3. Find the polar distance, by adding 90° to the declination, when the latitude and declination have different names, or by subtracting the declination from 90°, when the latitude and declination have the same names. 4. Correct the observed altitude for index correction, dip, semidiameter and correction in altitude, and thus get the true altitude. 5. Put down the latitude under the altitude and take their difference; under which put the polar distance; take the sum and difference. 6. To the log. secants of the two first terms in this form (omitting the tens in the index) add the halves of the log. haversines* of the two last, the result, rejecting 10 in the index, is the log. haversine of the true bearing or azimuth, which find from the table. 7. Mark the true bearing N. or S. according as the latitude is N. or S., mark it also E. or W., according as the heavenly body is E. or W. of the meridian. 8. With the true bearing thus found and the compass bearing, find the variation of the compass by Rule 57, p. 245. EXAMPLE. June 7, 1851, at 5h 50m A.M., mean time nearly, in lat. 50° 47′ N., and long. 99° 45′ W., when the sun bore by compass S. 92° 36′ E., the observed altitude of the sun's lower limb was 18° 35′ 20′′, index correction + 3' 10", and the height of the eye above the level of the sea was 19 feet; required the variation, the ship's head being N.E. *If the student have no table of haversines, the angle P z X, or true bearing, may be found in a similar manner to that pointed out in p. 171. When the ship is in harbour, or in any position where the sight of the horizon is intercepted by land, or obscured by fog, so that the altitude of the sun cannot be taken, the preceding methods are inapplicable. The following rule may then be used, in which it is supposed that the hour angle at the ship is known, or can be found by means of the chronometer, or the time at the place. Rule LX. Variation by azimuth (hour angle being known). Given the hour angle at ship and the compass bearing, to find the variation of the compass. Let x (fig. p. 250) be the place of the heavenly body when its compass bearing is observed: then in the triangle PZX are given two sides and the included angle, to find one of the remaining angles: namely, Pz colat. P X polar distance, and z P x the hour angle, to find the angle P Z X, the true bearing or azimuth. = 1. Get a Greenwich date. 2. Take out of the Nautical Almanac for this date, the equation of time and sun's declination. 3. Find the polar distance, by adding 90° to the declination, when the latitude and declination have different names, or by subtracting the declination from 90° when the latitude and declination have the same name. 4. Under the colatitude (found by subtracting the latitude from 90°) put the polar distance, take the sum and difference, and the half sum and half difference. 5. To find the hour angle at ship. Correct the time shown by chronometer when the compass bearing was observed, for its error on Greenwich mean time, and thus get the mean time at Greenwich; to mean time apply the equation of time to obtain apparent time; under this put the longitude in time, adding if east and subtracting if west; the result will be ship apparent time, and also the hour angle if P.M.; but if A.M. at ship, subtract the apparent time from 24 hours, the remainder will then be the hour angle required. 6. Divide the hour angle by 2. Then under heads (1) and (2) put down the following quantities. 7. Under both (1) and (2) put log. cotangent of half hour angle. Under (1) log. cosine} (2) log. sine. (1) log. sec. } of half difference of polar distance and colatitude. of half sum of polar distance and colatitude. 8. Add together the log. under (1) and (2) separately; and take out the angles corresponding to each as a log. tangent. Put one under the other, and take their sum, if the polar distance is greater than the colatitude, or their difference if the polar distance is less than the colatitude; the result will be the true bearing of the sun at the time of observation. 9. Then proceed to find the variation as in Rule LVII. EXAMPLE. June 23, 1847, at 10h 58m A.M., mean time nearly, in lat. 50° 48′ and long. 1° 6' W., when a chronometer showed 11h 3m 37s, the bearing of the sun was observed to be N. 173° 10′ E., the error of the chronometer on Greenwich mean time being 0m 54s fast; required the variation. (232) April 27th, 1847, at 1h 10m P.M., mean time nearly, in lat. 50° 48' N., and long. 1° 6′ W., when a chronometer showed 1h 15m 51s, the bearing of the sun was observed to be S. 51° 55′ W., the error of the chronometer on Greenwich mean time being 1m 18s fast; required the variation. Ans., 23° 46′ W. (233.) Dec. 14, 1847, at 10h 22m A.M., mean time nearly, in lat. 52° 10" N., and long. 1° 30′ W., when a chronometer showed 10h 30m 48s, the bearing of the sun was observed to be N. 179° 20' E., the error of the chronometer on Greenwich mean time being 3m 38s fast; required the variation. Ans.. 21° 23' 15" W. P.M., mean time nearly, (234.) Dec. 14, 1847, at 1h 55m in lat. 48° 50′ N., and long. 1° 30′ W., when a chronometer showed 1h 59m 55, the bearing of the sun was observed to be S. 51° 40′ W., the error of the chronometer on Greenwich mean time being Om 58 fast; required the variation. Ans., 22° 25′ 15′′ W. (235.) Dec. 14, 1848, at 11h 11m A.M., mean time nearly, in lat. 39° 40′ N., and long. 0° 40′ E., when a chronometer showed 11h 19m 43s, the bearing of the sun was observed to be N. 167° 50′ E., the error of the chronometer on Greenwich mean time being 3m 38s fast; required the variation. Ans., 2° 50′ 13" E. (236.) March 7, 1844, at 9h 59m A.M., mean time nearly, in lat. 49° 48′ N., and long. 1° 10′ E., when a chronometer showed 10h 24m 8s, the bearing of the sun was observed to be N. 164° 51′ 40′′ E., the error of the chronometer on Greenwich mean time being fast 20m 48s; required the variation. Ans., 20° 26' 40" W. (237.) May 26, 1851, at 9h 48m A.M., mean time nearly, in lat. 50° 48′ N., and long. 1° 6′ W., when a chronometer showed 9h 47m 37s, the bearing of the sun was observed to be S. 31° 7' E., the error of the chronometer on Greenwich mean time being 3m 17s fast; required the variation. Ans., 23° 35′ 15′′ W. |