Question on CB Roll for Physics Enthusiasts

Jal said:
Colin, here is the predicted ratio of distances in a nutshell. You might want to imagine a cut shot to the left with the y-axis aligned with the cueball's pre-impact direction and the x-axis pointing to the right as seen from above. Let the distances the cueball travels in the final roll direction during post-impact sliding, and then rolling, be Xcs and Xcr. Let the corresponding object ball distances be Xos and Xor. Then the ratio of distances Rd is:

Rd = (Xcs + Xcr)/(Xos + Xor)

In this ratio, the cueball's pre-impact velocity V and the gravitational acceleration g are common in the numerator and denominator and therefore cancel out. But they're included in the following formulas for the four distances.

Let Us and Ur be the coefficients of sliding friction and rolling resistance, respectively (approx. 0.2 and 0.01). Let Vcs be the cueball's initial post-impact velocity component in the direction of its final roll (as it just begins sliding along the tangent line), and Vcr be its velocity in the roll direction just as it reaches natural roll. For the object ball, let the corresponding velocities be Vos and Vor. Then, using a standard kinematic formula:

Xcs = (Vcr^2 - Vcs^2)/(2gUs)

Xcr = Vcr^2/(2gUr)

Xos = (Vos^2 - Vor^2)/(2gUs)

Xor = Vor^2/(2gUr)

The four velocities in the above are as follows. Let C be the cut angle and F the cueball's final direction measured with respect to the x-axis (positive counterclockwise). The angle F is:

F = arctan[sin(C)^2 + Bz/R)/(sin(C)cos(C))]

The quantity Bz/R is equal to 2/5 for a rolling cueball, but the equation holds for other values. That is, you can characterize its spin state as if it had been hit by an idealized squirtless cue anywhere above or below center by the fraction of its radius, Bz/R. According to the miscue limits, Bz/R ranges from +0.5 (max topspin) to -0.5 (max draw), and is 2/5 when at natural roll as just mentioned. The complement of the angle F, which is 90 - F, is close to 30 degrees for a wide range of cut angles (eg, the 30-degree rule). With this, the four velocities are:

Vcs = Vsin(C)cos(F - C)

Vcr = (5/7)Vsqrt[(1 + 2(Bz/R)sin(C)^2 + (Bz/R)^2]

Vos = Vcos(C)

Vor = (5/7)Vcos(C)

Note that sin(C)^2 denotes the square of sin(C). These can be corrected for throw (particularly the slight loss of spin during the collision) and inelasticity, but I'm not sure you're going to use them, so I'll leave it at that.

If you're going to be doing such analysis in the future, you might want to get comfy with the relation which gives the final vector velocity of a spinning ball as it reaches natural roll. If its spin is W and initial velocity V, its velocity Vr at natural roll is:

Vr = (5/7)[V - (2/5)W X R]

where R is the displacement vector from the center of the ball to the "point" of contact with the bed and X denotes the vector cross product. This is derived in one form or another by Dr. Dave and Ron Shepard in their technical articles. It can also be easily gotten by noting that 2/7'ths of the relative surface velocity between the ball and the cloth is subtracted from the ball's initial velocity (a fact which comes from the moment of inertia of a sphere = (2/5)m R^2):

Vr = V - (2/7)[V + W X R]

where [V + W X R] is the relative surface velocity.

For figuring the post-impact velocity of the cueball as it reaches natural roll, a more convenient form is:

Vr = (5/7)V[sin(C)cos(C)i + (sin(C)^2 + Bz/R)j]

where i and j are the standard unit vectors in the x and y directions (the cueball's pre-impact velocity is again aligned with the y-axis). Bz/R is as described earlier. Dividing the y-component by the x-component yields the formula for the tangent of the angle F used above, which is measured with respect to the x-axis and is positive in the counterclockwise sense. (Note: this latter form of the equation doesn't include a term for swerve, ie, if W has a component in the y direction.) You can use this to make a simple geometric construction that gives you the cueball's final direction at natural roll for any amount of draw or follow.

The magnitude of the cueball's velocity at natural roll, Vcr above, comes from taking the square root of the sum of the squares of the x and y-components, and a little algebra.

Jim
Click to expand...

I'd rather build a ramp. :D
 
To Jim and Dave,
Thanks so much. Great technical analysis. Well I assume so, cause I haven't grasped it all yet.

Now the problem is that even a physics enthusiast like myself struggles to interpret that info (great as it is), let alone the average mathematically adept pool player, not to mention the average pool player.

