by the late Dr Maskelyne, will be found very convenient for this purpose. PROBLEM II. Given the right ascension, the declination, and the obliquity of the ecliptic, to find the longitude and latitude. Let RA denote the right ascension, O the obliquity of the ecliptic, and D the declination; then using logarithms: Tan D-sin RA=tan A, North or South as the declination is. Call O in the first six signs of RA South or S., and in the last six signs North or N. Then A+0=B, regard being had to the algebraic signs. A being less than 45°, and using logarithms. Sec A+ cosec B+tan RA tan lon of the same kind as RA, unless B be more than 90°, when the quantity found of the same kind as RA must be taken from twelve signs. A being more than 45°. Tan A+cosec A+ cos B+tan RA=tan lon of the same kind as RA, unless B be more than 90°, when the quantity found of the same kind as RA must be taken from twelve signs. Lon being nearer III. and IX. signs than O and VI. signs. Lon nearer O and VI. signs, than III. and IX. signs. Tan lon+cos lon+tan B=tan lat of the same name as B. EXAMPLE. On Monday the 12th of June, 1826, the moon's R A at noon was found by observation to be 10h 39m 31s and her declination 2° 51′ 58′′ N.; required her longitude and latitude? D= 2° 51′ 58′′ N. tan 8.699533 RA-10h 39m 31s sine 9.536560 tan 9.563908 Given the longitude and latitude of a celestial object, and the obliquity of the ecliptic; to find the right ascension and declination. Tan lat-sin lon-tan A, North or South as the latitude is. Call O north in the six first signs, and South in the six last signs. A+0=B, as before. A being less than 45°, sec A+ cos B+tan lon=tan RA of the same kind as the longitude, unless B be more than 90°, when the quantity found of the same kind as the longitude must be subtracted from twelve signs. A being more than 45°, tan A+cosecant A+ cos B+tan lon= tan RA of the same kind as the longitude, unless B be more than 90°, when the quantity found of the same kind as the longitude must be subtracted from twelve signs. If RA be nearer III. signs and IX. signs, than O and VI. signs, sine RA+tan B-tan Dec of the same name as B. And RA being nearer O and VI. signs, than III. and IX. signs, tan RA+cos RA+tan B-tan Dec of the same name as B.* EXAMPLE. On the 1st of January, 1820, the mean longitude of the Star Fomalhaut was 111° 19′ 34′′, the mean latitude 21°6′ 45′′ S.; required the right ascension and declination, the obliquity of the ecliptic being 23° 27′ 46" ? Lat 21° 6′ 45′′ S. tan 9.586721 α 1. The mean longitude of a Arietis, on the 1st January, 1820, was 1 5° 8′ 48′′, and mean latitude 9° 57′ 34′′ N. when the obliquity of the ecliptic was 23° 27′ 46′′; what was the right ascension and declination? Ans.-R. A. 1 57m 35; Dec 22° 36′ 24′′ N. 2. Required the right ascension and declination of Pollux, when the longitude was 3° 20° 43′ 58′′, the latitude 6° 40′ 17′′ N. the obliquity of the ecliptic being 23° 27′ 46′′ ? Ans.-R. A. 7h 34m 17.5s; declination 28° 27′ 8′′ N. 3. The mean longitude of Spica Virginis is 6 21° 19′ 50′′, latitude 2° 2' 24" S. and the obliquity of the ecliptic 23° 27′ 46′′; required the right ascension and declination? Ans. R. A. 13h 15m 43.5; declination 10° 13′ 4′′ S. 4. The mean right ascension of a Aquilæ is 19h 42m, and declination 8° 24' 4" N. the obliquity of the ecliptic being 23° 27′ 46′′; required the longitude and latitude? Ans.-Longitude 9° 29° 14' 14", Latitude 29° 18′ 36′′ N. 5. Required the longitude and latitude of a Pegasi, of which the right ascension is 22h 55m 48, declination 14° 14' 24", the obliquity of the ecliptic being 23° 27′ 46′′ ? Ans.-Longitude 11° 20° 58' 47", Latitude 19° 24' 36" N. PROBLEM IV. Given the latitude of the place, and the sun's declination, to find his altitude and azimuth at 6 o'clock. These rules may, in general, be depended upon, except in peculiar circumstances, which a consideration of the figure will enable the computer to correct, as when the longitude, or RA, fall upon PP', or pp', &c. See Dr Abram Robertson's paper in the Phil. Trans. for 1816, which for want of room cannot be given here. 1. At Edinburgh, in latitude 55° 57′ 20′′ N. on the 21st of June, 1826, the sun's declination was 23° 27′ 36′′ N.; required his altitude and azimuth at 6 o'clock in the morning or evening, his declination being supposed to remain the same? r H P E n B G N Construction.