Question on CB Roll for Physics Enthusiasts

[Colin Colenso]

The image below is my rough estimate for how far a CB will roll for different cut angles, based on a natural rolling shot that has the velocity to hit the OB 10 diamonds on a straight on hit. Imagine the CB is being played from 1 diamond away and that the distance is measured from where it hits the OB.

For the straight shot I've started the curve at 1.5 diamonds. This is pretty close to what we see though it might be a little higher. Maybe 1.8 diamonds. Usually I work on about 6:1 for this shot.

I'm sure the curve works pretty much as drawn (asymtotal to 10), but not sure on the steepness. Any estimates on the values X, Y and Z at 30, 45 and 60 degrees? It's not as simple as 30 = 1.5 + (0.5[sin30] x 8.5)) or is that a reasonable approximation? I expect the value at 30 needs to be higher than 5.75 as that equation would produce.

Anyone want to offer the full solution or a good rule of thumb formula?

Colin

 

[mikepage]

a few comments:

1. Your 1.5 diamonds, the head-on cueball distance, depends on both the speed of the cloth and the slickness of the cloth. Ron Shepard has a formula for the distance the head-on cueball travels as a function of its initial spin. (see apapp problem 2.6)

2. Your 90 degree asymptote shouldn't be at 10 diamonds. A cueball that had enough energy to knock an object ball 10 diamonds AND roll another 1.5 diamonds itself had at least the energy to roll 11.5 diamonds had it merely grazed the object ball.

3. the distance a naturally rolling cueball travels is proportional to its kinetic energy. If you ignore the problem of each ball achieving natural roll (reasonable?), then it seems to me this suggests the distances of the two balls should sum to the same number regardless of cut angle. If this distance is 11.5 diamonds, then that suggests both that your asymptote should be at 11.5 diamonds and that your value of X should be 5.75. I say this because 30 degrees is a half-ball hit, and the half-ball hit pretty much splits the energy between the two balls.

The image below is my rough estimate for how far a CB will roll for different cut angles, based on a natural rolling shot that has the velocity to hit the OB 10 diamonds on a straight on hit. Imagine the CB is being played from 1 diamond away and that the distance is measured from where it hits the OB.

For the straight shot I've started the curve at 1.5 diamonds. This is pretty close to what we see though it might be a little higher. Maybe 1.8 diamonds. Usually I work on about 6:1 for this shot.

I'm sure the curve works pretty much as drawn (asymtotal to 10), but not sure on the steepness. Any estimates on the values X, Y and Z at 30, 45 and 60 degrees? It's not as simple as 30 = 1.5 + (0.5[sin30] x 8.5)) or is that a reasonable approximation? I expect the value at 30 needs to be higher than 5.75 as that equation would produce.

Anyone want to offer the full solution or a good rule of thumb formula?

Colin

 

[Colin Colenso]

a few comments:

1. Your 1.5 diamonds, the head-on cueball distance, depends on both the speed of the cloth and the slickness of the cloth. Ron Shepard has a formula for the distance the head-on cueball travels as a function of its initial spin. (see apapp problem 2.6)

2. Your 90 degree asymptote shouldn't be at 10 diamonds. A cueball that had enough energy to knock an object ball 10 diamonds AND roll another 1.5 diamonds itself had at least the energy to roll 11.5 diamonds had it merely grazed the object ball.

3. the distance a naturally rolling cueball travels is proportional to its kinetic energy. If you ignore the problem of each ball achieving natural roll (reasonable?), then it seems to me this suggests the distances of the two balls should sum to the same number regardless of cut angle. If this distance is 11.5 diamonds, then that suggests both that your asymptote should be at 11.5 diamonds and that your value of X should be 5.75. I say this because 30 degrees is a half-ball hit, and the half-ball hit pretty much splits the energy between the two balls.

I knew you'd find this Mike and help me think it out more clearly

I'll redo the diagram and re-post soon.

Colin

 

[Colin Colenso]

Ok, to make the figures easier, in this diagram the speed of shot is such that the OB would travel 6 diamonds on a straight in shot. This equates to about 1 diamond roll through on the CB.

On the 90 degree cut, the CB will roll 7 diamonds as it has the same total kinetic energy of the previous system (6 + 1). This is assuming kinetic energy varies relatively linearly with distance travelled.

On the half ball shot the CB will run about 3.5 diamonds, the same distance as the OB would for that shot.

Anyone want to make any observations or corrections?

