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 B 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 F = 1st 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 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,

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.]

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.

Required the bearing and distance of the port to which the ship is bound, from the place at which it has arrived.

Ans. Bearing = N. 57° 47′ W.

Distance = 40 miles.

CHAPTER III.

PARALLEL SAILING.

24. PARALLEL SAILING considers only the case where the ship sails exactly east or west, and therefore remains constantly on the same parallel of latitude. Its object is to find the change in longitude corresponding to the ship's track. [B. p. 63.]

25. Problem. To find the difference of longitude in parallel sailing. [B. p. 65.]

Solution. Let A B (fig. 18) be the distance sailed by the ship on the parallel of latitude A B. As the course is exactly east or west, the distance sailed must be itself equal to its departure.

The latitude of the parallel is ADA' or A A'. The angle AEB = A'DB', or the arc A' B, is the difference of longitude. Denote the radius of the earth A'D = B'D = AD by R, and the radius of the parallel AE = BE byr; also the circumference of the earth by C, and that of the parallel by c.

Since AB and A' B' correspond to the equal angles AEB and A'D B', they must be similar arcs, and give the proportion,

or

AB: A' B' = c : C,

Dep.: diff. long. = c : C.

But, as circumferences are proportional to their radii,

c:C=

r:R.

Hence, leaving out the common ratio,

Dep.: diff. long. =r: R.

Putting the product of the extremes equal to that of the means,

r. diff. of long. = R. departure.

But, in the triangle ADE, since

DAE=AD A' = latitude,

we have, from (22),

r

R X cos. lat.

which, substituted in the above equation, gives, if the result is divided by R,

26. Corollary. Since the distance is the same as the departure in parallel sailing, the word distance may be substituted for departure in (223) and (224).

27. Corollary. It appears, from (223) and (224,) that if a right triangle (fig. 18) is constructed, the hypothenuse of which is the difference of longitude, and one of the acute angles the latitude, the leg adjacent to this angle is the departure. All the cases of parallel sailing may, then, be reduced to the solution of this triangle.

28. Problem. To find the distance between two places which are upon the same parallel of latitude.

Solution. This problem is solved at once by (223).

29. The Table, p. 64, of the Navigator, which "shows for every degree of latitude how many miles distant two meridians are, whose difference of longitude is one degree," is readily calculated by this problem.