Preserving the angle

So as I read the curves, with six inches of separation you can get the cue ball to stop dead and cut the object ball four degrees. I usually demonstrate the challenge with a one-ball (2.25 inch) separation.
Yes Bob, assuming, of course, I did the math right.
Jim and Bob,

The analysis doesn't include the effects of swerve.

FYI, I just added a plot at the bottom of TP A.29 - Using throw to limit cue ball motion that shows that theoretically (neglecting swerve), it is possible to hold the CB and create an OB angle at any CB-OB distance. However, at the table, there is a practical limit to the CB-OB distances over which this can be done. The analysis by Jim and in TP A.29 ignores the effects of swerve. Swerve changes the effective cut angle of the shot, making it more difficult to hold the CB. And at larger distances, swerve becomes more of a factor. Also, stun (for maximum throw) is more difficult to control at larger distances. Also, it is much more difficult to judge squirt and swerve and be accurate with the hit at larger distances.

Therefore, Bob's challenge is safe. Nobody will be able to hold the CB at large distances between the CB and OB (although, if they cheat by adding chalk at the contact point, making larger throw possible, they can push the distance much farther than normal ;)).

Regards,
Dave
 
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So as I read the curves, with six inches of separation you can get the cue ball to stop dead and cut the object ball four degrees. I usually demonstrate the challenge with a one-ball (2.25 inch) separation.
Is there a video of this?
The first shot in NV B.21 - Straight shot squirt, swerve, and throw clearly shows both the CB and OB moving to the right with a straight initial alignment (at 2-3 balls of separation). Therefore, it is easy to make the CB stop and still create an OB angle with throw.

Regards,
Dave
 
I've been okay and good to see you posting every once in a while. Ah, for the good old days! Please forgive my long delay in responding, but I wanted to take a fresh look at the problem given the issues you raise below. I based my statement on the results of a program of several years ago where, as I recall, the CB speed changes seemed to correlate with the changes in the sine of the ghostball angle. However, I believe I must have been looking at using inside english to increase the cut angle and increase CB speed down the tangent line, rather than outside to slow it down. There are some cases where there is a somewhat strong correlation (at least, that's my story!).

Well, I think you're essentially right, though I believe the reduction in ghostball angle has something to do with it. But it alone does not provide an accurate characterization of the speed reduction, contrary to what I confidently exclaimed earlier! Here's a graph showing the fractional reduction in speed down the tangent line as compared to the "predicted curve" (red) using only the sines of the ghostball angles.

View attachment 387161

Not good! In fact the change in ghostball angle grossly overestimates the reduction (at least in this case where we're assuming 5 degrees of throw).

I'm sorry jsp, I'm just not grasping your argument, perhaps because of my general confusion concerning the whole issue. One thing I am sure of, I was wrong and I thank you for challenging my assumptions.

Since we're on the subject (or were almost a week ago ), here are some plots at different separations between the centers of the CB and OB. All are for stun shots. They indicate the CB's speed reduction down the tangent line when using outside english to "reverse throw" the OB by 5 degrees, as compared to using no english with normal throw. With the outside english, the geometric cut angle as well as the CB's speed are reduced so that the OB's velocity (speed and direction) is identical to the no english case. This makes for a fair comparison.

View attachment 387162

As the graphs indicate, the reduction is very significant, even at the larger separations, for very small cut angles. (Of course, you're reducing something very small to begin with, but reducing it a lot, nevertheless. :)) Where the curves drop down into negative territory, both the CB and OB would be traveling on the same side of their original line of centers (Bob Jewett's challenge).

Since travel distance is proportional to the square of a ball's speed, squaring the fractional speed gives you the fractional travel distance.

The opposite effect, boosting CB tangent line speed, is shown below. Here, inside or "subgearing" english is employed to increase the geometric cut angle. The OB's velocity is again identical for the no english and inside english cases.

View attachment 387163

Jim
Great stuff Jim! Took me a while to understand what's going on in those graphs. But now that I do, it's great info.

So the middle graph shows that for a given cut angle, the reduction in velocity (for supergearing english) gets smaller for larger CB/OB separations. The reason for this is that for larger distances, the CB has to be hit harder (the CB has to travel faster) in order for it to achieve stun (CB not rolling) at impact with the OB, and because the CB has a greater velocity the magnitude of throw gets reduced. Is this reasoning correct?

EDIT: But wait, the graph states that in each case the amount of throw remains constant at 5 degrees. So maybe I still don't understand what's going on. Hmmm.

