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ARTES
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SCIENTIA
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QA
77
C97
1729
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In-Me
10
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8/0
Lat:
10
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6018
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90
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70
70
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Cho:
Hou:
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210
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Lin
Cho
715 fec
Poly
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75
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18
Tho. Heath at the Hercules & Globe, next the Fountain Tavern in thes
Strand. Makes & Sells Wholesale & Retale, all Sorts of Mathematical Instru
ments, in Gold, Silver, Brass, Ivory, or Wood with Books of their use
8
9191
Tan
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£
20
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100
90
30
40
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50
60
70 80 90
Sin
70
75
tan
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tan
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40
50
an:
610
lan
8090 Sin
30, 40 30 ;
60
16
V: Sin:
70 Sin:
10 Num
OF
ICH
W
A NEW
TREATISE
OF THE
Conftruction and Ufe
OF THE
SECTOR.
CONTAINING,
The SOLUTIONS of the principal Problems by
that admirable Inftrument in the Chief·
Branches of MATHEMATICKS, viz.
ARITHMETICK,
MENSURATION,
Projection of the SPHERE
GEOGRAPHY,
Plain TRIGONOMETRY, ASTRONOMY,
Spherick GEOMETRY, DIALLING, &c.
Illuftrated with Variety of neceffary Obfervations,
and pleaſant Conclufions: Containing ſeveral
Applications intirely New.
Being a Work of the late Mr. SAMUEL CUNN's,
Teacher of Mathematicks, &c. Now carefully
Reviſed by EDMUND STONE.
LONDON:
Printed for JOHN WILCOX, at the Green Dragon,
in Little Britain, and THOMAS HEATH, Ma-
thematical Inftrument Maker, at the Hercules
and Globe, next the Fountain Tavern in the
Strand. M.DCC. XXIX.
Sothern
1-22-28-
12-13
THE
REVISOR
TO THE
READER.
MONG the Multitude of Mathe-
Amatical Inftruments that have been
invented, the SECTOR (as we call
it) or, Compafs of Proportion (as the
Foreigners) claims a principal Place, and,
ever fince the Invention thereof, has been
had always in the greatest Esteem by the
Ingenious of the Mathematical Kind, but
more especially with those who bufy them-
felves in the practical Parts of that Learn-
ing.
The Value of this Inftrument may
be ea-
fily gathered from hence; That feveral
eminent Mathematicians, as well Engliſh
as Foreigners, have thought it worth their
Pains to write Treatifes concerning it.
Such as Juftus Burgius, and Goldman, in
A 2
Latin;
PREFA C E.
Latin; Galileo, in Italian; Dechales, in
Latin; Ozanam, in French; Gunter,
Fofter, &c. in English; And now, at
length, the late Mr. Samuel Cunn, a Per-
fon whom all Men, capable of judging, muſt
allow, in juftice, to have been a Mathema-
tician of the first Rank, and confequently
would never have written now on a Sub-
ject that was trifling, or well handled by
Others before.
The Sectors made formerly (as well as
the Foreign ones now) differ very much
from thofe made by our Inftrument Makers
at this Time. I mean as to the Scales of
Lines drawn upon the Face of them, and
their Pofitions; Thefe latter having feve-
ral Lines omitted, viz. The Lines of Qua-
drature, the Lines of Segments and In-
fcribed Bodies, the Lines of Planes, the
Lines of Solids, the Lines of Metals, &c.
to make Room for others of far more exten-
five Ufe; and the Pofition of thofe (that
fue from the Center) are jo alter'd, that
the Sector, as now made, is not only a Scale
of Chords, Sines, Tangents, &c. to innu-
merable Radius's, but likewife by it you
may readily work Proportions in Lines,
Sines, Tangents, &c. feparately, or with
each other, and that to an Exactness fuffi-
cient enough for ordinary Practice, if the
Inftrnment be tolerably large.
From
PREFACE.
From whence it manifeftly appears, that
this Inftrumentis vaftly useful in Trigo-
nometry, Spherical Geometry, Projection
of the Sphere, Practical Aftronomy, Di-
alling, and, indeed, all Practical Parts of
Mathematicks where Scales of equal Parts,
Chords, Sines, Tangents, &c. are con-
cerned.
Now Mr. CUNN having examin'd the
Divifions on ſeveral Sectors, even Brafi
ones of 12 Inches, made by notable Work-
men, found egregious Faults in all of them;
upon which he himſelf, with Mr Heath,
Mathematical Inftrument Makr, in the
Strand, took the Pains of vividing anew
from Sherwin's Tables, a Brafs Pattern of
a Foot Sector.
Hence, if we contder Mr. CUNN's Skill
and Accuracy in Things of this Nature,
it will evidently follow, that Sectors care-
fully made by the faid Pattern, muſt equal,
or even exceed, any others whatsoever, in
anfwering the End defigned.
This Faultinefs of Sectors was the prin-
cipal Caufe of Mr. CUNN's writing the fol-
lowing little Tract of the Foot Sector,
which will ferve for any fized ones, wherein
he has profoundly handled the Matter, and
not so much as omitted one fingle Circum-
ftance of Ufe, in giving the Reader a thorow
Know-
PREFACE.
Knowledge of this admirable Inftrument,
both in its Construction and Uſe.
And, to compleat the Whole, tho' to
me there feems no abfolute Neceffity, he
has given the Conftruction and Uses of the
Artificial Numbers, Sines, and Tangents,
which are Lines properly belonging to Gun-
ter's Scale, but now put upon Sectors, as
well to fill up vacant Spaces, as for their
good Ufes. He gives you likewife the Con-
Aruction and Uses of the Dialing Lines of
Latrudes and Hours.
But, that he might not be altogether
filent with respect to the Lines formerly
put upon Secors, he at length finishes the
Piece with a brief and diftinct Account of
their Nature and Tfe.
Thus much in general; Not in the leaft
doubting but the following Sheets will be
read with abundance of Pleaſure by every
Practical Mathematician.
Edmund Stone.
THE
CONTENTS.
CHAP. I.
T
HE Definitions of Chords,
Tangents, &c. in a Circle.
Sines,
P. 1
Chap. II. Definitions of the Lines of
Chords, Sines, Tangents, &c.
Chap. III. Of the Radius on the Chords, Sines,
Tangents, and Secants.
8
Chap. IV. The Geometrical Conſtruction of the
Lines of Chords, Sines, Tangents, &c. 10
Chap. V. The Manner of laying down the Sines,
Tangents, &c. by Help of Tables.
14
Chap. VI. The Construction of the artificial
Lines on the Sector.
20
Chap. VII. The Numeration, Eftimation, or
Valuation of the feveral Divifions and Di-
ftances on the Scales or Lines of the Sector.
25
Chap.
CONTENT S.
Chap. VIII. Of the Sector in general, with gene-
ral Laws for working by the fectoral Lines. 33
Chap. IX. Of the Ufe of the Line of Lines 39
Chap. X. Of the Ufe of the Line of Chords and
Polygons.
52
Chap. XI. Some Ufes of the Line of Sines 64
Chap. XII. Some Ufes of the Lines of Tan-
gents and Secants.
81
Chap. XIII. The general Uſe of the Line of
Numbers.
96
Chap. XIV. Of the Ufe of the Sector in Tri-
gonometry.
109
Chap: XV. The Use of the Sector in Spherical
Geometry.
131
Chap. XVI. The Ufe of the Sector in Spheri-
cal Trigonometry.
148
Chap. XVII. The Ufe of the Sector in the
Projection of the Sphere.
171
Chap. XVIII. The Construction and Uſe of the
Lines of Latitudes and Hours.
180
Chap. XIX. The drawing of the Hour Lines
upon Dials by the Lines of Sines and Tan-
gents.
186
Chap. XX. A fhort Account of the Lines for-
merly put upon the Sector.
195
THE
CONSTRUCTION
OF THE
LINES upon the SECTOR.
CHAP. I.
Chords, Sines, Tangents, &c. defined.
NY Part X
1.
A
of the
Circum-
ference I
of a Circle is called
ân Arch; fo either
AB or AG (Fig. 1.)
AGL or AGLM is
an Arch: But the
Arch AK, becauſe
the fourth Part of
the Circumference or FH
Periphery, is called
a Quadrant; and the
Arch AKF is called
a Semi-circumference
and fometimes, tho'
improperly, a Semicircle.
B
W Ka TB
L
M
D
E
11. The
[ 2 ]
II. The ftrait Line which is drawn from
one End of an Arch to the other End of the
fame, is call'd the Chord of that Arch; fo
BM is the Chord of the Arch BAM, and
alfo the Chord of the Arch BFM, fince both
thefe Arches have the fame Ends, and the
ftrait Line BM is drawn thro' thofe Ends,
and terminated by them. In like manner
the ftrait Line BA is the Chord of the Arch
BA, and alfo of the Arch BFLA, and GL
the Chord of GAL and of GFL.
III. The Right Line, or (for its great
Ufes) the Sine of an Arch, is half the Chord
of double that Arch; fo if AM be taken
equal to the Arch BA, and the ſtrait Line
BM be drawn; BD the half of BM is the
Sine of BA, becauſe the Arch BAM is the
double of the Arch BA. In like manner,
the Sine of the Arch GA is GH, the half of
GL, which is the Chord of the Arch GAL,
the double of the Arch GA.
Otherwife thus, The Sine of an Arch is a
ftrait Line drawn from one End of that Arch,
perpendicular to the Diameter which paffes
thro' the other End of that Arch; fo BD is
the Sine of the Arch BA.
That this Definition of a Sine co-incides with
the former, will appear from the 3d Propofition
of the Third Book of the Elements of Euclid.
IV. If at the Extremities of a Circle's
Diameter paffing thro' one End of an Arch,
be raiſed two Lines perpendicular to that
Diameter;
[3]
Diameter ; thoſe two Lines will touch the
Circle in thoſe Extremities, by the 16th of
Euclid's 3d Book; and if from the Center of
the Circle be drawn a ftrait Line through
the other End of the forementioned
Arch, and this ſtrait Line be produc'd till
it meets one of the forefaid Perpendi-
culars; then that Part of the Perpendicular
which is between the Point of meeting
and the Diameter, is called the Tangent
of that Arch. And that Part of that Line
which meets the Tangent, and limits it; I
fay, that Part, which is between the Center
and the Point where it meets the Tangent, is
called the Secant of that Arch; fo AE is the
Tangent, and CE the Secant of the Arch BA.
and FI is the Tangent, and CI the Secant of
the Arch FG.
V. The verfed Sine of an Arch is that Part
of the Diameter which lies between that End
of the Arch which the Diameter paffes thro',
and the right Sine of the fame Arch. So AD
is the verſed Sine of the Arch AB; and AH
the verſed Sine of the Arch AG.
VI. Some neceffary Corollaries to the pre-
ceding Definitions.
1. While an Arch AB, lefs than a Qua-
drant is increafing, its Sine BD increaſes
alfo: But when the Arch hath increaſed till
it is become the Quadrant AK, the Sine KC
is the greateſt of all Sines; for if the Arch
ſhould ſtill increaſe, fuppofe till it becomes
AG,
B 2
[ 4 ]
E
AG, its Sine GH is less than KC. (Thefe
Things are evident from the 3d and 15th
Props. of Eucl. lib. 3.) that is, after the Arch
is become a Quadrant, if it continues to in-
creafe, the Sines will continually decreafe,
even till the Arch becomes AKF.
2. Since KC the Quadrant's Sine falls on
the Center of the Circle; KC is a Radius of
that Circle, and is equal to CA or CF, and
is fometimes called the whole Sine.
#
3. If the Arch FG be equal to the Arch
AB, the Sine of the former GH, is equal to
BD, the Sine of the latter. But when the
Arch FB is equal to the Arch AG, the Şine
of the former BD, is equal to GH the Sine
of the latter (Thefe Things are evident
from the forementioned 3d and 15th Props.
of Eucl. lib. 3.) And hence it appears, that
the 4 Arches AG, FG, AB, FB, have equal
Sines GH, BD, and of theſe the two former
have one and the fame Sine GH; and the two
latter one and the fame Sine BD. Hence it
appears, that any Arch AG, and the Arch
FG, which is its Supplement to a Semicircle,
have the fame Sine.
4. While an Arch AB increaſes to AT,
the Tangent AE will be increafed to AV,
and the Secant CE to CV. And if an Arch
leſs than a Quadrant continues to increaſe till
it becomes the Quadrant AK, the Secant
Line, or rather that which fhould be the
Secant Line, falls on KC, which is perpen-
dicular to AC, as well as AE; and fo KC
(28 p. of Euc. lib. 1.) is parallel to AE; con-
fequently that Line which fhould be the
Secant
1
[ 5 ]
Secant to determine the Length of the Tan-
gent and itself, doth not meet it. Therefore in
this Cafe, that is, when the Arch is a Qua-
drant, the Tangent and Secant are both in-
finite. But if the Arch AB had fo increaſed
as to want of being equal to AK, a very
fmall Quantity a K; the Secant Line C a pro-
duc'd would meet the Tangent, and deter-
mine that and itfelf: but at a Diſtance very
great. And fo AB continually increafing til
it becomes a A; the Tangent and Secant con-
tinually increaſe likewiſe.
5. The Arch being greater than AK,
and continually increafing, the Tangent and
Secant are continually decreafing, even till
the Arch becomes AKF; for the Tangent of
the Arch AW is FX, and its Secant is CX;
but the Tangent of the Arch AG is FI, and
its Secant is CI. But FI is evidently lefs than
FX and CI (24 Prop. of Euc. lib. 1.) is lefs
than CX.
6. In the Tangents and Secants, as in the
Sines, when two Arches together are juſt a
Semi-circumference, they have the fame Tan-
gents, and the fame Secants.
7. If the Arch AB increaſes, the verfed
Sine AD increaſes; if the Arch AB increaſes
to the Quadrant AK, the verfed Sine
becomes AC the Radius; if it ftill in-
creafes, till it becomes A G, the verfed
Sine becomes AH: So that the verfed
Sine always increaſes with the Arch, even
when the Arch is greater than a Qua-
drant, until the Arch become equal to AKF.
Therefore the fame verfed Sine doth not
anfwer
B 3
[6]
anſwer to an Arch, and its Supplement to a
Semi-circumference.
LORDAY
CHA P. II.
The Lines of Chords, Sines, Tangents,
I.
&c. defined.
N order to thefe Definitions, we
are to obferve, that the Circum-
ference of a Circle is commonly
divided into 360 equal Parts,
called Degrees; and theſe again divided, or
fuppofed to be divided, into 60 equal Parts,
called Minutes; and thefe again are fub-
divided continually by 60, if the Circum-
ference be large enough.
2. And when the Length of an Arch con-
tains any Number of thefe Degrees, it is
called an Arch of thofe Degrees; fo if an
Arch contains 57 of fo many Degrees, it is
call'd an Arch of 57 Degrees.
3. And the Chord, Sine, Tangent, or
Secant of that Arch, is called the Chord,
Sinc, Tangent, or Secant of (57) the De-
grees of that Arch.
4. Now a Line of Sines is a Line to which
the Sines of every Degree of the Quadrant
are transferred and number'd, as their cor-
refponding Arches of the Circle, and all of
them
[ 7 ]
them beginning from one Point. Such a Line
is CG in Fig. 2.
In that Line C 20, is the
Sine of 20; C 50 the Sine of 50; and CG
the Sine of 90, to the Radius CF or CA.
5. What hath been faid of a Line of Sines
may be applied to a Line of Chords; viz.
It is a Line to which the Chords of every
Degree, as far as 60, 90, 180, your de-
figned Length, are transferred. All of them
begin from the fame End of that Line; and
the other Ends of thefe Chords fo transferr'd
are number'd according to their Correfpon-
dent Arches. Such Lines are AE, AD,
Fig. 2.
6. Lines of Secants and Tangents are Lines
to which as many of the Secants and Tan-
gents are transferred, as the Length of the
Inftrument will receive, and number'd as
noted above. So FH, Fig. 2. is a Line of
Tangents to 70; and if the Divifions on CG
were ftruck out, CK would be a Line of
Secants to 70 Degrees.
B 4
CHAP.
لبيا
8 ]
:
味
​CHA P. III.
Of the Radius on the Chords, Sines,
Tangents, and Secants.
I.
1. PRESENEST hath been fhewn in the first
I
Chapter (Coroll 2.) that the
Sine of 90 is equal to the Radius
of the Circle, and therefore may
at all Times be taken for it, and called by
the fame Name.
2. A Side of an Equilateral 6 fided Figure,
infcribed in a Circle (that is a Chord of the
6th Part of the Circumference, or of 60 De-
grees) is equal to the Radius of that Circle
(15 Prop. of Book IV. of Euclid).
Of the equal Sides of the Hexagon (Fig. 3.)
any one of them AB, or BD, or DE, &c. is
K
D
P
C
L
FIL
སྙན
the
[9]
the Chord of 60 Degrees, and equal to the
Radius of the Circle CA.
3. The Tangent of 45 Degrees is alfo
equal to the Radius of the Circle, and may
therefore be taken for it.
For in the fame Figure, raife CK perpen-
dicular to CA, and KL perpendicular to CK,
and produce it till it meets AL. Then (28 Prop.
of Eucl. lib. 1.) is AL parallel to CK, and
KL to CA; and fo ACKL is a Parallelogram:
It is alfo a Square, fince KC is equal to CA.
Whence (34 Prop. Euclid, lib. 1.) KL is equal
to CA, and LA to KC, draw CL; then by
(8 Prop. Euclid, lib. 1.) the Angle KCL is
equal to the Angle ACL; and ſo (26 Prop.
Euclid, lib. 3.) the Arch KH is equal to the
Arch HA; that is, HA is an Arch of 45
Degrees, fince KA is one of 90 Degrees. LA,
which hath been fhewn to be equal to CA,
is the Tangent of the Arch HA; therefore
LA, the Tangent of 45 Degrees, is equal to
CA the Radius, and therefore may be taken
for it.
4 The Secant of o Degrees is equal to the
Radius, and therefore may be taken for it.
For (in Fig. 1.) if the Arch BA decreaſes,
the Secant CE will decreaſe alſo. Let the
Arch decreaſe to nothing, that is, let B come
to A; then will E come to A, and CE co-
incide with CA. Therefore the Secant of
o Degrees may be taken for the Radius.
玉
​CHAP.
[ 10 ]
F
CHA P. IV.
The Geometrical Conftruction of the Lines
of Chords, Sines, Tangents, &c.
XXXXROM the Second Chapter, it appears,
That the first Thing necessary to per-
form the Work propofed, is a Circle
divided into 360 equal Parts exactly.
But this Task is fo hard, that common Geome-
try pronounces it impoffible. It is true, indeed,
Euclid in his 4th Book bath fhewn us how to
divide the Circumference of a Circle into 3,
4, 6, 5, 10, 15, &c. equal Parts; and
bow. from them to gain other equal Parts :
And yet all these, and all those that may be
deduced from them, are infufficient for our
Purpoſe.
Dr. Halley (I ſuppoſe) ſeeing the Impof-
fibility of performing this Divifion of the whole
Circumference into 360 equal Parts Geometri-
cally, did, indeed, divide bis large Mural
Quadrant of 8 Feet Radius into 90 Parts;
but as a Check to thefe 90 Divifions, adds
another Number of Divifions; viz. 96. which
might be made equal Parts by common Geometry.
And then by a Table of Reduction compares the
two Orders of Divifions.
The Defign in this Chapter being to illuftrate
fully and Geometrically what thefe Lines are;
Ifball
[II]
I ſhall here take it for granted that a Circle is
fo divided, and leave the Manner of attaining
it by Approximation for a while.
A Circle then, being quartered, divided,
and number'd (as in Fig. 2.) I proceed, and
firft for the Chords.
1. Draw AD. Then fetting one Foot of the
Compaffes in A, extend the other to 10 on the
Arch; and with that Diſtance, the Com-
paſſes ſtill refting in A, deſcribe the Arch 10,
10, cutting the Line AD in 10: And in like
manner, one Foot ftill refting in A, extend
the other on the Arch fucceffively to the
Points;
20
20, 20
30
30, 30
40
and defcribe
40, 40
50
the Arches
cutting the
50, 50
60
60, 60
&c.
&c. &c.
20
30
Line AD in the Points
40
Do the like
50
60
&c.
with the intermediate Divifions of the Arch,
and number the Diviſions on the right Line
AD, ſo that they may correfpond with the
Numbers on the Arch, and you will have a
Line of Chords to 90 Degrees. If you want
it but to 60, you need not lay any Diviſions
on
[ 12 ]
on the Line beyond 60, or inſtead of AD
you might have ufed. AE.
2. Now for the Tangents draw from the
Center C, ftrait Lines to the Diviſions in the
Quadrant GF, and produce them till they
meet the Perpendicular FH; fo fhall FH be
a Line of Tangents divided, and you muſt
continue it to as many Degrees as will come
on the affigned Length. Thefe muft be
number'd correfpondent to the Arches.
3. For the Secants: Produce CG towards K;
then ſetting one Foot of the Compaffes in C,
extend the other to the feveral Divifions in
the Tangent Line, and defcribe Arches
meeting GK; and the Line CK will be a
Line of Secants divided between G and K,
which must be number'd as the Line of Tan-
gents is, from whence the Arches forming
the Divifions were defcribed. This Line
begins at C, and muſt be continued to aș
many Degrees as the Length affign'd will
permit.
Here obferve, that the fmalleft Secant iş
greater than the Radius; and that therefore
there are no Divifions between C and G.
4. For the Sines. From the feveral
Diviſions in the Quadrant FG, let fall Per-
pendiculars to CF; and thoſe Perpendiculars
may be transferred to one Line, and num-
bered to answer the refpective Divifions on
the Arch, and it will be a Line of Sines.
Becauſe the greateſt Sine is CG; and
becauſe there are no Divifions on the Secant
Line CK between C and G: if thofe Sines
are transferr❜d to CG; you have the Line of
Sines
[ 13 ]
Sines as you there fee. And this is the
way
they are ufually laid down on Gunter's Scale.
5. By the preceding Work you have
divided the Line FC; which, if number'd as
you there fee, is a Line of verfed Sines, as far as
a Quadrant or 90 Degrees. And if the like
be perform'd in the other Quadrant CAG;
or if C 100 be made equal to 80, C 110
equal to C70, &c. and fo on to 180; AF
will be a whole Line of verſed Sines.
6. But the Line of Sines may be otherwiſe
conftructed thus. The two Quadrants AG,
FG, being divided and number'd from
A and F towards G; draw Lines from 10 to
10, from 20 to 20, from 30 to 30, &c. and
alfo from the intermediate Divifions; and the
Line CG will be a Line of Sines divided,
which muſt be number'd according to the
Numbers on the correſpondent Arches.
In the next Place, to lay down theſe Lines
en the Sector, by the help of Mathematical
Tables, carefully calculated or corrected; the
Geometrical way by the oblique Interfections of
Lines being not fufficiently exact;
I fall take for an Example, a Sector whose
Legs were each one Foot. I mention the
Length on account of the Number of the fmaller
Divifions between the principal ones. I chufe
this Sector for an Example, becauſe I have
tried occafionally Brafs ones of the fame Size,
the Works of feveral notable Workmen, but
could find none without egregious Faults. As
to the Accuracy of this, I shall refer to the
Gentlemen who will be pleafed to confider the
Methods
[ 14 ]
Methods uſed in dividing it, to try it and com→
pare it with others.
CHA P. V.
To lay down the Sines, Tangents, &c. by
the help of Tables.
T
HE firft Thing abfolutely neceffary
to this Work, is a Line of equal
Parts; or at least of Parts fo nearly
equal, that the Differences from
Equality may be infenfible.
Mr. Edm. Gunter, formerly Profeſſor of Aſtro-
nomy, in Gresham College, London, lays fo
great a Stress on this Line, that he says it is
the Ground of all the reft. In his Treatife of
the Making of the Sector, he gives calculated
Tables for the laying down every Divifion on
bis Sector, by a Line of equal Parts.
Mr. Samuel Fofter, his Succeffor in the
Profefforſhip, treating of the Construction of the
Lines both Natural and Artificial, pews how to
perform the Work by a Line divided into equal
Parts, and calculated Tables; he never fo much
as once mention'd the Performance of any one by
The Interfections of Lines. From whence, by
their Judgments it appears, that this is the
only Method to be depended on, viz. by Tables.
If a Perfon has a mind to examine the Divi-
fions on any Line of these Inftruments, he will
Scarcely
[ 15 ]
fcarcely come near his Aim, unless he uses the.
Line of equal Parts, ufually called the Line of
Lines, perhaps for its Excellency; together
with Tables fitted to that Line you would
examine.
I
But now the principal Difficulty approaches;
we are well affured of the neceſſary Uſe of ſuch
a Line, but how to attain it is the Question.
From the 9th and 10th Propofitions of the 6tb
Book of Euclid, may be deduced a Method to
divide any Line given into any Number of equal
Parts, by dividing first another Line into a
like Number of Parts; or rather, by taking any
Part at pleaſure, and laying it as many times
on this Line, as the Line given is to be
divided into: This is certainly true in Theory,
but difficult to be practifed, especially when the
Divifions are very many, and cloſe together;
as will be evident to thofe who give themfelves
the Trouble of trying how exactly they can cut
500 Divifions in a Foot.
To prove whether your Line be cut into equal
Parts is eafy; for get a fliding Rule of Brafs,
that will move fmoothly, but not too loosely;
let the Slider be pinn'd faft to the fix'd Part.
Now by the beft Method you can think of,
divide both Parts together, but let the Divi-
fions be Hair-ftroaks; which done, then unpin
the Slider, and compare the Divifions of it
with thoſe of the fix'd Part; thus, move the
Slider till the first Divifion of it falls exactly
against the fecond on the fix'd Part; and then
carefully examine whether all the now corre-
fponding Divifions fit and agree. Then fide
the fame first Divifion of the Slider to the
fecond
[ 16 ]
fecond Divifion of the fix'd Part, and examine
as before. Then flide to the third, and fo on
to the last. Also repeat the fame the contrary
way, and you will find if your Divifions are
exactly equal; and if not, you may estimate
how much, and where the Errors are. This
to me feems to be an infallible Proof, and a
fure Method to correct minute Errors, especially
if you ufe Glaffes.
I
The Line of Equal Parts in this Example
was divided into 10 grand or primary Divifions 3
each of theſe were fubdivided into 10 others of
the first inferior Order, or fecondary Divifions 3
and thefe fecondary ones were again fubdivided
into 5 equal Parts, each of which reprefented
two tertiary Divifions, or two of the third
Order; thofe of the Fourth are to be estimated
by the Eye.
And fince we were at the Liberty to aſſign
this Line any Length fomewhat less than the
Leg of the Sector it was taken with, a very
great Convenience of fuch a Length, that of
thefe 500 Divifions, 42 were equal to one
Inch; and the whole Length of the Line was
11.32 Inches.
II
2
1 now proceed to fhew how this Line of
Equal Parts was made uſe of in laying down
the Sectoral Lines: I mean thofe Lines
which tend to the Center of the Sector, as
the Sines, Chords, Tangents, Secants, and
Polygons as alfo doth this Line of equal
Parts.
I fay, having tranfmitted this Line of
Equal Parts to the Sector becauſe every
Sine (in Sherwyn's Tables) confifted at moſt
;
of
[ 17 ]
of 7 Places, the three laft were eſteem'd as
Decimals, the other as Integers, in order to
introduce 4 Places in laying down theſe
Divifions. To do this, becauſe our fecon-
dary Divifions were divided but into 5 Parts,
the two laft Figures were halved, and theſe
Halves uſed in their ftead. Obferve, that by
the two laft, I mean the two laſt of thoſe I
eſteem as Integers. Obferve farther, that
the Divifions we are about to lay down, were
taken from the Original Line of equal Parts ;
and not from a Copy of it.
1. From the Sine of 90, down to the Sine
of 5 Degrees, 47 Minutes, all the Sines con-
fifted of 7 Places; and therefore we had four
to confider.
Example, The Sine of 41° 15′ is 6593 L458,
which when the two laft Figures are halved will
be 6546L 729. Therefore there was taken from
the equal Parts 6546, or rather 6543, becauſe
of the Fractions following; i.e. there were taken
6 of the principal Divifions, or thoſe of the ift
Order; 5 of the 2d; 4 of the 3d; and 3 of a
Diſtance of a Diviſion of the fame 3d Order.
But from 5°47′, down to o°L45′, the
Table conſiſts only of 6 Places, and the three
laft as before are Fractions. But the des
fective Place in the Beginning was fupplied
by a Cypher, to compleat the Number of 7.
And fo every where elſe.
Suppoſe we were to lay down the Sine of
3° 30': I find in the Tables the Sine is
610L485, or rather 0610L 485, and by
halving the two laft Figures I have 0605L 242.
Therefore I lay down none of the 1ſt Order;
C
6 of
[ 18 ]
6 of the Second; none of the Third; and 5
In like manner the other
of the Fourth,
Sines were laid down.
2. The Lines of Tangents, which are in
Length, as well as the Sines, equal to the
Lines of equal Parts; and fo, as was fhewn
in the third Chapter, would run to 45 De-
grees, were laid down in the fame manner as
the Sines.
3. The Chord's which run to 60 Degrees,
and fo (by the 3d Chapter) are equal in length
to the preceding Lines, were also laid down
by the Table: thus, The Sine of half their
Degrees was doubled, and this Double laid
down from the equal Parts with the like
Management as above.
And each Pair of thefe Lines makes equal
Angles with one another, that they may be
ufed conveniently together at the fame
opening of the Sector.
4. The Radius of the Lines of Secants and
of upper Tangents of above 45°, were taken
exactly equal to the 4th Part of the Length
of the Line of Lines.
Therefore in laying them down, we divided
the Tabular Secants and Tangents by 4, and
then used the Quotients as we did the Tabular
Sines.
5. The Lines of the Pohgons are from a
Radius equal to the Length of the Line of
Lines. Every Divifion of them was allo
taken from the Tables, nor only of the Poly-
gons, but of the Chords, Sines, Tangents, and
Secants alfo. Now if the laft Flace of the
Figures be fo eftimated, that you err not
above
[ 19 ]
I O
above Part of the Diſtance of the laft
Diviſions on the equal Parts; the Error will
be less than the 400th Part of an Inch in the
Line laid down from thence. If the Error
in the Eftimation be not greater than of
the laſt Divifions on the equal Parts, the
Error in laying down this Diſtance muſt be
leſs than the 200th Part of an Inch. But
whofoever will come and examine the Ori-
ginal, I believe, will be convinc'd that it is
eafy by that Line of equal Parts fo to efti-
mate theſe laſt Diſtances, that the Error fhall
fcarce exceed th Part of one of thofe
Diſtances.
The former Pair of Lines were all that
were laid down on the preceding Sector.
Of which fee the Figure.
6. Formerly it was ufual to lay other
Sectoral Lines on thefe Inftruments; viz.
Lines of Superficies, Lines of Solids, Lines
of infcrib'à Bodies, Lines of Metals, &c.
Of which you will meet with a Hint here-
after. Theſe laſt mention'd have been
omitted to make room for more uſeful ones,
without Confufion. I defign alfo a Word or
two on Mr. Samuel Forfier's Sector, when
the Example in Hand hath been a little
fet-forth.
Befides the Sectoral Lines (I mean, thofe
beforemention'd which run to the Center)
there are ſeveral others running parallel to,
and on the outward Edge, whoſe Ufes are very
great; and therefore, the laying them down
very exactly, requires a great deal of Care
and Caution. Thefe mot properly belong to
C 2
the
[ 20 ]
the Gunter's Scale; but to render this Inftru-
ment compleat are here infcrib'd.
Theſe are, the double Line of Numbers,
or Scale of Logarithms; the Lines of
Logarithmical or Artificial Sines; the Arti-
ficial Tangents; and the verſed Sines fitted
Back to Back with the right Sines.
Beſides a Line of Inches fubdivided into
8th Parts; a Line of Inches fubdivided into
10th Parts; a Line of Foot Meafure into 100th
Parts, the Dialling Lines, &c.
CHAP.
VI.
To Construct the Artificial Lines laid down
on the Sector.
F
IRST of the Numbers: The Sector
being quite open'd, and having
affign'd one Line to the Right-
Hand, perpendicular to the Side
of the Sector, and within 3 of an Inch from
the End; and having drawn parallel Lines to
receive the Divifions: the Length of the
forementioned Line of Lines was laid down
twice towards the Left-Hand on that deſign'd
for the Line of Numbers, which gave us the
Points mark'd I at the Left-Hand; I in
the Middle; and I at the Right-Hand;
which repreſent the Places of 1, 10, 100; or
10, 100, 1000; or 100, 1000, 10000.
Then
[ 21 ]
Then by the Tables you find the Loga-
rithm of 101 to be 2L0043214; and neg-
lecting the Characteriſtick, I have L0043214,
which manag'd as in laying down the Sectoral
Sines,becomes firfto043 L 214,then 0021L607.
