OB on the rail & CIT?

[CreeDo]

Some thoughts...

1. How much does CIT throw a ball off, with clean balls? In other words, if I hit it into the center of a pocket 5 feet away, will it still go in the hole (just a little on the 'full side')? Or will it miss entirely?

Is it possible that, on clean equipment, even with a full 9ish feet of travel, CIT won't throw a ball out enough to cause a miss, unless the ball was already heading towards the full side of the pocket?

2. Is a frozen ball combo an accurate way to test CIT? In other words, if you hit a dead frozen combo from angle X, does the object ball travel on the same path if you removed the first ball on the combo, and shot the cue ball from the exact same position?

It seems to me that frozen ball combos throw off quite a bit more than a cut from the same angle. But if so, why?

3. If you learn to make balls without the benefit of CIT knowledge... or even with it... isn't it fair to say that all shots you send to the center of the pocket already have CIT compensation built in? Unless you've been outside spinning balls in from day one, isn't it likely that any cut angle that makes you think "this is going to the center of the pocket" is actually a cut angle to thin side of the pocket?

 

[Bob Jewett]

Some thoughts...

1. How much does CIT throw a ball off, with clean balls? In other words, if I hit it into the center of a pocket 5 feet away, will it still go in the hole (just a little on the 'full side')? Or will it miss entirely?

Is it possible that, on clean equipment, even with a full 9ish feet of travel, CIT won't throw a ball out enough to cause a miss, unless the ball was already heading towards the full side of the pocket?

2. Is a frozen ball combo an accurate way to test CIT? In other words, if you hit a dead frozen combo from angle X, does the object ball travel on the same path if you removed the first ball on the combo, and shot the cue ball from the exact same position?

It seems to me that frozen ball combos throw off quite a bit more than a cut from the same angle. But if so, why?

3. If you learn to make balls without the benefit of CIT knowledge... [/i]or even with it[/i]... isn't it fair to say that all shots you send to the center of the pocket already have CIT compensation built in? Unless you've been outside spinning balls in from day one, isn't it likely that any cut angle that makes you think "this is going to the center of the pocket" is actually a cut angle to thin side of the pocket?

CIT is up to 6 degrees even with clean balls depending on what's on the surface. Hard wax may reduce throw. On a shot from the spot to a far corner pocket, you need to have your error down under a degree.

A frozen ball seems to give close to the same throw as a slightly separated ball. The hard thing to control when the balls are more separated is the follow on the incoming ball. As mentioned, a good stun shot is the worst offender for throw.

Ten years ago no one seemed to know that both follow and draw reduce CIT. They change the throw by more than enough to miss the shot mentioned above. Maybe that reality is what caused (and causes) a lot of players to use outside english even though the side spin complicates the shot in other ways.

 

[dr_dave]

1. How much does CIT throw a ball off, with clean balls?

Maximum throw is about 1" per foot for typical equipment. For more info, including a video demo, see:

maximum throw resource page​

In other words, if I hit it into the center of a pocket 5 feet away, will it still go in the hole (just a little on the 'full side')? Or will it miss entirely?

That depends on how "tight" the pocket is. it also depends on the cut angle, shot speed, and whether or not the CB has stun. With a firm follow or draw shot at small cut angles and decent size pockets, the ball should still go in with center-ball aim ... unless there is cling.

Is it possible that, on clean equipment, even with a full 9ish feet of travel, CIT won't throw a ball out enough to cause a miss, unless the ball was already heading towards the full side of the pocket?

Not for a 1/2-ball hit with slow stun. You will miss this every time, regardless of how new or clean the balls are.

2. Is a frozen ball combo an accurate way to test CIT? In other words, if you hit a dead frozen combo from angle X, does the object ball travel on the same path if you removed the first ball on the combo, and shot the cue ball from the exact same position?

It seems to me that frozen ball combos throw off quite a bit more than a cut from the same angle. But if so, why?

I think Bob already answered this fairly well. A frozen ball acts mostly like a stunned ball.

3. If you learn to make balls without the benefit of CIT knowledge... [/i]or even with it[/i]... isn't it fair to say that all shots you send to the center of the pocket already have CIT compensation built in? Unless you've been outside spinning balls in from day one, isn't it likely that any cut angle that makes you think "this is going to the center of the pocket" is actually a cut angle to thin side of the pocket?

If you are good at "spinning balls in" (i.e., judging the gearing amount of outside English for every shot), you don't need to worry about throw (or cling); however, if you don't do this, you do need to compensate (even if only subconsciously) ... especially on tight equipment.

Regards,
Dave

 

[dr_dave]

Ten years ago no one seemed to know that both follow and draw reduce CIT.

If people want proof of this (from both me and you), it can be found here:

draw and follow effects resource page​

Regards,
Dave

 

[CreeDo]

I guess where I was going with those questions... I was wondering if it would be possible to remain blissfully ignorant about CIT, and make the balls with only unknowing compensation (i.e. learning to slightly overcut center ball shots by experience, not realizing you were actually hitting balls to the thin side of the pocket and throwing them to the center).

