Visual Indicator of Maximum Throw/SpinXfr?

[Patrick Johnson]

A question for our scientifically-minded fellows:

We try to maximize the amount of throw and transferred spin from a CB/OB collision by making the CB hit the OB with just the right amount of side spin (ignoring the effects of speed and vertical spin). We estimate the right combinations, but how do we know immediately how successful we are? How can we visually confirm that we've maximized throw/spinxfr?

It seems to me that if we've produced the maximum amount of throw and transferred the maximum amount of spin, then the CB should have no side spin (in either direction) after the collision. Setting aside the difficulty of actually being able to see this in real time, am I right about what to look for?

pj
chgo

 

[dr_dave]

A question for our scientifically-minded fellows:

We try to maximize the amount of throw and transferred spin from a CB/OB collision by making the CB hit the OB with just the right amount of side spin (ignoring the effects of speed and vertical spin). We estimate the right combinations, but how do we know immediately how successful we are? How can we visually confirm that we've maximized throw/spinxfr?

It seems to me that if we've produced the maximum amount of throw and transferred the maximum amount of spin, then the CB should have no side spin (in either direction) after the collision. Setting aside the difficulty of actually being able to see this in real time, am I right about what to look for?

pj
chgo

Pat,

Check out the statements and results in:

TP A.27 - Spin transfer

The amount of spin transferred to the OB is the exact same as the amount of spin lost by the CB (from conservation of angular momentum), and the theoretical limit for the amount of spin that can be transferred is about 35%. So if you could somehow visualize that 1/3 of the CB spin is lost, then you would know you've transferred the most spin and gotten the most throw possible.

As I'm sure you know, many videos and other resources related to spin transfer can be found on the spin transfer resource page.

Regards,
Dave

 

[dr_dave]

How can we visually confirm that we've maximized throw/spinxfr?

How about a simpler answer to a simple question:

You can visually confirm when you've maximized throw and spin transfer when the OB gets thrown the most you expect for the given ball conditions. For more info, see the maximum throw resource page.

Regards,
Dave

 

[Patrick Johnson]

lThe amount of spin transferred to the OB is the exact same as the amount of spin lost by the CB (from conservation of angular momentum), and the theoretical limit for the amount of spin that can be transferred is about 35%. So if you could somehow visualize that 1/3 of the CB spin is lost, then you would know you've transferred the most spin and gotten the most throw possible.

Oh yeah, now that you mention it I think I remember that - probably too hard to see. I do try to gauge the results like you suggest, but always looking for a shorter learning curve.

So based on the rule of thumb that says about 50% of max CB spin is optimal (for a slow straight stop shot), the maximum spin that can be transferred to an OB is 50% of 35% or about 1/6 of maximum...?

Thanks,

pj
chgo

 

[Drop The Rock]

There are a lot of one pocket shots that come up that are only make-able by throwing the OB. I know how much throw I can get with a little spin at low speed (max) and I know how I can get at medium speed and high speed. My confirmation usually comes from how the ob reacts because my indication is whether it went into the pocket or not. Spin transfer is obviously different and for me personally I can only simplify in a rudimentary way with long and short rail banks. I know what banking with center results in at various speeds and I match the results of banking with spin. This might sound a little crazy but its helped a lot with understand banking with English.

 

[336Robin]

Maximum Spin

I understand exactly what you are talking about. I consider that Ive reached maximum spin when I pass the ball and object ball comes back straight off the rail. It can helpful to know how to get the results off the rail you need. You might need kiss into something while you hide the white.

There are a lot of one pocket shots that come up that are only make-able by throwing the OB. I know how much throw I can get with a little spin at low speed (max) and I know how I can get at medium speed and high speed. My confirmation usually comes from how the ob reacts because my indication is whether it went into the pocket or not. Spin transfer is obviously different and for me personally I can only simplify in a rudimentary way with long and short rail banks. I know what banking with center results in at various speeds and I match the results of banking with spin. This might sound a little crazy but its helped a lot with understand banking with English.

 

[Jal]

....
It seems to me that if we've produced the maximum amount of throw and transferred the maximum amount of spin, then the CB should have no side spin (in either direction) after the collision. Setting aside the difficulty of actually being able to see this in real time, am I right about what to look for?

