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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 : tang 9°::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 510.32' N. to find the time of day-break in the morning, and the end of twilight in the evening ?

Answer. The Los Nadir=730.35'.34" 4h.54'.22" the time from midnight. Hence, day breaks at 46.54.22" in the morning, and twilight ends at 76.5'.38" in the evening; admitting the sun's declination to be constant for one day.

2. Given the sun's declination 23o.28' S. and the latitude of the place 51o.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'=6.1.4" the time from midnight when the ois 18° below the horizon. Consequently day breaks at 5h.58'.56" and twilight ends at 65.1'.4", on the shortest day at London.

3. In the latitude of London, 519.32' N., what time does the day break when the sun has 119.30' N. declination?

Answer. 25.41', rejecting seconds.

4. Required the beginning and ending of twilight at London, lat. 510.32 N., when the sun's declination is 150.12 N.?

Answer. 2h.4'.12'' and 9h.55'.48".

5. Required the time the twilight begins and ends at London, latitude 510.32', when the sun has 119.30' South declination ?

Answer. 55.2' and.66.58', rejecting seconds.

6. At London, latitude 51°.32 Ñ., required the sun's declination, day of the month, and duration of twilight when it is the shortest?

Answer. Sun's declination = 70.7'.25" south, answering to

* Vide L'Hospit. Infinim. Petit. sect. iii. prop. 13. Dr. Gregory's Astron. page 328. Heath's Royal Astron. page 225. Emerson's Miscellanies, page 492. Vince's Astronomy, vol. i. page 18, &c. Also, Leybourn's edition of the Ladies' Diary, vol. iv. from page 314 to 339, wherein the subject is amply discussed.

March 2d and October 11th; between which days, the twilight increases, and from the latter to the former it decreases. The duration of twilight is 16.56'.32"; this is found by taking the difference between the time of sun-rise and day-break.

PROBLEM VIII. (Plate III. Fig. 6.) (Q) 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 1822=239.28'; the latitude of the place=510.32' N.; the horizontal refraction=33'; the sun's horizontal parallax =9"; 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 1' in an hour.

Let s be the true point of the sun's rising, and b that of his apparent* rising, then bo=33-9"-32.51" 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=NO.

In the oblique-angled triangle ZNO. 20 =90°.32,51" app. dist. of O's centre from the zenith.

. NO=66°32'. O' app. dist. of O's centre from the north pole. ZN = 380.28'. On the complement of the latitude.

2)195.32.51 ZN=38°.28' cosec

NO=660.32' cosec
Sum '97.16.251 Half sum=97°.46'.25" sine
ZO I 90.32.51 Remainder=70.13.341" sine

•20617 •03749 9.99599 9.09964

Rem. .7.13.343

2)19:33929

Cosine

620.8.15"

2

9.66964

[ ZNO=124.16.30=85.17.6" the apparent time from noon when the sun's centre appears in the western part of the horizon. Hence the apparent time of the * sun's rising is 32.42'.54", and the apparent time of setting is 8h.17'.6". The true time of the sun's rising and setting, on this day, has been determined (Prob. II. W. 286.) to be 3h.47'.32 and 8h.12.28"; hence the apparent day is 9'.16" longer than the astronomical day.

* The apparent time of rising and setting of the heavenly bodies always differs from the true time, for they are elevated by refractiun (S. 87, &c.), and depressed by parallax (S. 95, &c.).

All the heavenly bodies, when in the horizon, appear 33' above their true places, by the effect of refraction (sce Table IV.); but they are depressed by 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 9'' (see Table VI.) his centre will appear in the horizon when it is 32.51" below it, er 900.32.51" from the zenith. If to this zenith distance, the sun's semi-diameter 15'.47" be added, the zenith distance of the centre for the set. ting of the upper limb will be had, if subtracted, that for the lower limb will be obtained; the former being 90°.48'.38" and the latter 90°.17'.4", and the respec. tive times 8h.19'.20' and sh.14'.52".

(R) 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 át Glasgow, latitude 550.32' N., longitude 4o.15' W., on the 12th of October 1822 ?

The sun's declination at noon, for the meridian of Greenwich (Naut. Alm.), being 79.177.34" S. and on the 13th 70.40'.8" S.

