Figures, and more exact.
si
The Square of the Diameter is me
2200) 585.64 (
14564
Solopose
13640
$
4400
13.3100
Note, The Quotient is the Area or the
Content of one Foot in length of the
Tree
Suppoſe a Tree to be in Circumference
at one end 46 Inches, at the other 34,
and the middle 39; What's the Con-
tent in Square Feet, the length being 27
Inches.
F
You
68 Lux Stereometria
You are to find the Diameters of each
of the places by Moltiplying the Circum-
ference of each place ſeveraly, by. 318,
as before directed, and one third of the
Sum of the 3 Circumferences, is the
mean D'amerer, the Square whereof
divided by 2200 exhibits the Area, the
which multiply by the Length, and you
have the Content. de
Roqqo
gunsda olaraqua
NOY
69
?
Lux Stereometriæ.
Example.
318
318
318
39
Circumf. 46
34
1908
1272
1272
954
2862
954
14.628
10812
12402
10.812
I 2.402
3) 37 843
18
12.614
12614
0042
12
50456
.:
12614
75684
25228
12614
:
.
i
*
2200) 159012996 (072279
5012
6129
1799
18996
1396
9
:
Taare
70
Lux Stereometria.
Theſe few Examples are ſufficient for
the Learner, being there is ſo little Vari-
ety in Round Timber. Only obſerve
if it be a Growing Tree, to take the Cire
cumference about the Middle of the Boole
of it ; for they are commonly ſomwhat
Taper ; but if it be a Cut Tree, you need
not trouble your ſelf much with the Cir
cumference, but take the Diameter of each
end, (if the Tree be not Irregular) and
half the Sum is the Mean Diameter.
I ſhall hereunto annex a Table in which
by taking theCompaſs of the Tree, you may
find the Content, by infpe&tion of one
Foot in Length, the which multiply'd
by the wole Length gives you the COR-
tent of the whole Tree.
IQT
THE
-
Lux Stereometria.
70
THE
..
USE of the
.
TABLE
YO
ou will find the Compaſs
of the Tree under Com. and
over againſt it under Foot and
Parts, you will find the Content of
one Foot in length.
F
F
Example:
72
Lux Stereometria.
Example.
There is a piece of Timber 48 Inches
in Compaſs,and 20 Foot long ; , 1 find it
to be 25 Foot and 460 Parts, for 48
Inches in Compaſs, gives 1.273, which
Multwlyed by 20, gives 25.460 Foot
.
स I
.
fruit die
het doel vastatas, trabajo enero
On the batit Ilireanne Aroble
cigion is that are
Lux Stereometria.
73
A
Ś
3
TABLE
Which by the Compaſs of
any piece of RoundTim-
ber fbews the trueContent
of one Foot in Length
thereof.
Com.
F
IO
.055
066
fuduiog agi fo requun
12
079
13
093
108
::
14
Lux Stereometrice.
Com.
F.
Pa.
15
*124
q
16
go
141
17
159
179
18
0
1790
Th191 10 200
20 i So
o
2673
221 43
21
243
Inches of the Compaſs.
22
23
292
24
b
318
25
343
26
374
27
403
28
O
433
29
465
90
Lux Stereometria
75
.
Con.
E
&
30
*497
31
O
531
32
566
33
602
O O O O
*
* 34
639
35
677
36
O
716
Inches of the Compaſs.
37
756
38
O
798
:
39
840
:
40
537
* 41
929
1042
974
43
I
021
$8044
1 079
2
45
Lux Stereometria.
76
CAM.
Pa.
•119
.
46
I
169
47
I
220
.
48
273
49
I
327
So
T
381
ipduos aqi to samour
437
- 52
I
490
353
I
552
1
612
55
671
0:56
1
732
57
I
795
38
1
860
1959
923
21
Lux Stereometria.
77
Con.
F'
Pa
60
I
.988
61
056
:i
62
134
63
2
193
64
2
264
65
2
335
66
2
407
Laches of the Compaſs.
67
a
27 480
3
68
555
炎
​69
631
70
2
707
71
2
785
72
864
173
945
>
74
3
Lux Stereometriæ.
78
Com
F.
Pa.
3074
3
2026
75
3
108
76
3
ΙΟΥ
.
77
3
276
78
3
362
79
3
449
3
537
Incbes of the Compaſso
8
3
625
3
715
83
3
807
84
3
0866
2285
3
1990
86
4 $084
283
87
4
88
279
89
Lux Stereometria.
79
Com.
F
Pan
*
89
;
377
90
4
475
:
g!
4
4
576
677
789
*882
4
the
93
94
4
Inches
95
4
987
96
S
093
97
5
98
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6142
$
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THAS
:
81
mu
Lüx Stereometrie,
ASPART II.
CH A P. I.
The Art of Meaſuring Surfaces and
Solids, Practically and The-
orecically Demonstrated.
After a moft Exact Mechód.
Lanometry is that Part of the Ma-
ematicks, by which the Surface
or Planes of Things are Meaſored
and Superficial Content found, which is
dove, either by a Square inch, Foot, Tard,
Pace, Percb, &c.
2
1011
Girion So
82 Lux Stereometria.
So that I knowing how many Inches
Feet, Tards, &c. the Sides Diameter, Cir
cnmference, &c. of any Figure are, and u
ſing the Methods following, you ſhall find
how many Inches, Feet, or Tards, are con
cain'd in ſuch Figures.
(1.) To find the Superficial Content
of a Geometrical Square.
А
B
VIDEO
A
Geometrical Square is equal both in
Sides and Angles, and the Segments
by the Diameter A, are equal ; that is,
the Segment Ais equal to the Segment B,
otherwiſe it would be no Geometrical
Square.
There is Two ways to find the Superfi-
cial Content of this Figure, ( viz. ) Mnltiply
the
Lux Stereometria
83
and you
you have
the Side by its ſelf, and you have the Su-
perficial Content in Square Inches; or Multi-
ply the Diameter by it ſelf, and
double the Content: For, Euclids 47. Prop.
Book 1. fays, That the Square of the Two
Leſſer Sides of a Right Angle Triangle, ad-
ded together, is Equal to the Square of
the Greater Side.
.
Example, The Side 12. by 12. Produ-
ces 144, and the Diameter a Inevitably is
'16. 97. which, being Squared, Produces
287.9789, which wants not 100 Parts
: of 288, double the Square of 12,
Die on
(2.) To find the Superficial Content of 4
Paralellogram, in Love
25b12 on loated sit
A -Paralellogram is a Figure, whoſe op-
A
poſite Sides are Equal and Parallel,
and has Four Right-Angles, and the Seg-
ments Cut by the Diameter are Equal.
...
bre
G 2
84
Lux Stereometria.
b
slogo3
e
A
Fe
Parallellogram is equal to a Triangle
of double Baſe, and equal height For
if you Multiply the Side a. B. into B.c. you
have the Content of the Parallellogram, So
likewiſe, if you Multiply the Side b. r. by
the half of the Side e. d. you have the Con-
tent of the Triangle, bed, which alſo
proves that a Triangle is but the of a Pa-
Tallellogram, or Geometrical Square, the Baſe
of the Triangle being equal to one ſide of
the Square, alſo equal in height
(3.) TO
Lux Stereometriä. 85
*
but the Angle d. is Acute, being leſs than
( 3.) To find the Superficial Content
of a Triangle.
Triangles are either Right or Oblique-
Angled ; A Right- angle is Compo-
fed of Two Right-Lines, Perpendicular
b C
g
to each other, as a. b. c. the Angle c. is
Right, for the Line a. c. is Perpendicular
to the Line b. c. as the Line b. c. is to a. c.
i
f
Ve
*
a Right-Angles and the Lines which
Inclofe the Angle, not being Perpendicu-
lar to each other; and the Angle f. c b. is
Obtuſe. by the fame Reaſon, and being
greater than a Right-Angle, by the An-
gle e. for the Sides, or Shanks b. f. Incloſe
the Obtuſe Angle, and the sides f. g. the
Acute Angle.
G 3
The
86
Lux Stereometrie.
vegada i el
The Sides of a Triangle have ſeveral
Denominations, viz. The Longeſt Side of
a Right-Angled Triangle is call'd the Hy-
pothenuſal, the shorteſt the Perpendicular,
and the other the Baſe; but it matters not
whether of the two Shorteſt you call the
Baſe: but of an Acute, or Obtuſe Angled
Triangle, you are to find a Perpendicular by
letting fall a Line from the Greateſt An-
gle upon the nighteſt place of the oppo-
fite Side, as above.
For ſuppoſe the side oppoſite to the
Greateſt Angle, be 72 Inches, Feet, or
Yards,
Lux Stereometriz.
87
Yards, and the Perpendicular to be 60. you
muſt Multiply the of the Baſe into the
Whole Perpendicular, or the of the Peya
pendicular into the Whole Bafe, as 72. by
30. makes 2100. equal to the Content in
Inches, Fect, Yards, or whatſoever your Gi-
ven Meaſure were.
( 4.) To find the Superficial Content
of a Rhombus.
e
b
:
.
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​C
A
Rhombus is a Figure of T10 Obtufe,
and Two Acute Angles ; and all
the Sides are equal ; but a Rhomboides is a
Figure, whoſe oppoſite Sides are only
qual , having Length and Breadth, as a
Parallellogram, or
Oblong put
Square, as above, a. b. C. d.
