[ 14 ] .^. 



that the force of the fun is equal to F x by the produd of the 

 fine and cofine of his height above the plane of the annulus, 

 therefore the force of the fun at H is equal to F j-^ X i_j^~J^t. 

 But this force ads entirely in the plane P G tip, therefore we 

 muft refolve it into two forces, one ading in the plane PQ^A, 

 which is that we are looking for, the other in the plane PC/, 

 perpendicular to the former ; this latter force is deftroyed by 

 an eqtial and contrary force, when the fun is equidiftant on 

 the other fide of the line of the nodes ; but the other force 

 always ading in the fanae diredion, is that only by which the 

 ring is annually affeded. The Cos. G H : Cos. angle D C A : : 

 Rad. : Sin. angle H (Gas. 1 1. Sph. Trig.) and Rad : Sin. angle H : : 

 Sin. CH: Sin. CG (Cas. 2.) •.• Cos. GH (i—s'-y-^^): Cos. DC A 



(c) : : Sin. CYl (y): Sin. C G =: ~^ ' ;|^ Then, to find the part 



of the force ading in the plane PQ^A, Rad. (i.) : F s y/i — j*_y* 



cy 

 (the whole force) : : S. G C (— ,_ -- -^ =^ ) : F r,r^% the force in the 



\/i — j' J* 



diredion P G, And hence to find the mean annual force, we 



mufl find the fum of all the F csy' in the circle, or the fluent 



F c s y ^ y 

 of T c s y- z — —^=^-^— \ whofe fluent, found as before, is 

 V\ — y^ 



^ F c s% — 7 F csy -J i — y^ ; and when y = r, the fluent becomes 



~ F csp, and in the whole circle = ¥ csp; this divided by the 



v>'hole circumference 2 p, the mean force comes out ^Fcx, 



that 



