To find the angle Zenith N©, measured by the arc aÆ,—the time from noon. Conse 136 34 56=ZONZ=9h6m203, time from noon *when the sun is 18° below the horizon. quently the day breaks at 2h 53m 40s in the morning, and twilight ends at 9h 6m 20s in the evening, supposing the sun's declination to undergo no change between the beginning of twilight in the morning, and the ending thereof at night, being about 18 hours. = The same might have been found from the triangle OsNadir,for so=90°+10° 100°, Nadir=90°-18° 72°, and Nadir s = comp. lat. 38° 28′. Then, by the method above, find the angle Os Nadir (measured by the arc aq)= 43° 25′-2h 53m 40s as before, the time from midnight, when the sun is 18° below the horizon. (583) When the declination of the sun, the latitude and declination being of the same name, is greater than the difference between the complement of latitude and 18°, the parallel of declination ( sss) will not cut the parallel of 18° (TOW) below the horizon: consequently there will be no real night at these times, but constant day or twilight; as is the case at London from the 22d of May to the 21st of July. (584) Since the sun sets more obliquely at some times of the year than at others, it necessarily follows that he will be longer in descending 18° below the horizon at one season than at another. When the sun is on the same side of the equinoctial as the visible pole, the duration of twilight will constantly increase as he approaches that pole, till he enters the tropic, at which *The NZ represents the time from apparent noon, that is, from the time thecomes to the meridian, and not the time from 12 o'clock, as shown by a well-regulated clock or time-piece. If the equation of time, given in page I. of each month in the Nautical Almanac, be applied to this apparent time by addition or subtraction, as there directed, you will obtain the mean time, or that shown by a well-regulated clock or chronometer. time the duration of twilight will be the longest. It will then decrease till some time after the sun passes the equinox, but will increase again before he arrives at the other tropic; therefore there must be a point between the tropics, where the duration of twilight is the shortest, which point may be found by the following proportion; viz. rad : tan 9o :: sine of the latitude : sine of the sun's declination.* The declination must always be of a contrary name with the latitude. PRACTICAL EXAMPLES. 1. Given the sun's declination 10° south, and the latitude of the place 51° 32′ N. to find the time of day-break in the morning, and the end of twilight in the evening. Answer. The LOS Nadir=73° 35′ 34′′ = 4h 54m 22s the time from midnight. Hence day breaks at 4h 54m 22s in the morning, and twilight ends at 7h 5m 38s in the evening; admitting the sun's declination to be constant for one day. 2. Given the sun's declination 23° 28′ S. and the latitude of the place 51° 32′ N. to find the time of day-break in the morning, and the end of twilight in the evening. Answer. The LOS Nadir=90° 16′ =6h 1m 4s the time from midnight when the is 18° below the horizon. Consequently day breaks at 6h 1m 45, and twilight ends at 5h 58m 56s, on the shortest day at London. 3. In the latitude of London, 51° 32′ N., what time does the day break when the sun has 11° 30′ N. declination? Answer. 2h 41m, rejecting seconds. 4. Required the beginning and ending of twilight at London, lat. 51° 32′ N., when the sun's declination is 15° 12′ N. Answer. 2h 4m 11s and 9h 55m 49s. 5. Required the time the twilight begins and ends at London, latitude 51° 32′, when the sun has 11° 30′ south declination. Answer. 5h 2m and 6h 58m, rejecting seconds. 6. At London, latitude 51° 32′ N., required the sun's declination, day of the month, and duration of twilight when it is the shortest. Answer. Sun's declination=7° 7' 25" south, answering to March 2nd and October 11th. The duration of twilight is * Vide Dr. Gregory's Astron. page 328. Vince's Astronomy, vol. i. page 18.&c. Also, Leybourn's edition of the Ladies' Diary, vol. iv., from page 314 to 339, where the subject is amply discussed. 1 56m 32; this is found by taking the difference between the time of sun-rise and day-break. (585) Given the day of the month, the latitude of the place, the horizontal refraction, and the sun's horizontal parallax, to find the apparent time of his centre appearing in the Eastern and Western part of the horizon. Example. Given the sun's declination at noon on the 21st of June 1840=23° 28′, the latitude of the place=51° 32′ N., the horizontal refraction =33', the sun's horizontal parallax 8"; to find the apparent time of his rising and setting. This example being given for the longest day, the sun's declination may be considered the same at his rising as at noon; for the declination does not vary above 5′′ in 24 hours at this time, though near the equinoxes it varies about l' in an hour. * Let s be the true point of the sun's rising, and b that of his apparent rising, then bo=33′-8"-32′ 52" the distance of the sun's centre below the horizon; hence the true zenith distance zs will be increased, by refraction to zO, but the declination will remain the same, for NS=N O. In the oblique-angled triangle ZN O. z= 90° 32′ 52" app. dist. of O's centre from the zenith. NO 66 32 0 app. dist. of O's centre from the north pole. ZN = 38 28 0 the complement of the latitude. LZNO 124 16 30=8h 17m 6s the apparent time from noon when the sun's centre appears