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To find (he geocentric longitude of Mercury , first subtractOs. 11°23' 1",8, the true place of 0 corrected by the aberration oflight, from 4s. 3" 33'58", 1; and the remainder, 3s. 27° 10' 56",3,will be the commutation. Next, the logarithm of the distance of'£ from©, cArg. g’smean Anomaly,) is 9,5193S2; the reduc-tion, (Arg. $ from is —311)4 ; their sum, 9,510188, is the truelogarithm of £’ s distance from the bun : if this be subtracted, ac-cording to the rule, from 0,00.0105, the difference will be10,483917,to which logarithm corresponds the tangent of 71° 49' 57’'.Subtract 45° from this last quantity, and the remainder will be thetangent of 26" 49' 57". The log. of 90" 49' 57", 9,70409, being addedto the log. of half the commutation, 10,21423, the sum, 9,91825,(subtracting 10 from the index,) is the tangent-logarithm of 39° 38'2t",6. Take this last-found quantity from half of the commutation,Is. 28° 35' 28", 1, g being an interior planet, and the remainder,18° 57' 0",5, is the true elongation or distance of Mercury from theSun, as seen from the Karth. This elongation being added to theSun’s true place, (because li.c commutation is less than six signs,)gives Is. 0° 19' 48",3 for the true geocentric longitude of Mercury required. To find the geocentric latitude, first reduce the signsof the commutation and elongation into degrees ; and if the num-ber of degrees in either exceed 90°, subtract them from 180°, andthen work by logarithms as below :
The Sine of the complem. commutation, 62° 49'3",7 ... 9,94917
Is to Sine of elongation, 18°57'6'',5 . 9,51158
As tangent of heliocentric latitude, 6° 56' 22" N. 9,08533
Is to the tangent of the geocentric, latitude, 2°32' 40" N. 8,64774(87.) The right ascension of a celestial body is its distance from thefirst point of Aries counted upon the equator, and differs from itslongitude, the latter being its distance from the equinoctial pointcounted upon the ecliptic. The Sun’s right ascension may bereadily found by means of his meridian-altitude; for when itamounts to 38^- degrees, which is equal to the co-latitude of Lon-don , we are then assured that be is exactly in the equinoctialpoint where his right ascension is nothing or 12h. We may obtain theright ascension of a star, by observing the interval of time whichsucceeds its meridian-passage, and that of the Sun, for this quantitybeing added to the right ascension of the Sun will give that of thestar required. The declination of the Sun, a star, or planet, is itsnearest distance to the equator; and is denominated north orsouth, according to its position with respect to this great circle.If the meridian-altitude of the equator, which is equal to the co-latitude of the place, be known, the^leclination of a celestial bodymay he exactly determined by observing its altitude when duesouth, and talcing the difference of the altitudes, the remainderbeing the declination ; and is either north or south, as the meri-dian-altitude of the object is greater or less than that of the equator,if the place of observation be to the noith of the equinoctial line;and vice versa, if the place he to the south.
Example 1. If the Ram’s following horn be observed to passover the meridian of St. Paul’s, London , at the altitude of 60° 59'47", what is its declination?—Subtract 38° 29', the co-latitude of theplace, from 60° 59' 47", and the remainder, 22° 30' 47 ', is the decli-nation of the star, which is north, because the altitude of the staris greater than that of the equator ; the latitude of the place of ob-servation being also north.
Example 2. If the Rain’s following horn be observed to passover the meridian of Rio Janeiro at the altitude of 54° 35' 3", re-quired its declination.—Subtract 54° 35' 3" from 77° 5' 50" theco-latitude of the place, and the remainder, 22° 30' 47", is the star’sdeclination; and is qoith, because the altitude of the star is lessthan the co-latitude, the place being to the south of the equator.(88.) When the Sun’s longitude and the obliquity of the ecliplic are
5 ;ivcn, his right ascension may be found bv the followingproportion:ladius is to the co-sine of the obliquity of the ecliptic, as the tangentof the longitude of the Sun is to the tangent of his right ascension.
Example. On October 1, 1813, apparent noon at Greenwich,the Sun’s place is 6s. 7° 53' 28", and the obliquity of the ecliptic
23° 27'45",8; required his right ascension.