Patrick,

Consider a stun shot. There's one, and only one class of cases where the CB will end up with no side spin after the collision. That occurs when using inside english and the amount of inside english (measured as surface speed at its equator) is 5/14'ths of the total change in relative surface speed at the point of contact during the collision. (The total relative surface speed is the sum of the cueball's velocity component along the tangent line, plus the additional surface speed from the inside english.) However, this cannot produce maximum throw.

Maximum throw occurs when the balls end up gearing at the very end of the collision (i.e., the total relative surface speed is reduced to zero). In order for this to take place, the cueball's velocity component along the tangent line (pre-collision) must be 1/7'th of the total relative surface speed. In the above scenario where the inside english is 5/14'ths of the relative surface speed, the CB's velocity component along the tangent line must therefore be 9/14'ths of it (1 - 5/14) - very different from 1/7'th.

Also, there are no cases where outside english results in maximum throw AND the cueball ends up with no spin. So using post-collision CB spin (or lack of it) as a gauge is not very useful.

We try to maximize the amount of throw and transferred spin from a CB/OB collision by making the CB hit the OB with just the right amount of side spin (ignoring the effects of speed and vertical spin). We estimate the right combinations, but how do we know immediately how successful we are? How can we visually confirm that we've maximized throw/spinxfr?

I don't think there's any easy way. How much spin to use for a given cut angle is calculable, and you can find the results in one of Dr. Dave's massive technical articles on throw.

Jim

 

[duckie]

I really wish people would stop using the "we" word. Cause we includes me and I do not consider any of what was stated. I am no part of any "we" when it comes to pool and how it is approached. Thought that was kinda apparent.

I know I am successful with a shot, spin or no spin, when all the balls go where I want.

Its that simple, that is the visual indicator.....the balls go where you want.

 

[HawaiianEye]

I think TOI would fix it.

 

[Patrick Johnson]

So based on the rule of thumb that says about 50% of max CB spin is optimal (for a slow straight stop shot), the maximum spin that can be transferred to an OB is 50% of 35% or about 1/6 of maximum...?

If this is true, then a test can be devised based on how much a bank changes angle with transferred spin vs. how much a kick does with 1/6 of maximum tip offset. On my home room table that's about a half diamond cross table.

pj
chgo

 

[Jal]

...
So based on the rule of thumb that says about 50% of max CB spin is optimal (for a slow straight stop shot), the maximum spin that can be transferred to an OB is 50% of 35% or about 1/6 of maximum...?

True, but probably slightly more as in 1/5'th.

Jim

 

[book collector]

If this is true, then a test can be devised based on how much a bank changes angle with transferred spin vs. how much a kick does with 1/6 of maximum tip offset. On my home room table that's about a half diamond cross table.

pj
chgo

It seems like a robotic setup like Bob Meucci used, would probably work, at least it seems so in my mind.
What do you think?

 

[Patrick Johnson]

If this is true, then a test can be devised based on how much a bank changes angle with transferred spin vs. how much a kick does with 1/6 of maximum tip offset. On my home room table that's about a half diamond cross table.

pj
chgo

It seems like a robotic setup like Bob Meucci used, would probably work, at least it seems so in my mind.
What do you think?

I think it can be tested informally without special equipment. Just set up a shot 90 degrees into the rail and see how much angle you can get on the rebound (or how much throw into the rail). A practice technique.

pj
chgo

 

[Drop The Rock]

I think it can be tested informally without special equipment. Just set up a shot 90 degrees into the rail and see how much angle you can get on the rebound (or how much throw into the rail). A practice technique.

pj
chgo

This would be interesting two tips/max offset 90 deg. into the side rail with just the cueball is a little over two diamonds. It would be interesting to see which comes closer to that: Max spin transfer, max throw or the best of both.

 

[Patrick Johnson]

This would be interesting two tips/max offset 90 deg. into the side rail with just the cueball is a little over two diamonds.

That's how it works on my table too - if you define "two tips" as 2/3 of maximum side. See my writeup of that in this old thread: Calibrating Sidespin

It would be interesting to see which comes closer to that: Max spin transfer, max throw or the best of both.

