Jsp, for what it's worth, I agree with both of your observations.

Exactly the same amount, but in the opposite direction along the tangent line.

For instance, the cueball has a certain component of velocity in the direction of the tangent line, call it Vct. If the object ball is thrown in the same direction with velocty Vot, then the cueball's new velocity along the tangent will be Vct - Vot. If the object ball is thrown in the opposite direction along the tangent line because of excessive outside english, then the cueball's new velocity will be Vct + Vot.

This is just Newton's Third Law of action/reaction.

Jim

Another way to put it is that the object ball is thrown by the slightly sticky surface of the cue ball pulling it to the side. Since the cue ball is not attached to the table, it will have be pushed in the other direction. Depending on the situation, the cue ball will be going faster or slower along the tangent line right after the contact due to that frictional force against the object ball.

Thanks to all for considering this question. As a number of you have suggested, there is an equal and opposite reaction on the cue ball, the magnitude and direction, of which, depend upon the specific conditions, i.e., stun, draw, etc. Published papers on this concept appear to be in need of correction if they say that there is none or very little effect on the cue ball's motion. We would all benefit if someone would explore this effect in some mathematical way, and applied in a practical way.

The CB departs on the tangent line defined by conservation of momentum when throw is applied. To the tangent line is altered in exactly the same way as the line of departure for the OB is altered by the throw.

The answer starts pointing at tangents and the theory of tangents. I think that because of the velocity vectors, the throw for the cueball is in the same direction as its "tangent line path," but the throw for the object ball is perpendicular to its direction of line of centers creating a different angle.

Therefore, the vector of the cueball might be tangent to the contact point line of centers, but the path of the cueball is not perpendicular to the path of the object ball.

Freddie <~~~ now somebody set me straight

Hunger is the best seasoning
--------

Name: Freddie Agnir
Shooting Cue: Schuler
Breaking Cue:Gilbert
Playing time: zero
--------

Thanks to all for considering this question. As a number of you have suggested, there is an equal and opposite reaction on the cue ball, the magnitude and direction, of which, depend upon the specific conditions, i.e., stun, draw, etc. Published papers on this concept appear to be in need of correction if they say that there is none or very little effect on the cue ball's motion. We would all benefit if someone would explore this effect in some mathematical way, and applied in a practical way.

Well, in some sense there is less effect on the cue ball's motion than on the object ball. Suppose the cue ball lands full on the object ball at a speed of 10 inches per second and with enough spin to give the object ball a speed to the side of 1/2 inch per second (a "slope" of 20:1, or about 3 degrees of throw). Suppose the cue ball has no draw or follow. The cue ball will be moving to the other side at 1/2 inch per second. If the rolling friction on the cloth is such that the cue ball rolls one inch to the side before stopping, the object ball will roll about 400 times as far in total (assuming it doesn't hit a cushion) for a total distance of 400 inches roughly. With a slope of 20:1 off dead straight ahead, that means it will go to the side 20 inches in its 400 inches of travel. If you count "total distance to the side" as the important parameter, there was more effect on the OB than the CB.

Of course, it would have to be a pretty large table to have a 400-inch clear path, but if the object ball went a more reasonable 80 inches before a rail, it would have gone to the side 4 inches in that distance.

Published papers on this concept appear to be in need of correction if they say that there is none or very little effect on the cue ball's motion.

Exactly. Some of the published books and papers are so far off in this ignoring of the counter effect it's laughable. And then people will tell you they can hold a cueball dead still and throw the object ball some amazing sideways distance to, say, get a stop shot position on a break out shot in 14.1.

It's a shame.

Freddie

Hunger is the best seasoning
--------

Name: Freddie Agnir
Shooting Cue: Schuler
Breaking Cue:Gilbert
Playing time: zero
--------

Does anyone know how much reverse throw is applied to the cue ball when throw is applied to the object ball?

As others have pointed out, the effect is equal and opposite. The resulting effect on CB speed and angle are different depending on the type of shot. For example, the effect on stun and rolling-ball shots are described here:

For cut shots with English, the CB will pick up a little tangent line speed with outside English greater than the gearing amount, and will lose a little tangent line speed with inside English or outside English less than the gearing amount.

"Reverse throw" is also useful to help "hold" the cue ball. For more info, see:

......We would all benefit if someone would explore this effect in some mathematical way, and applied in a practical way.

