The Variation in Attraction Due to the Attracting Bodies. 221 



Per Conic Sections, 



Therefore b'^ = B^ + ( A'^ - B^) sin^ O cos^ ^ 

 a/ B;2 — A'^ b- = (A- - B2)'2 sin^ O cos^ O sin^ i 



Sin^ a (A^'- B^)^ sin^ O cos^ O sin^ ^ E-* sin^ O cns^ O sin^ | 



Dill \j <juo KJ aiiA. <^ xu oiia vy ^l^^'o ^^ dijj q 



a;^ B/ ^ (1-E- sin'^ O sin- I) (l-E^ cos- O)- 



Sin- tr, = E^ sin- O cos- O sin^ ^ + E^ sin- O cos^ O sin- ? + „ 

 Sin ex, = E- sin O cos O sin ? + |- E^ sin O cos'' O sin ^ + | E^ sin^ O cos O 

 sin^ I + i E« sin^ O cos^ O sin^ |-„. 



E,^ = ^"' ~^^'' ---E- cos- O - E- sin- O sin^ I + E^ cos^ O - E' sin^ O cos^ O 

 B,'- 



sin= § + E" cos" O — „ 



1^ cos^ O — 2 E^ sin- O cos- O sin- | + E^ sin* O sin* ? + 2 E'' cos« 



£'/ = E*cos*0 



Diagram 8. 



To find an expression for angle (a. + j,) for any two opposite wedges. 



Diagram 8 is constructed in accordance with diagram 7, having lower 

 part laid over so that point A,„ becomes a„,. A„ n and a„, m are dr^wn 

 perpendicular to P C. Angle a ,, R A,, equals 2 or, C b, R or b, P C equals 5, 

 A„ P R equals (S- + h) and a,„ PR equals (B — v). 



In triangle A„ Pa,„ , 



(a), sin A, Pa „ or sin (h + v)= sin (3- + >d sin (B — v). 



cos tX ^ ' 



Pb = D cos S ; PR = D cos -^ ; and b, R = D sin 5 cos 3^. 



^ 



A„ R =:a,„ R==^D sin 5 cos 3 = 1/1 + £^ D sin 3 cos 5 



A„ n=a,„ m=:= ^/l + E'^ D sin 5 cos 3 cos or, 



