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11-27-2017, 08:12 AM

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Originally Posted by denwhit View Post
Look at the 2 ball on the end rail. How would you get shape on it? Here is a little video that I learned from my instructor. Just a little bit of "force follow" CB, hitting the rail first, enough to get shape. https://youtu.be/Y7cyXUkm8L4 Just a reason why you should find and learn from a valuable pool instructor. You can always learn more than you know.
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11-27-2017, 09:02 AM

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Originally Posted by skipbales View Post
Kudos to the OP for bringing up a topic that generates comments and educates us all.
Thanks for that. Here is a little FF safety play. I'm hitting the one ball straight on, no cuts and If it was just follow, the CB would go on out to the right, but the FF (force-follow) makes it go down the rail and behind the two ball. https://youtu.be/UyVQiNndAAs


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11-27-2017, 09:02 AM

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Originally Posted by Neil View Post
Here's the def. I have always gone by- It is a force follow shot when the cb is forced farther down the tangent line before the follow takes over; or when it is forced farther from the rail before the follow takes over. This is accomplished simply by adding more force, or speed, to the cb.

Max follow occurs at about 70% above the center of the cb. If you have the stripe horizontal, 70% is at the top of the stripe.

Overspin, which is the cb rotating faster than forward momentum, dissipates almost immediately as Mike Page has shown. In BC21's video, he thinks it is showing overspin. What it is showing is that the cb will have overspin with only air friction. If one watches the video closely, one can see that the cb takes one bounce and then hits the ob while still airborne. (1:02 clearly shows it off the table at impact, which also is why the cb jumps up in the air after impact)

Any time the cb is struck above center, it will have two forces acting on it. Forward momentum, and rotational momentum. Due to friction force acting on the cb from the cloth, the rotational force (or momentum) will almost immediately match the forward force and the cb will be rolling across the felt with no overspin.

Immediately upon contact with the ob, some (partial hit) to all (full hit) of the forward momentum will be transferred to the ob. Only a small portion of the rotational force will be transferred.

On a soft hit at an angle on the ob, the cb initially travels down the tangent line. at this point the forward force is greater than the rotational force. Due to friction on the cloth, the forward force is almost immediately gone, and then the rotational force is greater and the cb then rolls forward. On a soft hit, this happens almost immediately. The harder the hit, the farther the cb will travel down the tangent line before the cb goes forward. As the two forces become slightly in favor of the rotational force, you can see the cb start to curve or bend on it's path. Shortly after the bend, the directional force will be gone, and all that is left is rotational force.

The rotational force still has a large amount of force as very little of it was lost in contact with the ob. But, at all times it is in contact with the cloth, and all that is left is rotational force, the cb will be a rolling ball. How far it travels depends on the force applied to it.

edit: Just to be clear here- even on a hard hit, with the cb travelling a ways down the tangent line before you see it go forward, the cb still has the same forward speed at the beginning of the travel down the tangent line as it does at the end of the tangent line. But, because the directinal speed is more than the rotational speed, the cb appears to follow the tangent line for a long distance. It just is moving sideways faster than it is going forward at that time.
You are correct. In my video the CB has overspin on it while it is headed toward the OB. The downward force generated by the above-center hit (extreme top with a very firm stroke) is why the CB is airborne, and why it retains overspin. This CB was only a couple of feet from the OB so it didn't have enough distance to begin rolling.

A little softer, yet still firm, the CB would've been rolling instead of flying when it hit the OB. It would bounce off the table regardless of striking the ob above it's equator. The reason why is because it would be rolling very fast (high torque) when it hits the OB, and that force must go somewhere. 100% of CB momentum transfers to the object ball in accordance with the full collision, but that torque on the CB does not transfer. It causes the CB to climb the back of the OB because it has no where else to go. And when it becomes airborne, losing the frictional hold from the cloth, the torque on the ball is seen as overspin. On a softer follow shot there is no overspin created because the CB doesn't leave the table. It just has more inertia than normal and rolls farther than normal.


