When the calculations are made for any other meridian than that of Greenwich, for which the quantities in the Nautical Almanac are calculated, we must take into consideration the change of the mean sun's place arising from the difference of longitude. For example, the tables of the Connaissance des Tems are computed for Paris, the long. of which is 9m 22s to the east of Greenwich: as in that time the mean sun moves to the eastward through an arc of 1$.53 in time (for 24h: 9m 22s :: 3m 56s.55 : 1.53), it follows that we must add 1.53 to all the right ascensions of the mean sun in the French tables to obtain those of the mean sun at mean noon at Greenwich. EXAMPLE. April 27, 1841, the right ascension of the mean sun at mean noon at Paris, by the Connaissance des Tems, was 2h 21m 10s-09, required the same for Greenwich mean noon. Right asc. mean sun at Greenwich 2 21 11.62 The longitude is usually found at sea by means of a chronometer showing Greenwich mean time at the instant the mean time at the ship is known. The mean time at the ship is deduced from the hour angle of a heavenly body, and this hour angle is calculated by means of the altitude of the body observed with a sextant and certain elements found in the Nautical Almanac. Rules for finding the hour angle of a heavenly body will be given hereafter. We will therefore suppose the hour angle known, and proceed to show how mean time might be found from it. F Rule XXI. Given, the hour angle of a heavenly body, to find mean time at the ship. Hour angle found by table of haversines.* When the heavenly body is west of meridian, take the hour angle out at top of page of haversines and add thereto the right ascension of the heavenly body: from the sum (increased if necessary by 24 hours) subtract the right ascension of mean sun (corrected for estimated mean time at Greenwich or Greenwich date), the result (rejecting 24 hours if greater than 24 hours) is the mean time required. When the heavenly body is east of meridian, take the hour angle out of the table at bottom of page of haversines, then proceed as directed above. If the heavenly body observed is the sun, the hour angle taken out will also be apparent time (p. 64, art. 24); the mean time will then be found by applying the equation of time with the proper sign as shown in the Nautical Almanac. Hour angle found by any other table. The angle taken out turned into time (if necessary) will also be the hour angle if the body be west of meridian; but if the body be east of meridian, subtract the hour angle from 24 hours, and then, as in the former case, add thereto the star's right ascension, and from the sum subtract the right ascension of mean sun as pointed out in the rule. * 24h Star's hour angle can be taken out of Inman's table of haversines by inspection: when, therefore, the longitude by chronometer is found by means of this table, the above rule applies whether the heavenly body is east or west of meridian. EXAMPLE. August 11, 1846, at 8h 50m P.M., mean time nearly, in long. 90° W., the hour angle of Arcturus was 3h 56m 55s west of meridian, required correct mean time at the place. This result is slightly incorrect, arising from the estimated mean time, 8h 50m, being different from the true time. When great accuracy is required, the operation should be repeated, using mean time last found, namely 8h 45m, instead of the one used before; thus, (74.) Nov. 22, 1853, at 7h 15m P.M., mean time nearly, in long. 22° 0′ W., the hour angle of Aldebaran (a Tauri) was 5h 10m 20s east of meridian, required mean time at the place. Ans., 17h 30m 54s. (75.) June 23, 1853, at 3h 15m A.M., mean time nearly, in long. 100° 40′ E., the hour angle of a Lyre was 3h 42m 40s west of meridian, required mean time at the place. Ans., 16h 11m 6s. (76.) June 15, 1853, at 10h 10m P.M., supposed mean time nearly, in long. 10° 42′ W., the hour angle of Arcturus was 2h 2m 30s east of meridian, required mean time at the place. Ans., 6h 30m first approximation; 6h 30m 35s more exactly. Elements from Nautical Almanac. Right ascension mean sun. Nov. 22, 1853 16h 5m 32s Right ascension star. Aldebaran 4h 27m 32s Rule XXII. To find at what time any heavenly body will pass the meridian. 1. Take out of the Nautical Almanac the right ascension of the given star, and also the right ascension of the mean sun for noon of the given day. 2. From the right ascension of the star (increased if necessary by 24 hours) subtract the right ascension of the mean sun, the remainder is mean time at the ship nearly. 3. Apply the longitude in time, and thus get a Greenwich date; with this Greenwich date correct the right ascension of mean sun for Greenwich date. 4. Then from the right ascension of the star subtract the right ascension of the mean sun thus corrected, the remainder is the mean time when the heavenly body is on the meridian. As in the last problem, the table of acceleration ought to have been entered with the correct mean time; but the error for all practical purposes is inappreciable. EXAMPLE. At what time will Sirius pass the meridian of a place in long. 68° 30′ W. on Nov. 20, 1845? Therefore the transit of Sirius is at 14h 37m 48s on Nov. 20, or at 2h 37m 48s A.M. on Nov. 21. To find at what time it will pass the meridian on the morning of Nov. 20, we must evidently begin one day back and take out the right ascension of the mean sun for Nov. 19. (77.) At what time will a Pegasi pass the meridian of Portsmouth, long. 1° 6' W., on Nov. 25, 1853 ? Ans., Nov. 25, 6h 38m 58s. (78.) At what time will the star Regulus (a Leonis) pass the meridian of Land's End, long. 5° 42′ W., on May 30, 1845 ? Ans., May 30, 5h 27m 45s P.M. (79.) At what time will Antares pass the meridian of Portsmouth, long. 1° 6′ W., on Aug. 20, 1845 ? Ans., Aug. 20, 6h 24m 11s. (80.) At what time will a Leonis pass the meridian of Lisbon, long. 9° 8' W., on June 4, 1846 ? Ans., June 4, 5h 9m 4s. (81.) At what time will the star Antares pass the meridian of Copenhagen, long. 12° 35' E., on Aug. 20, 1846 ? Ans., Aug. 20, 6h 25m 21s. |