English Transfer
- Thread starter Jal
- Start date
Hmm...I'm not quite sure. At first I voted that the amount of english transfer is independent of CB speed.Jal said:
But thinking about it again, I think the harder you hit it, the more english will transfer. This seems contrary to observable data, but let me explain...
The amount of english transfer is all dependent on the frictional force (Ff) tangent to the surface of the OB at the contact point. The frictional force is simply the coefficient of friction (mu) multiplied by the normal force (Fn), which is the force at the contact point perpendicular to the surface of the ball. So Ff = Fn * mu.
I think it's safe to assume the coefficient of friction remains constant for a fixed CB spin rate. However, the normal force is proportional to the velocity of the CB. That means, the frictional force is also proportional to the velocity of the CB, given the equation in the previous paragraph. So because the frictional force increases with CB speed, so does the amount of english transfer.
Now, I think the reason why this might sound contrary to what you observe on the pool table is because you can really only observe transferred english after the OB hits a rail. You might observe that the slower you hit a shot with english, the greater the deflection angle off the rail. It's because the spin "takes" more on the rails for the slower the shot. As Mr. Lucky said, "the slower hit allows to catch the cloth more."
So I would agree that if you hit the OB full with english, then in general the OB would rebound off the rail at a greater angle the slower you hit it. However, that doesn't mean that more english is transferred for the slower cases. Quite the contrary actually. It just means the slower you hit the shot, the more the spin will "take" on the rails, which would increase the rebounding the angle.
Jal, it's been a while, so how off am I? Right or wrong, it's nice to fiinally respond to a physics-related thread. Thanks.
TrickyRicky
happy, happy, joy, joy!
Oh yah, well e=mc squared
!!!!!
My opinion is based on simple (a few decades of) observation.
The quicker the (true) stroke is through the cue ball, the faster the cue ball will spin.
BUTTTTT, that is only part of the formula. The cue ball rotation will have to slow down some before it will allow the friction between it and the cloth to optimally grab it and fully result in the desired effect. During this "free spin time" the cueball will travel in a relatively straight path in the direction down table that it was hit until the ball rotation slows down enough for the optimal friction to grab and scoot it in the direction of the spin. No rocket science here--but it sure looks cool
Thus, a masse player may use any number of stroke speeds needed to attain the desired english which can "kick in" at different travel positions of the path of the cue ball, hence finally sending it in the desired direction, whether that be 2 inches or 2 feet out from the initial point of contact.
The same principle applies to rail and object ball "grab". If the cueball is spinning faster than optimal friction it won't "grab" as much, and if it's spinning slower than optimal it will "grab" more than if it was spinning too fast, but less than if spinning at the optimal speed. Thus, a slow to medium stroke will usually allow the cue ball to "grab" rather than "slip", and this will result in a more reliable shot.
So my answer to your poll, as it is to most of the forum polls, is that it depends on the desired result. Slow, fast, medium-- advanced players can use them all, hopefully as little as possible
And, for what it's worth, if you're finding yourself playing with a lot of left and right english, or two foot draw, you probably need to work harder on your cue ball speed control.
Rick P.
!!!!!My opinion is based on simple (a few decades of) observation.
The quicker the (true) stroke is through the cue ball, the faster the cue ball will spin.
BUTTTTT, that is only part of the formula. The cue ball rotation will have to slow down some before it will allow the friction between it and the cloth to optimally grab it and fully result in the desired effect. During this "free spin time" the cueball will travel in a relatively straight path in the direction down table that it was hit until the ball rotation slows down enough for the optimal friction to grab and scoot it in the direction of the spin. No rocket science here--but it sure looks cool
Thus, a masse player may use any number of stroke speeds needed to attain the desired english which can "kick in" at different travel positions of the path of the cue ball, hence finally sending it in the desired direction, whether that be 2 inches or 2 feet out from the initial point of contact.
The same principle applies to rail and object ball "grab". If the cueball is spinning faster than optimal friction it won't "grab" as much, and if it's spinning slower than optimal it will "grab" more than if it was spinning too fast, but less than if spinning at the optimal speed. Thus, a slow to medium stroke will usually allow the cue ball to "grab" rather than "slip", and this will result in a more reliable shot.
So my answer to your poll, as it is to most of the forum polls, is that it depends on the desired result. Slow, fast, medium-- advanced players can use them all, hopefully as little as possible
And, for what it's worth, if you're finding yourself playing with a lot of left and right english, or two foot draw, you probably need to work harder on your cue ball speed control.
Rick P.
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I agree that more spin is transferred to the object ball at a slow speed, but I am not sure that as much as 18% can be transferred from ball to ball. For that reason, 6% with a medium stroke sounds more accurate, although perhaps a tad high.
I'd love to hear some of the instructors weigh in on this!
