Which Points or Poles may be thus found. Point 1. For the Pole of the Hour-Circle PO QS, lay a Ruler from P to Q, it will cut the Primitive Circle in a, fet so Deg. of your Chords, from a to b; a Ruler laid from P to b, will cut the Equinoctial Circle in R, fo is R the Pole of the Circle POQS. 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 to c, a Ruler laid from P to c, will cut the Equinoctial 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 d to 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 D N. And in this manner may the Pole of any Great Circle be found. For the Centers of them, they always fall in the fame 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 Treatife of Projection, but of the Meofuring of Triangles, to which I now proceed. V. How to Measure the Sides and Angles of Spherical Triangles, they being thus 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 Triargle PO, Right-angled at O, there is given, (1.) The Perpendicular PO 40 Deg. equal to the Latitude given by the Propofition. (2.) The Hypotenuse PO, the Complement of the Sun's Declination 66 Deg. 30 Min. And let it be required to to find the Bafe O, whofe Measure is O, the Sun's Amplitude. Lay Fig. Lay a Ruler to Z, the Pole of the Horizon, and to the Angular Point, it will cut the Primitive Circle in the Point f, XLIV. fo the Distance from f to O, measured upon the Scale of Chords, will be found to contain 58 Deg. 38 Min. and that is the Quantity of the Side OO; and is the Amplitude of the Sun's Rifing or Setting from the North Part of the Horizon at 0; and its Complement, 31 Deg. 22 Min. is the Quantity of the Arch A, the Sun's Amplitude from A, the East and Weft Points. of the Horizon. 2. By the fame Things given to find the Angle at the Perpendicular PO, whofe Measure is the Arch of the Equinoctial Ta, 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 measured upon the Scale of Chords, will be found to be 68 Deg. 36 Min. And that is the Quantity of the Angle PO: Which 68 Deg. 36 Min. converted into Time, (by allowing 15 Deg. of the Equinoctial 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 Rifing in the Morning: Whofe Complement to 12 Hours, is 7 Hours, 26 Min. the Time of the Sun's Setting. As Radius The Canon for Calculation is, 3. By the fame Things Given, to find the Angle at the Base POO; whofe Measure is an Arch of a Great Circle intercepted between the Hour-Circle PS, and the Horizon HOO. Forafmuch as the Sides P, and O, of the Triangle POO, are less than Quadrants, we must measure the Angle POO, by the Angle HO S, which is equal to it. To perform which, Lay a Ruler to R, (the Pole of the Hour-Circle PQS) and O, the Angular Point, and it will cat the Primitive Circle in b; then fet go Deg. from b to a: A Ruler laid from the Pole R, to a, Fig. will cut the Hour-Circle P QS in k: Then lay a Ruler from LXIV. O to k, it will cut the Primitive Circle in, 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 POO required: And this is the Angle of the Sun's Pofition, (in refpect of the Pole P, and North Part of the Horizon O,) at the Time of his Rifing or Setting. As s. Po The Canon for Calculation is Radius SPOO. :: s. PO : go d. :: 40 d. : 44 d. 30 m. And thus are all the Sides and Angles of the Right-angled Spherical Triangle PO 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 C P. 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 P C, 30 Deg. the Horary Diftance of the Sun from the South Part of the Meridian, namely, at Ten or Two of the Clock. And from thefe 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 Queftion, viz. at Ten in the Morning, or at Two in the Afternoon. Which, to perform, Lay a Ruler to V, the Pole of the Azimuth Circle Z D N, 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. whofe 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 To find the Angle ZC P. Fig. This is the Angle of the Sun's Pofition at the time of the XLIV. Queftion, with Reference to the Pole and Zenith; and may be thus found. In regard the Two Sides C Z and C P, comprehending the enquired Angle ZCP, are both of them lefs than Quadrants, we must make ufe of the alternate and oppofite Angle SC N, which is equal to Z CP. And, whofe Measure is the Arch of a Great Circle intercepted between the Meridian or HourCircle SCP, and the Azimuth-Circle ZC N, at 90 Deg. Diftance from the Angular Point C: And to find the Quantity hereof, you must, Lay a Ruler to V, the Pole of the Circle ZCDN, and the Point C, it will cut the Primitive Circle in m, fet 90 Deg. from m, and it will reach ton: A Ruler laid from V to n, will cross the Circle ZDN in the Point X. Again, Lay a Ruler from T, the Pole of the Circle P B S, to C, the Angular Point, and it will cut the Primitive Circle in o, fet 90 Deg. from o top: Then a Ruler laid from T top, will cross the Circle PBS in the Point Y. -Laftly, A Ruler laid from C, to X and Y, will cut the Primitive Circle in r and s, fo the Distance between r and s, measured upon the Scale of Chords, will give 49 Deg. 46 Min. for the Quantity of the Angle YC X, which is equal to the Angle ZC P enquired. And is the Angle of the Sun's Pofition at the time of the Queftion. The Canon for Calculation.. s. ZPC:: s. Ꮓ Ꮲ 30 d. As s. ZC : s. 30 d. 7 m. : :: 50 d. : s. ZC P. : 49d. 45 m. To find the Vertical Angle CZ P. This Angle is the Sun's Azimuth from the North Part of the Meridian ZON, whofe Measure is the Arch of the Horizon DAO; and to find the Quantity of it, lay a Ruler from Z, the Pole of the Horizon, to D, it will cut the Primitive Circle in d; fo the Quantity of the Arch d N O, measured upon the Scale of Chords, will be found to be 113 Deg. 57 Min. And fuch is the Sun's Azimuth from the North Part of the Meridian: The Arch d H, 66 Deg. 3 Min. is the Sun's Azimuth from the South; and the Arch d N, 23 Deg. 57 Min. is the Sun's Azimuth from the Eaft and Weft. And thus are all the Sides and Angles of this Oblique-angled Triangle alfo Measured: And fo may any other, being thus Projected. And, to conclude, This Projedive Way will give great Light to Calculation, for by the true delineating of your Triangle upon the Projection, you fhall thereby difcover whether your Side or Angle be More or Less than a Quadrant, and fo be pofitive in your Refolution; which otherwife you must have Given, or render a double Solution: As in this laft Angle C Z P, whether it were 66 Deg. 3 Min. Or 113 Deg 57 Min. The fame Sine in the Canon answering to both. JANUA |