by the chronometer for the same instant will be the error of the chronometer. EXAMPLE. At a given place the altitude of the sun was the same at 9h 34m 20s A.M., and 2h 32m 26s P.M.; required the error of the clock, the polar distance being decreasing, the equation of equal altitudes 88-4, and the equation of time I'm 57s-6 to be added to apparent time. Clock is fast at apparent noon 0 1 36 Instead of only two observations being taken, several corresponding pairs may be taken, and the sum of the times of observation in the forenoon being divided by their number, and also the sum of the afternoon observations being similarly divided, the quotients are the mean times of observation, which are then to be treated as the two times of observation. The times of observation ought to be more than two hours distant from noon. EXERCISES. 1. At a given place the altitude of the sun was the same at 8h 4m 54s, and 4h 2m 36s; required the error of the clock, the polar distance being increasing, the equation of equal altitudes 125-4, and the equation of time 4m 16s-7 to be subtracted from apparent time. Ans. Clock fast 8m 148.1. 2. Suppose that at a given place the altitude of the sun was the same at 9h 40m 2s A.M., and 2h 10m 25s P.M.; required the error of the clock, the polar distance being decreasing, the equation of equal altitudes 14.5, and the equation of time 3m 50s-2, to be added apparent time. Ans. Clock slow 8m 51s.2. 473. PROBLEM XXX.-Given the latitude of the place, and the declination of a celestial body, to find its amplitude and ascensional difference. Z E Let HZR be the meridian of the place; EQ the equator, Pits pole; HR the horizon, Z the zenith, and B the body in the horizon when rising; then in the triangle ODB, D is a right angle; O is the colatitude; w BD the declination; OB the ampli- н tude; OD the ascensional difference; and EOOD the semi-diurnal arc. When the declination is south, OB′ is the amplitude; OD' the ascensional difference; and EO-OD' the semi-diurnal arc. the body is setting, the figure is exactly similar. N It is evident that if any two parts of the right-angled triangle OBD are given, the other parts can be found. Let angle BODC the colatitude, BDD the declination, ODN the ascensional difference, BW I the semi-diurnal arc. To find the amplitude M. L. Sin D = 9.3912057 P L. Sin M And M W B L. R L. Tan D To find the amplitude OB. R⚫sin BD = sin O · sin OB, or sin C: R= sin D: sin M. To find the ascensional difference OD. = R⚫sin OD cot O⚫ tan DB, or R : tan D = cot C: sin N. I=6h N. Then To find N. B EXAMPLE. When the declination of a celestial body is 14° 15′ N., what is its amplitude and ascensional difference in latitude 36° 45′ ̃N.? Here C = 53° 15', D = 14° 15′. D = 10· R When =9-4047784 9.8731668 = L. Sin N =9-2779452 9.4874356 17° 53′ 28′′ N. And N 10° 55′ 56′′. The semi-diurnal arc I = 6h + N = 6h 43m 438.7. = EXERCISES. = 1. The declination of a celestial body is 37° 34′ N.; what is its amplitude and ascensional difference in latitude 46° 8' N.? Ans. M 61° 37′ 5′′ N., and N = 3h 32m 38.4. 2. The declination of a celestial body is 26° 3′ 53′′ S.; what is its amplitude and semi-diurnal arc in latitude 55° N.? Ans. M 50° S., and I = 3h 2m 45s.4. 474. PROBLEM XXXI.-To find the apparent time at which the sun's centre rises or sets at a given place. 'Find the sun's zenith distance when his centre appears on the horizon; then, its polar distance and the colatitude being known, find the corresponding semi-diurnal arc, and this arc, converted into time, will be the time of rising or setting before or after apparent noon. The sun's zenith distance, when the apparent altitude of his centre is zero, is found by subtracting its parallax from the sum of the dip and horizontal refraction, and adding the remainder to 90°." The declination to be used is of course that at rising and setting, which can be found by first determining the semidiurnal arc, as in last problem, supposing the declination to be that at noon