348 
HERON OF ALEXANDRIA 
case suppose the circle AG to roll along AE till ODG takes 
the position 0'FE\ then D will be at F where AE — BF. 
And similarly if a whole revolution is performed and OB A is 
again perpendicular to AE. Contrary, therefore, to the prin 
ciple that the greater circle moves quicker than the smaller on 
the same axis, it would appear that the movement of the 
smaller in this case is as quick as that of the greater, since 
BH = AG, and BF = AE. Heron’s explanation is that, e.g. 
in the case where the larger circle rolls on AE, the lesser 
circle maintains the same speed as the greater because it has 
two motions; for if we regard the smaller circle as merely 
fastened to the larger, and not rolling at all, its centre 0 will 
move to 0' traversing a distance 00' equal to AE and BF; 
hence the greater circle will take the lesser with it over an 
equal distance, the rolling of the lesser circle having no effect 
upon this. 
The r parallelogram of velocities. 
Heron next proves the parallelogram of velocities (chap. 8); 
he takes the case of a rectangle, but the proof is applicable 
generally. 
The way it is put is this. A 
point moves with uniform velocity 
along a straight line AB, from A 
to B, while at the same time AB 
moves with uniform velocity always 
parallel to itself with its extremity 
A describing the straight line AC. 
Suppose that, when the point arrives at B, the straight line