Maltiply the Perpendicular e.d. by the ſide
C. d. and you have the Content in Inches ;
for if the Sidec, d. be 56. and Rerpen. 50.
the Product will be 2800.
(5) To
puc out of
G4
88
Lux Stereometrie.
er hins
*
(5) To find the Content of a Trapezia
,
or Unequal Many Sided Figure, fuck
As a Connexion of Triangles.com
ကို လည်း
D
Tvide the whole Figure into Triangles
by Lines, from one Angle to ano
ther, and let fall Perpendiculars, fb chi as
the Prickt-Lines, from the Greateſt in
gles to the nigheſt part of the made Lines;
then find the Content of the ſeveral Tri-
angles, as before directed and add them
all together, and you have the Content of
the Tråpezia.
12 5:
(6) To
Lux Stereometria.
89
bul
top
6) To find the Superficial Content of
a Regular Poligon
..
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2a
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​A
Regular Poligon is a Figure of more
then 4 Sides, and all equal: and
every Angle is greater than a Right An-
ple.
Multiply the half Sum of the Meaſure
of all the sides by the Perpendicular, let
fall from the Middle of any one side to
the Center, and you have the Content in
Superficial Inches : Example the Length of
each
go
Lux Stereomatrid.
each side is 6 Inches, and Perpendicular
U. P. 5. the of the Sum of the Sides 36. i
18. Multiply'd by S. produces 94. the
Content of the Hexagon.
To find the Center of a Poligon
; if it
hath even Sides,draw a ſtraight Line from
the Middle of any oppofite fide, to the
Middle of the other; and if it hąth an odd
number of Sides, as 7. or 5. draw a Line
from any Angle, to the Middle of the op:
poſite ſide, and ſo croſs that with another
ſuch Line, and where they crofs is the
Center.
:
V
sono
mietin TOISU
5700314
Ve (7) TO
32 sio yne 19 olabim odj mori !!ci
Ati friend or stopy bnb di
to rirgto oslonilaudan him and
Lux Stereomatrie.
91
7) To find the Superficial Content of a
Circle. Je
:
A
Circle, of all Plains, is the moſt Or-
dinate by 10.
There is Two ways to find the Super-
ficial Content of a Circle, one by Plato, and.
another by Euclid. But the Proportion
between a Circle and a Square, could ne-
ver be exaâly found : ſo that until we
find what Proportion the Diameter bears
to
92
Lux Stereometriæ.
to the Circumference, we can never for
the Quadrature of a Circle ; for Ram
ſays, That the Circumference is Thri
the Diameter, and almoſt one Seventh
the ſame Diameter s ſo that this bein
nigheſt the Truth, we fall take it for
granted. Example.
The Diameter is 18. divided by : give
2. 57. added to Thrice the Diameter, 54
produces 56. 57. for the True Circumfe
rence. Now having thus found the Cir-
cumference, Multiply the Circumference
by , the Diameter, and you have the
Quadrature of the Circle , or the Square
Inches in the Circle.
Euclid, as Hero Relates, Squares the Cir-
cle thus ; If from the Quadrature of the
Diameter you Subſtract parts of the
ſame Diameter, the Remainder ſhall be P
the Content of the Circle. Example, of
both ways. ti et constig
T
The Diam. is-18.
The; of the Diam. 2. 57
7.) 18 (2.57
40
50
I
om
Added to thrice the Diam. 54
2.57
12.57
Makes
56,57 the Circ.
The
Lux Stereometria.
93
The whereof is
28. 28
The Diam. 18. is.com domain -9
The Content of the Circle--254.52
This is Plato's Way.
::
.
Here follows Euclids way.
14) 324 (23. i
The Diam, is
18
44 3
The Square of the Diam.324
20 69:3
The Å of the Square 69.3
6
of the Dia,
Subſtr. from the
Squareof the Diam.
254:7 the Cont of the Circle
Produces 1915
Duis
70 harighe
Theſe Two ways differ not two tenths
of an Inch : Therefore I hold both ways
to be exact.
The Common way is by the Rule of
Proportion, or fingle Rule of Three di-
rect, to find the Circumference, having
the Diameter, or having your Diameter, to
find the Circumference, As 7. is to 22.
fo is the Diameter, given to the Circum-
ference Required: Or as 22. is to 7. ſo
is the Circumference to the Diameter.
:- 22 : 18: 56.57 Circumference.
22:7; 56.57 ; 18 Diameter.
(8.) TO
94
Lux Stereometria.
(8) To find the Content of a Sem
circle.
:
E
A
Rectangle made of the Semi-dian
and the Circumference, is equ
to the Superficial Content of the Semi
Circle. witor 104
bod hley .
For if the Diameter be 18. and Circuit
ference 56.57. the Quarter of the Circum
ference 14. 14. Multiply'd by 9. the D
anieter produces 127.26. equal to the
Content of the Semi-Circle.
HOS
Radiation
(9)
Lux Stereometrie
95
(9) To find the Superficial Content of
any Sector of a Circle, or Segment
thereof, extending to the Center..
:
M"
Ultiply · the Arch 0.0, by the Semi-
diam. of the Whole Circle, and
you have the Content of the Section 0.0.0.
whether it be above or under a Quadrant.
Suppoſe the Diam. of the whole Circle be
18. and the Arch o, o to 22. Multiply the
Arch 11, by the Diam. 9. and the
Produ& is 99. the Content of the Section
Required to start ague
Si hastas: 20.000
body lunos ojn won
Page neslo sai
og host 2008 (10) To
96
Lux Stereometriæ.
(101) To find the Content of ſuch
an Segment as a. b.f. having the Con
tent of the whale Circle ; alſo havin
the bare Segment given without it!
Meaſure, to find the Content of the
Segment, and of the whole Circle
which it is a part.
d/
mais it pd dora
DTIG storia lodvor 16b
0.0.0-41011110
3293b200. Toh
so stari slattw siin .
fm da
Na firſt to find the Content of the
A
Circle.
Segment, having the Content of the
0 2
Suppoſe the Content of the whole Circk
to be 254. 52, its granted the Diameter is
18. and Circumference, 56. 57.
Now you are to find the Content of the
Two Sections e. d. a. and f.d.e. by the
laſt foregoing Rule, and Subſtract the Sum
of theſe Two Sections from the Content
of
the
Lux Stereometriæ.
97
mains 65. 52.
But the more eaſie way, and near
the Circle, and the Remainder is the Setti-
on a. C. d.
The Arch a. e. is 21 Deg. and ſo is f.g:
the of both thoſe Arches is the whole of
one Arch 21. Multiply'd by the Cemidia-
meter 9. Produces 1 89. for the Content of
the Sections a, e. f. g. which being Sub-
Itracted from the Content of the whole
Circle, leaves the Content of the Section
4. d. c. Thus 189. from 254.52. Re-
Then to find the Content of the Segment
72. C. Subſtract the Content of the Trian-
gle a. d. c. from 65.52. the Content of the
Section a.d.c. and the Remainder is the
Content of the Segment a.c. Example, the
Baſe a. g. is 12. and Perpendicular d. 1.6.
the of 92 Multiply'd by 6. produces 36..
the Content of the Triangle, which being
Subſtracted from 65. 22. the Section leaves
25. 52. the Content of the Segment a. b. o.
nough the Truth in Practice, is to Multi-
ply the Chord a. c. by , the Sine h.b. and
you have the Content of the Segment very
near the Truth.
H
(11) TO
98
Lux Stereometria.
(mr) To find the Diameter of a Cir-
cle, having the Segment of a Circle.
Chid
3
.
D'
sa di wobnie
(vide the Square of the Chord of
the Segment,
by the Sine or Alci.
tude of the Segment, add the Quot, to the
Sine, and the Sum is the Diameter of the
Circle, of which the Segment is a part,
Suppoſe the Chord of the Segment al
be 20, and Sine c. p. 6, the Square of the
* Chord 10 is 1oo, divided by 6 Quotes
16,6, which, being added to the Sine, PTO-
duces 22.6 the Diameter fought,
(12) Having
Lux Stereometrie.
99
(12) Having the Content of a Circle,
to find the Diameter.
STER
Ippoſe the Content of a Circle be 148
Inches.
A$ 22, is to 28. fo is 148 to the Square
of the Diameter, the Square Root where-
of is the Diameter.
I fhall work an Example of this, and
by the fame Rule you may find all the O-
ther Proportions that follow.
22 : 28 : 148
28
:
396
1184
Diameter.
22 ) 4144 ( 188 (137
194
19.00
184 23)
8 | 267)
31
اش
I might have brought out more Deci-
mals, and the Diameter would have
been more exact ; but I only give this as
an Example
H2
(13) Ha.
.
100
Lux Stereometrice.
(13) Having the Content of a Cir
cle, to find the Circumference.
A
:
S 7 is to Four times 22, which is
88, ſo is 148 the Content to the
Circumference.
Having the Diameter of a Circle; to find
the side of a Square equal to the Circle.
As 1.oo is to .886 fo is the Diameter
of the Circle to the side of a Square,
which ſhall be equal in Content to the
Circle.
By the Circumference given , to find
the side of a Square Equal.
