Example.-If on May 7th and 8th, 1828, at Portsmouth, to four sets of equal altitudes of t sun's lower limb I find the times as under, required the error and rate of the chronometer! To find the longitude by a chronometer. Take an altitude of a celestial object, or rather a series of altitudes at short intervals of time, noting the time of each altitude. Take the mean of the times and the mean of the altitudes. To the mean of the times apply the last known error of the chronometer, adding if it was slow, and subtracting if it was fast. Multiply the rate by the number of days elapsed since the first error was determined, and add the product to the above corrected time if the chronometer is losing, but subtract it from it if gaining. To the result add the longitude of the place for which the error is found if west, but 22 15.6 log. 2-54900 log. 4-32933 4-4613 log. 1.3397 subtract it if east, and the sum or remainder wi be the mean time at Greenwich. For that a stant take the equation of time, and apply it a contrary sign, and the result will be the app rent time at Greenwich. Then, with the mean corrected altitude, th latitude of the place, and the polar distance a the object, find its meridian distance, and thenc: the apparent time at the place of observation: and the difference between that time and the apparent time found at Greenwich, found a above, will be the longitude of the place in tint, west if the Greenwich time is greater or before. but east if the Greenwich time is less or be hind the time at the place of observation. Example 1.-On June 5th, 1828, my chronometer was 5 m. 37 s. slow, and on June 15th 4 m. 27 s. slow, for mean time at Greenwich. On July 3d, in lat. 30° 25' N. at 6 h. 49 m. 435 P. M., by the chronometer the altitude of was 26° 48′, height of the eye fifteen feet; required the longitude? From June 15th to July 3d is h. m. S. 6 49 43 4 27 6 54 10 6 Mean time at Greenwich Equation of time with contrary sign Refr. par.. 42 26 1 Example 2.-If on May 10th, 1828, at Cape Town, long. 18° 23′ E., I find my chronometer 1 h. 30 m. 26 s. slow, and on June 3d, at James Town, St. Helena, long. 5° 43′ W. 5 m. 28 s. fast; and on July 12th, in lat. 20° 3′ N., on my voyage homeward to England, the altitude of be 29° 25′, at 7 h. 1 m. 25 s. by the chronometer, height of the eye twenty feet, required the longitude? The longitude of Cape Town, in time, being 1 h. 13 m. 32 s. east, if the chronometer were right for Greenwich time it would be 1h. 13 m. 32 s. slow for time at Cape Town. But it is 1 h. 30 m. 26 s. slow for time at that place, whence it is 16 m. 54 s. slow for Greenwich time on May 10th. In the same manner if the chronometer were right for Greenwich time it ought to be 22 m. 52 s. fast for time at James Town, whereas it is only 5 m. 28 s. fast for time at that place Consequently, on June 3d, it is 17 m. 24 s. slow for Greenwich time. From June 3d till July 12th is 39, and 39 × 9 35 s. loss from the rate. At 7 h. 16 m. P. M. July 12th, the sun's polar distance 68° 4′ 36′′. 7 14 9 True alt. 29 34 49 To find the longitude by lunar observation, that is by the distance of the moon from the sun or a star, with the altitudes of both objects; the latitude of the place of observation being known, as well as the time and longitude by account. With the time and the longitude by account, find the Greenwich time by account, and for that time take the moon's semidiameter and horizontal parallax from the Nautical Almanac, and to the semidiameter apply the augmentation corresponding to the altitude. Correct the altitudes for semidiameter and dip, and call the results the apparent altitudes. Correct them further for the parallax and refraction, and the results will be the true altitudes. If the sun is one of the objects observed, the distance observed will be that of the nearest limb; therefore, if the sum of the semidiameters be added to it, the apparent distance of the centres, as seen at the surface of the earth, will be obtained. If the observed distance is that of a star from the moon's nearest limb, add the moon's semidiameter to the observed distance; if it is from the farther or most remote limb, subtract the moon's semidiameter from the observed distance, for the distance of the star from the moon's centre as seen at the surface of the earth. From the altitudes and apparent central distance of the objects compute what the distance would have been if the observer had been at the centre. There are many methods by which this computation may be made. We give the following from the formula of Banda. See LoNGITUDE in this Encyclopædia. Place under each other, in order, the apparent distance, and the apparent altitudes of the objects, half the sum of the three arcs, and the difference between the half sum and the apparent distance. Below place the true altitudes and half their sum. Then add together the secants of the apparent altitudes, the cosine of half the sum of the apparent altitudes, and apparent distance, the cosine of the difference between that half sum and the apparent distance, and the cosines of the true altitudes, and from the sum of these six logarithms (rejecting twenty from the index), subtract twice the cosine of half the sum of the true altitudes, and half the remainder will be the sine of an arc. And the cosine of that arc added to the cosine of half the sum of the true altitudes (rejecting ten from the index of the sum) will be the sine of half the true distance, or that which the objects would have had if the observer had been at the centre of the earth. With this distance enter the Nautical Almanac, pp. 8, 9, 10, or 11, of the month, and take the two distances of the moon from the object between which the true distance falls, and write them under the true distance in the order in which they stand in the Almanac. Take the difference between the middle one of these three distances and each of the others, and subtract the proportional logarithm of the greater difference from that of the less, and the remainder will be the proportional logarithm of a portion of time, which, added to the time corresponding to the first distance taken from the Almanac, will be the Greenwich time. If the true distance be found in the Nautical Almanac, the apparent time at Greenwich will be found above it. Having now found the Greenwich time, find the time at the place of observation from the altitude of one of the objects in the latitude of the place; the polar distance and right ascension of the object; and the difference between that time and the Greenwich time found from the distance will be the longitude of the place in time, west when the Greenwich time is before, but east when it is behind, that at the place of observation. at Example 1.-On March 27th, 1828, in latitude 35° 10′ N., longitude by account 31° 30′ W., 10 h. 2 m. 12 s. P. M. per watch, the altitude of) was 60° 46'-, of Spica 25° 45′ 30′′ + ; distance of ✶ from 's farthest limb 54° 7′ 40′′, height of the eye sixteen feet; required the longitude? For this time the sun's right ascension is 0 h. 26 m. 29 s., the star's right ascension 13h. 16 m. 11 s., and polar distance 100° 15′ 49′′. Example 2.-On August 5th, 1828, in latitude 20° 3' N., longitude by account 20° E., at 7 h. 0 m. 20 s. A. M. by watch, the altitude of was 19° 20′ + of, 77° 37′+, distance of nearest limbs 59° 23′ 41′′, height of the eye eighteen feet; required the true longitude? |