(33.) May 20, 1850, at 5h 30m P.M. mean time, in long. 95° 30′ E., required the sun's declination. Ans. 19° 57′ 22′′ N. (34.) May 16, 1850, at 7h 50m A.M. mean time, in long. 120° 0' E., required the sun's declination. Ans. 18° 57' 48" N. (35.) March 23, 1850, at 3h 20m P.M. mean time, in long. 9° 0′ W., required the sun's declination. To take out the equation of time. 1. Get a Greenwich date. 2. Take out the equation of time for two consecutive noons between which the Greenwich date lies, and take their difference. 3. Add together the Greenwich date logarithm for sun and proportional logarithm of difference: the sum is the proportional logarithm of correction, which find from the table, and apply it with its proper sign to the equation of time first taken out; the result is the equation of time required. Or thus, by hourly differences. 1. Take out the equation of time for the noon of Greenwich date and the hourly difference opposite thereto. 2. Multiply hourly difference by the hours of the Greenwich date, and, if great accuracy is required, by the fractional parts of hour in the Greenwich date; the result will be the correction to be applied with its proper sign to the equation of time taken out. EXAMPLE. July 12, 1853, at 5h 8m A.M. mean time nearly, in long. 160° W., required the equation of time. Find the equation of time in the following examples : (36.) March 2, 1853, at 6h 10m P.M. mean time in long. 38° 42′ W. Eq. of time March 2 12m 22.1. March 3 12m 95.3 To take out the moon's semidiameter and horizontal parallax. The moon's semidiameter and horizontal parallax are put down in the Nautical Almanac for every mean noon and mean midnight at Greenwich: to find these quantities for any other time we may proceed as follows: First. To find the moon's semidiameter. 1. Get a Greenwich date. 2. Take out of the Nautical Almanac the moon's semidiameter for the two times between which the Greenwich date lies, and take the difference. To the Greenwich date logarithm for moon add the proportional logarithm of the difference just found; the result will be the proportional logarithm of an arc, which being found and added to the semidiameter first taken out, or subtracted therefrom (according as the semidiameter is increasing or decreasing), will be the semidiameter at the given time. Second. To find the moon's horizontal parallax. Proceed in a similar manner to that pointed out above for finding the moon's semidiameter. EXAMPLES. 1. August 3, 1853, at 4h 10m P.M. mean time nearly, in long. 48° 42′ W., required the moon's semidiameter and hori 2. July 14, 1853, at 6h 42m A.M. mean time nearly, in long. 30° W., required the moon's semidiameter and horizontal parallax. Find the moon's semidiameter and horizontal parallax in the following examples : * When the Greenwich date exceeds 12 hours, as in this example, look out the Greenwich date logarithm moon for the excess of the Greenwich date above 12 hours. It is better, however, in examples of this kind, where the differences are small, to estimate the correction at sight, without using logarithms; after some experience this is easily done; the above method, however, by means of logarithms, should be adopted by beginners. Rule XI. To take out the sun's right ascension. 1. Get a Greenwich date. 2. Take out the right ascension for two consecutive noons between which the Greenwich date lies, and take their difference. 3. Add together the Greenwich date logarithm for sun and proportional logarithm of difference; the sum will be the proportional logarithm of correction to be added to the right ascension for noon of Greenwich date. EXAMPLE. July 13, 1853, at 6h 31m A.M. mean time nearly, in long. 172° 10′ W., required the sun's right ascension. Find the sun's right ascension in the following examples:(42.) March 11, 1853, at 6h 42m P.M. mean time, long. 42° 41′ W. Elements from Nautical Almanac. Sun's right asc. March 11 23h 26m 26.3 March 12 23h 30m 68.6 Answers to (42), (43), (44) :— 23h 27m 53.3 Oh Om 10.0 22 0 6 36.4 Oh 4m 248.2 |