in the western part of the horizon. Hence the apparent time of The apparent time of rising and setting of the heavenly bodies always differs from the true time, for they are elevated by refraction (165), and depressed by parallax (184). All the heavenly bodies, when in the horizon, appear 33' above their true places, by the effect of refraction (see Table IV.); but they are depressed by * the sun's apparent rising is 3h 42m 54s, and the time of apparent setting is 8h 17m 65. The time of the sun's true rising and setting on this day, has been determined (page 287) to be 3h 47m 32 and 8h 12m 28; hence the apparent day is 9m 16 longer than the true day. (586) As the refraction causes an error in the rising and setting of the celestial objects, so it will cause an error in the amplitudes, as may be seen by comparing the triangle ABS with the triangle ADb. If the sun's amplitude be taken when his centre is in the visible horizon, an allowance depending on the horizontal refraction, parallax, height of the eye, and latitude of the place, should be applied to the observed amplitude, in order to obtain the true amplitude. This inconvenience may be avoided, by observing the amplitude when the altitude of the sun's lower limb is equal to 16'+the dip of the horizon. PRACTICAL EXAMPLES. 1. Required the apparent time of the sun's rising and setting at Glasgow, latitude 55° 32′ N., longitude 4° 15′ W., on the 12th of October 1840. The sun's declination at apparent noon, for the meridian of Greenwich (Naut. Alm.), being 7° 31′ 50′′ S., and on the 13th 7° 54' 19" S. Answer. The sun's declination, when it is noon at Glasgow, is 7° 32′ 6′′ S. (558). The O's declination being south, the triangle ZNO will fall on the left hand of the prime vertical za. Then ZO 90° 32′ 52′′ With these three sides, find the hour LZNO 79° 53′ 6′′ 5h 19m 32s, the time from noon when the sun sets; hence the sun rises at 6h 40m 28s. Call these the approximate times of rising and setting, calculate the sun's declination for these times of rising and setting, which you will find to be 7° 27′ 7′′ S., and 7° 37′ 5′′ S.; parallax. Hence, by these causes combined, a body becomes visible when its distance from the zenith is equal to 90° +33′-horizontal parallax, and as the sun's horizontal parallax is about 8" (see Table VI.) his centre will appear in the horizon when it is 32′ 52′′ below it, or 90° 32′ 52" from the zenith. If to this zenith distance, the sun's semi-diameter 15′ 45′′ be added, the zenith distance of the centre for the setting of the upper limb will be had; if subtracted, that for the lower limb will be obtained; the former being 90° 48′ 37′′, and the latter 90° 17' 7", and the respective times 8h 19" 20 and 8h 14m 52. A star has no sensible parallax, and, therefore, will appear in the horizon when it is 33' below it, or 90° 33' from the zenith. * The equation of time on this day at noon (see page I. of the Nautical Almanac) is 1m 24 at Greenwich; this, applied as directed in the Nautical Almanac, will give the mean time of the sun's rising and setting, as shown by a well-regulated clock or time-piece. then, with the polar distance N = 97° 27′ 7′′ repeat the operation, and the apparent time of apparent rising will be 6h 39m 58; with the polar distance NO 97° 37′ 5′′ repeat the operation a second time, and the apparent time of the sun's apparent setting will be 5h 20m 2s. To obtain the mean time of rising and setting, as shown by a well regulated clock, or time-piece, the equation of time must be applied as directed in the Nautical Almanac. 2. Required the apparent time of the rising and setting of the sun, in latitude 51° 32′ N., supposing his declination to be 23° 29′ S. Answer. 8h 8m 2s and 3h 51m 58o. 3. Required the apparent time of the rising and setting of the sun, in latitude 51° 32′ N. when his declination is 17°32′ N., supposing it to undergo no change from sun-rise to sun-set. Answer. 4h 22m 148 and 7h 37m 46s. 4. Required the apparent time of the rising and setting of the sun in latitude 56° N., supposing the sun's declination at his rising to be 22° 34′ 35′′ N., and at his setting 22° 30′ 7′′ N. Answer. 3h 22m 19s and 8h 36m 59s. PROBLEM IX. (Plate III. Fig. 7.) (587) Given the latitude of the place, the day of the month, the moon's horizontal parallax and refraction, to find the mean and apparent time of her rising. Required the time of the moon's rising at Greenwich, latitude 51° 28′ 39′′ N., on the 4th of August 1840. The horizontal parallax of the moon varying from 53' to 62', always exceeds the horizontal refraction; therefore, when the moon's centre appears in the horizon it is really above it, by a quantity equal to the horizontal parallax minus the refraction. The moon's declination must be found as near to the time of her rising as possible, because it is subject to a considerable variation in the course of a few hours. Let n be the place of the moon, when in the horizon, m the point where she becomes visible, and e her place on the meridian; then de will represent her change of declination from the time of her rising to the time of her transit. The time of the moon's passage over the meridian at Greenwich is given in page IV. of the Naut. Alm., and may be reduced to any other meridian by art. (562); reduce the declination to this time (558). With this declination, the latitude, &c. calculate the hour angle, which subtract from the time of the moon's passage over the meridian, and call the |