Therefore from the Left-Hand 1, and alfo
from the I
the I in the Middle, lay down
0021L607, or rather 002; that is, none
of the 1ft Order, none of the 2d, 2 of the
3d; and Part of the Diſtance between
thofe Divifions of the 3d Order, and each
of theſe repreſent 101.
ΙΟΙ.
Again the Logarithm of 102 is 2L 0086002,
which by neglecting the Characteriſtick is
0086002; and then ordering, as in the Sines,
you have 0043001; that is, none of the ift
Order, none of the 2d, 4 of the 3d, and 3
of the 4th (to be eſtimated by the Eye) to
be laid of the equal Parts from the preceding
Points mark'd 1 at the Left-Hand, and 1 in
the Middle, which gave us the Divifions
for 102.
And the Logarithm of 103 is 2L 0128372;
which without the Characteriſtick is 0128372,
and gives 0114186; none of the ift,
I of the 2d, I of the 3d, and 4 of the 4th
Order; which laid off on the equal Parts, from
the two ones as before, gives the Divifion
for 103.
Thus the Line was continued to the Divi-
fions mark'd 2, denoting 200. But now the
Divifions growing clofer together, there was
not room for every fingle one; therefore
every two only is laid down, e. g.
C 3
For
[22]
For the Divifion repreſenting 258: Its
Logarithm without the Characteriſtick is
4116197, which gives 4108L098; which
laid down from I gives the Place of 258.
Thus proceed till all the Divifions be-
tween the three ones are laid down;
diminiſhing the Number of the Divifions be-
tween the Primary ones as they came clofer
together, as you may fee on the Line on the
Sector mark'd Number at the End.
2. The Line of Sines was laid down from
this Confideration; viz. That the Logarith-
mical Co-fecant abating the Radius, is the
Complement Arithmetical of the Logarithmi-
cal Sine to the Logarithmical Radius or Sine
of 90°. Therefore the Logarithmical Co-
fecants, abating the Radius, being laid from
the Sine of 90°, which is placed under the
End of the Line of Numbers, gives the Sine.
For Example, Let it be required to lay
down the Sine of 10. Its Logarithmical
Co-fecant is 10L 7603298, and abating the
Radius, it is 7603298, and ordering as in the
natural Sines, you have 7601 649; that is,
of the 1ft Order, 6 of the 2d, none of the
3d, and I of the 4th; or rather 7 of the 1ft,
6 of the 2d, none of the 3d, but of a
Diſtance of the fame third Order; which
laid downwards from the Sine of 90° is the
Divifion repreſenting the Sine of 10°.
7
I
And for the Sine of 4°, its Logarithmi-
cal Co-fecant is 11L1564155, which by
abating a Radius becomes 11564155. And
by ordering as before; firſt eſteeming the
three laft Decimals, you have 11564-155;
and
[ 23 ]
and by halving the two laft Figures, it
becomes 11532L077: That is, 11 of the 1ft
Order; 5 of the 2d; 3 of the 3d; and 2 of
the 4th; which laid down, gives the Divifion
repreſenting the Sine of 4 Degrees.
3. The verfed Sines, with their Fiducial
Line, very near to the Fiducial Line of the
right Sines, juſt now laid down, were con-
ftructed from this Confideration; viz. That
the Diſtance of the Sine of any Number of
Degrees (fuppofe 40) from the Sine of 90,
being doubled, muſt give us the Divifion re-
preſenting twice the Difference (50) of thoſe
Degrees from 90 (viz. 100) on the Line of
the verfed Sines.
Therefore, half the Degrees of the verfed
Sine to be laid down, and the Logarithm
Secant, abating the Radius of thofe half
Degrees, being doubled; and this Double or-
dered as in the natural Sines, gives the equal
Parts to be laid down.
For Example, Let it be required to lay
down the Divifion reprefenting roo°: its half
is 50, whofe Log-Secant is IoL 1919325;
which abating the Radius and Doubling, is
3838650; and this ordered as in the Sines,
gives 3819 325; which laid down from the
3819-325
Right towards the Left, gives the Divifion
repreſenting the verfed Sine of 100.
4. The Artificial Tangents run and are
number'd the fame way with the Sines, till
they come to 45 Degrees, which Divifion
falls exactly under the Sine of 90 Degrees:
But above 45 Degrees the Divifions are the
fame with the former, but number'd back-
C 4
wards
[ 24 ]
wards with fmaller Figures inverted. So that
the Divifion which reprefents 40 Degrees,
alfo reprefents 50; that which reprefents 30,
alfo reprefents 60, &c. only the I runs
upwards as the Sines do, and the other the
contrary way.
Therefore the eafieft way will be to lay the
upper Tangents from 45 downwards, and
that will be the Line required.
Thus take any Degrees above 45, feek its
Logarithmical Tangent in the Tables; neg-
lect a Radius, then order as in the natural
Sines.
So for the Tangent of 50 Degrees, whoſe
Logarithm is 10L 0761865, which by abating
the Radius is 0761865, and ordering as in
the natural Sines, you have 07301932; or
rather 0731, to be taken off the equal Parts,
and from 45° downwards, and it gives the
Divifion repreſenting 50 Degrees which alſo
reprefents 40.
I fhould now proceed to fhew how the
other Lines on the Sector were conftructed:
but becauſe they may elegantly be laid down
by the Sectoral Lines already laid down, I omit
doing it in this Place, in order to make it one of
the Ufes of the Lines already laid down.
CHAP.
[ 25 ]
CHAP. VII.
To Read, Eftimate, and Value the ſeveral
Divifions and Distances on the Scales or
Lines already conftructed.
XXXXHE Divifions on ſome of thefe
T Lines
are
determined by the
Figures adjacent to them; and
theſe are all of the firſt Order, and
proceed by Tens, and are accordingly num-
bered by 10, 20, 30, 40, 50, &c. Such are
thofe of the Chords, Sectoral Sines, Tangents,
and Secants, and alſo the Artificial Sines, Tan-
gents, and verfed Sines.
The Diviſions on the Line of equal Parts,
and on the Line of Numbers, which are
diftinguiſh'd by 1, 2, 3, 4, 5, 6, &c. are
intirely arbitrary, and may fignify what
Value you pleafe to give them.
I. In the Sectoral Tangents, running to 45
Degrees between the numbered Divifions
which are of the firft Order; they are 9
longer than thofe adjacent to them, and con-
fequently they form 10 Diſtances between
every two Divifions of the firſt Order; and
each of theſe denote whole Degrees, every
5th Degree is repreſented by a Stroke a little
longer
[ 26 ]
longer than the other whole Degrees are
repreſented by.
The fhort Divifions between thofe of the
fecond Order, and which therefore are of the
Third, are valued by their Number between
thofe of the fecond Order. Thus, if there
be only one Divifion between the Divifions.
reprefenting Degrees, and confequently two
Diſtances; that Divifion anfwers to the
Middle of the Degrees, and then each
Diſtance anſwers to 30 Minutes. If there
had been two Diviſions, and fo 3 Diſtances;
each Diſtance would have been Part of a
Degree, or 20 Minutes.
If 3 Divifions, and fo 4 Diſtances; each
would reprefent of a Degree, or 15 Minutes.
And in this laft Manner are the Tangents
divided in this our Example. But if the
Number of the Diſtances had been
૬ ટ્
5
6
IO
12
each would reprefent
12
ΙΟ
6
Minutes.
If the Compaſs-Point falls between theſe laſt
Divifions, or if you want to take fome
Number of Minutes which does not exactly fall
on any one of theſe Divifions; the Diſtance
of thofe Divifions must be fuppofed to be
divided into thofe Minutes they reprefent.
So if there be wanted the Place which re-
preſents the Tangent of 32°. 35'; and there
be three Diviſions, or 4 Diſtances in the third
Order between thofe of the Second; then
each
[ 27 ]
each of theſe laſt Diſtances will reprefent 15
Minutes, and two of them 30'. But becaufe
I want 5 more, I eftimate by my Eye a third
of this Diſtance of 15.
Therefore from 30
towards 40 I count 2 long Divifions which
anfwer to my 32 Degrees; then I count
forward 2 fhort Divifions, and Part of the
next Diſtance, which Place will repreſent the
Tangent of 32° 35', as required.
This Explanation of the Tangents will be
nearly fufficient for the other Sectoral Lines.
II. You are only to obferve, That on the
Sines, from 80 to 85, the whole Degrees are
only mark'd; and this 85 is a long Stroak as
the other Fives are. But from 85 to 90
there was not room for the whole Degrees;
therefore from 80 to 85, you muſt by thę
Eye eſtimate the half Degrees; and from 85
the whole Degrees muſt be ſo eſtimated. The
Law which thefe Divifions on the Sines near
90 obferve, is this, 'Their Diſtances from the
Sine of 90 are proportional to the Squares of
the Degrees they want of 90; e. g. the
Diſtance of the Sine of 85 from 90, is to the
Diſtance of the Sine of 87 from 90, as the
Square of 5 (25) to the Square of 21 (6)
that is, as 4 to 1. And 85 is diſtant from
90, 25 times as far as the Sine of 89 is from
90. Hence it appears, That if you will
imagine the Diſtance between the Sines of
85 and 90, divided into 25 equal Parts; of
thefe Parts, 86, 87°, 88°, 89°, will be
16, 9, 4, 1, diftant from the Sine of 90
Degrees,
III. The
[ 28 ]
III. The Secants have no Divifions till
after the Radial Point, which is at a Hole
punch'd againſt 25 on the Line of Lines.
The first Divifion next after this punch'd
Hole is at 10 Degrees; for the Diſtance is fo
fmall, that there is not room for any other
Divifions. Between 10 and 20 every two
Degrees is laid down; there are 4 Divifions,
and 5 Diſtances, and the Middle of the
Diſtances denotes the odd Degrees. Between
20 and 30, and 30 and 40, every whole
Degree is laid down; and there you may
judge the Halves by the Eye. Between 40
and 50, the Diſtances of the Degrees are
greater, and the half Degree is laid down.
From so to the End, which is at about 76,
every Degree is divided into Quarters by
3 Divifions of the 3d Order, like the Chords
and Tangents; and therefore to be valued
like them.
IV. Of the Artificial right Sines, between
80 and 90 there is but one Divifion, and it
denotes 85. The Law to be obſerved in the
Eſtimation of the other intermediate Degrees
is nearly the fame with that laid down for
the natural Sines among the Sectorals.
From 80 to 60, every whole Degree. From
60 to 30, every half Degree. From 30 to 20,
every Quarter; and from 20 to 10, every
6th Part of a Degree; that is, every 10th
Minute. And from 10 down to the End of
the Scale, every Degree is divided into 12
Diſtances; and fo every Diſtance denotes
5 Minutes,
[ 29 ]
:
3 Minutes. All theſe intermediate Spaces
are to be fubdivided by the Eye; and in this
Part of the Line you may eſtimate the of
a. Minute.
V. Of the verfed Sines running Back to
Back with the Sines, but counted the con-
trary way; the firft 10 Degrees have no
Divifions, nor can there be any diſtinct, for
the verfed Sine of 5 Degrees, is but of the
verfed Sine of 10 Degrees. From 10 to 20
every fecond Degree is laid down, there
being not room enough to expreſs any more
diftinctly. From 20 to 90, every whole
Degree. From 90 to 130, every half Degree.
From 130 quite to the End, to every Quarter.
In this Part of the Line you may judge very
well to 2 or 3 Minutes.
VI. The Artificial Tangents have the
Divisions of the ift Order number'd 10, 20,
30, 40, 45, which 45 comes to the End
under the Sine of go: And on the fame
Divifions it is numbered backwards by ſmaller
Figures inverted, from 45 to 50, 60, 70, 80,
and continues on with Divifions of the zd
Order, to 81, 82, 83, 84, 85, 86, 87, 88,
89, as far beyond as to about 89°. 20'.
Sometimes every one of theſe Diviſions are
numbered by a fmall Figure, and fometimes
only the 85.
From the Beginning so' upwards to 10,
and from 80 downwards to 89°. 20', every
Degree is divided into 12 Parts, that is, into
every 5th Minute. From 10 to 20, and
from
I
[ 30 ]
30]
from 70 to 80; into 6 Parts, and ſo every
Diſtance denotes 10 Minutes. In the two
preceding Ranges of Divifions you will
fcarce err above one Minute. From 20 to
45, and back again to 70, the Degrees are
every where divided into 4 Parts; and fo
each is 15 Minutes: And here you will
fcarce err above 2 or 3 Minutes.
But you may obferve, That the Artificial
Sines, like the Tangents, are number'd back-
wards, with the Figures inverted; thus the
Io falls on the Divifion of 80; the 20 on
that of 70; the 30 on that of 60, &c. And
this by the Construction of the Sines is an
Artificial or Logarithmical Line of Secants,
and fo entitled in a fmall Letter engraven in
an inverted Pofition.
VII. The Sectoral Line of Polygons needs
no Explanation, for the punch'd Holes are
number'd with the Figures 6, 7, 8, 9, 10,
11, 12, denoting the Number of the Sides
of that Polygon.
And this admits of no
Sub-divifions.
We faid the Line of Lines, and the Line
of Numbers had their Divifions arbitrary.
VIII. The Line of Lines is divided into
10 equal Parts, numbered 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, and theſe I call Divifions of
the firſt Order. Each of theſe are again
fub-divided into 10 other equal Parts, which
are of the fecond Order; each of theſe laſt
are fubdivided into 5 equal Parts, which by
eftimating the Middles of the Diſtances may
be
[ 31 ]
be faid to be into ten, which are of the third
Order. And fo the whole Line would be
divided into 1000 equal Parts.
Now if the whole Line be conceived to be
divided into 1000 Wholes or Integers (which
in Practice it very well may, without the
leaſt Confuſion); the Divifions of the firſt
Order, that is the numbered ones, will repre-
fent Hundreds; fo the Figures 1, 2, 3, &c.
denote 100, 200, 300, &c.; the Lines of
the fecond Order muft denote Tens, viz. 10,
20, 30, &c. And thofe of the third Order
Units only.
When the whole Line reprefents 100;
then the Divifions of the firſt Order will be
Tens; and fo 1, 2, 3, &c. will reprefent 10,
20, 30, &c.
Thoſe of the fecond Order
will be Units; and thoſe of the Third will
be tenth Parts.
If the whole Line reprefents 10, the
Divifions of the firft Order will reprefent
what their Numbers annexed fhew; thoſe of
the Second, Decimal Parts; and thoſe of the
Third, Centefimal ones.
Laftly, 'When the whole Line muft repre-
fent 10000 Wholes or Integers; the Divifions
of the firſt Order denote Thouſands: fo the
Figures 1, 2, 3, &c. denote 1000, 2000,
3000, &c. The Divifions of the fecond
Order, will denote Hundreds, thoſe of the
Third, Tens. And if the half of one of
thefe fmall Diſtances be conceived divided
into ten Parts, thofe Parts will be Units.
Moreover,
I
[ 32 ]
Moreover, whatſoever Value you affix to
the Divifions on one Leg of the Sector, the
fame Value must be given on the like Divi-
fions on the other Leg.
IX. In the two Lines of Numbers the one
beginning where the other ends, the Divifions
of the firft Order are number'd thus, 1, 2,
3, 4, 5, 6, 7, 8, 9; 1, 2, 3, 4, 5, 6, 7, 8, 9,
10. Between every one of thefe Divifions
of the firſt Order, there are 9 Divifions of
the Second, and fo 10 Diſtances.
The
Divifions of the third Order are alike in both
Lines. From 1 to 2 there are 10 Diſtances
of the third Order between every two Divi-
fions of the fecond Order. From 2 to 3
there are but 4 Divifions, and fo 5 Diſtances,
the Middles of thefe Diſtances being eftimated,
you will have all the 9 Divifions of the third
Order. From 3 to 10 the Diſtances of the
fecond Order are only cut into two Parts,
each denoting 5 Diſtances of the third Order;
the other Diſtances of this third Order are to
be eſtimated by the Eye, and to be uſed with
very ſharp-pointed Compaffes.
When the Divifions in the firft Line
denote Units, thoſe in the ſecond denote
Tens; when thofe in the firft are Tens, thofe
in the fecond are Hundreds; when Hundreds,
thoſe in the ſecond will be Thouſands. And
the Value of the Divifions of the fecond
Order are regulated by thofe of the firft
Order, according to the Rules laid down for
the Line of Lines. That was, if thofe of
the firſt Order were Tens, thofe of the fecond
were
[ 33 ]
were Units, and thofe of the third, Decimals.
If thoſe of the firft were Hundreds, thofe of
the ſecond were Tens, and thoſe of the third
Units, &e.
Laftly, though the Examples here takeņ
relate only to the Foot Sector before men-
tioned (knowing that if the Sector had been
longer, more Divifions might have come on,
but if fhorter not fo many) yet the general
Rules herein deliver'd duly obferved, are fuf-
ficient to fhew the Ufe of all Sizes of Sectors.
CHA P. VIII.
Of the Sector in general, with general Laws
for working by the Sectoral Lines.
N refpect of its Ufe, it is appli-
cable to every Thing wherein
Proportions are made ufe of
The Ground of the Sectoral
Lines depends on the 2d and 4th Propofitions
of the 6th Book of Euclid's Elements.
The Ground of the Artificial Lines deperds
on the Nature and Properties of Logarithms.
In the Ufe of the Sectoral Lines, two
Openings of the Compaffes are always ne-
ceffary.
In the Artificial Lines feldom above one.
D
For
34 ]
For where there is but one Proportion,
and that wrought by one or two Lines at
moſt; this is the general Law.
Extend the Compaffes from the firſt Term
to one of the Middle ones; and the fame
Extent applied the fame way, will reach from
the other of the Middle ones, to the Anſwer.
In this Chapter, I fhall only confider the
Natural or Sectoral Lines, being thoſe that
meet together at the Center. And ſhall here-
after in a proper Place, particularly treat of
the Artificial ones.
All the Lines as AD, AE, meeting
B
G
C
>
D
I
in the Center, are of an equal Length;
and thoſe that bear the fame Name are cqual-
ly and alike divided and number'd. And
therefore in all the Operations wrought by
the Sectorals, the Points B, G, repreſenting
the fame Diviſions, are equally diftant from the
Center A. The fame is true of any other
two Points, as C and F. And therefore,
AB is to AC, as BG is to CF, by the 2d and
4th Propofitions of Euclid's 6th Book.
NB. The
[ 35 ]
N. B. The Lengths AB, BC, as alſo AG,
GF, are call'd Crurals, becauſe they lie, or
are meafur'd in the Legs of the Sectors.
The Lines BG, CF are call'd Parallels;
becauſe in all Proportions work'd by the
Sectorals, the two Extents between the Legs
will be two fuch Lines parallel.
If AC on the Line of Lines, meafures 67
Parts, and fo AF likewife; then is AC and
AF faid to be 67 Crural Parts, and CF 67
Parallel Parts.
If AC were the Chord of 39 Degrees; then
is AC the Crural Chord, and CF the Parallel
Chord of 39°.
In working Proportions by theſe Lines,
the firſt Term may in moſt Caſes be Crural or
Parallel. When the first is { parallel
Parallel} the laſt
or fought Term is {Parallel that is the laft
Crural
is in a Poſition different from the firſt.
Of the three Terms given, to find the
Fourth; the firſt Term, and that which is of
the fame Denomination, whether it be the
2d or 3d, I call Terms of the firſt Denomi-
nation: And the other Term, with that
fought, I call Terms of the ſecond Denomi-
nation.
N.B. To the Line of Lines run three others
parallel, to receive the Divifions and Num-
bers. But that is the Line to be uſed which
receives all the Divifions; it is that next the
inner Edge of the Sector, and it runs exactly
into the Center; whereas the others paſs by,
D 2
or
[ 36 ]
or ſtop before they come at the Center. The
fame is to be underſtood of all the other
Sectoral Lines.
The general Laws for working Proportions
by the Sectoral Lines are theſe three follow-
ing; according as you would have the firſt
Term Crural or Parallel.
First, When the firſt Term is to be Crural.
Let the three given Quantities be R (50),
Б
G
F
D
E
$ (60), T (70), to find a fourth Propor-
tional; and let R and S be Feet; and 60
and the Quantity fought be Shillings. Find on
both the Legs AD, AE, the Lengths of the
Terms of the firſt Denomination, all counted
from the Center; that is, take AB, AG,
each equal to R (50), and AC, AF, each
equal to T (70). Then open the Sector till
$ (60) of the fecond Denomination, reaches
from 50 to 50, the Ends of the firſt Term ;
fo fhall the Diſtance from 70 to 70 be (+);
the Anſwer, if meafur'd on the fame Line
the other was meaſur'd on.
Secondly,
[ 37 ]
Secondly, When the firft Term is to be
parallel, take (50) the first Term in the
Compaffes, and open the Sector till it reaches
(from 60 to 60) the Term given of the 2d
Denomination. The Sector being thus
open'd, take (70) the other given Term in
your Compaffes, and draw it along the Legs
of the Sector till the Points reft on the fame
Diviſions of both Legs; and thofe Diviſions
ſhew the Anſwer or Quantity fought (in this
Example as before. 4.
Here it may be obferv'd, that the Opera-
tion by the fecond Rule is perform'd by Ten-
tation; in the firft without.
The fecond Rule ought to be avoided
when the Sector is to be open'd to a very
fmall Diſtance.
Thirdly, If the firſt Term of the Propor-
tion be made Crural, and the Diſtance of the
Points repreſenting the firſt Term, when it is
the greateſt that the Sector will admit of, is
fhorter than the given Term of the fecond
Denomination; either the Term of the ſecond
Denomination muſt be made Crural, or taken
from a fmaller Scale, or fome of the Varia-
tions hereafter laid down muſt be made.
When the 4 Terms of the Proportion are
all of the fame Denomination; either the
ſecond or the third may be compar'd with the
firft.
The five following Variations may be al-
low'd when Conveniency requires it.
D 3
1. The
[ 38 ]
1. The firit Term, and either of the mid-
dle Terms may be multiply'd or divided by
any one and the fame Number.
2. If you multiply or divide the firft Term
by any Number, and work with the Refult;
you muſt accordingly multiply or divide what
arifes from that Work by the fame Number,
you will have the Solution.
and
{ multiply either of the mid-
3. You may divide
dle Terms by any Number, if you will
S divide
multiply the other by the fame Number.
4. If you { multiply either of the middle
divide
Terms by any Number, and work with the
divide }
Refult; you muſt {multiply
{multiply what arifes
from that Work by the fame Number, and
you will have the Solution.
5. The Parallels may be taken from any
one Scale whatſoever, as well as from thoſe
Scales laid down on the Sector.
N. B. Of Compaffes the Beam Compaffes
are fitteft to be uſed with the Sector; becaufe
their Points are always perpendicular to the
Plane of the Sector.
When the Sector is to be open'd by a
Crural Term which falls near to the Center;
multiply that Crural Term and its correfpon-
dent Parallel, till they produce Numbers at a
fufficient Diſtance from the Center. The
farther the better.
CHA P.
[ 39 ]
BIRKORGHEORGHES
I
CHA P. IX.
Of the Ufe of the Line of Lines.
25252
225T is impoffible to write all the Ufes
to which this Line is applicable. It
is fufficient in this fmall Tract, to
lay down fome that are eafy, and
fit for the common Practices of Life. Such as
are every Day required by the Architect, the
Working Builder, the Land Surveyor, the
Watch Maker, the Mathematical Inſtrument
Maker, &c.
I. To make a Scale of a defired Length, that
fo a Draught defign'd to be laid down by it,
may come within an affign'd Compafs.
I
Example. To make a Scale that may repre-
fent 100 Yards in 1 Foot; or, which is the fame
Thing, to divide a Foot into equal 100 Parts.
Take in your Compaffes (1 Foot) the
Length of the Scale, and open the Sector till
the Compafs-Points reach (100) the Parts to be
contain'd on that Scale, on both Legs. The
Sector thus open'd, Take the Diſtance from
90 to 90 in your Compaffes, and it will be
90 Yards on your Scale. And the Diſtance.
from 57 to 57 is 57 Yards on the Scale. So
every Diviſion may be thus transferr'd to your
Scale, or the Sector thus open'd is your Scale.
That
D 4
[ 40 ]
That you may not be at a Lofs in under-
ftanding other Treatifes on the Sector, and
perhaps, fome of the following Paffages in
this; obferve the Stile in the following
Words containing the preceding Example.
Make 1 Foot the Parallel of 100; and the
Parallels of 90 and 57, are 90 and 57 Yards,
&c.
Example II. To divide a Foot into 96 equal
Parts.
Take 1 Foot in your Compaffes, and open
the Sector till the Compafs-Points reach from
96 to 96; then the Diſtance from 50 to 50
gives so of thofe Parts required.
In the Sectoral Language, make 1 Foot
the Parallel at 96, and the Parallel at 50 is
50 Parts.
Example III. To divide 1 Foot into 120
equal Parts.
Here it is to be obſerv'd, that if the Divi-
fions of the firft Order be made Tens; 120
will be beyond the End of the Line of Lines:
And if I make the fame Divifion Hundreds,
the 120 cannot be open'd at the Diſtance
of a Foot, which is the Line to be divided.
Therefore divide or multiply the Foot, and
alfo the 120 Parts by any Number which will
fuit you beft. Here in this Example, if you
divide by 2, the Quotients are, or half
of a Foot, and 60 the Parts it is to be divided
into. Therefore by the preceding Doctrine,
divide half a Foot into 60 Parts, and it is the
Scale requir'd. Thus, open the Sector till
half
[ 41 ]
half a Foot reaches from 60 to 60; and your
Sector as far as 100 is the Scale you want.
Hence appears one Beauty of this Line of
Lines; viz. that it is a Scale of equal Parts
of all Lengths: And is therefore preferable
to any Scale where the Divifions are only of
one particular Length.
Example IV. To form a Scale of 1 Foot,
9 Inches, that may reprefent 145 Yards.
I
Here I cannot open the Sector wide
enough to receive 1 Foot, 9 Inches; there-
fore I multiply both by fome Number,
fuppofe 4, and there will arife 7 Foot, and
580 Yards; then I divide all by 7, and find
I Foot, and 82L85 Yards. Therefore
opening the Sector till 1 Foot reaches from
8285 to 82 85, it will be a Scale of 100
Yards of the Bignefs requir'd.
On or near the outer Edges of the Sectors
are uſually plac'd 3 Lines of equal Parts, viz.
a Foot Line in 100 equal Parts. The Manner
of laying this down is in the firſt Example.
The 2d is a Foot Line divided into Inches,
and each of thefe into 8 Parts. This is con-
ftructed in the fecond Example, for (12
times 8 is 96) and there the Foot was divided
into 96 equal Parts.
The 3d is a Foot Line divided into Inches,
and each of theſe into tenth Parts of Inches;
that is, into 120 Parts, for 10 times 12 is
I20. This therefore is perform'd in the 3d
Example.
And now I have in fome meaſure, or
partly perform'd the Promife I made in the
laft
[ 42 ]
laft Paragraph of the 6th Chapter. Theſe
Lines are alſo mention'd, and a little defcrib'd
in the laſt Paragraph of the 5th Chapter.
V. When you are about to take from the
Sector thofe Divifions which are near the
Center, in order to transfer them to the Scale
you are making; or if you uſe the Sector
open'd properly for your Scale, and you are
to lay down a fmall Number of Divifions,
ſuppoſe 3, you may be a little confus'd. For
there are two Lines which run to the Center
on each Leg; and to avoid Confufion, as much
as may be, the Lines parallel to the Fidu-
cial ones, are cut off (from one Line on each
Leg) two Inches from the Center, and
the Fiducial Lines only continu'd, with
prick'd Holes inftead of Divifions; and
yet even then, very near the Center, the
Diſtances are fcarce diftinguiſhable. Ob-
ferve the Remedy, If you are to lay from
A-
·科
​D
C
A towards D, 3 Parts; lay any larger Num-
ber, which is diſtinct, fuppofe 20, from A to
C; and then 17 Parts backwards from C to
E; fo AE will be 3 Parts as requir'd.
VI. If a Draught of Land, Buildings, &c.
be drawn without a Scale, and you would
add a Scale to it:
Meaſure
[ 43 ]
Meaſure on the Land or Building any one
Line not the ſhorteſt, fuppofe it to be 50
Poles, or Feet. And let that Line be re-
prefented in the Draught by HI.
Take the Line HI in your Compaffes, and
open the Sector till the Points reach from
50 to 50, the Meafure of HI; and fo
will the Sector thus open'd, be a Scale of 100
Poles or Feet for that Draught, to which it
may be transferr'd.
VII. To increaſe or diminiſh a Line in
any Proportion affign'd.
Let it be requir'd to diminiſh the Line HI
in the Proportion of 3 to 2.
Becauſe 3 and 2 fall near the Center, I
multiply them, and fo form 9 and 6, which
I uſe inſtead of 3 and 2.
Open the Sector
till 6 reaches from 9 to 9. Then laying the
given Line AB from the Center, keep the
upper Leg of the Compaffes in the Place to
which it reaches, and turn the other over to
the fame Place in the other Leg; and the
Diſtance now in the Compaffes is the Line
diminiſh'd as requir'd.
VIII. In like manner may the other Sides,
and alſo the Diagonals be diminiſh'd, even
without altering the Sector. For take IK in
the Compaffes, and lay it from the Center on
the Line of Lines; and keep faft the upper Point,
turn the other about till it falls on the fame
Part on the other Leg; and the Diſtance
of the Compaffes. will give BC diminish'd.
Hence any Figure may be diminish'd at
pleaſure.
[ 44 ]
pleaſure. But a Figure of
But a Figure of many Sides may
be reduc'd fhorter thus:
IX. Figure 7. Numb. 1. is to be reduc'd
to a leſs in a given Proportion.
G
L
9
K
H
h
I
Any where in the Figure take a Point O;
draw Lines OG, OH, OI, OK, OL, to all
the Angles. Diminiſh theſe Lines in the
Proportion affign'd, and let them be Og, Oh,
Oi, &c. and draw gh, hi, ik, &c. and the
Figure is diminish'd as requir'd.
X. If only one of thefe Lines had been
diminiſh'd, fuppofe OG to Og; and then the
Lines gb, bi, ik, &c. had been drawn paral-
lel to GH, HI, IK, &c. the Figure would
have been diminiſh'd as before. But the
Method laid down in the 8th Article is the
better way; becauſe the Points g, h, i, &c.
are found independant on any other. But
in
[ 45 ]
in the latter, the Points i, k, l, &c. depend
on the preceding ones.
XI. The fame had been equally true, if
the Point O had been taken without the
Figure, or on one of its Sides, or in one of
its Angles. (Fig. 7. Numb. 2, 3, 4.)
Fig. 7. Numb. 2.
k
l
I
h
K
H
I
T
Fig.
[ 46 ]
Fig. 7. Numb. 3, 4.
G
I
H
G
hi
K
I
I
H
K
4
I
6
XII. The
[ 47 ]
XII. The fame
had been alfo
true, if the Point
O had been taken
without, and the
diminish'd Lines
had been laid to-
wardsthecontrary
Parts from O, as
in Fig. 7. N°. 5.
And this is a
very excellent
Method, becauſe
you do not offend
the ift Draught.
the new
And
Draught is drawn
thus
clean without any
transferring. This
Problem,
manag'd,deferves
a much larger Ex-
planation, and to
be exemplify'd in
many Varieties;
but here 1 have
not Room. I
have fully fhewn
the Uſe of the Se-
Єtor herein; and
therefore leave it
for another Place.
H
h
G
9
k
K
I
Only obferve that there are particular Inſtru-
ments for this purpoſe.
2
XIII. And
[ 48 ]
XIII. And the fame Method might have
been obferv'd in increafing a Figure; but
this is feldom requir'd.
XIV. Two Numbers being given, and
one Line: to find another Line fuch; that,
as one Number to the other, fo may the Line
given be to the Line fought.
This is the fame with the foregoing, but in
other Words.
XV. Three Numbers being given; to find
a fourth Proportional.
This was doubly handled in the general
Uſe of the Sector. Turn back to Chap. 8.
XVI. Two Numbers being given; to find
a Third continually Proportional.
Supply a Third by writing the Second over
again. So if the Numbers 50 and 60 had
been given; fay, As 5o is to 60, ſo is 60 to
the Anſwer, which you will find to be 72.
The Operation in the Sectoral Stile is, make
60 a Parallel of 50; and then the Parallel of
60 is 72, the Anſwer.
XVII. When 3 Lines are given, to find a
fourth Proportional. Meaſure the firſt and
either of the other two on the Line of
Lines, and exprefs them in Numbers ; then
you have two Numbers, and a Line given:
to find another Line to which that given
bears fuch Proportion, as the Number ex-
preffing the Meaſure of the firſt Line given,
1
doth
[ 49 ]
doth to the Number expreffing the Meaſure of
the other meafur'd Line. And this is the
fame with Art. 13th of this Chap.
Let the 3 Lines be A, B, C. Let
A meaſure 70, and B 50; then make C a
A.
