clination. Deduce the horary angle, or the star's distance from the meridian. Add this distance to the star's right ascension, if at the time of observation it is west of the meridian, but subtract it if east, and the sum or remainder will be the right ascension of the meridian. From the right ascension of the meridian, increased by 24 hours if necessary, subtract the sun's right ascension, reduced to the time of observation, and the remainder will be the apparent time. If to this time the equation of time be applied, the result will be the mean time. If the table of reduced versed sines be used, the horary angle must be taken from the top of the page if the star be west of the meridian, but from the bottom if to the east, which being added to the stars R. A. gives the R. A. of the meridian. EXAMPLES TO THE PRECEDING RULES. Edinburgh, 30th March, 1829. In latitude 55° 57′ 20′′ N., longitude 3° 12′ W., the following series of observations were made on the star Arcturus, east of the meridian, and on Aldebaran, west of the meridian, in succession, to determine the error of the chronometer, the barometer being at 29.56 inches, and the thermometer at 54° Fahrenheit. By observing an object both to the east and west about the same time, and having the same altitude nearly, the errors of reading, the glasses, the latitude,,&c. will be very much lessened, if not entirely destroyed, and are consequently nearly equivalent to equal altitudes in point of accuracy, and are more convenient, as there is less risk of losing corresponding altitudes. Several variations may be made on the six things here proposed, that may serve as a useful exercise, which, by a reference to the theorems and rules already given, will be easily performed. PROBLEM II. Given the latitude of the place and the sun's declination, to find the time when twilight begins and ends. At what time will twilight begin and end at London in latitude 51° 32′ N., on the second of May, 1827, the sun's declination being 15° 14' N.? C In figure, (page 87), suppose a parallel n m to the equator EQ to be drawn at the distance of 15° 14′ above it, while another parallel to the horizon HO is drawn at the distance of 18° below it, these two would cut one another somewhere between c and m in S, forming the triangle ZPS, in which ZP, PS, and ZS, are given to find the angle ZPS, the angle between the meridian PEP and another meridian passing through the sun at the time he is 18° degrees below the horizon, his situation when twilight begins and ends. Z s or zenith distance 108° 0' Ps or polar distance 0.015534 0.206168 74.46 PZ or colatitude 38 28 cosecant PROBLEM III. Given the right ascensions and declinations, or the longitudes and latitudes of two celestial objects, to find their angular distance. In this problem there are given two sides and the contained angle to find its opposite side. The contained angle is the difference between their right ascensions or longitudes, and the containing sides are the complements of the declinations or latitudes. If the sun be one of the objects, as his latitude is very small, he may be supposed to be always in the ecliptic; then the triangle so formed will be right-angled if the longitudes and latitudes are used, and the computation becomes more simple. By means of this problem the lunar distances in the nautical almanac are computed. On the 1st of June, 1828, required the distance between the moon and a Pegasi, at noon, on the meridian of Greenwich, the moon's right ascension being 295° 23′ 46′′, and declination 16° 11′ 45′′ S., the star's right ascension being 22h 56m 13.85, or 344° 3′ 28′′, and north polar distance 75° 43′ 2′′, or declination 14° 16′ 58′′ N. 344° 3′ 28′′-295° 23′ 46′′-48° 39′ 42′′ the angle at the pole. Instead, however, of following the operation derived from the spherical triangle, a more simple practical rule may be derived from it according to theorem X. To twice the sine of half the contained angle add the cosines of the moon and star's declinations, and take half the sum of these three logarithms. From this half sum subtract the sine of half the sum of the declinations if they are of contrary names, or that of half their difference if of the same name, the remainder will be the tangent of an arc, the sine of which being subtracted from half the sum of the three logarithms already found will give the sine of half the required distance. 1. Required the distance between the moon and sun on July 2d, 1828, at noon, on the meridian of Greenwich, the longitude of the 9 sun being 3° 10° 28′ 44', the longitude of the moon 11' 17° 59′ 39′′, and latitude 2° 51′ 40′′ N.? Ans. 112° 27′ 19′′ east of her. 2. Required the distance between the moon and sun on the 20th January, 1828, at noon, the sun's longitude being 9° 29° 29′ 39′′, that of the moon 11s 17° 54′ 42′′, and latitude 3° 24′ 28′′ ? 3. Required the distance between the moon and Aquilæ, at noon, on the 10th of May, 1828, the right ascension of the moon being 6° 58′ 43′′, the declination 4° 44′ 48′′ N., the right ascension of « Aquila in time, being 19h 42m 25.62, and north polar distance 81° 34 41"? Ans.-70° 54′ 51′′ west of her. 4. Required the distance between the moon and Aldebaran, at midnight, on the 16th of December, the moon's R. A. being 32° 31' 30", the declination 11° 18′ 11′′ N., the R. A. of Aldebaran being 4h 26m 8.67, and N. P. D. 73° 50′ 37′′.4? Ans.-33° 21' 10" west of her. PROBLEM IV. On finding the Latitude by Observation. The most simple practical method of finding the latitude is from the meridian altitude of a celestial body whose declination is known. Should the object be the sun, moon, or some of the planets, the altitude or zenith distance of the lower or upper limb, or both, are observed, and by the application of several corrections, that of the centre is obtained. When reflecting instruments, such as the sextant, repeating circle, &c. with an artificial horizon, are employed, the arc read off must, from the principles of optics, be halved before the other corrections are applied.* A meridian altitude of the sun, moon, or a planet taken at land, must be corrected for refraction, parallax, and semidiameter, and at sea, in addition to these, for the dip of the horizon.† When great accuracy is required, a number of observations are taken near the meridian, and reduced to it by formulæ, or tables, such as XXVIII., to which the index error, when necessary, must be applied. Having found the true altitude, take its complement to 90°, which gives the zenith distance, denominated north or south, according as the observer is north or south of the object. Now, if the zenith distance and declination are of the same name, their sum is the latitude; if of contrary names, their difference is the latitude of the same name with the greater. Ex. 1.-Edinburgh Observatory, March 28th, 1825, with an artificial horizon and one of Troughton's best sextants, the vernier of which showed 10", Captain Pringle Stokes, R. N., found the meridian altitude of the sun's lower limb to be 73° 32′ 15′′, the index error being+2' 26", the barometer standing at 29.66 inches, and Fahrenheit's thermometer 56°; what was the latitude, employing the refractions in the table in the nautical almanac ? *See explanation of Table XXV. + Tables XIII, and XIV. combining the whole in one, have been computed for the sun and stars expressly with this view. They are sufficiently accurate for practice at sea, and diminish that liability to error which the corrections taken out separately are likely to occasion. Ex. 2.—On the same evening the altitude of Polaris to the north, when at the upper transit, was 115° 7' 44", the index error was+2′ 26", barometer 29.7 inches, the thermometer 50° Fah.; required the latitude? Observed altitude Index error Sum Half Refraction True altitude Polar distance Latitude by Polaris Mean 3. When the sextant is employed to determine the latitude, an object having the same altitude nearly should be chosen both to the north and south, in order to destroy the errors of the glasses, divisions, and reading, as far as possible. In these examples the mean is very near the truth, though either separately is erroneous to the amount of about 8". Edinburgh, 10th January, 1826. Ex. 3.-On the Calton-hill, near the Observatory, with one of Troughton's reflecting circles on a stand, and an artificial horizon, the author, at about ten o'clock, P. M., observed the following double altitudes of the polar star, when the sympiesometer stood at 29.86 inches, and thermometer at 42° Fahrenheit; required the latitude of the place of observation? |