So I must make some sense of this knowledge and how it relates to real play in simpler terms. I also think it is necessary to cut some corners, make some rules of thumb which are more easily understood and implemented. To bridge some of that billiard knowledge poverty gap;)

While I attempt to do that, can either of you tell me in simple terms why jsp and I made practical experiments that resulted in the OB travelling only about 1/2 the distance that the CB would have travelled without hitting the OB, when it hit the OB full on while rolling?

My best thinking, having reconsidered some of the information provided is that 2/5ths of the CB's energy was in angular momentum and of the remaining 3/5ths should have transfered as linear momentum to the OB.

But 16% of that 60% (~10% overall) is lost in friction through collision and overcoming cloth resistance from a standing start. Any insights into that loss? It seems higher than I would have expected.

Colin
 
Colin Colenso said:
... Anyone want to offer the full solution or a good rule of thumb formula? ...
Click to expand...
There is a simple graphical construction that gives the velocities and directions of both the cue ball and the object ball for any degree of cut and amount of draw or follow on the cue ball. The method gives both the initial speeds and directions (simple sine and cosine of the cut angle) and the final speeds and directions after smooth rolling sets in.

To get the distances traveled, you need to take the squares of the speeds, since distance traveled seems to be very close to proportional to the energy of the ball.

The method is illustrated in several of my articles in Billiards Digest including a recent one (July?). That article describes using the method on the table to determine the follow direction for any cut shot with a rolling cue ball. See http://www.sfbilliards.com/articles/BD_articles.html -- sorry, I don't have time now to pull out the pertinent articles.
 
Bob Jewett said:
There is a simple graphical construction that gives the velocities and directions of both the cue ball and the object ball for any degree of cut and amount of draw or follow on the cue ball. The method gives both the initial speeds and directions (simple sine and cosine of the cut angle) and the final speeds and directions after smooth rolling sets in.

To get the distances traveled, you need to take the squares of the speeds, since distance traveled seems to be very close to proportional to the energy of the ball.

The method is illustrated in several of my articles in Billiards Digest including a recent one (July?). That article describes using the method on the table to determine the follow direction for any cut shot with a rolling cue ball. See http://www.sfbilliards.com/articles/BD_articles.html -- sorry, I don't have time now to pull out the pertinent articles.
Click to expand...
Thanks Bob,

Will look through those articles.

Colin
 
graphical method for rolling ball speed and direction

Bob Jewett said:
There is a simple graphical construction that gives the velocities and directions of both the cue ball and the object ball for any degree of cut and amount of draw or follow on the cue ball. The method gives both the initial speeds and directions (simple sine and cosine of the cut angle) and the final speeds and directions after smooth rolling sets in.
Click to expand...
This is derived, illustrated, and described in TP A.4 and Bob's July '08 article.

Bob Jewett said:
To get the distances traveled, you need to take the squares of the speeds, since distance traveled seems to be very close to proportional to the energy of the ball.
Click to expand...
The relationship between speed and distance is presented in TP 4.1.

Regards,
Dave
 
Colin Colenso said:
...Now the problem is that even a physics enthusiast like myself struggles to interpret that info (great as it is), let alone the average mathematically adept pool player, not to mention the average pool player.
Click to expand...
Colin, you're obviously familiar with trig. From there it's a pretty short step to vector arithmetic, and in fact, the vector stuff usually greatly simplifies the geometry. On the other hand, I've got a long row to hoe before I could hope to shoot those 24 power shots in your video of a few years back (man, I'm still impressed). And let's not even bring up your break. :)

Colin Colenso said:
So I must make some sense of this knowledge and how it relates to real play in simpler terms. I also think it is necessary to cut some corners, make some rules of thumb which are more easily understood and implemented.
Click to expand...
I agree that the physics, as I presented it, is useless. I thought you might make some plots to see if there's an apparent pattern or easy reference points that could be established. I might do it, but I was sort of hoping you would. (I'm not sure I can fit all the math into my free graphics package.)

Colin Colenso said:
While I attempt to do that, can either of you tell me in simple terms why jsp and I made practical experiments that resulted in the OB travelling only about 1/2 the distance that the CB would have travelled without hitting the OB, when it hit the OB full on while rolling?

My best thinking, having reconsidered some of the information provided is that 2/5ths of the CB's energy was in angular momentum and of the remaining 3/5ths should have transfered as linear momentum to the OB.