B
C
D-
Parallel to A (70), and the Parallel of B (50),
will be D the Line fought.
XVIII. When two Lines are given, to find
a Third in a continual Proportion;
Confider the Second twice; and you have
3 Lines given, then work as in the laft. Let
the two Lines be E, F. Under F draw
E
F
F
G-
another equal to it, and call it alfo F. Then
meaſure the two firft, and let them be E (98)
F (84). Then make F a Parallel to E (98),
and the Parallel of F (84), will be G the
Line fought.
E
XIX. To
4
[ 50 ]
XIX. To divide a large Circle, Quadrant,
Sextant, Dodrant, &c. into Degrees.
Firſt the Circle is eafily quarter'd, by
raifing BA perpendicular to the Diame-
ter CH; then Sextants and Dodrants are
A
H
to
F
E
G
D
C
had, by laying the Radius from A to E,
and from C to F; or by biffecting the Arch
CF. Biffect the Arch CE in D, and the
Arches ED, DC, will be each of 15 Degrees.
you
Meaſure the Radius of this Circle by the
Line of Lines, and Parts thereof; and value
the Diſtances and Diviſions in the fame man-
ner as you did in dividing the Sector; fo
will have four Places in the Decimals, if
you want them. Let this Meaſure be (for
an Example) 5 times the Length of the Line
of Lines, and the following Parts; viz. one
of the Ift Order; 4 of the 2d; and 2 of the
3d; fo confider'd as the 2d, are divided into
5 Parts; and then Decimally, calling the
whole Line of Lines one Unite, the Length
is 5L 14.4.
Then
[51]
Then from the Tables of Sines calculate
the Sine of 30 Minutes, and double it; fo
the Chord of 1 Degree will be L089778,
where the whole Line is the Integer. But
when the Divifions of the ft Order are In-
tegers, the Chord of 1 Degree to that Radius
is L 89778, And this according to the
Method laid down in ufing the Sines, is
8938L9; which therefore may be called
8939, for it differs from it but Part
of an Inch.
I
I
Therefore open the Sector till 1 of the 1ft
Order will reach from 10 to 1o. And then
take the Parallel of 8939; and it ſhall be the
Chord of one Degree anfwerable to the given
Radius. And taking this Diſtance with hair
Compaffes, and ufing a Glafs, you may be
very exact; for the Sector thus open'd is a
Diagonal Scale of 50000 in the Foot, fup-
pofing each of the fmall Diftances divided
into 10 equal Parts by the Eye. Now as oft
as you err of theſe Diſtances, fo oft will
you err 2 Seconds of a Minute, and no more.
Lay this Chord from D to G; and the
Arch GC will confift of 16 Degrees; and
confequently by a continual Biffection may
be divided into fingle Degrees.
By repeating this Biffection you may have
the Halves and Quarters.
If from the Arch of 45° be taken the Arch
2 L 20', or from the Arch of 22° 30′ be
taken the Arch of 1° 10', &c. by a continued
Biffection you may attain every 10th Minute;
and fo confequently every fifth.
E 2
If
[
52 ]
If to the Arch of 7° 10' be added the
Arch of 62 Minutes, or to the Arch 3° 45′
be added the Arch of 31 Minutes; the Arch
of every fingle Minute may be had by Bif
fection.
Dr. Halley fo well approv'd of Divifions
by Biſſection, that he cauſed his large mural
Quadrant of 8 Feet Radius, to be divided
into 96 equal Parts, becauſe he could attain
thefe Diviſions by a continual Biſſection of
the Dodrantal Arches. And to reduce theſe
to the Degrees, hath a Table calculated,
whereby they appear, as it were, by In-
fpection.
MATOKEO
CHA P. X.
Of the Ufe of the Lines of Chords and
Polygons.
T
XXXXHE Chord in this Example runs
only to 60 Degrees, yet is fufficient
to lay down any Angle, or meaſure
one, tho' never fo great, and more
exactly than can be done by thoſe which run
to 90 or 180 Degrees.
By thofe you muft open the Compaffes
twice, by this you do no more; by thofe
you cut the Arch obliquely (which is a great
Dif
[ 53 ]
Difadvantage) by this your Compass Points
never make with the Arch an Angle leſs than
60 Degrees.
Beſides, if any one affects thefe Lines of
Chords to many Degrees, they will find them
fupplied in the Ufe of Sines, even to the 180
Degrees; tho' for my Part, I am not fond of
Chords much above 90°.
I. The Radius AB and Degrees
A-
C-
B
D
of an Arch, fuppofe 56, being given; to
find the Chord CD.
Take the Radius AB given, and open the
Sector till the Compaffes reach from 60 to 60.
Then the Diſtance from 56 to 56 is the Chord
required.
II. The Chord CD, and the Degrees 56
being given; to find the Radius.
Open the Sector till the Chord CD reach
from 56 to 56, and then the Diſtance from
60 to 60, is AB the Radius.
III. The Radius AB, and Chord CD
being given; to find the Degrees.
Make AB a Parallel Radius, that is, open
the Sector till it reaches from 60 to 60. Carry
the Chord CD along the Legs till they reft on
E 3
like
[ 54 ]
like Divifions in the Fiducial Lines, which
you will find to be at 56°.
The principal Ufe of the Sector is to make
Angles of affign'd Magnitudes, with given
Lines, and at a given Point thereof. And
alfo to meaſure thofe Angles, that are made,
&c.
IV. Let there be a Line AB and at
A, a given Point thereof, let it be required.
D
A
B
to make ah Angle (of 51°L 30') less than
of 60 Degrees.
This may be done two
without opening the Sector.
different Ways
By taking the
whole Radius of the Line of Chords to de-
fcribe the Arch with, or by taking the Di-
ſtances of the 60's to defcribe the Arch with.
And in the former Cafe to take the Crural
Chord of 51° 30'; and in the latter the
Diſtance of 51° 30' to 51° 30'. Which will
be eaſily underſtood after the Solution of this
firft Example, univerfally.
о
о
Open the Sector at Pleaſure, and in the
Compaffes take the Distance from 60 to 60;
and
[ 55 ]
and fetting one Foot in A, with the other
defcribe the Arch BC; the Sector remaining
thus open, take the Diſtance from 51° 30′
to 51° 30'; and lay it on the Arch from
B to D; then draw AD, and the Angle DAB
is of 51° 30', as requir'd.
If any one Radius would have been more
convenient than another; take that in your
Compaffes, and defcribe your Arch, and open
the Sector till that Radius reaches from 60
to 60; then as before, take the Diſtance from
51L 30 to 51-30, and lay it off upon the
Arch, and draw AD; it is done.
This in the Sectoral Stile is thus; make
the convenient Radius a Parallel at the Chord
of 60; and then the Parallel at 51°L 30' is
to be laid on the Arch.
Or thus; Make the convenient Radius a
parallel Radius; and the parallel Chord of
SIL 30 must be laid upon the Arch.
V. If the Angle to be made had been very
fmall (fuppofe of 3 Degrees) which to make
would be troublefome, on account of the ap-
proaching of the Lines towards the Center;
See Fig. 1. following.
From the given Point E, with a conveni-
ent Radius deſcribe the Arch FG; and lay
this Radius on the Arch from F to H (which
will be 60 Degrees): Then make this Radius
a Parallel of 60; fo will the Parallel of (57)
as much as your fmall Angle wants of 60,
reach backwards on the Arch from H to I.
Draw EI, and the Angle at E is of 3 Degrees,
as required.
VI. If
E 4
[56]
VI. If the Angle to be laid down be of more
than 60° (fuppofe of 94.) See Fig. 2. following
Firſt with any Radius deſcribe the Arch LO,
from the given Point K; and on it lay
the fame Diſtance from L to M; fo will
LM be 60 Degrees; now make the conve-
nient Radius a Parallel at 60; then take the
Parallel at (34), the Degrees which were
given above 60, and lay it from M to N,
and draw KN; fo will the Angle K be of
94 Degrees as required.
VII. If the Angle to be made be of more
than 120 Degrees (or more than two 60's)
ſuppoſe of 167 Degrees; See Fig. 3. following:
With any convenient Radius defcribe the
Arch QU; lay the fame on it from Q to R,
and from R to S; fo will QS be 120 Degrees,
too little by 47. Then make the Radius a
Parallel at 60; and lay the Parallel at
and draw TP; then will
QT be 167°: An Angle made by TP and
FQ, will be an Angle of 167°°
47 from S to T,
VIII. If in the two laft Cafes the Degrees to
be laid off after the 60's, had been very fmall,
they might have been managed as in the 2d
Example.
IX. In the fourth Example, Fig. 3.; if
the Arch WS had been defcrib'd, and as
many Degrees laid from W to T, as the
required Angle wanted of 180; in this
Example 13: And then the Line TP be
drawn ;
[ 57 ]
drawn; the Angle made by TP, and TQ would
be of 167°, as before.
X. Suppoſe the preceding, or any other
Angles were to be meaſured;
1. With any convenient Radius from the
angular Points A, E, K, P, (Fig. 1, 2,
3.) defcribe Arches meeting the Legs of
G
Fig. 1.
H
N
Fig. 2.
MX
E
I
F
Fig. 3.
R
K
I
W
P
the Angles; and where the Arch is greater
than 60, or twice 60 (which will be known
by applying the Radius to the Arch) turn
the Radius over the Arch once, as you will
find
[58]
I
find in Fig. 1; and twice, as you will find in Fig.
2. but never a time in the two Figs. p. 53, 54.
2. Now make your convenient Radius a
Parallel at 60, and leave the Sector at this
Opening.
3. Then in the 1ft Example, take BD in
the Compaffes, and draw them over the Legs
till they reſt at like Divifions on the Fiducial
Lines, and this you will find to be on 51, 30;
the Degrees of the Arch DB, and fo alfo of
the Angle A.
4. In the Second, becauſe very ſmall, lay
off the Radius from F to H, and then mea-
fure as in the laft Arch HI, by carrying
along the Legs till they reſt on like Divifions,
which you will find to be at 57; which taken
from HF (60°) leaves 3° for the Angle.
5. In the third Cafe meaſure NM the
falling beyond LM, and you will find it to
be 34°, which added to LM 60 gives 94°,
which is NL the Meaſure of the Angle K.
6. In like manner TS 47, added to SR
and RQ, 60 and 60, gives TQR 167°, the
Meaſure of the Angle TPQ
Had the Sector been fout, or bad you but
one Line of Chords as on the plain Scale,
any Angle might have been made or mea-
fured. By taking the Distance from the
"Center to 60° to defcribe the Arch; and the
Distance from the Center to the Degrees of
The required Angle to lay on the Arch.
But then the Radius is fix'd to one Length;
whereas on the Sector it may be any
Length between quite hout, and fo
open'd
[ 59 ]
open'd as to put the 60's at the greatest
Distance from one another; which differs
very little from being quite open'd.
Here again it appears how much more the Sector
is preferable to all other Inftruments for
thefe and the like Purpoſes.
XI. To open the Sector, fo that the Lines
of Chords may make any given Angle with
one another, not exceeding 6c Degrees.
Take the Ditance from the Center to the
given Degrees, and make it a Parallel at 60,
the Thing is done.
XII. To find what Angle the Lines of
Chords are open'd to.
Take the Diſtance from 60 to 60, and
meaſure it from the Center on the Chords,
and you will have the
required.
In the Ufe of the Sines
Degrees of the Angle
will be shewn bow
to open the Sector fo that Chords, Lines,
Sines, and Tangents, may form any Angle
affign'd, & contra.
The Lines of Polygons are very near to the
inner Edge of the Sector. At the End
ftands the Figure 6. And the other Small
punch'd Holes in it are numbered towards
the Center by the Figures 7, 8, 9, 10,
II, 12.
XIII. To
[ 60 ]
XIII. To infcribe a regular Polygon of 7
or any Number of Sides in a given Circle.
D
E
A
HA
费
​B
Open the Sector till (AB) the Radius of
the Circle reaches from 6 to 6, and then the
Diſtance from 7 to 7 will be BC the Side of
the Polygon fought; which will divide the
Circumference of the Circle into 7 equal
Parts, in the Points B, C, D, E, F, G, H.
Draw the Lines BC, CD, DE, &c. and you
have the Polygon required.
XIV. And thus you may work for a Poly-
gon of 8, 9, 10, 11 or 12 Sides.
XV. If you want a Polygon of 5 Sides,
divide the Circumference into 10 equal Parts,
and then every two will be a fifth Part.
XVI. If
[61]
VI. If you want one of 3 Sides, the
Radius will divide the Circumference into 6
Parts, and every two of theſe will be a
3d Part.
XVII. If the Polygon you want conſiſts of
more Sides than 12 (as for Example 15);
divide 360, the Degrees of the whole Cir-
cumference, by (15) the Number of the Poly-
gon's Sides; and to the Radius of the Circle
in which the Polygon is to be infcrib'd, find
the Chord of (24°) the Quotient: And it is
the Side of the Polygon fought.
XVIII. But if it had been required to
divide the Circle into many equal Parts (fup-
pofe 32) the Work is beft performed by
taking firſt the Chord of (11° 15') one of
thofe Parts; then of (22, 30) two of thoſe
Parts; then of (33, 45) 3 Parts; then of
(45) 4 Parts; of (56, 15) 5 Parts; of
(67, 30) 6 Parts; of (7845) 7 Parts; of
(90) 8 Parts: And then from one and the
fame Point, fuppofe C, lay on the Arch the
Chord of one Part; then from C lay the
Chord of 2 Parts; from C lay the Chord of
3 Parts, &c. till you arrive at or near the
Chord of 90 Degrees; call this laft D. Now
begin from D, and work as before you did
from C, proceeding onwards till you come to,
or near to 90 Degrees. And then leave off
and begin again.
The Reafon of this Proceſs will eaſily ap-
pear. For if you had taken the Chord of the
32d
+
[ 62 ]
32d Part of the Circumference, fo nearly
true, that a fenfible Error was not found;
yet by turning the Compaffes over 32 times,
it's very likely fome Error would have ap-
peared at laſt.
But by taking 6, 7, 8, or more of theſe
Parts before I begin again, I have no Depen-
dance on, and therefore do not communicate
the Errors of the intermediate Divifions ;
which Errors tho', fingly confider'd, are very
minute, yet when multiply'd by 32, produce
one very fenfible.
It is the beſt way to begin again when you
have about 90 Degrees; becaufe afterwards
fhould you go farther, the Arches ftruck with
the Compaffes would cut the Circumference.
of the Circle too obliquely.
If a Circumference were fo divided, and
Lines drawn from thofe Divifions to the
Center, it would reprefent the Mariners
Compaís. And if this be laid down in a Sea-
Chart; and thefe Lines continued every way
to the utmoſt Extent of the Chart, they would
be the Rhumbs.
In dividing the Periphery of a Circle into
many equal Parts (fuppofe 45) you will no
ways fall exactly on the 90, but frequently
fall fhort of it, or exceed it, as in the 45; for
II Parts would be fhort of the 90, and 12
would exceed it. Therefore in this Cafe it is
beft to renew the Work after you have laid
down the 11.
XIX. If the Side and Number of Sides
of any Polygon be given; the Radius of
the
[63]
the Circle in which it may be infcrib'd, may
be found.
Let the Polygons Side be BC in the Fig.
aforegoing, and the Number of Sides be 7.
Open the Sector till (BC) the Side given,
reaches on the Polygons from (7 to 7) the
Numbers of the Sides; then will the Diſtance
from 6 to 6, be (AB), the Radius fought.
The fame may be done by the Line of
Chords; for dividing 360 by 7, you find the
Quotient to be 51° 26' neareſt. Then you
have a Chord BC, and its Degrees 51° 26':
Therefore by the 2d of this Chapter you may
find the Radius.
The Ujes of these are frequent among Inginiers
in fortifying, and drawing the Plans of forti-
fied Towns. They know the Distance of the
Angles of the Baftions ought to be within
Musket-fhot; that is, they know the Diſtance
of the Baftions or Side of the exterior Polygon,
which fuppofe to be GH: to complect the Poly-
gon, first find the Radius of the Circle (by the
laft) which will circumfcribe fuch a Polygon.
Then from G and H with that Radius defcribe
Arches cutting one another; this will give the
Center of the Circle; which defcribe, and then
compleat the Polygon.
CHAP.
[64]
SACHIN ELSZER
CHA P. XI.
Some Ufes of the Line of Sines.
I. NY Radius A B being given;
A to find the Lines anfwering
to (35) any Number of De-
A
grees.
В.
で
​D
Make (AB) the given Radius a Parallel
at 90, and the Parallel at (35) the given
Degrees, is (CD), the Line fought.
II. A Sine (CD) and its Degrees (35)
given; to find the Radius.
Make the Sine (CD) a Parallel at (35)
its Degrees; and the Parallel at 90, is (AB),
the Radius fought.
III. A Sine (CD) and the Radius (AB)
being given; to find the Degrees to the
Sine.
Make
[ 65]
Make (AB) the Radius a Parallel at 90;
then move (CD) the Sine, along the Lines,
till the Points of the Compaffes fall on the
fame on each side; or, which is the fame Thing,
move (CD) the Sine till it becomes a Parallel;
which you will find to be at (35) the Degrees
anſwering to the Sine.
IV. If you are to deal with a Sine fo near
the Center, that it is not of more than 5
Degrees, inſtead of it you may take the
Chord as taught in the laft Chapter. For
under 6° 15′ the Chords and Sines fo nearly
agree, that their Differences are leſs than the
6000th Part of the Radius; and in a Sector
whofe Leg is a Foot, the Differences are lefs
than the sooth Part of an Inch.
N.B. In the first and third Examples, if
the given Radius had been equal to that on the
Leg, the Sine fought in the first would have
been bad by taking the Degrees from the Cen-
ter; and the Degrees fought in the 3d, by lay-
ing the Sine from the Center upwards. And
this without opening the Sector.
V. To find the verfed Sine of any Number
of Degrees, whether of (55) leſs than 90, or
of (125) more than 90; the Radius (EF)
being given.
+
+
H
E
h
ཀ་
F
Find
[66]
;
Find (EH or Eh) the right Sine of (35),
the Difference between (55 or 125), the
Degrees given, and 90 Degrees. And when
the Degrees given are
s fewer than
} than 90 De-
more
Eh
{fubtract }{E} the right Sine
grees, add
found {from? (EF) the Radius given;
to
and you will have the verfed Sine fought
hF
HF
{FFFF}
VI. If a verfed Sine, and the Radius were
given, to find the Degrees anſwering to it.
Make the Difference between the Radius
and the verſed Sine, a right Sine to the fame
Radius; and find its correfponding Degrees.
Then if the verfed Sine were { greater than
the Radius ;
s add to
2 lefs
2 S
fubtract from }
90
De-
grees; and thofe Degrees being found, you
have the Degrees fought.
VII. If the Degrees anfwering to a verſed
Sine, and that verfed Sine are given; to find
the Radius.
Take the right Sine of the Difference of
the given Degrees, and 90 from the Leg;
and if the given Degrees were
add to
S more
{fewer than
9°, {take from} the Crural Radius this
right
[ 67 ]
right Sine; and opening the Sector make
Sum
this Differénce a Parallel at 99 on the
Sines. Laſtly take the given verfed Sines in
the Compaffes, and carry them along the
Lines till they reft on like Sines; then from
the Sine where they reft to the Center is the
Radius fought.
VIII. The Radius (GH) or (its double
AH) the Diameter being given; to find the
Chord of (94), or any Number of Degrees,
even to 180.
A
Ꮮ
TH-
H
M
Make the given Radius (GH) a Parallel
at the Sine of 30; or the Diameter (AH) a
Parallel at 90. Then the Parallel at (47),
half (94) the given Number of Degrees, is
(LM) the Chord fought.
IX. A Chord (LM), and (94) its De-
grees being given; to find the Radius or
Diameter.
Make on the Sines (LM) the Chord a
Parallel at (47) half the Degrees given ;
then
will the Parallel 30 be (GH) the Radius,
and at 90 (AH) the Diameter.
F 2
X. The
[ 68 ]
X. The Radius (GH), or the Diameter
(AH) being given, and alfo (LM) a Chord;
to find the Degrees anfwering to the Chord.
Make (GH) the Radius a Parallel at 30,
or (AH) the Diameter at 90; then carry
(LM) the Chord along the Lines till it be-
come a Parallel; and the double of (47) the
Degrees where it is fo, gives (94) the De-
grees fought.
Tho' the three laft Articles are Geome-
trically true, even to 180 Degrees in find-
ing the Chord, & contra; yet in making
Angles much greater than 90 Degrees, it is
not fo proper to take the whole Chord as it
is half. And in measuring large Angles
ufe the Method laid down in the Ufe of the
Line of Chords.
XI. To open the Sector, ſo that the
Lines of Lines, Sines, Chords, Tangents, each
may make a right Angle with his like.
So open the Sector, that on the Lines 10
may reach from 6 to 8;
Ór that on the Sines 90, may reach from
45 to 45;
Or from 40 to 50;
Or from 30 to 60;
Or from any Degrees to their Comple-
ment;
Or that on the Sines 45, may reach from
30 to 30.
XII. To
[ 69 ]
XII. To open the Sector, fo that the
aforefaid Lines may make (86), any Angle
requir'd with their likes.
Take (43) half the given Degrees from the
Sines on the Leg, and make it a Parallel at
30 on the Sines; it is done.
XIII. The Sector being open'd, to find
the Angle that any of the aforemention'd
Lines makes with his like.
The Parallel at 30 meafur'd on the Lines,
gives half the Degrees of the Angle.
In the two laſt Examples; if the Angle to
be made or meaſured doth not exceed 60 De-
grees, the beſt way will be to uſe the Chords.
XIV. To defcribe an Ellipfis whofe Tranf
verfe and Conjugate Axes are given.
Let theſe Axes AC, de, cut each other
into two equal Parts at right Angles in B.
6
Hk
3
D
G
H
Of
Alxwvt
B
F
C
K F
kh g Fe
Make
F 3
[ 7༠ ]
Make AB a parallel Radius on the Sines
that is, open the Sector till AB reach from
90 to 90. And from that Point B lay to
wards A the Sine
ΙΟ
20
30
of/40
P
r
to
50
t
60
u
770
8c
W
; and at the fame time you
may do the like in the other half d Ge.
Thro' every one of theſe Points draw Perpen-
diculars to AC, both upwards and downwards.
Then make Bd, half the conjugate Axis, a
parallel Radius on the Sines; and lay the
complement Sines of the preceding Degrees,
anfwerable to the Radius Bd on theſe Per-
pendiculars both upwards and downwards;
that is, lay the Sine
80
p
70
60
9
1
of 50 from
S
to
and alfo down-
40
t
30
u
m
20
W
11
IO
wards on the fame Lines. Likewife do the
fame in the other half of the Figure from B to C.
Laſtly,
[71]
Laſtly, Do the like for as many of the in-
termediate Degrees as you fhall think fuf-
ficient; and thro' theſe Points f, g, h, k,&c.
draw a crooked Line fmoothly, and you will
have the Ellipfis required.
If the Diſtances p, q, r, s, &c. had been
the Sines of 15, 30, 45, 60, &c. anfwering
to the Meaſures of Time in Hours; and if the
Intermediates had been 7, 22, 37,
&c. anfwering to the Time in half Hours;
and others between thefe to Time in Quarters:
you would have had that Ellipfis in that
curious Problem, the Conftruction of a Solar
Eclipfe: In which cafe there would have been
no need of drawing the Periphery, the Points
thro' which it is to pafs are fufficient.
Points thro' which the Ellipfis is to paſs
may be found by the Line of Lines. For on
B, with the Diſtance AB, deſcribe the Semi-
circumference ADC. Take p, q, r, s, t,
&c. at pleaſure, and draw Perpendiculars
thro' them till produc'd they meet ADC:
Now make Bd a Parallel at BD; then will
pf, qg, rb, &c. be Parallels at p 1, q 2, r 3,
Ec. therefore the Points f, g, h, &c. may be
found.
Or Points may be found without any divided
Scale. For with the Diſtance BC or AB
from d; defcribe Arches cutting AC in
F and F. Break BF any how, fuppofe in
K; take AK and KC, and from F and F
defcribe Arches cutting one another in the 4
Points H, H, H, H; every one of theſe
4 Points will be in the Periphery of the
Ellipfis.
F 4
In
[72]
In like manner break BF in another Place,
and you may find 4 other Points: And fo as
many as you pleaſe.
But of theſe three Methods the firft is the
beft. It is better than either of the others
on account of the great Affinity there is be-
tween the Circle and Ellipfis; there being no
Property of the Ellipfis but what is alſo one
in the Circle.
Each of the Points F is call'd a Focus; and
they have this Property, viz. If to any Point
whatſoever, fuppofe G or H, two Lines be
drawn to thofe Points F, F; thoſe Lines GF,
GF together, or the Lines FH, FH together;
are every where equal to AC.
Therefore Gardeners when they would
ftrike an Ellipfis on the Ground, after they
have mark'd out the Tranfverfe and Conju-
gate Axes AC, de, cutting one another at
right Angles in B; and with the Diſtance
BC from d have defcrib'd Arches cutting AC
in F and F; they ftick Pegs in the Points
F, F, d, and tie a String gently ftrain'd
round them; this String will be repreſented
by the Lines FF, FG, GF.
Then taking up the Peg at d, move to-
wards H marking on the Ground the crooked
Line d H, the String being ftretch'd with the
fame Tenfion as it was at d. At length the
Peg will come to H; and by continuing the
Motion, will defcribe HC, and come to C.
And by proceeding till you come to e, A,
and at laft to d; and fo the Ellipfis will be
defcrib❜d.
The
[ 73 ]
The Builders uſe a Tool for this Purpoſe
call'd a Trammel, made by the Mathematical
Inftrument Makers.
If a Billiard Table be Elliptical, and if the
Balls and the Table's Rim were perfectly
elaftick; if a Ball lay in one Focus, and
there was a Net in the other, puſh the Ball
which way foever you pleaſe, hard enough,
it will come into the Net; and if this ſhould
ſtrike 1, 2, 3, or more Balls, hard enough,
no matter whether directly or obliquely,
they would all come into the Net.
If Heat, Light, or any other Vertue,
which will be fo reflected that the Angle of
Incidence be equal to that of Reflection;
and this Virtue proceeding every way from
one Focus, is reflected by the Periphery of
the Ellipfis; they will all be collected in the
other Focus.
Some Folio Volumes are inſufficient to fhew
all the Properties of Conick Sections; never-
theleſs I thought it might yield a little Plea-
fure to the Reader to know fome of them,
in hopes that he might farther pleaſe and
profit himſelf; moſt Mechanicks having Oc-
cafion fometimes, at leaft, to contemplate
them, in order to form juft Notions of what
they are at work on.
XV. To defcribe a Parabola, whofe Ver-
tex and Focus is given.
Let the Vertex be K (Fig. 21. N° 1.) and
the Focus D. Draw KD downwards, and it
is the Axis. Make DA equal to DK.
AK a parallel Radius at the Sine of 90.
Make
And
divide
[74]
divide AK as a Line of Sines to every 10
Degrees, in B, C, D, E, F, G, H, I, to
the Radius AK; and draw the Perpendiculars
Aa, Bb, Cc, &c. on both Sides of the
Axis. To the Radius AK make Ii, Hh,
Gg, Ff, &c. the Chords of 10, 20, 30,
40, &c. Degrees. And all the Points i, b,
g, f, &c. will be in the parabolick Curve.
And if other intermediate Degrees are laid
down, more Points may be found between
thefe; and fo the Curve more nearly exhibi-
ted to the Eye.
This is the Curve in which (abating the
Refiſtance of the Air) all Projectiles move,
as Cannon Balls, Bombs, Stones, &c. when
flung forward.
A Bomb flung from d fo as to go to e, would
pafs thro' the Points f, g, h, i, K, i, h, g,
f, e, d, if the Ground was Horizontal.
But
if the Ground (N°. 2.) defcended, the Ball
would continue to go on in the Curve to c₂
b, a, &c. till it ftruck the Ground. And if
the Ground afcended, it muſt meet the
Ground before it came to d, perhaps at e, f,
g, &c. (See N°. 3.)
But in every of the three Cafes the Bali
will move in a Parabola.
If a Part of half the Parabola GK g, ſhould
be turn'd round upon GK at reft, it would
form a Concave like a Bowl-difh. If fuch a
Concave be made of Metal, and very well
poliſh'd, and Gg be about 18 Inches; and
if the Sun fhining on this polifh'd Concave be
fo turn'd about, that the Axis KG points to
the Sun; all the Rays of Heat coming from
the
[75]
1
the Sun, and falling in this Concave, will be
reflected to the Focus D: And there the
Heat will be fo intenfe, as to vitrify hard
Stones, and to melt Steel, in lefs than a
Minute. But the Concave itſelf will be no
hotter than other the like Metal expos'd to
the Sun.
The Parabola might have been defcrib'd
by the Line of Lines, For drawing PK
perpendicular to the Axis KA. And taking
the Points L, M, N, O, &c. at pleaſure,
and raiſing the Perpendiculars Li, Mb, Ng,
3c. Find a third continual Proportional
(as in the 15th Example of the Ufe of the
Line of Lines) to the double of AK,
KL
Li
KM
and
Mb
and
KN
you will
get
N
KO, &c.
201, 80.5
&c.
L
M
which Diſtances laid from
b
to
N
g
20,00. S TEES
will give thoſe Points in the Curve.
The Parabola alfo may be laid down with-
out any divided Lines. But the firft Method
here propoſed I recommend as the beſt.
If
[76]
If thro' the three Points i Ki be defcrib'd
an Arch of a Circle, whofe Center is very
near to A; that circular Arch will have very
nearly the fame Properties with the fmall para-
bolical ArchiKi: And becauſe it is eaſier to
poliſh a Glaſs or Metal, that is in a ſpherical
Form than any other, the Philofophers
often uſe the Spherical one inſtead of that
formed by the Rotation of the Parabola on
its Axis.
XVI. To defcribe a Cycloid whoſe Height,
that is, the Diameter, of its generating Circle
is given.
Let its Height be AB77 (Fig. 22.); and let
the Line on which the Cycloid is to be defcrib'd,
be z BZ; then will the Semicircumference of
the Circle AHB be 121; which lay from
B to Z, and from B to z Divide the Semi-
circumference AHB into any Number of equal
Parts, the more the better, fuppofe 12, in the
Points BCDEFGHIKLMN; and divide
the half-Bafe BZ into the like Number of
equal Parts. Thro' B, C, D, E, &c. draw
Lines parallel to BZ on both Sides of AB,
and lay BO, BP, BQ, BR, &c. from N
to n, from M to m, from L to 1, from K
to k, &c. And do the fame on the other
Side of the generating Circle; fo will the
Points c, d, e, f, g, h, &c. be all in the Curve
of the Cycloid.
If this Curve be inverted, as in Numb. 2.
and a Body defcends from P, T or
S, by its own Gravity, to B the loweſt
Point
[ 77 ]
Point and Vertex of the Curve; the Times
of theſe Deſcents fhall all be equal.
Numb. 2.
P
T
W
B
If a Body deſcends from T to S in the Curve
TVS by its own Gravity; the Time of its
Defcent will be lefs than if it went in the
ftrait Line TWS; or lefs than if it went in
any other Line: therefore this is call'd the
Line of fwifteft Deſcent.
If the Curve QBP were cut into two equal
Parts in the Vertex B; and thofe Parts QB,
PB, were turn'd upwards above the Line
QP, fo that P and Q of P and Q become C;
and alfo the convex Parts of thefe Curve
Lines towards one another; and if Plates
were bent in the Form of thoſe crooked Lines,
and a Pendulum hung at C, of a length cqual
to the crooked Line BQ, counted from the
Point
[ 78 ]
Point of Sufpenfion to the Center of Ofcilla
tion I fay that Center of Ofcillation will
defcribe the Cycloid PBQ; and all the Vi-
brations, whether long or fhort, fhall be per-
form'd in the fame time; therefore it is
call'd the Ifochronal Curve.
This Curve may be conceiv'd generated by
a Nail in a Coach Wheel. Firſt let the
Wheel touch the Ground at z, and let it row!
towards great Z. By that time that is car-
ried forward fo that its Center be S, the Nail
7 will be raiſed to ; and have deſcrib'd the
Arch zb. And when the Center is come to
w, the Nail will be at the Top A, and
will have defcrib'd the Curve zb A. The
Wheel going ſtill on till its Center comes to v,
the Nail will touch the Ground at Z; and
have defcrib'd the Curve zb AZ. And the
Circumference of the Wheel or generating
Circle would have meafur'd out Zz the Bafe
of the Curve. And fo the Bafe of the Curve
is equal to the Circumference of the genera-
ting Circle.
If the Wheel fhould continue thus to rowl
on ad infinitum, an Infinity of ſuch crooked
Arches would be defcrib'd, which all con-
fider'd together, is but the fame Curve con-
tinu'd; and fo this Curve by Mathematicians
is faid to be a Line of the infinite Order; for it
may be cut by one right Line into an Infinity
of Points.
