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Fig. to %, it will cross the Axis of the World P AS, in the Point F; XLIV. so have you Three Points, o F %, through which you may draw.

the Circle s Fa; which will cut the Horizon HA O, in the Point O, the place where the Sun will rise that Day.

Fifıbly, This Point o being found, you have Three other Points given, viz. P, O and S; through which you may describe the Circle POS which is the Hour at which the Sun Riseth. And thus have you Projekted, fo far of the Problem, as

concerns the Time of the Sun's Rising. Now for his Place of being at Ten in the Morning, or 'Two in the

Afternoon. Sixthly, Tun in the Morning, or Two in the Afternoon, are (either of them) Two Hours, or 30 D.g. distant from the Meridian : Wherefore, take 30 Dag. from your Scale of Chords, and let them upon the Meridian, from A to G, then a Ruler laid from P to G, will cross the. Æquino&tial Circle # A & in the Point B; so have you Three Points, S, B and P, by which you may Project the Circle SB P, for the Hour Circle of Ten in the Morning, or Two in the Afternoon : And the.. Circle S BP, will cut the Tropick of Cancer % A % (the Line which the Sun traces that Day,) in the PointC; in which point the . Sun will be at Ten in the Morning, or at Two in the Afternoon. .

Seventhly, And now have you Three other Points, N O and Z, through which you may draw the Azimuih Circle ZCN, for the dzimuth, that the Sun will be upon at Ten and Two a Clock. III. Concerning the Spherical Triangles that are made by the Inter- . fe&tions of these Great Circles thus proje&ted.

Many are the Triangles, that are made by the Intersed ions of ihese few Circles upon this Proje{tion, but I shall exemplifie only in Two of them; namely, the Triangle P OʻO, Right-angled at 0, which is composed of Three Arches of Great Circles of the Sphere, viz. (1.) OF 0. O, an Arch of the Horizon, comprehended between 0, the North Point of the Horizon, and o, the Point in the Horizon, whereon the Sun Riseth that Day, or his Amplitude from the Norib. (2.) OF F 0; an Arch of a Meridian or Hour-Circle, passing through P, the North Pole of the World ; ard 0, the Place of the Sun's Rising that Day, equal to 66 Deg. 30 Min. the Com

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plement of the Sun's Declination that Day. And, (3.) OF PO,

(Fig. an Arch of the Meridian of the Place, comprehended between XLIV. O, the North Point of the Horizon, and P, the North Pole, equal to the Latitude given, 40 Deg.

The other Triangle shall be the Oblique-angled Triangle Z CP, constituted by the Interfe&tion of Three other Great Circles of the Sphere ; namely, Of (1.) Z P, an Arch of the Meridian of the Place, comprehended between Z the Zenith, and P, the North Pole, equal to 50 Deg. the Complement of the Latitude of the Place. (2.) Of P C, an Arch of a Meridian, or HourÇucle of Ten or Two a Clock, equal to 66 Deg. 30 Min. the Complement of the Sun's Declination; or, the Sun's Di. stance from the elevated Pole P. And (3.) OF Z C, an Arch of an Azimuth or Vertical Circle, passing through the Hour-Circle SC P, in the Point P, where the Sun is at Ten or Two a Clock that Day, and is the Complement of the Sun's Alsitude at that time. IV. To find the Poles of these Great Circles, they being thus :

Proješted.

The Pole of any Great Circle of the Sphere, is, always, 90 Deg. distant in all Places, from the same Circle, upon a Right Line, or Circle, which cuts it at Right Angles: So, 1. The Pole of the General Meridian, or Primitive Circle

ZHN O, is A, the Center thereof, in all Places distano

90 Deg. 2. The Poles of the Horizon H 0, are Z and N, the Zenith

and Nadir Points. 3. The Poles of the Æquinoctial FE 2, are P and S, the Polos :

of the World. 4. The Poles of the Aquinoctial Colure, or Axis of the World

PS, are Æ and . s. The Poles of the Prime Vertical Circle, or Azimuth of East

and West, ZAN, are H and O), the North and South Points

of the Horizon. And thus you see, that all Great Circles of the Sphere, which, when Projetted, become Strait Lines, their Poles fall all in the Periferie of the Primitive Circle: But for all other Great Circles, which consist of Circular Arches, as POS, PB S, and Z DN, the Poles of them fall within the Primitive Circle Ź HNO, upon those Grear Circles, which cut them at Right Angls: S.,

Ri

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Fig. XLIV.

Z DNY

che Line.

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T.
V.

R? The Pole of the SP will be upon S&A at the

R.
Great Circle.

RZ
Which Points or Poles may be thus found.
1. For the Pole of the Hour-Circle POQS, lay a Ruler

from P to R, it will cut the Primitive Circle in a, fet 50 Deg. of your Chords, from a to b; a Ruler laid from P to b, will cut the Aquinołtial Circle in R, fo is R the Pale of

the Circle POS 2. For the Pole of the Hour-Circle P B S, lay a Ruler from

P to B, it will cut the Primitive Circle in G, set 90 Deg. from G toc, a Ruler_laid from P to c, will cut the Æquinoftial in T, fo is T the Pole of the Hour Circle of 10

and 2, viz. PBS. 3. For the Pole of the Azimuth, or Vertical Circle Z DN,

lay a Ruler from Z to D, it will cut the Primitive Circle in d, fet 90 Deg. from déto e, a Ruler laid from Z to e, will cut the Horizon H A O in V, fo is V the Pole of the

Azimuth Circle Z DN. And in this manner may the Pole of any Great Circle be found. For the Centers of them, ihey always fall in the same Right Line, in which their Poles are, they being extended, if Need be: But no more of this in this place : For I intend not here a Treatise of Projeftion, but of the Meofuring of Triangles, to which i

I now proceed. V. How to Measure the sides and Angles of Spherical Triangles,

they being ihus Projected.

