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When the zenith of the spectator is to the north of the body it is said to have a meridian altitude, zenith north. When the zenith is south of the body, its meridian altitude is called a meridian altitude, zenith south. By constructing figures similar to the above to suit different cases we shall find that the latitude is equal to the sum of the zenith distance and declination when the declination and zenith distance have the same names, namely, both north or both south and of the same name as either; and that the latitude is equal to the difference of the zenith distance and declination when they are of different names, and the latitude will be of the same name (N. or S.) as the greater.

Hence the following rules for finding the latitude from the meridian altitudes of different bodies.

Rule XXVIII.

To find the latitude by the meridian altitude of the sun, and its declination.

1. Find a Greenwich date in apparent time.

2. By means of the Nautical Almanac find the sun's declination for this date (p. 74). Take out also the sun's semidiameter, which is to be added to apparent altitude when lower limb is observed, and subtracted when upper limb is observed.

3. Correct the observed altitude for index correction, dip, semidiameter, and refraction and parallax, and thus get the true altitude (p. 109), subtract the true altitude from 90°, the result will be the true zenith distance.

4. Mark the zenith distance N. or S. according as the zenith is north or south of the sun.

5. Add together the declination and zenith distance if they have the same names; but take the difference if their

names be unlike; the result in each case will be the latitude, in the former of the name of either, in the latter of the name of the greater.

EXAMPLE.

April 27, 1853, in long. 87° 42' W., the observed meridian altitude of the sun's lower limb was 48° 42′ 30′′ (zenith north), the index correction was + 1' 42", and the height of eye above the sea was 18 feet, required the latitude.

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(100.) January 14, 1853, in long. 72° 42′ W. the observed meridian altitude of the sun's L. L. was 32° 42′ 10′′ (Z. N.) the index correction + 2′ 10′′, and height of eye above the sea 14 feet, required the latitude.

Ans., lat. 35° 50′ 34′′ N. (101.) March 20, 1853, in long. 72° 42′ E. the observed meridian altitude of the sun's L.L. was 45° 4′ 20′′ (Z. S.),

index correction 3' 4", and height of eye above the sea 20 feet, required the latitude. Ans., lat. 44° 56′ 54′′ S.

(102.) July 4, 1853, in long. 100° 0' W. the observed meridian altitude of the sun's L. L. was 62° 8' 7" (Z. N.), index correction 3′ 0′′, and height of eye above the sea 15 feet, required the latitude. Ans., lat. 50° 34′ 59′′ N.

(103.) March 21, 1853, in long. 62° 0′ W., the observed meridian altitude of the sun's U. L. was 50° 10′ 5′′ (Z. N.), index correction + 7' 10", and height of eye 14 feet, required the latitude. Ans., lat. 40° 26′ 47′′ N.

(104.) Sept. 24, 1853, in long. 33° 0' E., the observed meridian altitude of the sun's U. L. was 42° 3′ 15′′ (Z. N.), index correction - 1' 4", and height of eye above the sea 18 feet, required the latitude.

Ans., lat. 47° 49′ 39′′ N.

(105.) June 3, 1853, in long. 178° 30′ W., the observed meridian altitude of the sun's U. L. was 16° 20′ 0′′ (Z. S.), index correction + 3' 30", and height of eye above the sea 20 feet, required the latitude.

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When the altitude of a heavenly body is observed by means of an artificial horizon, the reading off on the instrument will be the angular distance between the heavenly body and its image in the artificial horizon, and this will be double the altitude as observed from the true horizon. This will be easily seen by the following figure. Let s A, a ray

S

of light proceeding from the body at s, be reflected by means of an artificial horizon placed at a, in the line A E. Then, if the spectator's eye is in the line A E, as at E, the image of the body will appear in the direction E A coming from a point s' below the horizon A. Now the observer is supposed to be placed so near a that the distance EA is inappreciable when compared with the distance As of the heavenly body,

H

-H

A

that is, the angle observed between s and s', namely, SES' may be considered to be sa s' and this angle s A s' is manifestly double s AH, the altitude above the horizontal plane н н. For by the principles of optics it is proved that the angle SA H is equal to E A H, which is equal to the vertical or opposite angle s'A H, that is, the horizontal line AH bisects the angle observed. Hence the following rule for finding the true altitude from an observed altitude in the artificial horizon.

Rule XXIX.

Given, the observed altitude of a heavenly body in an artificial horizon, to find the true altitude.

1. Correct the observed altitude for index correction. 2. Half of the result will be the apparent altitude of the point observed.

3. Then proceed as in the preceding rules to find the true altitude.

EXAMPLES.

1. The observed altitude of the sun's lower limb in an artificial horizon was 98° 14' 10," index correction 4o 10" required, apparent altitude of sun's lower limb.

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2. Oct. 21, 1853, in long. 1° 6′ W., observed the meridian altitude of the sun's lower limb (in quicksilver horizon) to be 56° 14' 0" (Z. N.), index correction — 0' 10", required the latitude.

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(106.) Oct. 9, 1853, in long. 19° 20′ W., the observed meridian altitude of the sun's lower limb (in artificial horizon) was 44° 30′ 15" (Z. S.), index correction - 2' 10", required the latitude. Ans., lat. 73° 53′ 28′′ S. (107.) June 10, 1853, in long., 23° 40′ E. the observed meridian altitude of the sun's lower limb (in quicksilver horizon) was 72° 15′ 20′′ (Z. N.), index correction + 4' 5", required the latitude. Ans., lat. 76° 37′ 45′′ N.

(108.) Aug. 7, 1853, in long. 62° 11' E., the observed. meridian altitude of sun's lower limb (in artificial horizon)

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