I'd love to hear some of the instructors weigh in on this!
Thanks JSP and to all who have responded and voted thus far. I always enjoy reading your well thought out posts and am glad that you're still on the forum.jsp said:
I would like to give the poll a little more time so I'll probably post what I think is the right answer later tonight. If anyone feels that one of the choices resonates a little more than the others, please vote. Don't feel bad if you get it wrong. I believe I've had a fair understanding of throw/spin for some time now, but up until a few days ago would have picked the wrong one. Though the choices may seem kind of tough, three of them are way off.
Thanks again.
Jim
jsp said:Hmm...I'm not quite sure. At first I voted that the amount of english transfer is independent of CB speed.
But thinking about it again, I think the harder you hit it, the more english will transfer. This seems contrary to observable data, but let me explain...
The amount of english transfer is all dependent on the frictional force (Ff) tangent to the surface of the OB at the contact point. The frictional force is simply the coefficient of friction (mu) multiplied by the normal force (Fn), which is the force at the contact point perpendicular to the surface of the ball. So Ff = Fn * mu.
I think it's safe to assume the coefficient of friction remains constant for a fixed CB spin rate. However, the normal force is proportional to the velocity of the CB. That means, the frictional force is also proportional to the velocity of the CB, given the equation in the previous paragraph. So because the frictional force increases with CB speed, so does the amount of english transfer.
Now, I think the reason why this might sound contrary to what you observe on the pool table is because you can really only observe transferred english after the OB hits a rail. You might observe that the slower you hit a shot with english, the greater the deflection angle off the rail. It's because the spin "takes" more on the rails for the slower the shot. As Mr. Lucky said, "the slower hit allows to catch the cloth more."
So I would agree that if you hit the OB full with english, then in general the OB would rebound off the rail at a greater angle the slower you hit it. However, that doesn't mean that more english is transferred for the slower cases. Quite the contrary actually. It just means the slower you hit the shot, the more the spin will "take" on the rails, which would increase the rebounding the angle.
Jal, it's been a while, so how off am I? Right or wrong, it's nice to fiinally respond to a physics-related thread. Thanks.
NERRRRRRRRRRRD!!!!!
j/k
Actually I think that a ball rebounds from a cushion at a wider angle when hit slow is because the rubber doesn't compress as much and doesn't "catch" the ball. Have you ever seen a slow-mo video of a ball hitting a cushion even at a medium speed? The cushion compresses like crazy, the ball actually seems to go 'inside' the cushion a bit. The harder the ball is hit, the more the cushion would compress and "catch" the ball.
The most spin transfer is at slower speed and less than a tip of english. I didn't vote, don't want to calculate percentage even if I could. Thats for the rocket scientists to figure out. LOL
Can I throw and spin this ball enough? Yep, looks like I can. OOPS, hit it a little hard and it went short. LOL
Rod
Can I throw and spin this ball enough? Yep, looks like I can. OOPS, hit it a little hard and it went short. LOL
Rod
Slower CB speed creates increased transfer. I am no expert, but I do know that. An easy trial would be to place to OB's on the table with one in front of the other a few inches. Then place the CB approx 8" away, but in direct line with the other two balls. Hit the CB as you would if you were going to draw it. Doing this will of course impart top spin on the first OB. Do this at varying speeds. Measure which one allows the 1st OB to roll farther due to top spin. You will see that a slower speed (of course it can not be too slow) will create more top spin on the 1st OB, and make it roll farther. Have fun!
jsp said:
I toyed around a bit with this a while back, and discovered that no matter how slow or hard I played the cue-ball I couldn't get the objectball to rotate anywhere near half a rotation before hitting a rail, so I suspect the answer is nearest the 2%, I think it doesn't matter much how hard or soft you play the shot.
gr. Dave
Would seem to me that the 2% transfer is about right.
That accords with Robert Byrne's estimation as I recall.
I would have assumed that there would be more transfer at a slow forward speed and medium offset of CB spin....in the zone where one would expect maximum throw for a full ball shot.
Looking forward to hearing Jal's insights.
Colin
That accords with Robert Byrne's estimation as I recall.
I would have assumed that there would be more transfer at a slow forward speed and medium offset of CB spin....in the zone where one would expect maximum throw for a full ball shot.
Looking forward to hearing Jal's insights.
Colin
Again, thanks to everyone for taking a stab at it. Without further ado, and I realize some of you will probably find this very hard to believe, the last choice of 35% at any speed is the correct one (actually 35.7% according to theory). I'm amazed that this received as many votes as it did - I would have chosen 18% at slow speed since, unquestionably, you can get more throw at slower speeds and the induced spin is directly related to throw, as several posters have indicated.
I'll try to prove why this is so with a minimum of math.
