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9. The course of the ship, or the bearing of the two places from each other, is the angle which the ship's path makes with the meridian. [B. p. 52.]

10. The departure of two places is the distance of either from the meridian of the other, when they are so near each other that the earth's surface may be considered as plane and its curvature neglected. But, if the two places are at a great distance from each other, the distance is to be divided into small portions, and the departure of the two places is the sum of the departures corresponding to all these portions.

11. Instead of dividing the quadrant into 90 degrees, navigators are in the habit of dividing it into eight equal parts called points; and of subdividing the points into halves and quarters. A point, therefore, is equal to one eighth of 90°, or to 11° 15'. [B. p. 52.]

Names are given to the directions determined by the different points, as in the diagram (fig. 14), which represents the face of the card of the Mariner's Compass.

The Mariner's Compass consists of this card, attached to a magnetic needle, which has the property of constantly pointing toward the north, and thereby determining the ship's

course.

On page 53 of the Navigator a table is given of the angles which every point of the compass makes with the meridian, and on page 169, table XXV. the log., sines, &c. are given.

12. The object of Plane Sailing is to calculate the Distance, Course, or Bearing, Difference of Latitude and Departure, when either two of them are known. [B. p. 52.]

13. Problem. To find the difference of latitude and departure, when the distance and course are known. [B. p. 54.]

Solution. First. When the distance is so small that the curvature of the earth's surface may be neglected. Let AB (fig. 15) be the

distance. Draw through A the meridian AC, and let fall on it the perpendicular BC. The angle A is the course, AC is the difference of latitude, and BC is the departure. Then, by (21 and 22),

Diff. of lat. dist. X cos. course,

(211)

Departure =

dist. X sin. course.

(212)

Secondly. When the distance is great, as AB (fig. 16), then divide it into smaller portions, as A a, ab, bc, &c. Through the points of division, draw the meridians AN, an, bp, &c. Let fall the perpendiculars am, b n, cp, &c. Then, as the course is every where the same, each of the angles mA a, nab, pb c, &c. is equal to the angle A, or the course. Moreover, the distances, Am, an, bp, &c. are the differences of latitude respectively of A and a, a and b, b and c, &c. Also a m, bn, cp, &c. are the departures of the points A and a, a and b, b and c, &c. Therefore, as the difference of latitude of A and B is evidently equal to the sum of these partial differences of latitude; and as the departure of A and B is by § 10 equal to the sum of the partial departures, we have

Diff. of lat. = Am+an+bp + &c.

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But the right triangles m A a, n ab, pbc, &c. give by (211) and (212)

Am Aa X cos. course, am Aa X sin. course;

an ab X cos. course, b n = a b × sin. course;

=

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But

Hence,

A a + ab + b c + &c. АВ distance.

Diff. of lat.

dist. X cos. course,

Departure

dist. X sin. course;

precisely the same with (211) and (212).

This shows that the method of calculating the difference of latitude and departure is the same for all distances, and that all the problems of Plane Sailing may be solved by the right triangle (fig. 15). [B. p. 52.]

Tables of difference of latitude and departure are given in pages 1–6, Tables I. and II. of the Navigator, which might be calculated by (211) and (212).

14. Problem. To find the distance and difference of latitude, when the course and departure are known. [B. p. 55.]

Solution. There are given (fig. 15) the angle A and the side BC. Hence, by (23) and (24),

Distance

departure X cosec. course,

(213)

Diff. of lat. = departure X cotan. course.

(214)

15. Problem. To find the distance and departure, when the course and difference of latitude are known. [B. p. 55.]

Solution. There are given (fig. 15) the angle A and the side AC. Then, by (25) and (26),

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16. Problem. To find the course and difference of latitude, when the distance and departure are known. [B. p. 27.]

Solution. There are given (fig. 15) the hypothenuse AB and the side BC. Then, by (27) and (29),

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17. Problem. To find the course and departure, when the distance and difference of latitude are known. [B. p. 56.]

Solution. There are given (fig. 15) the hypothenuse AB and the leg AC. Then, by (27) and (29),

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18. Problem. To find the course and distance, when the departure and difference of latitude are known. [B. p. 57.]

Solution. There are given (fig. 15) the legs AC and BC. Then, by (30) and (32),

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1. A ship sails from latitude 3° 45′ S., upon a course N. by E., a distance of 2345 miles; to find the latitude at which it arrives, and the departure which it makes.

Ans. Latitude = 34° 35' N.

Departure 458 miles.

2. A ship sails from latitude 62° 19° N., upon a couse W. N. W., till it makes a departure of 1000 miles; to find the latitude at which it arrives, and the distance sailed.

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3. The bearing of Paris from Athens is N. 54° 56′ W.; find the distance and departure of these two places from each other.

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4. A ship sails from latitude 72° 3′ S., a distance of 2000 miles, upon a course between the north and the west, that is, northwesterly, until it makes a departure of 100 miles; find the latitude at which it arrives, and the course.

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5. The distance from New Orleans to Portland is 1257 miles; find the bearing and departure.

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6. The departure of Boston from Canton is 8790 miles; find the

bearing and distance.

Ans. Bearing N. 82° 31′ E.

Distance

8865 miles.

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