No Algebraick Equation can exprefs the
Relation of its Ordinate to its Abfcifs; there-
fore it is call'd a Tranfcendent Curve. And
notwithſtanding this, its Area is comparable
with
[ 79 ]
with the Area of its generating Circle, and is
indeed juſt 3 times that Circle. And there
is no Part or Segment of it, but what may
be compared with fome Part of the Circle.
Nay, feveral Portions of it are actually fquare-
able by right Lines, fuch is AbH, AkВ; Aw
being Radius, and AB with many other
Portions. And tho' right Lines can't be af
fign'd equal to the Arches of Circles; yet there
is no Arch of this Curve, notwithſtanding its
Superiority, which we cannot affign a right
Line equal to. For any Arch (as Ale) of
the Cycloid taken from A the Vertex, is
equal to twice AL the Chord in the generating
Circle, whofe End falls in the Ordinate e E
to that Arch of the Cycloid. And fo the
whole Cycloidal Curve Line ZAZ is equal
to 4 Diameters of the generating Circle.
XVII. To folve by the Lines of Sectoral
Sines the grand Geographical Problem; that
is, having the Latitudes of two Places, and
their Differences of Longitude, to find their
Diſtance.
Let the Example be of London in the
Latitude of 51° 32′ N, and of Pekin in China,
in the Latitude of 40° N; their Difference
of Longitude is 117–30.
With any Radius (See the following Fig.)
deſcribe the Circle NP A SQ, draw NP G,
and lay the Chord of 90 from NP to Æ, and
draw GQ Lay the Chord of $1L 32
(London's Latitude) from A to L and draw
LG; lay the Chord of Pekin's Latitude
(400) from A to P and from Q to F.
2
And
[80]
と​の
​Æ
I
B
NP
P
C
F
And draw pF; make Bp on the Sines a
parallel Radius. Then take the Sine of the
Difference of 90 Degrees and Difference of
Longitude (27°L 30) and lay it from B
to P; if the Difference of Longitude be
S lefs
than 90 Degrees from B towards
greater
P
{} call it P. That is the Place of Pekin.
F
From P draw PC perpendicular to LG; then
is LC the Diſtance fought, which meaſure by
the Chords.
!
The two Aftronomical Problems are per-
formed after the fame Manner; viz. when
the Latitudes of two Stars, with their Dif-
ferences of Longitude; or the Declinations
of two Stars with their Differences of Afcen-
fion are given; to find their Diſtance.
1
XVIII. To
[ 81]
XVIII. To cut a given Line AB into
mean and extream Proportion.
Parallel at the Chord of 60.
Make AB a
Then is the
Parallel at the Chord of 36 the greater Seg-
A-
C
B
TD.
ment AC.
Or make AB a Parallel at
the Sine of 54; then is the Parallel at
30 the greater Segment AC; and the Paral-
lel at 18 the leffer Segment BC. Et
vice verfa.
HETOUR FP85a
CHA P. XII.
Some Ufes of the Tangents and Secants.
T
HERE are two Lines of Tan-
gents on the Sector, one (like the
Chords) beginning at the Center,
and running to 45 Degrees, where
it is equal to the Radius, as is demonſtrated
in the 3d Chapter: The other which I call
the upper Tangents, beginning with 45
Degrees (which therefore is the Radius) at
the Diſtance of the Number 25 of the Lines
G
from
[ 82 ]
from the Center, and running to upwards of
70, viz. to as many Degrees as will come
on.
By the former may be found the Tangent of
any Number of Degrees not more than 45, to
any Radius within the Compafs of the Sector,
& contra.
I. For let the Radius given be (AB) any
Line; to find the Tangent of (37° 15') any
Number of Degrees.
A
C
-B
D
Make (AB) the Radius given a Parallel
at 45; and then the Parallel at (37° 15)
the Degrees given, is the Tangent (CD)
fought.
II. A Tangent (CD) being given of
(37° 15′) a given Number of Degrees; to
find the Radius.
Make (CD) the Tangent given a Parallel
at (3715) the Degrees given; and then
will the Parallel at 45 be the Radius fought.
III. A Tangent (CD) and the Radius
(AB) being given; to find the Degrees an-
fwering to that Tangent.
Make the given Radius (AB) a Parallel at
45°; then the Tangent (CD) carry along
the Lines till it becomes a Parallel; that is,
till
[83]
till the Points of the Compaffes reft on the
fame Tangents on fuch Line;
which you
will find to be at 37° 15.
IV. The Radius (EF) being given; to
find the Tangent of (57°) any Number of
L
E
G
F
H
Degrees above 45, as far as the Sector reaches
on the upper Tangents.
Make EF a Parallel at 45° on the upper
Tangents (commonly call'd the leffer Tangent,
on account of the leffer crural Radius) and
the Parallel at 57 is (GH) the Tangent
fought.
V. A Tangent GH and its correfponding
Degrees (57) above 45 being given; to find
the Radius.
Make CD the given Tangent a Parallel at
(57) the Degrees given, and then will the
Parallel at 45 be the Radius fought.
VI. A Tangent (GH) and the Radius
(EF) being given; to find the Degrees an-
fwering to this Tangent.
Make the Radius (EF) a Parallel at 45;
and then the Tangent GH carried along the
Lines till it becomes a Parallel, will give (57)
the Degrees.required.
G 2
VII. The
1
[ 84 ]
VII. The Radius (IK) being given; to
find the Secant of (57°) any Number of De-
I
L
K
M
grees as far as the Sector reaches on the Line
of Secants.
Make (IK) the Radius a Parallel at the
Beginning of the Diviſions; that is, at the
Secant of o Degrees; and the Parallel at
(57) the Degrees given, will give (LM) the
Secant required.
VIII. A Secant (LM) and (57) its cor-
refponding Degrees being given; to find the
Radius.
Make (LM) the Secant a Parallel at (57)
the Degrees given; and the Parallel at o De-
grees will be the Radius fought.
IX. The Radius (IK) and the Secant
(LM) being given; to find the Degrees an-
ſwering to that Secant.
Make (IK) the Radius a Parallel at o De-
grees; and carry (LM) the Secant along the
Lines until it becomes a Parallel; which will
be at (57) the Degrees fought.
It is to be obferv'd, that as the crural Ra-
dius of this Line of upper Tangents, and alſo of
the Secants, were made but to Part of the
Line
[85]
1
旭
​649
9
Line of Lines, in order to bring on as many of
the upper Degrees as we could conveniently;
the Radius is here but 2 Inches and 118; and
fo its double is 5 Inches and ½, which is the
greatest Radius that thefe Secants and upper
Tangents can fimply be applied to; and fo the
greatest Circle that they can fimply be applied.
to, is that whofe Diameter is equal to the
Length of the Line of Lines; viz. 11 Inches
and 23
32
Had it been made much larger, there had
been wholly loft the Use of the upper Degrees;
had it been made much less, the Detriment
would have been greater than the Gain.
But tho' the Tangents and Secants reach but
to little more than 75 Degrees, fuppoſe to 76,
yet by the fame Sector the Tangent and Secant
of any Number of Degrees may be bad.
XX. To find the Tangent of (82°) any
Number of Degrees above 76, not exceeding
(83) the middle Degree between 76 and 90.
From 90 fubtract (82) the Degrees, and
from (82) the Degrees given, fubtract (8)
the preceding Remainder; and there will be
left 74: Then the Sum of the Tangent and
Secant of (74), this laft Refult will be the
Tangent of (82) fought.
*
XI. A Tangent of (82) any Number of
Degrees being given; to find the Secant of
the fame.
To the Tangent of (82) a given Number
of Degrees, add the Tangent of half (8) the
Complement of (82) thoſe given Degrees to
G 3
× Jee Gunter 149
90
"
A
}
the secant of an Arch is equal to the
Tangent of it, & the Tangend of the Tangent
of halt it's comp Tee Emerson Jorg p 23.
7
}
[ 86 ]
90, (viz. the Tangent of 4); and you will
have the Secant of (82) the given Number
of Degrees.
And now the Sum of the Secant and Tan-
gent of 82, is (as in N. 10.) the Tangent of
86°. And if to this you add the Tangent of
half 4, i. e. 2; you have the Secant of 86
by N. 11. In like manner you may pro-
ceed to find the Tangent and Secant of 88,
89, 89 Degrees.
XII. To find the Tangent and Secant of
any Number of Degrees beyond the Length
of the Sector; fuppofe of 87.
Take 87 from 90, and there remains 3;
take 3 from 87, and there is left 84. Now,
if you had gotten the Tangent and Secant of
84, the Tangent and Secant of 87 would
have been gotten (by N. 10. and 11.)
But 84 is not on the Sector.
Take 84 from 90, and there remains 6;
take 6 from 84, and there is left 78. Now,
if you had gotten the Tangent and Secant of
78, the Tangent and Secant of 84 might be
had by the 10th and 11th. But if 78 be not
on the Sector;
Take 78 from 90, and there remains 12;
take 12 from 78, and there is left 66; whoſe
Tangent and Secant are on the Sector.
Therefore (by N. 10. and 11.) from the
Tangent and Secant of 66, you may find the
'Tangent and Secant of 78.
And from the Tangent and Secant of 78,
you may find the Tangent and Secant of 84.
And
[ 87 ]
And then from the Tangent and Secant of
84 you may find the Tangent and Secant of
87 fought. The like may be done of any
other Tangent and Secant.
XIII. The Sector cannot be fo open'd as
to make the Radius on the Secants or upper
Tangents quite 6 Inches. But this Defect may
be very eaſily cured by the Sines and the
lower Tangents. For let it be requir'd to
find the Tangent and Secant of 65 Degrees,
to a Radius of 6 Inches. Make 6 Inches à
Parallel at the Tangent of (25) the Comple-
ment of (65) the Degrees given: Then the
Parallel at 45 gives the Tangent fought.
Make 6 Inches a Parallel at the Sine of
(25) the Complement of (65) the Degrees
given. Then the Parallel at 90 is the Secant
fought. In like manner the Tangent and
Secant of Degrees beyond the Rule may be
had if a leffer Radius would fuffice. For to a
Radius of 3 Inches may be had the Tangent
and Secant to 82 Degrees, at one opening
of the Sector. And if 2 Inches will fuffice
for the Radius, you may have the Tangent
and Secant to 85 Degrees at one opening of
the Sector. For make 2 Inches a Parallel on
the Sines at 5 Degrees the Complement of
85; and then the Parallel at 90 is the Secant
fought. And make 2 Inches a Parallel on
the Tangents at 5 Degrees, then the Parallel
at 45 is the Tangent of 85 fought. And the
Converſe of theſe, viz. the finding of the De-
grees anfwering to a Tangent or Secant when
the Radius given is greater than the Radius
G 4
on
[88]
on the Secants or upper Radius is equally eafy.
For make the given Tangent a Parallel at
45 on the lower Tangents; and where the
given Radius is a Parallel on the Tangents,
you have the Complement of the Degrees
fought. Or make the given Secant a Parallel
on the Sines at 90; and where the given
Radius is a Parallel on the Sines you have the
Complement of the Degrees fought.
Therefore the Degrees anfwering to any given
Tangent or Secant not longer than 23 Inches
7 to a Radius given less than them; may
be found by the foremention'd Sector. Or,
16
Laftly, If the Secant or Tangent, and
their Degrees were given, the refpective
Radius would be found. For the Tangent
given being made a Parallel at 45 on the Tan-
gents; the Parallel at the Tangent of the
Complement of the Degrees is the Radius
fought. And the Secant being made a Paral-
lel at the Sine of 90; the Parallel at the Sine
of the Complement of the Degrees is the
Radius fought.
XIV. The Latitudes of two Places, toge-
ther with their Difference of Longitude being
given; to find their Diſtance at one opening
of the Sector.
Let the Example be for London and Rome,
whofe Latitudes are 51° 30' N, and 41
50' N ; and Difference of Longitude 12° 40'.
With
[89]
With any Radius defcribe a Circle NP EQ,
and quarter it by drawing NP S and
Qat
Æ R
1
C
f
N
Q
right Angles to one another. By help of the
Chords (the Opening of the Sector remaining
the fame) lay off the Latitude of London and
Rome 51 30 and 41 50 from Æ to 1 and R ;
and alfo of Rome from S to r, and draw lr.
Lay the Tangent of the Difference of Longi-
tude 12° 40′ from C to D, and the Tan-
gent of its half from C to d. And draw dl.
with the Diſtance DNP from D, deſcribe an
Arch cutting dl in L. Bifect the Diſtance
RL at right Angles with the ſtrait Line cF.
Produce F till it meet Cr in F. Meafure
CF on the Tangents, and lay the Tangent
of half thofe Degrees from C to f. Laftly,
a Ruler laid from f to L meets NPE in G.
And the Diſtance GR is the Diſtance fought;
which meaſure on the Chords.
XV. To defcribe an Hyperbola, the Af
fymptotes and Vertex being given.
Let
[ 90 ]
Let A B, AC, (Fig. 29.) be the Affymp-
totes, and E the Vertex of the Curve. From
E draw EO, EQ parallel to the Affymptotes.
Make AO or AQ a Radius on the upper
Tangents; and from A lay the upper Tan-
gents to F, G, H, I, &c. and to K, L, M,
&c. and thro' thefe Points draw Ff, Gg,
Hb, &c. Kk, LI, Mm, &c. parallel to the
Affymptotes. Make the fame AO the Radius
on the lower Tangents; and lay the Tangents
of the Complements of the preceding Tangents
from F to f, from G to g, &c. and alfo from
K to k, from L to 1, &c. Then will the
Points E, f,g,b,&c k,l,m,&c. be Points in the
Hyperbolick Curve. Confequently a crooked
Line fmoothly drawn thro' thofe Points,
(efpecially if many of them are taken, fuppofe
to every Degree) will exhibit the Curve very
beautifully to the Eye.
But Points may be found thro' which the
Hyperbola may pafs without any divided
Scale.
This Curve the Hyperbola continually ap-
proaches its Affymptotes, but never meets
them.
In any Part of the burning Zone, or in
any Part of the temperate Zones, if a Staff
be any how ftuck in a plain level Ground,
whether upright or leaning, the End of the
Shadow of the Staff ſhall deſcribe an Hyper-
bola every Day in the Year.
In all Dials whofe Stiles height is lefs than
66° 30', the Parallels of Declination are
Hyperbola's. And moft part of the beautiful
Ornaments drawn on Dials depend on them.
By
[91]
By Nature alfo in this Curve doth Fluids
rife to any heighth above the common open
Level.
Let RYXS, RYVT be two fquare Plates of
Glafs pinch'd cloſe on one Side of RY by two
R
T
Z
W
X
bits of bent Tin lapping round the Glaſſes; and
open'd on the oppofite Side to about a quarter
of an Inch by a fmall wooden Wedge; the
Wood gently giving way a little, the Wedge
foft not breaking the Glafs. If this Com-
bination be plac'd in a flat Bottom, with
Water about of an Inch deep, or more or
lefs at Pleaſure; you will fee the Water rife
up to the Top RST, at the cloſe Side RY,
in the Form of the Figure RXZWXYR;
and the Curve XZW will be an. Hyperbola;
RYX its Affymptotes; and Z its Vertex.
If the Liquor be tinged, and the Inſide of
the Glaſs Plates made very plain, and wetted
before you try the Experiment, the Phæno-
menon will be more pleaſant.
This
[92]
This Truth is proved by thoſe who fhew
the Philofophical Experiments; from the
rifing of Fluids in capillary Tubes.
If the wooden Wedge be drawn out a little
fo that the Plates come nearer together, the
Vertex and every Part of the Curve will be
farther diftant from the Affymptotes. But the
Curve will be fimilar to what it was before.
If RY be perpendicular to the Surface of the
Water in the Veffel; all the Hyperbola's will
be Equilateral. If the Plates are perpendicu-
lar to the Surface of the Water; but the Line
RY inclin'd to it; the Hyperbola's will be
Scalene. If RY be perpendicular to YZ; and
the whole Combination inclin'd to the Surface
of the Water; the Hyperbola's will be far-
ther remov'd from the Affymptotes.
XVI. To defcribe the true Sea-Chart,
fometimes call'd Wright's or Mercator's. And
let the Example be from the Latitude of 45
to the Latitude of 55 Degrees.
Open the Sector fo that the Diſtance of the
Beginning of the Secants may repreſent one
Degree of Longitude. And having drawn
AM (Fig. 31.) to repreſent the Parallel of
45 Degrees; lay on it the aforefaid Diſtance
(viz. the Parallel at 45) from A to B, from
B to C, from C to D, &c. as far as you pleaſe.
And draw A X at right Angles to A B.
The Sector opcn'd as before; take the Paral-
lel Secant of 45, and lay it from A to N ;
and the Parallel at 46, and lay it from N
to
[ 93 ]
to O. And in like manner the Parallel
47
48
49
✪ to P
P to Q
Q to R
50
R to S
at
and lay it from
JI
S to T
52
T to V
53
V to W
54
W to X
Now thro' N, O, P, Q, &c.
draw Lines
parallel to AB. And thro' B, C, D, &c.
draw others parallel to AX. Then in fome
convenient Place of the Draught lay down
the Rumbs, as taught in the 18th Article of
the 10th Chapter; and you will have the true
Chart requir'd, divided to every whole
Degree.
If greater Exactness be required, make
the parallel Radius on the Secants equal to
half a Degree of Longitude; that is, take the
Diſtance from 45 to 45, and lay from A
to B, from B to C, from C to D, &c.
3
I
Then take the Diſtances at the Middle of
every half Degree, viz. at 45, 45, 46
746 49
&c. and lay them from A to N, from N to
O, &c.
And thro' thefe Points draw Lines parallel
to AB and AX; and you will have the
Parallels of Latitude for every half Degree's
Diſtance; and the Meridians at every half
Degree of Longitude diſtant from one
another.
The
t
[94]
The Parts of a Degree lefs than a half,
fuppofe Quarters, are had in Longitude by
Bifection. The fame may be done for the
farther Divifion on the Meridian, near enough
for any Ufes.
If the Foot Sector beforementioned be here
made ufe of, a Degree of Longitude may be
made 11 Inches; and fo every League will
be reprefented by more than half an Inch.
Which is more than fufficient for any Degrees
of Exactnefs required in keeping an Account
of a Ship's Way, and pricking the Ship's Way
down on that Chart. For indeed there is no
Neceffity for laying down every half Degree;
unleſs it be for Voyages very near the Pole.
But divide every whole Degree of the Meri-
dian into 20 equal Parts, to reprefent
Leagues. And the fame in fome one Parallel
of Latitude; or on the Equinoctial.
This Chart differs from the plain Chart and
the Globe in theſe following Particulars.
The Meridian in this Chart is a ftrait Line
divided by unequal Diſtances reprefenting
Degrees; and thefe unequal Diſtances con-
tinually increaſe as they come nearer the
Poles. In the plain Chart the Meridian is a
ftrait Line equally divided into Degrees. On
the Globe the Meridian is a great Circle
equally divided into Degrees.
The Parallels of Latitude in this Chart,
are all ftrait Lines of the fame Bignefs, and
are all alike equally divided; but thefe Parallels
are at unequal Diſtances from one another.
But a Degree of one of thefe Parallels bears
fuch Proportion to a Degree of the Meridian
in
[ 95 ]
in the Latitude of this Parallel; as a Degree
of the fame Parallel of Latitude on the Globe,
is to a Degree of the Meridian on the Globe.
But tho' the Parallels of Latitude in the plain
Chart are all ſtrait Lines of the fame Bignefs,
and alike divided; they do not bear that Ana-
logy to their correfpondent Degrees of Eaſt-
ing and Weſting, as thoſe on the Globe do.
This Chart and the plain Chart agree only
in this, viz. the Rumbs in both are ſtrait
Lines; and the like Rumbs make equal
Angles with every Meridian they meet. But
on the Globe the Rumbs are crooked Lines of
the ſpiral Kind refpecting the Pole as their
common Eye; one of which they continually
approach: But cannot be faid to come quite
to it without an indefinite Number of Times
running round it.
This Chart at its firft Appearance was, and
ftill is look'd upon as the fittest of all others
for thoſe who fail long Voyages; efpecially
if they are oblig'd to be in high Latitudes.
-
The Problems ufually perform'd by this
Chart are theſe following.
Of any two Places laid down on the Chart,
to find their Latitudes, Difference of Longi-
tude, the Rumb from the one to the other;
and their true Diſtance in that Rumb.
And of thefe 4, any two being given, the
other two may be found; fave that when the
Difference of Longitude and Diſtance are the
two given; the Operation will be tentative.
And theſe 16 Problems are by this Chart
better folv'd than they are by the Globe
itſelf.
of
[96]
Of thefe Solutions, together with the Com-
pariſon of this Chart with the plain Chart,
I have not at prefent Room enough to give
Examples.
SAITTARD DESALON
CHA P. XIII.
The general Ufe of the Line of Numbers.
T
30000
233 HE Reading, Eftimating, and Valu-
ing the feveral Divifions and Di-
ftances on this Line were fully laid
down in the 7th Chapter; therefore
need not be repeated here.
The Operations perform'd by this Line are
theſe following; viz. Multiplication, Divifion,
Proportion, Extraction of Roots; and feveral
others depending on theſe.
I. The general Rule for working Propor-
tion is, Extend the Compaffes from the firſt
Number to the fecond; and the fame Extent
laid the fame Way, will reach from the third
to the Anſwer.
N. B. By the fame Way, I mean, that if
the Compaffes, open'd from the first to the
fecond, was from the Left-Hand to the
Right, that is upwards, then they ſhould be
applied from the Left to the Right for the
Anſwer. And on the other hand; if the
Extent
[97]
Extent of the Compaffes from the Firſt to the
Second, be from the Right to the Left; then
the Compaffes from the Third to the Anſwer,
muſt be alfo from the Right to the Left.
Example 1. To the three given Numbers 40,
55, 56; to find a fourth Proportional. Ex-
tend the Compaffes from 40 to 55; and that
Opening will reach from 56 to 17, the
Anfwer.
Example 2. To the three given Numbers
150, 90, 81, to find a fourth Proportional.
Extend the Compaffes from 150 to 90; the
fame will reach from 81 to 54, the Anſwer.
In the firſt of thefe Examples the Extent
was made to the Right-Hand, or upwards ;
in the laſt, to the Left, or downwards. But
we are to remember what hath been faid elfe-
where; viz. that of four proportional Num-
bers, either the Second and Third may be
taken for the other. Note alfo, in taking of the
Extent of the Firſt to the Second, one Value
may
be affign'd to the Divifions; and in lay-
ing it from the Third to the Anfwer, the
fame, or any other Value may be affign'd.
Example 3. To the given Numbers 4, 9, 63 2,
to find a fourth Proportional. Extend the
Compaffes from 4 to 9 (calling the Left-Hand
1 'an Unite) and the fame Extent (calling the
fame 1 an 100) will reach from 632 to 1422.
I
Example 4. To the given Numbers 12, 5,
420, to find a fourth Proportional. Extend the
Com-
H
[98]
Compaffes from 12 to 5 (calling the middle
I ten) and the fame Extent (calling the
middle 1 an 100) will reach from 420 to
175.
୮
If the firſt Extent be either too long for
your Compaffes, or falls beyond the Line, be
it upwards or downwards; vary (but let it be
Decimally) your given Numbers, as is taught
in the 4 firſt Variations in Chap. 8.
Example 5. To the given Numbers 2,98,50;
to find a fourth Proportional. Here I vary
them by multiplying the firft by 10; and
then they are 20, 98, 50. And the Anfwer
comes out 245, which by the 2d Variation is
2450.
29
L32
Example 6. To the given Numbers
L7; to find a fourth Proportional. Calling the
Middle an Unit, the Extent from 2 to L 3,
will reach from 7 to 1L05, the Anfwer.
II. Two Numbers being given; to find a
3d, 4th, 5th, &c. continually Proportional.
Extend the Compaffes from the firſt to the
fecond; and that Extent will reach from that,
2d to the 3d Proportional; the fame Extent
will reach from the 3d to the 4th; from the
4th to the 5th, &c.
Example 1. Let the Numbers be 4 and 8 ;
then the Extent from 4 to 8, will reach from
8 to 16, from 16 to 32, and from 32 to 64;
but from 64 on the first Line to 128, and
from 128 to 256, &c.
Example
[ 99 ]
Example 2. Let the Numbers be 10 and 7.
The Extent from 10 to 7 will reach from 7 to
4L9, from 79 to 443, &c.
III. Multiplication is perform'd from this
Confideration. Unity is to one Factor, as
the other is to the Product.
Example 1. Multiply 33 by 25. Extend
the Compaffes from 1 to 25, the fame Extent
will reach from 33 to 825 the Product.
In this Example, I chang'd the Proportion
by Variation of Chap. 8. viz. I made it as
I is to 2L 5, fo is 330 the Anſwer.
But the Products may be had without any
Confideration of thofe Variations thus: Find
both the Factors in the firft Line of Numbers,
without regard to the Value of the Units.
Extend the Compaffes from 1 to the leaft of
the Factors; and that Extent will reach from
the other to the Anfwer, which will always
conſiſt of as many Integral Places, as there
were Integral Places in both Factors, if the An-
fwer falls above the Middle 1. But will be a
Place lefs, if it falls fhort of it.
Example 2. Multiply 575 by 28. Seek
28 in the firſt Line of Numbers, and alfo
575 without any regard to the Value of the
Unit. Then the Extent from 1 to 28, will
reach from 575 to 16100; viz. to the Num-
ber of the Integral Places in both Factors,
becauſe the Product falls above the Middle 1.
H 2
Ex-
[ 100 ]
Example 3. Multiply 42L 5 by 36. Ex-
tend the Compaffes from 1 to 425; and the
fame Extent will reach from 36 to 1530Lo.
That is, to as many Integral Places, as there
were Integral Places in both Factors.
Example 4. But if the Factors had been
425 and 16, the Product falls below the
Middle 1; and therefore confifts but of 4
Integral Places; viz. one lefs than 5 the
Number of Integral Places in Factors.
The fame had been equally true if one or
both Factors had been pure Fractions; pro-
vided that the Cyphers, between the figni-
ficant Figures and the Place of Unity, be
efteem'd as Negatives. So if 22 were to be
multiply'd by Lo8. Extend the Compaſſes
from 1 to 8, the fame Extent will reach from
22 to 17L6.
IV. Divifion is perform'd from this Con-
fideration. The Divifor is to Unity, as the
Dividend is to the Quotient. Therefore the
Extent from the Divitor to Unity, will reach
from the Dividend, downwards to the
Quotient.
Example 1. Divide 825 by 25. Extend
the Compaffes from 25 to 1, the fame Extent
will reach from 825 to 33 the Quotient.
But the Quotients may be had thus. Find
both the Divifor and the Dividend in the
fecond Line of Numbers, without regard to
the Value of the Unit. Then the Extent
from
[ 101 ]
from the Divifor to the Middle 1; will reach
from the Dividend downwards to the Quo-
tient; which, if it falls below the middle 1,
will confift of as many Integers, as the Inte-
gral Places of the Dividend are in Number
more than thofe in the Divifor. If it reaches
not to the Middle, the Quotient will confift
of one more than the Exceſs.
Example 2. Divide 1530 by 36. Theſe
being both fought in the fecond Line, extend
the Compaſſes from 36 to the Middle 1, and
the fame Extent will reach from 1530 to
425. But becauſe it falls below the Middle
I,
1, the Integral Places are 2, equal to the
Difference between the Number of the Inte-
gral Places in the Dividend, and thoſe in the
Divifor.
Example 3. Divide 16100 by 28. Theſe
being found in the fecond Line, the Extent
from 28 to the Middle 1, will reach from
16100, down to 575; which becauſe it falls
below an Unit, confifts of 3 Integral Places
3 being the Difference between 5 and 2.
;
Thefe
Example 4. Divide 935 by 25.
being found in the fecond Line as before
the Extent from 25 to the Middle 1, will
reach from 935 to 374; which becauſe it
falls above the Middle 1, will confift of one
Integral Place more than the Difference of
the Places in the Numbers given; therefore
the Anſwer confifts of two Integral Places;
confequently it is 37L4.
H 3
What
[ 102 ]
What hath been faid in Multiplication
concerning the pure Decimals may be ap-
plied here.
V. To find a mean Proportional between
two Numbers given. When the Numbers
confiſt both of odd Numbers of Places, or both
of even; ſeek them both in the fecond Line
of Numbers. But when one hath an odd
Number of Places, and the other an even;
feck one in the firſt Line of Numbers, the
other in the ſecond. Then the Middle
between theſe two Numbers gives the mean
Proportional fought. And the Number of
the Integral Places of this Mean fhall
be half the Sum of the Number of Inte-
gral Places in both the given Numbers, if
that Sum can be halv'd. But if it be odd,
and fo cannot be halved exactly, it fhall be
half of one S more
2 lefs
} when the Anfwer falls
{belove} the middle Unit.
S
Example 1. Find a mean Proportional be-
tween 252 and 468.
Becauſe they are both of odd Places I feek
them on the 2d Line, and I find their Mid-
dle 336, which is the Mean fought, and doth
confift of 3 Integral Places, the half of the
given 6 Places.
Example 2. Find a mean Proportional be-
tween 8 and 128.
Becauſe
[ 103 ]
Becauſe both are of odd Places, I feek
both in the ſecond Line, and their Middle
gives me 32, which are both Integers, be-
caufe 2 is the half of 4 the Number of Inte-
gral Places given.
Example 3. Find a mean Proportional be-
tween 405 and 84L 50.
Becauſe both are of even Places, I feek
them in the fecond Line, and the Middle
between them I find to be 585. Becauſe
there was but 4 Integer Places in both Num-
bers, the Anfwer is 585.
Example 4. To find a mean Proportional
between 8 and 98.
2
I ſeek the 8 on the firſt Line, and the 98 on
the fecond, according to the Rule. And
becauſe the middle Point falls above the
middle Unit on 28; I add 1 to 3 the Num-
ber of Integers in the given Numbers, and
half the Refult 4 (viz. 2.) fhews the Num-
ber of Integers in the Mean fought.
Example 5. Find a mean Proportional be-
tween 544 5 and 121 5.
The Numbers being fought on both Lines ac-
cording to the preceding Directions, I find the
middle Point falls below the Middle 1 on 825.
Now becauſe the middle Point falls below the
Middle 1,I fubtract 1 from theSum of theNum-
bers of the Integral Places in the given Num-
bers, and conclude that 2, half the Remainder,
is the Number of the Integral Places in the
I
H 4
Mean
[104]
>
Mean fought.
is 82L 5.
Therefore the Mean fought
If one or both are purely Decimals, feek them
as tho' they were Integers, by counting the
Cyphers between the fignificant Places, and the
Place of Unity as negative Integers. And in
valuing the Mean confider the Number of theſe
Cyphers negatively.
VI. To extract the Square Root of any
given Number. Find a mean Proportional
between that Number and Unity. It is the
Square Root fought.
So the Square Root of 4225 is 65, the
mean Proportional between 4225 and 1.
VII. To extract the Cube Root. The
Integral Places of the Number given, con-
fifts of a Multitude which is either an exact
Multiple of 3, or exceeds a Multiple of 3 by 1,
or exceeds a Multiple of 3 by 2. If it exceeds
a Multiple of 3 by 1, feek it in the firſt Line;
if by 2 in the fecond. Divide the Diſtance
between the Number and the firſt 1 into 3
equal Parts; and that Point of Divifion
which falls next the firſt I is the Cube Root
fought.
So 3 is the Cube Root of 27, 2 of 8, and
4 of 64. When the Places are in Multitude
an exact Multiple of 3, feek it in the laſt
Line; and divide the Diſtance between it
and the laft 1 into 3 equal Parts. And the
Divifion next the laft i fhall be the Root
fought.
So
[ 105 -]
So the Cube Root of 512 is 8: The Cube
Root of 125 is 5.
The Multitude of Integral Places may be
thus known. If thofe given do not exceed 3,
there is but one Integral Place in the Root.
If they do exceed 3 but not exceed 6, there
will be 2. If they exceed 6 but not exceed
9, there will be 3, &c.
So the Cube Root of 1520L875 will be
found by the Line to be exactly 11L5.
VIII. Three Numbers being given; to
find a fourth in a duplicate Proportion.
The Extent from the Firſt to the Second,
doubled, will reach from the Third to the
Fourth.
Example and Ufe. The Content of a
Circle whofe Diameter is 7, is 38L5; what
is the Content of the Circle if the Diameter
be 10L 5?
Becauſe all fimilar Superficies are to one
another in a duplicate Proportion of any of
their Homologous Sides: It is,
As 7 to 10L 5 Duplicate, fo is 385 to
the Content fought. Therefore the Extent
from 7 to 1oL5 Duplicate, will reach from
38L5 to 86L 6, the Content of the Circle
fought. For the Extent from 7 to 10 5 will
reach from 385 to 57L75; and the fame
Extent will reach from 5775 to 86L 6 the
Content.
