To find the apparent time at Greenwich. True distance 540 0' 37" . . 3h 52 34 58 second difference 1 45 31 P. log 2320 24 0 0 To find the apparent time at the place of observation. 0 0 4 O's reduced R. A...... 12 13 6 + *'s R. A. Jan. 1, 1820, 9h 58m 47s + 3.21s Declination.... 12° 50'36" N - 17.23s 3.21s x 3 years = .. 12 1 4 *'s reduced R. A. 9 58 59 *'s reduced dec. 12 49 32 N 90 0 0 *'s polar dist. 102 49 32 *'s true altitude.. 10°31' 37" Latitude 28 37 0 sec 056583 *'s polar distance 102 49 32 cosec .010973 Sum..... 2)141 58 9 Half sum 70 59 4 cos 9.512985 Remainder 60 27 27 sin 9.939507 2)19.520048 Half hour angle 35° 8' sin 9.760024 8 *'s meridian dist. 4h41m 4s *'s R. A.. 9 58 59 R. A. meridian.... 5 17 55 O's R. A.... 12 13 6 17 4 49 apparent time at the place of obser on. 24 33 54 apparent time at Greenwich. Difference 7 295, longitude in time W, 112° 164 W. 1 EXAMPLE III. 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 Q was 30° 14' -, and at “h 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 difference of latitude made from noon till 4h P. M. is 13, and till 7h 40m, 24 miles, consequently the latitude, when the O'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 SW1W 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. 90 2 .. Time per watch when distance was observed ..... 7h 40m2 s Watch fast for time at the meridian on which the ship was at 4h P. M. 0 14 14 Apparent time at that meridian 7 25 48 Difference of longitude in time made since 4h P. M.... 0 14 Apparent time at the meridian on which the distance was taken 7 24 44 Longitude by account in time 2 40 0 Greenwich time by account when the distance was observed 10 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", 4 Apparent time.... 7h 24m 44s 18h 59m 44s 16 18 33 *'s R. A. R. A. mer... 18 59 44 *'s mer, distance 2 41 11 21 2 52 40° 18' To compute the moon's altitude. Latitude.... 28° 3' cot •273412 sin 9•672321 58 12 tan 10.207460 sec .278226 cos 9.833377 )'s hor. semid... 14' 52" D's true altitude. 37° 27' 0" sin 9.783924 Aug....... 0 0 9 Correction nearly 42 0 )'s true semidiam. 0 15 1 Apparent altitude nearly . 36 45 0 ) 's N. L. from * 66 30 12 True correction.. 42 24 App. central dist. 66 45 13 Apparent altitude 36 44 36 Difference of apparent altitudes 13° 14' 0" 60 17 50 Apparent distance.. 66 45 13 Sum of auxiliary arc and preceding one 73 31 50 suvers 83458 Difference of ditto 47 3 50 suvers 81147 Sum of arc first and following one 127 3 3 vers 02512 Difference of ditto .. 6 27 23 vers 06330 Difference of true altitudes 13 58 35 vers 29564 03011 Sum of parts for " 146 Vers true distance 03157 66°37' 011 03119 8 38 True distance .... 66 37 8 Parts for 46 36 12 12 40 vers true distance, To find the apparent time at Greenwich. 7488 True distance...... 66°37' 8". 1h 3m5ls prop. log 9 0 0 Apparent time at Greenwich 10 3 51 Apparent time at ship .... 7 24 44 Longitude in time 2 39 7 39° 463! W EXAMPLES FOR EXERCISE. In the following examples the time at the place of observation is deduced from O's altitude observed with the distance, the longitude being required ? 2 32 35 2 32 462 78 5 25 45 13 26+ 15 24+ 37 16+ 23 36+ 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 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 elapsed between the observations. (See p. 230.) LET Z or Z' be the zenith, ID CH the horizon, P the pole, A and B the planes of the sun at the two times of observation, A D and B C the true altitudes, A Z, BZ, or A Z, BZ' the zenith distances, A Pand B P the polar distances, which, in the practice of this problem, may be considered as equal. Join A B, and bisect it by the |