Rule.

Find the Greenwich date.

Correct the equation of

time according to Art. 55 and apply it to the given time as directed in the Nautical Almanac. The result is the time

required.

EXAMPLE 1.-The local mean time of a ship in longitude 16° W, April 27, 1899, is 9h 10m P. M. Find the corresponding apparent time.

App. time required = 9h 12m 29.24s. Ans.

Change in 1h

= 0.4s incrs. X 10.2h

Corr.

= 4.08s

According to the Nautical Almanac, the equation of time in this case is additive to local mean time; therefore, the sign (+) is placed to the right of the corrected equation of time, as shown in the solution.

EXAMPLE 2.-The local apparent time of a ship in longitude 16° 30 E, January 1, 1899, is 9h 26m A. M. Required, the corresponding

mean time.

EXAMPLE 3.-The local apparent time June 22, 1899, is 5h 42m P. M. Find the mean time, the longitude being 100° 30′ E.

SOLUTION.-L. App. T., June 22

Mean time required = 5h 43m 41.6s. Ans.

66. To Find the Sidereal Time Corresponding to a Certain Mean Time.-Let the circumference of each circle, Fig. 4, represent the celestial equator; mn the meridian; P the celestial pole; ' the first point of Aries, or the vernal equinoctial point; M the mean sun situated to the west of the meridian mn; and M' the mean sun when situated to the

east of the meridian m n. Then, the angle 'Pn will represent the sidereal time, as it is the hour angle of Y, or it will represent the time since I was on the meridian; the angle MPn will represent the given mean time, being the hour angle of the mean sun; and the arc I'M will represent the right ascension of the mean sun.

The sidereal time is therefore equal to the sum of the right ascension of the mean sun and the given mean time, 24 hours being subtracted from the result if over 24 hours. Hence, the procedure for finding the sidereal time when the mean time is known may be embodied in the following rule:

Rule. Find the Greenwich date. Find from the Nautical Almanac the right ascension of the mean sun according to Art. 57, and add to it the given mean time. rejecting 24 hours, if over 24 hours, will be the required sidereal time.

The sum,

EXAMPLE 1.--The mean time of a ship in longitude 84° 35′ W, January 23, is 4h 36m A. M. Find the corresponding sidereal time.

EXAMPLE 2.-A ship is in longitude 75° 20′ E. About 1:20 P. M., February 26, 1899, the ship's chronometer indicated 8h 14m 31s, its error on Greenwich mean time being 2m 19s slow. Required, the local sidereal time.

SOLUTION. First find the local mean time at ship according to Art. 49, and then the corresponding sidereal time. Thus,

67. If the given time is apparent, it must be reduced to mean time by the application of the equation of time, according to Art. 65. The sidereal time is then found according to the rule of Art. 66.

68. To Find the Mean Time Corresponding to a Given Sidereal Time.-There are several methods by which the mean time corresponding to a given sidereal time may be determined. Among these the following is selected as being simple and easily committed to memory.

Hence, when the sidereal time is known, the corresponding mean time is readily obtained by subtracting from the given sidereal time (increased by 24h if necessary) the right ascension of the mean sun.

But, as the right ascension of the mean sun cannot be obtained before the mean time is known, an approximate value of the mean time must be found by using in the preceding formula the right ascension of the mean sun for the given day as tabulated in the Nautical Almanac. By apply ing to this approximate local mean time the longitude in time, an approximate Greenwich date is found, and from this a more correct value of the right ascension of the mean sun is obtained. This new value of the right ascension of the mean sun subtracted from the given sidereal time will produce a more correct value of the mean time, by means of which a still more correct value of the right ascension of the mean sun may be obtained. This procedure may be repeated until a desired degree of accuracy is arrived at; the second approximation, however, is quite sufficient for all practical purposes.

EXAMPLE 1.-On July 6, 1899, in longitude 124° 40′ W, the sidereal time is 7h 24m 48s. Find the corresponding mean time.

The second value of the right ascension of the mean sun differs only by .2s from that previously found; the required mean time is therefore Oh 26m 21s. Ans.

EXAMPLE 2.-On January 18, 1899, in longitude 174° 30′ E, the sidereal time is 19h 26m 14s. Required, the corresponding mean time.

69. In the foregoing examples it will be noticed that the corrections used are taken from Table III, Nautical Almanac, just as was done in the method for converting mean into sidereal time. The same result, however, may be obtained by using Table II of the Almanac and a slightly different method, as follows:

Find from the Almanac the right ascension of the mean sun for the given local astronomical day. Apply to this a correction for the longitude in time taken from Table III, Nautical Almanac, adding for west and subtracting for east longitude. Subtract the result from the local sidereal time (increased by 24 hours if necessary), and apply to the