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30. EXAMPLES.

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

Ans. Longitude sought=48° 32′ W.

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°.

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CHAPTER IV.

MIDDLE LATITUDE SAILING.

31. THE object of Middle Latitude Sailing is to give an ap proximative 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),

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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,

Middle lat. =

sum of the lats. of the extremities of the

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or, by substituting (212),

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

(227)

diff. long. "This method of calculating the difference of longitude may be rendered perfectly accurate by applying to the middle latitude a correction," which is given in the Navigator, and the method of computing which will be explained in the succeeding chapter. [B. p. 76.]

33. By combining the triangle (fig. 15) of Plane sailing with that (fig. 18) of Parallel sailing, a triangle (fig. 19) is obtained, by which all the cases of Middle Latitude sailing may be solved.

34. Problem. To find the distance and bearing of two places from each other, when their latitudes and longitudes are known. [B. p. 68.]

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35. Problem. To find the course, distance, and 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).

36. Problem. To find the departure, distance, and 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)

diff. long.

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

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

Solution. The course is found by the formula.

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

Solution. The difference of latitude is found by the formula

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

Solution. The course is found by the formula.

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40. EXAMPLES.

1. A ship sailed from Halifax (Nova Scotia) a distance of 2509 miles, upon a course S. 79° 34′ E.; find the place at which it arrived.

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Ans. The place arrived at is one mile south of Cape St. Vincent in Portugal.

2. Find the bearing and distance of Canton from Washington.

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