's dec. April 3d 5° 24′ 50′′ N + 22′ 53′′. 24h. : 21m. 53s. : : 23h. 38m. (mid time): 22 32′′ Corr. Example 2.-On March 2d, 1828, in lat. 46° N., long. 150° E. by estimation, at 10h. 45m. 13s. A. M., the alt. of was 33° 26' + 2 bearing per compass S E E, and at 1h. 43m. 27s. P. M. the alt. of was 30° 4′-, course in the interval ES, seven miles per hour, height of the eye fourteen feet; required the latitude? The ship in this example is running nearer the sun, within three points of his bearing at the first observation. With this the distance and run between the observations (about twenty-one miles) the correction of the first altitude is about 17′ 5′′ : =or, 17′ 30′′ additive. The middle time corrected for longitude, is at about 14h. 14m. of March 1st, for which instant the sun's declination is 7° 14′ 37′′ S., and consequently the polar distance is 97° 14′ 37′′. To find the latitude from the altitude of two known fired stars.--Let Z (fig. 14, plate II.) be the zenith, P the pole, B and C two stars whose altitudes B Dand C E are measured, but not at the same time, that of B being measured first. Let B be at B' when the altitude of C is measured, and let PA be the meridian passing over the first point of Aries. Then AP B' is the right ascension of B', APC' the right ascension of C'; and BPB the elapsed sidereal time; which, being added to AP B', gives the right ascension of the point B; and we may then consider B. and C as two stars whose altitudes are taken at the same instant. Now in the triangle BPC are given the two polar distances of the stars BP and PC, and the angle BPC the difference of their right ascension to find B C and the angle PBC'. To compute these, drop C F a perpendicular from Con PB; then rad. : tan. PC:: cos. BPC : tan. PF. The difference of PF and PB is FB; and cos. B F: cos. FB:: cos. PC; cos. BC; also sin. PF: sin. FB:: cos. BPC: cot. PBC. Then in the triangle Z BC we have Z B and ZC the two zenith distances of the stars, deduced from their altitudes; and BC before computed, to find the angle Z BC, which may be found from ZBC this expression: cos. 2 ZBBC + ZC The difference between ZBC and PBC is Z BP. Then in the triangle ZBP are now given Z B the zenith distance, and PB the polar distance of B, and the contained angle Z BP to find Z P the colatitude. To make the computation drop Z G, a perpendicular from Z or PB, then rad. tan. ZB:: cos. ZBP tan. BG. The difference of BP and BG is PG and cos. ZC)cosect. Z B, cosect. BC. BG: cos. PG:: cos. BZ: cos. Z P, or sine of the latitude. Example.-On January 26th, 1828, in lat. 510 N., long. 20° W. by account, at 8h. 31m. 46s., the altitude of Sirius was 18° 15′ 48", and at 8h. 35m. 24s. that of Regulus was 18° 41′ 48′′, height of the eye sixteen feet, required the latitude? If one of the altitudes be increasing, and the other decreasing, the zenith, Z, will fall between BP and CP. To find the time at sea from the altitude of a known celestial object, at a place whose true latitude and longitude by account are known. Add together the altitude of the object, the latitude of the place, and the polar distance of the object, and take the difference between half the sum and the object's altitude. Then add together the secant of the latitude, the cosecant of the polar distance, the cosine of the half sum, 51 29 17 east of the meridian, or if its altitude is increasing, subtract the meridian distance from object's right ascension; if it is west of the meridian, or if its altitude is decreasing, add the meridian distance to its right ascension, and the result will be the sidereal time, or the right ascension of the meridian; from which subtra: the sun's right ascension, and the remainder w be the apparent time. To the apparent time apply the equation. time, and the result will be the mean time. Investigation of the method of computing the meridian distance. In fig. 15, plate II., call 2 the colatitude, l, the latitude l, A B, the altitude a, and A Z its complement a', and call A P the por Example 1.-If on April 19th, 1828, in lat. 43° 27′ N., long. by account 38° W. at 9 h. 4 m. it's A. M., the altitude of C be 39° 56′+, height of the eye ten feet, what is the true mean time! Example 2.—If on August 15th, 1828, in lat. 20° 10′ S. loug. by acct. 6o E., at 10 h. 27 m. 13s | P. M. per watch, the altitude of Antores be 38° 14', height of the eye 17 feet; required the tru mean time? For this time, per Nautical Almanac, the sun's R. A. is 9h. 41 m. 13s., the equation 4 m. 5s. add.; the polar distance of Antores 63° 57′ 33′′, and its right ascension 16 h. 18m. 56s. of time The nearer the object is to the east or west, when the altitude of it is taken for time, the better To find the error and rate of a chronometer from equal altitudes of the sun observed in the fore and afternoon of two successive days. Deduct the time by the chronometer for the altitude of the first forenoon from the time by the chronometer for the altitude of the second forenoon increased by twenty-four hours; the remainder will be the interval between the two observations. Do the same with the times of the altitudes of the two afternoons, in order to have the interval of time between them. The difference between half the sum of those intervals and forty-eight hours will be the rate of the chronometer, or its daily gain or loss; gaining if the half sum exceeds, but losing if it is less than forty-eight hours. servations, then n+1 the days between the corresponding forenoon or afternoon observations. Put m the seconds in twenty-four hours, A the apparent time of the first observation, B that of the second, and A+D = that of the third; then BD will differ from the time of the fourth by a quantity too minute to be worth attention. Hence (n+1. m + A + D) — A = n + 1 .m D the first interval; and (n + 1. m+BD) - Bn + 1. m D the second interval. The sum of these intervals is 2. n + 1. m, whence the difference between the sum of the intervals as measured by the chronometer, and 2 . n + 1 . m, is the gain of the chronometer in 2. n + 1 days. Ifn, or the observations are made on two successive days, then half the difference between the sum of the intervals and 2 m will be the rate or the gain or loss per day. Take the difference between the two intervals already found, and likewise the interval between the times of the first two observations of equal altitudes. Add together the logarithms of these two intervals, in seconds, and 4-46143, and the sum, rejecting the tens from the index, will be the logarithm of a correction to be added to, or Again, (n + 1 .m + D) (n + 1. m + D) subtracted from, half the interval between the+2 D; and if we put i the interval in first two observations, according as the interval between the observations on the forenoons is greater or less than the interval between those on the afternoons. The sum or remainder added to the time of the first forenoon will give the time of true noon by the chronometer on the first day of observation. Hence by applying the equation of time the error of the chronometer for mean time may be found. The following is the investigation of the rule, which was first published by J. de Mendoza Rios, but without demonstration, in his Nautical Tables. Let n the complete days between the ob seconds in 24 h. + BA, the time between the first and second observations, and ethe change in the hour angle resulting from the change of declination in the interval i, we have n+1.m:2 D::i: 2 e; or Di = e 2 2 n + 1. m the correction to be applied to the half interval, or to the middle time between this observation, to obtain the time of true noon. |