Colin

 

[Colin Colenso]

So a rule of thumb, when estimating the OB travel on rolling shots could be;
Full Ball = 1/6
3/4 Ball = 1/3
1/2 ball = Just over 1/2
45 degrees = Just over 3/4
60 degrees = Just under 1

That is, in comparison to the distance the OB would travel if it was hit center ball.

Or in relation to the distance the OB will take after the cut:
Full Ball = 1:6
3/4 Ball = 2:5
1/2 Ball = 1:1
45 Degrees = 2:1
60 Degrees = 4.5:1

Note: Personally I think it is easier to judge the disance the OB will travel when playing a shot than the distance a rolling CB will travel. Particularly on fuller angles.

Colin

 

[jsp]

2. Your 90 degree asymptote shouldn't be at 10 diamonds. A cueball that had enough energy to knock an object ball 10 diamonds AND roll another 1.5 diamonds itself had at least the energy to roll 11.5 diamonds had it merely grazed the object ball.

"At least" is the key phrase, because I think the asymptote in actuality should be closer to 15 (based on the Colin's numbers). More on this later...

3. the distance a naturally rolling cueball travels is proportional to its kinetic energy. If you ignore the problem of each ball achieving natural roll (reasonable?), then it seems to me this suggests the distances of the two balls should sum to the same number regardless of cut angle. If this distance is 11.5 diamonds, then that suggests both that your asymptote should be at 11.5 diamonds and that your value of X should be 5.75. I say this because 30 degrees is a half-ball hit, and the half-ball hit pretty much splits the energy between the two balls.

Is it reasonable to ignore the problem of the CB achieving natural roll after contact? I don't think it is, which is why I think in actuality the value at X would be significantly bigger than 5.75...closer to 7.5. The following is my explanation...

For a half-ball hit, if you assume a sliding CB on contact, then the energy actually does NOT split between CB and OB. After contact, the OB would actually have 3x the energy as the CB (square the cosine and sine of 30 degrees). So if we assume that on a direct hit the OB would travel 10 diamonds, then for a half-ball hit the OB should travel approximately 7.5 diamonds and the CB should travel only 2.5 diamonds (again assuming a sliding CB on contact).

Now, if we then consider that the CB was rolling on contact with the OB, then the actual CB distance would obviously be greater than 2.5 diamonds. If we assume that it attains the same added distance due to the roll in the case of a direct hit, then we simply should add 1.5 to 2.5, giving the result of only 4 diamonds.

4 + 7.5 does total 11.5 diamonds-worth of total energy, so energy does seem to be conserved. But does this result mesh with what our experience tells us? Not really, since our experience dictates that the CB and OB distances should be roughly equal (with a naturally rolling CB and a half-ball hit).

I think the flaw in the reasoning above is precisely what Mike said he'd ignore...that is, the problem of each ball (especially the CB) achieving natural roll after contact. We neglect the amount of energy that is lost due to friction between the CB and felt immediately after contact, and this lost energy is inversely proportional to the cut angle. We shouldn't assume that the CB in all cases attains the same amount of extra distance as in the case of a direct hit, since that isn't exactly accurate.

On a direct hit, it is given that the CB rolls a distance of 1.5 diamonds after contact due to the energy retained by its natural roll. Immediately after contact, the CB has zero linear veloctiy but still has angular velocity (the CB is still rolling). As the CB begins to pick up speed again due to its roll, energy is lost due to friction between the CB and felt, and the amount of energy lost gets greater the smaller the cut angle.

So I do believe that the distance the OB travels is about 7.5 diamonds, as the previous calculations show. From experience, the CB should travel roughly the same distance as the OB (with a half-ball hit and a naturally rolling CB). Therefore, it can be argued that the value of X in your graph should be closer to 7.5 diamonds.

Given these new numbers, this also suggests that the asymptote should be closer to 15 diamonds. Let me know if I'm terribly off.

 

[Colin Colenso]

"At least" is the key phrase, because I think the asymptote in actuality should be closer to 15 (based on the Colin's numbers). More on this later...


Is it reasonable to ignore the problem of the CB achieving natural roll after contact? I don't think it is, which is why I think in actuality the value at X would be significantly bigger than 5.75...closer to 7.5. The following is my explanation...

For a half-ball hit, if you assume a sliding CB on contact, then the energy actually does NOT split between CB and OB. After contact, the OB would actually have 3x the energy as the CB (square the cosine and sine of 30 degrees). So if we assume that on a direct hit the OB would travel 10 diamonds, then for a half-ball hit the OB should travel approximately 7.5 diamonds and the CB should travel only 2.5 diamonds (again assuming a sliding CB on contact).