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11-27-2017, 09:10 AM

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Originally Posted by skipbales View Post
I understand that part. It is the point of contention in this thread. I understand it is not overspinning as it rolls. Any overspin seems to be the result of the rotation continuing after the forward roll is interrupted. My question was if a single ball could provide adequate resistance to create significant overspin at distance. I know a rail or rack of balls can and maybe the definition of force follow does not need to result in any overspin?

What causes the cue ball to already be up in the air? Is it a bounce off the table from the force of the cue hitting the ball originally? Like a partial jump shot? I know the cue ball lifts slightly on most shots but not sure when and where.
High speed video shows that there is not significant overspin at contact with a single ball. or with a rack of balls.

Now with a full hit on a rack of balls, the balls actually push back on the cb. That is why you see the cb "bounce" back off the rack, hopefully to center table. If the cb was in contact with the felt at impact with the rack, the forward rotation of the cb stays nearly the same (only a little is transferred to the ball it hit full) and does not stop at impact.

At this point, you could technically say the cb has overspin, because while the amount of forward rotation has not changed except for what is naturally lost due to friction, the cb is actually moving in a direction opposite the forward rotation. This is because the forward momentum is greater in the opposite direction than the rotational momentum and friction can overcome for a distance.

Eventually, due to friction, the forward momentum dissipates, and the rotational momentum takes over. How much rotational momentum is left determines how far the cb will now travel forward. A rack is the same effect as a rail, but the rail imparts more forward momentum to the cb.

As far as the cb in the air, that is nothing more than " for every action, there is an equal and opposite reaction". If your cue is not level with the bed of the table and it seldom is, then you actually forcing the cb down into the table. The table pushes back and since it has much more mass than the cb, the cb gives and goes airborne. How high and how far depends on the angle of the cue, and the force applied. It will lift at impact.
  
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11-27-2017, 09:20 AM

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Originally Posted by BC21 View Post
... it should be considered a forced-follow shot if any top spin imparted on the CB kicks in after it contacts the OB. ...
This is true for all follow shots. That is very fundamental to the way follow works. For a full hit, the cue ball always stops dead (or bounces forward for very hard shots where the cue ball arrives off the cloth) and then gradually accelerates forward. It just happens a lot faster for slow shots and takes a lot longer for power shots.


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11-27-2017, 09:28 AM

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Originally Posted by BC21 View Post
You are correct. In my video the CB has overspin on it while it is headed toward the OB. The downward force generated by the above-center hit (extreme top with a very firm stroke) is why the CB is airborne, and why it retains overspin. This CB was only a couple of feet from the OB so it didn't have enough distance to begin rolling.

A little softer, yet still firm, the CB would've been rolling instead of flying when it hit the OB. It would bounce off the table regardless of striking the ob above it's equator. The reason why is because it would be rolling very fast (high torque) when it hits the OB, and that force must go somewhere. 100% of CB momentum transfers to the object ball in accordance with the full collision, but that torque on the CB does not transfer. It causes the CB to climb the back of the OB because it has no where else to go.
And when it becomes airborne, losing the frictional hold from the cloth, the torque on the ball is seen as overspin. On a softer follow shot there is no overspin created because the CB doesn't leave the table. It just has more inertia than normal and rolls farther than normal.
No. The cb would have stayed on the table with no lift at all. It does not climb the ob unless hit above center, and even then it does not climb it but deflects off it at the angle determined by the part of the balls that made contact. (think tangent line)

The cb and ob ball are only in contact for about a miilisecond. During that short amount of time, a little of the forward rotational force is transferred to the ob by what is called the gear effect. Which causes the ob to spin the opposite direction as the cb. (most of the transfer is lost due to slippage, that is, lack of friction.)

Since the ob separates from the cb so quickly, much faster than the rotational force is moving the cb foward, the cb does not and can not "climb" the ob. The cb only rotates a small amount during contact but is just long enough to impart a little spin to the ob. Every time you see the cb bounce up after contact with an ob, it is because the cb was higher off the felt than the ob was.
  
