Dif. of lat. 58.7 Is to dep. 26.9 To find the course. Difference of Lat. N. Latitude left 6'6 6.6 10. 1.76864 S. 19.6 8.5 15. ... 22.2 65.3 6.6 58.7 = = Latitude in = 50 26 The whole difference of latitude and departure being now known, namely, 58.7 S. and 26.9 W., the corresponding course and distance can be found, as in the sixth example of the last problem, or as in article 209. 3.9 Sin course 10° 12.3 9.2 35.4 8.5 26.9 To find the distance. 9.61966 10. 1.42975 To tan course 24° 37′ 9.66111 To dist. 64.6 1.81009 The course is therefore S. 24° 37′ W., and the distance 64.6 miles. The course and distance can also be found by inspection with sufficient accuracy. In the page under 24° of a course is found in the departure column 26; and opposite to it, in the columns of latitude and distance, are 59.4 and 65, which shows that the course is 24° and the distance 65. These two proportions can also be performed by Gunter's scale as formerly. Construction. The different courses and distances may also be drawn as in the annexed diagram, and the equivalent course and distance measured. C Describe a circle bed, and let OL represent the meridian; draw the radii Oa, Ob, Oc, Od, Oe, corresponding to the different courses; then on Oa lay off the corresponding distance OA=20; draw AB parallel to Ob, and = 12; BC parallel to Oc, and = = 18; CD parallel to Od, and 14; and DE parallel to Oe, and = 24. Draw EL perpendicular to OL, then OL is the whole difference of latitude, EL the whole departure, and (supposing O and E joined) angle EOL the equivalent course, and OE the equi E valent distance. B EXERCISES. 1. A ship takes her departure from the Lizard W. light, in latitude 49° 58′ N., which then bears NNW., its distance being 15 miles, and sails SE. 34 miles, W. b S. 16, WNW. 39, and S. 6 E. 40; what is the latitude in, and the bearing and distance of the Lizard? Ans. Latitude in 48° 53′ N., bearing of Lizard N. 12° 16′ E., and its distance 66.8. 2. A ship's place is in north latitude 50° 36', and it sails during 24 hours in the following manner :-SSW. 54 miles, W.US. 39, NW.6 N. 40, NE. 6 E. 69, and NNW. 60; what is the latitude in, and the equivalent course and distance from the former place? Ans. Lat. in 51° 45', course N. 33° 54′ W. or NW. ¿ W., and distance 83.7. 505. If the ship has sailed in a current during any time, its effect for that time is allowed for as a separate course and distance. For instance, if the ship has been sailing for 10 hours under the influence of a current setting NE. at the rate of 21 miles per hour, the effect is the same as if the ship had sailed NE. 25 miles. GLOBULAR SAILING. 506. In globular sailing the methods of calculation are derived on the supposition that the earth is of a spherical form, and they apply with sufficient accuracy for the determination of the ship's place at any time, and the bearing and distance of the port bound for or of that left. Case I.-When the ship sails between two places on the same meridian. The difference of latitude is just the distance sailed, and the course is due north or south, and there is no difference of longitude. Case II.-When the ship sails on the equator. The distance sailed is the difference of longitude, the course is due east or west, and there is no difference of latitude. Case III. When the ship sails on the same parallel of latitude. 507. To find the distance, when the latitude is given, and the longitudes of the two places. Radius is to the cosine of the latitude, as the difference of longitude to the distance." Rad: cos lat. = dif. long. : distance. 508. To find the difference of longitude, when the latitude and distance are given. Cosine latitude is to radius as the distance to the difference of longitude.' Cos lat. : radius=distance : dif. long. 509. To find the latitude, when the distance and difference of longitude are given. "The difference of longitude is to the distance as radius to the cosine of the latitude." Dif. long. : distance radius: cosine lat. This case is sometimes called parallel sailing. The proportions in this case can be represented by this construction: ABC is a right-angled triangle, of which B is the right angle, AB the distance, AC the difference of longitude, and angle A the latitude. Then, when AC is radius,... Dif. long. : distance = radius : cos lat. Lat. L. Radius L. Cos lat 56° L. Dif. long. 505 Diff.of Long = 10. = 9.74756 A Dist. B EXAMPLE. The longitudes of two places in the latitude of 56° S. are 140° 20′ and 148° 45′; find the distance. Dif. of long. =8° 25′505 miles. To find the distance. C L. Distance 282.4 = 2.45085 The proportion can be derived from the figure in art. 502. Let O be the centre of the earth's equator, and OQ its radius; P the centre of the parallel of latitude at B, and PB its radius; then the distance between two meridians, measured on the equator, is to their distance on the parallel at B as OQ: PB, that is, dif. long. : dist. = radius: cos lat. EXERCISES. 1. A ship, in latitude 49° 30', sails due E. till her difference of longitude is 3° 30'; what is the distance sailed? Ans. 136.4 miles. 2. A ship sails 136.4 miles due W. on the parallel of latitude 49° 30'; required the difference of longitude made. Ans. 210 miles. 3. A ship sails 136·4 miles due E., and her difference of longitude is then 3° 30′; on what parallel of latitude did it sail? Ans. Latitude 49° 30'. Case IV. -When the course is compound. 510. Method I.-The first method of solution is by middle latitude sailing. This method combines plane and parallel sailing; and in it, it is supposed that the departure made by a ship is equal to the meridional distance on the middle parallelthat is, the meridional distance EF (fig. to article 502) on the parallel of latitude in the middle between the latitudes of A and B, the latitude left and that arrived at, is equal to the sum of the elementary meridian distances Gg, Hh,...which it nearly is. There are two proportions used, namely 511. The difference of latitude is to the difference of longitude, as the cosine of the middle latitude to the tangent of the course." Dif. lat.: dif. long cos mid. lat. : tan course. 512. The cosine of the middle latitude is to the sine of the course, as the distance to the difference of longitude." Cos mid. lat. : sin course distance: dif. long. 513. These two analogies can be obtained by means of two right-angled triangles, ABC, DBC, D, having their right angles at C. In the triangle ABC, angle A is the course, AB the distance, BC the departure, and AC the difference of latitude; and in the tri- C angle BCD, BC is the departure, angle B the middle latitude, and BD the difference of longitude. ABC is a triangle in plane sailing, and BCD in parallel sailing. (See fig. to art. 509.) A The two proportions given above may be changed by alternation, inversion, &c., so as to make any term the last, or the unknown part, the other three being supposed to be given. Besides these two proportions, several others can be deduced from the triangle ABC, as in plane sailing, and from the triangle DCB, as in parallel sailing (509.) But those obtained from triangle ABC can be used only for a short distance. In comparing the triangle BCD to ABC in article 509, angle B in the former corresponds to A in the latter, and the departure BC in the former to the distance AB in the latter; also the differences of longitude in both correspond. D. Long. M.Lat Dep. Course. B Distance. |