But 16% of that 60% (~10% overall) is lost in friction through collision and overcoming cloth resistance from a standing start. Any insights into that loss? It seems higher than I would have expected.
Click to expand...
Well, the main reason the OB goes only half the distance is that it starts out sliding - analogous to hitting centerball on the cueball. By the time it reaches natural roll, its speed is reduced to 5/7'ths of its initial speed. That loss of 2/7'ths is 2/7'ths of its initial relative surface speed between it and the cloth, which happens to be equal to its initial translational speed (in this case):

Vr = V - (2/7)[V + W X R]

Since it has no appreciable spin to begin with except what little it picks up during the collision, W=0 (approximately). So:

Vr = V - (2/7)V = (5/7)V

Here, V is the velocity the cueball would have had, had it been allowed to pass through the object ball unscathed. But the cueball would have already been rolling whereas the OB has to lose 2/7'ths of its initial speed to attain roll. Since distance traversed is proportional to the square of velocity, it will travel about (5/7)^2 or 25/49'ths of the cueball's distance, or just about 1/2.

When the cueball stops after colliding with the OB, it attains roll by giving up spin to gain speed. Using the same formula, but with V now equal to zero and W X R equal to - V(approximately), its speed at the onset of roll should be:

Vr = V - (2/7)[V + W X R] = -(2/7)W X R = -(2/7)[-V] = (2/7)V

Squaring this, its travel distance should be (2/7)^2 or 4/49 as much as if it were allowed to pass through, which is very close to 1/12.

Congrats to you and jsp for performing careful experiments. You guys really nailed it! And thanks for verifying the math.

As far as energy goes, its spin energy is 2/5'ths of its translational energy, or 2/7'ths of its total energy (spin + translation). But to me, thinking in terms of energy is sort of confusing...using the formula is easier (for me). The formula simply says that 2/7'ths of the initial relative surface velocity will be subtracted (vectorally) from a ball's translational velocity by the time it reaches roll. The surface velocity can be due to translational speed, or spin, or both. In the case of a head on collision of a rolling cueball, the math is relatively easy as per above. Since all this is taking place in one direction, the vector arithmetic can be dropped, but I used it anyway just to provide an example. In the long run, it really does make things easier to figure, once you get used to it.

Jim
 
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Jal said:
Colin, you're obviously familiar with trig. From there it's a pretty short step to vector arithmetic, and in fact, the vector stuff usually greatly simplifies the geometry. On the other hand, I've got a long row to hoe before I could hope to shoot those 24 power shots in your video of a few years back (man, I'm still impressed). And let's not even bring up your break. :)

I agree that the physics, as I presented it, is useless. I thought you might make some plots to see if there's an apparent pattern or easy reference points that could be established. I might do it, but I was sort of hoping you would. (I'm not sure I can fit all the math into my free graphics package.)

Well, the main reason the OB goes only half the distance is that it starts out sliding - analogous to hitting centerball on the cueball. By the time it reaches natural roll, its speed is reduced to 5/7'ths of its initial speed. That loss of 2/7'ths is 2/7'ths of its initial relative surface speed between it and the cloth, which happens to be equal to its initial translational speed (in this case):

Vr = V - (2/7)[V + W X R]

Since it has no appreciable spin to begin with except what little it picks up during the collision, W=0 (approximately). So:

Vr = V - (2/7)V = (5/7)V

Here, V is the velocity the cueball would have had, had it been allowed to pass through the object ball unscathed. But the cueball would have already been rolling whereas the OB has to lose 2/7'ths of its initial speed to attain roll. Since distance traversed is proportional to the square of velocity, it will travel about (5/7)^2 or 25/49'ths of the cueball's distance, or just about 1/2.

When the cueball stops after colliding with the OB, it attains roll by giving up spin to gain speed. Using the same formula, but with V now equal to zero and W X R equal to - V(approximately), its speed at the onset of roll should be:

Vr = V - (2/7)[V + W X R] = -(2/7)W X R = -(2/7)[-V] = (2/7)V

Squaring this, its travel distance should be (2/7)^2 or 4/49 as much as if it were allowed to pass through, which is very close to 1/12.

Congrats to you and jsp for performing careful experiments. You guys really nailed it! And thanks for verifying the math.

As far as energy goes, its spin energy is 2/5'ths of its translational energy, or 2/7'ths of its total energy (spin + translation). But to me, thinking in terms of energy is sort of confusing...using the formula is easier (for me). The formula simply says that 2/7'ths of the initial relative surface velocity will be subtracted (vectorally) from a ball's translational velocity by the time it reaches roll. The surface velocity can be due to translational speed, or spin, or both. In the case of a head on collision of a rolling cueball, the math is relatively easy as per above. Since all this is taking place in one direction, the vector arithmetic can be dropped, but I used it anyway just to provide an example. In the long run, it really does make things easier to figure, once you get used to it.

