made good, will be at once understood from the following examples. EXAMPLES. 1. A ship sails from a place in lat. 24° 32′ N., and has run the following courses and distances, viz. 1st, S. W. by W., distance 45 miles; 2d, E. S. E., distance 50 miles; 3d, S. W., distance 30 miles; 4th, S. E. by E., dtstance 60 miles; 5th, S. W. by S., W.. distance 63 miles : required her present latitude, with the direct course and dis tance from the place left to the place arrived at. It appears from the results of this table that the difference of latitude made by the ship during the traverse is 149.2 S. =2° 29′ S. It appears also that the departures east are equal to the departures west, so that the ship has returned to the meridian she sailed from, consequently the direct course from the place left to that come to is due south, and the distance is equal to the difference of latitude, which is 149.2 miles. The construction of the traverse is as follows: With the chord of 60°, taken from the line of chords on the mariner's scale, describe the horizon circle, and draw the north and south line N. S. From the line of rhumbs take the chords of the several courses, and as these are all southerly they must be laid off from the south point S, those which are westerly to the left, and those which are easterly to the right, their extremities being marked 1, 2, 3, &c., in the order of the courses. This done, lay off from any convenient scale of equal parts, and in the direction of Al the distance AB sailed on the first course; then in the direction A N E parallel to 12, the distance BC sailed on the second course; in the direction parallel to A3, the distance CD on the third course; in the direction parallel to A4, the distance DE on the fourth course; and, lastly, in the direction parallel to 15, the distance EF on the fifth course; then F will represent the plane of the ship at the end of the traverse; FA, being applied to the scale of equal parts, will show the distance made good, and the chord of the arc included between this distance, and the meridian, being applied to the line of rhumbs, will show the direct course. In the present case the intercepted arc will be 0, showing that F is on the meridian of A. 2. A ship from Cape Clear, in lat. 51° 25' N., sails 1st, S.S.E. E., 16 miles; 2d, E.S.E., 23 miles; 3d, S.W. by W. W., 36 miles; 4th, W. & N., 12 miles; 5th, S.E. by E. † E., 41 miles required the distance made good, the direct course, and the latitude in? Then by the formulæ for the solution of right-angled plane 1.797635 ; tan, course 18°12′ 9.517010 : distance 62.74 therefore, as the difference of latitude is south, and the departure east, the direct course is S. 18° 12' E., and the distance made good 62.74 miles. To construct this traverse, describe, as before, the horizon circle, with a radius equal to the chord of 60°, and taking from the line of rhumbs the chord of the first course, 21 points, apply it from S. to 1, to the right of S.N., as this course is south-easterly; apply, in like manner, the chord of the second course, six points from S. to 2, also to the right of the meridian line; apply the chord of the third course, 5 points from S. to 3, to the left of the meridian, the chord of the fourth course, 7 from N. to 4, to the left of N. S., this course being north-westerly, and, lastly, apply the chord of the fifth course, 5 points, from S. 4. 36; in the direction parallel to A4, lay off DE 12 miles; and, lastly, in the direction parallel to A5, lay off EF = then F will be the place 41; of the ship at the end of E the traverse; consequently, AF will be the distance made good, and the angle FAs the direct course; applying, therefore, the distance AF to the scale of equal parts, we shall find it reach from 0 to 622; and applying the distance Sa to the line of chords, we shall find it reach from 0 to 180. 3. A ship from lat. 28° 32′ N., has run the following courses, viz. 1st, N.W. by N., 20 miles; 2d, S.W., 40 miles; 3d, N.E. by E., 60 miles; 4th, S.E. 55 miles; 5th, W. by S., 41 miles; 6th, E.N.E., 66 miles. Required her present latitude, the distance made good, and the direct course from the place left to that come to. The direct course is due east, and distance 70.2 miles, the ship being in the same latitude at the end as at the beginning of the traverse. 4. A ship from lat. 41° 12' N., sails S.W. by W:, 21 miles; S.W. S. 31 miles; W.S.W. S., 16 miles ; S. & E., 18 miles; S.W. W., 14 miles; and W. N., 30 miles : required the latitude of the place arrived at, and the direct course and distance to it. Lat. 40° 5' N.; course S. 52° 49' W.; distance 111.7 miles. 5. A ship runs the following courses, viz. 1st, S.E., 40 miles; 2d, N.E., 28 miles; 3d, S. W. by W., 52 miles; 4th, N.W. by W., 30 miles; 5th, S.S.E., 36 miles; 6th, S.E. by E., 58 miles: required the direct course and distance made good. Direct course S. 25° 59' E., or S.S.EE. nearly; distance 95.87 miles. These examples will, perhaps, suffice to illustrate the principles of plane sailing, in which, course, distance, difference of latitude, and departure, are the only things concerned. The determination of the difference of longitude made on any course which is the distance between the meridians measured on the equator, cannot be effected by these principles, for this element is not the same as if the meridians were all parallel to each other, as is the case with the other elements. The finding of the difference of longitude is the easiest when the ship sails due east or due west, that is upon a parallel of latitude; this is called parallel sailing. Parallel Sailing. 99. The theory of parallel sailing is comprehended in the following proposition, viz : The cosine of the latitude of the parallel is to the distance run as the radius to the difference of longitude. This may be demonstrated as follows |