-Describe the primitive HPON on the plane of the meridian. Let HO represent the horizon, ZN the prime vertical at right angles to the former. Make OP, from a scale of chords equal to the latitude of the place, North in the present instance; draw PP', the six-o'clock hour circle in this case, and at right angles to it draw the equator EQ; describe the small circle nm at the distance of 23° 27' 36" from the equator, representing the parallel of declination, and it will cut the sixo'clock hour circle PP' in F, the sun's place at the given time. Through Z, F, and N, describe the azimuth circle ZFN cutting the horizon in D, then FD is the altitude, FZ the zenith distance, and the angle FZP, or its measure, the arc DO, is the azimuth; consequently the things given and required fall in either of the triangles FZP, or FDA, which are supplemental to each other. For, since OP is the latitude, PZ is the colatitude, AF is the declination; consequently FP is the polar distance, DF being the altitude, FZ must be the zenith distance. Calculation. In the right-angled spherical triangle FPZ, rightangled at P, FP and PZ are given, to find the angle FZP and FZ; or in the triangle ADF, right-angled at D, there are given the angle FAD, equal to the latitude of the place, and AF, the sun's declination, to find DF, the altitude, and the side AD the azimuth. By the rule of the circular parts FP, PZ, and PZF, are all connected, therefore PZ is the middle part, and PZF and PF are the adjacent parts, where Rx sine ZP-tan PF x cot PZF, or cot azimuth To log cos lat cot dec X cot azimuth, therefore Sum-log cot az Again, to find FZ the coaltitude, and the same things being given, cos lat x tan dec cot dec Add log tan dec 23 27 36 76 20 38 Rx cos FZ-cos ZP x cos FP, or sine alt-sine lat x sine dec. Given the latitude of the place, and the sun's declination, to find the altitude and hour when the sun is due East or West. EXAMPLE. At Edinburgh, on the 21st June, 1826, what was the sun's altitude and hour when due East or West, the declination being 23° 27' 36" N? In the last figure let ZAN meet the parallel n m in K, and suppose a circle to be drawn through the points PKP', forming the triangle ZKP, right-angled at Z, then ZK is the coaltitude, and ZPK the hour from noon; hence From noon, that is, at 7h 8m 12 A. M., and 4h 51m 48 P. M. This problem is of considerable utility to the navigator and practical astronomer, for the purpose of determining time accurately when an altitude instrument is used. As the change of altitude, on which the accuracy of the determination of the time depends, is quickest when the object is on the prime vertical, the most proper time for observing an altitude for that purpose is, therefore, when the object is due East or West, as any small error in the observation has then the least possible effect on the time. Other errors are also in this case in a great degree avoided, or at least considerably lessened, particularly that arising from any small error in the estimated latitude at the time of observation. To facilitate its application, the following table, corresponding to the latitude and declination (which must be of the same name with the latitude), has been given. When the latitude and declination are of different names, the altitude must be as near the horizon as is consistent with accuracy, so far as depends upon the uncertainty of the horizontal refraction. Altitudes under 5° should not be used when great accuracy is required. Table, showing the Time from Noon and Altitude when a Star whose Declination is less than the Latitude is on the Prime Vertical. λ The change of altitude on the prime vertical in one second of time is=15" x cos a const log 1.176091+log cos λ, in which a is the latitude. This gives 8".4 at 56° N. PROBLEM VI. Given the latitude of the place and the sun's declination, required his amplitude and ascensional difference ?* At Edinburgh, on the 21st of June, 1826, from the data given in the last example, on what point, and at what time, did the sun rise and set? In the triangle ABC, in the last figure, there are given the angle BAC, equal to the colatitude, and BC the sun's declination; to find AC and AB. Rx sine BC=sine AC x sine BAC, or AC, CO, 45 19 33 sec 10.251939 sine 9.851941 44 40 27, in which case AC is the amplitude reckoned from the East or West, to the North or South, according to the name of the declination, and CO is that reckoned from the meridian, or from the North or South, according to the name of the declination. * By the ascensional difference is meant the time before or after six o'clock the sun rises or sets. By this problem, therefore, the lengths of day and night are determined, and the variation of the mariner's compass. M |