(129.) Feb. 10, 1853, the meridian altitude of a Argûs below the pole was observed to be 6° 41′ 15′′, index correction —2′ 10′′, and height of eye above the sea 14 feet, required the latitude. Ans., lat. 43° 50′ 18" S. (130.) January 11, 1853, the observed meridian altitude of a Ursa Majoris, below the pole, was 14° 14' 30", the index correction -4' 5", and height of eye 20 feet, required the latitude. Ans., lat. 41° 29′ 47′′ N. (131) April 20, 1853, the observed meridian altitude of ʼn Argûs, below the pole was 20° 14′ 15′′, the index correction 4' 5", and the height of eye 10 feet, required the latitude. Ans., lat. 51° 9′ 27′′ S. (132) June 1, 1853, in long. 30° 52′ W., the observed meridian altitude of the sun's lower limb, below the pole, was 10° 42' 0", the index correction + 2' 10", and height of eye 20 feet, required the latitude. Ans., lat. 77° 41′ 0′′ N. (133) June 10, 1853, at 2h 40m A.M., mean time nearly, in long. 30° W., observed the meridian altitude of the moon's lower limb, below the pole, to be 14° 30′ 10′′, index correction + 2' 45", height of eye 14 feet, required the latitude. Ans., lat. 81° 32′ 31′′ N. (134.) July 1, 1853, at 9h 30m P.M., mean time nearly, in long. 62° W., the observed meridian altitude of Mars below the pole was 10° 32′ 30′′, index correction 3' 0", and height of eye 18 feet, required the latitude. Latitude by observations off the meridian. In the volume of astronomical problems* will be found several methods for finding the latitude depending on some particular bearing or hour angle of the heavenly body: as, when it bears due east, or, when it is in the horizon, or when the hour angle is 6 hours, &c.; but since it is difficult to determine the precise moment when the heavenly body is in either of these positions, the methods referred to are of little use in practice. Problem 131, however, is one from which a useful rule may be derived, as it depends on the declination, altitude and hour angle of the heavenly body. The altitude and declination are easily obtained at sea; the hour angle is only known accurately when the ship time is given, and this is a quantity somewhat difficult to discover independently of an observation: the ship time, however, may always be considered to be known nearly. To render therefore a rule for finding the latitude, depending on the declination, altitude and ship time of practical value, we must ascertain in what position of a heavenly body an error of a few minutes in the ship time will produce the smallest error in the latitude deduced from it: and this we find will be the case, if the observed altitude is taken when the body is near the meridian. It is for this reason that single altitude observations taken off the meridian for finding the latitude are confined to bodies within half an hour of the meridian, when the time at the ship is uncertain to 3 or 4 minutes. * "Problems in Astronomy, &c., and Solutions,” pp. 33, 34, &c. X To find the latitude from the hour angle, altitude, and P Z M Let p z Q be the celestial meridian, è the pole, z the zenith, and x the place of a heavenly body. Then in the triangle P x z are given the hour angle P, the polar distance P X = 90 declination, and the zenith distance zx 90 altitude, to find P z the colatitude, and thence the latitude. = Investigation. Let fall a perpendicular xм upon the meridian PQ, thus forming two right-angled spherical triangles PX M, ZXM. Let PX = p, zx = a, p = h. also PM, ZM = y, XM=2. Then colat. PZ = x y, when the perpendicular x м does not fall between the pole and the zenith, and P z = x + y, when the perpendicular does so fall, a position easy to discover by observation. sec. p. cos. p cos. a COS. x = cos. y (2). = Cos. x. sin. alt. cosec. decl. Formula (1) and (2) determine x and y and thence the х latitude, since colat. = xy.. EXAMPLE. Given, hour angle = 3h 5m 36s, declination = 10° 54' 26" N., and altitude = 35° 4' 7", to find x the colatitude. Tan. x cot. d, cos. h. y Cos. y cosec. d, sin. alt. cos. x. A rule deduced from the above formulæ is open to the objection of a distinction of cases. For, first, if the perpendicular x м fall upon P z, between the pole and the zenith, then the colatitude PZ is equal to x + y or the two arcs must be added, instead of being subtracted as in the above example. Secondly, If the declination and latitude are of different names then tangent x is negative: this is evident by a figure. To find x in the latter case the arc taken out of the tables must be subtracted from 180°. If the student is able to discover these distinctions, the above formula is valuable, as he can derive from it a useful and practical rule. We will now proceed to give another rule, which is free from the objection mentioned above; this rule, however, will require the latitude to be known within a quarter of a degree of the truth, otherwise it may be necessary to repeat a part of the work, perhaps more than once. To find the latitude from the altitude of a heavenly body near the meridian, and its declination and hour angle. Р ૨ Let p be the pole, z the zenith, Pz Q the celestial meridian, PQ 90°, then z latitude of spectator. X Let x be the place of a heavenly body near the meridian. Draw the circle of declination 'P X and circle of altitude z x through x, then in the spherical triangle P zx are given the hour angle P, the polar distance PX, and the zenith distance zzx, to find the colatitude PZ. This may be done by dropping a perpendicular from x upon PQ, in the manner pointed out in Problem 131 of the Astronomical Problems: but the direct method of solving it being long and tedious an analytical formula is obtained for this purpose (see astronomical problems, p. 201), from which the following rule is deduced. Rule XXXIV. To find the latitude from an altitude of the sun near the meridian. 1. Find the Greenwich date in mean time. 2. Take out the declination and equation of time for this date. 3. To find the sun's hour angle. To the Greenwich mean time found as accurately as possible apply the longitude in time; subtracting if west, and adding if east; the result will be ship mean time: to this apply the equation of time with its proper sign to reduce mean time into apparent time; the result will be the sun's hour angle. 4. Add together the following logarithms,— Constant log., 6·301030 Log. haversine hour angle.* reject 30 in the index, and look for the result as a logarithm, and take out its natural number. *Or, instead of log. haversine, twice the log. sine of half the hour angle (rejecting in this case 40 from the index). |