The Variation in Attraction Due to the Attracting Bodies. 217 



Likewise the product of the thickness at a, into length r s into width at 

 a, is to that of tlie tliickness at a,„ into lengtli i f into width at a,„. 



( B, : B,„ ) 



As - 1 : cos 5 ^ :: A,B, cos a : A,„B^,, cos b, cos ^ :: 1 : cos 3^. 



( A, cos a: :A„,cosb, ) 



Mass-factor then at a, is to mass-factor at a,, or a„, as 1 : cos 3-. 



If in Diagram 7 conjugate diameters A, A, and B, B, were at right angles 

 the chord element of mass and the wedge mass would be cos^ a. to unity 

 greater. Regardless of the law of variations of the angle ex, from wedge to 

 wedge if we assume a homogenous solid made up of elliptic wedges having, 

 a common edge length of B, B,, and limited in width by an ellipse with 

 principal semi-axis A and A, having its plane in center C and at right angles 

 to diameter B, B , then we have a volume that can be divided into infini- 

 tesimal wedges of equal masses in accordance with Art. 14. If the ellip^ 

 soid be divided into wedges having the same angular thicknesses as those 

 of the assumed solid, then the mass of any wedge of the ellipsoid is to the 

 mass of the corresponding wedge of the assumed solid as the square of the 

 cosine of the a angle (cos- a) to unity. In the assumed solid let m repre- 

 sent the mass of the chord element, m that of the wedge and M, the roass 

 of the whole solid; then in the ellipsoid m cos'^ a, is the mass of the chord 

 element, m cos'^ or, that of the wedge. Let M be the mass of the ellipsoid. 



Draw k b„ and jb„, parallel to diameter B, B,. As it was'proven that a„d 

 equals a, cos 5 so, by similar construction of diagram with radms of cir- 

 cle a, C, with a point P, in extension of diameter A, A' and with points of 

 tangency corresponding to a,, and a,„ at b„ and b„ ,, it can be proven that 

 k b„ and j b,, , each equals b^ cos 3. 



By construction angle Cb„ k equals angle a„ P C, and Cb,,, j equals a,„ PC. 

 Angle b,, k C and b„,j C each varies from a right angle by the angle a. Let 

 angle a,, PC be represented by 3 + u, and a„,P C by 3— v, 



DsinaCP, 



a„ P = r 



cos b 



From a,, conceive a perpendicular drawn to diameter B, B,. In con- 

 struction then we have a riglit angled triangle with hypothenuse Ca,, or a, 

 and with perpendicular equal to a, cos 3 cos a, in which 



^ _, a' cos 9- cos a A, cos 3 cos a 

 sm a,, C P = — 



a, P= 



a„ A„ 



D A' cos 5 cos a 



A, cos b 



Likewise, 



A,„ P = D A, cos 3 cos a 

 A , cos b. 



Draw line P c making an angle (a±z) with P C, and let P c be the result- 

 ant direction of attraction for the two opposite chord elements. Draw 

 line V C u making angle u C k or v C w equal to angle z; and draw lines 



