To find the difference of longitude. By Mid. Lat. Sailing. 2.38561 Dif. long. 390.5 2-59162 = By Mercator's Sailing. Dif. lat. 346(a.c) 7·46092 Dif. long. 391.2 2.59239 Long. left = 36° 45′ W. Long. in 43 15 W. Long. in =43 16 W. The place arrived at is therefore in latitude 48° 38' N., and longitude 43° 16′ W. EXERCISE. A ship from latitude 37° N., longitude 48° 20' W., sails between the north and east till her difference of latitude and departure are 855 and 564 miles; required the latitude and longitude in. Ans. Latitude 51° 15' N., longitude in 35° 14′ W., by mid. lat. sailing, and 35° 8′ W. by Mercator's. 521. PROBLEM IV.-To perform a traverse or compound course by middle latitude and Mercator's sailing. "Form a traverse table, and find by it the whole difference of latitude and the departure; then find the latitude in and the course made good, as in article 504. Find then the middle latitude between that left and that arrived at, or find the meridional difference of latitude for these two latitudes; then, to find the difference of longitude, Cos mid. lat. radius dep. : dif. long. by mid. lat. sailing. : Or, rad: tan course Mer. dif. lat. : dif. long. by Mer. sailing. EXAMPLE. Find the longitude and latitude of the place. of the ship arrived at, after sailing the various courses and distances given in the example of a traverse in plane sailing in article 504, supposing the longitude left to be 23° 40′ W. Construct the traverse as in that example, and it will be found that the difference of latitude is 587 S., and the departure 26-9 W. Hence, Dif. long. 42.2 Long. in Then dif. lat. : dep. = radius : tan course, and, as found in that example, the course is S. 24° 37′ W. To find the difference of longitude. Mer. parts Mer. dif. lat. By Mid. Lat. Sailing. Cos mid. lat. 50° 25′ 9.80428 Radius 1.62547 = 23° 40′ W. = 0 42 W. In Plane Sailing. Course Dif. lat. By Mercator's Sailing. 10. Tan course 24° 37′ 9.66104 1.42975 Mer. dif. lat. 94 1.97313 Dif. long. 43.1 = 24 22 W. Long. in 3609 3515 94 22. Since the difference of latitude, distance, and course, are the same parts of a right-angled triangle in plane sailing that the departure, difference of longitude, and middle latitude, are in middle latitude sailing; and the difference of latitude, departure, and course, are the same parts of the triangle in plane sailing that meridional difference of latitude, difference of longitude, and course, are in Mercator's sailing; therefore the difference of longitude for the last two proportions can be found, by inspection, in the table of the difference of latitude and departure. corresponds to 1.63417 = 23° 40′ W. = 0 43 W. = 24 23 W. In Mid. Lat. Sailing. Mid. lat. In Plane Sailing. Course Dif. lat. In Mercator's Sailing. Mer. dif. lat. corresponds to ... Thus, for the proportion above by middle latitude sailing, in the table of difference of latitude and departure in the page for course = 50°, and departure 26-9 in the difference of latitude column, there is 42 in the distance column for the difference of longitude, as above; and for the proportion by Mecator's sailing, in the same table for course = 25° (for 24° 37′), and meridional difference of latitude 94 in the difference of latitude column, there is 44 for the difference of longitude in the departure column. 523. When great accuracy is required, or when sailing in high latitudes, it would be preferable to calculate the difference of longitude for each course and distance, supposing the distances not to exceed a few miles, instead of merely finding the difference of longitude on a whole day's sailing. This can be done by first constructing the usual traverse table, by means of which the latitude at the beginning and end of each distance sailed can be found; and hence also the middle latitude and the meridional difference of latitude, and then the difference of longitude for each distance can be found, as in the two last proportions, by both middle latitude and Mercator's sailing, either by calculation or inspection; and hence the latitude and longitude for the beginning and end of each distance is found, and hence also for the place arrived at. This method, however, is tedious, and is seldom used; the former method generally leads to no greater errors than the unavoidable errors of the methods of measuring the course and the distance by the compass and log. The former method may be called the plane traverse, and this method the globular traverse. EXERCISES. 1. A ship from a place in latitude 50° 6′ N., and longitude 5° 55′ W., is bound to a port in the island of St Mary's in latitude 37° N., and longitude 25° 6′ W., and steers the following courses:-S. 6 W. 24 miles, WSW. 32, NW. W. 41, SSE. E. 49, ENE. & E. 19, W. 21, NE. E. 36, S. 41, SSW. 92, and N. 36; what is the latitude and longitude in, and also the direct course and distance to the intended port? Ans. Latitude in 48° 9', and longitude in 7° 19', by mid. lat. and by Mercator's sailing. Or, Course 49° 35', and dist. 1032, by mid. lat. sailing. Course 49° 29', and dist. 1030, by Mer. sailing. In the following exercise, the difference of longitude is found on each course, as explained in article 523, that is, by the globular traverse : 2. A ship from latitude 68° 38′ N., and longitude 8° 40' E., is bound for the North Cape in latitude 71° 10′ N., and longitude 26° 3′ E., sails the courses and distances in the subjoined table; what is the latitude and longitude in, and the direct course and distance of the Cape? Courses. NE. 6 N. NNE. N. NW.6N. N. 6 E. NE. 6 E.E. SE. ENE. Or, 276.2 219 714-2 Ans. Latitude in = 73° 14′, longitude in 20° 34′ E. The course required is S. 39° 3′ E., distance 159.7, by mid. lat. sailing. The course is S. 39° 3′ E., distance 160, by Mercator's sailing. 3. A ship in latitude 67° 30′ N., longitude 8° 46′ W., sails NE. 64 miles, NNE. 50, NW.6 N. 58, WNW. 72, W. 48, SSW. 38, S. b E. 45, and ESE. 40; what is the latitude and longitude in? Ans. By the plane traverse, the lat. in is 68° 43′ N., and long. 11° 3′ W.; and by the globular traverse, the long. in is 11° 37′ W. by mid lat., and 11° 43′ by Mer. sailing. Departure of a Ship. The place of departure of a ship, that is, the place from which the beginning of a voyage is reckoned, is generally some promontory or other convenient object whose latitude and longitude are known; and as the vessel is usually some miles distant from it, observations must be taken to determine this distance. 524. PROBLEM V. Given the bearing of a headland from a ship at two places, and the distance and direction sailed between them, to find the distance of the promontory from the ship. a Let P be the promontory; S and H the two places of the ship; mn and ab parts of the meridians through H and S; then angle mSP, the bearing of P from H S, is given; and angle mSH or HS, the ship's course; and also angle PHb, the bearing of P from H. Therefore, in the triangle PHS, the angle at S= mSP+mSH is known, and that at H PHỏ - 6HS is also known, and the side HS; therefore the distances PH and PS can be found (p. 84, Part I.) EXERCISES. n m P 1. A headland was observed from a ship to bear NE. 6 N., and after sailing 7.5 miles on a NNW. course, the headland then bore ESE.; required the distance of the headland from both places of the ship. Ans. 5.4 and 6.3 miles. |