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

rx diff. long. RX departure.

But, in the triangle ADE, since

=

EADA'DA = latitude,

we have, from (22),

r=R× cos. lat.,

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

diff. long. X cos. lat. departure (or distance).

(223)

Hence, by (6),

dep. (or dist.)

diff. long. =

cos. lat.

= dep. (or dist.) X sec. lat. (224)

26. 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).

27. 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 formula.

28. Corollary. It appears from (223) and (224) that if a right triangle (fig. 18) is constructed of which the hypothenuse 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.

29. EXAMPLES.

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

Ans. Longitude sought 48° 20' W.

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

Ans. Distance 3250 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.

4. Find the difference of longitude which corresponds to a departure of one league, or three sea miles, in the latitude of 72°.

CHAPTER IV.

MIDDLE LATITUDE SAILING.

30. Middle Latitude Sailing is an approximate method of solving those problems of rhumb sailing which involve the consideration of the Difference of Longitude. It is properly applicable to cases in which the difference of latitude is small, and consists in calculating the difference of longitude from the departure or the departure from the difference of longitude by the formulas of Parallel Sailing, on the hypothesis that the departure is equal to the distance between the extreme meridians measured at the Middle Latitude; that is, at the latitude of the middle point of the rhumb. [B., p. 66.]

If A and B (fig. 71) are situated on the same side of the equator, A being in the higher latitude, their departure is less than the meridional distance BB' and greater than the meridional distance AA', since each of the partial departures, as ma, is less than the corresponding arc of BB' and greater than the corresponding arc of AA'. Hence, the departure of A and B must be equal to the meridional distance measured on some intermediate parallel, DD'; so that the departure and difference of longitude of A and B are the same as those of D and D'. Since the meridional distance regularly increases, as we go from AA' to BB', it is natural to take the middle parallel as an approximation to the position of DD'; and it is evident. that if the difference of latitude is small, little error can result from this assumption, especially if the places are near the equator, where the meridians converge but slightly.

It is evident that

mid. lat. of two places sum of their lats.

=

either lat. ± 1 diff. lat. (225)

31. Corollary. By combining the triangle (fig. 15) of Plane Sailing with that (fig. 18) of Parallel Sailing, and making the latitude in the latter equal to the middle latitude, we obtain a triangle (fig. 19) by which all the cases of Middle Latitude Sailing can be solved.

32. Problem. To find the difference of latitude, the departure, and the difference of longitude, when the course and the distance are known, and the latitude of one extremity of the ship's track. [B., p. 71.]

Solution. The triangle (fig. 19) gives at once, as in Plane Sailing dist. X cos. course

diff. lat.

The middle latitude may then be found by (225); and we have, as in (224),

dep.

diff. long. =

= dep. X sec. mid. lat.

(226)

cos. mid. lat.

diff. long. dist. X sin. course X sec. mid. lat.

(227)

or, by substituting the value of the departure,

33. Problem. To find the bearing and the distance of two given places from each other.

Solution. The places being

[B., p. 68.]

given, their latitudes and longitudes

are supposed to be known, so that the diff. lat., mid. lat., and diff. long. are easily found. Then we have (fig. 19), by the principles of the solution of right triangles,

departure diff. long. X cos. mid. lat.

(228)

(229)

(230)

34. Problem. To find the course, the distance, and the difference of longitude, when both latitudes and the departure are given. [B., p. 70.]

Solution. The difference of longitude is found by (226), the course by (229), and the distance by (230).

35. Problem. To find the departure, the distance, and the difference of longitude, when both latitudes and the course are given. [B., p. 72.]

Solution. The departure is found by the formula

departure diff. lat. X tang. course;

(231)

the distance by (230); and the difference of longitude may be found by (226), or by substituting (231) in (226), as follows,

diff. long.: 36. Problem. To find the course, the departure, and the difference of longitude, when both latitudes and the distance are given. [B., p. 73.]

= diff. lat. X tang. course X sec. mid. lat. (232)

Solution. The course is found by the formula

Cos. course =

diff. lat.
dist.

;

(233)

the departure by (212) or by the formula

departure = ✔[(dist.)? — (diff. lat.)2];

(234)

and the difference of longitude by (226) or (227).

37. Problem. To find the difference of latitude, the distance, and the difference of longitude, when one latitude, the course, and the departure are given. [B., p. 74.]

Solution. The difference of latitude is found by the formula

diff. lat.= dep. X cotan. course;

the distance by the formula

dist. = dep. X cosec. course;

(235)

(236)

the mid. lat. by (225); and the difference of longitude by (226).

38. Problem. To find the course, the difference of latitude, and the difference of longitude, when one latitude, the distance, and the departure are given. [B., p. 75.]

Solution. The course is found by the formula

the difference of latitude by (211) or by the formula

diff. lat. [(dist.)2- (dep.)2];

=

and the difference of longitude by (226).

(237)

(238)

39. Scholium. If two places are in opposite latitudes, the middle latitude may evidently differ considerably from the latitude of DD';