Now, if we then consider that the CB was rolling on contact with the OB, then the actual CB distance would obviously be greater than 2.5 diamonds. If we assume that it attains the same added distance due to the roll in the case of a direct hit, then we simply should add 1.5 to 2.5, giving the result of only 4 diamonds.

4 + 7.5 does total 11.5 diamonds-worth of total energy, so energy does seem to be conserved. But does this result mesh with what our experience tells us? Not really, since our experience dictates that the CB and OB distances should be roughly equal (with a naturally rolling CB and a half-ball hit).

I think the flaw in the reasoning above is precisely what Mike said he'd ignore...that is, the problem of each ball (especially the CB) achieving natural roll after contact. We neglect the amount of energy that is lost due to friction between the CB and felt immediately after contact, and this lost energy is inversely proportional to the cut angle. We shouldn't assume that the CB in all cases attains the same amount of extra distance as in the case of a direct hit, since that isn't exactly accurate.

On a direct hit, it is given that the CB rolls a distance of 1.5 diamonds after contact due to the energy retained by its natural roll. Immediately after contact, the CB has zero linear veloctiy but still has angular velocity (the CB is still rolling). As the CB begins to pick up speed again due to its roll, energy is lost due to friction between the CB and felt, and the amount of energy lost gets greater the smaller the cut angle.

So I do believe that the distance the OB travels is about 7.5 diamonds, as the previous calculations show. From experience, the CB should travel roughly the same distance as the OB (with a half-ball hit and a naturally rolling CB). Therefore, it can be argued that the value of X in your graph should be closer to 7.5 diamonds.

Given these new numbers, this also suggests that the asymptote should be closer to 15 diamonds. Let me know if I'm terribly off.

Thanks for your input jsp!

I don't think the 10 + 1.5 being added would be too far off the total travel of the CB if it didn't touch the OB. Perhaps around 12. Any more would surprise me. This could be tested with a ramp.

As for the 1/2 ball hit, it does make sense that the CB has more energy due to its rolling. I think angular momentum works out at about 2/7 of total momentum. A 5:2 ratio. (Or is this 7:2 ratio?) So if the OB rolled half (5 diamonds in the first diagram) the CB would have about 2 extra diamonds worth of energy. Another interesting test. In the 2nd graph that would make the CB be at about 7/12ths of 7 diamonds = 4.1 diamonds, (with the OB travelling about 2.9 diamonds) instead of 3.5 which would give a steeper curve which seems to make more sense intuitively.

I'm only interested at this stage of working this out for a CB that is rolling initially. I think shooting distance judgements shots with stun isn't common enough to be worth pondering, other that for comparative purposes.

Colin

 

[jsp]

Thanks for your input jsp!

No prob, I always enjoy your threads.

I don't think the 10 + 1.5 being added would be too far off the total travel of the CB if it didn't touch the OB. Perhaps around 12. Any more would surprise me. This could be tested with a ramp.

You think? I'm not sure, but my gut feeling is that the asymptote would be closer to 15. I think the energy lost due to friction between the CB and cloth immediately after contact (for a direct hit) is much more than we think.

For example, think of how lightly you need to tap the CB for it to roll 4 diamonds. Now, with that same stroke, imagine putting an OB directly in front of its path (once the CB achieves natural roll). Then add the resulting distances the OB and CB travel. I think the total would be closer to 3 diamonds than 4.

But you're right that this can be tested with a ramp. I'll try it out on my home table after work and I'll report the results.

 

[JoeW]

Seems to me that the math is useful as it gets you in the ball park (so to speak). Then its a matter of determining what happens on two tables with different cloth and the same relative humidity.

Someone somewhere must have a table with Simonis 860 sitting next to a table with "X" cloth in a semi-controlled environment.

These numbers would give closer estimates to what "really" happens and would hopefully include all of the variables that have not been considered. Whne the numbers are in, the scientific types can attempt to explain the differences between theory and observation and may find "other" variables to consider.

Of course there will be error due to human factors but these can be randomized with a sufficient number of trials.

If nothing else an ipsative (based on your personal characteristics) data set and Colin's initial estimates are exremely useful -- thank you.

I like Colin's last initial estimates as a place to start one's expectations.

