Required the true longitude, the eye being 18 feet above the level of the sea. D's semidiameter at noon, March 22d (Naut. Alm.) = 14′ 48′′, and at midnight 14′ 47′′; horizontal parallax at noon 54′ 18′′, and at midnight 54′ 13". Equation of time (subtracted from mean time) on March 22d is 6m 57.6, and on March 23d it is 6m 39s-1. Answer. The apparent time at Greenwich is 4h 3m 463, D's true semidiameter=14′ 54′′, horizontal parallax= 54′ 16′′, apparent altitude of the D's centre 18° 56′ 36′′, and her correction 48′ 35′′. = App. altitude of the star 20° 10′ 56′′, correction 2′ 34′′. (Naut. Alm. distance at III"-30° 37′ 24′′, and at VI132° 6' 27".) = (611) In all the preceding examples the watch is supposed to have been previously regulated; when that is not the case, the error of the watch must be found from observations of the altitudes of the sun or of a star, taken either before or after that of the distance, as directed in Problems XIV. or XV., pages 316 and 319. Or if the sun or star be sufficiently distant from the meridian, the mean of the sun's or star's altitudes, taken at the same time as the distance is taken, together with the latitude of the place and the sun's declination, &c. may be used to correct the watch; with this corrected time proceed as before. EXAMPLE V. At sea, April 7th, 1840, in latitude 47° 39′ N. and longitude 184° 45′ West from Greenwich by account at 3h 56m P. M. apparent time, per watch not previously regulated; suppose the observed altitude of the sun's lower limb to be 25° 57′ 10′′, and that of the moon's lower limb 46° 22' 32", and at the same time the distance of the sun's and moon's nearest limbs to be 75° 33′ 37"; required the longitude, the eye being 18 feet above the level of the sea. ▷'s semidiameter at midnight, April 7th (Naut. Alm.)= 16' 9', and at noon April 8th=16′ 4′′; horizontal parallax April 7th at midnight=59′ 15′′, and at noon April 8th= 58 58; also the O's semidiameter 15′ 59′′. Equation of time (subtracted from mean time) April 7th is 2m 7-0, and April 8th it is 1m 49s-9. Answer. The apparent time at Greenwich is April 7416h 15m; D's true semidiameter 16′ 18′′; horizontal parallax 59′ 9′′; apparent altitude of her centre 46° 34′ 47′′, and correction of her altitude 39′ 45′′. The apparent altitude of the sun's centre is 26° 9′ 6′′; correction l′ 47′′; apparent distance of centres 76° 5′ 54′′; true distance 75° 51′ 24′′. Apparent time at Greenwich 16h 12m 19s. The sun's declination at apparent noon (Naut. Alm.) on the 7th, is 6° 58′ 3" N., and hourly difference 56"-1, which reduced to the apparent time at Greenwich is 7° 13′ 15′′ N.; with this declination, the C's true altitude 26° 7′ 19′′; and the co-latitude 42° 21′, find the correct apparent time at ship-3h 54m 43; and hence the true longitude is 184° 24' West. (Naut. Alm. distance at XV 75° 10′ 44′′, and at XVIII =76° 49′ 21′′".) CHAP. XIII. ON THE DIFFERENTIAL ANALOGIES OF SPHERICAL TRIANGLES.* PROPOSITION I. (Plate IV. Fig. 3.) A preparatory Proposition. (612) CONSTRUCT a general figure as at Prop. XIX. Book III. Chap. I. Thus, let a be the pole of the circle HGFE; F the pole of ABH; E the pole of CGI; and c the pole of EDI. See Simpson's Fluxions, vol. ii. page 278. Traité de Trigonométrie, par M Cagnoli, chap. xix. page 310. Hind's Trigonometry, 2nd edition, page 286. Young's Trigonometry, page 203. Suppose these circles to be invariable whilst another great circle DFCB revolves about the pole F, and let cn be at right angles to the great circle mnorsd; then the three triangles ABC, CGF, and EDF will be variable, viz. I. In the right-angled triangle ABC, the A will be a fixed quantity, and the other parts will be variable; viz. вm will be the increment of AB; no the increment of BC; co the increment of AC; and Ds the increment of the arc ID which measures the c. II. In the right-angled triangle CGF, the side FG will be a fixed quantity, and the other parts will be variable; viz, co will be the decrement of CG; no the decrement of FC; Bm the decrement of the CFG; and Ds the increment of the ▲ c. III. In the right-angled triangle FDE, the hypothenuse EF will be a fixed quantity, and the other parts will be variable; viz. sd=no, will be the increment of FD=BC; SD the decrement of ED; and Bm the decrement of the EFD=/ CFG. PROPOSITION. II. (Plate IV. Fig. 3.) (613) In any right-angled spherical triangle ABC, right-angled at B, suppose one of the angles as a to remain constant, it is required to find the ratio of the differentials of the other parts. 1. In the triangles FBM, FCn, having the same acute at F, sin FB: sin FC:: tan Bm; tan cn (382). But FB 90°, Fc is the complement of BC, and вm and cn, being very small arcs, have the same ratio to each other as their tangents. 2. In the triangles DCI, con, where the DCI may be supposed equal to the con, (382). tan DI : sin CI :: tan cn: sin no But DI is the measure of the c, ci=90°, and the tangent of cn and the sine of no, have the same ratio to each other as Again, sin DI: sin DC :: sin cn: sin co (382). But DIC, DC=90°, and the very small arcs cn and co, have the same ratio to each other as their sines, 3. In the triangles BFM, DFS, where FS FD ultimately, sin FB: sin Bm :: sin Fs: sin Ds (382). But FB 90°, FS=FD=BC, and the small arcs are as their sines, But no represents the differential of BC, co of AC, ds of di =c, and Bm of AB (269). Therefore, sin BC dab. C (614) Hence, in any right-angled spherical triangle ABC, right-angled at B; by denoting the sides opposite to the angles A, B, C, by a, b, c, respectively, we derive by substitution and reduction, the following general equations, which include all the varieties that can possibly happen wherein the LB is 90° and one of the other angles (viz. a) constant. b B (615) Any of the foregoing equations may be turned into proportions, or varied in the expression, by reference to pages 104, 105, &c. Thus, from the fourth set of equations dc. cos a=db. sin c. that is, db: dc :: cos a: sin c. Simpson's Fluxions, page 280. :: cos b sin c : (384). COS C sin b :: cos b. sin b: cos c. sin c. :: sin 26 sin 2c+ (228). + Vince's Trigonometry, 2d edition, page 139.; Traité de Trigonométrie, par M. Cagnoli, art. 677. page 329 PROPOSITION III. (Plate IV. Fig. 3.) (615) In any right-angled spherical triangle CGF, right-angled at G, suppose one of the sides as FG to remain constant, it is required to find the ratio of the differentials of the other parts. By the foregoing proposition no = COS BC Bm; co= Bm. But no is the decrement of FC; Bm the decrement of the LF; co the decrement of CG; and Ds the increment of the c, also Fc is the complement of (616) Hence, if the fixed side FG be represented by c, the hypothenuse FC by b, the side co by a, and the angles opposite to these sides by C, B, and A respectively, as at (614), we derive the following general equations or formulæ, which comprehend all the different cases that can possibly happen wherein the angle B is 90°, and one of its adjacent sides (viz c) constant. (617) The preceding formulæ may be varied in the expression by reference to pages 104, 105, &c. or they may be turned into proportions thus, from the first set of equations da. sin b da. sin C, that is, da: da:: sin b sin c.+ * Simpson's Fluxions, page 280. + Vince's Trigonometry, 2d edition, page 140. |