As 1.oo is to .282 fo is the Circumfe-
rence to the side of a Square equal to the
Circle.
By the Diameter to find the side of a
Square Inſcribd in the Circle.
As 1.00 is to.7071, ſo is the Diameter
of the Circle to the Side of a Square in-
ſcrib'd.
By the Circumference to find the Sides
of a Square Inſcribd.
As 1.00 is to .225, ſo is the Circumfe-
rence to the ſide of a Square equal to
the Circle.
(14) TO
Lux Stereometria.
10l
:..
(14) To find the Superficial. Con-
tent of an Elipſis.
Multiply the Tranſverſe Diameter
of an Elipſis by the Conjugate,
and Extract the Square-Root of the Pro-
duct, and you have the Diameter of the
Elipſis equal to the Diameter of a Circle,
which will be equal in Content to the
Elipſis. Suppoſe the longeſt Diameter be
62. and the ſhorteſt 32. the Product of
theſe are 3224. Extract the Square-Root
of this and you have the Mean Diameter
South
H 3
(15) TO
of the Elipſis.
TO2
Lux Stereometriæ.
:.::
(15). To find the Superficial Content of
a Spherical Triangle.
b
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Sun
Uppoſe the Triangle be a. b. c. the
Spherical Triangle being made of the
3 Arches of a great Circle, as all Spheri-
cal Triangles are, you are to cut off the
Segments a. b. b. c. and 6. a. by the Prick
Lines in the Triangle, and meaſure them
as is directed in Propoſition 10. Chap. !.
Then find the Content of the Plain Tri-
angle, 4. 6. b. which is made, by the
prickt Lines, by Propoſition 3.Chap. 1. and
add to the Content of the Plain Triangle,
the Content of the Segments, a. c. and 6
b. and ſubſtract from the Triangle the
Segment a. c. and you have the Content of
the Spherical Triangle a.b.c.
Example
Lux Stereometriæ.
103
Example.
Let the Content of the Plain Triangle
be 108. and the Content of the Segntent
a. c. 30. and b. 6. 22. all theſe added
together amounts to 160. from that take
the Segment a. 6. 20, and you have 140.
Equal to the Content of the Spherical
Triangle, a, boc
CHAP. I I.
Menſuration of Solids, Regular
and Irregular ; Alfo of Imboſſed
Solids.
R
Egular Solids are ſuch whoſe
Bounds are either ſtreight Lines,
or Circular. And Irregular Solids
are ſuch Solids, whoſe bounds are
mixt Lines, Streight and Crooked. And
Imbofs'd Solids are comprehended of an
Imboffed Sarface, ſuch as an Egg-Shell.
[
(I TO
104
>
Lux Stereometrice.
**
(1) To find the Solid Content of a
Cube.
в у
A
6
D
E
G
F
... sont
A
.
of
$ Length is only proper to a Line,
Length and Breadth to a Surface ſo
a Solid Body is a Compoſition of Lines and
Surfaces, therefore conſiſts of Length
Breadth, and Thickneſs, for every part of
a Body is alſo a Body.
So that as a Line, Multiply'd in it felf
,
produces a Surface or Geometrical Square,
to that Square being Multiply'd by its
firſt Original or Root, produces a Cube
,
or Solid Body, of equal height to the Root
of the Square Surface.
For
Lux Stereometria.
105
For if a. b. be 6 Inches, that Line Mul-
tiply'd by it's ſelf, produces the Square
Surface A. B. Ć. whoſe Superficial
Area is 36. and Sides all equal to the
Side a.b. given, and that Surface being
Multiply'd by 6 again, gives Thickneſs and
Height, and ſo becomes the Solid Body
AB.C.E.F.G. of Length, Breadth, and
Thickneſs,
So that to find the Solid Content of a
Cube, Multiply the Side by it's ſelf, and
the Product by. the side again, and you
have the Solid Content in Inches, then to
know how many Solid Foot is herein :
divide by 1728. the folid Inches in a Foot
and you have your deſire. !
Suppoſe the Side a.b. be 30 Inches, Square
30 and it Produces 200 Multiply that by
30 and the Product is 27000 divided by
128, quores 15.6 Foot : But for thoſe
that underſtand the Rule of Practice it is
done more eaſy thus,
f.
:
106
Lux Stereomatriæ.
f. 8
.
2.63
3.11
2.6
5.0
1.31
ber
6.32
Side
2.6
12.6 Duo
..
15.71
.
The Sum of all the Segments of a Cube
are equal to the Cube. For if the side of
a Cube be 30 Inches, the Cube of 15, the
of the Side is but 3375, equal to of
the Cube. So that in a Cube of a Foot
w Solid, there is 8 Solid Feet and 64
Quarter-feet, c. So that a Cube is in
aquadruple Ratio to a Square.
.
(2) Ta
Lux Stereomatrie, .
107
Sagt
-Y
( 2 ) To find the Content of a Pa-
ralellapipedon, or Solid Oblong.
id
Ind the Superficial content of any one
of the Plains and Multiply that into
the height, and you have the Solidity.
Suppoſe the side a. be 60.2 and Side b.
23. the product of that plaine is 1384.6
and that Multiply'd by the height, or
Side a. produces 83352.92 Solid Inches,
which may be brought into Solid Yards,
Feet, Gallons, &c, by dividing the number
of Solid laches, by the Solid Inches, in
either of theſe reſpedively.
.
(3) TP
108
Lux Stereometrix.
!
( 3.) To find the Solid Content of
*
Bruido Cronogiqslla
Cylinder.
1
d
f
توجی
--
:
PE
е с 2
Et. Ramus ſays, That a Cylinder is
made by the turning about of a Right
angled Parralellogram, the one ſide ſtand-
ing ſtill. It is a Solid, whoſe Diameters
in all parts are equal, and Baſes Parrallel
.
The Superficial Area of the Baſe Mul.
tiply'd into the height produces the Solid
Content,
So that if the Diameter of the Baſe a.de
be
40, the Circumference (by the Com-
mon-Rule as 7 is to 22 ) will be 125.7.
Now you may omit the way of finding the
Area of a Circle, practiced by Plato, &c
and take the common way, by Multiply.
ing the - Circumference by the Diame-
tefs
Lux Stereometria,
109
1.
at
ter, and Multiply that Product by the
Depth.
4. :) To find the Content of the Section,
or Cemi-Cylinder d. b. e.
Y this
Ou are only to Multiply the Area of
the Baſe by the height,and you have
the Solid Content, but to find the Content
of the Section d.b.c. it is harder ; for you
muſt find the Content of the Priſm a.c.d.
by Multiplying the Area of the Baſe by
Altitude ( but the Priſm is a Cylin-
der ) and Sabitract the Content of the
Priſm. from the Content of the Cylinder,
and, the remainder is the Content of
the Segment dib.c. but d.b.c. is the Cy-
linder, by the 16 Propofition 1. B. Euclid.
riangles of equal Baſe and equal height
are equal. So the Triangle d.b.c. is equal
to die.c. being both on the fame Baſe and
between the fame Parallels; therefore d.
kis
, is of the Cylinder of which it is a
Segment f.b.c. would have been the of a
Cylinder of the fame height with the Seg-
ment.
(5) TO
saigon.
:
G
So you are to Multiply the Area of the
Multiply'd into the Area of the Bale
ES 401983.8. the Content
IIO Lux Stereometriæ.
(5) To find the Solid Content of
Pyramid.
A
Pyramid is of a Parallellapipedor
or Priſm ; or of any of the Regular
Poligons of equal Baſe and Altitude.
Pyramids Baſe by the Altitnde and you
have the Content.
Suppoſe the Area of the Baſe to
37923. and height 32, the of 32 is 10
of the Pyramid.
be
There
Lux Stereometria. III
There is ſome difficulty in finding the
(pendicular height of a Pyramid or Cone,
but it is nothing but a Confectary out of
the
he 47. E. 1. B. from the Square of the
laint height Subſtract the Square of of
one of the ſides of the Baſe, and the ſquare
Root of the difference is the Perpendicular
altitude or Axis.
6) To find the Solidity of a Cone
.
.
-
cis
A
Cone is of a Cylinder, of Equal
Baſe and Altitude, and is made by
the turning about of a Rect-angled-Tri.
one lide ſtanding ſtill.
The Content of a Cone is found after the
line inanner as the Pyramid, only the one
has a Circular Baſe and the other a Square,
Triangular, Pentagonal, Oxogonal, or a
ſtreight
112
Lux Stereometriæ.
.
ſtreight Linc-Baſe ; ſo that the Area of
their Baſes muſt be found, according to
the ſuperficial propoſitions foregoing.
G) To find the Solid Content of the
Fruſt um of aCone ſuch as a Brewers
Tun.
nebo ... ..
He moſt Practical way is to reduce
it into a Cylinder,by finding a mean
betwixt the greater and leſſer Diameters,
thus ſubſtract the leſſer Diameter from the
greater, and add the difference to the
leſler Diameter, and that is the mot
Practical mean. Multiply the 1 of this
mean by the Circumference of that mean
Daneter, and multiply the Product by
the Depth or Length, and you have the
Content in Solid Inches and Parts.