 

[CueAndMe]

I'm barely keeping up with you guys on the technical stuff here, but I think I notice a problem with this whole exercise. You are beginning with the assumption that the ratios you come up with for a 10 diamond shot are going to translate for other speeds. In the straight pool forum, Bob Jewett recently posted a way of calculating the OB and CB distances for any shot with no spin. Once you add natural roll to the mix, I assume it wouldn't work across the board.

 

[Colin Colenso]

Not really useful for firmer shots.

I'm barely keeping up with you guys on the technical stuff here, but I think I notice a problem with this whole exercise. You are beginning with the assumption that the ratios you come up with for a 10 diamond shot are going to translate for other speeds. In the straight pool forum, Bob Jewett recently posted a way of calculating the OB and CB distances for any shot with no spin. Once you add natural roll to the mix, I assume it wouldn't work across the board.

I realize some of these rules of thumb are going to slide with longer distances and when rails become involved but I am actually more interested in the softer side of the game when using these ratios.

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.

But when there is an angle I have been guessing. I think a more predictive formula for 7/8th, 3/4, 1/2 ball etc angles would come in very useful for the soft roll through game.

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

Colin

 

[CueAndMe]

That's true. It could be useful for softer shots.

 

[jsp]

For example, think of how lightly you need to tap the CB for it to roll 4 diamonds. Now, with that same stroke, imagine putting an OB directly in front of its path (once the CB achieves natural roll). Then add the resulting distances the OB and CB travel. I think the total would be closer to 3 diamonds than 4.

But you're right that this can be tested with a ramp. I'll try it out on my home table after work and I'll report the results.

I just performed this experiment on my home table.

For the ramp, I butted two house cues side by side with the butts on the rail and the tips on the cloth. I marked the point on the ramp where the CB would roll exactly 4 diamonds (once achieving natural roll). I then put an OB at the starting point directly in front of the ramp (but off a few inches away from the end of the ramp to ensure a naturally rolling CB when it contacts the ball). The cloth I have is simonis 860, about 10 months worn in.

The results were fairly surprising. The total distance the CB and OB traveled was only 2 diamonds. The results were very repeatable. After contact, the OB would travel slightly less than 2 diamonds, while the CB would only roll forward another revolution or two after contact. The total distance traveled by both balls came out to be only 2 diamonds long.

 

[Da Poet]

Ok, to make the figures easier, in this diagram the speed of shot is such that the OB would travel 6 diamonds on a straight in shot. This equates to about 1 diamond roll through on the CB.

On the 90 degree cut, the CB will roll 7 diamonds as it has the same total kinetic energy of the previous system (6 + 1). This is assuming kinetic energy varies relatively linearly with distance travelled.

On the half ball shot the CB will run about 3.5 diamonds, the same distance as the OB would for that shot.

Anyone want to make any observations or corrections?

Colin

This is only my intuition, but I'm wondering if the graph should be "S" shaped. For example the change in resistance between a straight on shot and a five degree shot may not be as great as the change in resistance in a 30 degree to 35 degree shot?

I would think in five degree increments, starting from straight on, the change might start out more slowly and then as the angle increased, the change would increase and perhaps the middle of the "S" might be around a 1/2 ball hit? (Just guessing) Then going from there towards the sharpest cuts, I couldn't imagine there would be a significant change in resistance from an 80 degree cut and an 85 degree cut.

Does this make sense or am I way off the reservation.

Edit- This more dramatic difference around a half ball hit may be why those simple looking slow roll 1/2 ball safeties are so touchy?

 

[Colin Colenso]

I just performed this experiment on my home table.

For the ramp, I butted two house cues side by side with the butts on the rail and the tips on the cloth. I marked the point on the ramp where the CB would roll exactly 4 diamonds (once achieving natural roll). I then put an OB at the starting point directly in front of the ramp (but off a few inches away from the end of the ramp to ensure a naturally rolling CB when it contacts the ball). The cloth I have is simonis 860, about 10 months worn in.

The results were fairly surprising. The total distance the CB and OB traveled was only 2 diamonds. The results were very repeatable. After contact, the OB would travel slightly less than 2 diamonds, while the CB would only roll forward another revolution or two after contact. The total distance traveled by both balls came out to be only 2 diamonds long.

jsp,
Well done, you were right. At least it predicting much differing ratios.

I did some tests also tonight which correlate pretty closely with your data.

I found the OB travelled approximately one half the distance that the CB, left rolling, travels with the same velocity. After impact, the CB rolls about 1/6th the distance of the OB, which concurs with earlier tests I've conducted.

