Variation of the sun's declination between the given and following noons (the longitude being west) is 22'6". Sun's declination at noon of the given day, per Nautical Propl. part to long. 75:0 and var. 22: 0" 4:35% 0" ditto 0. 60. 1.15 Ditto 75.0 Correction of declination, additive = . Sun's reduced, or corrected declination = Observed altitude of the sun's lower limb = 7:56:42 N. 4:36:15 + 4:36% 8: 1:18 N. 57:40:30 S. 4.30 } difference, add 11:29" Parallax 0:5 refrac. 0:35", diff. 0:30% subtractive = 57:51:59 S. 0:30: Sun's true central altitude = 57:51:29 S. Sun's meridional zenith distance = 32: 8:31 N. . 8. 1.18 N. Apparent altitude of the sun's centre. 40: 9:49 N. September 21st, 1825, in longitude 60° east, the meridian altitude of the sun's lower limb was 56:26 north, and the height of the observer's eye above the level of the sea 26 feet; required the latitude of the place of observation? Variation of the sun's declination between the given and preceding noons, the longitude being east, is 23:22′′. Sun's declination at noon of the given day, per Nautical Propl. part to long. 60:0 and var. 23: 03:50% 0% 0:43:34 N. 60.0 ditto Correction of declination, additive = 3:53:40 + 3:54" 0:47:28% N. Observed altitude of the sun's lower limb = Sun's semidiameter 15:58 dip of the horizon for 26 feet 4:52" difference = 56:26 0 N. +11: 6 Parallax 0:5" refrac. 0:37" diff. = 0:32" subtractive = 56:37! 6" N. 0:32" Given the difference of Longitude between two Places, both under the same Parallel of Latitude; to find their Distance. RULE. To the logarithmic co-sine of the latitude, add the logarithm of the difference of longitude, in miles; and the sum, abating 10 in the index, will be the logarithm of the distance. Example. Required the distance between Portsmouth, in longitude 1:6 west, and Green Island, Newfoundland, in longitude 55:35! west, their common latitude being 50:47 north? PROBLEM IV. Given the Distance between two Places, both under the same Parallel of Latitude; to find their Difference of Longitude. RULE. To the logarithmic secant of the latitude, add the logarithm of the distance, and the sum, abating 10 in the index, will be the logarithm of the difference of longitude. Example. A ship from Cape Clear, in latitude 51:25 north, and longitude 9:29 west, sailed due west 1040 miles; required the longitude at which she then arrived? Note. The above two Problems are essentially useful when a ship sails upon a parallel of latitude; that is, when she steers either due east or due west. PROBLEM V. Given the Latitudes and Longitudes of two Places; to find the Course and Distance. RULE. From the logarithm of the difference of longitude, the index being augmented by 10, subtract the logarithm of the meridional difference of latitude; the remainder will be the logarithmic tangent of the course then, to the logarithmic secant of the course, thus found, add the logarithm of the difference of latitude, and the sum, abating 10 in the index, will be the logarithm of the distance. - Example. Required the course and distance between Cape Bajoli, in latitude 40:3 north, longitude 3:52: east, and Cape Sicie, in latitude 43°2: north, and longitude 5:58: east? Lat. of C. Bajoli 40: 3 N. Merid. pts. 2626.6 Longitude 3:52. E. Diff. of latitude 2:59: Merid. diff. lat.=239.2 Diff.long. 2: 6: Hence the true course is N. 27:47:53 E., or N.N.E. E. nearly, and the distance 2024 miles. Note. The true course is to be reduced to the magnetic or compass course by Problem V., page 576. PROBLEM VI. Given the Latitude and Longitude of the Place sailed from, with the Course and Distance; to find the Latitude and Longitude of the Place come to. RULE. To the logarithmic co-sine of the course, add the logarithm of the distance; the sum, abating 10 in the index, will be the logarithm of the difference of latitude; which being applied to the latitude left by addition or subtraction, according as the latter is increasing or decreasing, the sum, or difference, will be the latitude come to. Now, to the logarithmic tangent of the course, add the logarithm of the meridional * The course steered, per compass, is to be reduced to the true course by Problem VI, page 577. difference of latitude; the sum, abating 10 in the index, will be the logarithm of the difference of longitude; which being applied by addition or subtraction to the longitude left, according as the latter is increasing or decreasing, the sum or difference will be the longitude come to. Example 1. A ship from Cape Ortegal, in latitude 43:47 N. and longitude 7:49′ W., sailed N.W. N. 560 miles; required the latitude and longitude of the place come to? Latitude of Cape Ortegal 43:47 N. Meridional parts Latitude come to . . . 51: 0 N. Meridional parts To find the Difference of Longitude, and hence the Longitude come to. Remarks. When a ship decreases her latitude; that is, when the difference of latitude made good is of a different name to the latitude sailed from; then, if the difference of latitude, expressed in degrees, be greater than the latitude left, their difference will be the latitude come to; which will be of a contrary denomination to that sailed from; because, in this case, it is evident that the ship must have crossed the Equator. And, when a ship decreases her longitude; that is, when the dif |