Example 1. April 10th, 1825, in longitude 75: W., the meridian altitude of the sun's lower limb was 57:40:30" S., and the height of the eye above the level of the sea 22 feet; required the latitude? Observed altitude of the sun's lower limb = 57:40:30 S. Sun's declination at noon, April 10th = 7:56:42 N. + 4.36 Note. The meridional zenith distance and the declination are added together, because they are both of the same name: hence, the latitude is 40:9:49" N. Example 2. October 24th, 1825, in longitude 90: east, the meridian altitude of the sun's lower limb was 27:31:20 S., and the height of the eye above the surface of the sea 23 feet; required the latitude? Observed altitude of the sun's lower limb = Sun's semi-diameter = Dip of the horiz. for 23 feet 27:31:20% S. Sun's declination at noon, Oct. 24th = 11:45:42" S. Correction for longitude 90: east = Sun's reduced declination = 5.15 Sun's meridional zenith distance = Latitude, as required = Note. The difference between the meridional zenith distance and the declination is taken, because they are of contrary names: hence, the latitude is 50:38:21 N. PROBLEM II. Given the Moon's Meridional Altitude, to find the Latitude of the Place of Observation. RULE. Reduce the moon's passage over the meridian of Greenwich, on the given day, to the meridian of the place of observation, by applying thereto the correction in Table XXXVIII., by addition or subtraction, according as the longitude is west or east ; as explained in examples 1 and 2, pages 101 and 102. To the time of the moon's passage over the meridian of the place of observation, thus found, let the longitude of that meridian, in time, be added if it be west, or subtracted if east; and the sum, or difference, will be the corresponding time at Greenwich: to which let the moon's declination, horizontal parallax, and semi-diameter, be reduced by Problem VI., page 302, (or by means of Table XVI., as explained in page 30,) and let the moon's reduced semi-diameter be corrected by the augmentation contained in Table IV. Find the true altitude of the moon's centre, by Problem XV., page 323, and call it north or south, according as it may be situate with respect to the observer at the time of observation; which, subtracted from 90°, will give the moon's meridional zenith distance of a contrary denomination to that of its altitude. Then, if the meridional zenith distance and the declination are of the same name, their sum will be the latitude of the place of observation; but if they are of contrary names, their difference will be the latitude, of the same name with the greater term. Note. In strictness, the moon's declination should be corrected by the equation of second difference contained in Table XVII, as explained between pages 33 and 37. Example 1. January 27th, 1825, in longitude 55: W., the meridian altitude of the moon's lower limb was 58:40, S., and the height of the eye above the level of the sea 26 feet; required the latitude? Time of D's passage over the meridian of Greenwich = = Time of D's pass. over the merid. of the place of observation = 6 123: Longitude 55 W., in time = + 3.40. 0 9:41 23: 55:20% Moon's horizontal parallax at noon, Jan. 27th = Moon's reduced horizontal parallax = = Moon's semi-diameter at noon, Jan. 27th = Moon's true semi-diameter = + 0.17 55:37" 15: 5% + 4 +12 15:21" . Moon's declination at noon, Jan. 27th = 18:19:18" N. Example 2. February 3d, 1825, in longitude 65: E., the meridian altitude of the moon's upper limb was 62:45 north, and the height of the eye above the level of the sea 29 feet; required the latitude? Time of D's passage over the meridian of Greenwich = 12:25 0: 9.44 Time of D's pass. over the merid. of the place of observation=12:15 16: Longitude 65: E., in time = Greenwich time = 4.20. 0 Moon's horizontal parallax at noon, February 3d = 60:49" + 5 60:54" Remark.-Although this method of finding the latitude at sea is strictly correct when the longitude of the place of observation is well determined; yet, in some cases, it is subject to such peculiarities as to render it inconvenient to the practical navigator: this happens in high latitudes, and when the variation in the moon's declination is very considerable; because, under such circumstances, the moon's altitude sometimes continues to increase after she has actually passed the meridian. To provide against this, the observer should be furnished with a chronometer, or other well-regulated watch, to show the instant of the moon's coming to the meridian of the ship or place; at which time her altitude should be taken, without waiting for its ceasing to rise or beginning to dip, as it is generally termed at sea: then this altitude is to be considered as the observed meridional altitude of that object, and to be acted upon accordingly. PROBLEM III. Given the Meridional Altitude of a Planet, to find the Latitude of the Place of Observation. RULE. To the apparent time of observation (always reckoning from the preceding noon,) apply the longitude, in time, by addition or subtraction, according as it is west or east; and the sum, or difference, will be the corresponding time at Greenwich, to which let the planet's declination be reduced, by Problem VII., page 307. Find the true altitude of the planet's centre, by Problem XVI., page 325; and hence its meridional zenith distance, noting whether it be north or south then, if the meridional zenith distance and the declination are of the same name, their sum will be the latitude of the place of observation; but if they are of contrary names, their difference will be the latitude, of the same name with the greater term. Example 1. February 3d, 1825, in longitude 80: W., at 11:28:30: apparent time, the meridional central altitude of the planet Jupiter was 58:22. S., the height of the eye above the level of the sea 24 feet, and the planet's horizontal parallax 2 seconds; required the latitude? Apparent time of observation, February = 3:11*28*30: + 5.20. 0 3:16:48:30: |