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ASTRONOMY.

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The following examples'will sufficiently illustrate the use of thepreceding Table:

Example 1. Required the apparent time of the setting ofVenus at London , on October 10, 1810; her passage over themeridian being at 3h. 7m. her declination at noon 25° 58' S. andher motion in declination + 10' per day.

First Approximation. Find the semidiurnal arcs corres-ponding to the two tabular consecutive declinations between whichthe planet’s declination is at noon: thus, the declination beingin this case 25° 58' south, the semidiurnal arc, in the Table, cor-responding to 25° 50' of south declination, is found to be 3h. 35m.5s. and for 26° O', 3h. 33m. 49s. the difference of which is lm.16s. And 10' is to lm. 16s. the change of the semidiurnal arcfor 10' of south declination, as 8', the quantity by which theplanet’s declination at noon exceeds 25° 50', is to 6ls. Thisbeing subtracted from 3h. 35m. 5s. because the planet is to thesouth of the equinoctial, the remainder 3h. 34m. 4s. is the planet’ssemidiurnal arc at noon. And3h. 34m. 4s. -f- 3h. 7in. = 6h. 41m.4s. the time of Venus ’s setting nearly.

Second Approximation. Find the planet’s declination atthe approximate time of her setting, by saying, 24 hours is to 10',the planet’s daily variation in declination, as 6h. 4lm. 4s. is to 2'47". This quantity added to 25° 58', because the planet is in-creasing in declination, gives 26? 0' 47", for her declination atthe time of setting. The tabular semidiurnal arcs correspondingto 26° 0' and 26° 10'south declinations, (that of the planet beingbetwixt these,) are 3h. 33m. 49s. and 3h. 32m. 31s. the dif-ference of which is 78s. To find the true semidiurnal arc ofVenus at the approximate time of her setting, say, 10' is to 78s. as47", the excess of the planet’s declination above 26° 0' at theapproximate time of her setting, is to 6 seconds, which subtractedfrom 3h. 33m. 49s. for the reason before assigned, leaves 3h. 33m.43s. for the correct semidiurnal arc of Venus at the approximatetime of her setting. And 3h. 33m. 43s. -j- 3h. 7m. = 6h. 40m.43s. the true time of her setting required.

Example 2. Required the time of Jupiter’s rising at London oh November 7, 1810, his passage over the meridian being at 12h.53m. his declination at noon 18° 37', north, and his daily variationin declination — 1' 40".

First Approximation. The tabular semidiurnal arcs for 18°30' and 18' 40° north declination, between which the planet’s isat noon, are 7h. 43m. 41s. and 7h. 44m. 43s. the difference ofwhich is 62s, And 10': 62s.: : 7', the planet’s declination atnoon above .18° 30': 43s. This quantity added to 7h. 43m. 4ls.because the declination is north, the sum, 7h. 44m. 24s. will beJupiter ’s semidiurnal arc at noon; and 12h. 53m. — 7h. 44m.24s. = 5h. 8m. 36s. the time of his rising very nearly.

Second Approximation. The planet’s declination at theapproximate time of his rising, is found by the following propor-tion: 24 hours: 1' 40", his daily variation in declination,:: 5h.8m. 36s, : 22". This quantity subtracted from 18° 37', becausethe planet is decreasing in declination, leaves 18° 36' 38" for Ju-piter ’s north declination at the approximate time of his rising.And 10' : 62s. : : 6'38", the excess ot the planet’s declination above18° 30' at the approximate time of his rising, : 4ls. which addedto 7h. 43m. 4ls. gives 7h. 44m. 22s. for his semidiurnal arc at thesame time. And 12h. 53m. — 7h. 44m. 22s. = 5h. 8m, 38s. thetrue time of his rising required. N. B. The first approximation issufficiently exact for the planets Jupiter, Saturn , and Ilerschel,on account of their very small diurnal variation in declination ; andeven in the other planets the difference of the first and second ap-proximations will in mod cases be verv inconsiderable.

Example 3. Required the apparent time of the Sun ’s rising atLondon on August 26, 1810; his declination north the preced-ing noon being 10° 55' 44'', and his motion in declination in thesucceeding 24 hours — 20' 45".

First Approximation. The semidiurnal arc in the table cor-responding to 10° 55' 44", north declination, after making the ne-cessary proportions as instructed in the two last examples, is 6h.59m. 58s. which subtracted from noon, the time of the sun’s pass-ing the meridian, leaves 5h. 0m. 2s. on the morning of the givenday for the first approximate time of the Sun ’s rising.

Second Approximation. Find the Sun ’s declination at theapproximate time ol his rising thus: 24 hours is to 20'45”, theSun ’s diurnal decrease of declination, as 17 hours, the time of theSun ’s rising alter the noon of August 25, nearly, to 13' 19". This

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subtracted from 10° 55' 44”, because the Sun is approaching theequinoctial, gives 10° 42' 25” for his declination at the requiredtime; and the tabular semidiurnal arc corresponding to this, 6h.58m. 45s. after making the necessary proportion, subtracted fromnoon, leaves 5h. lm. 15s. for the correct time of the Sun ’s risingrequired.

