equal to -20".06: when the right ascension of the star becomes 6 hours, cos. a, and consequently the precession in north polar distance, vanishes. The variations afterwards increase till the right ascension is 12 hours, when they attain their greatest positive value, being then about +20".06: they afterwards diminish till the right ascension is 18 hours; and they again increase till the right ascension is 24 hours.

227. The annual precession in right ascension is the difference between Ym and En: in order to find it let fall re perpendicularly on Eq'; then in the right angled triangle Ere, which may be considered as plane, we have E = 50′′.2, EY the angle at E equal to pP (=) and the right angle at e; therefore (Pl. Trigo., art. 56.)

ET cos Ee (in seconds).

Now, because of the smallness of the angle subtended at the pole P or P' by the interval in right ascension between m and n, that part of the precession in right ascension may be represented by nn'; and this may be found from the two spherical triangles Pst and nsn', which have an equal angle at s, and in which the angles at t' and n may be considered as right angles. PS and Sn are known, the former being the given polar distance, and the other being supposed to be equal to its complement, that is, to the given declination: also in the triangle PP't,

Pt(PP' sin. YP's) 50.2 sin. ◊ sin. a. Therefore, in the spherical triangles Pst and ns n' (art. 60. (e)), r sin. Pt sin. S sin. PS and r sin. nn' sin. s sin. S n.

Consequently

sin. PS: sin. Pt:: sin. sn: sin. nn';

or substituting the small arcs or angles for their sines, sin. PS: Pt: sin. sn nn';

sin. sn

sin. PS

nn'

whence nn'=' sin. d cos. d

Pt; or (d being the declination)

50".2 sin. ◊ sin. a — 50′′.2 sin. ◊ sin. a tan. d.

Therefore the whole annual precession in right ascension (Ee + nn') is equal to

50".2 (cos. + sin. ◊ sin. a tan. d).

And these formula for the annual precession in declination and right ascension are found to express very nearly the differences between the declinations, and between the right ascensions when observed at considerable intervals of time.

228. About the year 1750, Dr. Bradley, while observing certain stars which passed the meridian near the zenith of his

station, discovered that, after having made the necessary corrections in the right ascensions and declinations on account of the effects of general precession, such stars apparently described in the heavens the circumferences of very small circles, or the peripheries of small ellipses each returning to the place in which it was first observed at the end of one year. Thus, at the time of the vernal equinox, the star y Draconis was observed on the meridian near the zenith about the time of sun-rise, when, consequently, the plane of the meridian must have passed through a tangent to the earth's orbit; and the star then appeared to be to the south of that which, according to computation, should have been its true place in declination, while its apparent right ascension was unchanged. At the summer solstice the star was on the meridian about midnight, at which time the plane of the meridian is nearly at right angles to a line touching the earth's orbit; and then the star was nearly in its true place with respect to declination, but its transit took place at a time later than the computed time, as if its right ascension had increased, or the star had moved eastward. Again, at the time of the autumnal equinox, y Draconis was on the meridian about sun-set, when the plane of the meridian must have passed through a tangent to the orbit; and the star appeared to the north of its computed place, while the right ascension became the same as at first. Lastly, at the winter solstice, the star passed the meridian above the pole at noon, at which time the declination was the same as at first, but the time of the transit took place earlier than the computed time, as if the right ascension had diminished, or the star had moved westward. The like phenomena were exhibited by every other star on which observations were made. These variations of position were found to correspond to such as would result from the movement of the light transmitted by the star, and that of the earth in its orbit; and this cause of the phenomena being admitted, it is evident that similar variations must result from the movement of light combined with that of the earth on its axis, but these are found to be very small.

229. The apparent deviation of a star on account of the combined movements of light and of the earth, is called the aberration of the star; and the manner in which the phenomena take place may be thus explained.

Let s be the place of a star, and ME the position of the axis of a telescope when the eye is at E, and when a particle of light in a ray proceeding from the star s is at M; then, while the particle moves from M to E', the eye of the spectator is carried from E to E' by the motion of the earth; so that

N

E's' becomes the position of the axis of the telescope when the star is visible, and s' becomes the apparent place of the star.

Now the ratio of ME' to E E' is equal to that of the velocity of light to the velocity of the earth in its orbit; and (Pl. Trigon., art. 57.)

ME: EE: sin. MEE': sin EME'.

But the angle MEE' may be considered as equal to SE'T, and EME' is equal to SES'; therefore, substituting the small angle SE's' for its sine, the velocity of light is to that of the earth as sin. SE'T is to SE'S'. Again, the ratio of the velocity of light to that of the earth being constant, it follows that the aberration in the plane SE'T varies with the sine of the angle SET. Here, from the smallness of the earth compared with the celestial sphere, E and E' may be considered as either at the surface or at the centre of the earth, and ET as a tangent to the earth's orbit: therefore the angle SE's', or the aberration, is always in a plane passing through the star, the centre of the earth and a tangent to the orbit. Hence, on the hypothesis that the earth revolves annually about the sun, it is evident that the effect of the combined motions of light and the earth may be represented by supposing each star to describe annually in the heavens a curve similar and parallel to the earth's orbit; the star being always in advance of the point corresponding to the place of the earth's centre by the value of the angle SE's'. But, in the region of the fixed stars, the magnitude of the earth's orbit scarcely subtends any sensible angle; and therefore the star appears to describe, about the place which it would seem to occupy if light passed instantaneously to the earth, a circle or ellipse at a distance equal to the measure of the angle SE'S': the visible form of this curve will evidently be different for different stars according to the position of the plane in which such curve appears to be projected on the celestial sphere.

