6o 
Mr. Robertson on the Precession 
: QP. Consequently, during the generation of the angles 
BCH, FEM, the angular velocity of CB is to the angular 
velocity of EF as KL to QP. 
4. If ACB (Fig. 3.) be any plane angle, and from any 
point E, EF be drawn perpendicular to AC and EH perpen- 
dicular to CB, then the angle FEH is equal to the angle ACB. 
For let HK be drawn perpendicular to AC, and let EH be 
produced to G. First let ACB be an obtuse angle, and then 
< ACB = < KHC + < HKC-f < KHC + < CHG = < KHG 
= ( 29. I. ) < FEH. Secondly, ACB being an acute angle, 
the right angle GHC = < GHK -J- < KHC = < KHC -f- < 
HCK = ( 15. 1 . ) < KHC 4* < ACB. Consequently the angle 
ACB == < GHK = < FEH. 
5. Let G ( Fig. 4. ) be the centre of gravity of a body, and 
AB, DC two axes passing through G ; and, while the body 
revolves round AB, let AB and consequently the whole body 
revolve round DC, the periodical times of these revolutions 
not being necessarily equal ; it is required to determine the 
direction and angular velocity with which any particle of the 
body revolves in consequence of this compound motion. 
Suppose the simple motion of the body about AB to be such, 
that during the revolution the parts towards D from AB 
would rise above the plane, on which the figure is drawn, and 
the parts towards C sink below it. And suppose the simple 
motion about DC to be such, that during the revolution the 
parts towards A from DC would rise above the plane of 
the figure, and those towards B sink below it. Let P (Fig. 5.) 
be a particle of the body, above the plane, and let PR be a 
perpendicular to the said plane.* With the centre G suppose 
* The axis DC, and the line RM are intentionally omitted in Fig. 5, with a view 
to prevent confusion in the figure. 
