In fact, more than half maximum spin always produces less than maximum throw/spin transfer.
pj
chgo
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Any ideas on something that's been bugging me?
- Thread starter CeaseLess
- Start date
pj
chgo
most likely its you for a reason. no tip or shaft is going to noticeably change the spin hitting the cue ball in the exact same spot.
besides if you are a bank player you should be hitting almost all shots the same speed and more often using only a slight amount of inside english
which make most banks easier and easier to determine the spot to aim for.
the best bankers that ever lived used a plain old wooden shaft and whatever tip was available.
besides if you are a bank player you should be hitting almost all shots the same speed and more often using only a slight amount of inside english
which make most banks easier and easier to determine the spot to aim for.
the best bankers that ever lived used a plain old wooden shaft and whatever tip was available.
Stupid question time. Is there a wind up time for a cue ball struck with english? If so is it significant? Just trying to understand the point of transfer on those banks.
If I understand you, there's no "wind up time"; the cue ball has the most spin immediately after being struck - it's all downhill from there.
pj
chgo
The physics says something else. The distance off-center will determine the spin/speed ratio. A better (more efficient) tip/shaft combination will have more spin but it will also have more speed. To make that efficient shaft play like a "deader" shaft, you should hit the same distance from center but slightly softer.
It's not clear what he means by wind up time. The time the tip is on the ball? The time the two balls are in contact? The ball on the cushion?
WobblyStroke
Well-known member
Can always count on u and PJ to fix any misunderstandings I may have on the physics of pool. Edited. Ty. But wouldn't the same speed closer to center axis also put less spin on the cue ball?
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WobblyStroke
Well-known member
Thank you.
Always nice when the physics backs up what commentators suggest. Certainly not always the case.
You have two things going on there. The closer you hit to center, the more speed you get, but you also have a lower spin/speed ratio. You do have less spin as a result, but it is nearly always the ratio you care about for spin transfer and other uses of side spin.
I was thinking more along the lines of miss hitting the intended contact point on the object ball. If you hit the ball fuller than you are intending it won't react the way you expect.
I understand throw, kind of
CeaseLess
Active member
I Did not know that, I will keeping that in mind from now on.
I'm not Pat, but I am online....
If you have super, extra spin on the cue ball, it slips more on the surface of the object ball. Shooting harder has the same effect.
You want to use just enough side spin that the cue ball and object ball just achieve "meshing" as they leave contact. That is, you don't want any excess side spin on the cue ball at all.
For bank shots when you are "twisting" the object ball, sometimes you should use plain stun if you have about a half-ball cut angle.
This is the reason that inside spin can actually decrease how much the object ball is thrown -- in effect all that spin makes the ball slippery.
CeaseLess
Active member
Already got that taken care of.
There should be a rise time though, right? Also on transfer there should be latency as the two balls hook up. (?) Yeah I know - insignificant but the <less english = more transfer> seems to indicate something along those lines.
Nuther thing that occurs to me is the ball might have higher forward speed when struck with less english bringing into play ball elasticity and impact moosh - increasing transfer. (?)
so how can more than half of the maximum spin produce less than the maximum spin
more than the maximum spin can and certainly would produce less than the maximum spin.
i think i get what he is saying and understand naturally but it doesnt sound right at the start.
but no one will think about that. all they will and should, imop, think about where to hit on the cueball to get the spin they want.
however if hitting directly on the object ball the more spin on the cueball will send the object ball off its straight course. it wont be less when you use more spin than half of the maximum you can apply. in theory it may be, for slippy surfaces, but practically the cueball doesn't slip more so you get less spin. that has to break down proportionately as the surfaces become contaminated.
more than the maximum spin can and certainly would produce less than the maximum spin.
i think i get what he is saying and understand naturally but it doesnt sound right at the start.
but no one will think about that. all they will and should, imop, think about where to hit on the cueball to get the spin they want.
however if hitting directly on the object ball the more spin on the cueball will send the object ball off its straight course. it wont be less when you use more spin than half of the maximum you can apply. in theory it may be, for slippy surfaces, but practically the cueball doesn't slip more so you get less spin. that has to break down proportionately as the surfaces become contaminated.
I think too much spin just "glances off" - doing nothing.
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