In like manner, if the Content of one of
any two fimilar Superficies be given, to-
gether with Homologous Lines in each; the
Content of the other may be found.
IX. Three
[106]
IX. Three Numbers being given; to find
a fourth in a fubduplicate Proportion.
This is the Converfe of the foregoing;
therefore half the Diſtance of the Firſt to the
Second, will reach from the Third to the
Fourth.
Example and Ufe. There is a Circle
whofe Content is 866, and its Diameter is
IOL 5; there is another I would have contain
385: I demand its Diameter ?
Becauſe the fubduplicate Ratio of fimilar
Superficies are as their like Lines: The fub-
duplicate of 86L6 to 38L5, is as 1oL5 to
the Anſwer. Therefore bifect the Distance
between 8616 and 38L5, which will be at
5775; I fay the Diſtance from 57×75 to
385, will reach from 10L 5 to 7, the Diame-
ter fought.
X. Three Numbers being given; to find
a fourth in a triplicate Proportion.
The Triple of the Extent of the Firſt to
the Second, will reach from the Third to
the Fourth.
Example and Ufe. The Weight of an
Iron Bullet, whofe Diameter is 4 Inches, is
9 lb. what will a Bullet weigh whofe Diame-
ter is 8 Inches?
Since all fimilar Solids are to one another,
in the triplicate Proportion of their like
Lines: The Triplicate of 4 to 8, is as 9 to
the Anfwer. Therefore,
The
[ 107 ]
The Triple of the Extent from 4 to 8, will
reach from 9 to 72 the Weight fought. For
the Extent from 4 to 8 will reach from 9 to
18, from 18 to 36, and from 36 to 72.
The like of all other fimilar Solids.
XI. Three Numbers being given; to find
a Fourth in a ſubtriplicate Proportion.
One third Part of the Extent of the First
to the Second, will reach from the Third to
the Fourth.
Example and Ufe. An Iron Bullet of 4
Inches Diameter weighs 9.; what is the
Diameter if the Weight be 727b.? The Sub-
triplicate of 9 to 72, is as 4 to the Anſwer.
Therefore divide the Diſtance between 72
and 9 into 3 equal Parts, and the laſt Divi-
fion will fall on 18; then the Diſtance from
9 to 18 will reach from 4 to 8.
XII. What hath been faid in this Chapter
of Numbers, is applicable to the Lines,
Superficies, Bodies, &c. that theſe Numbers
may repeſent. And this too, to a much
greater Purpoſe than could be done by the
Sectoral Lines of Superficies and Solids. For
Inftance, two Lines A, B being given; to find
a mean Proportional: Whether you are to
work by the Sectoral Lines, or the Line of
Numbers; thefe Lines A, B, given, muft
be meaſured on fome Line of equal Parts.
And the Middle between the Points on the
Line of Numbers repreſents the Length of the
mean Proportional in Numbers. By the
Sectoral Lines, the mean Proportional Line
itfelf
[ 108 ]
itſelf is fought; and therefore it muſt neither
be too long nor too fhort for your Sector.
For if it or the given Lines fhould be fo,
you muſt prepare them or their Parts to fit
them for the Work; which is not required
on the Numbers. And whatſoever hath
been faid of finding two mean Proportional
Numbers, between two Numbers given, may
be applied to two given Lines, if they are
meaſured by fome Line of equal Parts.
CHAP.
[ 109 ]
BRXONGIFOLCATE
CHA P. XIV.
Of the joint life of the preceding Lines in
the Solution of right lin'd Triangles.
1. From the Chords and Lines by Protraction.
2. From the other Sectoral
Lines
3. From the Artificial Lines
by Calculation.
I. N a right lined Triangle ABC,
right lined at C, are given the
Hypothenuſe AB (65) and the
oblique Angles A (30°. 31) B
(59. 29); to find the Legs AC, BC.
B
A
30 31
65
59.29
33
56
*
C
Ift. By
[ 110 ]
First, By Protraction.
Draw AC at pleaſure; make the Angle A
30°31' as taught in the 10th and 11th
Chapters; lay 65 equal Parts on AB: At B
make the Angle B 59° 29', or from B let
fall BC perpendicular to AC; and fo will the
Triangle ABC be protracted: Then AC and
BC may be meafur'd by the fame Scale that
AB was laid down by.
2dly, By Calculation from the Sectoral Lines.
Since the Radius, is to the Sine of either
Angle (A), as the Hypothenuſe (AB) is to
(BC) the Leg oppofite to that Angle.
* Take AB (65) from the Lines or any
equal Parts; and make it a Parallel at the
Sine of 90. Then the Parallel at the Sine of
(30° 31') the Angle A, meafur'd on the
fame equal Parts, gives (33) the Leg BC.
And the Parallel at (59° 29′) the Degrees
of the Angle B, meafur'd on the equal Parts,
gives 56 the Leg AC.
Thirdly, By the Artificial Lines.
By the preceding Proportion and the gene-
ral Law laid down in Chap. VIII. it follows,
* See Chap. VIII.
That
[ III]
That the Extent from the Sine of 90 De-
grees to the Sine of 30° 31, will reach on
the Numbers from 65 to 33.
And the Extent from the Sine of 90 to the
Sine of 59° 21 will reach from 65 to 56 on
the Numbers.
Or thus, The Extent from the Sine of 90°
to the Hypothenufe AB (65) on the Num-
bers, will reach from the Sine (30° 31′) the
Angle A, to BC (33) on the Numbers.
And the fame will reach from the Sine of
(59° 29') the Angle (B) to (56) AC on the
Numbers.
General Obfervations.
This latter Method requires but one open-
ing of the Compaffes, to find both the Legs.
It is called crofs Work. It is fometimes very
convenient, fometimes very improper. When
the Extent of the Compaffes by crofs Work
is taken from Points almoft over one another,
the Refult will be very doubtful and imper-
fect. This Inconvenience may fometimes
be remedied by counting on the other 5
Numbers. But when the Points to be taken
in the Compaffes by crofs Work, are at a
confiderable Diſtance, it may be practis'd
with Safety; provided that theſe Extents are
nicely taken from, and applied to the Fiducial
Lines. Thefe Fiducial Lines are known,
becauſe all the Divifions ftand on them, as
hinted in the VIIIth Chapter.
When
I
[112]
When feveral Proportions are the fame,
and fo may be wrought by croſs Work at
one opening of the Compaffes, as in the
Solution of this Example; or if the Diſtance
to be taken off the fame Line be too great
for the Compaffes, uſe croſs Work if it may
be remedied thereby.
It was hinted (Chap. VIII.) that the Extent
from the 1ft to one of the middle Terms,
will reach from the other, applied the fame
way, to the Anfwer. That is, if you ex-
tend from the 1ft, upwards, or towards the
Right-Hand, the Application of that Extent,
to find the 4th, fhall be made from the given
Quantity upwards. And if the Extent be
made downwards, the Application fhall be
made downwards alfo.
The Lines of Sines and Tangents begin at
a little leſs than 1 Degree. And in fome
Examples, the Extent being applied, will
fall below the Divifions on the Sines or Tan-
gents. In this Cafe, keeping the Compaffes
with the fame Extent, bring 1 Foot of the
Compaffes to 1 Degree, and where the other
falls, hold it faft till you have fo fqueez'd the
Legs together, that the other Point which
was plac'd at 1 Degree, falls on the Term to
which the Application was made. Then this
Diſtance laid from 60 on the Numbers down-
wards, will reach to the Minutes, and Parts
of Minutes requir’d.
Example 1. Let the Proportion be, as 91
to 22, fo is the Sine of 2 Degrees to a fourth
So the Extent from 91 to 22, will
Sine.
reach
[ 113 ]
reach from the Sine of 2 Degrees downwards,
beyond the End of the Sector. Therefore
with the fame Extent one Foot placed in one
Degree, and the other will reach upwards to
fomewhat more than 4°, where hold the
Point faft; then fqueeze the Compaffes toge-
ther till the other Point falls on 2, the Point
to which the Application was made. This
Diſtance applied downwards on the Numbers
from 60, will reach to 29, which is the
Anfwer in Minutes.
I
Example 2. As 95 is to 19, fo is the Sine
of 1° 30' to another Sine. So the Extent
from 95 to 19 will reach from the Sine of
1° 30′ below the End of the Sector.
о
This
Diſtance applied upwards from one Degree,
reaches to above 5, where being held faſt, and
the other Foot brought to 1° 30, you have a
Diſtance, which laid on the Numbers, reaches
from 60 downwards to 18, for the Minutes
fought.
II. In a right-lin'd Triangle ABC, right-
angled at C, there is given the Angles A
(30°31) and B (59L29) and one Leg
AC (56): To find the other Leg and the Hy-
pothenufe.
Firſt, By Protraction.
Draw AC at pleaſure; make the Angle
A 30° 31'; lay 56 equal Parts from A to C,
and at C raiſe BC perpendicular to AC; and
produce AB, CB till they meet in B; fo
I
will
[114]
will the Triangle be conftructed: And the
Hypothenufe AB and the Leg BC may be
meafur'd from the Scale AC was laid down
by.
Secondly, By the Sectoral Lines.
Since as the Sine of the Angle (B) op
pofite to the given Leg (AC) is to that Leg,
fo is the Radius to (AB) the Hypothenufe;
and fo is the Sine of the Angle A to (BC)
the Side oppofite to it. Therefore,
Open the Sector ſo that AC (56) on
the Line of Sines, may be a Parallel at the
Sine of (59° 29') the Angle B; and then a
Parallel at the Sine of (30°31′) the Angle
A will, when meafur'd on the Lines, give
(33) the Leg BC; And the Sector remain-
ing with the fame Aperture, a Parallel at
the Sine of 90, will give AB (65) on the
Line of Sines.
Thirdly, By the Artificial Lines.
From the preceding Proportion and the
general Law in Chap. VIII. The Extent
from the Sine of 59°L 29' to the Sine of 90°
will reach from 56 on the Numbers to AC 65.
And the Extent from the Sine of 59°L 29' to
the Sine of 30°L 31 will reach from 56 on
the Numbers to BC 33.
Or, at one opening by the Croſs Work
thus. The Extent from the Sine of 59° 29′
to 56 on the Numbers, will reach from the
Sine
[ 15 ]
Sine of 90° to 65 on the Numbers; and from
the Sine of 30°31' to 33 on the Numbers.
III. In a right-lin'd Triangle ABC, right-
angled at C; there is given the Hypothenuſe
AB (65) and one Leg BC (33): To find the
other Leg AC, and the Angles A and B.
Firſt, By Protraction.
Draw AC at pleafure, and at C raiſe the
Perpendicular BC, and lay thereon 33 from
any Line of equal Parts from C to B. Pro-
duce AC towards A. With 65 in your Com-
paffes taken from the fame Line of equal
Parts, one Foot in B with the other defcribe
an Arch cutting AC in A, and draw AB:
So will the Triangle be protracted. And
the Leg AC as well as the Angles may be
meaſured.
Secondly, By the Sectoral Lines.
Since the Hypothenufe (AB) is to the
Radius or Sine of 90°, as the Leg (BC) to
the Sine of its oppoſite Angle A. And fince
the Radius or Sine of 90° is to the Hypothe-
nufe AB, fo is the Sine of the Angle B to
its oppofite Side AC. Therefore,
Make the Sine of 90 a Parallel at the Hypo-
thenufe AB (65) on the Line of Lines, and the
Parallel at the Leg BC (33) will be the Sine of
(30° 31′), the Angle A. And fince (3c°L 31′
taken from 90, leaves 59°L 29) for the AngleB;
the Sine of 59° 29′ carried along the Line
I 2
of
[ 116 ]
1
of Lines will be found to be a Parallel at 56,
which is the Leg AC.
But the Leg AC might have been found
without finding the Angles, by the Line of
Lines only. For (by the 11th Article of the
11th Chap.) open the Sector fo that the Line of
Lines may form a right Angle, and count the
given Leg BC (33) on one Leg; and with
the Hypothenufe AB (65) in the Compaffes,
from 33 as a Center turn the Compaffes
about till the other Leg falls on the other Line
of Lines in the Fiducial Line, which you will
find to be at 56, which is AC the Leg
fought.
Thirdly, By the Artificial Lines.
From the preceding Proportion, the Ex-
tent from 65 to 56 on the Numbers will
reach from the Sine of 90 Degrees to the
Sine of 30° 31' the Angle A. And fo as
before the Angle B will, by fubtracting
the Angle A (30°L 31') from 90, be found
to be 59° 29'. And then the Extent from
the Sine of 90° to the Sine of 59° 29′, will
reach from 65 on the Numbers, to (56) the
Leg AC.
Or, by the Cross Work at one opening of
the Compaffes. The Extent from 65 on the
Numbers to the Sine of 90 Degrees, will
reach from 33 on the Numbers to the Sine of
(30° 31') the Angle A.
the Angle B is 59° 29′:
Extent will reach downwards from the Sine
And as before
Then the fame
of
[ 117 ]
of (59°L29) the Angle B to AC (56) on
the Line of Numbers.
IV. In a right-angled right-lined Triangle
ABC, there are given the Legs AC (56)
and BC (33); to find the Angles A, B,
and the Hypothenuſe AB.
First, By Protraction.
Draw AC, and perpendicular thereto raiſe
BC: Make AC 56, and BC 33 (from any Scale
of equal Parts, and draw AB); fo is the
Triangle protracted, and the Hypothenufe
and the fought Angles may be meafur'd.
Secondly, By the Sectoral Lines.
Since one Leg is to the other as the Radius.
(here the Tangent of 45°) is to the Tangent
of the Angle oppofite to the other Leg. And
then the Hypothenufe may be found by the
fecond Example. Therefore,
Make the Tangent of 45° a Parallel on
the Lines, at (56) the Length of AC one
of the Legs; and then the Parallel at 33 on
the Numbers meaſur'd on the Tangents,
will give (30°31') the Angle A. Then,
as in the fecond Example, may be found the
Hypothenufe AB.
But the Hypothenufe might have been
found without the Angles by the help of the
Lines of Lines only. For open the Sector
fo that the Lines of Lines may form a right
Angle; then count AC (56) on one Line,
I 3
and
[ 118 ]
and BC (33) on the other; and the Diſtance
from 56 to 33 meafur'd on the Lines, will be
AB (65) the Hypothenufe.
Thirdly, By the Artificial Lines.
From the preceding Proportion, the Ex-
tent from AC (56) one Leg, to BC (33)
the other Leg on the Line of Numbers, will
reach from the Tangent of 45° to the Tan-
gent of (30° 31') the Angle A appofite to
the latter Leg. This Angle being found,
the Hypothenufe may be found as in the
fecond Example.
The Crofs Work here can never affift
you.
V. In any right-lined Triangle DEF;
the Angles D, E, F, being given (55, 20,
105) to find the Ratio of the Sides.
Meaſure the Sines of thoſe Angles on the
Line of Sines, and you will have the Propor-
tions of the Sides; viz. 81L9, 34L 2, 96L 5;
always obferving that the greater Side is
oppofite to the greater Angle.
This might be perform'd by Protraction
or by the Artificial Lines; by affuming any
one Length for either of the Sides.
then it would change into the following,
which fee.
But
VI. In
[ 119 ]
VI. In any right-lined Triangle DEF.
The Angles D (55) E (20) F (105) and
one Side DE 145; to find the other two
Sides DF, EF.
F
55
5113
105
123
· D
145
First, By Protraction.
20
E
Draw DE at pleaſure, and make it 145
from any Line of equal Parts. At D make
an Angle of 55°, and at E one of 20 (per
Chap. 1o.) and produce the Lines till they
meet in F. Then is the Triangle protracted,
and the fought Lines DF, EF, may be
meafur'd by the fame Scale that DE was
protracted by.
Secondly, By the Sectoral Lines.
Since the Sine of any Angle (F) is to its
oppofite Side DE, as the Sine of any other
Angle D to its oppofite Side.
Make DE (145) a Parallel at the Sine of
(105°, that is, by the 3d Obferv. in Chap. I.
the Sine of 75°) the Angle F; and then the
Parallel
I 4
[ 120 ]
Parallel at the Sine of (55°) the Angle D, will,
when meafur'd on the Lines of Lines, give
(123) the Side FE. And the Parallel at the
Sine of (20) the Angle E, will give 5113.
Thirdly, By the Artificial Lines.
From the preceding Proportion, the Ex-
tent from the Sine of 75° to the Sine of 55°,
will reach on the Numbers from 145 to 123
the Side EF. And the Extent from the Sine
of 75° to the Sine of 20°, will reach on the
Numbers from 145 to 51 3 the Side DF.
Or, by the Croſs Work at one opening of
the Compaffes. The Extent from the Šine
of 75° to 145 on the Numbers, will reach
from the Sine of 55° to (123) the Side FE ;
and alfo from the Sine of 20° to (5IL 3) the
Side DF.
VII. In any right-lined Triangle DEF, two
Sides DF (513), FE (123), and the Angle
D (55°) oppofite to one of them, being given;
to find the other Angles F, E, and the
other Side DE.
First, By Protraction.
Draw DE, make the Angle D 55°, and
make DF from a Scale of equal Parts 51L 3.
Then take FE 123 from the fame Line of equal
Parts; and with one Foot in F defcribe an
Arch cutting the Line DE in E, and draw
FE: So you will have protracted the Triangle.
And
[ 121 ]
And the Angles F, E, and the third Side
DE may be meafur'd.
Secondly, By the Sectoral Lines.
Since any Side (FE) is to the Sine of its
oppofite Angle (D), as any other Side DF
is to the Sine of (E) the Angle oppoſite to
the other Side.
Make the Sine of (55°) the Angle D
a Parallel at (123) the Length of FE; and
then, &c.
But we are to obſerve that 123 falls fo
near the Center, that the Sine of 55° is too
great to be made a Parallel at 123. There-
fore make 123 a Parallel at the Sine of 55,
and then (513) the Line DF carried along
the Sines till it becomes a Parallel, will give
us the Sine of (20°) the Angle E. Now
becauſe the two Angles D and E are known,
therefore the third F is alfo known; for
all the Angles of every right-lined
Triangle taken together will be equal to two
right ones; whofe Meaſure is 180 Degrees.
Therefore the two Angles D and E, 55 and
20, that is, 75 taken from 180, will leave
105 for the Angle F. And the Parallel at
the Sine of (105°, that is) 75° gives on the
Lines DE 145.
But the third Side DE may be found
without the fought Angles, by the Line of
Lines. For fo open the Sector (by the 12th
of the 11th Chapter) that the Angle may be
55°; and count on one Leg 51L3: then
with 123 in the Compaffes from 55, turn the
other
[122]
other Point about till it falls on the Lines in the
other Leg of the Sector, which you will find
to be at 145, which is the third Side fought.
Thirdly, By the Artificial Numbers..
From the preceding Proportion, the Ex-
tent from 123 on the Numbers to 51L3,
will reach from the Sine of 55° to the Sine
of 20° the Angle E. And then as before,
the Angle F will be found to be 105; which
hath the fame Sine that 75 hath. And then
the Extent from the Sine of 55° to the Sine
of 75°, will reach from 123 on the Numbers,
to 145 the Side DE.
Or, by the Cross Work, at one opening of
the Compaffes. For the Extent from 123
on the Numbers to the Sine of 55° will reach
from 513 on the Numbers to the Sine of
20°. And the fame opening will reach the
contrary way from the Sine of 75 Degrees,
to 145 on the Numbers.
VIII. In any right-lined Triangle DEF.
Two Sides DF (51L3), FE (123), and the
Angle F (105°) contain'd by thofe Sides
being given; to find the Angles D and E, and
the third Side DE.
Firſt, By Protraction.
Draw DF at pleaſure, and make the Angle
F 105°. Then from fome Scale of equal
Parts make FD 51L3 and FE 123; and
draw DE: So is the Triangle protracted:
And
[ 123 ]
And the Angles D, E, and the Side DE
may be meafur'd.
Secondly, Since the Sum of the given Sides
is to their Difference, as the Tangent of half
the Sum of their oppofite Angles is to the
Tangent of half their Difference.
And in this Example the Angle F being
105°, and fo the Sum of the Angles at D
and E 75°, and the half Sum 37°30'.
Make the Sum of the Sides 174L 3 a Paral-
lel at the Tangent of 37° 30'. And then
71L6, the Difference of the Sides carried
along the Tangents, will be a Parallel at
17° 30', which is the half Difference. But
the half Difference (17°L 30') added to the
half Sum (37° 30') gives (55°) the greater
Angle (D). And the half Difference
(17° 30′) taken from the half Sum (37°L 30′)
leaves (20°) the leffer Angle E. And now by
the 6th Cafe, the third Side DE may be
found. But you may obferve, that when
the half Sum of the oppofite Angles is more
than 45 Degrees; and the half Difference
proves lefs than 45 Degrees, it will be
troubleſome to perform it by the Sectoral
Lines. But this Inconvenience is very well
cured by help of the Artificial ones.
But the third Side may be found without
the unknown Angles very eaſily by the Line
of Lines; and then the Angles may be found,
as in the following.
For (by the 12th of Chap. XI.) open the
Sector till the Lines form an Angle of 105°;
count 65 on the Lines on one Leg, and 123
on
[124]
on the other, and the Diſtance from this 65
to 123 is 145 the third Side.
Thirdly, By the Artificial Lines.
From the preceding Proportion, the Ex-
tent from 174 3 on the Numbers to 71L6,
will reach from the Tangent of (37°L 30′)
the half Sum of the unknown Angles, to the
Tangent of (17° 30') their half Difference
as before. And fo as before the Angle D is
55°, and E 20°. The Third now is found
by the laſt.
The Crofs Work cannot here do Service.
If the Sum of the unknown Angles is
more than 45 Degrees, and their Difference.
proves to be fewer, then the Extent from
the Sum of the Sides to the Difference of the
Sides, will reach from the half Sum of the
unknown Angles upwards beyond the Tan-
gent of 45 Degrees. In which Cafe, the
Compaffes keeping the fame Opening, bring
the upper Point back to 45, then keep the
other where it refts, and bring the Point
which was plac'd at 45, to the Tangent of
the half Sum. Now lay this laft Extent from
45 downwards, and you will have the Tan-
gent of the Degrees fought: e. g. Let it be
as 90 to 30, fo the Tangent of 55° to ano-
ther Tangent. Then the Extent from 90 on
the Numbers to 30, will reach from the
Tangent upwards beyond 45. But this
Distance being laid from 45 downwards,
keep the lower Point fix'd, and bring the
Point to (the Tangent of 55°) the Place
from
[ 125 ]
from whence the Application of the Extent
was to be made; and this laſt Diſtance will
reach from 45° downwards, to the Tangent
of 25° 28', the Anfwer.
In like manner, if it had been as 30 to 90,
(or any other Proportion) fo the Tangent of
25°L 28, less than 45°, to a Tangent
greater than 45°.
The Extent from 30 to 90, will reach
from the Tangent of 25° 28′ beyond 45°,
that is, to a Tangent greater than 45°: With
this opening bring one Point to 45; where this
falls hold it faft, and bring the other to
25° 28′: Apply this laſt Extent from 45,
and it will reach in the upper or inverted
Tangents, to the Tangent of 45, the Anſwer.
The Reaſon of this Operation is this:
The upper Line of Tangents ſhould have
been conceiv'd as continued beyond 45 right
forwards. But becauſe the Divifions would
have fallen beyond 45 upwards, in the fame
manner that they fall under 45 the contrary
way; to avoid Incumbrance and Charge,
the fame Line is number'd backwards, but
with the upper Number. But from hence
it appears, that the Figures above 45 increaſe
towards the Left-Hand, and fo they must be
accounted upwards from the Right to the
Left, and downwards from the Left to the
Right.
Again, in our laft Example the Extent
from 30 to 90 reaches from 25°L 28′ beyond
45. Let (in Fig. following) 25° 28' be
denoted by A, 45 by C, and the Extent by
Ab. Now it's evident that if the Tangents
had
[126]
had been continued beyond 45, b would have
given the Anſwer. But the Tangents are
not continued beyond 45, for inſtead thereof
they are folded backwards; confequently
the Point where the Anfwer will be found,
L
d
A
B
C
b
Therefore
is as far below C as b is above it.
Cb laid from C to B gives at B 55°, the
Anfwer. But becauſe I cannot conveniently
meaſure Cb, b falling off the Rule, I lay
Ab from C to d, and meaſure d A which is
equal to Cb, and lay it from C to B; and
fo gain the Anfwer. I faid Ad was equal
to Cb, which will thus appear: By the
Operation Cd was made from the fame Ex-
teht that Ab was, whence Cd is equal to
Ab; and by taking away the common Part
AC, there remains A d equal to Cb. There-
fore Ad may be ufed inftead of Cb.
It may be alfo obferv'd, that if the
Extent taken from the Numbers be greater
than that from 10 to 1, and perhaps too
long for the Compaffes; and this is to be
applied from the Tangent of 45° upwards or
downwards, or the Sine of 90°, that then
the Diſtance from 10 to I may be neglected;
and the remaining Diſtance applied from
5°43′ on the Tangents, and from 5°45′
on the Sines towards the Left-Hand, will
give the Tangent or Sine fought: e. g. Let
9o be to 8 as the Sine of 90° to another Sine:
о
Then
[127]
Then becauſe the Extent from 90 to 9, is
equal to the Extent from 10 to 1; I take
the Extent from 9 to 8, or, which is the
fame Thing, I feek both Numbers on one
Line, and then the Extent will reach from
the Sine of 5° 45′ to the Sine of 5° 07', or
on the Tangents from 5° 43′ to 5°L05.
In like manner, if you are to take the
Extent from 90° on the Sines, or 45° on
the Tangents, to a Sine below 5°L45', or a
Tangent below 5°L43'; extend the Com-
paffes from 5°45' or 5°43′ to the Sine or
Tangent given; and this Extent will reach
from the third Term, counted on the upper
Line of Numbers, downwards to the 4th
fought, to be efteemed as tho' it were in a
Line of Numbers below where it falls.
If the Proportion had been from a Sine
below 5° 45' to the Radius; take the Extent
from the given Sine to 5°45', and it will
reach from the third Term counted on the
lower Numbers upwards to the 4th fought,
to be eſteem'd as on a Line of Numbers next
above it.
And if it were from the Sine of 90 to a
Sine of Degrecs, a few more than 5°L45, as
6°Loo, and the Compaffes too fhort; then
the Extent from 545 to 6°Loo', will
reach from the third Term counted on the
lower Line of Numbers upwards to the 4th
fought, to be efteem'd as on a Line of Num-
bers next below where it falls.
If it be as a Sine of 6ºLoo to the Sine of
90°, fo a Number to another fought. It
may be as 6°Loo' to 5°L45', fo is the third
on
[128]
on the upper Numbers downwards to the
4th fought, to be eſteem'd as on a Line of
Numbers next above where it falls.
If the Extent on the Numbers be almoſt
the Length of one of the Lines of Numbers,
and be afterwards to be apply'd to the Sine
of 90°: Find both Numbers on one Line,
and the Extent will reach from 5°45′ on
the Sines, to the Sine fought.
Example 1. As the Sine of 90° to the Sine
of 3°; fo is 57 to 2L98.
As the Sine of 4° to the Sine of 90°; ſo
is 85 to 122.
fo
As the Sine of 90° to the Sine of 6°; ſo
is 59 to 6L 13.
As the Sine of 6° to the Sine of 90°; fo
is 63 to 605.
As 575 is to 6L 3; fo is the Sine of 90°
to the Sine of 6°L03'.
What hath been faid in thefe Obfervations
of the upper and lower Tangents, when both
are in uſe together, may be applied to a Sine
and a Secant when ufed together.
And what hath been faid of comparing the
Sine of 90 Degrees with a Sine leſs than, or
very near to the Sine of 5° 45'; and of com-
paring the Tangent of 45° with a Tangent
lefs, or little more than the Tangent of
5°43', may be applied to a Tangent greater
or little lefs than the Tangent of 84°L17',
changing the Word upwards for downwards.
Underſtand the fame of the Secant of
84°L 15'.
IX. The
[129]
IX. The three Sides of a right-lined Tri-
angle being given, to find either of the An-
gles.
Let the three Sides be D E 145, DF 65,
FE 123: To find the Angle F.
First, By Protraction.
Draw DE at Pleafure, and make it 145
equal Parts. Take DF 65, and FE 123 of
thofe equal Parts in your Compaffes, and
making D and E Centers, defcribe Arches,
cutting one another in F. Draw DF, E F,
fo is the Triangle protracted, and the Angle
F may be meaſured as taught in the 10th and
11th Chapters.
65
F
123
D
145
E
But the Triangle may be repreſented on
the Legs of the Sector; for count DF 65
on one of the Lines of Lines, and 123 on the
other; take 145 in the Compaffes, and open
the Sector 'till this Extent reaches from 65
to 123; and then the Angle that the Lines of
Lines
K
[130]
Lines make, is the Angle F, which may be
meaſured by the Precepts of Chap. XI.
And the Angle may be found by the two
following Proportions. As either of the Sides
adjacent to the fought Angle (which call the
Baſe) is to the Sum of the other two Sides
(which call Legs); fo is the Difference of
thoſe two, to a fourth.
Then add and fubtract the firſt and laſt
Terms in the preceding Proportion; and it
follows, That if the Angle fought be adjacent
to the {leffer
greater Leg, then the {
leffer
}
greater
Leg, is to the Radius, or Sine of 90°, as
the Half {Difference
Sum
of the firſt and laſt
'Terms in the preceding Proportion, to the
Sine of a Number of Degrees, which taken
from 90, fhall leave the Angle fought; un-
lefs when the fourth Term, in the firft Pro-
portion, is greater than the firſt Term, and
the Angle adjacent to the leffer Leg is fought:
For in that Cafe the Degrees found in the
laft Proportion, muſt be added to 90°, and
the Refult will be the Angle fought.
In the preceding Example,
As the Bafe 123, is to the Sum of the
Legs 210; fo is their Difference 80, to a
fourth 136/1/
Therefore the Sum of the firſt and laſt
Terms, is 259-5, and their Difference is
135. Whence the half Sum is 12975,
and the half Difference 6L75.
And becauſe the Angle fought is adjacent
to DF, the leffer Leg, it is, As DF (65) is
to
[ 131 ]
to the Radius; fo is the half Diff. (6175)
of the firſt and laft Terms in the preceding
Proportion, to the Sine of 6°; which, be-
cauſe the fourth Term in the firſt Proportion,
is greater than the firft, and the Angle adja-
cent to the leffer Leg is fought; add it to
90°, and you have 96°, the Angle fought.
But, by the Help of the Line of Numbers,
and the artificial verfed Sines running cloſe to
the right Sines, the fought Angle may be
found much eaſier: Thus,
The Sides containing the fought, DF 65
Angle F
SEF 123
The Side oppoſite to the fought DE 145
Angle
The Sum of all the Sides
Half the Sum of all the Sides
333
166/2/2
145
Half the Sum of the Sides wanting the 21
Oppofite
The Extent on the Numbers from the half
Sum (166) to (EF 123) one of the con-
taining Sides, will reach from the other con-
taining Side, to a fourth; where hold the
Point unmov'd, and extend the other Point
to (21) half the Sum of the Sides leffened
by the oppofite Side: And this fecond Extent
will reach from the Beginning of the verſed
Sines to (96°) the Angle F required.
K 2
As
[132]
As Trigonometry is an Art ufeful in al-
moſt all the practical Parts of Mathematicks,
and fince, by the Sector, the various Cafes
(as evidently appears from the Examples here
laid down) are concifely and elegantly folved,
I have been more particularly full to exem-
plify and explain the Ufe of the Sector herein.
I believe alſo, by this Time the Reader
does perceive, that the Sector, as it is now
made, is capable of performing all the more
ufeful Problems required in common Life,
with lefs Confufion, more Speed and Exact-
nefs, than by the old Sectors, loaded with
Multitudes of Lines: And that the artificial
Lines, not dreamt of before the great In-
vention of Logarithms, are fitter for
thefe Operations, in all Cafes where propor-
tional Numbers are concerned, than any
other Lines that have been laid down on
Inftruments; and therefore were, with very
great Judgment, cauſed to be cut on the
Sector: And this will again appear in the
Application of theſe artificial Lines, to Pro-
blems in Geography, Navigation, Dialling,
Aftronomy, &c.
CHA P.
3
[133]
EASY-SWEL BEBED
CHA P. XV.
The joint Ufe of the Chords, and the natu-
ral Tangents and Secants, in protracting
and meafuring the Reprefentatives of
Spherical Arches and Angles, on a Plane.
I
N the Management hereof, I beg
leave, for Brevity's fake, to call theſe
plain Repreſentatives, the ſpherical
Angles, Arches, and their Poles,
themſelves. Every where the Radius is the
Radius of the Primitive Circle, unleſs other-
wife expreffed.
I. A Point being given: To find another
diametrically oppofite to it.
