How much force is generated between the tip and the cue ball during a break shot, and what would it take to generate a “ton” of force at the tip?

See: TP B.20 – Peak forces during a break shot and TP B.22 – How peak tip contact force and contact patch size vary with shot speed, and drop tests.

As another example, and to keep things simple, let’s use a cue weight of 18 oz and assume a perfect tip with a center-ball hit. For this case, TP A.30 predicts that the outgoing CB speed is about 3/2 (1.5) the incoming cue speed. Let’s also assume that the average force during tip contact is about half the peak force. And let’s assume the tip is in contact with the ball for 0.001 sec, which is typical.

For any CB speed (vb), given the CB mass (mb), the momentum delivered to the CB is:

mom = mb * vb

For a given duration of contact (dt), this momentum must equal the impulse delivered from the cue:

imp = 1/2 * Fmax * dt

So to find the peak force for a given CB speed:

Fmax = 2*mb*vb/dt

And for a given peak force, the CB speed is:

vb = Fmax*dt/2 / mb

And the cue stick speed required to create this is about:

vs = 2/3 vb = Fmax*dt / 3*mb

For a 20mph break, with a 6oz pool ball, the Fmax equation gives a peak force of:

Fmax = 683 pounds

To achieve a 1 ton (2000 pound) peak force, the vs equation gives a required cue speed of:

vs = 39mph

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