TO FIND THE DEVIATION BY AN AMPLITUDE OBSERVATION 10. Directions. - Find the latitude in. Be ready at the compass a few minutes in advance of rising or setting and have the sight vanes, attached to the compass, pointed in the direction of the body to be observed (which is usually the sun). When the sun's center appears to be. on the horizon note its bearing. Refer this bearing to the east or west point. To the bearing apply a correction taken from Table III; apply this correction to the right at rising, and to the left at setting. This will give the correct compass bearing or compass amplitude, as it is sometimes called. From the Amplitude Table take out the true amplitude, according to the ship's latitude and the declination of the observed body. Draw a diagram (no matter how rough) representing the cardinal points, and from the center of this diagram lay off, respectively, the true amplitude and the compass bearing. The angle subtended by these bearings will be the whole error of the compass, and is named east when the true amplitude falls to the right of the compass bearing, but west when it falls to the left of the compass bearing. Then, in order to find the deviation, use the same diagram but turn around to the north point (which represents true north) and lay off from this point the whole compass error according to its name, and from the same point also lay off the variation. The deviation is, then, the difference between the error and the variation if both have the same names, but their sum, if of different names, and is named east when the compass north falls to the right of the magnetic north, but west when the compass north falls to the left of the magnetic north. The deviation thus found is the deviation for the point on which the ship is heading when the bearing is taken. 11. Illustration. -On July 11, 1901, at sunrise, the bearing by compass of the sun's center when on the horizon was N 80° E. The ship's course by compass when bearing this case the compass bearing is N 80° E, which, when referred to the east point, is equal to E 10° N. To find the is taken at sunrise, the cor- Comp.N W FIG. 4 S rection must be applied to the right. This gives the correct compass bearing as E 9.3° N. Then take from the Amplitude Table the true amplitude of the sun, which in this case is found to be E 30.3° N. According to the preceding directions the deviation is then found as follows: In Fig. 4, the compass bearing is represented by the line OB, and the true amplitude, or true bearing, by the line OS'; the angle BO S' subtended by these lines is the total error of the compass, and is named west because the true amplitude falls to the left of the compass bearing. The error and variation laid off, respectively, from the true north N will then show a deviation of 14°, which is westerly, because the compass north falls to the left of the magnetic north. = EXAMPLE 1.-On June 20, 1901, the bearing by compass of the sun's center at rising was ENE. Latitude of ship 44° 10' N. Find the error of the compass and the deviation for the point on which the ship is heading, assuming the variation by chart to be 15° E. SOLUTION. - Follow directions of Art. 10. Thus, In this case the sum of the total error and the variation (see Fig. 5) is equal to the deviation, because they are of different names; and it is named west since the compass north falls to the left of the magnetic north. EXAMPLE 2.-On November 27, 1901, the sun's center, at setting. bore S W by compass. Latitude in = 48° N. Find the deviation of the compass, assuming the variation by chart to be 6° W. SOLUTION. - Proceed according to directions in Art. 10. Thus, Sun's Decl., Nov. 27, 1901 = S 21° 6′ 7′′ In this case the total compass error is easterly (see Fig. 6). since the true amplitude falls to the right of the compass bearing. The deviation is easterly, also, because compass north falls to the right of the magnetic north. Ans. = EXAMPLE 3.-On May 20, 1901, at sunset, the bearing by compass of the sun's center was NW by W. Latitude in 42° N. Find the deviation of the compass, assuming the variation by chart to be 12.7° W. SOLUTION. - Proceed as in the preceding examples. Thus, In this case the total error is westerly, since the true amplitude falls to the left of the compass bearing (see Fig. 7); and the resulting deviation is easterly, because compass north falls to the right of the magnetic north. W B Mag. N E 12. From what has been shown, it is evident that by this simple method-by an amplitude of the sun- the error and deviation of the compass may be conveniently found any clear morning and evening at the rising or setting of the sun with sufficient accuracy for all practical purposes. At every observation for amplitude, it is important, that the ship's head by the compass, by which the bearing is taken, should be noted. The student should remember that the deviation thus found applies only to the compass used in taking the bearing, and is the deviation for that point on which the ship is heading at instant of taking the bearing. In each and every case of determining the deviation by this method, the student should make use of a diagram as shown in the preceding examples. By making it an invariable rule to use such a diagram in this manner, no mistakes can occur, whereas mistakes in determining and naming the deviation may frequently occur in cases where verbal rules are used without the aid of an illustration. |