A
B
H
E
+D
Thro' the given Point, and the Center of
the primitive Circle, draw a right Line.
Then if the given Point be A on the primitive
K 3
Circle,
[134]
1
Circle, the Point fought will be C alfo on the
primitive Circle. But if the given Point be
B, not on, but in the primitive Circle, the
Sector being opened to the Radius E A,
meaſure E B on the Tangents: Then lay the
Tangent of the Complement of E B, from
E to D, and then D is the Point. ×
II. To draw a great Circle through H, B,
two given Points. By the preceding find a
Point (D) diametrically oppofite to (B) either
of thofe given; and then thro' the three
Points A, B, C, defcribe the Periphery of a
Circle, and it is done.
III. A great Circle being given, to find its
Poles And a Pole being given, to defcribe
the great Circle. 1. If the Circle given be
the Primitive, its Center is its Pole.
2. If
the Circle be a right one, as A DC, its Poles
are found by laying of the Chord of 90°
from A or C, the Points where the right
Circle meets the Primitive, to the Points H
and G on the Primitive. 3. Therefore, if
a Point G or H is the Pole of a Circle, lay
the Chord of 90° from G or H, to A and
C, and draw the right Line A DC; it is the
Repreſentative of the Circle required. 4. If
the Circle be an oblique one, as ABC ;
S
}
meaſure BD on the Tangents fubtract
add
5 from } 45, and lay the Tan-
the Degrees
S
{
to
gent of the {Remainder
Sum
from D to
E
{2}}
KS
fo
[ 35 ]
fo will E, K be the Poles fought. 5. If the
Pole E be given to defcribe the Circle, draw
D E, and at right Angles to it, through the
Center, draw ADC: Meafure DE on the
TK
G
}
B
D
EL
H
ML
Q
F
N
Tangents, and lay the Tang. of twice the De-
grecs of DE, to DF; fo will F be the Center
of the Circle fought, and it muft pafs thro' C
and A fo the Circle ABC may be defcribed,
which is that fought. 6. If the Tangent had
fallen off the Sector, or if the Radius of the
Circle had been greater than the upper Tan-
gents could be opened to; from C, with the
Radius CD, defcribe the Arch D L, and lay
on it the Degrees of the Tangent from D to
M; and draw CM, and produce it 'till it
meets D G in N; then will DN be the Tan-
gent, and CN the Secant of D M. 7. If
the Pole given had been (K) out of the
K 4
Pri-
[136]
Primitive: Then by n. 1. find E, another
Point diametrically oppofite to it, which is
the other Pole to this Circle; and then the
Circle may be defcribed, as in the 5th of
this III.
IV. To meaſure any given Arch; and on
a given Circle, and from a given Point there-
in, to lay off any affigned Degrees, towards
affigned Parts. 1. If the given Circle be the
Primitive, Arches are meaſured and Degrees
laid off by the Chords. 2. If the given Circle
be (GK) a right one, and one End of the
Arch to be meaſured is G, the Center of the
N HO
A
D
L
R
E
F
ic
K
Primitive; then GH and GK meaſured on
the Tangents (with the Radius of the Primi-
tive) will give Half the Degrees of G H and
GK, and thefe doubled the Degrees of G H
and GK: But neither of the Extremities
were the Center of the Primitive (as of the
Arches
[137]
Arches { FK
FH
5 the fame Side
ર
contrary Sides
} then when the Ends are on
of the Center; then mea-
fure on the Tangents, the Diſtances of thoſe
Ends, F and K and H, from the Center G,
and the Difference of thofe Degrees, give
Half the Degrees of {FH
FK } and fo the Dou-
ble of the {Difference? gives the Degrees
Sum
requir'd. 3. Therefore, to lay off any Num-
ber of Degrees: If from the Center, lay off
the Tangent of half thoſe Degrees. If the
given Point be not the Center, meaſure its
Diſtance from the Center on the Tangents in
Degrees: And the Sum and Difference of
thefe Degrees, and half the Degrees to be
laid off, are to be laid off from the Center by
the Tangents, towards the given Point, if
the Part F K, to be laid off, be requir'd con-
trary ways to the Center; or if F E, the Part
to be laid off, be less than FG, the Diſtance
of F from the Center. But, on the other
Side of the Center, if FG, the Part to be
laid off towards the Center, be greater than
the Diſtance of the given Point F, from G
the Center. If half the Degrees to be laid off
}
from F, be {more than the Degrees of
fewer
the Tangent FG; then FG is
lefs
greater)
than the Arch to be laid off 5. To meafure
(NL, or NO) Arches of an oblique Circle,
or
[138]
or of any Circle in general: Find its Pole F
within the Primitive, and lay a Ruler from
F (to N, L,O) the Extremities of the Arches,
and it will meet the Primitive in the Points
R, Q, S; then will R Q, and QS on the
Chords, meaſure LN and NO. 6. On A D,
or any Circle whatfoever, from a given Point
N, to lay off any affigned Number of De-
grees, fuppofe 56. Firſt find F, its Pole,
lay a Ruler from F to N, it will meet the
Primitive in Q; then by the Chords lay 56
from Q to R, and from Q to S; and lay a
Ruler from F to R and S, it will crofs A D
in L and O; and fo will NL and NO be
each 56°, as requir'd.
V. To deſcribe a leffer Circle, parallel to
any great Circle given, and at any affigned
Diſtance from its Pole. 1. If the great Cir-
cle be the Primitive (and the leffer Circle is
at 56°, or any other affigned Diſtance from
its Pole) take the Tangent of (28) Half the
affigned Diſtance, and from A, the Center of
the Primitive, defcribe GHIK, and it is the
Parallel required. 2. If the great Circle be
(BD) a right one, whofe Pole is E, from
whence the leffer Circle is to be (30°, or)
any affigned Diſtance, lay from the Chords
(30°) the affigned Diſtance from (E) the
Pole, to P and Q. Then with the Tangent
of P E, or Q E, defcribe Arches cutting one
another in O; (or the Point O may be found
by laying the Secant of 30 from A) and from
O, with the fame Diſtance, defcribe the Cir-
cle P Q, and it will be that required. 3. If
the
[ 139 ]
the great Circle be (BFD) an oblique one,
let it be required to defcribe a leffer Circle
parallel to it; and at (70°) a given Diſtance.
From L, its Pole, lay (by the IV. of this
Chapter) 70° the given Diſtance both Ways,
n
S
S
F
G
H
N
B
A
D
K
I
L
R
P
S
E
S
M
on EF from L, viz. from L to M, and
from L to N. Then bifect MN in R; and
from R, with the Diſtance RM, or RN,
defcribe the Circle MSN, and it is the leffer
Circle
[ 140 ]
Circle required. In like manner the Circle
Sn S is deſcribed parallel to BFD, at 105°
Diſtance from that Pole L; for Ln, L m, are
each made 105°, and from r, the Middle
between n and m, SnS is defcribed.
VI. To a given great Circle, to draw ano-
ther at right Angles, thro' a Point given, ei-
ther on it, or off it. 1. Let the given great
Circle be (AE BK) the Primitive, and the
given Point be (E) on the Primitive, or (N)
off it. Draw a right Circle (NC, or EC)
thro' the given Point, and the Center of the
Primitive; and NB, or EC, is the great
E
D
G
I
M
A
B
F
N
H
I
K
Circle required. 2. If the given Circle be
ACB, a right one, and the given Point be
(M) on it, or (W) off it. (By the III. of this
Chapter) find (E, K) its Poles; and through
thofe Poles, and the given Point, draw the
Circumference of a Circle, and it is that re-
quired. 3. If the Circle be (AGB) an ob-
lique one, and the given Point be (L) on it,
or (F) off it: Find (H) the Pole of (AGB)
the
[ 141 ]
the given Circle. Then (by the II. of this
Chapter) defcribe (DHI) a great Circle thro'
(H) the Pole, and (F, or L) the given Point,
and it will be the Circle required.
VII. To meaſure any Angle. 1. If the
Angle be (GCB) at the Center, it is mea-
fured as a plain Angle, by applying GB to
the Line of Chords. 2. If the Angle be at
the Primitive, and made by the Primitive
and a right Circle, it is a right one.
3. If
the Angle be (CBH) at the Primitive, and
formed by (CB) a right Circle, and an ob-
lique one (HB); then draw CE at right
Angles; meaſure CH on the Tangents, and
E A
Va
DH
K
A
LC
0
N
P
B
GK
the Double of the Degrees gives the Meaſure
of the Angle. 4. If the Angle be EBH at
the Primitive, and form'd by (EB) the Pri-
mitive, and (HB) an oblique Circle. Mea-
fure CH on the Tang. and fubtract the Deg.
from 45, the then Double of the Remainder
gives the Degrees of the Angle (HBE)
fought; and fo the Sum of the Degrees of
the
[142]
the Tangents HC, CO, doubled, is the
Meaſure of the Angle HBO; and the Dif
ference of the Degrees of the Tangents C O,
CP, gives half the Meaſure of the Angle
OBP. 5. If the Angle be (F DB) neither
at the Center, nor at the Primitive, whether
both are oblique, or one a right Circle.
Make DQ, DL any how equal; and from
Land Q, with any Diſtance, deſcribe Arches
cutting one another in M, and draw DM.
In like Manner make DK, DI, equal, and
from D and I, with any Diſtances, defcribe
Arches cutting one another in N, and draw
DN. Laftly, Meafure the Angle MDN
by the Chords, and its Meaſure will be alfo
the Meaſure of the Angle B D F. 6. When
the angular Point is fo near the Primitive that
you cannot with Convenience lay off the
equal Diſtances from the angular Point: Then
G
F
H
K
.I
draw from (F) the angular Point (FG, FH)
Tangents to the Circles, forming the Angle;
and by the Chords meaſure the right-lined
Angle GFH, and you have the Meaſure of
the
[ 143 ]
the Angle formed by the Circles. To draw
a Tangent to a Circle, without finding the
Center, fuppofe at A; take A B, BC, any
how equal, draw A C, and from A defcribe
the Arch DBE; laſtly, make BE equal to
B D, and draw A E, and it is a Tangent, as
required. 7. Univerfally find the Poles
(I, K) of thoſe Circles, and lay a Ruler from
the angular Point (F) to thoſe Poles, it will
meet the Primitive in L and M, and L M
meaſured on the Chords, gives the Meaſure
of the Angle.
VIII. With a given great Circle, and at a
Point therein, to draw another that ſhall
make any affigned Angle with that given.
1. If the great Circle be (AB) a right one,
A
E
G
B
and the Point given in it be (C) the Center
of the Primitive, the Angle (BCG) will be
made as a plain one. 2. If the great Circle
be the Primitive, and the Point given in it
be B, to make an Angle of (40) a given
Number of Degrees, lay the Tangent of it
from
[ 144 ]
;
from C to F, or lay the Secant from B to F
fo fhall F be the Center, and B a Point thro'
which the Circle BDA is to be defcribed
and then the Angle DBE will be that re-
quired. As in the upper Tangents the Ra-
dius CB was fitted to its Radius 45°; fo
in the Secants, the Radius C B muſt be fitted
to the Beginning of the Secants.
3. If the
Angle had been to be made with the right
Circle CB, the Tangent Complement of the
required Angle CBD, muſt have been taken
to find the Center. 4. If the given Point be
neither at the Center, nor at the Primitive,
(as E) whether the given Circle be right, or
F
E
A
B
IK
D GH
oblique; from E (as in the 5th of VII. of
this Chapter) raife GE perpendicular to
(E B) the given Circle: And by the Chords
make the Angle GEH equal to the Angle
required to be made; draw from E a right
Line ECI thro' the Center; meaſure EC
on the Tangents, and lay the Tangent of the
Complement of its Double, from C to I,
and draw IH perpendicular to CI; fo fhall
H
[145]
H be the Center of the Circle DEF, which
defcribed thro' E, will form an Angle as re-
quired. And if it had been required to make
another Angle of a Magnitude different from
the former: Make a plain Angle GEK equal
to the required Angle, and in K, where it
meets with HI, is the Center of the Circle
required. For all the great Circles which paſs
thro' the Point E, have their Centers in the
Line IH..
IX. With a given great Circle, and from
a Point of it, to defcribe a great Circle, which
ſhall form any poffible Angle. ft. If the
given Circle be the Primitive, and the Point
given be D. From the Center, with the Tan-
gent of the given Angle, and from D, with
XD
E
E
the Secant, defcribe Arches cutting one ano-
ther in (E) the Center of the Circle required,
which, becauſe it is to pass through D,
may be deſcribed. 2. If the given Circle
(See Figure of the next Page) be not the
Primitive, let it be ABC, whofe Pole is D,
and let the given Point be E. Then E is
L
the
[146]
the Pole of fome Circle which may be de
fcribed by the III. of this Chapter, which
ſuppoſe to be FGH. From D, the Pole of
the given Circle A BC, defcribe a leffer Cir-
cle at the Diſtance of the Meaſure, or the
:
L
E
M
is
Supplement of the required Angle. If this
meets the Circle FGH, the Problem is poffi-
ble, otherwife not; but it meets it in two
Points L, M. Then make either of them a
Fole of a great Circle EB, which, by the III.
of this Chapter, defcribe, and it will form the
Angle required.
X. With two given great Circles to defcribe
a third that fhall make Angles with them
equal to any affigned Angles poffible, and
both on the fame Side of the required Circle.
1. Let one of the affigned Angles be a right
one, and the other an oblique one. Find G,
the Pole of (AE) one of the given great Cir-
cles
[ 47 ]
cles (by the III. of this Chapter). By the V.
describe HFI, a leffer Circle, at a Diſtance
from the Pole (G) equal to the Meaſure of
the Angle to be made, or its Supplement
with (AE) that Circle: If this leffer Circle
meets (AD) the other given great Circle,
the Problem is poffible, otherwiſe not: But
M
LE
F
A
D
H G
it meets it in two Points (H and I). Then
make either of theſe a Pole of a great Circle,
which (by the III. of this Chapter) defcribe.
If you take H for its Pole, it will be the
right Circle L. M; the Angle at M will be a
right one, and that at L equal to that
required. 2. If both Angles are oblique,
(See Fig. of the next Page) find G, the
Pole of AD, one of them (by the III.
of this Chapter) and at a Diſtance from
it, equal to the Angle that required Circle is
to make; with that (AD) defcribe (by the
V. of this Chapter) a leffer Circle (KHF).
Then at a Diſtance from (C) the Pole of
(A E) the other Circle, equal to the Supple-
ment to 180° of the Angle to be made by the
required Circle, defcribe (by the V. of this
Chapter) a leffer Circle (HNI). If this
L 2
meets
[ 148 ]
meets the former (KHF), the Problem is pof-
fible, otherwife not; but it meets it in two
Points (H and I). Then make either of
them a Pole of a Circle, which (by the III.
of this Chapter) defcribe. If you take H
for the Pole, the Circle will be LM, and it
will make the Angles at L and M as requir❜d.
L
D
M
E
K
HING
RESIDABRIOL TEND
CHA P. XVI.
The Use of the SECTOR in Spherical Tri-
gonometry, both by Protraction and
Calculation.
I
N the Management of this Chapter,
I fhall fhew the Protraction of all
the various Datas from the natural
Lines. Give the Solutions to every
Cafe, and the Operations of fome, particularly
of the more difficult; and leave the reft for
the Exercife of Beginners. I farther hint,
that
[149]
that when Tangents are concerned, it is im-
poffible to work every Example by the natu-
ral Lines, without an Expence of a great deal
of Time; and then it will ſcarce be done to
the Satisfaction of the Quærift: But the arti-
ficial Tangents being always ready, and fit
for every Operation, I fhall uſe them for the
moſt Part, and the natural ſparingly.
Of Right-angled spherical Triangles. In every
Cafe of which the right Angle is at D.
I. Given (A E 50 Lo') the Hypothenufe,
and an Angle (A 23-30): To find (DE)
the Leg oppofite to this Angle.
A
B
E
F
D
G
Firſt, By Protraction.
Make (by the VIII. of the laſt Chap.) the
Angle FAG of 23° 30′. And, by the IV. of
the laſt, AE 50°, and draw the right Circle
CD; then the Angle at D is a right one,
and the Triangle is protracted. Therefore
the other Parts A D, E D, and the Angle E,
may be meaſured by the IV. and VII. of the
L 3
laft.
[ 150 ]
K
laft. The Proportion to find DE is, As the
Radius is to the Sine of (A E) the Hypothe-
neufe; fo is the Sine of the Angle given (A)
to the Sine of (DE) its oppofite Side.
Therefore, by the natural, or fectoral Sines,
make the Sine of 23° 30', a Parallel at the
Sine of 90°, and the Parallel at the Sine of
50°, is the Sine of (17°L47) the Anſwer.
Secondly, By the Artificial Lines.
The Extent from 90°, to 23°30', will
reach from 50°, to the Anfwer 17°L47.
II. Given (ED 17°L4%) one Leg, and
(E 70° 0') the Angle adjacent: To find
(AD) the Leg oppofite to it.
i
B
E
F
A
D
G
First, By Protraction.
Make (by the VIII. of the laft) the Angle
FEG 70° 0'; and by the Chords make DE
1747. Thro' D (by the VI. of the laſt)
draw a great Circle perpendicular to DE;
it will be the right Circle CD. So the Tri-
angle
[151]
angle ADE, right-angled at D, will be
conftructed as required; and now the Hypo-
thenuſe A E, the Leg A D, and the Angle
A, may be meaſured (by the IV. and VII.
of the laſt Chapter). The Canon to find
AD is, As the Radius is to the Sine of
ED (17°47); fo is Tangent of (70°)
the Angle E, to the Tangent of (40°L2')
the Leg fought A D.
2dly, It is worth while to obferve, that if
you would folve this Queſtion by the fecto-
ral Sines and Tangents on the modern Sector,
or on Gunter's, as he made it, you are thus
to proceed; Open the Sector to the full Ex-
tent, and on the Tangents take the Diſtance
from 70° to 70°: Make this a Parallel at
the Sine of 90 Degrees, and then take the
Parallel at the Sine of 17° 47' in your
Compaffes, and lay the Compaffes by: Again,
open the Sector quite to its greater Opening,
and with another Pair of Compaffes, take
on the upper Tangents, the Diſtance from
45 to 45, and now make it a Parallel at
45 on the lower Tangents: Laftly, Take
the firſt Compaffes, having the Opening you
laft gave them, and draw them along the
lower Tangents 'till they become a Parallel,
which will be at (40°) the Anſwer. Nor is
the Matter any whit the better, if with the
former Sectorifts
you fo
vary the Proportion,
that the Co-Tangents are uſed inſtead of the
Tangents; and this will always happen when
two Tangents are to be compared with one
another, whereof the one is greater than of
L4.
45°,
[ 152 ]
45°, the other of lefs. Nor is it mended by
the Addition of Gunter's natural Tangent on
the Edge of the Sector. Laftly, Neither by
Gunter's, Fofter's, Collins's, &c. can the
Solution be wrought out at all, if the Num-
ber falls beyond the Tangent on the Sector,
without conftructing a Tangent, de novo.
And the fame may be faid of comparing a
Sine with a Secant.
Thirdly, By the Artificial Lines.
The Extent from the Radius to the Sine
of 17° 47', will reach from the Tangent of
70, upwards beyond the Tangent of 45;
therefore (fee the Remarks in the 8th Article
of Chap. XIV.) I bring the upper Point to
45, and the other will reach below 70, where
hold it faft, and bring the other Point to 70;
this Opening laid from 45, will reach to the
Tangent of 40°, the Anfwer. I think this
one Example may fatisfy the curious Enquirer
why the Sector was altered, and the artificial
Lines introduced.
III. Given (A D 40°) one Leg, and its
oppofite Angle (70°): To find the other
Angle A.
Firſt, By Protraction.
Draw (by the VIII. of the laft) FE, fo
that it makes the Angle FEG, 70°; and fince
AD is to be 40°, CA will be 50°; therefore
at the Diſtance of 50° (by the V. of the laſt)
defcribe
[ 153 ]
deſcribe H AK, a leffer Circle, cutting FE
in A. Laftly, thro' A (by the VI. of the
laft) draw AD perpendicular to EG; but
this will be a right Circle, and the Triangle
ADE will be conftructed. And then (by
the IV. and VII. of the laſt Chapter) the
other Angle A, the other Leg ED, and the
Hypothenufe A E, may be meaſured, and
fo known.
H
C
E
F/A
D
K
G
Secondly, By the Artificial Lin e
Becauſe the Cofine of AD is to the Radius,
as the Cofine of the Angle E, to the Sine of
the fought Angle A. Therefore the Extent
from the Sine of (50°), the Complement of
AD (40°), to the Radius, will reach from
the Sine of (20°), the Complement of the
Angle E, to the Sine of (26° 31') the Angle
A fought.
IV. Given both the oblique Angles (A 48°,
E 56°): To find the Hypothenufe.
Firſt,
[ 154 ]
ľ
First, By Protraction.
Make (by the VIII. of the laft Chap.) the
Angle FAG 48°; and then (by the IX. of
the laſt Chap.) draw a Circle, fo that it may
make an Angle with AF of 50°, and with
A G, a right Angle; that is, at the Diſtance
of 56° from P, the Pole of AF, deſcribe a
leffer Circle HK I, which meets the primitive
Circle AG in the Points H and I; then
make H (or I) the Pole of a great Circle,
which deſcribed (by the III. of the laft Chap.)
will be, if you take H, the right Circle LCD ;
and fo the Triangle ADE will be that re-
quired. And then the Hypothenuſe (A E),
and the Legs AD, ED, may be meaſured
by the IV. and VII. of the laft Chapter.
I
H
P
K
D/E
#3
Secondly, By the Artificial Lines.
Becauſe the Tangent of (E 56) one of the
Angles, is to Tangent of (32) the Comple-
ment of (48) the other Angle, as the Radius
is to the Sine of the Complement of (A E)
the Hypothenufe. In order to get in your
Compaffes
[ 155 ]
Compaffes the Diſtance from the Tangent of
56°, which is above 45°, to the Tangent of
42°, which is lefs than 45°; take the Diſtance
from the Tangent of 56°, one of the two, to
the Radius, and lay it from 42°, the other,
downwards; then take the Extent from the
lower Point of the Compaffes to 45°, and it is
that Diſtance fought which will reach from
the Radius on the Sines, to the Sine of (37°
24') the Complement of (52° 36°) the Hy-
pothenufe.
V. Given (AE 50°) the Hypothenufe,
and (A D 40°L 2) one Leg: To find ED
the other Leg.
G
TO
FA
D
B
First, By Protraction.
Make A D (by the IV. of the laft Chap.)
40° 2′; and thro' D (by the VI. of the laſt
Chap) draw (DEG) at right Angles to AD.
Then at the Diſtance of the Length A E (by
the V. of the laſt Chap.) defcribe a leffer
Circle meeting DG in E: Laſtly, draw AE,
and the Triangle is protracted. And now:
the
[ 156 ]
the Leg DE, and the Angles A and E, may
be meaſured by the IV. and VII. of the laft
Chapter. In this and the following, I have
protracted the Triangles at the Center of the
Primitive, in the four preceding I have pro-
tracted at the Primitive: But you may ob-
ferve, that, by the Doctrine of the forego-
ing Chapter, you might have protracted all
of theſe, either at the Center, or at the Pri-
mitive, or within it, or without it, or partly
within, and partly without.
Secondly, By the Artificial Lines.
Becauſe the Cofine of (A D) a Leg, is to
the Radius, as the Cofine of (AE) the Hy-
pothenufe, to the Cofine of (DE) the other
Leg. The Extent on the Sines from (49°
L 58') the Complement of 40°L 2′, will, on
the Sines, reach from (40°) the Complement
of 50°, to the Sine of (57°05′) whoſe
Complement is 32 L 55', DE fought.
VI. Given (AD, 40L 02; DE, 17°47′)
the Legs: To find the Angle A.
Firſt, By Protraction.
Make (by the IV. of the laſt Chap.) A D
40°L02', and (by the VI. of the laſt Chap.)
draw DEG at right Angles to AD; then
(by the aforefaid IV. of the laſt Chap.) make
DE 17°47′, and draw AE: So is the Tri-
angle protracted. Whence (by the IV. and
VII. of the laft Chapter) the Hypothenufe
(A E)
[ 157 ]
(AE) and the Angles (A and E) may be
meaſured: And fince, as the Sine of one
Leg (AD) is to the Radius, as the Tangent
of (ED) the other Leg, to the Tangent of
(A) its oppofite Angle.
E
A
D
Secondly, By the Artificial Lines.
The Extent on the Sines from 40°L 2', to
the Radius, will, on the Tangents, reach
from 17°47', to 26° 31'. Thefe are all
the various Datas in the 16 Cafes of right-
angled Sphericks, and confequently all the
Varieties of Protraction. In the other 10
Cafes, whofe Protractions are already laid
down, I fhall only propofe them, and give
the Anfwers, and leave the Operation for the
Reader's Practice.
N. B. The Figure referred to in the Laft So-
lution, is to be referred to in every one of
the following.
A Synopfis
[ 158 ]
A Synopfis of the Ten remaining Cafes of right-
angled fpberick Triangles.
VII. & VIII. Given the Hypothenufe AE
43°9′, and one Angle A 26°L 31′: To
find the other Angle E, and AD, the Leg
adjacent to this Angle.
I. As the Radius is to the Sine of the
Complement of the Hypothenufe A E; fo is
the Tangent of the Angle A, to the Tan-
gent of (20°) the Complement of (70°) the
Angle E.
2. As the Radius is to the Sine of the
Complement of the Angle A; fo is the Tan-
gent of the Hypothenufe A E, to the Tan-
gent of the Side A D 40°L 2'.
IX. & X. Given one Leg ED 17° 47',
and the adjacent Angle 70°: To find the
other Angle A, and the Hypothenuſe AE.
1. As the Tangent of the Angle E, is to
the Tangent of its adjacent Leg ED; fo is
the Radius to the Sine (63° 29′) the Com-
plement of (26°L 31) the Angle A.
2. As the Sine of the Complement of the
Angle E, is to the Radius; fo is the Tangent
of the Leg ED, to the Tangent of (43°19′)
the Hypothenuſe.
ว
XI. & XII. Given one Leg A D 40°L 2',
and its oppofite Angle E 70°: To find the
other Leg DE, and the Hypothenufe AE.
1. As
[159]
1. As the Tangent of the Angle E is to
the Tangent of the oppofite. Leg AD; fo is
the Radius to the Sine of (17°47′) the other
Leg.
2. As the Sine of the Angle E, is to the
Sine of the Leg AD; fo is the Radius to
the Sine of 43°L9' the Hypothenuſe AE.
XIII. Given the Angles A 26°31′, E
70°: To find a Leg A D.
As the Radius is to the Sine of the Com-
plement of the Angle E oppofite to the Leg
fought; fo is the Tangent of the other Angle
A, to the Tang. of (40°L 2) the Leg AD.
XIV. & XV. Given the Hypothenufe AE
43° 9', and one Leg AD 40°L 2: To find
the Angles A and E.
1. As the Tangent of the Hypothenufe
A E, is to the Tangent of the Leg AD; fo
is the Radius to the Sine of (63° 29′) the
Complement of (26L 31) the Angle A adja-
cent to the given Leg.
2. As the Sine of the Hypothenuſe is to
the Radius; fo is the Sine of the Leg AD,
to the Sine of (70°) the Angle E oppoſite to
this Leg.
XVI. Given the Legs A D 40°L 2′, DE
17°47': To find the Hypothenufe AE.
As
[160]
As the Radius is to the Sine of (49°L 58′)
the Complement of (40°L 2) A D one Lég;
fo is the Sine of (72°L15) the Complement
of (17°L47) DE the other Leg, to the
Sine of (46° 15′) the Complement of (43°
L9') the Hypothenuſe A E.
Of oblique angled fpherical Triangles.
XVII. Given the three Sides of a ſpherical
Triangle (viz. AB 40°, AC 46°, BC 76°);
To find (A, or) either of the Angles,
G
B
F
E
G
First, By Protraction.
By the Chords make A B 40°; at the Di-
ſtance BC (76°, from B deſcribe a leffer Cir-
cle DCE (by the V. of the laſt Chapter);
and by the fame, at the Diſtance of AC 46°,
from A deſcribe a leffer Circle FCG, cutting
the former in C. Laftly, thro' B and C,
and alfo thro' A and C, draw the great Cir-
cles BC, AC (by the II. of the laft Chap.)
and the Triangle is protracted; and then
either
[ 161 ]
either of the Angles may be meaſured by the
VII. of the laſt Chapter.
As to the Solution by the Sector, an Ex-
ample confidered in general Terms, will, in
this Cafe, perhaps, be more inſtructive than
a dry Rule, tho' that Rule be afterwards
exemplified.
The Sides containing the
required Angle are
The Side oppofite to the re-
quired Angle
The Sum of all the Sides
AB 40
AC 46
{BC 76
162
Half the Sum of all the Sides
Half the Sum of all the Sides
?
81 call it S
leffened by the Side oppofite os call it D
to the required Angle
Now the Extent of the Compaffes on the
Sines, from (81) S, to (46) A C, one of the
containing Sides, will reach from (40) A B,
the other containing Side, to another Sine,
where hold the Point faft, and carry the
other to D. This fecond Opening of the
Compaffes will reach on the verfed Sines from
the Beginning to (128° 52') the Angle A
fought. And thus this most difficult Problem
in Numbers, is readily and easily folved by the
artificial right Sines, and verfed Sines. Had
there been nothing else to recommend thefe two
Lines, but the Solution of this Problem alone,
it were fufficient for them to claim a Place on the
Sector; not only because they fo neatly and expo-
ditiously folve the Query, but also on account
M
Of
[162]
of the great Use of this Problem in Aftronomy;
and the daily Demand for its Solution by the
Navigators, in order to find the Variation of
the Compafs; without the Knowledge of which,
it would be impoffible for them to steer their
Courſe aright.
XVIII. Given two Sides (AB 40°, CB
76°) and their contained Angle (B 35° 16):
To find the third Side A C.
B
A
E
D
First, By Protraction.
Make (by the VIII. of the laft Chapter)
the Angle B 35°16′; and on the Primitive
lay AB 40°, and on CB lay 76° (by the IV.
of the laſt Chap.) Laftly, Thro' AC (by
the II. of the laſt) ſtrike the great Circle AC,
and the Triangle is protracted. And there-
fore the third Side A C, and the Angles A
and C, may be meafured by the IV. and VII.
of the laft Chap.
Secondly,
[163]
Secondly, By the Artificial Lines.
The Extent from the Radius, to the Sine
of (54°44) the Complement of (35°L 16'′)
the given Angle B, will reach from the Tan-
gent of (40°) AB, one of the given Sides,
to the Tangent of a fourth Arch. This
fourth fhall be leſs than 90°, if the given
Angle be lefs; but greater, when greater,
provided that the preceding given Side be
leſs than 90°. And then the Extent from
the Sine of (55°L35) the Complement of
(34°25') the preceding fourth, to the Sine
of (50°) the Complement of (40°) the pre-
ceding Side, will reach from the Sine of
(48°25) the Complement of (41° 35')
the Difference between the fecond Side (76°)
and the fourth Arch (34° 25′) before found,
to the Sine of (44) the Complement of
(46°) A C, the Side fought.
XIX. Given as above: To find the two
Angles A and C.
The Protraction is as before, becauſe the
Data are the fame.
The given Sides containing the given § BC 76
Angle are
Their Sum is
Their Difference is
Half their Sum is
AB 40
116
36
58
18
M 2
The
Half their Difference is
[164]
The Extent on the Sines from (58°) the
half Sum, to (18°) the half Difference, will
reach on the Tangents from (72° 22′) the
Complement of (17° 38') the Half of (35°
16') the contained Angle B, to (48° 54′)
half the Difference of the oppoſite fought
Angles A and C. And,
The Extent on the Sines from (32°) the
Complement of (58°) half the Sum of the
given Sides, to (72°) the Complement of
(180) half their Difference, will reach on the
Tangents from (72°L 22) the Complement
of (17° 38') the Half of (35°16') the
given contained Angle B, to (79° 58′) half
the Sum of the oppofite fought Angles.
An-}
1-3 79 °
The half Sum of the oppofite An-
gles A and C
The half Difference of A and C
The greater Angle A
The leffer C
58'
48 54
128 $2
31 04
But note, that as half the Sum of the Sides
AB, CB, is greater or lefs than 90°; half
the Sum of the oppofite Angles A and C,
fhall accordingly be greater or leſs than 90°.
Et vice verfa.
XX. But if the fame Things are given,
and all the three unknown Parts (viz. the
Bafe A C, and the two Angles A, C) are
fought; which is very frequent in Aftro-
nomy. Firſt (by the XIX.) find the two An-
gles A and C, and then, the Extent on the
Sines
[ 165 ]
Sines from (48° 54) the half Difference of
thofe Angles, to (79° 58′) their half Sum,
will reach from the Tangent of (18°) half
the Difference of the Legs BC, AC, to the
Tangent of (23°) half the Bafe.
