index correction - 1′10′′, and height of eye above the sea 30 feet; required the longitude. On May 19, the chronometer was fast 5m 16s on Greenwich mean time, and its daily rate was 38.5 gaining. Ans., 156° 20' E. (179.) July 8, 1849, at 1h 40m P.M., mean time nearly, in latitude 50° 48' N., and longitude by account 1° 1' W., the observed altitude of the sun's lower limb, taken by the artificial horizon, was 109° 54′ 44′′ the chronometer showed 1h 44m 14s, the index correction + 1' 25"; required the longitude. On July 1, the chronometer was slow on Greenwich mean time 8m 18s-4, and its daily rate was 35.5 losing. Ans., 1° 6' 0" W. (180.) January 20, 1846 at 6h 40m A.M., mean time nearly, in latitude 56° 20′ S., and longitude by account 83° 10′ W., when a chronometer showed Oh 14m 50s, the observed altitude of the sun's lower limb was 20° 20′ 30′′, the index correction 1' 30", and the height of the eye above the sea 20 feet; required the longitude. On Jan. 2, the chronometer was fast on Greenwich mean time 5m 20s, and on Jan. 6 it was fast 4m 52s, from which may be found its mean daily rate. Ans., 83° 5' 45" W. (181.) Feb. 10, 1846, at 7h 50m, A.M., mean time nearly, in latitude 50° 48′ N., and longitude by account 170° 30′ E., when a chronometer showed 9h 59m 25s, the observed altitude of the sun's lower limb was 51° 9' 10", the index correction 3' 20", and the height of eye above the sea 16 feet, required the longitude. On Jan. 31, at Greenwich noon, the chronometer was fast 1h 34m 43s, and its daily rate was 20.6 losing. Ans., 170° 34′ 15′′ E. To find the longitude by star chronometer. Object observed, a star. 1. Get a Greenwich date. 2. Find Greenwich mean time by bringing up the error of the chronometer to the instant of observation by Rule XLII. 3. Take out of the Nautical Almanac the right ascension and declination of the star, and also the right ascension of the mean sun (called in Nautical Almanac sidereal timé), for mean noon of the Greenwich date. 4. Correct the right ascension of the mean sun for Greenwich date. 5. Correct the observed altitude for index correction, dip, and refraction, and thus get the true altitude, which subtract from 90° to obtain the zenith distance. 6. To find the star's hour angle (using log. haversines*). Under the latitude put the star's declination; add if the names be unlike, subtract if like; under the result put star's zenith distance, and take the sum and difference. Add together the log. secants of the two first terms in this form (omitting the tens in each index), and the halves of the log. haversines of the two last, the sum rejecting ten in the index, will be the log. haversines of star's *If the student have no table of haversines, he may proceed as directed in the note to p. 171, and Ex. p. 174, to find the sun's hour angle or apparent time, using the star's declination instead of sun's, so as to get the star's hour angle. hour angle, to be taken out at top of page, if heavenly body be west of meridian, but at bottom, if east of meridian. 7. To find mean time at ship. To the hour angle thus found, add the star's right ascension; and from the sum, increased if necessary by 24 hours, subtract the right ascension of the mean sun; the remainder is mean time at the place at the instant of observation. 8. To find the longitude. Under ship mean time put Greenwich mean time as known by the chronometer; the difference is the longitude in time, west, if Greenwich time is greater than ship time, otherwise east. EXAMPLES. 1. Sept. 10, 1844, at 7h 15m P.M., mean time nearly, in latitude 48° 20′ N., and longitude by account 32° E., when a timekeeper showed 5h 1m 28s, the observed altitude of a Bootis (Arcturus) W. of meridian was 31° 5′ 40′′, the index correction 4' 10", and height of eye above the sea 20 feet; required the longitude. On Aug. 25, the chronometer was slow on Greenwich mean time 2m 40s, and its daily rate was 4.3 gaining. |