1873.] of a Body about a Fixed Point. 237 



Eliminating cos AGN and tan GF, we have 



AT Gr 2 



resolved velocity of G= — — — cot AG. 



Alx 



Hence the velocity of G resolved perpendicular to its central radius vector 



varies as the cotangent of that radius vector. 



If GM, G-M' be perpendicular arcs drawn from Gr on the axes, we 



may show that the velocity of Gr resolved along its central radius vector 



n B-C sin G-M . sin GM' mi . , 

 is = — It . ^ AG ' result may be deduced from 



one of Euler's equations. 



10. The relations of the motions of the points G, H, I to one another 

 may easily be deduced from the following simple formulae. 



If we eliminate to v w 2 , w 3 from the expressions for the direction 



T 



cosines of OG, OH, 01, and w by Lagrange's theorem, w cos GI = — , we 



G 



find cos AI _ cos AG _ cos AH 



G 2 cos GI AT GVAT' 



If P be any point on the sphere, we have, by Napier's rules, 



cos BP cos BAP t PAB. 

 cos CP cos CAP 



Prom this we easily deduce 



tan 6' tan 9 tan 6" 



B ~ C ~ VBC' 



where 0', 6, 0" are the angles the central radii vectores AI, AG, ATT 

 make with the major axis. 



11. Let us apply these to find the motion of the extremity I of the 

 instantaneous axis along its ellipse. 



B 



Differentiating the expression tan 6' = tan 6, we get 



C 



dd' B cos 2 6' dd 



dt C cos 2 6 dt 



If (a, (3, y), (a', /3', y) be the direction angles of OG and 01 referred to 

 the principal axes at O, we have, by Napier's rules, cos 6 = CQS ^ and 



sin a 



cos 6' = cos ^ . Also ^ has been already found. Hence 

 sin a dt 



dd' ^ cos 2 /3 ' sin 2 a AT— G 2 1 B 



dt ~ sin 2 a ' cos 2 /3 ' AG tan a . sin a C ' 



. ,dd' G AT — G 2 , , nT 

 .'. sin a — = — — — . cot a . COS GI. 



dt T BC 



This gives the velocity perpendicular to AI. The velocity along its 



