ASTRONOMY.
Example 1. Required the true time of the apparent rising andsetting of the singular star in the neck of the Whale marked o, byBayer, on Oct. 3d, 1813, at Greenwich; its passage over themeridian being at llh. 51m. 16",'', night, fand its declination3° 40'31" south.
We must first find the ascensional difference as follows :
Is to tang, of star’s declination, 3° 49' 31".8,8251857
As tang, of the latitude of the place, 51° 28' 40".10,09.00491
Is to sine of the ascensional difference, 4" 49' 5".8,9242348
The ascensional difference turned into time, at the rate of 15degrees per hour, gives 19m. 18s.; this subtracted from six hours,because the declination is of a dilieiont name with the latitude ofthe place, there remains 5h. 40in. 42s. for the semidiurnal arc,without regard to parallax or refraction. The stun of the star’smeridian-passage and semidiurnal arc, 17h. 31m. 5S,2s. gives thetrue time of the star’s real selling; but if the latter l>e subtractedfrom the former, the remainder, 5h. 18m. 34,2s. is the true timeof its real rising. But on account of the refractive nature of theatmosphere, the rising of the stars and planets are continually ac-celerated a few minutes, and their settings retarded as much. Theparallax of the stare, and most of the planets, bring very small, cannever sensibly alter the effect of refraction. In order to find theeffect of refraction in our present example, enter the above Tablewith the declination and latitude of the place, and after makingthe necessary proportions, you find 3m. 38s. forthe ellect of refrac-tion, if the declination had been north ; but it being south in thepresent instance, reference mint be made to the last column ofthe Table for the quantity to be subtracted from 3m. 38s. whichaccording to it, is 2s. so the whole ellect is 3m. 36s. This cor-rection subtracted ’ from the real time of the star’s rising, gives I5h. 14m. 58,2s. for the true time of its apparent vising, and addedto the time of its real setting gives 17h. 35m. 34,2s. for the truetime of its apparent setting required.
Example^. Required the true time of the Moon ’s apparentsetting at Greenwich, on Jan. 10, 1813 ; her passage over the me-ridian being at 6h. 36m. her declination at apparent noon 6° 3'north, and the difference of the latter in the succeeding 12 hours+ 2° 14, and in the succeeding 24 hours -(- 4° 22'.
First Approximation. To obtain this we must find thesemidiurnal arc corresponding to the Moon ’s declination at noon,
as follows:
Is to tangent of D’s declination, 6° 3'.9,0232510
As tangent of the latitude, 51° 28'40"..'.10,0990491
Is to sine of the ascensional difference, 7° 39' 3".9,1243001
This being turned into time is 30m. 36s. which added to sixhours, because the latitude of the place and the lunar declinationare of the same name, gives 6h. 30m. 36s. for the semidiurnalarc. This quantity added to 6h. 36m. the meridian-passage of theMoon, the sum is 13h. 6m. 36s. the time of the Moon ’s seltin"-nearly. °
Second Approximation. Find the Moon ’s declination atthe time of her setting according to the first approximation, ln-saying, 12 1 lours is to 2° 14', her increase of declination in 12hours, as 13h. 6m. 36s. the first approximate time of the Moon ’ssetting, is to 2° 20' 23". Add this last quantity to fi 11 3', because
the declination is increasing, and the sum, S 1 ’ 29' 23", will be theMoon ’s declination at the lime of her setting nearly. Next findthe semidiurnal arc answering to this declination as follows:
Is to tangent of 8° 29' 23".9,1739650
As tangent of the latitude, 51° 28'40", .10,0990491
Is to the sine of the ascensional difference, 10°4S'26", 9,2730141This quantity turned into time amounts to 43m. 1 is. which-,added to six hours for the reason before assigned, gives 6h. 43m.14s. for the semidiurnal arc of the Moon at the time of her settingaccording to the first approximation ; and 6h. 3i>m. -)-6h. 43m. 14s.= 13h. 19m. 14s. the apparent time of the Moon ’s setting verynearly.