When friction is applied to the surface of a ball, say by another ball rubbing against it, both balls will undergo a change in linear speed (movement as a whole) and spin rate such as to reduce the relative surface speed (rubbing action) between them. If one ball is stationary to begin with, then "change" means that it will acquire a non-zero linear speed and spin. Because they're spheres, for every unit of reduction in surface speed due to a change in linear speed of either ball, there will be 2.5 units of reduction due to a change in spin. (It's easier to spin a sphere than to move it as a whole when trying to alter surface speed.)
So the total reduction in surface speed from the changes to both ball's linear speeds and spins will be 1 + 2.5 + 1 + 2.5 = 7 units. Looking at it another way, the change in linear speed of either ball will contribute to 1/7 'th of the reduction in surface speed, while the change in spin will contribute 2.5/7 'ths (= 5/14) of it.
Okay we're just about there. If the intial relative surface speed isn't too great to begin with, it will be brought to zero during the collision and the balls will end up rolling across each other. At this point the friction is reduced to zero (or extremely close to zero). Both balls will have undergone a change in speed and spin so as to exactly cancel the initial surface speed. By the previous paragraph, we know that the stationary ball will now have 5/14 'ths of this initial surface speed as spin. And since in the case of a full ball hit the initial surface speed is due only to the spin on the cueball, the object ball will acquire 5/14 'ths or 35.7% of its spin.
This is true for any case where the balls end up rolling across each other. It also happens to be true in the case of maximum throw, because this occurs when the balls reach the rolling state just at the very end of the collision period.
Notice that no mention was made of the cueball's speed in the above argument. As far as the percentage transfer goes, it's irrelevant. However, it should be mentioned that the higher the cueball's speed, the more spin that can be imparted to the object ball. To get maximum throw, a slow speed shot requires a tip offset of about 1/2 of maximum (it varies some with speed). On a high speed shot, say 4X the slow speed one, it's roughly 1/4 of maximum. Both will transfer 35.7% of the cueball's spin, but this will be about 2X greater for the high speed shot (four times the speed but one-half the tip offset yields a 2X spin rate). But the spin/speed ratio of the object ball will be 1/2 that of the slow speed shot.
One practical application and test of this, similar to mantis99's suggestion, is when two object balls are very close to each other, say an inch or less, and you want the first object ball to travel as far forward as possible after the in-line combo is executed. Hitting hard with about a quarter of maximum draw should get the best results (no, I haven't tried it, sorry). With the object balls that close, the first one won't have much time to pick up additional topspin from the cloth even when hit slow.
I think both Bob Jewett and Dr. Dave will or already have done articles on this in Billiards Digest (see chefjeff's post). Unlike me, they will probably have tested it. One or more of Dr. Dave's videos might show this but I haven't done the leg work.
Jim
(Great to see Colin back!)
I'll try to prove why this is so with a minimum of math.
When friction is applied to the surface of a ball, say by another ball rubbing against it, both balls will undergo a change in linear speed (movement as a whole) and spin rate such as to reduce the relative surface speed (rubbing action) between them. If one ball is stationary to begin with, then "change" means that it will acquire a non-zero linear speed and spin. Because they're spheres, for every unit of reduction in surface speed due to a change in linear speed of either ball, there will be 2.5 units of reduction due to a change in spin. (It's easier to spin a sphere than to move it as a whole when trying to alter surface speed.)
So the total reduction in surface speed from the changes to both ball's linear speeds and spins will be 1 + 2.5 + 1 + 2.5 = 7 units. Looking at it another way, the change in linear speed of either ball will contribute to 1/7 'th of the reduction in surface speed, while the change in spin will contribute 2.5/7 'ths (= 5/14) of it.
Okay we're just about there. If the intial relative surface speed isn't too great to begin with, it will be brought to zero during the collision and the balls will end up rolling across each other. At this point the friction is reduced to zero (or extremely close to zero). Both balls will have undergone a change in speed and spin so as to exactly cancel the initial surface speed. By the previous paragraph, we know that the stationary ball will now have 5/14 'ths of this initial surface speed as spin. And since in the case of a full ball hit the initial surface speed is due only to the spin on the cueball, the object ball will acquire 5/14 'ths or 35.7% of its spin.
This is true for any case where the balls end up rolling across each other. It also happens to be true in the case of maximum throw, because this occurs when the balls reach the rolling state just at the very end of the collision period.
Notice that no mention was made of the cueball's speed in the above argument. As far as the percentage transfer goes, it's irrelevant. However, it should be mentioned that the higher the cueball's speed, the more spin that can be imparted to the object ball. To get maximum throw, a slow speed shot requires a tip offset of about 1/2 of maximum (it varies some with speed). On a high speed shot, say 4X the slow speed one, it's roughly 1/4 of maximum. Both will transfer 35.7% of the cueball's spin, but this will be about 2X greater for the high speed shot (four times the speed but one-half the tip offset yields a 2X spin rate). But the spin/speed ratio of the object ball will be 1/2 that of the slow speed shot.
