and the different Motions of the Earth's Axis. 29

triangles DEe and D/T will also be equal and alike, in allrespects ; and so, DE -f De being = DE + DF — a semi-circle, both DE and De may be taken as quadrantal, or arcsof 90° each : whence, if 'VSR, the measure of the angle Thbe supposed to meet aED in r, it will be, as Jin. ED (radius)

: fin. <? (E) :: Ee ) : ED* = x ;

J v 'I' co-sin.^ 1 cn-l. <v X rad.

also, as Jin. a (qf>) : fin. <pD (co-sin. ‘Y’E) :: E De : op a =

«tA v fill. E X co-sin. E X co-fm. ^ re q U i re d quantity of the•l lin. v x co-iin. <r x rad. A

precession :

And, as fin. r (radius) : fin. DR (qpE) : : EDc (RDr) : Rr

wtA sin. E x co-sin. E x sin. <rE , r

= -77— X --, the corresponding quantity or

» 1 co-sin. v x rats]

the nutation, or the decrease of the inclination of the equatorto the ecliptic. Us. E. I.

COROLLARY.

It is evident from hence that the quantity of the nutation isto that of the precession, as —' - J-- to co '^”‘ ^ E , or as sin.qp to

rad.

rad. x co-sin. vE

sin. -v

-—--, that is, as the sine of C Y > to the co-tangent of

sin. rt 0

qpE- It appears moreover (because fin. E : fin. : ’.fin. N

: fin. op E, p. spherics) that the former of these quantities is

. r ^ wtA sin. N x sin. vN x co-sin. E , . ,

also truly explicable by — r - x-=rr-• which

r ■'I co-sin. -v X rad. |

expression will be of use in the following Problem.

PROBLEM VII.

7 0 determine the precejjion of the equinox , and the quantity ofithe nutation of the earth’s axis, caused by the moon, during dietime ofi half a revolution of the node of the moon’s orbit.

Things being supposed as in the preceding Problem, let the Fig-14distance cpN of the node from the equinoctial point be denot-ed by z, its sine by x, and its co-sine by y ; let also the sineol the angle Nq^E be put = a, its cosine = b, the sine of