Is to co-sine of 23° 27' 45",8.9,9625206
As tangent of Sim’s longitude (rejecting 180°).9,1417735
Is to tabgent of his right ascension (adding 180°)
187° 14' 45"-J .
VOL. i.— no. 18.
The Sun’s right ascension in parts of the ecliptic, 187° 14' 4Vj-may be reduced into time at the rate of 15 degrees per hour;thus the 0’s right ascension in time is 12h. 28m. 59s.
(89.) When the Sun’s place and the obliquity of the ecliptic aregiven, his declination may be found by the following proportion:Radius is to the sine of the obliquity of the ecliptic, as the sine ofthe Sun’s longitude is to the sine of the declination. N. B. Thedeclination is north in the first six signs, and south in the other six.
Example. On April 1, 1813, apparent noon at Greenwich, theSun’s longitude is 0s. 11° 22' 37", and the obliquity of the ecliptic
23° 27' 44",7 ; required his declination.
Is to sine of the obliquity of ecliptic, 23" 27' 44",7.9,6000433
As sine of 0’s longitude 11° 22'37".9,2950452
Is to sine of 0’s declination north, 4° 30' 17", (© in «p) 8,8950890
Rut in case the Sun’s right ascension and the obliquity cf theecliptic be given, the following proportion must be made use ofto find the Sun’s declination: Radius is to the tangent of theobliquity of the ecliptic, as the sine of the right ascension is to thetangent of the declination.
Example. On March 21, 1813, apparent noon at Greenwich,the Sun’s right ascension in parts of the ecliptic, is 27' 55"A, andthe obliquity of the ecliptic 23° 27' 44",7 ; required the Sun’s de-clination.
Is to tangent of the obliquity of the ecliptic, 23° 27' 44",7 9,6375224As sine of ©’s right ascension, 27' 55"A.7,9090206
Is to tangent of declination north, 12' 7", (© in qp) .7,5471430
(90.) To find the time of the southing of a star, it is sufficient, bymeans of the Nautical Almanac, to know how much its right ascen-sion exceeds that of the Sun, at the noon preceding, and the excessbeing turned into time, at the rate of 15 degrees per hour, and thetlie Sun’s right ascension subtracted from it, the remainderwill be the approximate time of the star’s southing, or coming to*the meridian. This first approximation for the fixed stars andplanets will lie exact enough for ail common purposes; but whenwe want to find the southing of the Moon , we must make use ofother approximations, which will be explained in the followingexamples:
Example 1. Required the apparent time of the first meridian-passage at Greenwich of that very remarkable star in 1 ho neck ofthe Whale which periodically varies in magnitude, marked byBayer e, that takes place alter the noon ot October 30, 1813’;tin* right ascension of the star for that time being 2h. 1 Dm. 36", 4, t hatof the bun at noon 1 111. 17m. 24",6, and the daily variation of thelatter in right ascension 3ni. 54", 1.
First Approximation. The right ascension of the Sun atapparent noon, 14h. 17m. 24",6, subtracted from 2h. 10m. 36",4,the right ascension of oCeti; the remainder, 11 h. 53m. 11",8, isthe southing of the star nearly. This first appioximation, in ex-treme cases, may differ from the truth between four and liveminutes.
Secox’d Approximation. Subtract one minute from thetime found by the first approximation, if the meridian-passage l.eafter three hours ; two minutes, if after nine hours; three minutes,it after fifteen hours; and four minutes, if the star’s southing beafter twenty-one hours. In our present example two minutesmust be subtracted from 1 111. 53m 1 F',8, and llie remainder, 1 111.
5 I m. 11”,8, is the apparent time of the star’s passage over the me-ridian, and can never deviate from the truth so much as oneminute, because the right ascension of the Sun varies about fourminutes every day at an average, the daily diifert nee being 3m.35s. when least, and 4m. 27s. when greatest. This approximationis of no use but for the fixed stars. 11 may also serve for the planets,when the daily difference of their right ascension is not veryconsiderable.
T hird Approximation. If we are not satisfied with having aminute’s uncertainty in the m vidian passage of a star, we mustmake use of the following proportion : 24 hours are to 3m. 54",1,tlie diurnal variation of the bun's right ascension, as ill:. 51 in.
11",8, the time of the star’s southing nearly, to 1 in. 55",6, whichbeing subtracted from the first approximate time of the stal ls ; a---4 C sage
,9,1042941