Example
The greatel Diameter is 62.3. the leſſer
is 57. the difference is 5-3, the of which
is 2. 65 added to the leſſer Diameter pro-
duces 59. 65 for a mean Diameter the
Circunference thereof is 187.47. Now
thes of this Circumference, viz. 93.73
Multiply'd by the Diameter 29.82 pro
duces
Lux Stereometrice.
113
duces 2795.0286. for the Superficial Area,
Multiply'd by the depth 22 produces
61 4.90.6292, the Solid Content.
But the moſt Geometrical way is Having
the Cones Fruſtum to find the Altitude of
the whole Cone, and ſubſtract the Altitude
of the Fruftum from the Altitude of the
whole Cone, and the difference is the Alti-
tude of the top or ſmall Cone c.d.e. the
Content of the ſmall Cone being taken from
the great Cone leaves the Content of the
Fruſtum a.b.c.d. Thus, let the height of
the Fruſtum be 22 the greateſt Diameter
62. 3. the leſer 57, the difference of Dia-
meters is 5.3. Multiply the greater Dia-
meter by the Fruſtum's altitude and
divide by the difference of Diameters and
you have the whole Cones Altitude
Example.
62.3. Greater Diam.
22 Fruſt. Altitude
.
1 24.6
124.6
-
5.3) 1370.6 (258.0
310
456
32,0
2
298,6
114
Lux Stereometrie.
258.6 The Altit. of the whole Cone.
22.0 The Altitude of the Fruſt.
236.6 The Altit.of the Leller Cone.
:
Now find the Solidity of the leſſer Cone,
and ſubſtract from the Solidity of the
whole Cone, and it leaves the Solidity of
the Fruftum a.b.c.d.
Note, That the Leſſer Diameter of the
Fruſtum is the Baſe of the ſmall Cone,
be
cauſe the Fruſtum is a part of the Great
Cone.
(8) To find the Solidity of a Sphere.
E
::
Very Sphere is equal unto Two Cones
,
whoſe height and Diameter at the
Bale is the ſame with the Axis of the
Sphere, or a Sphere is two thirds of a Cy.
linder, whoſe Diameter and height is equal
to the Axis of the Sphere, as Arebimedes
manifeſts in his firſt Book of the Sphere
and Cylinder.
Therefore, the Superficial Area of the
Sphere's Axis, being multiply'd' by twice
,
the
Lux Stereometria.
115
the Sphere's Axis is the ſolid Content of
the Sphere.
Or, as 2 is to 3, ſo is the Content of
a Globe to the Solid Content of a Cylinder
of the ſame Baſe and Altitude. So that
when you have the Dimenſions of your
Globe, compute the Content of it as it
were a Cylinder of that Diameter and Al-
titude, then as 2 is to 3 fo is the Content
of the Suppoſed Cylinder to the Content
of the Globe.
Suppoſe the Globe's Axis bei 2 Inches,
and greateſt Arch or Circle be 37.7.
the Circumference is 18.85.
the Diameter is
6.
Superficial Content II 3.1.
the Double Axis is 8.
The Content of the Sph. 904.8.
1 2
AS
A
116
Lux Stereometria.
::
the Altitude of the other Segment, by the
a Circle : having the Segment, Divide
As 21 is ta 11, ſo is the Cube of the
Globes Diameter to the Solid Content
Or as 42 is to 22, fo is 1728 the Cube
of the Globes Diameter to the Solid Con-
tent of the Globe.
Or, having the Circumference of the
Sphere, to find the Solid Content.
As 1.0000
Is to 0.01688
So is the Cube of the Globes Circunt-
ference to the Solid Content.
(9) To find the Solidity of a
Segment.
His muſt be done by the Rule of
Proportion double. Thus, firſt find
Propoſition for finding the Diameter of
..
Sphere's
1E
the
Lux Stereometria.
the Square of the Diameter of the Seg-
ments Baſe by the Altitude of the Seg-
ment, and add the Quot, to the Segments
Altitude, and the Product is the Axis of
of the Sphere. Then Subſtract the Seg-
ments Altitude from the Axis of the
Sphere, and the remainder is the Altitude
of the Great Segment.
Then by Proportion, As the Altitude
of the Great Segment is to the Altitude
of the Leſſer Segment given, ſo is the
Altitude of the Greater Segment added
to the Axis unto a fourth Number.
Then Multiply the Quadrant of the
the Chovd of the Segment by the fourth
Number, and you have the Solidity of the
Spberes Segment.
Example.
Suppoſe the Diameter at the Baſe of the
Segment a.b.c. be 10, and the Altitude 4
Inches.
Half the Baſe
5
S
Š
The Altitude is 4) 25 (6.35
10
29
13
6.25
118
Lux Stereometria.
6.25
08:25
.
the Segments Altitude.
10.25 the Diam,or Axis of the Sph
.
4
.
::
6.25 The Altit.of the great Seg.
5.12 the Axis.
11.37 Sum
6.25.411.377.2
:
100
7.2
200
700
720.0
The Solidity of the Spheres
. Segment
720
.
( 10.) The Superficial Content of a
Sphere is found by Multiplying the whole
Diameter by the whole Circumference;
for the Superficial Content of a Spbere is
equal to 4 times the Superficial Content
of its greateſt Circle.
The
Lux Stereometria.
119
The Superficial Content of the Segment
of a Globe is found by Multiplying the
height by the Circumference of the whole
Sphere:
( 11 ) How to find what Diameter the
Shell of a Boom was, having any part of
the Shell.
b
f
Apply a pair of Calloppers to the Arch
ing part of the Broken Piece, as near to
the Edge that was joyned to the whole
as you can, and ſet it off upon a Sheet
of Large Paper, and let it be the Line
a.c. then take the height of the Broken
Piece from the utmoſt extremity of the
Fracture to the higheſt part of the Arch,
and let that be the Line b.d. divide the
Line into Two equal Parts at d. and
raiſe the Line b.d. (being equal to the
14
height
1
I20 Lux Stereometrie.
height of the piece of the Boom ) Per-
pendicular to a.c. draw the Lines a.b. and
b.c. then raiſe the Perpendiculars e. and
f. and draw them out at length until they
meet at g. and where they Croſs is the
Center of the Boom, of which a.b.c. was
a part, and the Lines.e.g. and f.g. are the
Semi-diameters of the Boom, or theſe two
added together are equal to the Diameter
of the Boom.
( 12 ) Suppoſe a Brewer to have an
Oblong-Back, 240 Inches in Length, and
Breadth 104 Inches, and the Room where
he is to erect another Back ( which he
deſigns to be of equal Content with the
former ) will allow the Breadth to be but
68 Inches, what Length muſt this Back
be to be equal in Area and Content with
the former, being of equal Depth.
Divide the Superficial Area of the
Given Back by the Breadth of the de-
fign’d Back, and you have the Anſwer.
Example
The Superficial Area
24960.
The Breadth of the deſign d Back 68.
68) 24960 (367 the Length requir’d.
456
480
The
Lux Stereometrie.
HI
The Proportion holds thus. As the
Breadth of the deſigned Back is to the
Breadth of the Given Back, ſo is the
Length of the Given Back to the Length
of the Back requir'd.
( 13. ) Having the Length and Breadth
of the Given Back, and Length of the
Back requir’d, what ſhall the Breadth be.
As the Length of the required Back is
to the Length of the Given, fo is the
Breadth of the Given to the Breath of
the required.
367-240--104-68
(14.) A Square Back being 100 Inches
each ſide, and the Breadth of an Oblong
being given 52, what Length muſt the
Oblong be ſo as it may be equal in Content
to the given Square.
As the Breadth of the Oblong required
is to the side of the Square Given ; ſo is
the side of the Square to the Length of
the Oblong. Example.
tresnakutanen 1 00sportartending I 00-serien
10000
480
:
120
16,0
1
.
عدم معه
(15.) TO
A
Lux Stereomatria.
(15.) Tofind the Dimenſions of Cone
that ſhall be equal to a Sphere.
.
Uch a Cone as has its Axis equal to
the Radius or Semi-Axis of the Sphere,
and the Diameter of its Baſe twice the
Diameter of the Sphere is equal in Content
to the Sphere.
Or ſuch a Cone as hath its Axis equa
!
to twice the Diameter of the Sphere,
its Diameter at the Baſe equal to the Dia
meter of the Sphere, is equal to the Sphere
.
and
Example.
Let the Sphere be a.c. b.d. the Diameter
is a.b. or c.d. 30 Inches; twice the Dią.
meter
Lux Stereometria
123
2827.5
meter e.f. 60. the Semi-Diameters a.e. 152
the Content of the Sphere is 14142.
The Superficial Area of the Baſe of the
Cone c.o.f. is
Multiplyed by the Height is 14137.5
Which wants but 4.5 of the Spheres
Content, which deficiency proceeds from
the want of more decimals in the finding
the Content of the Baſe.
The Baſe of the Conc a.e.b is a.b
30
Whoſe Superficial Area is 707.10
Multiply by the height is 1414.2.0
Efore I enter upon Gauging I ſhall give
you an Account of the Area of Unity
and Gauge-Points together with their uſe,
and how they are found.
(16.) To find the Area of Unity, or
Superficial Content of a Circle whole
Diameter is one Inch.
The Diameter being 1, the Circum-
ference is
3.14159
Circumference.