One practical application and test of this, similar to mantis99's suggestion, is when two object balls are very close to each other, say an inch or less, and you want the first object ball to travel as far forward as possible after the in-line combo is executed. Hitting hard with about a quarter of maximum draw should get the best results (no, I haven't tried it, sorry). With the object balls that close, the first one won't have much time to pick up additional topspin from the cloth even when hit slow.
Sometimes instinct is better than reason JSP.jsp said:
As usual, no. This is how I like to get myself into trouble. But there is indirect supporting evidence from tests of throw itself. Dr. Dave Alciatore, who I'm sure you and many are familiar with, derives a figure of 25% for the transfer here. But he uses a tip offset which is more "typical" (ie, 1/2 of maximum on a slow shot) rather than the exact value at a given speed which produces the absolute maximum amount of throw and spin on the object ball.jsp said:
I think both Bob Jewett and Dr. Dave will or already have done articles on this in Billiards Digest (see chefjeff's post). Unlike me, they will probably have tested it. One or more of Dr. Dave's videos might show this but I haven't done the leg work.
Jim
(Great to see Colin back!)
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Wow...did I miss the boat on this this poll!
I wasn't even paying attention to those percentage numbers next to each option (2%, 8%, 16%, 35%). For some reason, my brain just thought those were the calculated percentages of the pollers picking that option. I guess a red flag would have been raised if I bothered adding up those percentages.
I thought the poll was asking which shot gives you the most english transfer, which is evident by my ramblings in my initial post. Ooops.
Finally, looking at the english transfer percentages for what they really do mean, then I would never have defended the 2% option...hehe.
When I have time, I'll look into your derivation a bit more Jal. Sorry for the goof up.
jsp <~~~~ Probably not the only one that made such a gaff.
I wasn't even paying attention to those percentage numbers next to each option (2%, 8%, 16%, 35%). For some reason, my brain just thought those were the calculated percentages of the pollers picking that option. I guess a red flag would have been raised if I bothered adding up those percentages.
I thought the poll was asking which shot gives you the most english transfer, which is evident by my ramblings in my initial post. Ooops.
Finally, looking at the english transfer percentages for what they really do mean, then I would never have defended the 2% option...hehe.
When I have time, I'll look into your derivation a bit more Jal. Sorry for the goof up.
jsp <~~~~ Probably not the only one that made such a gaff.
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Handsumm
Banned
There are a lot of factors involved in trying to calculate induced spin due to contact with a spinning cue ball ( throw).
Have you ever considered how clean, polished, dirty, chipped, or scratched balls affect the coefficient of friction of balls?
The best answer I have seen to this question are answered in Byrne's Complete Book of Pool Shots. In this he explains that the coefficient of friction between two (clean) balls is ~0.02. This means that roughly 2% of the cue balls spin is transferred to the second ball at contact.
For the OP's question (if I understood it correctly) the amount of spin transferred is dependent on the coefficient of friction, the spin (rotational velocity),and Impulse (how long the cue ball is in contact with the object ball and its force in that direction).
http://en.wikipedia.org/wiki/Impulse
... the impulse of a force is the product of the force and the time during which it acts. Although momentum is conserved within a closed system, individual parts of a system can undergo changes in momentum. Impulse has the same units and dimensions as momentum (kg m/s or N·s = Huygens Hy). The impulse of a time-varying force is calculated as the integral of force with respect to time:
Therefore, if a cue ball is spinning at a certain rotational velocity, it should impart MORE spin on the object ball if it is in contact with it longer, which would be true if it was moving at a slower linear velocity.
Have you ever considered how clean, polished, dirty, chipped, or scratched balls affect the coefficient of friction of balls?
The best answer I have seen to this question are answered in Byrne's Complete Book of Pool Shots. In this he explains that the coefficient of friction between two (clean) balls is ~0.02. This means that roughly 2% of the cue balls spin is transferred to the second ball at contact.
For the OP's question (if I understood it correctly) the amount of spin transferred is dependent on the coefficient of friction, the spin (rotational velocity),and Impulse (how long the cue ball is in contact with the object ball and its force in that direction).
http://en.wikipedia.org/wiki/Impulse
... the impulse of a force is the product of the force and the time during which it acts. Although momentum is conserved within a closed system, individual parts of a system can undergo changes in momentum. Impulse has the same units and dimensions as momentum (kg m/s or N·s = Huygens Hy). The impulse of a time-varying force is calculated as the integral of force with respect to time:
Therefore, if a cue ball is spinning at a certain rotational velocity, it should impart MORE spin on the object ball if it is in contact with it longer, which would be true if it was moving at a slower linear velocity.