Lay a Ruler to Z, the Pole of the Horizon, and to the An- Fig. gular Point, it will cut the Primitive Circle in the Point f, XLIV. To 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; and is the Amplitude of the Sun's Rising or Setting from the North Part of the Horizon at O; and its Complement, 3.1 Deg. 22 Min. is the Quantity of the Arch A, the Sun's Amplitude from A, the East and West Points of the Horizon. Ascs PO : १० d. :: 230.30m.: 31 d. 22 m. 2. By the fame Things given to find the Angle at the Perpendicular PO, whose Measure is the Arch of the Æquinoctial Te, 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 Æquinoctial 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: Whose Complement to 12 Hours, is 7 Hours, 26 Min. the Time of the Sun's Setting. As Radius : The Canon for Calculation is, :: : cs.OPO. 3. By the fame Things Given, to find the Angle at the Bafe POO; whose Measure is an drch of a Great Circle intercepted between the Hour-Circle PS, and the Horizon HOO. Forasmuch as the Sides P, and O, of the Triangle PO, are less than Quadrants, we must measure the Angle POO, by the Angle HS, which is equal to it. To perform which, Lay a Ruler to R, (the Pole of the Hour-Circle PQS) and 0, the Angular Point, and it will cat the Primitive Circle in b; then fet 90 Deg. fromh toa: A Ruler laid from the Pole R, to Fig. a, will cut the Hour-Circle P QS ink: Then lay a Ruler from LXIV. tok, it will cut the Primitive Circle in 1, and the Distance from / 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 O2) at the Time of his Rising or Setting. S. POSPOO. And thus are all the Sides and Angles gled Spherical Triangle PO measured: 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 ZPC, 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 Afternoon. 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 DC, 59 Deg. 53 Min. is the Sun's Altitude at Ten or Two of the Clock, that Day. : To find the Angle Z C P. Fig. This is the Angle of the Sun's Position at the time of the XLIV. Question, with Reference to the Pole and Zenith; and may be thus found.. In regard the Two Sides C Z and CP, comprehending the enquired Angle ZCP, are both of them less than Quadrants, we must make use of the alternate and opposite Angle SCN, which is equal to Z CP. And, whose Measure is the Arch of a Great Circle intercepted between the Meridian or HourCircle SCP, and the Azimuth-Circle ZCN, at 90 Deg. Distance 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 PBS, to C, the Angular Point, and it will cut the Primitive Circle ino, set go Deg. from o top: Then a Ruler laid from T top, will cross the Circle PBS, in the Point Y. -Lastly, A Ruler laid from C, to X and Y, will cut the Primitive Circle in rand s, so the Distance between r and s, meafured upon the Scale of Chords, will give 49 Deg. 46 Min. for the Quantity of the Angle YCX, which is equal to the Angle ZCP enquired. And is the Angle of the Sun's Position at the time of the Question. 30 d. 7 m. : The Canon for Calculation. :: S. ZPC : To find the Vertical Angle CZ P. This Angle is the Sun's Azimuth from the North Part of the Meridian ZON, whose 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, so the Quantity of the Arch d NO, 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 East and Weft. 1 The And thus are all the Sides and Angles of this Oblique-angled Triangle also Measured : And fo may any other, being thus Projected. And, to conclude, This Projetive Way will give great Light to Calculation; for by the true delineating of your Triangle upon the Projection, you shall thereby discover whether your Side or Angle be More or Less than a Quadrant, and so be positive in your Resolution; which otherwise you must have Given, or render a double Solution: As in this last Angle CZ P, whether it were 66 Deg. 3 Min. Or 113 Deg 57 Min. The fame Sine in the Canon answering to both. JANUA |