Third Approximation. Find the Moon ’s declination at13h. 19m. l4s. bv saying ; I? hours U to 2° 1-f', as 13h. !9m. 14s.is (o 2° 28'44". This being added to 6 1 ’ 3', gives 8" 31' 44" for
Is to tangent of 8° 31' 44"..n...0,1759941
As the tangent ef the latitude, 51° 28' 40'.iO,<h>904‘)l
Is to the sine of ascens. diff. 10° 51' 31"...9,2730432
This quantity turned into time is equal to 43m. 26s. whichadded to six hours gives Oh. 43m. 26s. forthe semidiurnal are otthe Moon at the time of her setting very nearly ; ami Oh. 36m.-f- 6h. 43m. 26s. = 13h. 19m. 26s. the true time of the Moon ’ssetting, without having regard to refraction and parallax. Thetime of the Moon ’s real rising or setting as obtained by this thirdapproximation can never vary from the truth more than a veryfew seconds of time, even in extreme cases, and therefore is sm-ficiently exact for most purposes. But we will make use ot afourth approximation in order to shew how the truth may be ar-rived at within much less than a second of time.
Fovrtii Approximation. First say, l't hours is to T’ 14',as 13h. 19m. 26s, the time of the Moon ’s real setting exlremeiynear the truth, to 2° 28' 47". Add this quantity to (5^ 3', and thesum, 8° 31'47", is the Moon ’s declination at the time of the thirdapproximation very nearly. To obtain it more exactly, say 12hours is to 6 minutes of a degree, the abatement of the progressivemotion of the Moon ip north declination in the 12 Hours succeed-ing midnight, compared with her motion during the preceding12 hours ; as lh. 19:n. 26s. is to 40". This quantity subtractedfrom 8° 31'47", because five progressive motion of {lie Moon indeclination is decreasing, gives 8° 31' T for the'dcclinalion exactly.Then find the semidiurnal arc corresponding to 8° 31' 7", as
follows :
Is to tangent of 8° 31' 7"..9,1754627
As tangent of the latitude, 51° 28' 40".10,0990491
Is to sine of the'ascen. differ. 10° 50' 42"...9,2745118
This turned into time is equal to 43m. 23s. which added to sixhours, gives 6h. 43m. 23s. for the exact semidiurnal arc of theMoon at the time of her setting ; and 6h. 36m. 6h. 43m. 23s.
= 13h. 19m. 23s. the true time of the Moon ’s real setting. Thislast result is as exact as the data whence the above calculationshave been made. On account of refraction and parallax the Moon ’sapparent rising or setting will for the most part vary a few minutesfrom the real time of her centre being in the horizon. To findthe combined effect of refraction and parallax upon the rising andsetting of the Moon or a planet, make use of the following methodas illustrated in the case of the Moon ’s setting : To the logarithm-cosine of the latitude of the place (/) -J- the Moon ’s declination,(d) add the logarithm-cosine of l—d, and subtract half the sum ofthe logarithm from the logarithm of the difference of the refractionand parallax, and the remainder will he the logarithm of the quan-'tity required. Ohs. If the parallax exceed the refraction, as isthe case in the Moon , the quantity found must be added to therising and Subtracted from the setting ; but if the refraction be,greater than the parallax, as is the case with the planets, this quan.tity must be applied vice versa. In our present example the alti-tude of the pole, i. e. the latitude of the place, is 51° 28'40',the declination of the Moon 8° 31' 7", the horizontal refraction33' 30”, Hie horizontal parallax of the Moon 59 ' 21", and die dif-ference of the parallax and refraction 23' 51", which reduced totime is 103”2. From these data we obtain the quantity required as
follows:
I.ogarilhm-cosine of (l-\-d) 59° 59' 47 '.9,6990174
Logarithm-cosine of (l — d) 42 u 57' 33".9,7644159
Sum of the above logarithms.259,5634333
The half-sum is.9,78! 7166
Lich subtract from the log. of 103”4......2,0149403
And the remainder is the log. of 9m. 51s. required.2,2332237
If this quantity be subtracted irnm 13h. 19m. 23.,. because theparallax is greater than the refraction, and the selling consequentlyaccelerated, the ivm-dmler, 13h..lom. .12s. wiil he {lie true time of| file Moon's apparent selling, required ; ?. t. the Moon 's cent in(I will be appur-iitly in the hor'/on of Greenwich, at lorn. 32s: pa-.t
one;