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

1. Find a Greenwich date in mean time.

2. By means of the Nautical Almanac find the planet's declination for this date; and when great accuracy is required take out the planet's semidiameter and horizontal parallax.

3. Correct the observed altitude for index correction, dip, refraction (and if necessary for semidiameter and parallax in altitude), and thus get the true altitude. Subtract the true altitude from 90° to get the true zenith distance.

4. Mark the zenith distance north or south according as the zenith is north or south of the planet.

5. Proceed then as in Rule XXVIII., Art. 5.

EXAMPLE.

November 20, 1853, at 6h 18m A.M., mean time nearly, in long. 62° 42′ E. observed the meridian altitude of Mars' lower limb to be 52° 10′ 45′′ (Z. N.), the index correction +4' 0", and height of eye above the sea 16 feet, required the latitude.

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50 34 29 N.

If the small corrections of the planet's semidiameter and parallax in altitude are neglected, the above example will be worked thus:

Ship, Nov. 19
Long. in time

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Greenwich, Nov. 19

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Obs. alt.

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12° 55' 36" N.

In. cor.

52° 10′ 45′′
4 0 +

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Planet's declin. 12 44 40 N.

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EXAMPLES.

(123.) May 4, 1853, at 2h 45m A.M., mean time nearly, in long. 42° 10' W., the observed meridian altitude of Jupiter's

centre was 16° 42′ 10′′ (Z. N.), index correction + 11′ 42′′, and height of eye above the sea 20 feet, required the latitude. Ans., lat. 50° 30′ 38′′ N. (124.) July 12, 1853, at 9h 36m P.M., mean time nearly, in long. 30° 30′ E., the observed meridian altitude of Jupiter's centre was 10° 10′ 50′′ (Z. N.), the index correction 4' 4", and height of eye above the sea 10 feet, required the latitude. Ans., lat. 57° 45′ 37′′ N.

(125.) November 27, 1853, at 6h 3m A.M., mean time nearly, in long. 100° 0' W., the observed meridian altitude of Mars' centre was 32° 40′ 10′′ (Z. S.), index correction 8′ 10′′, and height of eye 16 feet, required the latitude. Ans., lat. 45° 45′ 0′′ S.

(126.) Sept. 15, 1853, at 4h 20m A.M., mean time nearly, in long. 10° 6' W., the observed meridian altitude of Saturn's centre was 19° 42′ 10′′ (Z. N.), index correction 6′ 45′′, and height of eye 12 feet, required the latitude. Ans., lat. 88° 55′ 24′′ N. (127.) Jan. 12, 1853, at 7h 9m P.M., mean time nearly, in long. 32° 0' W., the observed meridian altitude of Saturn's centre was 62° 42′ 10′′ (Z. S.), index correction and height of eye 20 feet, required the latitude.

8' 10",

Ans., lat. 14° 36′ 41′′ S.

(128.) June 7, 1853, at 5h 40m P.M., mean time nearly, in long 72° 30′ E., the observed meridian altitude of Venus was 30° 40′ 10′′ (Z. S.), index correction + 4′ 20′′, and height of eye 24 feet, required the latitude.

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To find the latitude by the meridian altitude of a heavenly body below the pole, and the declination.

Let x be the place of a heavenly body on the meridian,

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1. Find the declination of the heavenly body at the time of observation.

2. From the observed altitude get the true altitude.

3. Add 90° to the true altitude, and from the sum subtract the declination; the remainder will be the latitude.

EXAMPLES.

1. April 27, 1853, the meridian altitude of a Crucis below the south pole was observed to be 14° 10′ 30′′, the index correction was + 4' 4", and the height of eye 20 feet, required the latitude.

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2. June 18, 1853, at apparent midnight, in long. 100° W., the observed meridian altitude of the sun's lower limb below the north pole was 8° 42′ 10′′, the index correction -3', and height of eye above the sea 14 feet, required the latitude.

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