True distance To find the apparent time at Greenwich. .... 54° 0' 37" Distance at noon, Sept. 27, 54 20 29 first difference 0°19′ 52" P. log. 9571 Distance at .3h 52 34 58 second difference 1 45 31 P. log 2320 September 27.. 0 33 54 P. log 7251 24 0 0 Greenwich time, September 26, 24 33 54; for the astro nomical time at the place of observation being on the 26th, the time at Greenwich must be estimated from the noon of the same day. To find the apparent time at the place of observation. O's R. A. September 27, at noon, Greenwich time 12h 13m 2s + 3m 37s 0 0 4 12 13 6 On September 16, 1823, at noon, we were in latitude 28° 27′ N, and longitude by account 40° W ; at 4h 1m 2s P. M. per watch, the altitude of was 30° 14' and at 7h 40m 2s the distance of 's nearest limb from Antares was 66° 30′ 12′′, the ship having run from noon S WW 5 miles an hour, height of the eye 12 feet, required the longitude when the distance was measured ? The distance run from noon till 4h P. M. is 20 miles, and till 7h 40m 38 miles, and with the course 4 points and these distances, the difference of latitude made from noon till 4h P. M. is 13, and till 7h 40m, 24 miles, consequently the latitude, when the 's altitude was observed for the time, was 28° 14′ N, and when the distance was observed 28° 3′ N. Again, from 4h P. M. till 7h 40m the ship had run SW W 18 miles, whence the departure which she had made in that time is 14 miles nearly, with which, and the middle latitude, nearly 28°, the difference of longitude which she has made is 1m 4s W. To find the error of the watch from O's altitude. Time per watch ... 4h 1m 2s 0 2 O's declination at noon 2°51′58′′N-23' 11" 0 6 27 2 45 31 90 87 14 29 Time per watch when distance was observed .... Watch fast for time at the meridian on which the ship was at 4h P. M. 0 Apparent time at that meridian 7h 40m2s 14 14 ........ Difference of longitude in time made since 4h P. M.... Greenwich time by account when the distance was observed The right ascension of the sun, and the right ascension, declination, &c. of the moon, and the star being reduced to this time, we have O's R. A. 11h 35m Os, )'s R. A. 21h 2m 52s, )'s P. D. 105° 15′, 's R. A. 16h 18m 33s, 's P. D. 116° 2′, )'s horizontal semidiameter 14′ 52", and parallax corrected by Table 14, 54′ 30′′, Sum of auxiliary arc and preceding one 73 31 50 suvers 83458 .... Difference of ditto Parts for " 46 47 3 50 suvers 81147 36 Difference of ditto... Difference of true altitudes True distance .... 66 37 8 To find the apparent time at Greenwich. True distance...... 66°37' 8" +7488 Distance at .... 9h 66 5 2 first difference .... 0°32′ 6" prop. log... .... In the following examples the time at the place of observation is deduced from 's altitude observed with the distance, the longitude being required? In the following examples the time at the place of observation is deduced from X's altitude observed with the distance, the longitude being required? In the following examples the true apparent time at the place of observation is given, and the altitudes for clearing the distance are to be computed, the longitude being required? INVESTIGATION OF THE TRIGONOMETRICAL FORMULÆ Method of finding the latitude from two altitudes of the sun, and the time LET Z or Z' be the zenith, I DCH the horizon, P the pole, A |