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11-27-2017, 09:46 AM

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Originally Posted by Bob Jewett View Post
This is true for all follow shots. That is very fundamental to the way follow works. For a full hit, the cue ball always stops dead (or bounces forward for very hard shots where the cue ball arrives off the cloth) and then gradually accelerates forward. It just happens a lot faster for slow shots and takes a lot longer for power shots.
I understand. What I'm suggesting is that anytime the CB is struck with enough force to cause an obvious forward spin after contacting the OB it should be considered a force follow. I mean, if it's observable without a slow motion video, then it's obvious. On soft follow shots the spin may only amount to a tiny fraction of rotational torque and almost immediately the CB starts rolling along the cloth. I would discount these shots as a force follow because it is typical of a normal rolling collision. But once the speed is sufficient to cause an obvious overspin to occur, it would be considered a force follow.


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11-27-2017, 09:49 AM

Here is a little bit of force follow to hit the 9 ball into the hole. It would be very hard to run out three rails to get it done as the obstruction balls in the way. Just use a bit of FF and watch the OB bend to take it out. Force follow is an amazing thing to learn! https://youtu.be/xnv_zAXU0Bc


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11-27-2017, 10:14 AM

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Originally Posted by denwhit View Post
Thanks for that. Here is a little FF safety play. I'm hitting the one ball straight on, no cuts and If it was just follow, the CB would go on out to the right, but the FF (force-follow) makes it go down the rail and behind the two ball. https://youtu.be/UyVQiNndAAs
Something about that shot bothers me --- there doesn't seem to be a whole lot of top spin on the cb after it hits the rail. Not sure if it qualifies as a force follow or just a top spin shot with some speed. I know what you're trying to say but I think if it were really a force follow shot, based on the angle I'm seeing, the cb would hit the same rail at least once more. I'm not seeing that second kick that you get with a force follow shot when cb stops sliding and the top spin kicks in.


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11-27-2017, 10:27 AM

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Originally Posted by FranCrimi View Post
Something about that shot bothers me --- there doesn't seem to be a whole lot of top spin on the cb after it hits the rail. Not sure if it qualifies as a force follow or just a top spin shot with some speed. I know what you're trying to say but I think if it were really a force follow shot, based on the angle I'm seeing, the cb would hit the same rail at least once more. I'm not seeing that second kick that you get with a force follow shot when cb stops sliding and the top spin kicks in.
I am hitting it pretty softly, so that it hits the rail and bounces out a little bit so that it won't scratch. If I hit it harder it would do like you are writing about. Why would it go along the rail if I'm hitting the one ball directly if not force follow?


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11-27-2017, 10:36 AM

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Originally Posted by Neil View Post
No. The cb would have stayed on the table with no lift at all. It does not climb the ob unless hit above center, and even then it does not climb it but deflects off it at the angle determined by the part of the balls that made contact. (think tangent line)

The cb and ob ball are only in contact for about a miilisecond. During that short amount of time, a little of the forward rotational force is transferred to the ob by what is called the gear effect. Which causes the ob to spin the opposite direction as the cb. (most of the transfer is lost due to slippage, that is, lack of friction.)

Since the ob separates from the cb so quickly, much faster than the rotational force is moving the cb foward, the cb does not and can not "climb" the ob. The cb only rotates a small amount during contact but is just long enough to impart a little spin to the ob. Every time you see the cb bounce up after contact with an ob, it is because the cb was higher off the felt than the ob was.
The friction between the CB (with its forward momentum and stored torque) and the stationary OB does allow the CB to slighty leave the cloth. You don't see it because it is very slight, but that CB will climb the back of that OB, especially with enough impact force. In that millisecond of contact time, the balls actually mash together and create quite a large contact patch, which increases the friction between the balls and allows the torque on that CB to find a momentary escape route. Then the OB is gone and the CB is spinning when it hits the cloth again. It may only leave the table by a height of a thousandth of an inch, but it happens all the same, whether we observe it or not.