XXI. Given two Angles (A 128°L 52′,
B 35° 16′) and the interjacent Side ( A B
40°): To find (A C, BC) the other two
Sides. (See the last Figure).
Firſt, By Protraction.
Make by the Chords A B 40°; and (by
the VIII. of the laſt Chap.) defcribe B CE,
to make the Angle B 35° 16', and AD to
make the Angle BAD 128° 52', or the
Angle DAE 51°L8', and the Triangle
ABC is protracted; and confequently the
Legs AC, BC, and the Angle C, may be
meaſured by the IV, and VII. of the laft
Chapter.
The Computation by the artificial Lines in
this and the following, is in all refpects the fame
with Article XIX. and XX. by changing the
Angles into Sides, and the Sides into Angles,
fave that inſtead of the interjacent Side, and
its oppofite Angle, you muſt take their Sup-
plements to 180°. For,
1
M 3
The
[ 166 ]
The given Angles
Their Sum
Their Difference
Half their Sum
Half their Difference
A 128° 52′
C
35 16
164 08
93 36
8204
46 48
Then the Extent on the Sines from (82°
4)
half the Sum of the given Angles, to (46°
48') half their Difference, will reach on the
Tangents from (20°) the Half of (40°) the
given Side, to (15°) half the Difference of
the fought Legs ACB. And,
The Extent on the Sines from (7°L 56′)
the Complement of (82°L4′) half the Sum
of the given Angles, to (43°12′) the Com-
plement of (46°48′) half the Difference of
the given Angles, will reach on the Tangents
from (20°) the Half of (40°) the given Side,
to (61°) half the Sum of the fought Legs.
Half the Sum of the fought Legs
61°
Half their Difference
15
The greater Leg BC
76
The leffer AC
4.6
N. B. Obferve the Note in the XIXth.
XXII. Given
[167]
XXII. Given as before: To find all the
three unknown Parts, viz. the Angle C,
and the Legs AC, BC.
Firſt, By the former find the Legs AC,
BC: And then the Extent of the Compaffes
on the Sines from (15°) half the Difference
of thoſe Legs, to (61) half their Sum, will
reach on the Tangents from (46°L 48′) half
the Difference of the Angles, to (74.°L 28')
the Complement of (15° 32') the Half of
(31°4') the Angle C fought: But the An-
gle C might have been found without the
Legs (as in the XIX) if you had changed
the Side AB, and its oppofite Angle C,
for their Supplements, to 180°; and then
called the Angles Sides, and the Sides
Angles.
XXIII. Given two Sides AB 49, AC 46,
and an Angle B 35° 16', oppofite to AB,
one of them: To find the other three un-
known Parts, viz. the Side BC, and the
Angles A, C. (See the following Figure).
First, By Protraction.
Make by the Chords A B 40°; draw BG
fo that the Angle B may be 35°16′, by
the VIII. of the laft. At the Diſtance of
AC 46°, defcribe about the Pole A, the
leffer Circle DCE meeting BG in C. Laftly,
thro' A and C draw the great Circle ACF,
and the Triangle is laid down; and then the
unknown
M 4
[168]
unknown Parts may be meaſured by the IV.
and VII. of the laft. Now,
D
G
B
E
F
Becauſe Sines of Angles are as the Sines
of their oppoſite Sides, the Extent on the
Sines from (46°) the Side AC, to (35°16')
its oppofite Angle B, will reach from (40°)
the Side A B, to (31°L4) the Angle C.
The other Side BC, and its oppofite An-
gle A, may be found by the latter Parts of
the XX. and XXII,
XXIV. Given two Angles, A 128°L 52′,
B 35° 16' and BC 76°, a Side oppofite
to one of them: To find the other three un-
known Parts.
Make the Angle B 35°L 16' (by the VIII.
of the laſt Chap.) and cut off BC 76° (by
the IV. of the laft Chap.). Then (by the
IX. of the laſt Chap.) from C fo draw the
great Circle ACE, that it may make the
Angle CAB 128 L 52', or the Ångle DAE
51° 8'; and fo will the Triangle be con-
°18';
ftructed;
[169]
ftructed and what's unknown may be mea-
;
fured (by the IV. and VII. of the laſt Chap.)
D
B
A
E
By the Artificial Sines.
The Extent from (51°L8) the Supple-
ment of (128° 52') the Angle A, to 180°
to (76°) the Side B C oppofite to it, will
reach from 35° 16') the Angle B, to (46°)
its oppoſite Side B C.
BC
The Angle C, and its oppofite Side A B,
may be found by the latter Parts of the
XX. and XXII.
XXV. Given all the Angles, A 128° 52',
B 35° 16′, C 31°L4: To find a Side AC.
(See the following Figure).
First, By Protraction.
Make (by the VIII. of the laſt Chap.) the
Angle B 35° 16'; and (by the X. of the
laft Chap.) draw A K fo that it may make
the Angle A 128° 52′, and the Angle C
31°L4; then the Triangle will be laid down,
and
[ 170 ]
and the Sides may be meaſured by the IV.
of the laſt Chapter.
B
A
K
Secondly, By the Artificial Lines.
The Solution is the fame with the XVII.
provided that you change one of the Angles
adjacent to the Side fought, no matter which,
into its Supplement to 180°; and then call
the Angles Sides, and the Sides Angles.
Here I change the Angle A.
The Angles adjacent to the 551° 08′ A
fought Side are
The third Angle
The Sum of all the Angles
Half the Sum of all the Ang.
Half the Sum leffened by the
Angle oppofite to the re-
quired Side is
31
04 С
35
16 B
117
28
58 44 S
23 28 D
Now the Extent of the Compaffes on the
Sines from (8°L44′) S, to (51°L8′) A,
one of the adjacent Angles, will reach from
(31°
[ 171 ]
(31°4') C, the other adjacent Angle, to
another Sine, where hold the Point faſt, and
carry the other, Point to D. This fecond
Opening of the Compaffes will reach on the
verfed Sines from the Beginning, to (46°)
the Side fought.
BRYONIC STED
CHA P. XVII.
The Use of the Sector in the Projection
of the Sphere.
W
Firſt, In the Orthographick.
[See Fig. of Orthographick Projection.]
ITH any convenient Radius de-
fcribe a Circle reprefenting the ge-
neral Meridian, which divide into
four equal Parts by the Diameters.
EQ, PS. EQ reprefenting the Equinoctial
Circle, PS reprefenting the 6 o' Clock Hour
Circle, and the Axis of the Sphere. Divide
each of thefe 4 Quarters of the Meridian in-
to 90°, and number every 10th from E to-
wards the Poles, and from the Poles towards
Q, by the Figures 10, 20, 30, &c. From
E and Q towards the Poles, count 23
and draw the ftrait Lines marked s F, G;
thefe will reprefent the Tropicks of Cancer
and Capricorn: And parallel to thefe, draw
ftrait Lines thro' every whole Degree between
them; and theſe will reprefent the Parallels
ΟΙ
20
of
[172]
of the Sun's Declination. The fame may be
done for all the other Parallels, and they will
divide the Lines CP, CS, into Lines of
Sines; draw C, and it will reprefent
the Ecliptick. Make CE, or CQ, Radius
on the Sines; and by it, from the Sector,
make CE, CQ, Cs, C, Lines of Sines
divided into every Degree, if there be Room
enough. Then, making the Half of each of
thefe Parallels of Declination a Radius, make
them Lines of Sines from SP, both upwards
and downwards: And thro' thefe Divifions,
at every 15th Degree of the Equinoctial,
draw fmoothly a crooked Line, and they will
repreſent the Hour Lincs; which, towards
P, the North Pole, are numbered to anſwer
the Morning Hours; thofe towards S, the
Evening ones. The Semicircle SEP denot-
ing the Mid-day Hour Circle, or 12 at Noon.
Produce the Tropicks to H and K, to L and
I; fo fhall the Spaces between K and L, H
and I, be for the Kalender; K and H de-
noting the 10th of December, and L and I
the 10th of June.
Seek in a Table what is the Sun's Declina-
tion at the End of December; and againſt
that Declination draw a Line, or Divifion,
in the Kalender; and alfo feek the Sun's
Declination for the laft Moment of January,
and againſt the Parallel of this Declination,
draw a Divifion in the Kalender, and the
Month of January falls wholly between thefe
two Divifions: And the like may be done
for every other Month; and alfo for every
Day, or every fifth Day, according to the
Room
[ 173 ]
Room in the Inftrument. Thus far the In-
ftrument is univerfal.
,
If you would reſtrain it to a particular La-
titude, count the Latitude in the Meridian
SEP from E, according as it is North or
South In this Example it is for London, in
the Latitude of 51° 32′ North, reprefented
by Z. Draw the prime Vertical Z CN, and
at right Angles to it, the Horizon HC,
and divide them as the Equinoctial is divided;
and parallel to the Horizon, at 18° below it,
draw T W, it will be the Line between the
enlightened and dark Hemiſphere; divide it
as the 18th Parallel of Declination is divided.
This is ufed in fhewing the Time of Day-
break. Laftly, If the Lines ZC, HC,
TW, were Rulers of Brafs fix'd to one
another, and another is made to flide on CZ,
the whole would be univerfal, and capable
to folve all ſpherical Problems relating to the
diurnal Motion of the Sun, and of as many
of the Stars alfo, as you pleafe to put on it.
Some of the more obvious Ufes of this
Projection.
By the Day of the Month, the Sun's Place,
or right Afcenfion, may be found the De-
clination; by the right Afcenfion may be
found the Flace, Et vice verfa. When the
Latitude is (51° 32') known as well as the
Day of the Month, which let be the 29th of
July, when the Sun is in 16° of Leo, and fo
has 15° Declination, and his right Afcen-
tion is 8 54. Then the Meridian Altitude is
HA
[174]
HA 53°30', the Midnight Depreffion is
Bb 23°30', the afcenfional Difference is
CG 126, the femidiurnal Arch is AD
h
h
h
26′ (and fo the Length of the Day 14
52), the feminocturnal Arch is DB 4 34',
(and fo the Length of the Night is 9" 8'),
the Amplitude CD is 27°L 50'; and from
the End of Twilight, to Midnight, is 1" 56′,
and fo the whole Length of real Night is
3h 52'; and from Day-break, to the End of
Twilight, is 2oli 8'.
If the Sun's Altitude were given 17°, draw
a Line thro' 17° of C z, parallel to the Ho-
rizon Hb; and where it cuts the Parallel of
the Declination, as in O, you will find by
Inſpection, the Hour of the Day; and the
Diſtance R. O laid on the Parallel of the La-
titude of 17°, fhews the Azimuth. And if
the Inftrument had been fitted as before di-
rected, theſe latter, with others, had been
done eaſier and exacter. This, with the
Projection uſed in the Conftruction of folar
Eclipfes, are the Principal in the Orthogra-
phick.
1
Of the Stereographick Projection.
[See Fig. of the Stereographick Projection.]
The two following are the moſt remark-
able; the one on the Plain of the general
Meridian, and the other on the Plain of the
Horizon; tho' the Sphere may be projected
on the Plain of any Circle.
Deſcribe a Circle to reprefent the general
Meridian, quarter it, divide it, and number
it as the foregoing. Lay the Tangent of
half
[ 175 ]
half 75° from C towards E, which gives a
Point, thro' which, and the Poles S, P, a
Circle is to paſs to reprefent the Hour Circle
of XI. and I.; lay the Tangent of (15°) its
Complement, from C towards Q, it gives
the Center; and therefore you may defcribe
the Circle. Again, Lay the Tangent of half
60°, from C towards E, which gives a Point,
thro' which, and the Poles, the Hour Circle
of X. and II. paffes: And its Center is had by
laying the Tangent of 30°, the Complement
of 60°, downwards from C towards Q; and
therefore the Circle may be defcribed. And
in like Manner you may proceed to lay the
Tangents of half 45, 30, 15 upwards; and
the Tangents of (45, 60, 75) their Comple-
ments downwards, to find the Centers, and
then deſcribe the Circles: And, as you have
drawn the Circles in one Semicircle, turn the
Paper and draw them in the other Semi-
circle. If the Plain be large, and you have
Room enough, you may bring in every 5th
Degree, and you will have the Circles to
every 20th Minute in Time; and every two
Diviſions will anfwer to 10° in Motion. Laſt-
ly, Subdivide by the Tangents of the half
Degrees, 'till you have divided the Equino-
ctial ECQ into 90°. The Parallels of La-
titude have their Centers in the Six o' Clock
Hour Circle produced. So, would I defcribe
the Parallel of 40°, I take the Tangent of
50°, its Dift. from the Pole; and one Foot
placed on 40 in the Meridian, I turn the other
about 'till it falls on the 6 o' Clock Hour Cir-
cle produced, that is its Center, therefore de-
fcribe
[ 176 ]
fcribe it. In like Manner defcribe the reft
of them with the two Tropicks and polar
Circles. The Tropick of Cancer is mark'd
I; that of Capricorn w H. Laftly, draw
HC the Ecliptick, and divide it from C
to, and from C to H by the Tangents of
half the Degrees: And do the like from C
to P, and from C to S.
you
If would reſtrain it to a particular
Latitude, draw the prime Vertical and the
Horizon, and divide them as C P is divided;
18° beneath the Horizon, draw a Parallel to
determine the Beginning and End of Twi-
light; which, if you pleafe, you may divide
into 18 equal Parts, reprefenting every 10th
Degree of Azimuth. As to the Ufes of this
Projection, they are fo much like the former,
that I fhall refer you to them: But take no-
tice, that this, as well as the preceding,
might be made univerfal.
The other Projection I fhall fhew how to
lay down by the Sector, is very frequently
ufed in Dials, both fix'd and portable. It is
a Projection on the Plain of the Horizon.
[See Projection on the Plain of the Horizon.]
Deſcribe a Circle, quarter it, and divide
it, and number it from A to B and D, and
from a to B and D. A a is the Meridian,
and BC and DC the prime Verticals. Lay
the Tangent of half (51° 30′) the Latitude
of the Place, from C to E; then is E the
Point where the Meridian and Equinoctial
interfect each other. Lay the Tangent of
half (38° 28′) the Complement of the Lat.
from
[ 177 ]
from C towards a, it will give the Center to
defcribe the Equinoctial, and it will pafs
thro' E, B, and D. Lay the Tangent of
half (38°28') the Complement of the La-
titude, from C towards a, viz. to P, and it
is the Pole; and therefore a Point thro
which all the Hour Circles are to pafs. Lay
the Tangent of the Latitude from C, on
CA, and it will give CF, the Center of the 6
o' Clock Hour Circle, which will paſs thro' B,
P, and D. At right Angles to FC, at F draw
GFH, in which are the Centers of all the
Hour Circles. Make PF the Radius, and
from F, both Ways, lay off the Tangents of
15, 30, 45. Then make PF a Radius on the
upper Tangents, and lay off, both Ways,
from F, the Tangents of 60° 75' ; and
theſe ſhall be the Centers to strike the Hour
Circles thro' the Pole P. If theſe Tangents
had been laid down to every 7° 30', the
half Hour Circles would have been defcribed:
And, if they had been laid down to every 10°
they would have reprefented Meridians to
every 10°.
Now make again the Radius of
the Primitive, a Radius on the Chords.
The Parallels of Declination may be
drawn (by the 5th Prop. of the 15th Chap.)
But for this Work the Parallels are very eaſily
defcribed: For, take their Diſtance from the
elevated Pole, and take alfo the Complement
of the Latitude, and find their Sum, and
alfo their Diff. Then lay the Tangent of
half their Diff. from C towards A, it will
give the Point where this Parallel cuts the
Meridian; alfo lay the Tangent of the half
N
Sum
[178]
Sum from C´on Ca; and from this Point to
the former, bifect the Diſtance, and it fhall
be the Center of the Parallel, which therefore
may be deſcribed. The Examples fhall be
the two Tropicks. Firſt, The Tropick of
Cancer is 66° 30' diftant from the Pole,
and the Complement of the Latitude 38°
30'; their Sum is 105°, and their Difference
is 28°. Therefore lay the Tangent of half
28° from C to 5, and the Tangent of
half 105°, from C to K. Then find the
Middle of K, which is L; from L, with
the Diſtance L, defcribe MN, and it is
the Parallel required. Alfo the Tropick of
Capricorn is diftant from the elevated Pole
113° 30', and the Complement of the La-
titude is 31° 30': Their Sum is 152°, and
their Difference is 75°. Therefore lay the
Tangent of half 152° from C, beyond a, to
R, and the Tangent of half 75° from C to
о
. Then find the Middle, which is 1, and
with the Diſtance I, from I defcribe S v T,
and it is the Parallel required. In like man-
ner the other Parallels of Declination were
defcribed; But you may obferve, that, if
this Projection had been made for any Place
in the Torrid Zone, fuppofe for the Latitude
of 23° North, the half Tangent of C, as
well as of CK, muft be laid on Ca: And
the fame muſt be obferved of every Parallel
that falls between the Zenith and the elevated
Pole; but where the Diſtance of the Parallel
is 180° from the Zenith, that Parallel will be
a ftrait Line.
The
[179]
The two Arches reprefenting the two
Halves of the Ecliptick, were thus defcribed.
The Diſtance from C to, was the Tan-
gent of half 28°; and therefore (by the IV.
of the XVth Chap.) lay the Tangent of 62°,
its Complement, from C to V, and it gives
the Center from whence B D may be de-
fcribed. The Diſtance from C to v was the
Tangent of half 75°; therefore lay the Tan-
gent of 15°, its Complement, from C to W,
and it is the Center to defcribe B ✨ D.
Theſe two reprefent the two Halves of the
Ecliptick.
For the Poles of the Ecliptick, fince is
diffant from C by the Tangent of half 28°,
lay the Tangent of half 62°, its Complement
from C to X; then a Ruler laid from X to
the feveral Divifions in the primitive Semi-
circle BAD, will divide B D. And in
like manner, fince is diftant from C by
the Tangent of half 75°, lay the Tangent of
half 15°, its Complement from C to y; and
then a Ruler laid from y to the feveral Divi-
fions in BAD, will divide B D into the
like Divifions. Put on a Kalender between
M and S, N and T, as taught in the firſt
Example of this Chapter. Laftly, divide C a
into a Line of half Tangents running to 90°
and alfo a Line of Shadows; or rather fit a
divided Scale to turn about on C.
N 2
CHAP.
[180]
SAINOSEOBIUS?
CHA P. XVIII.
The Conftruction and Ufe of the Lines of
T
Latitudes and Hours.
HESE Lines are placcd near the
Edges of the Sector, and properly
belong to the plain Scale, but are
often laid down on the Sector for
their Uſe in readily drawing the Hour Lines
on any plain Dial.
The Line of Hours is thus conftructed:
Having refolved on your Radius, make it
a Parallel at the Tangent of 45. Take the
parallel Tangent of 15°, and lay it both
Ways from III to II. and IV. And lay the
parallel Tangent of 30° from III. to I. and V.
And of 45° from III. to o and VI. And fo
the whole Hours will be laid down.
If you would draw it to every 5th Minute,
or nearer, you may do it by allowing 15°
to every Hour, or 15' a Quarter of a Degree,
in Motion to every Minute in Time. And
it is evident, that this Line is two Lines of
Tangents, running upwards and downwards.
from the Middle, numbered III. but reduc'd
to Time.
The
[181]
The Line of Latitudes may be conſtructed
thus:
Make the whole Line of Hours the Radius
of a Semicircle ABDC. At the Ends of a Plate
the Diameter raiſe the Perpendiculars AE, F: 63
DF, and make them Lines of Sines to the
fame Radius.
From the Center, to the feveral Divifions
in theſe Sines, draw ftrait Lines, meeting the
Periphery, in as many Points, which number
with the Numbers correfpondent to them on
the Lines of Sines.
Raiſe CB perpendicular, and lay a Ruler
crofs it from the like Numbers in the Arch,
and it will cut CB into a Line of Latitudes.
See Fig. 63.
The Line of Latitudes may be thus con-
ftructed:
Take the Line of Hours and make it a
parallel Sine of 90°. Then, for any Divi-
fion, fuppofe that of 35°. Take the crural
Sine of 35°, and lay it on the crural Tan-
gent, and
you will find it give on the Tan-
gents 29°50′; the parallel Sine of which
gives the Diſtance of 35° from the Beginning
of the Line: And fo for any other Degree.
But the fame Line may be thus conftru-
cted by the Tables of natural Sines, and the
Line of Lines on the Sector.
ว
The natural Sine of 35° is 5735764, which
fought among the Tangents, gives 29 L 50',
and the natural Sine of 29° 50' is 4974787.
Open the Sector 'till the Line of Hours is a
Parallel to the whole Line of Lines. And
then the Parallel of 4974787, will be the
N 3
Diſtance
[182]
Diſtance to be laid down from the Beginning
of the Line, to the Divifion repreſenting
35 And in the fame Manner may any
other Divifion be laid down.
The Ufe of thefe Lines will be manifeft
from the two following Examples.
ift. To draw an horizontal Dial for any
Latitude, fuppofe 54 N. Draw C, XII,
to reprefent the Meridian, and it is the Hour
2
VI VII
N V
B
VI
6
VI
5
VIL
IV
چی
a
D
3
VIII
IX
I
Line of 12.
I XII XI X
Croſs it at right Angles by AB,
and it gives the Hour Lines of 6 and 6. Take
from the Line of Latitudes, the Latitude of
the Place, fuppofe that of London 51N; and
lay
[183]
lay it on the 6 o' Clock Hour Lines from C
to A and B. Take the whole Line of Hours
in your Compaffes, and one Foot in B, or
A, turn the other about 'till it falls in the
Meridian, as here in D; and draw DA,
DB; and on DA, DB, transfer the Divi-
fions from the Line of Hours. Then Lines
drawn from C thro' thofe Divifions, will be
the Hour Lines fought. The Stile meets the
Plain in C, and ſtanding over the Meridian
C, XII, makes an Angle with it of 511,
the Latitude of London. The Hours 4, 5,
before 6 in the Morning, and 7 & 8 after 6 in
the Evening, are drawn by continuing their
oppofite Hour Lines 4, 5 in the Afternoon,
and 7, 8 in the Forenoon, beyond the Cen-
ter. In like manner may a direct South or
North Dial be drawn, omitting thofe Lines
that would be uſeleſs.
2dly. To defcribe a vertical declining Dial.
Let the Example be for a South Dial, de-
clining 20° Weftward, in the Latitude of
51 / North.
In order to this, there are three Things
neccffary, which may very readily be ob-
tained by the artificial Sines and Tangents.
Ift. The Extent from the Sine of (51)
the Latitude to the Sine of 90°, will reach
from the Tangent of (20°) the Declination,
to the Tangent of the Plain's Difference of
Longitude 24° 57'.
N4
2dly.
[ 184 ]
2dly. The Extent from the Sine of 90°,
to the Sine of 65°L 3', the Complement of
24° 57', the Plain's Difference of Longi-
tude, will reach from the Tangent of 38° 1,
the Complement of (51°) the Latitude, to
the Tangent of 35°48′ the Stile's Height.
3dly. The Extent from the Radius to the
Sine of the Stile's Height, will reach from
the Tangent of (24°L57) the Plain's Dif-
ference of Longitude, to the Tangent of
(15°13') the Subftile's Diſtance from the
Meridian.
Now, becauſe the Plain's Difference of
Longitude is 24°57', which reduced to
'Time, is 1h 40' neareft, I perceive that the
Subftile falls between the one and two
o' Clock Hour Lines: And one Hour being
taken away, in order to find the Hour' Line,
next to the Subftile you have left 40'.
Draw A C perpendicular to the Horizon,
to repreſent the 12 o' Clock Hour Line.
With it from C make an Angle of 15°13′,
by the Line of Chords, by drawing the Line
CB; and this muſt be to the right Hand,
becauſe the Declination is Weftward. This
CB is the fubftilar Line. Draw DCE per-
pendicular to CB; and from the Line of
Latitudes, take 35° 48', the Stile's Height,
and lay it from C to D. Take the whole
Line of Hours in your Compaffes, and with
one Foot in D, turn the other about 'till it
falls on the Subftile CB, as here in B, and
draw
[185]
draw DB. Make the Angle CBE equal to
the Angle CBD. Transfer the Divifions
of the Line of Hours, on DB and BE:
Then count from B towards D; and alfo
D
MII 8
C
11
IX
E
7
VII
6
VI
M
A
1
B
X XI XI II I
N
from E towards B, 40', 1h 40', 2h 40', 3h 40',
4h 40', 5h 40', &c. and thro' thefe and C,
fhall the Hour Lines be drawn: And in like
manner might the Halfs and Quarters be laid
down.
For Example; The Plain's Difference of
Longitude is 1h 40', from which take away
th
all the Hours, Halfs, and Quarters, and
there will be left 10'; to this add continu-
ally 15', and you have 10', 25', 40′, 55',
1h 1o, 1h 25, 1h 40', &c. to prick from the
Line of Hours on the Lines BD, E B.
I
And
[186]
And then Lines drawn thro' thefe Points
and C, are the Lines for the Hours, Halfs, and
Quarters. Laftly, the Stile muſt meet the
Plain at C, and ſtanding on CB, muft make
an Angle with it of 35°48'.
CHAP. XIX.
To draw the Lines on the Dials in the laft
Chapter, by the Line of Sines, and Lower
Tangents, &c. And firft for the hori-
zontal Dial.
AVING drawn the Meridian Line,
Hand crofs'd it at right Angles by
the Line AB; take any conveni-
ent Diſtance CD, and make it a
parallel Sine of 90°. Then lay the Sine of
the Latitude from C to A and B: And com-
pleat the Rectangles A CDE, BCDF; the
Sector ftill retaining the fame Opening, lay
the Tangent of 15°, 30°, 45°, from A and
B, to E and F. Now, making DE, or DF,
Radius, lay the Tangent of 15°, 30, 45,
from D. on DE and DF. Laſtly, from C,
thro' theſe Points, draw Lines, and they
will be the Hour Lines. In like manner the
Halfs and Quarters, or Minutes, may be
laid down, allowing according to the Pro-
portion
[187]
portion of 15° to an Hour, 7°30′ for
half an Hour, and ſo on.
VIII
N
VIII
C
VI
VI
A
B
15
15
V
N
VIII
30
30
VIII
E
FIX
3/0 15 D15 30
II I XII XI
XIX
2dly. For the declining Dial. Having, as
in the laſt Chapter, prick'd down the ſubſti-
lar Line, take any convenient Diſtance from
C thereon, as CB for the Radius; and croſs it
at right Angles in B and C. Then make
CB a parallel Radius on the Sines, and take
the parallel Sine of 35°48′, the Stile's
Height, and lay it from B to D and E, and
from C to F and G, and draw FD, GE,
compleating the Rectangle D EGF. Make
DB the Radius of a Line of Tangents. The
Plain's Difference of Longitude being 24
57', take from it 15° as often as poffible, and
there will be left 9°L 57; to this add 15° as
°L
often
[188]
often as the Reſult will be leſs than 45″, and
you will have 9° 57′, 24° 57′, 39° 57'.
To the Radius B D, take the Tangents of
theſe, and lay them from B towards D.
G
F
VIII
VII
VI
E
B
IX
D
IV
X XI XII I II I
O
3dly. Take 9° 57′ from 15°, and there
is left 5°3'; to this add 15° continually,
as often as the Refult will be less than
45
and you will have 5° 3', 20° 3', 35°3':
Of thefe, to the former Radius BD, take
the Tangents and lay them from B towards E.
4thly. Make FD, or GE, Radius, and
lay the Tangents of the former Degrees,
viz. of 9°57', 24° 57', 39° 57', from
G towards E: And the Tangents of the lat-
ter Degrees, viz. of 5°L3, 20°L 3', 35°
L3', from F towards D.
Laftly,
[ 189 ]
Laftly, Thro' thefe Points from C, ſtrait
Lines drawn are the Hour Lines.
If you had a Mind to bring in the Quar-
ters, you muſt, from 24°57', the Plain's
Difference of Longitude, take 15°, 7°L 30′,
°45', as often as poffible, and then you
have left 2° 27′; to this add 3°45′ con-
tinually, as often as the Reſult will be lefs
than 45°, viz. 6° 12′, 9°57′, 13 °L 42′,
3
17
L27, &c. and lay their Tangents from
B to D. Then fubtract 2°L27 from 3°L
45', there is left 1°L18; to which add 3°
L45' as before, and lay the Tangents of the
Sums from B to E. The former alfo, with
GE Radius, muſt be laid from G to E, and
the latter from F to D, and you will have
Points to draw the Hours, Halfs, and Quar-
ters thro'.
The fame may be done from the Meridian,
and without the Ufe of the fubftilar Line:
But first you muſt find the Angle made by
the 6 and the 12 o' Clock Lines; thus the
Extent on the Sines from the Radius, to the
Sine of the Declination, will reach from the
Tangent of the Latitude, to the Tangent of
(23° 16') the Complement of (66°L44)
the Angle fought.
Having drawn CE the Hour Line of 12,
make with the Chords the Angle E CD (66°
44') equal to that made by the 12 and 6 Hour
Lines. And lay CD to the right Hand,
becauſe the Declination is South-Weft. With
any convenient Radius, make CE the Sine
of
[ 190 ]
of (70°) the Complement of (20°) the De-
clination: And CN the Sine of (38°L 30')
the Complement of (51) the Latitude.
And draw N G parallel to CE, cutting CD
in D.
1
K
IXH
E
D
N
HVII
VI
G
IV
X XI XII I II III
Compleat the Parallelogram CDGE; con-
tinue G E 'till EH be equal to it; and draw
from Ha Line parallel to CE. Make GE,
or EH, a Radius on a Line of Tangents; and
lay from E towards G and H, 150 and 30',
for the Hours of 10, 11, 1, 2. Alfo make
DG Radius, and lay the Tangents of 15°
and 30°, from D and H towards G and N,
and divide H K, as G N is divided, as far as
the
T
*
[ 191 ]
the Plain will bear it. And draw Lines from
C to theſe ſeveral Diviſions, and alſo to the
Angles G and H, and you have the Hour
Lines required. The Stile is to form an An-
gle, and to be feated as in the laft Operation.
You might have proceeded to lay down the
Halfs and Quarters, or, indeed, the Minutes,
as in the firſt Example of this Chapter.
Here hath been confidered the Ufe of the
Sector in drawing the Hour Lines on Dials,
having the Pole confiderably elevated above
them: It remains to fhew how, by the fame.
Lines, to draw Dials that are polar, or have
the Pole
very little elevated above them.
The Example ſhall be for a full Eaſt Dial,
in the Latitude of 51 1.
I
IV V VI VIVIX X
N
V
VI
VII
VILL
IX C
X
10
XI
F
B
XI
Draw
[ 192 ]
Draw the horizontal Line CB, and, by
the Chords, make the Angle CBA (38)
the Complement of the Latitude, and A B
will reprefent the Equator. Chufe two Points,
E and F, near the Extremities of A B, thro'
which draw the Lines XI, XI, IV, ÍV, at
right Angles to AB: And they may repre-
fent the Hours of 4 and 11 in the Morning.
Make, by the Help of the Chords, the Angle
EFG 15°, and the Angle FEG 60°. Make
GE a Radius on the Sines, and lay the Sine
of 30° from E to D, and draw DG. Make
GD a Radius on the Tangents, and lay the
Tangent of 15° from 6 to 5 and 7; and the
Tangent of 30° from 6 to 4 and 8: Of 45°
from 6 to 9. Make GD the Tangent of
30°, and lay the Radius correfponding to it,
from D to 10. Thro' thefe feveral Points
draw right Lines at right Angles to the Equa-
tor, and they will be the Hour Lines. The
Stile muſt ſtand on DG, the 6 o'Clock Hour
Line; it muſt be parallel to it, and at the
Diſtance D 9. If you had ufed the Tangents
of 7°30', 15°, 22°L 30', &c. inftead of
22°30',
15°, 30°, &c. you would have drawn the
Hours and Halfs.
Laftly, Let it be required to draw an up-
right Dial for the Latitude of 51° 1, having
80° of Weſt Declination from the South,
whofe Requifites, by the preceding Doctrine,
will be found to be as follow, viz. the
Plain's Difference of Longitude 86°L 5', the
Subftile's Diſtance from the Meridian (38°
L24) the Height of the Stile 3°L 7.
If
[ 193 ]
If the Meridian Line be introduced on this
Dial, almoſt all the reft of the Hour Lines
will fall fo very near the Subftile, that they
will be of very little Ufe. It is therefore the
beſt Way to bring in fome of the more ufe-
ful Lines at a competent Diſtance, tho' you
lofe one or two, which would difcommode
the Dial in general.
In this Example, let it be propofed to bring
on all the Lines, after one o' Clock.
