Sorry for the delay an let's do forget about who's being defensive (you).FLICKit said:
Let's first look at a center ball hit which sends the cueball off at 20 mph and assume a one millisecond contact. The time averaged force is the momentum of the cueball divided by the contact time, so:
F = MV/T = (,012 slugs)(20 mph)(1.47 ft/sec/mph)/(.001 sec) = 352 lbs
At a large tip offset where the cueball takes off at 12 mph, let's assume a contact time of two milliseconds. It might be more like 1.3 or 1.5, but 2 is about the outside limit and gives us the least amount of force:
F = (.012)(12)(1.47)/(.002) = 106 lbs
To get the average force applied by the hand/arm for a center ball hit to bring the cue up to speed, we can use:
FX = (1/2)MV^2
where F is the spacially averaged force and X is the bridge length, assumed to be 8.5". V is the stick's speed just before impact and is about 2/3 of the cueball's post impact speed (ignoring energy losses). This formula is valid so long as the applied force is symmetric about the midpoint of the acceleration period, otherwise a correction needs to be made. But according to accelerometer recordings of actual strokes, the symmetry condition is roughly realized.
So:
F = (1/2)MV^2/X = .5(.035 slugs)[(2/3)(20 mph)(1.47 ft/sec/mph)]^2/(8.5/12 ft) = 9.5 lbs
I used a figure of 15 lbs in an earlier post but this is roughly the peak force. When you allow for energy losses during the collision, the 9.5 lbs might be more like 12 lbs.
Jim