Erample.-If on May 7th and 8th, 1828, at Portsmouth, to four sets of equal altitudes of te sun's lower limb I find the times as under, required the e.ror and rate of the chronometer? Times in the Forenoon. Times in the Afternoon, b. h. 35 2 28 12 S. m. S. 48 m. 31 May 7th . NN To find the longitude by a chronometer. subtract it if east, and the sum or remainder . be the mean time at Greenwich. For that Take an altitude of a celestial object, or rather stant take the equation of time, and apply it to a series of altitudes at short intervals of time, a contrary sign, and the result will be the app noting the time of each altitude, rent time at Greenwich. Take the mean of the times and the mean of Then, with the mean corrected altitude, the the altitudes. To the mean of the times apply latitude of the place, and the polar distance i the last known error of the chronometer, adding the object, find its meridian distance, and the if it was slow, and subtracting if it was fast. the apparent time at the place of observative; Multiply the rate by the number of days elapsed and the difference between that time and the since the first error was determined, and add the apparent time found at Greenwich, found 3 product to the above corrected time if the chro- above, will be the longitude of the place in tini , nometer is losing, but subtract it from it if gain- west if the Greenwich time is greater or before ing. To the result add the longitude of the but east if the Greenwich time is less or be place for which the error is found if west, but 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 1sc. 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 Chron. slow, June 5th 5 eighteen days. 15th 4 27 7 X 18 = 126 = 2 6, gain from rate. Gain in ten days 1 10 = 70 s. At 6h. 48 m. July 3d, the sun's declination is 22° 56' 36' N. = 7 s. rate gaining. whence his polar distance is 67° 3' 24". Time by chronometer, July 3d 6 49 Obs. alt. O 26° 48' 0" Chronometer slow, June 15th 27 Dip m. S. 37 m. S. 70 10 h. m. S. 43 4 3 49 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 1 h. 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. Loss in 34 days 30 rate 9 s. losing. h. 29 25 Obs. alt. O 17 24 Dip S. m. . 0 24 4 19 3 15 29 34 49 . Mean time at Greenwich 7 19 24 29 Equation of time with contrary sign 5 Semid. Alt. 29° 34' 49" Lat. 20 0 sect. Polar dist. 68 4 36 cosect. 3 . 027152 To find the longitude by lunar observation, that is Then add together the secants of the apparent by the distance of the moon from the sun or a altitudes, the cosine of half the sum of the appastar, with the altitudes of both objects; the lati- rent altitudes, and apparent distance, the cosine of tude of the place of observation being known, as the ditlerence between that half sum and the well as the time and longitudi by account. apparent distance, and the cosines of the true With the time and the longitude by account, altitudes, and from the sum of these six logafind the Greenwich time by account, and for rithms (rejecting twenty from the index), subthat time take the moon's semidiameter and ho. tract twice the cosine of half the sum of the true rizontal parallax from the Nautical Almanac, altitudes, and half the remainder will be the sine and to the semidiameter apply the augmentation of an arc. And the cosine of that arc added to corresponding to the altitude. the cosine of half the sum of the true altitudes Correct the altitudes for semidiameter and (rejecting ten from the index of the sum) will be dip, and call the results the apparent altitudes. the sine of half the true distance, or that which Correct them further for the parallax and refrac- the objects would have had if the observer had tion, and the results will be the true altitudes. been at the centre of the earth. If the sun is one of the objects observed, the With this distance enter the Nautical Almadistance observed will be that of the nearest nac, pp. 8, 9, 10, or 11, of the month, and take limb; therefore, if the sum of the semidiameters the two distances of the moon from the object be added to it, the apparent distance of the between which the true distance falls, and write centres, as seen at the surface of the earth, will them under the true distance in the order in be obtained. If the observed distance is that of which they stand in the Almanac. Take the a star from the moon's nearest limb, add the difference between the middle one of these three moon's semidiameter to the observed distance; distances and each of the others, and subtract if it is from the farther or most remote limb, the proportional logarithm of the greater diffesubtract the moon's semidiameter from the ob- rence from that of the less, and the remainder served distance, for the distance of the star from will be the proportional logarithm of a portion the moon's centre as seen at the surface of the of time, which, added to the time corresponding earih. to the first distance taken from the Almanac, From the altitudes and apparent central dis- will be the Greenwich time. If the true distance tance of the objects compute what the distance be found in the Nautical Almanac, the apparent would have been if the observer had been at the time at Greenwich will be found above it. centre. There are many methods by which this Having now found the Greenwich ne, find computation may be made. We give the fol- the time at the place of observation from the lowiny from the formula of Banda. See Lon- altitude of one of the objects in the latitude of GITUDE in this Encyclopædia. the place; the polar distance and right ascenPlace under each other, in order, the appa- sion of the object; and the difference between rent distance, and the apparent altitudes of the that time and the Greenwich time found from objects, half the sum of the three arcs, and the the distance will be the longitude of the place in difference between the half sum and the appa- time, west when the Greenwich time is before, rent distance, Below place the true altitudes but east when it is behind, that at the place of and half their sum. observation. Erumple 1.-On March 27th, 1828, in latitude 35° 10' N., longitude by account 31° 30' W., at 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 b. 26 m. 29 s., the star's right ascension 13b. 16 ra. :: 11 s., and polar distance 100° 15' 49". *'s true altitude 25 39' 36 35 10 0 sect. 10.087023 15 49 cosect. 10.007005 Erample 2.-On August 5th, 1828, in latitude 20° 3' N., longitude by account 20° E., at 7 h. O m. 20 s. A. M. by watch, the altitude of 0 was 19° 20' + of ), 77° 37' +, distance of nearest limbs 59° 23' 41", height of the eye eighteen feet; required the true longitude? h. par. 55 O Long. by acct. E. Aug. 15 1 20 |