Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

the sun's rising is 'sh.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 3b.47'.32" and gh.12'.28"'; hence the apparent day is 9.16" longer than the astronomical day.

(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 at 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 70.17.34" S. and on the 13th 7°.40'.8" S.

Answer. The sun's declination, when it is noon at Glasgow, is 7o.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 =°S NO=970.17.50") LZNO=800.14.3"=51.20.57". the time

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

sun rises at 65.39'.?". Call 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

by the effect of refraction (see 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, cr 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 gh.14'.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 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.

79.22.51" South,; then, with the polar distanceNO 97.12.49" repeat the operation, and the true apparent time of rising will be 65.38.33"; with the polar distance NO=97°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 23°.29' S?

Answer. 8h.8.2" and 3h.51.58".

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

sun, in latitude 51°.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 220.34'.35", and at his setting 220.30.7" ? Answer. 3h.22.20" and 8b.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 parallaz. 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 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.

Find the time of the moon's passage over the meridian (E. 275.), and calculate the declination to that time (B. 270.).

e

* 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 suf

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.

SOLUTION.

1. Find the time of the moon's passage over the meridian. The moon passes over the meridian at 13h.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=40.44'S.

Decrease of declination in 12 hours

3.

[ocr errors]

12h. : 30.0':: 15.32'.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 L2Nm. 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 IH.32.51" (C.271.), the time past midnight when the D is on the meridian, the horizontal parallax at that time will be 575.6". Hence we have

zm=90°.733' — 57'. 6"=890.35'.54" Given

To find the 70.20'.48"=970.20.48" NM=90°. + ZN =900.- 510.28'.40"-380.31'.20" °

=S LzNm. The X 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' 56.20'.5"=86.12.46" 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.

4. To find the D's declination at the estimated time of rising. D's decl. at noon 4th Aug. 1822. =100.371 Horizont. parallax=56'.49" D's decl. at midnight 4th Aug. 1822 = 79.44' Horizont.parallax = 57'. 4"

Decrease of declination

12 hours

20.53

Incr. in 12

09.15"

12h. : 20.53'::8h.12.46" : 10.58'.24" 12" : 15"::8h.12'.46" : 10" D's decl. at ris. = 109.37' –10.58'.24" =80.38'.36"S. 1) 's hor, parallax =56'.59!

[ocr errors]
[ocr errors]

:

5. Again, in the triangle znm, we have zm=90° +33-56'.59" =890.36'.1".

To find the Given NM=90° +8°.38.36" = 989.38.36" zn=90°-51°.28'.40" — 38°.31.20". S

LZNM. The LZNm will be found=78°.20'.4" = 54.13.20", the time the moon rises before she comes to the meridian; whence 13h.32'.51"-55.13.20"=8h.19'.31"P.M.thetimeofthe moon's rising, nearly.

To find the correct time. The ) passes the meridian on the 5th of October (Prob. viii. page 276.) at 145.18'.12", then 14".18'.12" – 13.32.51"=45.21" daily variation of transit; hence, 24h: 45'.21":: 8h.19'.31" : 15'.44", and 8h.19'.31"15'.43"=8h.3'.47" P.M. true time of the moon's rising..

(T) Should more examples be thought necessary, they may be formed and solved in a similar manner; likewise the rising of the different planets may be found by the same method. The times of the rising and setting of the moon and of the planets, are given in White's Ephemeris, and in the Connaissance des Tems. These works will serve as a check

upon

the calculation. The rising and setting of the moon, and of the planets, are not inserted in the Nautical Almanac.

PROBLEM X. (Plate III. Fig. 1.) (U) The latitude and longitude of a fixed star, or planet, being given, to find its right ascension and declination, et contra. *

In the oblique triangle nnc. Nn=obliquity of the ecliptic, or distance between the poles of the equator and ecliptic.

nc=the complement of the star's latitude vc. Nc=the complement of the star's declination RC. The angle nnc, measured by the arc ve, is the complement of the star's longitude from aries.

The supplement of the angle nnc, measured by the arc RQ, is the complement of the star's right ascension from aries.

Or, In the oblique triangle smx.
sm=the obliquity of the ecliptic.
mx=the complement of the star's latitude vx.
sx=the complement of the star's declination rx.

* A neat method of solving this problem may be seen at page 44 and 45, Vol. I. third edition, of Dr. Mackay's useful treatise on the theory and practice of finding the longitude, or in Dr. Maskelyne's introduction to Taylor's logarithms. The problem may be varied so as to admit of several cases, but the two above specified are the only useful ones.

X

The supplement of the angle xms, measured by the arc vv is the complement of the

star's longitude from aries. The angle xsm, measured by the arc rg, is the complement of the right ascension from aries.

EXAMPLE.

[ocr errors]

mx

2

76106

The latitude of Aldebaran in Taurus being=5°.28'.27" S.; its longitude 25.6°.56:24"*; required its right ascension and declination; the obliquity of the ecliptic being 230.28'?

In the triangle smx.
Sm = 230.28'. 0"=the obliquity of the ecliptic.
xms = 1569.56'.24" =star's longitude +90°.
mx = 849.31'.33" = the complement of the star's lat.

1. To find the co-declination sx. sm=230.28'.00" log sine 9•60012|| Nat. sine 220.22

38053 = 840.31'.33" log sine = 9.99802 jxmg=789.28'.12" log sine=

S 9.99114 29.99114

mx-sm=619.3'. 33'' Nat. cos = 48891 229.22' log sine Nat. cos 730.541.37''

27715

sx = 1069.5'.23" obtuse. Hence 106°.5'.23"-90°-16.5'23" =RX the declination N.

2. To find the co-right ascension xsm.
Sine sx=1069.5'.23" : sine xms=1569.56'.24"::

Sine mx=840.31'.33" : sine xsm=230.56.92".
Hence Ar=660.3ʻ.28", the right ascension.

OR THUS:

Draw the perpendicular XL.
Rad : cos Lmx::tang mx : tang Lm, and Lm +=LS.
Cos mL : cos mx:: cos sl : cos sx.
Sine ml : sine se::cot Lmx : cot xsm.

9.58042

[ocr errors]
[ocr errors]

PRACTICAL EXAMPLES.

1. Required the right ascension and declination of Regulus in the Lion's heart; latitude being 0°.27'.30” N., and longitude 48.26°.57'.5", obliquity of the ecliptic being 230.28'?

Answer. Right ascension=1490.19ʻ.54" and declination= 120.58'.21" N.

2. Required the right ascension and declination of Spica Virginis, in the sheaf of Virgo; latitude 20.2o.23" South, and longitude 68.209.571.10"; obliquity of the ecliptic 230.28'?

The latitudes and longitudes of the stars, in this, and many of the following examples, in the first edition of this work (published in the year 1801) were adapted to the year 1796 ; any change of data is at present unnecessary, as the method of solution remains the same, being independent of an ephemeris, and the answers, when obtained, would only differ a few seconds from those now given.

« ΠροηγούμενηΣυνέχεια »