Whence the error of chronometer is fast 23m 20s.5 on Greenwich mean time. (165.) May 20, 1847, at 5h 20m P.M., mean time nearly, in lat. 47° 20′ N., and long. 94° 30′ E., when a chronometer showed 11h 5m 20s, the observed altitude of the sun's lower limb was 20° 0' 15", the index correction 4' 10" and height of eye above the sea 20 feet, required the error of chronometer on Greenwich mean time. Ans., fast 0m 41s.4. (166.) Feb. 3, 1847, at 10h 30m A.M., mean time nearly, in lat. 49° 30′ N., and long. 22° W., when a chronometer showed Oh 2m 30s, the observed altitude of the sun's lower limb was 19° 21′ 30′′ the index correction + 3′ 20′′, and height of eye above the sea 18 feet, required the error of chronometer on Greenwich mean time. Ans., fast 10m 78.4. (167.) March 25, 1847, at 3h 20m P.M., mean time nearly, in lat. 52° 10′ N., and long. 36° 58′ 15′′ W., when a chronometer showed 5h 40m 58', the observed altitude of the sun's lower limb was 25° 10′ 20′′, the index correction 6' 10", and height of eye above the sea 20 feet, required the error of chronometer on Greenwich mean time. Ans., 9m 25s.2 slow. (168.) May 19, 1847, at 3h 0m P.M., mean time nearly, in lat. 49° 50′ N., and long. 21° 4′ 45′′ E., when a chronometer showed 1h 23m 20s, the observed altitude of the sun's lower limb was 42° 50′ 30′′, the index correction + 4′ 10′′, and height of eye above the sea 20 feet, required the error of chronometer on Greenwich mean time. Second. When the object observed is a star. Rule XLV. To find the error of chronometer on mean time at a place by a single altitude of a star. 1. Get a Greenwich date. 2. Take out of the Nautical Almanac the right ascension and declination of the star, and also the right ascension of the mean sun for mean noon of the Greenwich date. 3. Correct the right ascension of mean sun for Greenwich date (p. 83). 4. Correct the observed altitude for index correction, dip, and refraction, and thus get the true altitude, which subtract from 90° for the true zenith distance. 5. To find star's hour angle (using haversines*). Under the latitude put the star's declination; add if the names be unlike, subtract if like; under the result put the true zenith distance of star, 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 halves the log. haversines of the two last; the sum, (rejecting 10 in the index,) will be the log. haversine of hour angle, to be taken out at top of page if heavenly body be west of meridian, but at bottom if east of meridian. 6. 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 mean sun; the remainder is mean time at the place at the instant of observation. 7. Under mean time at place put the time shown by chronometer; the difference will be the error of chronometer on mean time at place. To find the error of chronometer on Greenwich mean time. Proceed as in the corresponding rule for the sun, p. 172. EXAMPLE. June 3, 1842, at 12h 9m P.M., mean time nearly, in lat. 50° 48' N., and long. 1° 6' 3" W., observed the altitude of a Bootis (west of meridian) to be 89° 53′ 30′′ in artificial horizon, when a chronometer showed Oh 14m 22-3, the index correction was 10", required the error of the chronometer on mean time at the place, and also on Greenwich mean time. * If the student have the table of log. sines, &c., only, the hour angle may be found in a similar manner as in note, p. 171, and example, p. 174. To find the error of chronometer on Greenwich mean time. (169.) May 4, 1847, at 4h 40m A.M., mean time nearly, in lat. 40° 10′ 20′′ N., and long. 81° 47' 15" E., when a chronometer showed 11h 13m 50s, the observed altitude of a Bootis (west of meridian) was 20° 45' 4" 5, the index correction 2′ 10′′, and the height of eye above the sea was 18 feet, required the error of the chronometer on Greenwich mean time. Ans., 0m 35-3 slow. was (170.) Feb. 10, 1847, at 9h 22m P.M., mean time nearly, in lat. 28° 30′ N., and long. 27° 15′ W., a chronometer showed 11h 17m 20s, when the observed altitude of a Leonis (east of meridian) was 42° 10' 0", the index correction. -3′ 20′′, and height of eye above the sea 20 feet, required the error of the chronometer on Greenwich mean time. Ans., 4m 575.3 fast. (171) April 18, 1848, at 0h 40m A.M., mean time nearly, in lat. 46° 32′ N., and long. 43° 36′ 15′′ E., when a chronometer showed 10h 13m 45s, the observed altitude of the star a Aquila was 14° 45′ 15′′ (east of meridian) the index correction + 4′ 5′′, and height of eye above the sea 18 feet, required the error of the chronometer on Greenwich mean time. Ans., 19m 318-7 fast. (172.) Aug. 11, 1848, at 8h 10m P.M., mean time nearly, in lat. 50°20′ N., and long. 29° 53'′ 15′′ E., when a chronometer showed 6h 6m 20s-0, the observed altitude of a Bootis (Arcturus) was 39° 5' 10" (west of meridian) the index correction 2′ 10′′, and height of eye above the sea 18 feet, required the error of the chronometer on Greenwich mean time. Ans., 11m 175.8 slow. |