Answer. The sun's declination, when it is noon at Glasgow, is 70.17'.50”S. (B. 270.). The O's declination being south, the triangle zno will fall on the left hand of the prime vertical za. Then

With these three sides find the hour ZO=90°.32.51"

S

LZNO=800.14'.3=55.201.57". the time NO=970.17.50"

from noon when the sun sets, hence the ZN = 340.28'

sun rises at 65.39'.?". Çall these the approximating times of rising and setting; calculate the sun's declination to the approximating time of rising and setting, which you will find to be 70.12'.49' S., and

A star has no sensible parallax, and, therefore, will appear in the horizon when it 33' below it, or 90°.33' from the zenith. * The equation of time on this day at noon (see page II. of the Nautical Almanac) is 1.15" at Greenwich ; this applied as directed in the 2d page, and page 149 of the Nautical Almanac, will give the true time of the sun's rising and setting, as shown by a well-regulated clock or time-piece.

70.22.51"South.; then, with the polar distance NO 97°.12.494 repeat the operation, and the true apparent time of rising will be 65.38'.33", with the polar distance NO=970.22'.51" repeat the operation a second time, and the true apparent time of the sun's setting will be 56.20'.27". To obtain the true, or 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 page 2, and page 165 of the Nautical Almanac for 1822.

PRACTICAL EXAMPLES. 1. Required the apparent time of the rising and setting of the sun, in latitude 510.32' N. supposing his declination to be 230.29' S?

Answer. 89.8.2" and 3h.51'.58".

2. Required the apparent time of the rising and setting of the sun,

in latitude 510.32 N. when his declination is 170.32 N. supposing it to undergo no change from sun-rise to sun-set ?

Answer. 4h.22.16" and 76.37.44".

3. 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 22o.34.35', and at his setting 220.30.7" ? Answer. 31.22'.20% and 8h.37.

PROBLEM IX. (Plate III. Fig. 7.) (S) Given the latitude of the place, the day of the month, the moon's horizontal parallax and refraction, to find the time, of her rising.

Required the time of the moon's rising at Greenwich, latitude 51°28'.40" N., on the 4th of August 1822.

The horizontal parallax, of the moon being 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 inust 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.

Find the time of the moon's passage over the meridian (E. 275.), and calculate the declination to that time (B. 270.). With this declination, the latitude, &c. calculate the hour angle, as in the preceding problem, and call this the estimated time of the moon's rising; correct the moon's declination, and her horizontal parallax, to the estimated time of rising; repeat the operation; and correct the result for the daily retardation of the moon's transit.

* The moon's declination at any hour may be found by interpolation or second differences, see Naut. Alm. 1822, page 174; but in general it may be found suf4. To find the D's declination at the estimated time of rising. D's decl. at noon 4th Aug. 1822. = 100.37 Horizont. parallax=56'.49" D's decl. at midnight 4th Aug. 1822 =

SOLUTION,

1. Find the time of the moon's passage over the meridian.

The moon passes over the meridian at 136.32.51" (by Prob. viii. page 276.)

2. To find the D's declination when she is on the meridian. D's declin. at midnight (Naut. Alm.) 4th Aug. 1822=70.44 S. D's declin. at noon ditto 5th Aug. 1822=4o.44' S.

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12h. : 30.0':: 16.32'.51": 23'.12" the change of the D's declination from midnight on the 4th of August, to the time of her transit; hence 70.44! — 23.12'' = 70.20.48“ S. the moon's declination when on the meridian.

3. In the triangle znm, find the LZNm. The horizontal refraction=33' (Table IV.). The D's horizontal parallax at midnight (Naut. Alm.)=57.4"; at noon on the 5th=57'.19", and by calculating the proportional part for Th.32:51" (C. 271.), the time past midnight when the V is on the meridian, the horizontal parallax at that time will be 57'.6". Hence we have zm = 90°. 733'— 57'. 6"=890.35'.54'

To find the Given NM=90°. + 70.20'.48"=970.20'.48"

zn=90°.- 51°.28'.40"=38°31'.2005 ZzNm. The L znm will be found = 80°.1'.12"=54.20'.5' nearly the time the moon rises before she comes to the meridian. Whence 13h.32.51"-55.20'.5” – 85.12.46" P.M.the estimated time of the moon's rising; and the time of setting will be nearly 65.52.56" A.M. on the 5th of August.

Horizont.parallax = 57'. 4"

79.44'

Decrease of declination in 12 hours

20.53

Incr. in 12 hours

0.15"

12h, : 20.53':: 84.12.46" : 10.58'.24" ||12): 15:: 84.12.46" : 10" D's decl. at ris. = 100.37'-19.58'.24" = 89.38'.36S. 1) 's hor. parallax = 56'.59''

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