A S T R O N O M Y.

0^

their Orbits were exterior to the Orbitthe Earth.

X. 1*

and on the Point V with the Radius V X describe 4Quadrant TA ; from T let fall the Perpendicular 1 Jand put B V^ d\ then AB =:at — d z=.v, the V er iSine of the Angle T V A. Now (by Euclid II-we hare a z = b z -f- x z -f- 2 b d, whence a z — 2 b"h x z ; and by adding 2 b x on each Side, a z -f- 2 j,2 &d — b z 2 b x x z \ that is, a z -f- 2 b v = ^ p—p. Then r — a z —2bvszs-, and w"tiplying by 2 x we have \bxv = 2 x s, whence 4 ^ fr :: 2 x : D ; that is, 4 b x : b‘ -j- 2 b x -f- * x —',j2 T V : A B the Diameter of A Circle to the ^ eI eSine of the exterior Angle T V A. , g

12. But in any Situation B of the Planet Rentes j

Arch of Illumination <2/ is equal to the Arch « 'which measures the exterior Angle b B d. And st ^been shewn, that the Area of the v hole Dijk of the P^eis to the Area of the enlightened Part as the Diatnetp ya Circle to the Versed Sine of the Arch of Illuming* *and therefore as 4 b x to b z -f- 2 b x -f- x z — d 1 . 5

13. But the Area of the whole Disk is every wh ese *

therefore, as 4 b x : V -}- 2 b x -f- x %■A which in

' xf

b 1 -f- 2 b x -f- .

4 ^ x3 , -.. ..11 Cafes will be P r °

portioned to the enlightened Area of the Dislc. ^o determine this a Maximum, its Fluxion must be ^or the negative Parts thereof be equal to the astst* 1 ^that 2bx-^-2xx'Xe r bx*zz 12 bx."

; and dividing all by 4 b x z *

tive

D

that is,

the

i X 2 x * 2, b z b x f 3

X $

b^ -f- 2 b x -{- x x — a

Equation becomes 2 p x ~f 2 x —^ ^ -p v </ «■ , - ," 3 aZ ‘ Consequently 2 > lb-f\bx-\-xxsxs 3 awhence we get * — f-faafHTb —2 b = 427. ^

14. If therefore we take 427 from the Scale of eC 3 ll ,,[Parts S T, and set from T to the Orbit of Venus, st ' ii }

intersect it in the Point*; and drawing T *, it w

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