Examples.

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

Differences of longitude in parallel sailing.

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 AB. As the course is exactly east or west, the distance sailed must be itself equal to its departure.

The latitude of the parallel is A D A' or A A'. The angle А Е В = A' D B', 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 A E B E by r; 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'DB', they must be similar arcs, and give the propor

Difference of longitude.

But, as circumferences are proportional to their radii, c: Cr: R.

Hence, leaving out the common ratio,

Dep. diff. long. = r : R.

Putting the product of the extremes equal to that of the

is divided by R,

r=RX cos. lat.

which, substituted in the above equation, gives, if the result

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.

Differences of places on the same parallel.

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.

30. EXAMPLES.

1. A ship sails from Boston 1000 miles exactly east; find the longitude at which it arrives.

2. Find the distance of Barcelona (Spain) from Nantucket (Massachusetts).

Ans. Distance 3252 miles.

3. Find the distance between two meridians, whose difference of longitude is one degree in the latitude of 45°.

Ans. Distance 42.43 miles.

Middle latitude.

CHAPTER IV.

MIDDLE LATITUDE SAILING.

31. The object of Middle Latitude Sailing is to give an approximative method of calculating the difference of longitude, when the difference of latitude is small. [B. p. 66.]

32. Problem. To find the difference of longitude by Middle Latitude Sailing, when the distance and course are known, and also the latitude of either extremity of the ship's track. [B. p. 71.]

Solution. The difference of latitude and departure are found by (211) and (212),

Diff. lat. = dist. X cos. course

Departure dist. X sin. course.

The difference of longitude may then be found by means of (224). But there is a difficulty with regard to the latitude to be used in (224); for, of the two extremities of the ship's track, the latitude of one is smaller, while the latitude of the other extremity is larger than the latitude of the rest of the track. Navigators have evaded this difficulty by using the Middle Latitude between the two, as sufficiently accurate, when the difference of latitude is small. Now the middle latitude is the arithmetical mean between the latitudes of the extremities, so that we have,