Examples.

Solution. There are given (fig. 15) the legs AC and BC.

1. A ship sails from latitude 3° 45′ S., upon a course N. by E., a distance of 2345 miles; to find the latitude at which it arrives, and the departure which it makes.

Ans. Latitude = 34° 35′ N.

2. A ship sails from latitude 62° 19′ N., upon a course W. N. W., till it makes a departure of 1000 miles; to find the latitude at which it arrives, and the distance sailed.

3. The bearing of Paris from Athens is N. 54° 56′ W.; find the distance and departure of these two places from each other.

Ans. Distance = 1135 miles.

4. A ship sails from latitude 72° 3′ S., a distance of 2000 miles, upon a course between the north and the west, that is, northwesterly, until it makes a departure of 1000 miles; find the latitude at which it arrives, and the course.

5. The distance from New Orleans to Portland is 1257 miles; find the bearing and departure.

Ans. Bearing N. 49° 24′ E.

Departure

954 miles.

6. The departure of Boston from Canton is 8790 miles; find the bearing and distance.

Traverse.

CHAPTER II.

TRAVERSE SAILING.

20. A traverse is an irregular track made by a ship when sailing on several different courses.

The object of Traverse Sailing is to reduce a traverse to a single course, where the distances sailed are so small that the earth's surface may be considered as a plane. [B. p. 59.]

21. Problem. To reduce several successive tracks of a ship to one; that is, to find the single track, leading to the place, which the ship has actually reached, by sailing on a traverse. [B. p. 59.]

Solution. Suppose the ship to start from the point A (fig. 17), and to sail, first from A to B, then from В to C, then from C to E, and lastly from E to F; to find the bearing and distance of F from A. Call the differences of latitude, corresponding to the 1st, 2d, 3d, and 4th tracks, the 1st, 2d, 3d, and 4th differences of latitude; and call the corresponding departures the 1st, 2d, 3d, and 4th departures. Then we need no demonstration to prove that

Diff. of lat. of A and F1st diff. of lat. 2d diff. of lat.

+3d diff. of lat. 4th diff. &c.;

or that the difference of latitude of A and F is found

To reduce a traverse to a single course.

by taking the sum of the differences of latitude corresponding to the northerly courses, and also the sum of those corresponding to the southerly courses, and the difference of these sums is the required difference of latitude.

By neglecting the earth's curvature, we also have,

Dep. of A and F 1st dep. 2d dep.- 3d dep. + 4th dep.

or the departure of A and F is found by taking the sum of the departures corresponding to the easterly courses, also the sum of those corresponding to the westerly courses; and the difference of these sums is the required departure.

Having thus found the difference of latitude and departure of A and F, their distance and bearing are found by $18.

22. The calculations of traverse sailing are usually put into a tabular form, as in the following example. In the first column of the table are the numbers of the courses; in the second and third columns are the courses and distances; in the fourth and fifth columns are the differences of latitude, the column, headed N, corresponding to the northerly courses, and that headed S, to the southerly courses; in the sixth and seventh columns are the departures, the column, headed E, corresponding to the easterly courses, and that, headed W, to the westerly courses. [B. p. 59.]

Examples.

23. EXAMPLES.

1. A ship sails on several successive tracks, in the order and with the courses and distances of the first three columns of the following table; find the bearing and distance of the place at which the ship arrives, from that from which it started.

2. A ship sails on the following successive tracks, South 10 miles, W. S. W. 25 miles, S. W. 30 miles, and West 20 miles; it is bound to a port which is at a distance of 100 miles from the place of starting, and its bearing is W. by S.