= cos L cos D Let now, H = ZPA, and x = APB, when AP is greater than 90°, that is, when the declination is of a different name from the latitude, and H' ZPA, and x' APB, for an equal declination, when of the same name with the latitude. Then, it is evident from the expression for cos ZPA, that H will be less than 90°, and that H' will be the supplement of H. Hence, we have, − x') + cos (H + x) H = cos H = 2 cos H. cos (H+ x), —cos (H — x) = 2 cos H (1. cos x) 4 cos H Consequently, sina sin (H — or, (App. 18), cos (H — x') —cos or, x') = sin x sin (H + x), cos (HBut, (App. 22), cos (Hx) + cos (H + x) = 2 cos H cos x. Hence, cos (H— x') sin2r. Now, since H is less than 90°, the second member is affirmative. Consequently, cos (Hx) is greater than cos (H - x), and therefore, x' is greater than ; that is, the twilight is longer when the latitude and sun's declination are of the same name, than when they are respectively of the same values, but of different names. And it is easy to perceive from equations (A) and (B), that, the longest twilight at a place, occurs when the declination is greatest and of the same name with the latitude. = For the shortest twilight. Let the triangle BPC, having the side BP AP, have also, PC PZ, and BC AZ: = 90°; then its three sides being respectively equal to those of the triangle APZ, we have the angle CPB = ZPA. Taking CPA from each, we have APB = ZPC. Hence, the twilight will be shortest when the angle ZPC is least. Let, ZD be a vertical circle through C, and PE an arc of a great circle bisecting the angle ZPC. Since, PZ and PC are constant, the angle ZPC will be least when ZC is least; and since BZ and BC are constant, ZC will be least when the angle ZBC is least, that is, when it becomes Zero, or when BZ and DZ coincide. But, when BZ and DZ coincide we must have DC BC 90°. Hence, as DZ when the twilight is shortest, ZC = 2a. 90° 2a, it follows, that Now, as the triangle ZPC is isosceles and PE bisects the vertical angle, it must also, bisect the base ZC and be perpendicular to it. Hence, in the right-angled triangle ZEP, we have, (App. 47), Twice the angle ZPE converted into time, gives the duration of shortest twilight. From the right-angled triangles ZEP and DEP, we have, As cos PD is negative, PD, the sun's distance from the elevated pole, must be more than 90°, and consequently, the declination must be of a name contrary to that of the latitude. When PD has been found from the above expression, the sun's declination is known; and by a Nautical Almanac, the times in the year when the sun has this declination are easily found. 54. To find the Annual Variations of a star in Right Ascension and Declination. Referring to Fig. 20, described in previous Articles (126 and 128), as it is evident that mg does not sensibly differ from EG, and as the difference between the complements of two arcs is the same as the difference between the arcs themselves, we have. Draw P'r perpendicular to the declination circle PSG. Then, as PP' and P'r are very small, we may, without material error, regard PpP as a spherical triangle, right angled at P and P', PrP' and E'mE as right angled plane triangles, and sP' = sr. Put, right ascen. of the star = declination = do. =pP = QEC obliq. of the ecliptic. Then, taking small arcs or angles instead of their sines, and observing that PpP' = = CC' EE' = 50′′.2 (126. Cor.), and P'Ps = 90° — GPQ EQGQ EG = = = A, we have, from the triangles PpP' and PrP', = PP': PpP' sin pP 50".2 sin ε, PP' sin A 50".2 sin sin A, Consequently, Ps P's = Ps―rs = Pr 50."2 sin cos A. (C) We have, also, from the right angled spherical triangles P'rs and gG's, the latter of which may be regarded as right angled at g as well as at G', we have, But, since the quantity which is multiplied by tang G's is small, we may, without sensible error, put tang. Gs or tang D, instead of tang G's. We (D) (E) Substituting in formulæ (A and B), the values of E'm, gG' and Ps—P's, (E, D, and C), we have, Ann. var. in R. Ascen. = 50".2 cos + 50".2 sin sin A Ann. var. in Declin. 