EDIT2: So given the same cut angle and the same amount of throw, doesn't the graph show that at larger CB/OB distances the OB has to be cut more full?
 
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all these guys can make up all these charts and shit about what ever and ramble on about them but cant figure out how to make a simple pro1 shot hahahaha

Have to love how bitter AZ billiard people can be lol
I think it's much more likely that you don't understand what they say about it. In fact I know it's the case.

pj
chgo
 
...So the middle graph shows...
Thanks for your kind words, Jsp. I'm afraid my response will be sluggish once again. I'd like to make a diagram or two in order to make things as clear as possible. But that will have to wait until late this evening (maybe longer, but hopefully not).

My apology.

Jim
 
...So the middle graph shows that for a given cut angle, the reduction in velocity (for supergearing english) gets smaller for larger CB/OB separations....
Jsp, on the off chance you're still looking for some clarification, I hope this might help.

The reason the CB's post-impact tangent line speed is reduced less at larger CB-OB separations has to do entirely with the ghostball angles. To see this let's step back a moment. First, a reminder of a basic theorem from geometry: an exterior angle of a triangle is equal to the sum of the opposite interior angles.

View attachment 388784

The angles are labeled to reflect shot geometry: C is the geometric cut angle (cut angle sans throw); P is the corresponding impact angle (Dr. Dave's term) between the line of centers of the CB/OB and the direction the OB would take without throw; and G is the ghostball angle.

Note that if you vary P while keeping the lengths 2R and d fixed, then G co-varies with P; increasing P increases G and vise versa. This is true up to where C (= P + G) reaches 90 degrees.The important point here is that C changes by a larger amount than P because of this covariance of G.

These angles are shown below for the two cases. Case 1 (C1, P1, G1) represents using no english, while Case 2 (C2, P2, G2) represents using outside english and throw to send the OB down the same path and at the same speed as in Case 1.

View attachment 388785

Throw angles T1 (varies with cut angle) and T2 (always equal to 5 degrees for the graphs) are also indicated.

Only two assumptions are made regarding the CB's pre-impact speed. One is that for Case 1, it lies in the range of slow to moderate such that the balls end up gearing during the collision. This makes it both easier to calculate the throw angle T1, and, more importantly, T1 is independent of the specific speed of the CB at any particular cut angle. (The throw velocity component added to the OB and subtracted from the CB is always 1/7'th of the CB's velocity component in the tangent line direction.)

Second, for Case 2, the CB's speed is adjusted so that the OB's speed is the same as in Case 1. Adjustments are required because of the generally thicker hit with Case 2 (but actually thinner at very small cut angles around a couple of degrees and less) and the different throw velocities imparted to the OB between Cases 1 and 2 (T2 versus T1). However, these adjustments are quite small/negligible.

There are four things then that enter into the CB's post-impact speed down the tangent line in Case 2. Two are the just mentioned CB speed adjustments to equalize the OB's speed, one is the added tangent line velocity component from the throw T2 (equal but opposite forces), and the last is the change in the cut angle from C1 to C2.

In the last diagram above, the impact angle P is altered (reduced) from P1 to P2 to compensate for the throw angles T1 and T2. In the course of doing this, because of the covariance of the change in the ghostball angles (G2-G1), the cut angle changes by a larger amount than the impact angle. The closer the balls are together and the smaller the cut angle, the greater is this disparity between a change in P and a change in C. This, plus the fact that the CB's added tangent line velocity from T2 doesn't vary much with P, results in the general shape of the curves. (It would take a little math here to back those statements up.)

If the CB and OB are an infinite distance apart, the ghostball angle is always zero no matter where the CB strikes the OB, as is the change in the ghostball angle from Case 1 to Case 2. Thus, the change in the cut angle C is exactly equal to the change in the impact angle P. Below is plot of the CB's post-impact speed reduction (sans the small speed adjustments to equalize the OB's speed):

View attachment 388786

Without the ghostball angles coming into play, the reduction is nil.

Next, and finally (yay!), is a graph showing the separate factors mentioned above which contribute to the CB's post-impact velocity (the green line combines the cut angle reduction with the added throw velocity from T2).

View attachment 388787

I hope all of this sheds a little light on the subject and doesn't totally lie in the realm of "more than I wanted to know." Before seeing the last graph above, I thought the CB's pre-impact speed adjustments for the varying throw (T2 versusT1 across the range of cut angles) would be much larger and a major contributor to the shape of a curve. Obviously, that's not the case.