I plan to find out. By the way, I think max spin transfer and max throw are achieved together in the same shot - because they're caused by the same thing. And it can be accomplished with pure cut angle (half ball with stun), pure side spin (straight shot with 50% side spin) and the right combinations of both (between straight and half ball).

pj
chgo

 

[app4dstn]

a question for our scientifically-minded fellows:

We try to maximize the amount of throw and transferred spin from a cb/ob collision by making the cb hit the ob with just the right amount of side spin (ignoring the effects of speed and vertical spin). We estimate the right combinations, but how do we know immediately how successful we are? How can we visually confirm that we've maximized throw/spinxfr?

It seems to me that if we've produced the maximum amount of throw and transferred the maximum amount of spin, then the cb should have no side spin (in either direction) after the collision. Setting aside the difficulty of actually being able to see this in real time, am i right about what to look for?

Pj
chgo

patrick,

consider a stun shot. There's one, and only one class of cases where the cb will end up with no side spin after the collision. That occurs when using inside english and the amount of inside english (measured as surface speed at its equator) is 5/14'ths of the total change in relative surface speed at the point of contact during the collision. (the total relative surface speed is the sum of the cueball's velocity component along the tangent line, plus the additional surface speed from the inside english.) however, this cannot produce maximum throw.

Maximum throw occurs when the balls end up gearing at the very end of the collision (i.e., the total relative surface speed is reduced to zero). In order for this to take place, the cueball's velocity component along the tangent line (pre-collision) must be 1/7'th of the total relative surface speed. In the above scenario where the inside english is 5/14'ths of the relative surface speed, the cb's velocity component along the tangent line must therefore be 9/14'ths of it (1 - 5/14) - very different from 1/7'th.

Also, there are no cases where outside english results in maximum throw and the cueball ends up with no spin. So using post-collision cb spin (or lack of it) as a gauge is not very useful.

I don't think there's any easy way. How much spin to use for a given cut angle is calculable, and you can find the results in one of dr. Dave's massive technical articles on throw.

Jim

man, i was confused by the OP. then confused by the reply. then i figured out you guys were actually geared up, as jal knew where to substitute "maximum" and "optimal" and still stay in conversational step..... jal, do my taxes!

so, borrowing and extending, or re-factoring the original question.... how do you know you've delivered the optimal gearing English to cancel throw? more specifically, how can i tell if i delivered the amount i intended, if i'm trying to follow the "40% of..." rule?

what can i look at, to discern whether i A) undercut and overthrew, B) overcut and underthrew, or C) delivered as intended? (the extent of my precision is a binary determination of spinning or not using a ball with a stripe).

now having digested jal's post.... for this case, there's nothing i can learn from observing the cue ball with my own eyes (CB spinning w outside before collision -> CB spinning w outside after collision).

and in thinking about it further, there's no way to differentiate between A,B,C based upon OB observation. (assuming the OB is headed to *exactly* the same spot each time, not within pocket tolerances. no tangential force == no sidespin). tell me i'm wrong (please!).... i'd love to know how.

great original question. what signals are there? besides a pre-aligned striped ball that may/not spin when moving?

 

[Drop The Rock]

That's how it works on my table too - if you define "two tips" as 2/3 of maximum side. See my writeup of that in this old thread: Calibrating Sidespin


I plan to find out. By the way, I think max spin transfer and max throw are achieved together in the same shot - because they're caused by the same thing. And it can be accomplished with pure cut angle (half ball with stun), pure side spin (straight shot with 50% side spin) and the right combinations of both (between straight and half ball).

pj
chgo

Hmmm you may be right, however I think there is variation with speed. Max throw and spin transfer may occur at the same amount of speed and spin (50% spin) however the spin retained after contact with the rail will be dependent up speed/distance from the rail. Is it possible that CIT can increase spin transfer theoretically if you cross bank with a 15 deg. or less angle?

It would be interesting to experiment with this as well. I do know that some short rail banks while producing negligible throw, will produce significant spin transfer.

There is also banking a ball with inside in order to shorten the angle. The results may be the same but they have always seemed more relative to speed just as much as ideal spin.