If the given civil time be P. M., the astronomical and civil time will agree; but if the civil time be A. M., add 12 hours to it, and the sum will be the corresponding astronomical time after the noon of the preceding day. For example, October 4th, at 2h 10m 3s P. M., civil time, is also October 4th, 2h 10m 3s astronomical time; but September 1st, 8h 4m 20s A. M. is August 31st, 20h 4m 20s astronomical time. From the time at any place, and the longitude of that place, to find the corresponding Greenwich time. Reduce the longitude into time by Problem 1; and if west, add it to the astronomical time at the given place, but subtract it if east, and the sum or remainder will be the Greenwich time. If in adding, the sum should exceed 24 hours, the excess will be the Greenwich time past noon of the following day; and if in subtracting, the longitude in time should exceed the astronomical time at the place, subtract the longitude in time from the given time, increased by 24 hours, and the remainder will be the Greenwich time past noon of the preceding day. EXAMPLES. 1. In longitude 21° 4' W, on September 13th at 8h 40m 3s A. M., required the astronomical time at Greenwich? h m S A stronomical time, September 12th 20 40 3 1 24 16 Astronomical time at Greenwich .. 22 4 19 2. In longitude 85° 24′ E, on July 20th at 3h 35m 7s P. M., required the Greenwich time? To find the time at which any celestial object culminates, or passes the meridian of a given place, on a given day. From the object's right ascension for the noon of the day, increased if necessary by 24 hours, subtract that of the sun, and the remainder will be the required time nearly. Find the corresponding Greenwich time by Problem 4, and for that time compute the sun's right ascension, (see the use of Table 30,) and also that of the given object, and from the object's corrected right ascension subtract the sun's, and the remainder will be the required time nearly. If still greater exactness is required, the operations may be repeated; but this will seldom be necessary. The time of the moon's passing the meridian of Greenwich is given in the Nautical Almanac to the nearest minute, for every day, and it may be found for any other meridian by adding the correction from Table 17, in west, but subtracting it in east longitude. EXAMPLE I. At what time will Altair pass the meridian of Petersburg, October 4th, 1823 ? Required the time at which the moon will pass the meridian of 40° W, September 10th, 1823? h m Per Naut. Alm. (see Tab. 31) D on the meridian of Greenwich 4 32 And on the following day 5 23 Difference 51 To 51m and long 40° W, in Table 17, corresponds 5m, which added to 4h 32m gives the required time 4h 37m. EXAMPLES FOR EXERCISE. At what time will the following celestial objects pass the meridian of the places respectively named in each question on the given day? Object. Date. Answer. Place. Time of culminating. h m S Fomalhaut.. April 14th, 1839.. Cape of Good Hope 21 17 21 From the observed altitude of a fixed star to find its true altitude. Correct the observed altitude for the index error of the instrument, subtract the dip, (Table 7,) if the altitude be taken by a fore observation, but add it if by a back observation, and the sum or remainder will be the apparent altitude, from which subtract the refraction, (Table 9) corresponding to the altitude, correcting the refraction by Table 11, when necessary, and the remainder will be the true altitude. Remark. When it is not otherwise stated, the altitudes in every example in this book are to be understood as taken by a fore observation. EXAMPLE. If the altitude of a star be 43° 12′, height of the eye 18 feet, required its true altitude, the index error of the instrument being 3'46"-? The true altitude of the star is required in each of the following examples? Answer. *'s observed Alt. Index Error. Height of the Eye. True Alt. From the observed altitude of the sun's lower or upper limb, to find the apparent and true altitude of his centre. Correct the observed altitude of the limb for the index error of the instrument and the dip, as in the last Problem, and the result will be the apparent altitude of the limb; to which add the semidiameter, if the lower, but subtract it if the upper limb is observed, and the sum or remainder will be the apparent altitude of the centre. From this subtract the difference between the parallax, (Table 10,) and the refraction (Table 9,) corresponding to the altitude of the observed limb, and the remainder will be the true altitude. Remark. The difference between the parallax and refraction corresponding to the altitude of the centre, is the correction of altitude which must be used in computing the true altitude, when it is required for clearing the apparent distance between the moon and the sun or a star from the effects of parallax and refraction. EXAMPLE. If the altitude of Q be 24° 2′ 40′′ on May 17th, index error of the instrument 2′ 23′′ +, height of the eye 15 feet, required the true altitude of his centre ? + 15 50 semidiameter, (Nautical Almanac or Table 16.) 24 17 4 2 00 cor of alt for ref and par (Tables 9, and 10.) 24 15 4 true altitude EXAMPLES FOR EXERCISE. In each of the following examples the true altitude of the sun's centre is required? From the observed altitude of the moon's upper or lower limb, her semidiameter and horizontal parallax as given in the Nautical Almanac, and the latitude of the place of observation, to find the true altitude. To the semidiameter from the Nautical Almanac, apply the augmentation corresponding to the altitude from Table 12; and from the horizontal parallax as given in the Nautical Almanac, subtract the reduction corresponding to the latitude, Table 14. Then correct the altitude for index error and dip, as in the preceding problems, and add or subtract the augmented semidiameter according as the lower or upper limb is observed, and the sum or remainder will be the apparent altitude of her centre, to which add the correction of altitude from Table 24, and the sum will be the true altitude. |