tan. 90 tan. 45 =b1 = 0, correct results. cos. Ľ' 869. Let the inclination of the plane to the horizon be a, and its inclination to the meridian B. Then taking a place whose horizon is parallel to the plane, it is evident the sun will begin to shine upon the plane when it rises at this assumed place. Let L be the given co-latitude, L' the assumed one. Then from the A, whose sides are the distance of the zeniths of the places, (viz. c), and L, L', we get cos. L.cos. a cos. ß = sin. L. sin. a and .. cos. L' = cos. L. cos. at sin. a. sin. L. cos. B, which gives (when adapted to logarithms) L'. Hence, if D be the co-declination of the sun, we get, from the right angled A, whose I and hypothenuse are 90° - L', and D, cot. H = cot. L' x cot. D, where H is the hour angle from the meridian of the assumed place. But inclination of the meridians of two places = inclination of their horizons. Consequently Ha the time required is known. 870. Let L be the co-latitude of the place, D the declination of the sun on one day, D' that on another, and d the difference of the lengths of the meridian shadows of the tower, whose altitude is x. Then since the meridian altitude of the sun is LED, we have x = 1 tan. (L+D) = (1 + d) tan. (LED) 2 being the actual length of one shadow. Hence, d x tan. (L +D) 1= tan. (L ID) – tan. (LED) and :. 2 = tan. ( LD. tan. (L + D) tan. ( LD) – tan. (LED) 871. Let L be the latitude of the place, D the co-declination, and A the right ascension of the star. Also let D' be the codeclination of the sun, which is known from the tables for the given day. Then from the right-angled A, whose I and hypothenuse are L and D', we have cos. H = tan. L. cot. D', H being the hour Z from midnight, and indicating the hour of sunrise. The time when the star rises is HŁA. 872. A writer, in Leybourn's Mathematical Repository gives the following solution of this problem : “ The greatest apparent distance of a star from the north pole, arising from aberration and nutation together, is when the sun's longitude, and also that of the D's ascending node, are three signs before the *, reckoning according to the order of the signs; and that apparent distance will be greatest of all when the star is in the solstitial colure ; but it does not appear that the period can be accurately determined when such an event will take place, for any proposed star. We can, however, calculate the time which seems to be required by the question, if it falls within the limits of some proposed revolution of the D's nodes, or if it is restricted to a given year; thus, for instance, let the year be 1809, and suppose the star's right ascension = 108°, and its declination 60° north. Let a great o be drawn to the E from the star I to the meridian through the star, and it will meet the ecliptic in 25° 41' of Libra, which is the O's place, when the aberration in declination southward is a maximum, (obliquity of ecliptic being 23° 27' 44') ; this answers to October 19, at London. On that day the star comes to the meridian at 22} before six in the morning. The longitude of the D's node at that time is 6* 23° 45', and the other is 6* 25° 26'; the star is therefore very ne at its maximum distance from the north pole, which distance will not sensibly vary during three or four days at that time.” 873. Let D, D' be the co-declinations of the stars, L the latitude of the place, and H, H + h the angular distances from the meridian at the moment of rising, h being the difference of the given right ascensions; then from the right-angled A, whose I and hypothenuses are D, L; D' L, we get cos. (H + h) = tan. L. cot. D cos. H = tan. L. cot. D', and eliminating H, we have tan. L. cot. D= cos. I tau. L cot. D – sin. 1 x {1 tan. L. cot. D'} which gives cot. L= cot.? D + cot.'D— 2 cos. h.cot. D.cot. D' sin.' 874. Let A, D, be the right ascension and co-declination of the star; A', D' the right ascension and co-declination of the sun ; also let L be the latitude of the place, and H, H' the hour angles for the star and sun when rising; then from the rightangled A, whose hypothenuses are D, D, we have cos. H = tan. L. cot. D cos. H' = tan. L. cot. D' But H' H= A A, :. cos. H'. cos. H + sin. H . sin. H' = cos. (A' - A) .. cos. (A' A) = tan. L. cot. D. cot. D' + {(1 tan. L. cot. D).(1 – tan. I cot. D}). Again, if O be the obliquity of the ecliptic, it is easily shown that cot. D' = tan. O sin. A' ...... (1) which being substituted in the above equation, gives cos. (A' - A) = tan. * L cot. D. tan. O x sin. A' + N {(1 tan.' L cot. D) (1 - tan.' L tan. O x sin.'A')}, which being reduced and resolved, will give A. Hence, by Equat. (1) is known D and therefore by the tables the day of the year is known. The time of rising on that day is known from cos. H' = tan. L. cot. D'. 875. “ To delineate the phases of the moon," (see Woodhouse's Astron., p. 552.) “ To find the < which the line joining the moon's cusps make with the horizon at any time, we will premise the following, LEMMA. Required the inclination of the plane passing through the centres of the earth, sun, and moon, to the ecliptic. Let S, E, m (Fig. 112,) be the intersection of these planes, S, M, E being the positions of the sun, moon, and earth. Draw Mm I ecliptic, mm 1 SE, and join Em, EM, m'M; then the inclination required is measured by Z Mm'm. Em mm. tan. MEm tan. () 's geocentric latitude) sin. () 's elongation in longitude) which gives the _ Mm'm. Again, let MM' (Fig. 113,) the line joining the moon's cusps be produced to meet the horizon in H', and let Mm', the plane passing through the earth, sun, and moon, intersect the ecliptic M'm', which also cuts the horizon in H, in the point m'. Then, by the nature of phases, (see Woodhouse,) the M = 90°; the si is given by the Lemma, and the Z M'HH' being measured by the altitude of the nonagesimal, is found from Woodhouse, p. 741; consequently, LH' = 180° M' - H H (M + m) 90° - (H + m') is known. = 900 876. Let A be the sun's right ascension, L the latitude of the place, D the sun's declination, and a the ascensional difference. Also let O be the obliquity of the ecliptic then, in the first place, from the right-angled A (D, A) we have tan. D= tan. O x sin. A which gives D. Again, from the right-angled A (D, x) we have sin. x = tan. D x tan. ( opposite x) tan. D x tan. L - tan. O sin. A x tan. L which is a general investigation of the ascensional difference. 877. From the A (p, 90° – 1, 90° — a), we have sin. I cos. p cos. t = Pcos. / a) cosec. p sec. |