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11-27-2017, 10:53 AM

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Originally Posted by BC21 View Post
The friction between the CB (with its forward momentum and stored torque) and the stationary OB does allow the CB to slighty leave the cloth. You don't see it because it is very slight, but that CB will climb the back of that OB, especially with enough impact force. In that millisecond of contact time, the balls actually mash together and create quite a large contact patch, which increases the friction between the balls and allows the torque on that CB to find a momentary escape route. Then the OB is gone and the CB is spinning when it hits the cloth again. It may only leave the table by a height of a thousandth of an inch, but it happens all the same, whether we observe it or not.
Well, now you have me wondering if you are just trying to be contrary. Earlier, and what I responded to, you were stating that the cb would climb the ob and go off the table. Now, you are saying it will only, maybe, climb a thousandth of an inch. Which you have no evidence of.

As far as the contact point increasing substantially, take a piece of carbon paper and lean it in front of the ob. Shoot it soft, then hard. Observe the results.
  
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11-27-2017, 11:14 AM

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Originally Posted by Neil View Post
Well, now you have me wondering if you are just trying to be contrary. Earlier, and what I responded to, you were stating that the cb would climb the ob and go off the table. Now, you are saying it will only, maybe, climb a thousandth of an inch. Which you have no evidence of.

As far as the contact point increasing substantially, take a piece of carbon paper and lean it in front of the ob. Shoot it soft, then hard. Observe the results.
I've consistently said the CB will leave the table because it climbs up the back of the OB. It could be a 16th of inch or a thousandth of an inch....it doesn't matter. The point is, it climbs the ob and goes vertical if struck firm enough with top. This is basic physics, not sure how or why you chose to argue with me about it. Here's the evidence,
which clearly shows a rolling CB leaving the cloth after contacting the OB. You can also find slow motion clips showing the compression of the balls at impact, which increases contact area.

http://www.billiards.colostate.edu/h...w/HSVA-143.htm


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11-27-2017, 12:07 PM

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Originally Posted by BC21 View Post
I've consistently said the CB will leave the table because it climbs up the back of the OB. It could be a 16th of inch or a thousandth of an inch....it doesn't matter. The point is, it climbs the ob and goes vertical if struck firm enough with top. This is basic physics, not sure how or why you chose to argue with me about it. Here's the evidence,
which clearly shows a rolling CB leaving the cloth after contacting the OB. You can also find slow motion clips showing the compression of the balls at impact, which increases contact area.

http://www.billiards.colostate.edu/h...w/HSVA-143.htm
When you stated "leave the table" I read it as going off the table to the floor. Actually leaving the table. Climbing that ob, growing a pair of wings, and saying "You ain't gonna hit me no mo. I'm outta here, Adios, arrivederci, sayounara, vaarwel, and goodbye."
  
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11-27-2017, 12:33 PM

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Originally Posted by Neil View Post
High speed video shows that there is not significant overspin at contact with a single ball. or with a rack of balls.

Now with a full hit on a rack of balls, the balls actually push back on the cb. That is why you see the cb "bounce" back off the rack, hopefully to center table. If the cb was in contact with the felt at impact with the rack, the forward rotation of the cb stays nearly the same (only a little is transferred to the ball it hit full) and does not stop at impact.

At this point, you could technically say the cb has overspin, because while the amount of forward rotation has not changed except for what is naturally lost due to friction, the cb is actually moving in a direction opposite the forward rotation. This is because the forward momentum is greater in the opposite direction than the rotational momentum and friction can overcome for a distance.

Eventually, due to friction, the forward momentum dissipates, and the rotational momentum takes over. How much rotational momentum is left determines how far the cb will now travel forward. A rack is the same effect as a rail, but the rail imparts more forward momentum to the cb.

As far as the cb in the air, that is nothing more than " for every action, there is an equal and opposite reaction". If your cue is not level with the bed of the table and it seldom is, then you actually forcing the cb down into the table. The table pushes back and since it has much more mass than the cb, the cb gives and goes airborne. How high and how far depends on the angle of the cue, and the force applied. It will lift at impact.
Good point. With absolutely no angle the cue ball does not go forward and off the table, right? It goes up and straight back down then forward. It is only with angle that it flies off the table onto the floor. I watch players do this all the time trying an 8 ball break with topspin usually coupled with an indirect hit.
  
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