514. Method II.-The second method of solution is by Mercator's sailing. In this method the surface of the earth is considered to be plane, the meridians being parallel lines, and also the parallels of latitude, as in plane sailing; and since by this hypothesis the distance between the meridians is increased, the lengths of the arcs of the meridians are increased in the same proportion, so that the distances between the parallels of latitude for every successive minute are continually increasing with the latitude; and the relative bearings of places are thus preserved. This method is so accurate, that it may be used without sensible error for any distance on the earth's surface. The lengths of the meridians from the equator to any latitude are thus increased, and the increase is greater the higher the latitude. For instance, the increased distance of the parallel of 10°, instead of being 600 miles, is found to be 603 miles; and that of the latitude of 50°, instead of 50 x 60 3000 miles, is 4527 miles. The increased lengths of the meridians, from the equator to any latitude, are called the meridional parts, from the manner in which they are computed; and their numerical values are contained in tables. = 515. The difference of the meridional parts for any two latitudes, is called the meridional difference of latitude; and the true difference of latitude is sometimes, for distinction, called the proper difference of latitude. The analogies peculiar to this method are :— 516. The difference of latitude is to the departure, as the meridional difference of latitude to the difference of longitude." Dif. lat. : departure mer. dif. lat. : dif. long. 517. The meridional difference of latitude is to the difference of longitude, as radius to the tangent of the course.' Mer. dif. lat. : dif. long. = radius: tan course. These two proportions may be varied so as to have any term last; and the first three being known, it could be found. Some of these proportions would, however, be practically useless. These proportions can be obtained from two right D Dif. Long. E angled triangles, ABC, ADE; in {Mer Dif of Lal Dif. Lat. From the triangle ABC, other proportions can be obtained; but as they are the same as in plane sailing, they are not to be depended on for any distance exceeding a few miles, nor in high latitudes, even for small distances. A. Dep. 518. PROBLEM I.-Given the place left and that bound for, to find the course and distance. 1. By Middle Latitude Sailing. Dif. lat. : dif. long cos mid. lat. : tan course. To find the distance. Cosine course: radius = dif. lat. : distance. Mer. dif. lat. : dif. long. -radius: tan course. By Mid. Lat. Sailing. Liverpool, lat. = 53° 25′ N. N. York, lat. = 40 42 N. Dif. lat. = 12 43 Mid. lat. To find the distance. Cosine course : radius dif. lat. : distance. The second proportion in the two methods is the same. EXAMPLE.-Required the bearing and distance of New York from Liverpool. Distance.} J - Dif. long. = 71° = 4260 mls. To find the course. 711748 3.62941 Dif. lat. 763 (a.c) Tan course 75° 16′ 10·58027 Mer. dif. lat. 1128 3·05231 3.62941 Tan course 75° 10′ 10.57710 To find the course. 10. Cos course 75° 16′ 9.40538 Cos course 75° 10′ 9.40825 10. Dif. lat. 763 2-88252 Dif. lat. 763 2.88252 Distance 3000 3-47714 Distance 2980 3.47427 EXERCISES. 1. What is the bearing and distance of a place in latitude 71° 10' N., longitude 26° 3′ E., from another place in latitude 60° 9′ N., and longitude 0° 58′ W.? Ans. Course N. 45° 19' E., distance 940 miles, by mid. lat. sailing. Course N. 44° 48′ E., distance 938.5 miles, by Mer cator's sailing. 2. A ship having arrived at a place in latitude 49° 57' N., longitude 5° 14′ W., and is bound for another place in latitude 37° N., longitude 25° 6′ W.; required the bearing and distance of the latter place from the former. Ans. Course S. 48° 4′ W., distance 1163 miles, by mid. lat. sailing. Course S. 47° 54' W., distance 1159 miles, by Mercator's sailing. 519. PROBLEM II.-Given the place sailed from, the course and distance, to find the place arrived at. By Middle Latitude Sailing. To find the difference of latitude. To find the difference of longitude. The analogy is the same as in the preceding method. Or, By Mercator's Sailing. To find the difference of longitude. By the first proportion in these two methods, the difference of latitude is found, and by the second, the difference of longitude; and hence the latitude and longitude of the place in are known. Lat. left Dif. lat. EXAMPLE.-A ship from a place in latitude 25° 40′ S., and longitude 35° 12′ W., sails SW. b S. 246 miles; required the latitude and longitude in. To find the difference of latitude by both methods. 10. Lat. in Rad Cos course 3 points 2-18722 Dif. long. 154.4 1594 1825 231 Long. in = 37 46 W. The two methods give the differences of longitude to within less than a mile of each other. The place arrived at is in latitude 29° 5' S., and longitude 37° 46′ W. 1. A ship from a place in latitude 50° 30' N., and longitude 145° 20′ W., sails 450 miles SSW.; what place is it in? Ans. Latitude 43° 34' N., longitude 149° 33′ W., by mid. lat. sailing, and 149° 33′ W., by Mercator's sailing. EXERCISES. 2. A ship from latitude 51° 15′ N., and longitude 9° 50′ W., sails SW. b S. till the distance run is 1022 miles; what are the latitude and longitude in? Ans. Latitude 37° 5' N., longitude 23° 2′ W., by mid. lat. sailing, and 23° 8′ W. by Mercator's sailing. 520. PROBLEM III.-Given the latitude left, the difference of latitude and departure, to find the difference of longitude. By Middle Latitude Sailing. Cos mid. lat. radius dep. : dif. longitude. : = Lat. left Lat. in By Mercator's Sailing. Dif. lat. : dep. = mer. dif. lat. : dif. long. EXAMPLE.-A ship on a course between the south and west from latitude 54° 24′ N., and longitude 36° 45′ W., has made 346 miles of difference of latitude, and 243 miles of departure; what is the latitude and longitude in? = 54° 24′ N. = 48 38 N. = 51 31 Mer. parts Mer. dif. lat. 3906 3349 557 |