8. A ship sails from Nantucket upon E., until its departure is 2274 miles; sailed, and the place arrived at. 131 miles. a course S. 62° 11′ find the distance Ans. Distance 2571 miles. The place arrived at is 261 miles north of Santa Cruz. 9. A ship sails southwesterly from Land's End (England), a distance of 3461 miles, when its departure is 3300 miles; find the course and the place arrived at. The place arrived at is Charleston (South Carolina). To find the difference of longitude. CHAPTER V. MERCATOR'S SAILING. 41. The object of Mercator's Sailing is to give an accurate method of calculating the difference of longitude. [B. p. 78.] 42. Problem. To find the difference of longitude, when the distance, the course, and one latitude are known. Solution. Let AB (fig. 16) be the ship's track. Divide it into the small portions A a, a b, bc, &c., which are such that the difference of longitude is the same for each of them, and let Let also d = this small difference in longitude. L = the latitude of A, L' — the latitude of B, = 7 = the latitude of one of the points of division as b, 'the latitude of the next point c, C: the course. The distance bc may then be supposed so small, that the formulas of middle latitude sailing may be applied to it; and (232) gives Difference of longitude. d = (l') X tang. CX sec. (+1), (239) by (14) But,(' -1) is a small arc expressed in minutes, and and (243) may be written in the usual form of a proportion sin. (l' — 7) : cos. ≥ (l' + 7) =m: 1; whence, by the theory of proportions, cos. (l' + 1) + sin. † (l' — 1) cos. (l' + 1) sin.( But if in (47) we put we have (244) m = A = 90° — (l' + 1), B = ¦ (l' — 1), A+B = 90° — 1, A and (47) becomes cos.(+)+sin.(-1) (246) Hence, the successive values of cotan. (45° — § 7) at the points A, a, b, &c., form a geometric progression; and if D= the difference of longitude of A and B, n the number of portions of AB; (251) cotan. (45° L') cotan. (45°- L) M", (252) and by logarithms = log.cotan.(45°-L')—log.cotan. (45°—3 L)=log.M" (253) which, substituted in (253), gives by a simple reduction Meridional difference of latitude. Now the value of cosec. 1' log. e ―log. cotan. (45° — 1L) has been calculated for every mile of latitude, and inserted in tables. [B. Table III.] It is called the Meridional Parts of the Latitude, and the method of computing it is given in the following section. The difference between the meridional parts of the two latitudes, when the latitudes are both north or both south, is called the Meridional Difference of Latitude; but when one of the latitudes is north and the other south, the sum of the meridional parts is the meridional difference of latitude. Hence (257) gives D = diff. long. mer. diff. lat. X tang. course. (258) 43. Corollary. The difference of longitude is as in (fig. 20) the leg DE of a right triangle, of which AD is the meridional difference of latitude, and the angle A the course; and by combining this triangle with the triangle ABC of plane sailing, all the cases of Mercator's Sailing are reduced to the solution of these two similar right triangles. 44. Problem. To calculate the Table of Meridional Parts. I. In finding the value of e, the portions of the distance are supposed to be infinitely small, hence m is by (243) also infinitely small, and its reciprocal is infinitely great. |