50".2 sina sin cos A. tang D. In applying these formulæ, the declination is to be regarded as affirmative or negative, according as it is north or south. When the mean right ascension and declination of a star is known for a given time, and the annual variations have been computed by the above formulæ, its mean right ascension and declination at a subsequent time, not distant from the former, more than 20 or 30 years, or in case of the pole star or other star near the pole, 10 or 15 years, are obtained by adding to the given right ascension and declination, the product of the corresponding annual variation by the number of years, and parts of a year in the interval. In like manner, the right ascension and declination may be obtained for a a prior time, only subtracting the products instead of adding them. 55. To find the Aberration of a fixed star in Right Ascension and Declination, for a given time. Lets, Fig. 60, be the star, E the earth, BQ the equator, P its pole, LC the ecliptic, V the vernal equinox, N the pole of the declination circle P&G, and 8NM an arc of a great circle, which, passing through the pole N, must be perpendicular to PSG. Also, let D be the point of the ecliptic towards which the earth is moving at the given time, and &D an arc of the great circle, in which the plane sED cuts the celestial sphere. Then, (132) the direct effect of aberration is to make the star appear to be at a point s' between s and D, such that ss' 20.36 sin Ds. Then, as ss' is very small, we may Let s'd be perpendicular to PG. regard the triangle s'ds as rectilineal. error, G's': Gd, Pe Ps, and s'd es. = sd, it is evident the effect of aberration on the star s, which is in the first quadrant, is to diminish the right ascension by the quantity G'G, and the declination by sd. Consequently, 20".36, A = VG star's right ascension, D = Put, a declination, S = sun's longitude, and GVF obliquity of the ecliptic. Then, using small arcs and angles instead of their sines, we have, also, ess'd ss' sin DsF 20′′.36 sin Ds sin DsF Hence, G'G cos D = a sin Ds sin DsF. =α a sin Ds sin DSF. But, sin DsF: sin F:: sin DF : sin Ds, or, sin Ds sin DsF = sin F sin DF; and in the right angled spherical triangle VGF, But, sin MSD sin M :: sin MD: sin Ds, or, sin Ds sin MSD = sin M sin MD; and, sin MV: sin NV :: sin MNV, or, sin GNs sin M sin NV sin GNs cos A sin D Then, (B), n = sin MD. a cos A sin D sin MV Aber. in declin. — — sd = n sin MD Now, in the triangle MNV, we have, (App. 36), (G) (H) sin cot D+ cos sin A sin NV cos A (I) 90+ 6.* Then, sin MD + 0), sin MV = cos 0, and cot MV Hence, (I, G, and H), * When, MV is less than 90°, as in the figure, it is 90° +360° that is to be regarded as equal to MV. The increase or diminution of an arc by 180°, changes the sign of its sine or cosine, but does not affect its numerical value. It is, therefore, evident that if the value of the arc o be increased or diminished by 180°, the expression for the aber. in right ascen. (F) will still be true; for the signs of both factors m and sin (S+) being thus changed, there will be no change in the sign of the product. Hence, we may always take o affirmative, and not exceeding 180°. Similar observations apply to the arc 0. The quantities, 0, m and n change but little for a number of years, and therefore, when once computed for any star, they serve for a long time in computing the aberration of that star. Table IX, contains the values of these quantities for 30 principal fixed stars. The values of m and n are all made affirmative by increasing the values of and by 180°, when requisite. 56. To find formula for the Lunar Nutation in right ascension and declination. To obtain these formulæ we have recourse to certain results established by Physical Astronomy. It has been proved that the phenomena of nutation may be explained on the supposition that the pole of the equator, instead of moving strictly in a circle about the pole of the ecliptic (126), moves in a small ellipse about the mean place of the pole, that is, around that point in the circle, at which the pole would be if the nutation did not exist, and in a period equal to that of the moon's nodes. The major axis of this ellipse is situated in the solstitial colure, and is to the minor axis in the ratio of the cosine of the obliquity of the ecliptic to the cosine of twice the obliquity. The major axis has been found to be equal to 18".44;* and hence, the minor axis is 13′′.73. Let ELF Fig 61, be the ecliptic, N its pole, NLM the solstitial colure, EMF the mean equator, P the mean place of the pole, AbCd the ellipse in which the pole is assumed to move, and ABCD a circle about the centre P. Then, according to the investigation in Physical Astronomy, if the arc ABO be made equal to the longitude of the moon's node, and Oc be drawn perpendicular to NL, the pointp in which it meets the ellipse, will be the true place of the pole, and the great circle E'G'F', described about the pole p, will be the true equator. Let s be the situation of a star, and let pr be drawn perpendicular to the declination circle PG, and Em perpendicular to * This is the value given by Struve in No. 426 of the Astr. Nach. as recently obtained from a series of observations made by him at Dorpat. |