Energy (proportional to speed squared) is conserved, not speed.
WHY DO THE HEATHEN CONTINUE TO RAGE....Part Two
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Energy (proportional to speed squared) is conserved, not speed.
Look at it like this: For stun shots, the cb's post impact speed is reduced based on the same proportion of cb-ob overlap. (Full ball hit reduces cb speed by 100%, halfball reduces its speed by 50%, a quarter ball by 25%, etc...). The ob post impact speed, however, isn't so proportionally tied to the cb-ob overlap.
On a full hit (0° cut) the ob takes on 100% of the cb's speed while the cb loses all of its speed. Friction is negligible. But on a 30° cut the ob takes on more than 80% of the cb's speed, regardless of the fact that the cb loses half (50%) of its speed. You would think if the cb loses half, then the ob should get that half. But not so.
I'm sure Dr Dave will say I'm wrong about this, lol, but I think it's because the friction on a cut shot acts like a brake for the cb, but has an insignificant effect on the ob. And this is the case up to a certain fractional overlap. At some point the friction just isn't enough to hold the cb back as much, and the ob post impact speed will end up being equal to cb post impact speed. For a stun shot, this occurs at a 45° cut (29.3% cb-ob overlap). And every cut thinner than that results in the cb retaining more of its speed and the ob taking on less and less speed.
The two extremes are a 0° cut (100% cb-ob overlap) and a 90° cut (zero overlap). At 0° the ob has a post impact speed of 100% and the cb's is 0%. At 90% the cb speed is 100% and the ob's is 0%. Halfway between 0° and 90° we have 45°. So it makes sense that 45° is the crossover point, the point of equality, where the two balls swap rolls as far which ball has has the faster post impact speed.
That is correct, but momentum is a vector (mass times the velocity vector). If you add the post-impact velocity (or momentum) vectors, they add to the original CB velocity (or momentum) vector. Both energy (scalar) and momentum (vector) are conserved in the collision.
You are correct about me saying you are wrong on this point. Again, the analysis in TP 3.2 assumes a perfectly elastic collision with no friction.
Lol.... Again?
TP A.16 mentions "perfect" balls and "imperfect" balls. I guess in reality there's not really a perfect elastic collision between the balls. Close enough that you consider the friction between the balls to be negligible, but the two different tables of values (using perfect and imperfect balls) show significant differences at different shot angles.
And nowhere in TP 3.2 does it state "perfect" or "imperfect" balls are being considered. It simply refers to "stun" shots. Are you saying the results are based on perfect balls? Does this means the numbers aren't what we can expect with real life "imperfect" balls?
Since there is a difference between post-impact speeds and rolling distances when accounting for and not accounting for friction between a rolling cb and an ob, as shown in TP A.16, and since friction is higher on stun shots, wouldn't there have to be even more of a significant difference between "perfect" and "imperfect" balls when looking at the type of data in TP 3.2?
Assuming perfectly elastic balls and no friction makes the math/physics analyses much easier, and the results are usually very close to reality. Pool balls don't lose much energy in collisions and they have fairly low friction.
I guess that depends on what you mean by "significant differences."
TP 3.2 is based on TP 3.1, where the assumptions are stated clearly.
I think my previous posts and the linked resources cover this fairly well.
Again, inelasticity and friction do change the results slightly, but not by much. Equal separation speed for a rolling CB shot still requires a hit slightly thinner than a 1/2-ball hit, whether you include the effects of inelasticity and friction or not.
NO SH**. That’s progress duckie. congratulations.