As Dr. Dave indicated, swerve tends to vitiate all of this; the father the balls are apart, the worse it gets (even converting a reduction in CB speed to a boost!) So the graphs are only strictly valid where a cue's intrinsic pivot point is at the same height as the center of the cueball. In that case, vertical squirt keeps the applied force level and the vertical spin axis truly vertical (i.e., no swerve).

Jim
 
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Good work Jim. Thank you for sharing it.

Catch you later,
Dave

Jsp, on the off chance you're still looking for some clarification, I hope this might help.

The reason the CB's post-impact tangent line speed is reduced less at larger CB-OB separations has to do entirely with the ghostball angles. To see this let's step back a moment. First, a reminder of a basic theorem from geometry: an exterior angle of a triangle is equal to the sum of the opposite interior angles.

View attachment 388784

The angles are labeled to reflect shot geometry: C is the geometric cut angle (cut angle sans throw); P is the corresponding impact angle (Dr. Dave's term) between the line of centers of the CB/OB and the direction the OB would take without throw; and G is the ghostball angle.

Note that if you vary P while keeping the lengths 2R and d fixed, then G co-varies with P; increasing P increases G and vise versa. This is true up to where C (= P + G) reaches 90 degrees.The important point here is that C changes by a larger amount than P because of this covariance of G.

These angles are shown below for the two cases. Case 1 (C1, P1, G1) represents using no english, while Case 2 (C2, P2, G2) represents using outside english and throw to send the OB down the same path and at the same speed as in Case 1.

View attachment 388785

Throw angles T1 (varies with cut angle) and T2 (always equal to 5 degrees for the graphs) are also indicated.

Only two assumptions are made regarding the CB's pre-impact speed. One is that for Case 1, it lies in the range of slow to moderate such that the balls end up gearing during the collision. This makes it both easier to calculate the throw angle T1, and, more importantly, T1 is independent of the specific speed of the CB at any particular cut angle. (The throw velocity component added to the OB and subtracted from the CB is always 1/7'th of the CB's velocity component in the tangent line direction.)

Second, for Case 2, the CB's speed is adjusted so that the OB's speed is the same as in Case 1. Adjustments are required because of the generally thicker hit with Case 2 (but actually thinner at very small cut angles around a couple of degrees and less) and the different throw velocities imparted to the OB between Cases 1 and 2 (T2 versus T1). However, these adjustments are quite small/negligible.

There are four things then that enter into the CB's post-impact speed down the tangent line in Case 2. Two are the just mentioned CB speed adjustments to equalize the OB's speed, one is the added tangent line velocity component from the throw T2 (equal but opposite forces), and the last is the change in the cut angle from C1 to C2.

In the last diagram above, the impact angle P is altered (reduced) from P1 to P2 to compensate for the throw angles T1 and T2. In the course of doing this, because of the covariance of the change in the ghostball angles (G2-G1), the cut angle changes by a larger amount than the impact angle. The closer the balls are together and the smaller the cut angle, the greater is this disparity between a change in P and a change in C. This, plus the fact that the CB's added tangent line velocity from T2 doesn't vary much with P, results in the general shape of the curves. (It would take a little math here to back those statements up.)

If the CB and OB are an infinite distance apart, the ghostball angle is always zero no matter where the CB strikes the OB, as is the change in the ghostball angle from Case 1 to Case 2. Thus, the change in the cut angle C is exactly equal to the change in the impact angle P. Below is plot of the CB's post-impact speed reduction (sans the small speed adjustments to equalize the OB's speed):

View attachment 388786

Without the ghostball angles coming into play, the reduction is nil.

Next, and finally (yay!), is a graph showing the separate factors mentioned above which contribute to the CB's post-impact velocity (the green line combines the cut angle reduction with the added throw velocity from T2).

View attachment 388787

I hope all of this sheds a little light on the subject and doesn't totally lie in the realm of "more than I wanted to know." Before seeing the last graph above, I thought the CB's pre-impact speed adjustments for the varying throw (T2 versusT1 across the range of cut angles) would be much larger and a major contributor to the shape of a curve. Obviously, that's not the case.

As Dr. Dave indicated, swerve tends to vitiate all of this; the father the balls are apart, the worse it gets (even converting a reduction in CB speed to a boost!) So the graphs are only strictly valid where a cue's intrinsic pivot point is at the same height as the center of the cueball. In that case, vertical squirt keeps the applied force level and the vertical spin axis truly vertical (i.e., no swerve).

Jim
 
Jsp, on the off chance you're still looking for some clarification, I hope this might help.

