PROBLEM V. Given sidereal time at any instant, to find mean time at the same instant. Let PQ represent the celestial meridian, AQ the celestial equator, A the first point of Aries, and m the mean sun. Then QA sidereal time, Qm=mean time; and Am=right ascension of the mean sun. By the fig., Qm=QA-Am, or mean time sidereal time-right ascension of mean sun. = m Q Hence it appears, that to find mean time we have only to subtract the right ascension of the mean sun for that instant of mean time from the given sidereal time, and the result will be mean time at that instant. But how shall we find the right ascension of the mean sun for that instant of mean time, since in the Nautical Almanac the right ascension of the mean sun is only recorded for every day at mean noon at Greenwich; and as we know not the time elapsed from mean noon at Greenwich, we cannot proportion for the change in the right ascension of the mean sun due to that elapsed time? We may proceed as follows. Let x=mean time required. Subtracting from sidereal time the right ascension of the mean sun at mean noon at the place, we get an approximate value of the mean time. Let this first approximation be denoted by t. Now this time t is manifestly too great by the motion of the mean sun, expressed in time in the interval x; and this quantity is equal to 0027379 xx, since the motion of the mean sun in the equator in 24 hours=3m 56.555$='0027379 of a day. ... x=t-0027379x, or 1.0027379x=t, ... x=9972696t. And this coefficient of t is the same as the factor which is used to reduce an interval of sidereal time into mean time (p. 312). Hence it appears, that to find the mean time x it will be sufficient to correct the approximate time t (obtained by using the right ascension of the mean sun at noon) as if we were about to reduce sidereal time into mean time; i. e. we must subtract from t the acceleration of sidereal on mean time for the interval t, which quantity may be taken out of the table. 40. Given sidereal time=3h 40m 45s, and the right ascension of the mean sun at mean noon at the place=12h 35m 24.14; required mean time. Mean time sidereal time-RA mean sun. Sidereal time.... 3h 40m 45.00s (+24h) Cor. from table, 15h..2m 27.85$ 5 20.86=t This result, however, is not quite correct, although nearer the truth than the quantity t; for we ought to have entered the table with the correct mean time, instead of the approximate time, 15h 5m 20.863. A nearer approximation, however, may now be got by repeating the work, using the last esti mated mean time, 15h 2m 52.14$, instead of 15h 5m 20.863: thus, If we approximate a third time, by using this last result instead of 151 2m 52∙143, we shall find no difference in the correction; we may therefore conclude that the correct mean time is 15h 2m 52.54s. (6). In almost every problem in Nautical Astronomy in which are used quantities taken from the Nautical Almanac, we shall find the results obtained are only approximate values of the quantities sought; and this arises from using an approximate time (called a Greenwich date) instead of the correct Greenwich time, which is seldom known. If the object of the problem is to find the correct time, we can make a second approximation, similar to the one above; but this, in the practical problems of Nautical Astronomy, is very seldom required. EXAMPLES FOR PRACTICE. 41. Given sidereal time=12h 10m 10s, and the right ascension of the mean sun at mean noon at the place=1h 42m 14.5s; required correct mean time. Ans. 10h 26m 12.4s. 42. Given sidereal time=6h 32m 40-5s, and the right ascension of the mean sun at mean noon=7h 37m 42.48; required correct mean time. Ans. 22h 51m 12.78. PROBLEM VI. To find at what time any heavenly body will pass a given meridian. Let X be the given heavenly body on the meridian PQ, A the first point of Aries, and m the mean sun. Then Qm mean time required, Am=right ascension of the mean sun at that time, and by the figure, Qm=AQ-Am, or mean time-star's RA-RA mean sun. From which expression the mean time of the star's transit may be found, as in the last problem. First. Let the given meridian be that of Greenwich, for which the quantities in the Nautical Almanac are calculated. 43. Find at what time Antares passed the meridian of Greenwich on October 3, 1846; the star's right ascension being at that time 16h 20m 1s, and the right ascension of the mean sun at mean noon at Greenwich 12h 47m 13.8s. Mean time=star's RA-RA mean sun. RA mean sun at noon..12 47 13 8 ...16h 20m 10s Second approximation. Cor. for 34.95. 0'09s Cor. mean time.3 32 12.34 EXAMPLES FOR PRACTICE. 44. Find at what time a Canis Majoris passed the meridian of Greenwich; the star's RA being 6h 38m 52.2s, and the right ascension of mean sun at Greenwich mean noon being 11h 6m 2-3s. Construct the figure, to show the positions of the first point of Aries and the mean sun with respect to the meridian. Ans. 19h 29m 38s nearly. 45. Find at what time a Aquila passed the meridian of Greenwich, having given the right ascension of the star=19h 43m 51.5s, and the right ascension of the mean sun at Greenwich mean moon Oh 6m 40·4s; and construct the figure. Ans. 19h 33m 58.3s. 46. Find at what time a Leonis will pass the meridian of Greenwich, when its right ascension is 10h 0m 49.763, and the right ascension of the mean sun at Greenwich mean moon= 4h 32m 4.68; and construct the figure. Ans. 5h 27m 51.24s. Second. When the calculations are made for any other meridian than that of Greenwich, we must take into consideration the change of the mean sun's place corresponding to the difference of longitude between the two places (see Navigation, Part I. p. 97). In practice, the correction of the RA of mean sun on this account is made in getting a Greenwich date (Navigation, Part I. p. 101). PROBLEM VII. Given the altitude and declination of a heavenly body, and the latitude of the place of observation; to calculate the hourangle of the heavenly body. Let PZQ be the celestial meridian, P the pole, Z the zenith of spectator, and X the place of the heavenly body. Let AQ be the celestial equator; and through X draw PXR a circle of declination, and ZX a circle of altitude. Then, in the spherical triangle ZPX, the three sides are given, to calculate P the hour-angle (Trigonometry, Part I. p. 55). |