Jal

01-21-2006, 11:03 PM

The model for squirt that's most often cited uses the concept of "endmass" at the tip portion of the shaft. The ball must push it aside as it begins to rotate, and the tip pushes the ball back with an equal amount of force. One reason cited that the entire mass of the cue is not involved is the relatively slow propagation of the transverse (shear) wave down the shaft.

But the cue also acts like a cantilever beam, providing some spring force resistance. If you look up "cantilever beam stiffness" on Google, you'll get endless references to the subject of Atomic Force Microscopy. Here can be found the formula:

k = (3/4)E[r^4/L^3]

where E is Young's modulus, r is the radius of the rod, and L its length. This is for a rod fixed at one end and of uniform radius, and for the equilibrium case.

While I think a cue can be treated as if it is fixed at one end during the brief contact period, and the formula can be modified for taper, it's not clear what its effective length is in relation to the distance the shear wave has traveled down the shaft.

Without a background in continuum physics, it's a tough nut to crack, so I'm looking for a relatively easy way out (although an extended discussion would be most welcome too).

If you're not well versed but have some thoughts, I'd be interested in hearing them. They cannot be any more bizarre than the ones I've entertained (such as thinking its effective length is the distance the shear wave has traveled).

Jim

But the cue also acts like a cantilever beam, providing some spring force resistance. If you look up "cantilever beam stiffness" on Google, you'll get endless references to the subject of Atomic Force Microscopy. Here can be found the formula:

k = (3/4)E[r^4/L^3]

where E is Young's modulus, r is the radius of the rod, and L its length. This is for a rod fixed at one end and of uniform radius, and for the equilibrium case.

While I think a cue can be treated as if it is fixed at one end during the brief contact period, and the formula can be modified for taper, it's not clear what its effective length is in relation to the distance the shear wave has traveled down the shaft.

Without a background in continuum physics, it's a tough nut to crack, so I'm looking for a relatively easy way out (although an extended discussion would be most welcome too).

If you're not well versed but have some thoughts, I'd be interested in hearing them. They cannot be any more bizarre than the ones I've entertained (such as thinking its effective length is the distance the shear wave has traveled).

Jim