subtract the lesser sum from the greater; the remainder will be the total difference of latitude made good with the same name as the greater. Proceed similarly with the departures. The result by the traverse in this case shows that the whole difference of latitude made good is 41.9 miles south, and the departure 39 miles east. From these data we then compute the direct course and distance made good according to the formulas tan C Dep.÷D. Lat.

The course and distance made good may also be found directly by inspection of the traverse tables. It should be noted that the course takes its name from the difference of latitude and departure made good.

19. By construction, the solution is effected as follows: On a sheet of paper draw two lines NS and WE that intersect at right angles, to represent the direction of the cardinal points, Fig. 9. From the center C lay out the first course S W by S and the distance 24 miles according to an adopted scale; then, from a lay off the second course N N W, 57 miles; thence SE by EE, 84 miles; and, finally, the last course south, 35 miles. The point b is then the place at which the ship has arrived. Connect with C; also draw db perpendicular to N S; Cd is then the difference of latitude; db, the departure; Cb, the distance; and the angled Cb, the course made good by the ship. By measuring the different parts of the triangles thus formed, they will be found to agree exactly with results obtained by the traverse table.

EXAMPLE 1.-A ship has sailed the following courses, namely: SSEE, 16 miles; E S E, 23 miles; S W by W W, 36 miles; W &N, 12 miles, and S E by EE, 41 miles. Find the course and distance made good.

SOLUTION.-Construct a traverse table similarly to the one in the preceding example and take out from the table the corresponding D. Lat. and Dep. for each course; then calculate the course and distance made good according to the proper formulas, thus:

Course made good = S 18° 12′ E. Ans. Dist. made good=62.7 mi. Ans.

SOLUTION BY CONSTRUCTION.-Proceed as in the preceding example, by drawing lines representing the direction of north, south, east, and west, Fig. 10. From their point of intersection c lay off the successive courses with their respective distances; the point d will then mark the

position of the ship. Connect d with c and draw a perpendicular from d to the line NS; fd is then the departure;, the difference of latitude; and cd, the distance made good

EXAMPLE 2.-A ship from latitude 24°32′N sails the following courses: SW by W, 45 miles; ESE, 50 miles; SW, 30 miles; SE by E, 60 miles; and SWS, 63 miles. Required her latitude in, the departure, and the course and distance made good.

SOLUTION.-Arrange a traverse table as in the preceding examples.

In this case the departure made good is zero, hence; the course made good is south, and the distance equal to the D. Lat., or 149.2 mi.

Ans.

EXAMPLE 3.-A ship in latitude 1° S is bound to a port in latitude 1°10′ N, distance 220 miles in the northwest quarter; but, meeting with contrary winds, runs the following courses: NE by N, 63 miles; NWW, 85 miles; north, 96 miles; and NN W, 87 miles. Required the latitude of the place arrived at, and the course and distance from there to the point of destination.

SOLUTION.-To begin with, find the difference of latitude eg, Fig. 11, between the two places, which, in this case, is equal to the sum of their latitudes, or 60+70=130 miles. Then compute the departure g c according to formulas of Plane Sailing. The result will be 177.5 miles. Next find the departure df and the difference of latitude made good by the ship, by arranging a traverse table, or by laying off from the starting point the courses and distances sailed.