Example 4. Required the apparent time of the Sun ’s setting atLondon Tin January 20, 1812; iiis declination at noon being 20° 17 '59” south, his daily variation in declination — 12' 54", and his in-crease of velocity toward the ecpiator in 24 hours, 23".

First Approximation. The tabular semidiurnal arc corres-ponding to 20° 17' 59", after making the necessary proportion, is4h. 13m. 30s. the first approximate time of the Sun ’s setting afternoon.

Second Approximation. Find the Sim’s declination at 4h.13m. 30s. by the proportion; 24 hours: 12'54'::4h. 13£m. : 2'16", the quantity of the decrease of the Sun ’s declination, at thefirst approximate time of his setting, supposing it to be uniformlyat the rate of 12' 54" through the 24 hours. But by the question,his increase of motion towards the equator in the 24 hours, is 23”rT o find the quantity of this correction, say, 24h. : 23":: 4h. 134m.:4". The complement of this to 23", i. c. 19", subtracted from 2'16", gives 1' 57", the true quantity to be taken from 20° 17' 59",the Sun ’s declination at noon; and 20° 17' 59" — 1' 57” = 20°16' 2", his declination at 4h. 134m. The corresponding semidi-urnal arc to this last quantity, 4h. 13m. 42s. is the correct timeof tlie Sun ’s setting required".

Example 5. Required the true time of the Moon ’s rising atLondon on September 25, 1812, the time of her passing the meri-dian being at l6h. 46m. 29s. her declination at noon 16° 5' north,at 12 hours 17° 16', and her decrease of motion from the equatorin 24 hours 16'.

First Approximation. The moon’s passage over the meri-dian having been obtained to the accuracy of a second of time bythe method before explained, (90,) we may find the time of herrising nearly by subtracting 7h. 29m. Is. the semidiurnal arc cor-responding to her declination atnoon from 1 Gh. 46m. 29s. the ap-parenttime of the Moon ’s southing, and the remainder, 9b. 17m.28s. is the Moon ’s rising nearly.

Second Approximation. Find the Moon ’s declination atthe time of her rising nearly, by saying, 24 hours : 1° 11', her in-crease of declination in 12 hours :: 9h. 17m. 2Ss. : 54'56'. Addthis quantity to 16° 5', and the sum, 16° 59' 56", is the Moon ’sdeclination at 9h. 17m. 28s. The corresponding semidiurnal arc,7h. 34m. 28s. subtracted from l6h. 46m. 29s. gives 9h. 12in. Is.for the Moon ’s rising very nearly.

Third Approximation. I* ind the Moon ’s declination at 91).12m. Is. thus : 12 hours: 1° 11':: 9h. 12m. Is. : 54'26”. Thisquantity is obtained on the supposition of the Moon ’s increase ofdeclination through the 12 hours after noon being uniform ; butaccording to the question the quantity of the Moon ’s motion fromthe equinoctial is decreased 16' in the 24 hours, which gives a cor-rection of -j- 9' 52" to be applied to 54' 26", w'hich makes it in all1° 4' 18". This added to 16° 5' gives 17° 9' 18” for the Moon ’sdeclination at 9h. 12m. Is. The semidiurnal arc correspondingto this last-found declination is 7h. 35m. 25s. and 1 Oh. 46m. 29s.— 7h. 35m. 25s. = 9h. 11m. 4s. evening, the apparent time qf theMoon ’s rising required, without having regard to her jxirallax.This is sufficiently exact for all common purposes; but in caseswhere great accuracy is wanted, the following method must bemade use of instead of the preceding.

First Approximation. Find the Moon ’s semidiurnal arcexclusive of refraction for 16° 5' north declination, by means ofthe ascensional difference in the 'Fable as follows: Tile ascensionaldifference corresponding to it, after making the necessary pro-portion in the same manner as for the semidiurnal arc, is 21“ 15'7", in parts of the ecliptic, or 111. 25m. 0s. in time. If to this lieadded 6 hours, because the declination and latitude of the placeare of tlie same name, the sum, 7h. 25m. will be the semidiurnalarc exclusive of refraction. And l6h. 46m. 29s. — 71). 25m. =Oil.21m. 29s. tlie Moon’s rising nearlv.

Second Approximation.' Find the Moon ’s declination at<)h. 21 in. 29s. as follows: 12 hours : 71':: 9h. 21m. 29s. : 55'22". And 16° 5' + 55'22"... 17° O' 22". 'File tabular ascen-sional difference corresponding to tins is 22° 36' 47" iii parts of tlieecliptic, or 111. 30n). 27s. And Hi. 30m. 27s. -j- 6h. = 71). 30m. 27s.

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