of the compass 2° 20' W. (see Table).

The compass

bearing reckoned from the same point as the true bearing

The true bearing is E. 10° N., when the compass bearing is E. 8° S.; required the true variation, the ship's head being S. W.

The true bearing is S. 80° W., when the compass bearing is N. 108° W.; required the true variation, the ship's head being S.W.b. W., and therefore the deviation by Table 710 W.

(220.) The true bearing of the sun was N. 36° E., when the compass bearing was N. 24° E., the ship's head being W. N.; required the variation of the compass.

Ans., 2310 E.

(221.) The true bearing was N. 110° 42′ W., when the

compass bearing was N. 90° 24′ W., the ship's head being S.S.W.; required the variation of the compass.

Ans., 15° 18' W.

(222.) The true bearing was S. 48° 30' W., when the compass bearing was N. 132° 33' W., the ship's head being S.W.b. W.; required the variation of the compass.

Ans., 8° 30′ E. (223.) The true bearing of the sun was E. 20° 20′ N., when the compass bearing was E. 32° 45' N., the ship's head being W.; required the variation of the compass.

Ans., 21° 15' E. (224.) The true bearing of the sun was W. 12° 32′ S., the compass bearing was W. 2° 10′ N., the ship's head being W.b.S.; required the variation of the compass.

Ans., 6° 22' W. (225.) The true bearing of the sun was W. 30° 10′ N., the compass bearing was W. 20° 42′ N., the ship's head being N.b.E.; required the variation of the compass.

Ans., 4° 31' E.

The variation of the compass is found at sea by either of the following problems.

1. Given the latitude of the ship and the sun's declination when in the horizon, to find the bearing or amplitude.

2. Given the latitude of the ship, and the altitude of the sun and declination, to find the true bearing or azimuth.

3. Given the latitude and time at the ship and the sun's declination, to find the true bearing or azimuth.

The compass bearing being observed at the time of observation, the difference of compass and true bearing, that is, the variation of the compass, is readily found, by the preceding rules.

2. Take out of the Nautical Almanac the sun's declination for this date.

3. Add together the log. sin. of the declination and log. secant of latitude; the sum, rejecting 10 in the index, is the log. sin of amplitude, which take from the tables.

4. If the body is rising, mark it east, if setting west: mark it also north or south according as the declination is north or south.

5. The result is the amplitude or true bearing.

6. Under the true bearing put the compass bearing, and determine the variation of the compass by the preceding rule.

EXAMPLE.

September 19, 1851, at 5h 51m A.M., mean time nearly, in latitude 47° 25′ N., and long. 72° 15′ W., the sun rose

by compass E. 12° 10' N.; required the variation, the ship's head being E.b.S.

(226.) May 6, 1846, at 5h 30m A.M., mean time nearly, in lat. 50° 48′ N., and long. 47° 12′ E., the sun rose by compass E. 2° 10' S.; required the variation, the ship's head being S.b.W. Ans., 24° 21' 30" W.

(227.) Nov. 14, 1846, at 6h 45m P.M., mean time nearly, in lat. 32° 14′ S., and long. 100° E., the sun set by compass W. 15° 40′ S.; required the variation, the ship's head being N.E. Ans., 16° 1' 30" W. (228.) January 10, 1846, at 6h 58m A.M., mean time nearly, in lat. 31° 56′ N. and long. 75° 30′ W., the sun rose by compass E. 30° 10' S.; required the variation, the ship's head being N.E.b.E. Ans., 67° 14′ 45′′ W.

(229.) March 21, 1846, at 6h 0m A.M., mean time nearly, in lat. 42° 13′ N., and long. 90° E., the sun rose by compass E. 11° 40′ S.; required the variation, the ship's head being W.b.S. Ans., 3° 20′ 15′′ W. (230.) March 31, 1850, at 6h 0m P.M., mean time nearly, in lat. 42° 13′ N. and long. 124° W., the sun set by compass W. 11° 30′ S.: required the variation, the ship's head being N. Ans., 14° 38′ 15′′ E.

(231.) Dec. 4, 1851, at 7h 50m A.M., mean time nearly, in lat. 50° 40′ N., and long. 94° W., the sun rose by compass E. 10° 42′ S.; required the variation of the compass, the ship's head being N.E. Ans., 57° 20′ 45′′ W.

Given the altitude of the sun, and the compass bearing, to find

2. Take out of the Nautical Almanac the sun's declina

tion for this date, and also the sun's semidiameter.