1.57079
Diameter
S
The Arca of Ulnity. 785395
Multiply the Square of any Diameter
by
124
Lux Stereometria.
1
by the Area of Unity, and the Product is
the Square Inches contain'd therein.
The Gauge-Point for Ale-Gallons is
18.95
The Gauge-Point for Wine-Gallons is
17.15.
The Square of the Gauge-Point for Ale
is
359.95
The Triple Square of the Gauge-Point
for Ale is
-1077
The Square of the Gauge-Point for
VVine is
294
The Triple Square of the Gauge-Point
for VVine is
882
The Gauge-Point is the Diameter of a
Circle which holds a Gallon on an Inch
in Depth. For if the Diameter of a
Circle be 18.95, and one Inch deep, its
Content will be/282 Inches, that is the
Solid Inches in an Ale-Gallon.
If you divide the Square of any Diameter
by the Square of the Gauge-Point, you
have the Gallons contain'd on an Inch in
depth ; Or if the Triple Square of a
Circle be divided by the Triple Square of
the Gauge-Point you have the Area : For,
Since the Single Square of any Diameter,
divided by the Single-Square of the Gauge
Point, produces the Area, conſequently
the Square of three Circles, divided by
thrice,
.
Lux Stereonretria.
125
thrice the Square of the Gauge-Point,
muſt needs give the Area.
CHA P. II.
Stereometry of Gauging.
L
Et none enter into Gauging until
he is ſufficiently qualified in Vul-
gar and Decimal Arithmetick,
and very dexterous and ready in
the ſingle and double Rules of Proportion
direct, for that is the moſt uſeful Rule
alluſive to Gauging. And likewiſe let
him be very perfect in the Geometrical
Prob. foregoing, and meaſuring of various
Surfaces.
(1) To find the Area of an Oblong,
or Geometrical Square in Ale or
Wine-Gallops.
TH
He Proportion is, as the Cubical
Inches in an Ale or VVine-Gallon,
126
Lux Stereometria.
is to tlie Breadth in Inches and Parts, ſo
is the length to the Area.
.*.*..
Example.
B L
282-47.259.39.92 Area.
For a Geometrical Square; as 282 is to
the ſide, ſo is the ſide to the Area,
Example.
fide fide
35.435.4: 4.44 Area.
Or as 231 is to the ſide, 10 is the fide
to the Area in VV'ine-Gallons.
·
282-35.4
( 2 ) To find the Area of a Triangle
in Ale or Wine-Gallons.
F"
hind the Superficial Area as is directed
in the 3d. Propoſition of Surfaces ;
and divide by 282 or 231, and you have
the Area in Ale or VVine.
Or by Proportion thus,
As 282 is to the Bafe, ſo is the , Per-
pendicular to the Area in Ale- Gallons.
Example. Suppoſe the whole Baſe be
12, and of the Perpendicular 30.
Baſe 3 Perp
282—72-30-7.6 Area.
(3.) T.
Lux Stereometrie.
127
( 3.) To find the Area of «Rhombus
in Ale or Wine Gallons.
A
S 282 is to one of the Longeſt fides,
ſo is the perpendicular to the Area
in Ale Gallons.
282
-5650-_-99 Area.
( 4.) To find the Area of a Trapezia
in Ale or Wine Galons.
Ivide it into Triangles, as is directed
in Prop. sth. and work the ſeveral
Triangles, as is directed in the foregoing
Theorem of a Triangle, and add the
ſeveral Areas of the Triangles together,
and the Sam is the Area of the Trapezia.
Example of one Triangle.
282.-03-29,2-
4.SI
..
(5.) To
128
Lux Stereometriæ.
(5) To find the Area of any Regular
Poligon.
Et fall a Perpendicular as in Prop. 6.
L
tlien by the Rule of Proportion,
As 282 is to the ſum of all the ſides
,
fo is the perpendicular to the Area in Ale
Gallons.
3
Example.
Suppoſe the Sum of all the ſides be
8o and the Perpend. 16.
282-80--16-4.5 Area.
(6.) To find the Area in Ale Gallons
of a Cylinder:
S 359 is to the Diameter, fo is the
fame Diameter to the Area ſought
.
A
Example.
3594-40
40-4.0
-404.45 Area:
Or, as 282 is to half the Circumference
fo is half the Diameter to the Area in Ale
Gallons.
Example
Lux Stereometrie.
129
,!!
20.
Example.
The Circumference is
125.7
The Diameter is
40.
Half the Circumference is 62.85
Half the Diameter is
282-62.85--20-4.45
Note, If you would know how many
Wine Gallons the Cylinder contains on
an Inch in Depth, inſtead of 282 you
muſt uſe 231, and to know the Content
of the Cylinder in Ale or Wine Gallons,
you muſt multiply the Area, or what it
holds upon one Inch, by the Namber of
Inches and parts the Cylinder is in
depth.
(6.) To find the Content of a Semi-
Cylinder, ſuch as the figure is.
L
K
Suppoſe
130
Lux Stereometrie.
Sum
Altitude 105.2.
Uppoſe the Baſe a.b. be 40 Inches, and
As 359 is to the Square of the Dia-
meter of the Baſe, fo is the half of the
Altitude to the Content in Ale Gallons,
The Square of the Diameter of the Baſe
is found to be 1600 half the Altitude is
52.6.
359---1600-_-52.6---234.4. Gallons, .
2.0.b. is likewiſe the of a Cylinder of
the ſame Baſe and Altitude by the 3d
Prop, of the 2d Chap.
(7.) To find the Area, or Content of
Such a Segment of a Cylinder, which
hath its ptain Surface cut parallel
to the Axis, not touching the Center.
d
flagga
Lux Stereometria.
131
FR
Urlt find the Superficial Content of the
Baſe a.0.c.d. in manner following ,
and Multiply that into the height.
Múltiply the Chord c.d. by the verſed
fine 2.0, and you have the fuperficial Area
in Inches, then divide by 282,and theQuot.
is the Area in Ale Gallons.
Example.
The Chord cd. is 20.2. and the verred
line is 9. and Depth 48.
282) 121,2 (42
84.0 48
Volgens een initi
27 6-
336
168
net
Ale Gallons
2 0.16
Or by the Rule of Proportion, as 3,82
or 231 is to a Rectangle made of the
Chord Line, and the verſed line, To is
the Length to the Content in Ale or Wine
Gallons
282---121.2---48---20,6 Ale Gallop.
K 2
(.8.) TO
132
Lux Stereometria.
( 8.) To find the Solid Content of a
Cone in Ale and Wine Gallons.
Ou are firſt to find the Cones Axis
by ſubſtracting the Square of the
Semi-diameter of the Baſe from the Square
of the Cone's flant height, that is a Line
drawn from the Extent of the Baſe's Dia-
meter to the Vertex; and the Square
Root of the difference is the Axis or Per-
pendicular Altitude of the Cone.
Example
The ſlant height is
The Diameter is
200
106
37191
192
The Square of the ſlant height is 40000
The Square of the Semi-diam. is 2809
TILOR
The Difference is
Square Root, or Perpendicular
The Perpendicular is
64
ei Then, as 359 or 294 is to the Square
of the Diameter of the Baſe, ſo is the the
Axis to the Content in Ale or Wine
Gallons
DOMINO D d
359
Lux Stereometrie.
133
2003
359_1123664
64
44944
67416
719104
Or, As 1077 is to the Square of the
Diameter at the Baſe, ſo is the whole
Altitude to the Content.
1077
-11236--192
2003
*
( 90 ) To find the Content in Ale or
Wine Gallons of any Fruftum of 4
Conc
.:
I
Thall only make uſe of the moſt Geo-
metrical Rule, and Conſequently Ra-
tional, and leave the moſt Practical ways
to thoſe who have a mind to make uſe of
theni.
ITp on
an
b K3
This
134
Lux Stereometria.
P R
AF
E
IN
KA
OSC
B G с
This figure is deſign'd only as a Geo-
metrical Demonftration, not to be uſed
in the Practice of Gauging, but to ſatisfy
the Curious Artiſt of the Truth of the
work and that the true mean is found. .
poll: DOV
Example.
Let Q.C.B.E, repreſent the Fruſtum of
a Cone ſtanding on its leſſer Baſe, whoſe
top Diameter is
** Bottom Diameter
2 Draw the Line C.R. Perpendicular to
C.B. and continue the Line C.B. to 0.
making c.0. cqual to Q.R. let the Line
Q.R. and 0.c. be divided into two equal
parts at the points P. and S.then draw the
Parrallel Line P.S. and where it Cuts the
40.4
32.2
Depth
38.0
fide
Lux Stereometria
135
3
be added to the Lefler Diameter
ſide of the Fruſtum is the Mean-Diameter
of the Fruitum.
Proof. Extend the Line C.B. to G. at
the difference of Drameters, and from
thence raiſe a Perpendicular, and the Lines
G. A. and S.P. make a Cylinder equal to
the Cone's Fruftum
For the Triangles M.and N. are equal,
and ſo are the Triangles K.and L.by the 29
of Euclid Lib. 1. ſo that as the Fruſtum is
more than a Cylinder at the Top by the
Triangles M.K. ſo it is leſs at the Baſis
by the Triangles N.L.
