Required the true longitude? the eye being 21 feet above the level of the sea.

Answer. The reduced time 4h.18'.10"; D's true semi-diameter 16.3"; horizontal parallax 58'.27"; apparent altitude of her centre 19°.42′.41", and correction of her altitude 52′.22′′. The apparent altitude of the sun's centre is 45°.32′.29"; correction 50"; apparent distance of centres 106.47′.24′′; true distance 106.1' and the true longitude 113°.10' W.

EXAMPLE IV.

Suppose on the 9th of May 1813, in longitude 16°.30' west, by account, the following observations to be taken:

Required the true longitude? the eye being 18 feet above the level of the sea.

Answer. The reduced time is 13h.47′19′′, and being so near to midnight, the D's semi-diameter and horizontal parallax may be taken out of the Nautical Almanac without using proportion, being 15′.8" and 55'.8".. Hence the apparent altitude of the D's centre is 18.56'.50", and her correction=49′.25". App. altitude of the star's centre 20°.10'.56", correction 2'.35"; App. distance of centres 31°.17.52"; true distance 31°.11'.52′′. The true longitude 17°.6'.30" west.

(M) 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. 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 6th 1813, in latitude 47°.39'. N. and longitude 57.16'. W. from Greenwich, by account, at 3h 55'.30′′ P. M. 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°.27′.52", required the longitude? The eye being 18 feet above the level of the sea.

Answer. The reduced time is 7.44' 36"; D's true semidiameter 16.15"; horizontal parallax 58.54"; apparent altitude of her centre 46°,34'.44", and correction of her altitude 39'.35".

The apparent altitude of the sun's centre is 26°.9'.7" ; correction 1.48"; apparent distance of centres 76°.0'.7′′; true distance 75°.45' 44',

The sun's declination reduced to the time and place of observation is 6°.32'.12" N.; with this declination, the O's true altitude 26.7.19", and the co-latitude 42°.21', find the correct apparent time at ship 3h.51',24"; and hence the true longitude is 56°.56'.45" west,

CHAPTER IX.

OF THE FLUXIONAL ANALOGIES OF SPHERICAL TRIANGLES †. PROPOSITION I. (Plate IV. Fig. 3.)

(N) A preparatory Proposition.

CONSTRUCT a general figure as at Prop. XIX. Book III. Chap. I. (L. 147.). 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.

This example, and its solution, were given to me by Mr. Kirby, the Editor of Bowditch's Navigation, who likewise worked the four preceding examples, by different methods, in order to check any errors which might have been committed in the solutions.

+ See Cote's "De Estimatione Errorum in mixtâ Mathesi." Cambridge, 1722. Simpson's Fluxions, vol. II. page 278, et seq.-Crakelt's translation of Mauduit's Trigonometry, Chap. V. page 164, &c.-Traité de Trigonométrie, par M. Cagnoli, Chap. XIX. page 310, &c.-La Lande's Astronomy, Paris, 1792, vol. III. page 588, et seq; or art. 3997 to 4051, &c. &c.

Suppose

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 mno Fsd; then the three triangles ABC, CGF, and LDF 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 . LC.

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 decremént of the CFG; and DS the increment of the

c.

III. In the right-angled triangle EDF, 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=4CFG (N. 131.).

PROPOSITION II. (Plate IV. Fig. 3.)

(0) In any right-angled spherical triangle ABC, rightangled at B, suppose one of the angles as A to remain constant, it is required to find the ratios of the fluxions of the other parts. 1. In the triangles FEM, FCn, having the same acute at F. sine FB sine FC :: tang вm : tang cn (M. 163.). But FB 90°, FC is the complement of BC, and Bm 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.

Tang DI: sine ci tang cn sine no (M. 163.) But.DI is the measure of the Zc, cr=90°, and the tangent of cn and the sine of no, have the same ratio to each other as the arcs,

Again, sine DI: sine DC :: sine cn sine co (M. 163.) 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 FD=FS extremely near. sine FB sine вm :: sine FD sine DS.

But FB=90°, FD=BC, and the small arcs are as their sines.

But no will represent the fluxion of BC; co the fluxion of ac; Ds the fluxion of DIC; and Bm the fluxion of AB. (B. 120.) Therefore,

(Q) Any of the foregoing equations may be turned into proportions, or varied in the expression, by reference to pages 98, 99, &c.

sin C

× b

cot a

(0.98.)

Simpson's Fluxions, p. 280.

Thus,

Thus, from the first equation ex cos a = 6x sin 4 c.

(R) In any right-angled spherical triangle CGE, right-angled at G, suppose one of the sides as FG to remain constant, it is required to find the fluxions of the other parts.

By the foregoing proposition n o =

COS BC

tang c

Bm; co =

But no is the decrement of

FC; Bm the decrement of the F; co the decrement of CG; and Ds the increment of the c, also FC is the complement of BC. Therefore

(S) Hence, if the fixed side FG be represented by c, the hypothenuse FC by b, the side CG by a, and the angles opposite to these sides by C, B, and a respectively (as at P. 310.) we derive the following general equations, or formulæ, viz.

* Vince's Trigonometry, 2d edition, page 139; Traité de Trigonométrie, par M. Cagnoli, art. 677, page 329.

↑ Simpson's Fluxions, page 280.

IV.