(93.) July 12, 1848, at 9h 18m P.M., mean time nearly, in long. 44° 40′ W., the observed altitude of the moon's L. L. was 27° 56′ 40′′, the index correction + 2′ 20′′, and height of eye above the sea 20 feet, required the true altitude. Ans., 28° 56′ 9′′. (94.) May 15, 1848, at 10h 25m P.M., mean time nearly, in long. 55° 40′ W., the observed altitude of the moon's L. L. was 21° 14' 10", the index correction + 2′ 20′′, and height of eye above the sea 15 feet, required the true altitude. Ans., 22° 15' 17". (95.) May 15, 1848, at 10h 22m P.M., mean time nearly, in long. 41° 30′ W., the observed altitude of the moon's U. L. was 45° 20′ 30′′, the index correction + 4' 10", and height of eye above the sea 20 feet, required the true altitude. 113 SECTION II. RULES FOR FINDING THE LATITUDE, LONGITUDE, ERROR AND RATE OF CHRONOMETERS, AND VARIATION OF THE COMPASS. CHAPTER VI. RULES FOR FINDING THE LATITUDE. To find the latitude by the meridian altitudes of a heavenly body above and below the pole. Let NW SE represent the horizon of the spectator, z the zenith, N z 8 the celestial meridian, N P the altitude of the pole, w Q E the celestial equator. Then N P (the altitude of the pole) latitude of spectator.* Let A B A' be a parallel of declination described by a heavenly body about the pole P, P Ꮓ W E Q above pole star's polar distance and ▲ N = PN + A P = lat. + star's polar distance adding A' N+AN = 2 lat. or lat. = (A'NAN) half sum of latitudes. *For ZNPQ (each being 90°) or, PN + PZ = PZ + ZQ .. PNZQ= latitude of spectator. See "Problems in Astronomy," by the Author. If the heavenly body when passing the meridian above and below the pole, is on different sides of the zenith, so that the altitudes are taken from opposite sides of the horizon, subtract the greater altitude from 180°, so as to reduce it to an altitude taken from the same point of the horizon as the other altitude (see Exercise 2, p. 115). Hence this Rule XXVII. To find the latitude by the meridian altitudes of 1. Correct the altitudes for index correction, height of eye, refraction and parallax (or as many of these as are applicable to the case), and thus get the true meridian altitudes. 2. Add together the true meridian altitudes (reckoning from the same point of horizon), and half the result will be the latitude of the spectator. EXAMPLES. 1. The meridian altitudes of a Ursa Majoris were observed above and below the north pole to be 74° 10′ 10′′ and 32° 42′ 15′′ respectively (zenith south at both observations), index correction-2′ 10′′, and height of eye above the sea 20 feet, required the latitude. 2. The meridian altitudes of a Auriga (Capella), were observed above and below the north pole to be 81° 10′52′′ (zenith north of star), and 3° 42′ 52" (zenith south), index correction 3′ 10′′, and height P (96.) The meridian altitudes of a star were observed above and below the north pole to be 69° 20′ 45′′ and 6° 14′ 30′′ respectively (zenith south at both observations), index correction 1'45", and height of eye 16 feet, reAns., lat. 37° 37′ 35′′ N. quired the latitude. (97.) The meridian altitudes of a star were observed above and below the north pole to be 85° 10′ 10′′ and 10° 10' 10" respectively (zenith south at both observations), index correction, -2' 40", and height of eye 20 feet, required the latitude. Ans., lat. 47° 30' 24" N. (98.) The meridian altitudes of a star were observed above and below the north pole to be 77° 8′ 10′′ (zenith north of star), and 3° 40′ 45′′ (zenith south), index correction + 1' 42", and height of eye 12 feet, required the latitude. Ans., lat. 53° 10′ 7′′ N. (99.) August 12, 1850, the meridian altitudes of a star were observed above and below the south pole to be 85°14′15′′ (zenith south), and 4° 52' 0" (zenith north), index correction 8′ 14′′, and height of eye above the sea was 30 feet, required the latitude. Ans., lat. 49° 43′ 39′′ S. To find the latitude by the meridian altitude of a heavenly body above the pole, and the declination. Let NW SE represent the horizon, N z s the celestial P W 7. N Q meridian, p the north pole of the heavens, and w Q E the celestial meridian. Then zo is the latitude of spectator (p. 113, note). First. Let the declination of E the body and the latitude be of the same name, as when the body is at A; then A Q is the declination of the body, A s is its meridian altitude, and z A its zenith distance, and z Q = Z A + A Q; or, lat. = meridian zenith distance + declination. Second. Let the declination and latitude be of different names (one north and the other south); as when the body is at A'. |