BOOK VI. APPLICATIONS OF SPHERICAL TRIGONOMETRY TO ASTRONOMY AND GEOGRAPHY. 187. The CELESTIAL SPHERE is the spherical concave surrounding the earth, in which all the heavenly bodies appear to be situated. 188. The ZENITH is that pole of the horizon which is directly overhead. 189. The ALTITUDE of a heavenly body is its distance above the horizon, measured on the arc of a great circle passing through that body and the zenith. 190. The DECLINATION of a heavenly body is its distance north or south of the celestial equator, measured on a meridian. 191. The altitude of the celestial pole is equal to the latitude of the place where the observer is located. For the distance from the zenith to the celestial equator is the latitude of the place, and the distance from the zenith to the pole is its complement; but the distance from the zenith to the pole is also the complement of the altitude of the pole; hence the latitude of the place and the altitude of the pole are equal. 192. To find the time of the RISING AND SETTING OF THE SUN at any place, the sun's declination and the latitude of the place being given. Let P represent the celestial north pole, EAQ the celestial equator, HAO the rational horizon, S the place of the sun's rising, S' the posi E P S' S H A 0 B 9 f tion of the sun at 6 o'clock, PEP' the meridian of the given place, PBP the meridian passing through S, and PAP the meridian 90° distant from PEP', passing through S. From the time of the sun's rising to 6 o'clock, it will pass over SS, the arc of a small circle, corresponding to B A, the arc of a great circle. The length of BA, expressed in time (Art. 147), will then give the amount to be taken from or added to 6 o'clock, to give the time of the sun's rising or setting. BS is the sun's declination, PO is the latitude of the place (Art. 191), and QO, which measures the angle BAS, is its complement; hence, in the right-angled spherical triangle ABS, there are known the side BS and the angle BAS, from which, by Art. 175, sin BA = tan BS cot BAS, or, log sin B A = log tan sun's decl. + log tan lat. of place. After reducing the arc BA to time, at the rate of 15° to an hour, or 4m. to a degree, it must be added to 6 o'clock for the time of the sun's setting, and subtracted for its rising, when the declination and latitude are both north or both south; but subtracted for its setting and added for its rising, when one is north and the other south. The preceding reasoning rests upon the assumption that the sun's declination does not change between sunrise and sunset, which, although not strictly true, is accurate enough for our present purpose. The time obtained is apparent time, and a correction must be applied if we wish to find mean time, or that indicated by the clock. Another correction is necessary for refraction. Neither of these corrections has, however, been applied to the answers that follow. EXAMPLES. 1. Required the time of the sun's rising and setting in Edinburgh, latitude 55°57′ N., when the sun's declination is 23°28′ N. Ans. Rises, 3h. 20m. 7s.; sets, 8h. 39m. 53s. 2. What is the time of the sun's rising and setting in latitude 60° 3′ N., when the sun's declination is 23° 28′ S.? Ans. Rises, 9h. 15m. 33s.; sets, 2h. 44m. 27s. 3. Required the time of the sun's rising and setting in places whose latitude is 48° S., when the sun's declination is 15° S. 193. To find the HOUR OF THE DAY at any place, the latitude of the place and the sun's declination and altitude being given. Let. Z represent the zenith, and ZD an arc of a great circle drawn through the zenith and the sun's place, S; PB P', a meridian drawn through the sun's place, &c., as in H the last article. Z E P S A 0 D P As before, the arc A B, added to or subtracted from 6 o'clock, will give the time when the sun is at S; but it will be more convenient to use its complement, E B, which is the time before or after 12 o'clock. EZ is the latitude of the place, and PZ is its complement; BS is the sun's declination, and PS is its complement; SD is the altitude of the sun, and ZS is its complement; hence, in the spherical triangle ZPS, the three sides are known, and the angle ZP S, or the arc E B, may be found by Art. 185. If either the sun's declination, or the latitude of the place, is south, it must be considered negative in taking its complement. unless the south pole is taken as a vertex of the triangle, when north will be negative. EXAMPLES. 1. Required the apparent time of day in the morning, at a place in latitude 39° 54′ N., the sun's declination being 17° 29′ N., and its corrected altitude 15° 54′. Ans. 6h. 25m. 30s. A. М. 2. In latitude 36° 39′ S., when the sun's declination was 9° 27′ N., its corrected altitude was observed, in the afternoon, to be 10° 40′; what was the apparent solar time? Ans. 4h. 36m. 10s. P. М. 3. Required the apparent time of day in Boston, latitude 42° 21′ N., when the sun's declination is 20° S., and its corrected altitude 15° 15', the sun being east of the meridian. 194. To find the SHORTEST DISTANCE between two places on the earth's surface, and the BEARING of one from the other, their latitudes and longitudes being given. Z P E S B A Let Z and S represent the two points on the earth's surface, and P the north pole of the earth. PZ and PS are the complements of the latitudes of the two places, and the arc EB, or the angle ZPS, is the difference of their longitudes; hence, in the spherical triangle SPZ, the two sides P PZ and PS, and their included an gle P, are known, from which the side ZS, and the angles ZSP and SZP may be found by Art. 183. The distance ZS can easily be reduced to miles by allowing 69.16 statute miles, or 60 nautical miles, to a degree. The answers which follow are given in statute miles. If one place is south and the other north of the equator, the south latitude must be considered negative in taking its comple ment. EXAMPLES. 1. What is the distance and bearing of Jerusalem, lat. 31° 47′ N., long. 35° 20′ E., from London, lat. 51° 30′ N., long. 6' W.? Ans. Distance, 2248 miles; bearing, S. 66° 31′ Ε. 2. Required the distance and bearing of Cape Horn, lat. 55° 58′ S., long. 67° 21′ W., from London. Ans. Distance, 8363 miles; bearing, S. 36° 59′ W. 3. Required the distance and bearing of Quito, lat. 0°, long. 78° 45′ W., from San Francisco, lat. 37° 49′ N., long. 122° 14′ W. A TABLE, CONTAINING THE LOGARITHMS OF NUMBERS FROM 1 TΟ 10,000. Numbers from 1 to 100 and their Logarithms, with their Indices. No. Log. No. Log. No. Log. No. Log. No. Log. 1 0.000000 21 1.322219 41 1.612784 61 1.785330 81 1.908485 2 0.301030 22 1.342423 42 1.623249 62 1.792392 82 1.913814 3 0.477121 23 1.361728 43 1.633468 63 1.799341 83 1.919078 4 0.602060 24 1.380211 44 1.643453 64 1.806180 84 1.924279 1.397940 45 1.653213 65 5 0.698970 25 1.812913 85 1.929419 11 1.041393 31 1.491362 51 1.707570 71 12 1.079181 32 1.505150 52 1.716003 72 13 1.113943 33 1.518514 53 1.724276 73 14 1.146128 34 1.531479 54 1.732394 74 1.869232 94 1.973128 1.851258 91 1.959041 1.857332 92 1.963788 1.863323 93 1.968483 1.995635 2.000000 191.278754 39 1.591065 59 1.770852 79 1.897627 99 20 1.301030 40 1.602060 60 1.778151 80 1.903090 100 NOTE. - In the following part of the Table the Indices are omitted, as they can be very easily supplied by the directions given in the Author's NEW HIGHER ALGEBRA, p. 315, or Book I. of his TRIGONOMETRY |