764 PEOFESSOE SYLVESTEE ON THE MOTION OP A EIGID BODY 



Then XPF being a quadrantal triangle, 



cos XP'=8in XP cos XPI=^jj^pj (ccs XI-cos XP cos PI) 



_ 1 /a)) a)| M\_ a), /I M\ ' 



— sin PI V?L ~ » !;<;/— L sin PI V a"* ~ "*/ ' 

 where 



u'=6/l-\-ul-\-ul. 



Again, for gi'eater simplicity, making §=1, i. e. considering the motions of the rolling 

 body and the free nucleus to absolutely coincide in time, we have from the Eulerian 

 equations. 



Hence if [F] be the couple due to F the friction force, 



[F]=S(J cos X cos XF) = L^i^^Jgpj^ S(c-=-5=)(«^i'+«V-iV)(.;'-M«^) 



~ La«6VsinPI 



~ a-iVv'I?«.*-M« 



And as the arm at which the friction acts, i. e. the distance of the fixed centre from 



the point of contact between the ellipsoid and the fixed plane is -^j: — sec PI, i. e. -7W, 



we have 



the mass of the ellipsoid throughout being treated as unity. 



"We might, in like manner, through the algorithm of spherical triangles, proceed to 

 calculate the value of the pressure couple [P] which is equal to the sum of the compo- 

 nents J cos X, J cos fi, J cos V multiplied respectively by the sines of the perpendicular 

 arcs dropped upon PI from X, Y, Z. But it will be obtained more expeditiously in its 

 simplest form by first calculating J itself, the value of the entire couple, and then using 

 the equation [P]'=J^-[F]l 



For brevity, in place of a', ¥, c^ u^, L- write /, g, h, Q, A respectively, and let 

 f+g+h=p, fg+(jh+hf=q, fgh=r. Then 



