282
ASTRONOMY.
sage over tin* meridian, gives IIh. 51m. 16”,2, the exact appa-rent time of the star’s southing. This last result is as accurate aspossible, since we have found by the preceding proportion thedifference of the right ascension of the Sun and star for 11 h. 51m.
11",8, and the apparent time of the star’s southing is only 4”,4afterwards; which small quantity of time can only accelerate themeridian-tran.'it about the -^ths of a second of lime.
Example ‘2. Required the apparent time of the Moon ’s passageover the meridian of Greenwich on December 30, 1813 ; the Sun’sright ascension being 18h. 36m. 58”,8, and the Moon ’s 30m. 48s.the daily diflerence of the former in right ascension being 4m. 25”, 6,and of the latter 12° 41' in parts of the ecliptic, or 50m. 44s. in time.
First Approximation. Subtract 18h. 36m. 58",8, the Sun’sright ascension at noon, from Oh. 30m. 48s. the Moon ’s do. andthe remainder, 5h. 53m. 49”,2, is the apparent time of the Moon ’spassing the meridian nearly.
Second Approximation. Subtract 4m. 25”,6, the daily dif-ference of the Sun in right ascension, from that of the Moon , 50m.44s. and the remainder, 46m. 18”,4, is the relative dad v differenceof Sun and Moon in right ascension. Then say, 24 hours is to 46m.18”,4, as 5h. 53m. 49”,2 is to 11m. 22",7, which added to the timeof the Moon ’s passing the meridian as found by the first approxi-mation, because her progressive motion in right ascension isgreater than the Sun’s, gives 6h. 5m. 11”,9, for the Moon ’s southingvery nearly.
Third Approximation. Take the proportional part of thequantity found by the second approximation, thus: 24hours is to46m. 13",4, as 11m. 22",7 is to 21",9, which added to 6h. 5m.
11",9, for the reason given in the second approximation, gives 6h.5m. 33”,S for the apparent time of the Moon ’s passing the meridian,which is very exact, and can never vary from the truth morethan a second or two at the most.
Fourth Approximation. Take the proportional part of thequantity found by the third approximation for 24 hours, thus: 24hours is to 40m. 18”,4, as 21",9, is to 0",7, which also added to (ill.5m. 33", 8, for the reason above assigned, gives 6h. 5m. 34",5 forthe apparent time of the meridian-passage of the Moon required.The result obtained bv this last approximation can never deviatefrom the truth more than about the T ^ c th of a second of time,which is a much greater degree of accuracy, than the tableswhence the right ascensions of the celestial bodies are calculateddare pretend to. lienee it is useless, and even absurd, to makeuse of a fifth approximation.
(91.) 'l lie rules which we have just given to find the meridian-passages, suppose the right ascensions of the Sun and star calcu-lated for the apparent noon of some particular meridian, as c. g.Greenwich; but if with the same data wc wish to find the pas-sage of a star over any other meridian, it is necessary to calculatethe same right ascensions for the meridian of the given place.The time of a meridian-passage would be the same in all places,if the Sun and star in the space of 24 hours remained at the samedistance from each other, and preserved also the same relative mo-tion in right ascension. If the Moon passed the meridian of Green-
77: r Suantiti/ Schick He fraction accelerates the Rising, or retards
33"
wich at 10 hours to day, and at 11 hours to-morrow, it would con-sequently pass the meridian of its antipodes at 10,j hours of theirtime, for when she reaches the opposite meridian, it is evidentthat she has just completed half of an intire revolution to the me-ridian of Greenwich again. To find the passage of a celestial bodyover the meridian of a given place, say, 24 hours is to the dif-ference of the meridians, as the retardation or acceleration o astar over the meridian of Greenwich from one day to another, isto the difference between the passage at Greenwich and the givenplace. This result must be added to tbe passage at Greenwich, ifthe place be to the west, otherwise subtracted, provided thepassages are later every day, as is the case with the Moon . But inthe case of the stars, and most of the planets, the latter clause ofthis rule is to be applied vice versa.
Example. Required the time of the Moon ’s passing over themeridian of Quito on December 23, 1813, her meridian-passageat Greenwich being at Oh. 34m. and the daily difference 4Sminutes.—First say, As 24 hours is to 5h. 1 lm. 40s. the longitude ofthe place in time to the west, so is 48 minutes to 10m. 23”,3.This quantity being added to Oh. 34m. because the place is lothe west, &c. the sum, Oh. Aim. 23”,3, will be the apparent timeof the Moon’s passing the meridian of Quito requited. This in-stant is the same with. oh. 56m. 3”,3 at Greenwich.
(92.) When a planet or star is in the horizon, its distance from themeridian is what is called its semidiurnal arc, and the quantity ofthis angle must first be known before we can ascertain the precisetime of a star or planet’s rising and setting. The most expeditiousmode of finding the semidiurnal arc is by means of the ascensi-onal difference. In an oblique sphere, the stars all obliquely riseand set. And as the right ascension of a star is the arc of theequator contained between the first point of Aries and that pointwhich comes to the meridian with the star, or that point which ina right sphere rises with the star : so the oblique ascension is thearc of the equator between the first point of Aries and that pointof the equator which rises together with the star in an obliquesphere; and reckoned from west to east, which according to theobliquity of the ecliptic is various. The difference between theright and oblique ascension is what is termed the ascensional dif-ference. To find the quantity »f this in any latitude, and forany declination, the following proportion may be used: Radius is to tiie tangent of the declination, as the tangent of the latitudeof the place is to the sine of the ascensional difference. Thisbeing turned into time, and six hours added to it, if the latitudeof the place be of the same name with the declination, otherwisesubtracted from six hours, the sum or remainder will be the semi-diurnal arc, without having regard to refraction or parallax. Butin order to obtain the apparent semidiurnal arc we must apply thesmall quantity which results from refraction or parallax, or fromboth conjointly, as in the case of the Moon . The following tablecalculated by M. de la Lande, is designed to shew the quantityof the equation arising from refraction at any place, not exceed-ing 60 degrees of latitude, and for :tny declination, not greaterthan 30 degrees:
the Setting of the Stars, the horizontal Refraction being supposed30”.
s North Declination, in
>lat
■es having
north latitude.
l>eg,
0°
5°
10°
15°
20°
23°
28'
25°
30°
Quantity
found.
Subtractfor south deck
0
2'
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22"
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26"
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