230. Since the angle of aberration SE's' varies with sin. SE'T, it will evidently be a maximum when SE'T is a right angle, in which case the star would be in the direction of a line drawn from the earth perpendicularly to the plane of its orbit, that is, on account of the great distance of the stars, in the pole of the ecliptic. Now the angle SE'T being known for some particular star, and the angle SE's' being found by observation, the amount of aberration for a star in the pole of the ecliptic may consequently be obtained from the proportion sin. SET radius :: SE's': A.

This last term is generally called the constant of aberration, and, from the best observations, it is found to be equal to 20".36. Therefore the formula 20".36 sin. SE'T gives, for any star, the value of the aberration in a plane passing through the star, the earth and a tangent to the earth's orbit. The position of the last-mentioned plane for any star at a given time being found, the effects of aberration in longitude and latitude, in right ascension and declination, may be computed by the rules of spherical trigonometry.

B

W

A

231. Let the primitive circle ETA, of which a part only is drawn, represent the ecliptic in the heavens, let p be its E pole, and s the place of a star: let r be the equinoctial point, and for the given time, let ro be the longitude of the sun; then E, at the extremity of a

T

a

Αν

M

N

diameter passing through o, will be the place of the earth when projected in the heavens by a line drawn from the

sun.

Since the earth's orbit may be considered as very small, the place both of the earth and of the sun may be conceived to be at the projection p of the pole of the ecliptic; and pT, at right angles to OE, may be considered as a tangent to the earth's orbit. Imagine a great circle to be drawn through s and T, and also a circle of longitude pa through s; then the plane psT is that which passes through the star, the earth and a tangent to the earth's orbit, and is consequently the plane of aberration, or that which above is designated SET. The arc OET (reckoning from o in the order TOET, that is of the signs) is equal to three quadrants, YO (= L) is the longitude of the sun, and YA (=) is the longitude of the star; therefore AOET is equal to L+270°-7. Now, in the plane spT, the aberration is equal to 20".36 sin. ST: let this be represented by ss', and imagine pa to pass through s'; also let s'q be drawn through s' perpendicular to pa: then s'q, when reduced to the ecliptic, on dividing it by cos. s'a (art. 70.) becomes A a, the aberration in longitude, and sq is the aberration in latitude.

232. To find the aberration in longitude.

N 2

In the triangle SAT, right-angled at A, we have (art. 60. (e)) sin. AT (= sin. AOET) = sin. AST sin. ST:

therefore

sin. ST

sin. (L + 270° — 1)

= sin. AST, or = cos. ss'q.

sin. (L+ 270° -1) sin. ST

But s'q ss' cos. ss'q = 20′′.36 sin. ST

= 20′′.36 sin. (L + 270° − 1);

Or (Pl. Trigon., art. 32.)

s'q=20′′.36 { sin. (L + 270°) cos. 1— cos. (L + 270°) sin. 1}; and supposing, as in the above figure, that l is less than 90°, also L+ 270° less than 360°, we have

-

sin (L+ 270°) = cos. L, and cos. (L+ 270°) = sin. L; therefore s'q=20".36 (cos. L cos. + sin. L sin. 1),

=

20".36 cos. (L-();

which being reduced to the ecliptic, we have (art. 70.) λ representing the latitude As of the star,

Aa (the aberration in longitude)

= · 20".36

cos. (L-1)

cos. A

233. To find the aberration in latitude. Imagine a great circle s'I (of which s'q may be considered as a part) to pass through s' perpendicularly to pa and to cut the ecliptic in 1; the point I will be the pole of PA and the angle at I will be measured by SA, the star's latitude. Then, in the triangle SIT (art. 61.),

sin. ST : sin. I :: sin. IT : sin. IS'T.

But sin. 1 = sin. λ; and since IA = TO, each of them being equal to a quadrant, we have IT = YO — YA = L−1; also s'T may be considered as equal to ST;

therefore sin. IS'T (sin. ss'q)

sin. A sin. (L—1)

sin. ST

and sq (the aberration in latitude) = ss' sin. ss'q

=20".36 sin. ST sin. ss'q =20".36 sin. A sin. (L-1).

234. To find the aberration in right ascension. Let WYM be the projection of the celestial equator, and P its pole: draw the declination circles through PS and PS', cutting the equator in M and N, and the former cutting the ecliptic in A'; also let s't be drawn from s' perpendicular to PS: then MN will represent the required aberration in right

ascension.