at the given place; and then the approximate time of rising and setting are known, and the longitude being also known, the reduced time, and hence also the reduced declination, can be found. EXAMPLE. Find the mean time of the apparent rising of the sun's centre, on the 24th of May 1841, at a place in latitude 55° 57' N., and longitude 25° 30′ W., the observer's eye being at the height of 24 feet. Approximate apparent time of rising Longitude in time on 23d = 0 15h 38m 1 42 = 17 20 20° 44′ 27′′ N. 0 33 51 4 47 0 0 9 0 37 29 By 467, 2 L. cos + H =L. sin S+L. sin (S-Z) + L. cosec P+L. cosec C. Here Z = 90° 37' 29" L. Sin S P= 69 15 33 L. Sin (S-Z) C= 34 3 0 L. Cosec P L. Cosec C 2)193 56 9.9967814 9.0430038 2 1 S= 96 58 S-Z-6° 20' 32" And H = 62° 46′ 29′′ Hence apparent time of rising on 24th is 3h 37m 488.2 A.M. Equation of time 0 3 30.9 Mean time of rising 3 34 17.3 EXERCISE. Find the mean time of the setting of the sun, on the 20th of July 1841, in longitude 35° 45′ E., and latitude 55° 57′ N., the eye of the observer being 20 feet high, its registered declination on the 20th and 21st being 20° 40′ 38′′ and 20° 29′ 12′′, and the equations of time 5m 588-7 and 6m 28.1. Ans. At 8h 27m 59s.9. = 2)19.3207615 9-6603807 L. Cos H = And H = 8h 22m 118.8. 475. The time of a star's rising or setting may be found thus: Compute the star's semi-diurnal arc, and it will be the sidereal interval from its rising to its culmination, which is to be reduced to the mean solar interval by 448; then find the mean time of the star's culmination by 456, and apply to it the preceding interval by subtraction for the mean time of rising, and by addition for the time of setting. The time of the moon's rising or setting may be found thus:-Find approximately its semi-diurnal arc, considering its declination and horizontal parallax to be that at the nearest noon; and find the time of its meridian passage, then the approximate time of its rising or setting is known. Compute its declination and parallax for the reduced approximate time of rising, and find again its semi-diurnal arc; then 24 hours is to the semi-diurnal arc found, as the daily retardation to a fourth term, which, added to the preceding arc, will give the interval between the rising or setting and the meridian passage in mean time. For a sidereal day is to any sidereal arc as a lunar day (expressed in mean time) to the corresponding lunar arc (expressed also in mean time.) Then the sum or difference of this interval, and the time of transit, will be the time of rising or setting. Let H H' semi-diurnal arc, in sidereal time, R r = moon's retardation for arc H', in mean time, the mean time of transit, t' t = the mean time of rising or setting; then 24: RH:r, or P.L. r = P.L. R+ P.L. H, H'=H+r; then t=t'+H', and where the upper sign refers to the time of setting, and the lower to the time of rising. The same method applies in finding the rising or setting of the planets; but when vv, or when v' is negative (461), r is negative, and H'H-r. EXERCISE. Find the mean time of the rising of the moon for the data of the example in article 458; having also given the moon's declination on 2d May at Oh = 5° 20' 7"-4 S., its horizontal parallax on the 2d at noon and midnight 57′ 28′′ and 57′ 14′′, and its declination on the 2d at 77° 4′ 18′′, and at 8h 7° 19′ 2′′, the horizontal refraction being 33′ 51′′, and the latitude of the place 54° 30'. Ans. 4h 19m 278.3. METHODS OF FINDING THE LATITUDE. 476. PROBLEM XXXII.-Given the declination of a celestial body, and its meridian altitude, to find the latitude of the place of observation. "Call the true zenith distance of the object north or south, according as the zenith is north or south of the body; then, when the zenith distance and declination are of the same name, their sum is the latitude also of the same name; but when of different names, their difference is the latitude of the same name as the greater. When the body is at the lower culmination, the latitude |