You must understand. That the Quantities of the sides and Angles of Spherical Triangles are measured upon the Periferie of the Primitive Circle.

Then, 1. In the Right angled Spherical Triargie Poo, Right-an

gled at 0, there is given, (1.) The Perpendicular PO 40 Deg. equal to the Latitude given by the Proposition. (2.) The Hypotcnuse Po, the Complement of the Sun's Déclination 66 Deg. 30 Min. And let it be required to

to find the B.ise ou, whose Measure is 0 (), the Sun's Amplitude.

Lay

:

sod.

:

go d.

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Lay a Ruler to Z, the Pole of the Horizon, and to the An- Fig. gular Point o, it will cut the Primitive Circle in the Point f, XLIV. fo the Distance from f to 0, measured upon the Scale of Chords, will be found 10 contain 58 Deg. 38 Min. and that is the Quantity of the Side 0 0; and is the Amplitude of the Sun's Rising or Setting from the Norih Part of the Horizon at (); and its Complement, 3.1 Deg.: 22 Min. is the Quantity of the dich A O, the Sun's Amplitude from A, the East and West Points of the Horizon.

The Canon for Calculation is,
As cs PO : Radius : : стро

CsOO.

: : 23 d. 30 m. : 31 d. 22 m. 2. By the same Things given to find the Angle at the Per

pendicular O PO, whose Measure is the Arch of the Aqui

noctial T &, or the Hour from Midnight,
Lay a Ruler to P, the Pole of the World, and T, it will cut
the Primitive Circle in a, and the Distance a é measured upon the
Scale of Chords, will be found to be 68 Deg. 36 Min. And
that is the Quantity of the Angle O PO: Which 58 Deg. 36
Mir. converted into Time, (by allowing 15 Deg. of the Æqui-
noctial for One Hour, and 4 Deg. for One Minute of Time) makes
4 Hours, and 34 Min. Which is the Hour from Midnight, or
the Time of the Sun's Rising in the Morning: Whofe Comple-
ment to r2 Hours, is 7 Hours, 26 Min. the Time of the Sun's
Setting

The Canon for Calculation is,
As Radius t. PV : : ct.Po

::

:

cs, OPO. god.

23 d. 30 m. 2id. 24 m.
3. By the fame Things Given, to find the Angle at the Base

POO; whose Measure is an drch of a Great Circle inter-
cepted between the Hour-Circle Po S, and the Horizon

HOO.
Forasmuch as the sides o P, and o 0, of the Triangle
POV, are less than Quadrants, we must measure the Angle
POO, by the Angle HOS, which is equal to it. To perform
which,

Lay a Ruler to R, (the Pole of the Hour-Circle P QS) and o; the Angular Point, and it will cot the Primitive Circle in bi then set go Deg. from h to a: A Ruler laid from the Pole R, tó

:

40 d.

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2

:

Fig.

a, will cut the Hour-Circle P Q S in k: Then lay a Ruler from LXIV. Ó to k, it will cut the Primitive Circle in l, and the Distance

from 7 to H, measured upon the Scale of Chords, will be found to contain 44 Deg. 30 Min. for the Quantity of the Angle HOS, which is equal to the Angle Po required: And this

is the Angle of the Sun's Position, (in respect of the Pole P, and North Part of the Horizon 0,) at the Time of his Rising or Setting.

The Canon for Calculation is
As s. Po ; Radius :: s. PO : SPOO.
660.30 m. :
so d.

40 d. 47d. 30 m.
And thus are all the sides and Angles of the Right-an.
gled Spherical Triangle Po O measured: And the like may
be done in any other Right-angled Triangle ; of which, in this
- Scheme there are many.
VI. And now for the Oblique-angled Spherical Triangle Z CP.

In this Triangle there is given, (1.) The Side Z P, 50 Deg. equal to the Complement of the Latitude given. (2.) The Side CP, 66 Deg. 33 Min. equal to the Complement of the Sun's Declination given. And, (3.) The Angle Z PC, 30 Deg. the Horary Distance of the Sun from the South Part of the Meridian; namely, at Ten or Two of the Clock. And from these Things given, let it be required to find the third Side Z C, the Complement of the Sun's Altitude at the Time of the

Question, viz. at Ten in the Morning, or at Two in the After. noon. Which, to perform,

Lay a Ruler to V, the Pole of the Azimuth Circle Z DN, and to the Angular Point C, it will cut the Primitive Circle in m, and the Distance from Z to m, by the Scale of Chords, will be found to 30 Deg. 7 Min. whose Complement D C , 59 Deg. 53 Min. is the Sun's Altitude at Ten or Two of the Clock, that Day.

The Canons for Calculation is (1.) As c t. PZ : Radius

:: cs. P

cs. P: PA.
60 d.:

45 d. 56 m. Which substra£ted from PC, 66 d. 30 m. there remains

20 Deg. 34 Min.
(2.) As c s. PA : SC. ZP :: cs. Rem. :

cs. Z c.
44d.4 m.
:: 69 d. 26 m. : 598.53 m.

To

40 d.

:

90 d.

::

:

40 d.

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