Rule, As 2 is to i, fo is the difference
of Diameters to the Number which muſt
Or,
the difference of Diameters added to
the Leffer Diameter is the true mean.
Top: Diameter 40.4
Bottom Diameter 32.24
Difference
8.2
.
Difference
Bottom Diameter
4.1
32.2
36.3
...:
136
Lux Stereometria.
As 359 is to the Mean Diameter, fo is
the Mean Diameter to the Content of one
mean Inch in Depth ; that Multiplyed
into the Depth Produces the Content.
Example.
359-36.336.363.667
Depth 38
29336
II OOI
Content 139.346
109. A Pyramid has a ſquare or right-line
Baſe, and has ſo many flat Surfaces as
there is fides at the Baſis, but decreaſes
gradually to the Vertex; fo that the dif-
ference between a Pyramid and a Cone is
in the Formation of their Baſis, and may
be meaſured by the ſame General Rule,
having the Areas of their Baſis, according
to the Rules preſcribed in the sth and 6th
Propoſition of the 2d Chapter. -
cho
C H A P.
Lux Stereometri&.
137
alike Scicuate, the Diameter of the
СНА Р. IV.
To Gauge Brewers Tuns of Vari.
ous Forms and Scituation.
Uppoſe a Tun to have a Circular
and an Elliptical Baſe Parallel and
Circular Baſe 72. the Length of the Ellipti-
cal 72.5 and Breadth 53 Inches and Depth
86 ; how many Ale Gallons will this Tun
contain.
Find the Area of the Circular Baſe and
then of the Elliptical, and multiply the
of the Sum of thoſe two by the Depth,
and you have the Content.
o
Example
138
Lux Stereomatriæ.
Example.
The Area of the Circular Baſe
Area of the Elliptical Baſe
14.4
10.6
Bananesinin insan
Sum of both Areas
25.0
12.5
Sum
Depth of the Tun is
86
750
1000
Content
10750
A Tun whoſe Baſes are unequal, one
being, Square and the other Oblong; the
Side of the Square Baſe is 40. 2 and Long
Side of the Oblong Baſe 40.5 and Thort
Side 32 and Depth 50; what's the Con-
tent,
Find the Area of both Baſes and mul-
tiply the Sum of both theſe Areas by
the depth of the Tun, and you have the
Content in 'Gallons.
Example
Lux Stereometrie.
139
Example
The Area of the Square Baſe is 5.73
The Area of the Oblong Bafe is 4.57
Sum
Sum
Depth
10.30
5.15
50
Content in Ale Gallons.
257.50
Of Cask-Gauging
A
E
go
F
D
Б
Efore a
B
Man goes abont to Gauge any
Vellel or Cask, he is to have reſpect
to the form of it, and to conſider under
what denomination the Figure of the Cask
ought to be ; whether Cylindrical, Parara-
bolical,
140 Lux Stereometriæ.
bolical, Hiperbolical, Spheroidal, or Co-
noical ;' But to wave all niceties in Cask-
Gauging, being that ſuch variety as is
here mentioned cannot well be diſtin-
guiſhed by a mans Eye, and ſome thereof
may be denied, we ſhall only make ſuch
diſtinctions here as are moſt obvious, viz.
the midle Fruſtum of a Spheroid, the
midle Fruſtum of a Parabola, and the
midle Fruftum of two Cones abutting on
a Common Baſe, as are demonſtrated in
the following Figure, the Outmoſt Arch-
line forms a Spheroid, the midle Arch
forms a Parabola, and the innermoſt de-
ſcribes two Cones abutting on a Common
Baſe.
Let A.B. be the Bong Diameter, and
C.D. the Head Diameter of either of theſe
Casks and E.F. the Length.
Now we ſhall Gauge theſe three Casks
ſeverally by the Rules proper to their
Variety, and ſhall begin firſt with a
Spheroid.
Suppoſe the Bong Diameter of a Spher-
oid to be 26 and head Diameter 22, and
length 34, how many Ale Gallons will this
Contain.
I ſhall here mention ſeveral ways pro-
poſed by ſeveral Authors and leave every
man to his own Choice in uſing any one
of
hem.
Firſt,
Lux Stereometrik.
141
i
Firſt, Mr. Everard in his Stereometry
makes uſe of a very Accurat way, viz.
To once the Sum of the Squares of the
Bong and Head Diameters, add the Sum
of the Squares, and to that add the
difference of the Squares of Head and
Bong, Multiply that Sum by the Casks
Length, and divide by 1077 and you have
the Content in Ale Gallons, or by 882
for Wine Gallons.
Secondly, Other Authors direct that to
the double Square of the Bong Diameter
you add the Square of the Head, and
Multiply the Sum by the Length, and
divide by 1077.
Thirdly, Another way is, to ſubſtract
the Head Diameter from the Bong, and
Multiply the difference by 1, add that
Product to the Head Diameter, and you
have the mean Diameter of the Cask,
which Squared and divided by 359 Pro-
duces the Area, which Multiply by the
Length, and you have the Content in Ale
Gallons
Example
143
Lux Stereometria.
@ Example.
Bong Diameter
27
Head Diameter 24
::
3
2. I
24
Mean Diameter 26.1
Squared and Multiplyed by the Length
40, and divided by 359 produces 76 Ale
Gallons.
I ſhall here ſhow another way which I
have not only took pains to prove by
Geometrical Lines of Proportion, but by
filling the Cask with Water by a ſtatute
Gallon I had made for the ſame end
which was 16.79 Inches each ſide and
one Inch in depth : and 77 of thoſe fill’d
the Cask of the before-mentioned Dimen-
fions, only there was about a Pint over
when the Cask was full.
I ſhall here ſhow how the mean Dia.
meter is found by Geometrical Lines
and
Lux Stereometrie.
142
BE
9
,
and leave it to the Judgment of the better
Artiſt to Judge on the truth of it.
Let A.B. be the Bong Diameter, 27
mean between thoſe, by Multiplying by
:7, and adding to the Head Diameter, will
be 26.1, equali to the Line ses, for if
the Line A.B. be 27 Equal parts, the
Line s.s. will be 26.1, but the true mean
is the Line 0.0.
B
Demonſtration, Continue the Line D.C.
to m. fo that the Line m.C. may be equal
to An. draw the Line Am, parallel to
1.C. then divide the Lines A.n, and m.c.
into four equal parts ; draw the Line 2 9.
and it will cut the Curved Line at o, which
is the mean between Head and Bong
viz.
144
Lux Stereometria.
viz. 26.25. for q.C. is three fourths of
the half difference of Diameters m.C.
therefore you are to add three fourths of
the difference of Diameters to the Head
Diameter, and that is the Mean-diameter
0.0. for ſince the Arch riſes gradually be-
tween Head and Bong; "the proportion
muſt be as 4 is to 3, and not as to is to
7. for 0.B. is of the difference of Dia-
meters.
Example.
Bong 27
Head 24
2) 3.00 775
3
20
2.25
24
The mean Diam.26.25
The Square of this Multiplyed by 40
the Length, and divided by 359 quotes
76.88.
Let there be a Cask of the ſame di-
menſion as to Bong, Head, and Length,
but leſs Curved than the former, ſuch as
the middle Line in the laſt Figure but one,
which we ſhall call a mean Cask.
Multiply
Lux Stereométriæ.
1451
.
Multiply the difference of Diameters by
by .64, and add the Product to the Head
Diameter, and the Sum is the mean Dia
meter, which being Squared and Multi-
ply'd by the Casks Length, and the Pro-
duct divided by 359, quotes the Content
in Ale Gallons, or by 294 for Winery
(2.) To find the Vacuity of the Spher-
oidal Cask poſited with its Axis
parallel to the Horizon. o xorit 2013
pd coniu
I
ſhall not go about to preſcribe any
new way, there being ſo many ways
already, which do to a niçety agree in
the main, but ſhall adviſe you to make
uſe of Mr. Everard's way in his Stereome-
try, by his ſliding Rule, where the Seg-
ments are fitted to a Spheroid, there
being none more eaſy; however I hall
not leave you altogether deſtitute here,
but ſhall fet down ſuch plain ways as my
brevity will allow.
Suppoſe the Dimenſions of a Spheroid
to be as follows, viz.
to be
Content
57
Bong
Wet
17.
L
Divide
1
25.8
1462 Lux Stereosnetrix.
Divide the wet. Inches by the Bong
Diameter, adding a Competent Number
of Cyphers to the wet Inches, and if the
Quot. Exceedi.go, add to it a third part
of what it is above,50, but if the Quotient
want of so, then ſubſtract a third part
of what it wants of .50, and Multiply the
remainder by the Content of the Cask,
and you have the Quantity of Liquor re-
maining in the Cask, as, near the Truth
as Common Practice doth require, buc
r 'would not have any perſwade them-
felves that they can do it exactly by any
Rule; Curved Lines being ſo Various and
the difficulty of finding a mean Segment
and Verſed Line either Geometrically
or Arithmetically between Head and
Bong, being ſo hard.
26W 2
gue si la
$ obou 2
* 500
erori zaidi 0 1983 gota Dovoza
Ym er 275* malu dabing-sasa un
biozole a la enoiloma
Pat Here
Lux Stereometria.
147
Here follows an Example.