Jim
Click to expand...
Jim,
That really helped me to clarify all the formulas and how it relates to what I have tested and what I am trying to work out.

I think I can work out the translational figures for the different angles and put together a pretty accurate plot.

It's been a while since I spent some time on vectorial analysis. Did a BSc with 2nd year physics and sometimes get my mind in the groove to work through equations, but my head's just not in the mood to focus on that stuff at the moment.

The part I bolded above was particularly illuminating. The 5/7ths squared giving close to the 1/2 ratio and the 2/7 squared giving 1/12, or 1/6th of OB travel. That is fantastic!

I'll work on a new plot and repost it here when done.

Thanks again to you and all who provided info.

Colin
 
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Colin Colenso said:
I've found the 6:1 ratio for rolling straight in shots to be very valuable.

For example, I'll see a straight in that is 3 diamonds to the pocket. I figure to dribble it in means I'l hit it 4 diamonds, hence the CB will roll 4/6ths of a diamond or 8 inches. I'll check that position for suitability of position.
In the past I would overcompensate to get past snookers, not being sure how far the CB would travel if the OB reached the pocket. Now I can regularly get within a ball width of the position I want on these roll through shots.


btw: I've found the 1:6 ratio for straight roll through shots works well on tables of all speeds.

Colin
Click to expand...

My head was starting to hurt by reading these posts and then this gem fell out of the mix....THAT, I can use.

Jeff Livingston
 
distance graphs and useful summary statements

Colin, Jim, and others,

FYI, I've added ball travel-distance plots, and some useful conclusion statements (e.g., the 6x factor mentioned by Jim and Colin), to my TP analyses.

For stun shots, see the graph and summary statements on page 2 of:

For roll shots, see the graph and summary statements on page 3 of:

Colin, if you still plan to create your own plots, let me know how they compare to mine.

Regards,
Dave

Colin Colenso said:
Jim,
That really helped me to clarify all the formulas and how it relates to what I have tested and what I am trying to work out.

I think I can work out the translational figures for the different angles and put together a pretty accurate plot.

It's been a while since I spent some time on vectorial analysis. Did a BSc with 2nd year physics and sometimes get my mind in the groove to work through equations, but my head's just not in the mood to focus on that stuff at the moment.

The part I bolded above was particularly illuminating. The 5/7ths squared giving close to the 1/2 ratio and the 2/7 squared giving 1/12, or 1/6th of OB travel. That is fantastic!

I'll work on a new plot and repost it here when done.

Thanks again to you and all who provided info.

Colin
Click to expand...
 
Ouch!

My head hurts. I just know that if I hit the cue ball into the object ball, it's going to head towards a pocket faster than the object ball seems to. :-)

Thanks for the insight
 
dr_dave said:
Colin, Jim, and others,

FYI, I've added ball travel-distance plots, and some useful conclusion statements (e.g., the 6x factor mentioned by Jim and Colin), to my TP analyses.

For stun shots, see the graph and summary statements on page 2 of:

For roll shots, see the graph and summary statements on page 3 of:

Colin, if you still plan to create your own plots, let me know how they compare to mine.

Regards,
Dave
Click to expand...
Geez, you really got to work on that Dave :-)

Thanks heaps.

I'll get a graph together a bit later. I think I'll stick with my current format.

It seems your curve runs the opposite way to the way I expected.

Cheers,
Colin
 
distance vs. cut angle and ball-hit fraction

Colin,

It only took me 5-10 minutes to add the distance graphs because I already had the speed plots and equations in my files.

Concerning the curve trends, be sure to look at the horizontal scale. I have two distance plots in each analysis, one for cut angle (phi) and one for ball-hit fraction (f). A small cut angle is a large ball-hit fraction.

Regards,
Dave

Colin Colenso said:
dr_dave said:
Colin, Jim, and others,

FYI, I've added ball travel-distance plots, and some useful conclusion statements (e.g., the 6x factor mentioned by Jim and Colin), to my TP analyses.

For stun shots, see the graph and summary statements on page 2 of:

TP 3.2 - Ball speeds and distances after stun-shot impact

For roll shots, see the graph and summary statements on page 3 of:

TP A.16 - Final ball speeds, distances, and directions for natural roll shots

Colin, if you still plan to create your own plots, let me know how they compare to mine.

Regards,
Dave
Click to expand...
Geez, you really got to work on that Dave :-)

Thanks heaps.

I'll get a graph together a bit later. I think I'll stick with my current format.

It seems your curve runs the opposite way to the way I expected.

Cheers,
Colin
Click to expand...
 
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