difference of latitude is 139. 6, under distance 220; twice the latter, or 440 miles, is, therefore, the distance sailed; and twice 139. 6 = 279.2 miles, or 4:39, is the difference of latitude: whence the latitude in, is 27:36 N., and the meridional difference of latitude 322. 1. Now, to course 4 points, and half the meridional difference of latitude =161 miles, in a latitude column, the corresponding departure is 196.3; the double of which, or 392.6 miles, is the difference of longitude: hence, the longitude come to is 71:19 W. PROBLEM IV. Given both Latitudes and the Course; to find the Distance and the This comes under the 6th analogy, page 237; therefore, This comes under the 10th analogy, page 237; therefore, To find the Distance, and the Difference of Longitude, by Inspection. To the course 3 points, and one third of the difference of latitude 180, the corresponding distance is 233; which multiplied by 3, gives 699 miles, the distance.-And, to course 3 points, and one third of the meridional difference of latitude = 225.4, in a latitude column, the corresponding departure is 185. 2; now, this being multiplied by 3, gives the difference of longitude =555.6 miles. PROBLEM V. Given both Latitudes and the Distance; to find the Course and Difference of Longitude. Example. A ship from Urris Head, Broad Haven, in lat. 54:21 north, and longitude 10:2? west, sailed 900 miles upon a direct course between the south and west, and then by observation was found to be in latitude 43°14' north; required the course steered, and the longitude come to? 1018.2 Difference latitude 11: 7667 miles. Merid. diff. lat. = This falls under the 4th analogy, page 237: hence, This falls under the 10th analogy, page 237: hence, As radius = 90:0: Log. co-secant=10.000000 3.007834 Is to mer. diff. of lat. 1018. 2 miles, Log..= To the diff. longitude 922.4 miles, Log. = . 2.964921 Longitude of Urris Head = = To find the Course, and the Difference of Longitude by Inspection. One-fourth of the distance = 225, and one-fourth of the difference of latitude = 166. 7, are found to agree at 33 points; which, therefore, is the course. Now, to course 33 points, and one fifth of the meridional difference of latitude = 203. 6, the departure nearest agreeing is 184.7; this being multiplied by 5, gives 923. 5 miles; which, therefore, is the difference of.longitude :-differing about one mile from the result by calculation. PROBLEM VI. Given one Latitude, Distance, and Departure; to find the other Latitude, the Course, and the Difference of Longitude. This comes under the 4th analogy, page 237; hence, As the distance = 1300 miles, Log. ar. compt. 6.886057 10.000000 2.963788 This comes under the 1st analogy, page 237; hence, 90:0: Log. co-secant = 10.000000 As radius= Lat. at which the ship arrived = 21:44 N. Meridional parts = 1336.4 Meridional difference of latitude = To find the Difference of Longitude = DE: This comes under the 10th analogy, page 237; hence, As radius = 90:0: Log. co-secant = 10.000000 3.025306 So is the course=45:2:52" Log. tangent. 10.000724 1060.0 3.026030 The course steered is S. 45:3 W., or S. W., very nearly. 26:42. W. To find the Course, and the Latitude and Longitude by Inspection. One-fifth of the distance = 260, and one-fifth of the departure = 184, are found to agree nearest at 45; which, therefore, is the course; and the corresponding difference of latitude is 183. 8, which, multiplied by 5, gives 919 miles, the difference of latitude: whence, the latitude come to is 21:44: north, and the meridional difference of latitude 1060 miles. Now, to the course 45, and one fifth of the meridional difference of latitude = 212, in a latitude column, the corresponding departure is 212.1; and this being multiplied by 5, gives the difference of longitude 1060.5 miles; hence, the longitude at which the ship arrived is 26:41. west. = PROBLEM VII. Given both Latitudes, and the Departure; to find the Course, Distance, and Difference of Longitude. This comes under the 5th analogy, page 237; hence, |