h will be smaller than the value assumed in the figure, and more and more so the greater d becomes, h varying as the square of d; consequently the path of the ray will fall below the tangent, TA, and continually depart from it. III. For the refraction. We will suppose an observer provided with an astronomical clock, carefully regulated to sidereal time (24h. = 360°), and an altitude and azimuth instrument, accurately leveled so that the upper and lower meridian transits of a circumpolar star shall take place at intervals of twelve hours as exactly as possible, and small, unavoidable errors allowed for. The data, both observed and calculated,* requisite for determining the law of atmospheric refraction, may be tabulated as follows. [Last of Aug., 1847.] Ap. zen. dist. of polaris. at upper transit = 45° 33′ 50′′, do. = 48° 34' 10' = 47° 4. do. do. at lower transit = .. ap. zen. dist. of pole [calculated] Note. The fixed stars are so distant that they suffer no displacement on account of parallax, appearing in the same direction whether seen from the earth's centre, or from any point on its surface. Now, the azimuths and hour angles of a Cygni being observed at the successive zenith distances, 15°, 30°, 45°, 60°, 75°, 80°, 85°, 90°, in the triangle ABC, we have (fig. 135.) A = observed hour angle, a2 = ob. zen. dis., a = calcul. zen. dis. b = cal. zen. dis. of pole = 47° 4′, C2 = ob. azimuth, C = cal. az. c = polar dis. of star = 45° 15' 41'5", [Naut. Al.] Barometer 30 inches, internal Thermometer 50°, external 47°. The values of C2 are found the same with those of C, which, therefore, the student may calculate and verify by employing them to find the values of A. * The student should find the calculated numbers for himself. It appears from the above that the effect of the atmospheric refraction is to elevate a heavenly body, without changing its azimuth, by an angle which varies, to within 10° or 15° of the horizon, nearly as the tangent of the zenith distance. With this limitation, therefore, we may assume (a− a): tana2 = 58'' (640) But the refraction is found to vary with the density of the atmosphere, at 45° zenith distance increasing about 2" for every additional inch of the barometer [B] above 30 inches [B — 30], and decreasing '12" for each additional degree of Fahrenheit's thermometer over 50° [T-50]; hence (640) becomes (641) a = a2+[58+2( B — 30) — ‘12( T —– 50)]′′ tana. Scholium. On account of the fluctuating nature of the atmosphere, observations within 15° or 20° of the horizon should be avoided. For the variation of the magnetic needle, however, the azimuth of the sun may be taken when the lower limb appears above the horizon by its diameter. IV. For the parallax. Let S be the position of a heavenly body seen from a point, A, of the earth's surface, O the centre, Z the zenith, and S, the position of the same body in the horizon;* let ASO = p, ZAS = a, ZOS = a; AS¿O = Pr‚ ZAS 90°; then P, the difference of a and a,, is called = Fig. 140. the parallax in altitude, p, the horizontal parallax, and we have, putting OS = d, OA = r, sinp sinar: d, and sinp: sin90° r: d; sinp sinp, sina, or p = pr sina, = when the parallaxes are small. = aca-Ph sina. (642) (643) Scholium. The horizontal parallaxes are given in the Nautical Almanac. V. For the semidiameter. Let S be the point of contact of the line SA and a heavenly body, as the sun, whose centre denote by S.; there results, d sina.] sina sinSAS, sinAS,S SS, AS = *These positions are supposed to be corrected for refraction. but, if we denote by s what s becomes when the point S falls a little below S, so as to make SAO = SOA (in which case it is obvious that s may be taken for the horizontal semidiameter), we have (645) In (643), replacing a = ZOS, and a by ZOS, and a — s, we find ZOS, = a -s—Pr sin(a — s), = αc s, nearly. Scholium I. The refraction for the ray S.A is less than that for SA; therefore, denoting this difference by dif. ref. (S,,S), when great accuracy is required, as in finding the longitude, (645) be comes True zen. dist. ZOS,—a—s—PÅ sin(a — s)+dif. ref. (Sc,S). (646) Scholium II. Except for the moon, we may assume (647) Example. On the 20th of August, 1847, at 6h. Greenwich mean time, the eye of the observer being 24 feet above the level of the sea, the altitude of the moon's lower limb above the apparent horizon was found to be 40° 20'. At the same time the barometer indicated 29 inches, and the thermometer 40°, geocentric zenith distance of the moon's centre. d (638) Ap. alt. from tr. horizon, Required the true 40° 20' 00'0" 4' 49'3" 40° 15' 10'7" Tr. geocentric zen. dis. of moon's Ir. limb a,, 49° 2′ 53'2′′ = 15' 35'0" True geocentric zen. dis. of moon's centre 48° 47′ 16′2′′ The student may make further corrections (646). ADDENDUM I. The ingenious theorem contained in the following communication is published by permission. PROFESSOR WHITLOCK, G. C. W. RICHMOND, June 22d, 1848. Dear Sir:-I have modified the demonstration of my theorem as you requested, and hasten to send you the result. I shall, at the earliest opportunity, attend to the problem which you suggested, to find the area of a polygon in terms of its sides and angles, excepting two sides and one angle. If I arrive at anything of importance, I will send it to you. Yours respectfully, DASCUM Green. To find a general expression for the area of a polygon. the double area of a triangle (a, b, c), in terms of its three angles and one side is we shall have, if we designate the area of the polygon a, b, ..., l, by P, POLYGON. (j+j) sin(k,j) sin(j,l) ___ (i +¿')2 sin(j,î) sin(i,l) 2P = sin(l,k) sin(j,l) sin(b,l) (b+b')2 sin(c,b) sin(b,l) a2 sin(b,a) sin(a,1) a2 sin(a,b) sin(a,l), (b+b')2 sin(b,c) sin(b,l) sin(c,l) + sin(b,l) sin(c,l) in which it remains to determine b', c', ..., j'. ... 313 + + + [o+ a sin(a,l)b sin(b,l) sin(c,l) sin(j,l) 2 a sin(a,1)]*sin(b,c) sin(b,l) sin(b,l) sin(c,l) in(b,l)] *sin(c,d) sin(c,l) sin(d,l) +... a sin(a,1)+b sin(b,?)+...+i sin(¿,l)]*sin( j,k) sin( j,l) .. 2P = [a sin(a,l)]2 sin(a,l) sin(b,l) sin(b,c) +[a sin(a,1)+b sin(b,7)]2 • sin(b,l) sin(c,l) |