Method 3rd.-By the altitude of the pole star, at any time of the day*. If the altitude of the pole star can be taken when on the meridian, its polar distance, either added to, or subtracted from, the altitude, gives at once the latitude; and when observed out of the meridian, as at the point S or S' in the figure, the latitude can be easily obtained, as follows: S' Let Z PO represent the meridian, Z the zenith, P the pole, and a Sa' the circle described by the polar star S, at its polar distance P S. The star's horary angle Z P S, or Z P S', is evidently the difference between its right ascension and the sidereal time of observation; and in the spherical triangle Z PS (or ZPS') we have ZS, PS, and the angle ZPS, to find ZP, the co-latitude. The result may be obtained with almost equal accuracy by considering PSc as a plain right-angled triangle, of which Pc is the cosine of the angle cPS to radius PS; the distance Pc thus found is to be added to, or subtracted from, the altitude HS, according as the star is above or below the pole, which is thus ascertained :-If the angle ZPS' be less than 6, or more than 18 hours, the star is above the pole, as at S'; if between 6 and 18 hours, it is below the pole, as at S. By the tables given in the Nautical Almanac, the solution is even more easy, and has the advantage of not requiring any other reference. The rule is as follows: 1st. From the corrected altitude subtract 1'. *This of course is only applicable to northern latitudes. In the southern hemisphere there is no star sufficiently near to the south pole to be made available in thus determining the latitude. 2nd. Reduce the mean time of observation at the place to the corresponding sidereal time. 3rd. With this sidereal time take out the first correction from Table I., with its proper sign, to be applied to the altitude for an approximate latitude. 4th. With this approximate latitude and sidereal time take out from Table II. the second correction; and with the day of the month and the same sidereal time take from Table III. the third correction. These are to be always added to the approximate latitude for the latitude of the place. EXAMPLE. On Oct. 26, 1838, the double altitude of Polaris, observed with a repeating circle, at 11h 55m 30s mean time, was 105° 44′ 63′′, the barometer standing at 29.8; thermometer, 50°. latitude of the place of observation. By the method given in the Nautical Almanac,— Required the Correction 1st for sidereal time By considering SPc as a plane rightangled triangle, in which P c, the correction to be subtracted, is the cosine of P to radius PS, the latitude is found by plane trigonometry within a few seconds of the above results. Method 4th.-By an altitude of the sun, or of a star, out of the meridian, the correct time of observation being known. By reference to the figure, it will be seen that this method simply involves the solution of the spherical triangle ZPS already alluded to, formed by the zenith, the pole, and the object at the time of observation; of which ZS, the zenith dis 2 tance, PS, the polar distance, and the angle at P are known, and Z P, the co-latitude, is the quantity sought. The formula given by Baily, for finding the third side, when the other two sides and an angle opposite to one of them are given, is tan a' = cos given angle × tan adjacent side and (aa"), which formula is used in the following examples : EXAMPLE I. On May 4, 1838, the observed altitude of the sun's upper limb at 5h. 47m. 15s. by chronometer was 14° 44′ 58′′. The index error of sextant being 28", and the watch 3m. 34s4 too fast. Barometer 29.9; thermometer 61; required the latitude. |