So by my calculations, the total travel, at least on relatively short shots which take cushions out of play is OB 6/12ths + CB 1/12 (1/6th of OB travel) = total 7/12ths of total ball travel.

Note: I trialled this on pretty slow cloth.

Hence to move an OB 10 diamonds, it needs to be hit with the CB with natural roll at a speed that would send that CB 20 diamonds past that point if it didn't hit any other ball. (Assuming no loss from cushions)

That is quite a surprising revolution to me. Any physicists care to explain why there is such an apparent loss in kinetic energy?

Colin

 

[Colin Colenso]

This is only my intuition, but I'm wondering if the graph should be "S" shaped. For example the change in resistance between a straight on shot and a five degree shot may not be as great as the change in resistance in a 30 degree to 35 degree shot?

I would think in five degree increments, starting from straight on, the change might start out more slowly and then as the angle increased, the change would increase and perhaps the middle of the "S" might be around a 1/2 ball hit? (Just guessing) Then going from there towards the sharpest cuts, I couldn't imagine there would be a significant change in resistance from an 80 degree cut and an 85 degree cut.

Does this make sense or am I way off the reservation.

Edit- This more dramatic difference around a half ball hit may be why those simple looking slow roll 1/2 ball safeties are so touchy?

I doubt it would be S shaped. But I was wrong earlier.:embarrassed2:

I think one curve with similar trend to that shown in the graphs above.

btw: I did some 1/2 ball experiments and found the distance travelled by both balls was quite similar. It was hard to be sure I was hitting exact 1/2 ball though.

This did cause me to rethink some basic trig. At 30 degrees (half ball) the OB will move forward significantly quicker than the CB, but the CB does have angular momentum. The scenarios mentioned in earlier posts are more relevant for the 45 degree hit (slightly thicker than 3/4 ball).

Colin

 

[mikepage]

[...]
That is quite a surprising revolution to me. Any physicists care to explain why there is such an apparent loss in kinetic energy?

Colin

Seems to me there are three sources of loss

1. inelasticity of the ball-ball collision
2. object ball rubbing on the cloth as it achieves natural roll
3. cueball rubbing on the cloth in the other direction as it achieves natural roll

 

[dr_dave]

ball speeds for rolling and stunned CB

Colin,

The full solution and graphs for a rolling CB can be found in TP A.16. The solution for a stunned CB can be found in TP 3.2. Post-impact ball speeds are directly related to how far the balls will travel, but the exact distances will depend on conditions.

Regards,
Dave

The image below is my rough estimate for how far a CB will roll for different cut angles, based on a natural rolling shot that has the velocity to hit the OB 10 diamonds on a straight on hit. Imagine the CB is being played from 1 diamond away and that the distance is measured from where it hits the OB.

For the straight shot I've started the curve at 1.5 diamonds. This is pretty close to what we see though it might be a little higher. Maybe 1.8 diamonds. Usually I work on about 6:1 for this shot.

I'm sure the curve works pretty much as drawn (asymtotal to 10), but not sure on the steepness. Any estimates on the values X, Y and Z at 30, 45 and 60 degrees? It's not as simple as 30 = 1.5 + (0.5[sin30] x 8.5)) or is that a reasonable approximation? I expect the value at 30 needs to be higher than 5.75 as that equation would produce.

Anyone want to offer the full solution or a good rule of thumb formula?

Colin

 

[JoeW]

Using ramps is a great idea JSP. As soon as I get some time I am going to set up this study on my 9' GC II with Simonis 860.

I have read (and apparantly it is true) that there are some differences in the roll obtained from the head of the table and from the foot of the table so I will look for differences here too.

Seems this would also be a good way to determine the natural roll return off a rail as well.

Just Read Dr Dave's paper. The predictions are interesting and useful. Now I want to see it happen in "real" time. Dave did you ever verify your equations to determine how far off reality is from the predictions?

 

[Jal]

...Anyone want to offer the full solution or a good rule of thumb formula?

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.

**********************************

Edit: The equation for Xcs, the distance traveled by the cueball in the final roll direction (but while still sliding) should be:

Xcs = (Vcr^2 - Vcs^2)/(2gUscos(B-F))

where the angle B is the direction of the cloth friction force measured from the x-axis as with F. It is:

B = arctan[cos(C)/(-sin(C))]

The implied plus sign on cos(C), and the minus sign on sin(C), place this direction in the correct quadrant when evaluating the arctan function. Sorry about that.

Jim