When the balls are close enough to each other and/or you're hitting hard enough such that the cueball doesn't lose any significant backspin on the way to the object ball (or gain more topspin), there is a method of determining the cueball's direction once it reaches natural roll after the collision. I call it the Bottom-Center-Arrow method, or B-C-A for short, in that it's easy to remember.

Imagine a circle centered on the ghostball with the bottom of the circle running through the center of the cueball. This circle represents the face of the cueball from the shooter's perspective. To determine the CB's roll direction after the collision for any vertical offset (no sidespin applied), draw a line from the center of the real cueball parallel to the line of centers between the ghostball and the object ball. This will intersect the tangent line at 90 degrees, call it point A. Thus, we have a triangle with the CB at vertex B (bottom of the circle), the ghostball at C (center of the circle) and point A from which we'll draw an arrow such that it intersects the vertical axis of the large circle. This yields the CB's direction once roll sets in, given that vertical tip offset on the face of cueball. Here's a diagram:

The relevant point here is that friction, amongst other things, has an effect on this idealized geometry. Below is plot of the deviations from this ideal due to CB "throw" at various vertical offsets and cut angles.

(Note:the discontinuities in some of the curves are transition points from partial to full sliding during the collision.)

As you can see, the largest departures happen at close to stun at smallish cut angles. But these deviations can be essentially eliminated by applying an amount of inside english determined as follows. Swing the vector going from (3/5)R below center to the desired vertical tip offset, parallel to the direction of the line of centers between the ghostball-object ball. The location of the tip of this vector yields the amount of inside english to use (as seen on the large circle epesenting the face of the CB) in order to negate the effects of friction. A couple of diagrams should help, one for draw, one for follow:

I should note that the above correction assumes that a certain component of the CB's spin (i.e., along the ghostball-object ball line of centers) is unaffected by the collision. While this is certainly not exactly true, I believe it's true enough compared to the effect of the friction on the CBs other spin component along the tangent line. Also, the use of the inside english will generate another deviation: post-impact swerve.

There are other deviations due to mismatched ball weights and the less than perfect elasticity of the collision, but I'll leave it at that for now, since we're discussing CB throw.

Does anyone know how much reverse throw is applied to the cue ball when throw is applied to the object ball?

Depends on what kinda reverse throw you use really.What kinda cb to.And you need to find out if the object ball is the kind that will accept reverse throw.

Im confused ..

Everyone has photographic memory; some just don't have the film.

Excellent post. I plan to look into this more closely when I can find some time. I'll also quote it on one of my resource pages at some point.

... remainder of the message deleted until I find time to think through the results ...

Thanks,
Dave

Quote:

Originally Posted by Jal

When the balls are close enough to each other and/or you're hitting hard enough such that the cueball doesn't lose any significant backspin on the way to the object ball (or gain more topspin), there is a method of determining the cueball's direction once it reaches natural roll after the collision. I call it the Bottom-Center-Arrow method, or B-C-A for short, in that it's easy to remember.

Imagine a circle centered on the ghostball with the bottom of the circle running through the center of the cueball. This circle represents the face of the cueball from the shooter's perspective. To determine the CB's roll direction after the collision for any vertical offset (no sidespin applied), draw a line from the center of the real cueball parallel to the line of centers between the ghostball and the object ball. This will intersect the tangent line at 90 degrees, call it point A. Thus, we have a triangle with the CB at vertex B (bottom of the circle), the ghostball at C (center of the circle) and point A from which we'll draw an arrow such that it intersects the vertical axis of the large circle. This yields the CB's direction once roll sets in, given that vertical tip offset on the face of cueball. Here's a diagram:

The relevant point here is that friction, amongst other things, has an effect on this idealized geometry. Below is plot of the deviations from this ideal due to CB "throw" at various vertical offsets and cut angles.

(Note:the discontinuities in some of the curves are transition points from partial to full sliding during the collision.)

As you can see, the largest departures happen at close to stun at smallish cut angles. But these deviations can be essentially eliminated by applying an amount of inside english determined as follows. Swing the vector going from (3/5)R below center to the desired vertical tip offset, parallel to the direction of the line of centers between the ghostball-object ball. The location of the tip of this vector yields the amount of inside english to use (as seen on the large circle epesenting the face of the CB) in order to negate the effects of friction. A couple of diagrams should help, one for draw, one for follow:

I should note that the above correction assumes that a certain component of the CB's spin (i.e., along the ghostball-object ball line of centers) is unaffected by the collision. While this is certainly not exactly true, I believe it's true enough compared to the effect of the friction on the CBs other spin component along the tangent line. Also, the use of the inside english will generate another deviation: post-impact swerve.