Let the Plain on which the Dial is to be
drawn, be ABCD. In the horizontal Line
CI IV V VI VII VIR D
2
4
8
,
EXG
A
H
M
V
IV
I IB
AB, take the Point E at Pleaſure towards the
left Hand, becauſe it is a South-west Decliner;
O
and,
[194]
and, with a Line of Chords, make the
Angle FEB of (38° 24') the Subftile's
Diſtance from the Meridian: So will E F be
the Equator. Affume any two Points, G and
H, in it to denote the Place of the extream
Hour Lines in this Equator. From (86° 5)
the Plain's Difference of Longitude, fubtract
15° continually, and you will have the Re-
mainders 71°5', 56°L 5′, 41°L 5', 26°L 5',
11°5', anfwering to the Hours 1, 2, 3, 4,
5. And 11°L5 taken from 15°, leaves
3° 55'; to which add 15° as often as there
is Occafion, and you will have 3°55′, 18°
L55', 33°55', for the Hours of 6, 7, 8,
Make with the Chords, the Angle FG I, of
18° 55', the Complement of 71°L 5', which
reprefents one o' Clock, and the Angle EHI
of 56°L 5', the Complement of 33°55′, an-
fwering to 8 o' Clock. And from I, where
thefe Lines meet, let fall to EF the Perpen-
dicular IK, and it is the Subftile. Make IK
a Radius on the Tangents, and lay the Tan-
gents of 3° 55', 18° 55', 33° 55', from
L
K to 6, 7, 8. Alfo lay the Tangents of
11°L5′, 26°L5', 41°L5', &c. from K to
5, 4, 3, &c.
Lay the Tangent of 3°L7', the Stile's
Height, from K to L, and draw IL. To
this let fall a Perpendicular from K, and on
it lay IK from K to M. And draw M N
parallel to IL; fo will MN be the Stile.
Take any other Point (as O) in the Subftile,
and thro' O draw IR parallel to EF. Make
the neareſt Diſtance from O to the Stile MN,
a Radius on the Tangents. And then lay
the
[ 195 ]
the Tangents of 3° 55', 18° 55', 33°55',
from 0 to 6, 7, 8; and the Tangents of
11°L5′, 26°15', 41°L5', &c. from 0 to
5, 4, 3, &c. Laftly draw right Lines thro'
1, 1; 2, 2; 3, 3; 4, 4, &c. and they will
be the Hour Lines required. The Stile muft
ſtand right over IK; its neareſt Diſtance
from K, muſt be K M equal to IK; and its
neareſt Diſtance to O, muſt be OR, the
Radius ufed in laying down the Divifions
on PQ. In this, as in other Dials, the Halfs
and Quarters might have been brought in by
allowing 7° 30' for half an Hour, and
3°45' for a Quarter.
If the Line KI be too great for the Radius
of the upper Tangents, uſe the lower Tan-
gents, as taught in the XIIth Chapter, at
Sect. XIII.
CHA P. XX.
A bort Account of thofe Lines which
were formerly put on the Sector, but are
1.
now omitted.
0
T
HE Line of Quadrature; it is
known generally by the Letter
Q at the End of the Sector, and
its punch'd Holes mark'd Q, 90,
5, 6; S, 7, 8, 9, 10, downwards.
0 2
From
[ 196 ]
From the Center to the Hole Q, is the
Side of a Square equal to the whole Length
of the Line of Lines. And from the Center
to the Holes 5, 6, 7, 8, 9, 10, give the
Sides of 5 fided, 6 fided, 7 fided, &c. regu-
lar Polygons, each of which contains exactly
as much as the preceding Square. The Di-
ftance from the Center to the Hole S, is the
Radius of a Circle, whofe Area is equal
to the aforefaid Square. And the Diſtance
from the Center to 90, is the Part of the
Periphery of the Circle to the fame Radius.
I
+
The Problems performed on this Line,
are ufually theſe four.
1ft. The Radius (AB) of a Circle being
given; to find the Side of a (Pentagon, or
B
AOMO G
-D
F
Բ
G-
F
other) regular Polygon of equal Area with
the Circle: Make A B a Parallel at S. And
then the Parallel at the Hole numbered (5)
anfwerable to the Number of the Sides of
the Polygon, is (CD) the Side of the Poly-
gon fought.
2dly. Given CD the Side of (a Pentagon,
or) any regular Polygon; to find the Side of
(a Hexagon,
[197]
(a Hexagon, or) any other regular Polygon,
or the Radius of a Circle, having Area's,
each equal to the Area of the Polygon
given. Make (CD) the Side of the given
Polygon, a Parallel at (5) the Number de-
noting its Sides; then will the Parallel at (6
or 8) the Number denoting the Sides of the
Polygon, be (EF, ef) the Side fought;
and the Parallel at S will be (A B) the Ra-
dius fought.
3dly. Given (AB) the Radius of a Circle;
to find the Length of a quadrantal Arch of
that Circle. Make (A B) the given Radius,
a Parallel at S; and then the Parallel at 90
fhall be (GH) the quadrantal Arch fought.
4thly. Given (GH) the Length of a qua-
drantal Arch of a Circle; to find the Radius.
Make (GH) the Length of the quadrantal
Arch, a Parallel at 90; and the Parallel at
S will be (A B) the Radius fought.
II. The Line of infcribed Bodies. It is
known by punch'd Holes mark'd T, O,
C, I, S, D (downward towards the Center)
denoting by their Diſtances from the Center,
the Proportions of the Sides of the 5 regular
Bodies infcribed in a Sphere, whofe Ra-
dius is equal to the Distance from the Center
to S. So T, O, C, I, S, D, denote the Te-
trahedron, the Octohedron, the Cube, the
Ifcofahedron, the Semidiameter of the Sphere,
the Dodecahedron.
The Problems performed on this Line, are
theſe two.
0 3
ift. The
[ 198 ]
If. The Radius (IK) being given; to
find the Side of (an Octohedron, or of) any
of the 5 regular Bodies infcribed in it. Make
IK a Parallel at S; and then a Parallel at
(0) the Letter denoting the regular Body,
will be (LM) the Side of it.
I
L
P
hd hd d
K
Q
M
zaly. The Side (LM) of (an Octohedron,
or) any regular Body being given, to find
the Side of (an Ifcofahedron) any other of
the regular Bodies; or the Semidiameter of
the Sphere, in which they may be both in-
fcribed. Make LM a Parallel at (0) the
Letter denoting the regular Body, whofe Side
is given, and then will the Parallel at (1) the
Letter denoting the regular Body, whofe
Side is fought, be (PQ) that Side fought;
and the Parallel at S will be the Radius of
that Sphere, in which they may be both in-
fcribed.
III The Line of equated Bodies; It is
known by punch'd Holes, mark'd T, O, S,
C, I, D (downward towards the Center) de-
noting by their Diſtances from the Center,
the Proportions of the Sides of the five regu-
lar Bodies, to the Diftance of S from the
Center,
[199]
Center, which is the Diameter of a Sphere,
to which each of theſe Bodies is equal.
The Problems performed on this Line,
are two.
ift. Given (AB) the Diameter of a Sphere;
to find the Side of (an Octahedron, or of)
A
C
F
B
D
F
any one of the five regular Bodies, equal to
the Sphere, whofe Radius is given. Make
(A B) the Diameter of the Sphere, a Parallel
at S, and then the Parallel at (0) the Letter
denoting the regular Body, will be (CD) its
Sice fought: And this regular Body will be
equal to the Sphere, whofe Diameter is
A B.
adly. Given (CD) the Side of (an Octo-
hedron, or) any regular Body; to find the
Side of (an Ifcofahedron, or) any other of
the regular Bodies; or the Diameter of a
Sphere, in fuch Sort, that each of thefe three
fhall be equal to one another. Make CD a
Parallel at (0) the Letter denoting the regu-
lar Body, whofe Side is given; and then
will the Parallel at (1) the Letter denoting
the regular Body, whofe Side is fought,
be (EF) that Side fought, and the Parallel
at S, will be the Diameter of that Sphere,
0 4
to
[200]
to which each of theſe regular Bodies is
equal.
IV. The Line of Segments. It is known
by being divided into 5 unequal Diſtances,
numbered 10, 9, 8, 7, 6, 5, from the End
of the Lines downwards towards the Center;
each of theſe Diſtances are again fubdivided
into other 10.
The principal Problems performed on this
Line, are theſe two.
Ift. Given (GH) the Diameter of a Cir-
cle, and (IK) the Depth of the greater Seg-
ment; to find the Proportion that the whole
G-
I-
K
-H
Circle bears to the greater Segment. Make
(GH) the Diameter, a Parallel at 100, and
carry (IK) the Depth of the greater Seg-
ment, along the Lines 'till it becomes a Pa-
rallel; which will be at 80: to which 100
bears fuch Proportion, as the whole Circle
doth to the greater Segment.
2dly. Given (GH) the Diameter of a
Circle; to cut it fo that the whole Circle
ſhall be to the greater Segment, as 100 to 80.
Make GH a Parallel at 100, and the Paral-
lel at 80 will be IK, the Depth of the
greater Segment.
V. The
[201]
V. The Line of Superficies; From the
Center it is numbered at the grand Divifions
by 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, reaching
as far as the Line of Lines reaches; and each
of theſe are again fubdivided to fuit with the
decimal Order, according to the Length of
the Sector, and may be read thus, 1, 10,
20, 30, 40, 50, 60, 70, 80, 90, 100; or,
100, 1000, 2000, 3000, &c. or, LoI, LI,
L2, L3, L4, L-5, &c.
The principal Problems performed by this
Line, are thefe.
pft. To find the Proportion between
(A, B) two fimilar Superficies. Make on the
AA
Line of Superficies, any Side (CD) of the
greater Superficies (A) a Parallel at 100;
then carry (EF) a like Side of the other
Superficies (B) along the Line of Superficies.
'till it becomes a Parallel, which you will
find to be at 441-4. Therefore A is to B,
as 100 to 444.
edly. A
[202]
2dly. A Superficies (A) being given; to
find (B) another like it; fo that (A) the
former, may be to (B) the latter, in a given
Proportion (viz. as 4 to 9). Make (CD)
any Side of (A) the given Superficies, a Pa-
rallel at 9, the Number anfwering to (A)
that Superficies; and (EF) the Parallel at
(80) the Number anſwering to the fought
Superficies, will be a like Side of that fought
Superficies.
3dly. Two like Superficies (A, B) being
given; to find a third like to them, and equal
to their
{Difference}. Seek, as in the firſt
Example, Numbers repreſenting their Pro-
portion to one another: This you will find to
be as 100 to 444, which is as 9 to 4.
Then the Sum of theſe Numbers are
5372
Diff.
{3}. Laftly, As in the ſecond Example,
213
find a Superficies like to A; fo that A may
be to it, as 25 to{37} and the Superficies
Sum
thus found, will be equal to the {m} of
Diffs
the Superficies.
4thly. Between two given Lines (A and C)
to find a mean Proportional: Find the Num-
bers expreffing their Proportion by meaſuring
them on the Line of Lines, which you will
find to be 75 and 48; then I make, on the
Super-
[203]
Superficies, C a Parallel at 75; and the Pa-
rallel at 48 is the mean Proportional fought.
AC
sthly. To find a mean Proportional be-
tween two given Numbers. Take from the
Line of Lines, two Lines anfwering to the
Numbers; and, by the laft, find a mean
Proportional between theſe Lines; this mean
proportional Line meaſured on the Line of
Lines, gives the mean proportional Number
fought.
6thly. To find the fquare Root of a given
Number. If the given Number confifts of
odd
an even
Number of Places feek its
;
Place on the Superficies between { the ift I
the 2d r
& the 2d 1 }, And the Diſtance of this Place
and 19
from the Center, applied from the Center on
the Line of Lines, will give the fquare Root
fought, which will confift of half as many
Places as the given Number confifts of, if
the Multitude of thofe Places be even; but
if odd, of half as many Flaces as the given
Number increafed by one, confifts of. So the
fquare Root of 25 is 5; of 250 is 15L8
neareft; of 2500 is 50; of 25000
is 158.
This
[204]
This is perform'd without opening the Sector.
Befides theſe already mentioned, there are
others; as three Numbers given; to find a
fourth in a duplicated, or fubduplicated Pro-
portion, &c. But all theſe are better done
by the Line of Numbers.
VI. The Line of Solids; From the Center
it is numbered at the grand Divifions by 1, 1,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10; and theſe are
again fubdivided to fuit with the decimal
Order, according to the Length of the Se-
ctor; and may be read thus, 1, 10, 100,
200, 300, 400, 500, 600, 700, 800, 900,
1000; or, 1000, 10000, 100000, 200000,
300000, 400000, &c.
The principal Problems performed by this
Line, are theſe.
Ift. To find the Proportion between (A,B)
two fimilar Solids.
2dly. A Solid (G) being given; to find
(H) another like it; fo that the former may
be to the latter, in Proportion of two given
Numbers (of 36 to 80).
3dly. Two fimilar Solids (A, B) being
given; to find a third like to them, that
fhall be equal either to the Sum, or the Dif
ference of thofe given.
4thly. Between (A and C) two given
Lines; to find two mean Proportionals.
Sthly. To
[205]
sthly. To find two mean Proportionals
between two given Numbers.
Theſe five Problems are performed exactly
as thoſe in Sect. V. in this Chapter, if you
ufe the Line of Solids, inſtead of the Line of
Superficies. And fo will the following, which
is to extract the Cube Root of a given Num-
ber, provided that you feek the given Num-
Ift. 1 & the 2d. 12
2d. 1 & the 3d. I
3d. 1 & the 10
ber between the
that given Number conſiſts of
when
I or 4 or 7
2 or 5 or 8
3 or 6 or 9
Places; always obferving, that the Root
will confift of
S: 2
235
1 2 3
Places, when the given
S
I or 2 or 3
Number confifts of 4 or 5 or 6 Places.
7 or 8 or 9
Alfo three Numbers being given, a fourth
may be found in a triplicated, or fubtripli-
cated Proportion. Let the three Numbers
given be 4, 8, 9. Make 9 taken from the
Solids, a Parallel at 4 on the Line of Lines,
and then will the Parallel of 8 be, when
meaſured on the Solids, 72, the Anfwer.
Ex. II. Let there be propofed the three
Numbers 27, 125, 6; to find a fourth in a
fubtriplicated Proportion. Make from the
Line of Lines, 6 a Parallel at 27 on the
given
[ 206 J
]
Solids; and then the Parallel at 125, will
give, when meaſured on the Line of Lines,
10, the Anſwer.
VII. The Line of Metals.
It hath 7
punch'd Holes mark'd h♂ 4, de-
noting Gold, Quickfilver, Lead, Silver,
Copper, Iron, Tin. By their Diſtances
from the Center of the Sector, is fhewn the
Proportions of their ſpherical Bodies (or of
their homologous Lines) when they have
equal Weights.
The Problems ufually performed by this
Line, in Conjunction with the Line of Lines,
and the Line of Solids, are thefe.
Ift. To find their fpecifick Gravities; that
is, to find the Proportions that the Weights
of thefe Metals bear to one another, in Bo
dies like, and equal in Magnitude. Suppofe
a Body of Gold weighs so Ounces, I demand
the Weight of a like and equal Body of Sil-
ver. Take from the Solids (50) the given
Weight of the Gold, and make it a Parallel
at the Points belonging to (Silver) the other
Metal; then will the Parallel at the Point
belonging to (Gold) the firſt Metal, mea-
fured on the Solids, give (27L 2) the Weight
fought of the Silver Body. And fo the
Weights of the others will be as follow,
Quickfilver 3547, Lead 303, Copper
2347, Iron 2141, Tin 3445.
2dly. Of two like Bodies of equal Weights,
but of different Metals (Gold and Silver)
(40)
[207]
(46) a Line of the (Gold) one being given;
to find a like Line of (Silver) the other.
Make the given Side of the Body a Parallel
at the Points belonging to his Metal, and the
parallel Diſtance of the other Body's Metal,
meaſured on the Line of Lines, will give the
Line fought, 4643.
3dly. In like Bodies of equal Weight,
(Gold and Silver) having A the Magnitude
of the golden one; to find the Magnitude
of the other of Silver. Take A from the
Line of Solids, and make it a Parallel at the
Point of (Gold) the Metal to which A be-
longs. And then the Parallel at the Point of
Silver, meaſured on the Solids, will give the
Magnitude of the Silver. And fo the Paral
lel at the Lead, will give the Magnitude at
the Lead, &c. So if Bodies were of equal
Weights, and the Magnitude of the Gold be
389, that of the Quickfilver will be 5492
that of Lead 643, that of Silver 716, that of
Copper 822, of Iron 925, and of Tin 1000.
4thly. Of two like Bodies, A and B, of
given different Metals, the Weights of both,
C, D, being given, and a Line E of one of
them; to find a like Line of the other. Firſt
find, by the fecond of this, a Line (F) for a
Body (G) of equal Weight (C) with the firſt,
but of the fame Metal with the fecond.
Then are there known the Weights C and D
of two Bodies, C and B of the fame Metal,
and a certain Line (F) of the former (G)
is given, to find a like Line of the latter (B).
This
[ 208 ]
This may be done by the ſecond Example of
the VIth Section of this Chapter.
VIII. The Tangent Line, Secant Line,
Rumb Line, and Meridian Line, are Lines
properly belonging to the plain Scale, but
are fometimes laid down on the Sector.
The Tangents and Secants are fometimes
uſed in Projections, and fometimes in the
Calculation of Triangles, in Conjunction
with the Lines of Sines and Lines: But thefe,
as hath been declared, are better done by
other Methods. The Line of Rumbs is ufed
in Navigation, in plotting a Traverſe, and
thereby exhibiting to the View, a Reprefen-
tation of the Ship's Way. It generally runs
to 8 Points of the Compafs, where it is equal
to the Chord of 90° of that Circle from
whence it was made; and is no other than
a Line of Chords of the Arches of a Qua-
drant, divided into 8 equal Parts. There
is a fmall Hole punch'd in it at the Beginning,
and another a little beyond 5 Points and
The Diſtance of thefe Holes, is equal to the
Radius of the Circle anfwering to this Line
of Rumbs. Its Ufes are thefe two, viz. The
Meridian being given, to draw a required
Rumb; or, The Meridian and Rumb being
both drawn, to determine what Rumb it is.
Ift. The Meridian AB being given; to
draw from it the Rumb called S WW, that
is the 5th from the South-Wefterly. With
the Radius of the Rumbs, defcribe from C,
the Circle NESW. Let S repreſent the
South,
I
[209]
South, and take from the Line of Rumbs 5
Points, and lay it from S to R, and draw
CR; it is the Rumb required.
B
N
R
E
A
zdly. If the Rumb be drawn, deſcribe, as
before, a Circle, and apply the Diſtance RS
to the Line of Rumbs; it will reach from the
Beginning to the 5th. The Ufe of this may
be fupplied by the Line of Chords, by allow-
ing 11°L15 for every Point. The Ufe of
the Meridian is principally, either to draw a
Mercator's Chart, or to folve the three fol-
lowing Problems.
Ift. Both Latitudes, and the Difference of
Longitude, being given; to find the Courſe.
From the Lizard, in the Latitude of 49°
55′ North, a Ship is bound to Barbadoes, in
the Latitude of 13°L10' North, the Diffe-
rence of Longitude is 53°. Generally, on
the Back of this Meridian Line, runs a Line
of equal Parts, on which the Longitude is
counted; therefore call it the Longitude Line.
Extend the Compaffes on the Meridian Line,
from
P
[210]
from 49° 55', to 13°10', and make this
Extent a parallel Radius on the Tangents.
Then take the Difference of Longitude in
your Compaffes, and apply it on the Tan-
gents 'till it becomes a Parallel, which you
will find to be at 50°: And this is the Ship's
Courſe from the South-weft, which is al-.
moft S WW.
2dly. Given both Latitudes, as before, and
the Courſe S WW; to find the Difference
of Longitude. Extend the Compaffes on the
Meridian Line, from 49°L 55', to 13°L10',
49°55′,
and make this Extent a parallel Tangent of
45°; then the Parallel of 50° on the Tan-
gents, meaſured on the Longitude Line,
gives 53°, the Difference of Longitude.
3dly. Given one Latitude 49° 55' North,
the Courſe South 50° Wefterly, and the
Difference of Longitude 53°; to find the
other Longitude. Make 53°, taken from the
Longitude Line, a parallel Tangent of the
Courſe 50°; and then the parallel Tangent
of 45°, laid on the Meridian Line from 491
55' downwards, will reach to 13°L10', the
Latitude fought.
If the Diſtance taken on the Meridian Line,
be meaſured on the Longitude Line, the
Operation will be performed by the artificial
Numbers and Tangents very elegantly. So
in the firſt Example, the Diſtance on the
Meridian Line will meafure 44°. There-
fore on the Numbers, the Extent from 44° ±
ΟΙ
I
[ 211]
to 53°, will reach on the Tangents from 45°
to 50°, the Courſe.
the Courſe. But, laftly, there is no
Occafion for this Meridian Line in theſe
Days; for the artificial Tangents counted
backwards from 45°, calling every 5 Deg. 10,
is really that Meridian Line: This was firft
demonſtrated by that happy Man Dr. Halley.
But it feems, at firft Sight, that we ftill
want the Longitude Line; but this is thus
ſupplied; Take from the artificial Tangents,
the Diſtance from the Radius to the Tangent
of 22°40′: And this Extent being made a
Parallel on the Line of Lines to 1000, will
reprefent Leagues to meafure the Longitude
in Leagues by; or made a Parallel at 50°,
will reprefent 50° of Longitude. Then the
firſt of theſe three Examples wrought by the
modern Sector, is thus; Make the Extent
from 45°, to 22° 40′ on the artificial Tan-
gents, a Parallel on the Lines at 50°; then
the Diſtance of the two Latitudes (viz. 13°
10' and 49° 55', counting as taught above)
from the Meridian Line (that is the Tan-
gents) meaſured on the Lines, will give 44
Then on the Numbers, the Extent from
44° to 53°, will reach on the Tangents
//
from 45° to 50°, the Courſe.
By this Time I believe it appears that the
Meridian Line was left out with Judgment.
There are three other Cafes in Navigation
where the Courſe, Diſtance, and Difference
of Latitude, are concerned: And this is
called Sailing by the plain Chart. Any two
of theſe being given, the other may be found;
P 2
and
[212]
and that too by the artificial Sines and Num-
bers, at one Opening of the Compaffes; for,
1st. The Extent from the Radius, to the
Cofine of the Complement of the Courſe,
will reach from the Diſtance run down, to
the Difference of Latitude: Or,
2dly. From the Difference of Latitude, up
to the Diſtance. And,
3dly. The Extent from the Diſtance, to
the Difference of Latitude on the Numbers,
will reach from the Radius, to the Sine of
the Complement of the Courfe.
The following is very ready for Practice,
and fufficiently exact in all Caſes, when the
Navigator cafts up his Log every Day,
which he ought not to fail of. The Extent
from the Sine of the Complement of half the
Sum of the Latitudes, to the Radius, will
reach from the Departure (that is, the Sum
of all the Increments of the Eaſtings and
Weftings) to the Difference of Longitude.
From what hath been faid, it is abundantly
evident, that the Sector is fufficient for all
the Calculations neceffary for Navigation,
particularly if you call to mind how eafily
and pleafantly the aftronomical Problems
are folved by it. I mean that of finding the
Amplitude and the Azimuth. For the for-
mer no more than one Opening of the Com-
paffes is neceffary, and the latter but two.
The Extent from the Sine Complement of
the
[213]
the Latitude, to the Radius, reaches from
the Sine of the Declination, to the Sine of
the Amplitude. And if the Co-Altitude, and
Co-Latitude, are two Sides of a ſpherical
Triangle, whofe contained Angle is the Azi-
muth fought. And the third Side is made
of 90°, and the Declination
the
'Sum
Diff.
when the Latitude and Declination are of
Ja different? Kind, you are to work as
the fame S
taught in the 17th Section of the 16th Chap.
where are required no more than two Open-
ings of the Compaffes.
I could have proceeded to fhew the Ufes
of this admirable Inftrument in many more
Particulars; as in Dialling, the Calculation of
the Hour Diſtances, at one Opening of the
Compaffes; the Infcription of the ornamental
Furnitures; and, in Aftronomy, the Conftru-
ction of folar Eclipfes. In Architecture the
Conſtruction of the Catena and Curves, of
the more abftrufe Kind. In Menfuration,
the Management of the moſt difficult Curves
that may be compared with the conickSections:
But theſe, and ſeveral others, muſt, at pre-
fent, be laid afide, I not having Leifure,
nor Room, here to do it. However, Reader,
if you take but as much Pleaſure in practiſing
the Directions herein contained, as I did in
penning them, you will not repent your
reading thefe few Lines. Farewell.
FINI S
HOFTIMADASEDIK
BOOKS Printed for J. WILCOX,
and T. HEATH.
I.
A
STRONOMY; or, The true Syftem of the
Planets demonſtrated. Wherein are fhewn by In-
ftrument, their Anomalies, Heliocentrick and Geocentrick
Places, both in Longitude and Latitude; their Aphelions,
Perihelions, Retrogradations and Elongations, Parallaxes
and Diſtances from the Sun and Earth: With the Method
of computing the Times when Venus and Mercury may be
feen in the Sun's Disk. Alfo the Moon's Phafes, and
Eclipfes of the Luminaries, for any time paft, prefent, or
to come. With proper Cuts to each Planet: By which
any Perfon may, in a few Hours, and with great Eafe,
attain to a perfect Knowledge of the Planetary, or Solar
Syſtem. Likewiſe the Places of the Heavenly Bodies' and
Motion of the Earth are not only fhewn, but plainly and
fuccinctly demonftrated to the meaneft Capacity, by short
and eafy Rules, and New Aftronomical Tables. With the
Places of 130 principal fixed Stars, 33 of which lie in the
Moon's Way: Defign'd as a Help towards difcovering the
Longitude at Sea. To which is prefix'd, An Alphabetical
Catalogue of as many Terms in Aftronomy, as are moſt
ufeful, and neceffary to be underſtood. A Work entirely
New, and in a Method hitherto unattempted. By Charles
Leadbetter, Teacher of the Mathematicks.
2. The Modern Navigator's compleat Tutor: Or, A
Treatife of the whole Art of Navigation in its Theory
and Practice, Curiofity, and Utility. Containing, 1. Dome-
ftick, or Confting Navigation, fundry Notes of the Moon's
Motions, Calender, Tides, ésc. 2. Theoretical Naviga-
tion, by various Methods, viz. Geometrical, Logarithme-
tical, Inftrumental, Tabular, and by the Pen only (without
Tables or Inftruments). Together with Practical Aftro-
nomy, the Deſcription, Conftruction, and Ufe of the Plain
and Mercator's Charts, and other nautical Inftruments,
3. Practical Navigation, the Application of the Theory to
Practice; in keeping a Sea-Journal, the working a Day's
Log at Sea, in all Cafes of Winds and Weather, making
c.
of
Books Printed for J.Wilcox & T. Heath.
or fhort'ning Sail, Trying, Hulling, c. And, the Man-
ner of making Allowance for all known Impediments
whatſoever (with rational Corrections); Illuftrated with
full, proper, and practical Examples of the fame. The
whole more amply and plainly demonftrated, than in any
Treatife of this Kind extant. To which are added, All
neceffary Tables, Nautical and Aftronomical, for the Mari-
ner's Practice, by Infpection. Alſo an Appendix touching
Pleafant and Critical Queftions in Navigation; with divers
other Things, hoth Curious and Uſeful in this Art. By
Joshua Kelley, Mariner, and Teacher of the Mathematicks.
3. The Defcription and Ufe of a Portable Inftrument,
vulgarly known by the Name of Gunter's Quadrant. By
which is perform'd moft Propofitions in Aftronomy; as,
The Altitude, Azimuth, right Aſcenſion and Declination
of the Sun, &c. Alfo his Rifing, and Setting, and Ampli-
tude, together with the Hour of the Day or Night, and
other Conclufions exemplified at large. To which is
added, The Uſe of Napier's Bones in Multiplication, Divi-
fion, and Extraction of Roots: Alſo the Nocturnal, the Ring-
Dyal, and Gunter's Line, in many neceffary and delightful
Conclufions, fitted to the Underſtanding of the meaneſt
Capacities. Collected and digeſted into this portable Vo-
lume for the Ufe of young Practitioners. By William
Leybourn.
4. The Builder's Cheft-Book: Or, A compleat Key to
the Five Orders of Columns in Architecture. Where, by
Way of Dialogue, in Nine Lectures, the Etymology, Chá-
racters, Proportions, Profiles, Ornaments, Meaſures, and
Difpofitions of the Members of their ſeveral Columns and
Entablatures, are diftinctly confider'd and explain'd with
reſpect to the Practice of Palladio. Together with the
Manner of drawing the Geometrical Elevation of the Five
Orders of Columns in Architecture, and to meaſure the
feveral Parts in general. The Whole exemplified by Way
of Dialogue, in a very conciſe and familiar Manner, illu-
ftrated on Seven Copper Plates: Being a neceflary Compa-
nion for Gentlemen, as well as Mafons, Carpenters, Joyn-
ers, Bricklayers, Plaifterers, Painters, &c. and all others
concerned in the feveral Parts of Buildings in general. By
B. Langley, of Twickenham.
5. A
Books Printed for J.Wilcox, T. Heath.
5. A compleat Syftem of Aftronomy. In Two Vò-
lumes. Containing the Deſcription and Ufe of the Sector;
the Laws of Spherick Geometry; the Projection of the Sphere
Orthographically and Stereographically upon the Planes of
the Meridian, Ecliptick, and Horizon; the Doctrine of the
Sphere; and the Eclipfes of the Sun and Moon for Thirty-
feven Years; Together with all the Precepts of Calcula-
tion. Alfo New Tables of the Motions of the Planets,
fix'd Stars, and the firſt Satellite of Jupiter; of Right and
Oblique Afcentions, and of Logiſtical Logarithms. To
the Whole are prefix'd, Aftronomical Definitions for the
Benefit of young Students. By Charles Leadbetter, Teacher
of the Mathematicks.
6. The Practical Surveyor; Shewing, ready and certain
Methods for Meaſuring, Mapping, and Adorning all Sorts
of Lands and Waters, by the ſeveral Inftruments uſed for
this Purpoſe: Particularly, of a New Theodolite, very
convenient to be uſed by thoſe who are refolved to be
Accurate, as well as Expeditious. Together with the Uſe
of the fame Theodolite, in drawing the perfpective Ap-
pearance of a Gentleman's Seat, without Meafuring one
fingle Length, at one ſetting down of the Inftrument, the
Picture having any given Pofition. Alfo, Its Ufe in Le-
velling, Meafuring Timber ftanding; and, by a Sliding
Rule improv'd, all Timbers, Shrubs, c. Extracted from
the Works of the moſt experienc'd Artiſts, by John
Hammond.
7. The Deſcription and Uſe of an Architectonick Sector:
And alfo of the Architectonick Sliding Plates, whereby
Scales of all Sizes are moſt readily and univerfally obtain'd
for Fluting Pillafters and Columns, and Drawing the Geo-
metrical Planes and Uprights, in any of the Five Orders,
according to the given Diameter of a Column.
feveral other Scales, very convenient and ready for the
Practice of the Ingenious Defigners of Building. By
T. Carwitham.
With
IHGFO
OKLM
Ꮒ Ꮛ Ꮅ
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Fig: 29 Page 90
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Fig: 22 Page. 76
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T
H
W
S
G
h
S
E
D
TS R Q P O B
द
H
K
70
60
Seçants-
80
170
11
Fig: 2 Page 1
40
30+
nes.
20
10
Tangent's
Z
R
F 90
E
Fig: 63
IB
Page..
180
181
40
510
40
B
30
310
20
10
1501401:50 120 110 100 Jo
1501401.30 120 110 100 90 80 70 60 50 40 30 20F
20
30
Chords
20 30 40 50 60 70 80
50
60 70
E
80 90
A
I
90
170
+50
10
50
20
20
-20
M
Hours
II H
VI
Page 176
X
L
P
W
69
24
10/10
P
D
N
A
B
C
N
Z
T
A
on the Plain of the Horizon.
Projection on
D
F
D
E
XI
XI
X
D
P
Page 174
I H
XX VI VI VI VI
VỊ
४
H
XII
II III
m
A VI VIII X
Stereographick
M
XI
XX XII
N
Projection
F
1
25
70
60
08
06
20
10
50
440
T
H
PONML K
Fig: 21 Page 73
€
B
a
Numb: 3
Numb: 2
K
I
F
30
20
10
M
О
R
E
10
Page 171
с
40
30
09
IIIIIN
A
VII
HA XI XIX
Crthographick
с
OL
N
NO
H
L
A M
20
30
40
50
гр
&
08
იგ
08
S
V
I
I
IXX XI
VII
•
III II
1
09
05
80
90
60
P
от
20
30
10
Projection
THE UNIVERSITY OF MICHIGAN
GRADUATE LIBRARY
TARK BOOK
DEC 1 7 1974
DATE DUE
1
J
201
DO NOT REMOVE
OR
MUTILATE CARD
UNIVERSITY OF MICHIGAN
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