25.8) 17.200 66
1 720 0:50
172
.16
sima
$).167.053 one third the difference.
1.00.66
113
$7 the Content
.
이
​4991
3565
40.641 Remaining Liquor
3.) To find the Vacuity of a Stand-
ing Cask poſited with its Axis Per-
pendicular to the Horizon
Tuppoſe the Dimenſions of the Cask
to be
Content
- - - -
148
Lux Stereometric.
Content
Bong Diameter
Head Diameter
Length
Wet Inches
| 1
95
28.6
21.8
48.5
13
Subſtract the Area of the Head Dia-
meter from the Area of the Bong, and
divide the difference by 1.78, then add
the Quotient that proceeds from that
diviſion to the Area of the Head-dia-
meter, and Multiply the Sum by the Wet
Inches, and you have the Quantity of the
remaining Liquor.
Example.
The Area of the Bong is
The Area of the Head is
2. 28
1.32
96
The difference is
21:
1978).9600 (-53 iniwal
700
950 0 10 ** 166,091
181
1.32
0.53
1.85
1.85
Lux Stereometriæ.
149
1.85
555
185
:
24.05
Whoever will be at the pains to Com-
pare this work with Mr. Everard's Slid-
ing Rule will find it agree exactly to the
400 part of a Galion,
( 4.) Some uſeful Rules in Gauging,
whereby a Man may not only inform
his Judgment, but at ſome time or
other facilitate his Work.
1. TF 282 be divided by the Area of Unity
-7854, the Quotient will be 359,the
Square of the Gauge-point for Ale. Or,
Multiply 282 by 1. 2735, the Square of
the Diameter of a Circle whoſe Area is
1.00, and the product is the ſame.
If 1.00 be divided by 3.14, the Quo-
tient will be .318, a Multiplicator to
Multiply the Circumference by to find
the Diameter.
L 3
If
150
Lux Stereometrie.
If 1.00 be divided by .318, the Quoti-
ent will be 3.14, a Multiplyer to Mul-
tiply the Diameter to find the Circum-
ference,
2. The Area of a Circle is 2.47,what's
the Diameter ?.
Multiply the Area by 359, and the
Square Root of the product is the Dia-
meter.
3. Having the Circumference of a Circle,
to find the Area in Inches.
Square the circumference and Multiply
the produa by .07958.
4. Having the Area of a Circle, to find
the Circumference, Multiply the Area by
12.5064.
5. If a Tub, at any given Number of
Inches deep, hold any Number of Gallons
ſuch as 270, or the Like, what was the
Diameter of that Tub.
Multiply the Number of Gallons it
holds at the given Number of Inches
deep by 359, and divide the product
by the given number of Inches deep ,
and the Square Root of the Quotient is
the Diameter ſought.
6.If the Gauge-point,or 18.95Diameter
give one Gallon upon an Inch in depth,
what Diameter will give a Gallon on .5
of an Inch.
Multiply
Lux Stereometrie. 150
Multiply 359 by 1.0 and divide the
product by :5, and the Square Root of the
Quotient is the Diameter of
Circle
which will hold a Gallon, or 5 Tenths
of an Inch in depth. Or add a Cypher
decimal ways to 359, and divide by .59
and Extract the Root.
Lobivia
notion 91). Otsi
Foro 901
(5.) How to Inch a Swelling Cask, ſuch
as the midle Fruſtum of a Spheroid,
ſo as you may know the Content of
every Inch or Inches from Head to
Head, being erected with its Axis
Perpendicular to the Horizon.
Vle. Subſtract the Area of the Head
from the Area of the Bong , and
divide the difference bym # the Casks
Length, and of the Quotient is the
Common addend below the Bong, and
ſnbducend above the Bong
..
10 tort srusinov
-Trious sii Example. The origh
Suppoſe the Length of a Cask tó be4o
Inches, and Bong. Diameter to be 28, and
Head- Diameter to be 24, the Content of
this Cask is 8o Ale-Gallons.
tent
L4
The
152
Lux Stereometrie.
it obivih bris
or The Area of the Bong-Diam. 28 is 2.18
The Arca of the Head-Diam. 24 is 1.60
115
The difference of Bong and Head Areas
is
Divided by the Casks Length, 40,
viz, 10, the Quotient divided by 3, and
the Quotient Multiply'd by 2, gives the
Common addend,
board 710).58 (058
80
358 (019
aixA 25028 2
58
038 the Com. addend.
Si
Having found my addend, I now begin
to delineate my Inch Table thus; In the
firft Column ander. I, is the Length of
the Cask from 1. Inch to 40. In the
ſecond Column under 1, is the Areas of
every Particular Inch from i to 40. In
the third Column, under Ad, is the Com-
mon Addend, or Subducend : and in the
fourth Columa, under C, is the Content
of
any Number of Inches from the firſt
Inch.
Now
Lux Stereometriæ.
133.
the fourth
the third Inch
Now having Lined out my Inch Table,
oppoſite to the firſt Inch I ſet the Area
of the Head, and to it I add the Addend
,038, and it makes 1,638, the Content of
the firſt Inch, to that I add the Addend,
and it makes 1.676, the Content of the
ſecond lach, add theſe two Areas together,
and it makes 3,314, which ſet oppoſite
Content of the firſt two Inches add the
Addend to the Ayer of the ſecond Inch
1.714, add that to the Content of the
firſt two Inches, and it makes 5.028, the
Content of the firſt three Inches, and ſo
continue adding the Addend to the Laſt
Area, and adding that Area to the Lalb
Content, until you come to the Bong-
Diameter, and you have the Semi-Content:
of the Cask, viz. 39.98.
Now fince the Cask above the Bong
doth decreaſe gradually, I uſe the Add-
end as a Subducend, by Subftra&ing it
from every Area, as before I added it to
every Area : The Area at 21 Inches
muſt be the ſame as at 20, for the Dia-
meter an Inch above the Bong, is Homo
geneal to the Diameter an Inch below the
Bong, ſo it muſt be 2.360 to the Con-
tent above the Bong, viz, 39.98, and it
is
154
Lux Stereomatria.
is equal to 42.34 Gallons, and ſo much
the Cask holds at 21 Inches deep: Then
Subſtract the Subducend from the Area
at 21 Inches, and the remainder is 2,322,
the Area at 22 Inches deep ; add that to
the Content at 21 Inches deep, and the
Sum is 44.662, the Content at 22 Inches
deep, and fo proceed to the laſt Inch 40,
and the Content at 40 Inches deep is
80,06, equal to the Content of the whole
Cask. The aſe of the Table is plain
for ſuppoſe you dip your Röle into the
Cask and it wet 29 Inches, you look for
29, and over againſt 29,
and under Cont.
you will find 59.8s, which is equal to
the Content at 29 Inches deep; but if
your Rale wet 29.5, Then add the Con-
tent at 29, and the Content at 30 together
and half the Sum is the Content at 29.5;
or if it wet 29. 2, you are to ſuppoſe .2
to the 4 of an Integer, therefore you are
to add the Area, at 30 Inches, to the
Content at 29, and the Sum is the Con-
tent at 29.2, and ſo in all other dips. Kant
wolun
Lax Stereometrie.
135
.
L.
A.
Ad.
Cont.
2
038
.038
5.028
6.780
98
7
8
19
10
1.638 1.038 1.638
1.676 .038
3.314
1.714
1,752
1.790
.038
8.579
1.828.038
.038 10,398
866 1.038 1 12,264
1.904 038 | 14.168
1.942 038 16.110
1.980.038
.038 18.090
2,018 1.038
20.108
2.056.038
,038 22,164
2.094.038 24.258
2,132.038 26.390
2.170.038 | 28.560
2.208 .038 30.768
2.246
,038 | 33.014
2.284 .038 35.298
2.322 1.038 37.620
2.360.038
.038 39.980
II
13
Krk
14
16
.
.
19
156
Lux Stereometria.
L.
Ad.
Cont
28 2.094
21 2.360 038 | 42.340
22 2.322 .038 44.662
23 2.284
2.284 | .038 46.946
24 | 2.246.038 49. 192
25 | 2.208
.038 51.400
26 2.170 | .038 53.570
27 2.132 .03855.702
.038 57, 796
29 | 2.056 1.038
.038 59.852
30 2.018 .038 | 61.870
31 1.980 .038 | 63.850
32 1.942 .038 | 65.792
33 1.904 .038 | 67.690
34
1.866
.038 | 69.562
351.828.038 71.490
36
1.790
.038 1. 73.280
37 1.752
75.032
1.714 1.038 76.746
39
1.676 .038 | 78.422
40 1.638.038 80.000
mm
lo ho
.038
38
(6.) To
į
Lux Stereometrie.
2
D'
(6.) To Inch a Round or Square Tun,
As the Fruſtum of a Cone, or Square
Pyramid.
Ivide the difference of Diameters
or Sides, by the depth, and Mul-
tiply thé Leſſer Diameter, or Side of the
Leſfer Baſe, by that Quotient, and add
the Product to the Square of the Leſſer
Diameter, and to that Sam add of the
Square of the firſt Quotient, and divide
the Laſt Sum by 359, or 282, and the
Quot, is the Content of the firſt Inch,
:
Example
3
Depth
8
14 Great Diameter 102-10
Leller Diameter 08
98 911
9 8
8) 4.00. S
kn
1
49.00
98
158
Lux Stereometrie.
