Distance.

Course.

Departure.

Longitude is reckoned East and West of the first meridian from 0° to 180°.

7. The difference of longitude of two places is the angle contained between the planes of the meridians passing through the two places; or it is the arc of the equator comprehended between these two meridians.

The difference of longitude of two places is equal to the difference of their longitudes, if they are on the same side of the first meridian; and to the sum of their longitudes, if they are on opposite sides of the first meridian, unless their sum be greater than 180°; in which case the sum must be subtracted from 360° to give the difference of longitude. [B. p. 50.]

8. The distance between two places in Navigation is the portion of a curve which would be described by a ship sailing from one place to the other in a path, which crosses every meridian at the same angle. [B. p. 52.]

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,

Point.

Mariner's compass.

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 A B

Given distance and course.

(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 departThen, by (21 and 22),

ure.

Secondly. When the distance is great, as A B (fig. 16), then divide it into smaller portions, as A a, a b, bc, &c. Through the points of division, draw the meridians A N, an, bp, &c. Let fall the perpendiculars am, bn, cp, &c. Then, as the course is every where the same, each of the angles m A a, n ab, p b c, &c. is equal to the angle A, or the course. Moreover, the distances, A m, a n, b P, &c. are the differences of latitude respectively of A and a, a and b, b and c, &c. Also a m, b n, c p, &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.

Departure = am + b n + cp+ &c.

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

Am A a X cos. course, a m—A a X sin course;

ana bXcos. course, b n = a b × sin. course;

bp b c X cos. course, c p = b c X sin. course.
&c. &c.

Given course and departure.

The sums of these equations give

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

(Aa+ab+b c +&c.) × cos. course,

=

(Aa+ab+be+ &c.) X sin, course.

But

Hence,

Aa+ab+be+ &c. AB distance.

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.]

A table of difference and 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),

Cases of plane sailing.

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),

Distance diff. of lat. X sec. course.
Departure = diff. of lat. X tang. course.

(215)

(216)

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),

Diff. of lat. [(dist.2) (departure)2].

(217)

(218)

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),

Departure [(dist.)2- (diff. of lat.)2].

(219)

(220)

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