The reason the CB's post-impact tangent line speed is reduced less at larger CB-OB separations has to do entirely with the ghostball angles. To see this let's step back a moment. First, a reminder of a basic theorem from geometry: an exterior angle of a triangle is equal to the sum of the opposite interior angles.

View attachment 388784

The angles are labeled to reflect shot geometry: C is the geometric cut angle (cut angle sans throw); P is the corresponding impact angle (Dr. Dave's term) between the line of centers of the CB/OB and the direction the OB would take without throw; and G is the ghostball angle.

Note that if you vary P while keeping the lengths 2R and d fixed, then G co-varies with P; increasing P increases G and vise versa. This is true up to where C (= P + G) reaches 90 degrees.The important point here is that C changes by a larger amount than P because of this covariance of G.

These angles are shown below for the two cases. Case 1 (C1, P1, G1) represents using no english, while Case 2 (C2, P2, G2) represents using outside english and throw to send the OB down the same path and at the same speed as in Case 1.

View attachment 388785

Throw angles T1 (varies with cut angle) and T2 (always equal to 5 degrees for the graphs) are also indicated.

Only two assumptions are made regarding the CB's pre-impact speed. One is that for Case 1, it lies in the range of slow to moderate such that the balls end up gearing during the collision. This makes it both easier to calculate the throw angle T1, and, more importantly, T1 is independent of the specific speed of the CB at any particular cut angle. (The throw velocity component added to the OB and subtracted from the CB is always 1/7'th of the CB's velocity component in the tangent line direction.)

Second, for Case 2, the CB's speed is adjusted so that the OB's speed is the same as in Case 1. Adjustments are required because of the generally thicker hit with Case 2 (but actually thinner at very small cut angles around a couple of degrees and less) and the different throw velocities imparted to the OB between Cases 1 and 2 (T2 versus T1). However, these adjustments are quite small/negligible.

There are four things then that enter into the CB's post-impact speed down the tangent line in Case 2. Two are the just mentioned CB speed adjustments to equalize the OB's speed, one is the added tangent line velocity component from the throw T2 (equal but opposite forces), and the last is the change in the cut angle from C1 to C2.

In the last diagram above, the impact angle P is altered (reduced) from P1 to P2 to compensate for the throw angles T1 and T2. In the course of doing this, because of the covariance of the change in the ghostball angles (G2-G1), the cut angle changes by a larger amount than the impact angle. The closer the balls are together and the smaller the cut angle, the greater is this disparity between a change in P and a change in C. This, plus the fact that the CB's added tangent line velocity from T2 doesn't vary much with P, results in the general shape of the curves. (It would take a little math here to back those statements up.)

If the CB and OB are an infinite distance apart, the ghostball angle is always zero no matter where the CB strikes the OB, as is the change in the ghostball angle from Case 1 to Case 2. Thus, the change in the cut angle C is exactly equal to the change in the impact angle P. Below is plot of the CB's post-impact speed reduction (sans the small speed adjustments to equalize the OB's speed):

View attachment 388786

Without the ghostball angles coming into play, the reduction is nil.

Next, and finally (yay!), is a graph showing the separate factors mentioned above which contribute to the CB's post-impact velocity (the green line combines the cut angle reduction with the added throw velocity from T2).

View attachment 388787

I hope all of this sheds a little light on the subject and doesn't totally lie in the realm of "more than I wanted to know." Before seeing the last graph above, I thought the CB's pre-impact speed adjustments for the varying throw (T2 versusT1 across the range of cut angles) would be much larger and a major contributor to the shape of a curve. Obviously, that's not the case.

As Dr. Dave indicated, swerve tends to vitiate all of this; the father the balls are apart, the worse it gets (even converting a reduction in CB speed to a boost!) So the graphs are only strictly valid where a cue's intrinsic pivot point is at the same height as the center of the cueball. In that case, vertical squirt keeps the applied force level and the vertical spin axis truly vertical (i.e., no swerve).

Jim
Jim, thanks for the taking the time and effort to explain this to me. I think I finally get it. Till our next conversation in which you can school me again. ;)
 
Wow! I'm glad to see this thread really go off. There's a lot of great info brought up here.

I played snooker the other night, and figured out pretty quick that it center ball is the way to go. Maybe it's the smaller balls, but throwing the ob with English was way too inaccurate for me, unlike on a pool table.
 
Here's a shortcut to preserving the angle:

viagra.jpg



You're welcome.

Jeff Livingston
 
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