98
98
784
882
9604 Sq, of the Leſs Diam.
49
To
Firſt Quot. :S
9653
.08
}).25 (.08
359) 9653.08 (26.91 Content of the
2473
firſt Inch.
319,0
598
39
Multiply the Leſſer Diameter by the
Quot. as before, and to the Product add
of the Square of that ſame Quot, and
Multiply the Sum by 2, and to that Sum
add the Square of the Leſſer Diameter,
ånd Multiply that Sum by 2, divide the
laſt Sum by 359, and you have the Con-
tent of the firft two Inches.
Example,
Lux Stereometrie.
159
Example
98
5
3).25 6.08
49.0
o 8
49.08
98.16
.
98
98
H
784
882
9604
08.16
9702.16
1940432
160
Lux Stereometriæ.
359) 19404.32 (54.05 Cont. of the
1454
two firſt Inches.
1832
37
Go on as in the laſt Paragraph, only
Multiply by 3 inſtead of 2.
98
49 o
.08
49.08
3
147.24
98
Lax Stereometrie.
161
:
98
98
784
882
9604
147.24
K
VE
9751.24
3
11
359) 29253.72 (81.48
333
1747
240
M
1,62
Lax Stereometria
The 3 firſt Inches
The 2 firſt Inches
81.48
-
- $4.05
The ift Inch -
26.91
The 20 Inch
| 11
27.14>81.48
The 3d Inch
27.43
siirre
23.18) $5.cz ces (ez:
napi 2d. 27.14
cute ift. 26.91
.23
90T
To
..
Lux Stereometriæ.
183
Deptb. Cons.of every imeh. ift Differ. 2d Differ.
26.917 o Donna
,23
27.14
.06
.29
27.43
.06
.35
27.78
06
ol
06
28.19
47
60028.66north
.53
7 29.19
59
8 29.78
.06
M 2
Haring
164
Lux Stereometria.
sid
Having ſet down the Content of the
Three firſt Inches ſeverally, as in the
Table, Subſtract the Firſt Inch from the
Second, and ſet down the difference, as
23 ; then Subftra&t the Second from the
Third, and ſet down the difference, as
29. Then Subſtract the difference of
the firſt and Second from the difference
of the Second and Third, which will be
106, and ſet down as in the Table; then
add.06 to 29, and that makes .35, which
muſt be added to the Second Inch, and
that makes the Content of theFourth Inch,
and ſo on.
bas un
A ni
.cc anorama
hodan
oo.
otto
( 7.) TO
Lux Stereometrik.
165
Suo
..
Sub
(7) To Inch the Middle Fruſtum
of two Cones abutsing on 4 Common
Baſe.
übſtract the Head-Area from the
"
Bong Area, and divide the differ
ence by the Semi-Length, and add that
Quotient to the Head-Area, and that is
the Content of the Firſt Inch then add
that fame Quotient to the Content of
the Firft Inch, and the Summ is the Con-
tent of the ſecond Inch, and ſo continue
adding that Quotient to every Inch, even
to the Semi-Length, and you have the
Casks Semi-Content by adding all the
Inches together. Then, when you come
above the Bong, inſtead of adding you
muſt Subſtract that Quotient from the
Bong Area, or rather from the Area of
the Diameter, i an Inch below the Bong,
for chele Area's are in the Middle of e
very Inch, from Head to Bong.
M 3
Length
166
Lux Stereometria.
166
Bong
O
Length 40
28 2.18 Area.
I Head I alb 24
241.60 Area.
KO
20) 58 (.029 (The com-
18.0 mon addend
below,or ſubducend above the Bong.
JETLI SI
ei Jahr 1912 ofit o, or
bborong iba
to 100 di
ho asi
sunt nobre
novo oni
volan So gabbs
boy
29.0-imad on
sil: lis gniste voorste zde
smoo hap monn daorsogos en
DO (- anilbi to u beplan 2008 9:1 bo podle
ori mort gripitolo di Fasilitas
jo 17A off out 1-100
brod or's wolod om te
The
sto olbb?n ons
nie
aon of her moil and 191
tirgas,
Lux Stereometria.
167
The Inch-Table for a Cask in the Form
of two Cones abutting on a Common
Baſe.
Leng. Cont. offCom. add. Cont.from Co.in B.E.G.
Levery In.lor Subduc. In.to Inch. from In.toIn.
1.6
2
3.287
8.435
7
1.861
1 11629 1,029 1.629
1.098.029
3
1.687,029
4.974
4 1.716
,029
6.690
5 1.745 1.029
6
1.774 1019 10.209
1.8031.029
12.012
8 1.832.029 13.844
.029 15.705
10 1.890.029
.029 17.595
II 1.919 1.029 | 19.514
1.948.029 21.462
13
1.977 1.029
231439
14 2.006 . 029 25.445
IS
21035 1.029 27.480
16
2,064.029
29.544
17 2,093
029 31.637
18 2,122 1.029 33.759
192 ISI
.029
351910
2.180 | .029
38 090
M4
Bong.
168
Lux Stereometria.
Bống. 4,
Leng.(Cont. of Com. add.Cont.fromCo. in B.E.G.
every In.for Subduc.In.to Inch from In.to In.
O' (1.6
23
24
O
2.18
029
40.270
22 2,151 ,029 42.421
2.1 22 029 44.543
2.093 ·029 | 46,636
25
2.064 .029 | 48.700
26 2.035 ,029 50.735
27
2.006 - 929 52.741
28 1.977 .029
54.718
29
1.948 | .029 56.666
30 1.919 .029 | 58.58s
31
1.890
.029 | 60.175
32
1861 .029 162.336
1.832 029 64.168
34 | 1.803
.029 165.971
35 1.774 | .029 | 67.745
36
029 | 69.490
37 1.716 .029 71.206
38
1,687 029 | 72:893
39 1.658
.029 1 74.551
40
1,629
76.180
mm
BOOKS
BOOKS Printed and Sold by H.
Nenman, at the Graſhopper in the
Poultry.
EY
Very Man his own Gauger ; wherein
not only the Artiſt is ſhown a more
ready and exact method of Gauging than
any hitherto extant. But the moſt Igno-
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twenty in Figures, is taught to find the
Content of any Veſel, either full or in
part full ; and to know if they be right
liz'd. Alſo what a Pipe, Hogſhead, &c.
amounts to at the common rate and mea-
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ful Tables to know the Content of any
Veſel by. Likewiſe a Table fhewing the
price of any Commodity, from one Pound
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To which is added, the true Art of Brew
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Diſtilling
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sid the Richeſt Fineſt Cordials, of
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Coopers, Diſtillers
Strongwátermen, Coffeemen, and all other
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Duodecimal Arithmeriok viz. Nota
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portion Dirce and Reverſe: Duodeci:
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Gauging of all ſorts of Brewers Tuns and
Ciske To find the whole Content, or
the Vanity or Remaining Liquor of
either and that with more Eak and Esa
pedition, than by Vulgar or Decimal
Arithmerick- Vory uſeful for all the
of Men, as well Gentlemen as outiers, bor
cfpecially for Kerchants, Viting M **,
sind all 3e slaning tuticers And al 31€
Roles macke Plain, and Eaſic for the meant
eft Canseity. By Tobaalfondase of E ,
Philo. Acomptant.
mlo 290W to
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in Foor Parts. 1. An Abſtract of her
Sentiments, and a Character of her Writ-
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raiſed againſt them. 3. The Evidences
ſhe brings of her being led by the Spirit
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added a Diſſertation of Dr. De Heyde, on
the ſame Subject. 4. An Abſtract of her
Life. To which are added Two Letters
from different Hands , containing Re-
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alſo ſome of her own Letters, whereby
her True Chriſtian Spirit and Sentiments
are farther juſtified and vindicated ; par-
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and Satisfaction of Jeſus Chriſt.
Dr. Sydenham's Compleat Method of
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on of their Symptoms. To which are
now
added, five diſcourſes of the ſame
Author, concerning the Pleuriſy, Gout,
Hyſterical Paſſion, Dropſy and Rheuma-
tiſm. Abridg’d and faithfully Tranſlated
out of the Original Latin. With ſhort
and uſeful Notes on the former part,
Written by a late Learned Phyſician, and
never Printed before, Twelves, 1 s. 6 d.
Advice
Advice to a Phyſician - Containing
particular directions relating to the Cure
of moft Diſeaſes: With Refleaions on
the Nature and Uſes of the molt Cele-
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Done from the Latin.
mei gilt
An Account of the Nature, Cauſes $ym-
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are incident to Seafaring People. With
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Medacina Magnetica : Or, the rare and
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The Folly of Love a new Satyr againſt
Women, Quarto.
The Pleaſures of Love and Marriage,
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The Comical Hiſtory of Don Quixot,
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Noah's Flood, or, the deſtruction of the
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Reflections upon Mr. Fobnfons Notes on
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Publiſhed for the Awakening a ſleepy Age,
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The Good Houſe-wife made a Doctor
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a plain way of Natures own preſcribing,
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The Ladies Diverſion : containing, I.
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FINIS
...
o
*****
..
.
A 546335
UNIVERSITY OF MICHIGAN
3 9015 06389 4904