There are other deviations due to mismatched ball weights and the less than perfect elasticity of the collision, but I'll leave it at that for now, since we're discussing CB throw.

Excellent post. I plan to look into this more closely when I can find some time. I'll also quote it on one of my resource pages at some point.

I really appreciate the kind comment Dr. Dave. (I realize at this point you haven't verified my results). I do hope you come up with the same conclusions!

Yes, I did mean inside english. It doesn't cancel the throw and will increase it in some cases, as you well know. But it balances its effects on the various components that go into the CB's final roll direction....according to my math of which I'm fairly certain. But, of course, it would be great if you get the time to confirm it (or otherwise!) and add your insights.

Note that the method suggests one unnecessary correction: when using pure stun. While harmless in a way, this is the only vertical offset (0) in which the method doesn't really apply.

Swing the vector going from (3/5)R below center to the desired vertical tip offset, parallel to the direction of the line of centers between the ghostball-object ball. The location of the tip of this vector yields the amount of inside english to use (as seen on the large circle epesenting the face of the CB) in order to negate the effects of friction. A couple of diagrams should help, one for draw, one for follow:

That's really interesting. Can you post your math for this as well? Based on your diagrams I believe I can ascertain the formula which leads to your conclusion, but I am having trouble quantifying the CB spin based on the tip location. I don't believe the relationship between tip position and CB spin is entirely linear, due to the inelastic nature of the tip and CB collision. In addition I think many of the factors in the tip/CB collision would make quantifying this spin extraordinarily difficult, such as tip end mass, tip shape, and tip frictional co-efficient.

Even if we assume a perfectly elastic collision for the stroke, and a linear relationship between tip position and CB spin, it seems in your diagrams that CB speed is not a factor either. This is definitely a departure from the standard thinking on throw, unless I have been dreadfully misled

If your math checks out, it will definitely change my game. I was expecting the line depicting the tip position for zero throw to be parabolic (or perhaps circular), returning to the vertical centre at 3R/5 above the horizontal centre. Clearly that kind of thinking would have me adjust in the wrong direction for follow cut shots.

That's really interesting. Can you post your math for this as well?

I don't have the math for the tip offset graphical interpretation, but the foundation physics and math along with a graphical interpretation for the rolling CB case can be found here:

See Equation 24 on page 4 and Equation 39 on page 6. Also see the explanation and illustration on page 7 based on Bob Jewett's July '08 Billiards Digest article (see page 13 here).

Maybe Jim can post a summary for us for the general case. I also plan to work through and eventually post something for this (with credit to Jim). Jim has made a nice contribution here by extending the graphical approach to all draw and follow shots (not just roll shots).

Quote:

Originally Posted by OnTheMF

Based on your diagrams I believe I can ascertain the formula which leads to your conclusion, but I am having trouble quantifying the CB spin based on the tip location. I don't believe the relationship between tip position and CB spin is entirely linear, due to the inelastic nature of the tip and CB collision. In addition I think many of the factors in the tip/CB collision would make quantifying this spin extraordinarily difficult, such as tip end mass, tip shape, and tip frictional co-efficient.

In the ideal case, the spin-to-tip-offset relationship is definitely linear. See:

Even if we assume a perfectly elastic collision for the stroke, and a linear relationship between tip position and CB spin, it seems in your diagrams that CB speed is not a factor either.

The final CB trajectory angle depends only on the spin-to-speed ratio (AKA spin-rate factor, percentage English, "tips" of English). For more info and illustrations, see:

This is definitely a departure from the standard thinking on throw, unless I have been dreadfully misled

If your math checks out, it will definitely change my game. I was expecting the line depicting the tip position for zero throw to be parabolic (or perhaps circular), returning to the vertical centre at 3R/5 above the horizontal centre. Clearly that kind of thinking would have me adjust in the wrong direction for follow cut shots.

Object ball throw is a different question. Zero throw occurs at the "gearing" amount of English; and you are correct ... this does vary nonlinearly with cut angle. For details, see:

Diagram 2 in my January '07 BD article shows how the amount of English required for "gearing" varies with cut angle. The relationship is nearly linear over a fairly wide range of cut angles typical in good pool play. A 1/2-ball hit requires 40% English.