Given the true altitude of the Sun 28° 16', the latitude 66° 12′ S., and the polar distance 81°, to find the true azimuth, the time of observation being A.M. True azimuth N 38. 50 E., as the altitude is increasing. Problem VI. 2. WHEN THE ALTITUDE IS NOT TAKEN. In this case (see last fig.) SP, PZ, and angle P are given, from which, by Napier's Analogies, we find the angle Z. Given, the Sun's hour angle 64° 35', the declination 2° S., and the latitude 64° 10′ S., to find the azimuth. Tan (Z+S) = cos(p—l') sec (p + l') cot P 2 Tan † (Z — S) = sin † (p − l') cosec † (p+1') cot Р 2 It increases towards the north from March till June, decreases from June till September; increases southward from September till December, and decreases from December till March. On the days of the equinoxes, the Sun travels on the equator, and travels on a parallel of north declination from March till September, and again till March on a parallel of south declination. It is evident, therefore, that the higher his declination, the greater will be his amplitude north or south. This will be shown from the following formula. In the figure, let Z be the zenith; P, the pole ; AB, the horizon; W, the east or west point. Then, if the object be at E when rising, EW is its amplitude. And in the right-angled triangle EBP, by Napier's rules :cos EB cos PB; Cos EP sin EW. cos PB; and sin EW = cos EP. sec PB, i.e., sin amp. = sin dec. sec lat. Now, since the latitude may be considered as constant, sin amp. varies as sin dec., or the amplitude varies as the declination. Example. In latitude 25°. 30' S, if the declination of the Sun be 24° 20′ N, required the rising amplitude. The amplitude is reckoned contrary to the latitude, and E since the object is rising, W CHAPTER XI. Problem VIII. OF FINDING THE LATITUDE BY IVORY'S METHOD. Two observations are taken, either both a.m. or p.m., or one a.m. and the other p.m. In the figure, the two positions of the object are A and B, both on the same side of the meridian NS. ZL and EZ K are circles of altitude passing through the two positions of the object. Then AL and BK are the altitudes; ZA and ZB, the zenith distances; PA and PB, the polar distances; and the angle BPA mea sures the elapsed time between the two observations. Bisect the angle BPA by PC. Then, if the polar distances PA and PB are supposed equal, AB will be bisected in C. From Z, draw ZD perpendicular to PC. Let the equal polar distances = p, the angle APC = h, half the sum and half the difference of the altitudes = S and D respectively. Let the arcs AC, CP, ZD, CD, and DP be represented by (A), (E), (I), (O), and (E — O). 1. In the triangle APC by Napier's rules. Sin AC sin AP sin APC, or, sin (A) = sin p sin h. 2. Cos AP cos AC. cos CP. Therefore cos CP= cos AP sec AC, or cos E = cos p . sec (A). 3. Since PCA and PCB are right angles, sin ZCD. Cos ZCA sin ZCD and cos ZCB = = Sin AK=cos ZA=cos ZC. cos CA+ sin ZC . sin CA sin ZCD. Sin BL=cos ZB=cos ZC. cos CB-sin ZC. sin CB sin ZCD. Therefore, by subtraction, observing that CB CA, or, But sin ZC sin ZCD = sin ZD. Therefore sin AK · sin BL = 2 sin CA sin ZD, 2 cos(AK + BL) sin † (AK – BL) 2 sin CA = sin ZD. or, sin ZD = cos S. sin D cosec A. Again, by addition, Sin AK + sin BL = 2 cos ZC. cos CA. Therefore sin AK + sin BL = 2 cos ZD cos DC cos CA. or, sin S. cos D . sec (A) sec (I) = cos (O). Again, PD = PC — CD = (E — O), and cos PZ = cos PD cos DZ ; i.e., sin lat. = cos I . cos (E-0). Example. On March 9th two altitudes of the Sun were taken in the southern hemisphere 21°. 47'. 5" and 38°. 42'. 38". The true declination was 4°. 13'. 51′′ S., and the time elapsed between the observations was 3h. 45m. 8s.; required the latitude. |