It's been several years since I did the math. Forgive me, but in my lazy dotage I just don't feel like looking up the details. (Actually, the fewer details the better, I guess.) But here's the gist of it.
If you apply a specific force over a certain distance (e.g., bridge length) to an initially stationary object, that object will end up with a specific kinetic energy. It doesn't matter how heavy or light the object happens to be; it will have that much energy, and no more and no less. If it then collides square on (e.g., center ball) with another object of exactly the same mass, it stops and the second object will "acquire" all of the first object's kinetic energy, as I think we all know. If the masses aren't equal, the first object will either rebound somewhat or continue forward after the collision, thus "retaining" some of that kinetic energy in either case. As a result, the second object will come away with less. So having the masses equal will ensure that the second object ends up with the most energy (speed).
The reason we don't play with 6 oz cues is that the force our arms generate must move not only the cue but our arms themselves. Their mass prevents us from generating the cue speed we would get by applying the same force directly to the cue, bypassing our arms. Given this unavoidable load (our arms), and the fact that once the cue makes contact with the ball that load is irrelevant (decoupled) from the collision, it pays to use a cue heavier than the cueball because the increase in the cue's kinetic energy from the added mass outstrips the reduction in speed from adding that mass to the combined (and preloaded) arm/cue system. Some of that energy is wasted in the form of non-zero post-collision cue velocity, but there's still a net gain. That's true up to a point, that is, and that point is the optimal cue mass for a center-ball hit...at least from an efficiency point of view. Add more weight and the cue/arm combination slows down too much. If you substituted a lighter ball, the optimal mass would shift to a lighter weight accordingly. (End of long preamble which I'm know you already understand).
When you hit off center, the cue, in effect, sees less of the cueball's mass. The farther off center, the less of the mass it sees. Analogous to a center-ball hit with a lighter cueball, the optimal cue weight is correspondingly reduced.
I don't know if that's at all convincing. It's kind of hard (for me) to translate the math into the vernacular (even if I did revisit the details).
To repeat an earlier point, though, that's the simple physics of it. When you add bio-mechanics, things aren't so clear. You can probably generate more force against a heavier cue, so the above assumption of a single value for the force, regardless of cue weight, doesn't rest on very solid ground. While a lighter cue might possibly still be the way to go for a power draw shot, the best weight likely wouldn't represent as great a reduction as the "naive" physics would suggest. Of course, if its weight is already below optimum for a center-ball hit, that would make any benefit from a reduction even less likely.
Jim