I was honestly expecting the effect to be small enough that even from long distances, a ball will still go in if you shoot with this subsconscious "pre-adjusted" aim. But if a ball can go off 6 degrees, you can miss by a mile without adjustment.

So, though it's kind of tedious, I guess it's better to have the knowledge and be aware that you're adjusting, vs. learning to judge the shot from hitting it a million times, and being left with this nagging feeling that all your shots into a distant pocket feel like you're overcutting them.

 

[Cornerman]

I guess where I was going with those questions... I was wondering if it would be possible to remain blissfully ignorant about CIT, and make the balls with only unknowing compensation (i.e. learning to slightly overcut center ball shots by experience, not realizing you were actually hitting balls to the thin side of the pocket and throwing them to the center).

I was honestly expecting the effect to be small enough that even from long distances, a ball will still go in if you shoot with this subsconscious "pre-adjusted" aim. But if a ball can go off 6 degrees, you can miss by a mile without adjustment.

So, though it's kind of tedious, I guess it's better to have the knowledge and be aware that you're adjusting, vs. learning to judge the shot from hitting it a million times, and being left with this nagging feeling that all your shots into a distant pocket feel like you're overcutting them.

I think this is a very important point to consider. First, Bob and others have said that it's "up to 6 degrees." Also consider that swerve and squirt can be more than 6 degrees. A lot more.

This is why I don't bother considering CIT (from a black box, input/output point of view) on most shots. The combo of squirt and swerve (squirve / absolute squirt/ effective squirt) is in my opinion the more important way to look at english effects.

Personally, I look at it as "thicker or thinner" depending on the shot.

Freddie

 

[dr_dave]

I think this is a very important point to consider. First, Bob and others have said that it's "up to 6 degrees." Also consider that swerve and squirt can be more than 6 degrees. A lot more.

This is why I don't bother considering CIT (from a black box, input/output point of view) on most shots. The combo of squirt and swerve (squirve / absolute squirt/ effective squirt) is in my opinion the more important way to look at english effects.

Personally, I look at it as "thicker or thinner" depending on the shot.

I agree 100%. Squirt and swerve are much bigger problems (or "opportunities") than throw. To be a good player, one must be aware of (even if only subconsciously) all English and throw-related effects.

Regards,
Dave

 

[Robert Raiford]

I think this is a very important point to consider. First, Bob and others have said that it's "up to 6 degrees." Also consider that swerve and squirt can be more than 6 degrees. A lot more.

This is why I don't bother considering CIT (from a black box, input/output point of view) on most shots. The combo of squirt and swerve (squirve / absolute squirt/ effective squirt) is in my opinion the more important way to look at english effects.

Personally, I look at it as "thicker or thinner" depending on the shot.

Freddie

For reference, 6 degrees is around 10.5" (i.e. almost a full diamond) difference across the length of the table. That's roughly 1.3" error for each diamond the OB is away from a pocket. An OB from the center spot shot to a corner pocket will be off by an inch for every degree of throw (6" in this case). It may not sound like it at first, but even 6 degrees is big difference to compensate for.

Worrying about sidespin consequences like squirt and swerve is certainly important, but I think you're exaggerating about it being "a lot more" than 6 degrees for anything resembling a normal level-ish stroke. Squirt and swerve can have a canceling effect since they work in opposite directions, so their combined practical effect is usually much smaller than that unless you have an extremely high deflection cue and the CB and OB are very close together where swerve can be ignored. Masse shots are another matter, of course.

Regardless, the "up to 6 degrees" for CIT we're referring to is for tip contacts strictly on the vertical center line of the CB and doesn't involve sidespin-related complications. You can get a 4-5 degree variation in throw just by changing between natural roll and stun for a half-ball hit. This is surprising to many players.

To experiment with it, place the CB on the head spot and an OB on the center spot. Under typical conditions, aiming the center of the CB to the edge of the OB (i.e. 1/2-ball hit) with straight follow (NR) at medium speed will pocket the OB in either corner pocket. However, slowly shooting a STUN shot (no follow or draw at contact) using the same 1/2-ball center-to-edge aiming will cause you to completely miss the shot into the short rail. To compensate, you have to significantly overcut the ball towards the long rail for it to throw into the pocket. It's a large correction, but it's very repeatable and easy to see that it's not a random cling/skid effect. To see how much alignment has to change for the stun version, set it up as a frozen combination. (Frozen balls act throw-wise like STUN shots.)

Robert

 

[Patrick Johnson]

Robert Raiford:
Frozen balls act throw-wise like STUN shots.

I thought static friction (frozen balls) is significantly greater than kinetic friction (unfrozen balls). Maybe the difference isn't as great as I imagine...?

Worrying about sidespin consequences like squirt and swerve is certainly important, but I think you're exaggerating about it being "a lot more" than 6 degrees for anything resembling a normal level-ish stroke.

I'm guessing that Fred meant the effect squirt/swerve can have on the OB's direction.

pj
chgo

 

[Bob Jewett]

I thought static friction (frozen balls) is significantly greater than kinetic friction (unfrozen balls). Maybe the difference isn't as great as I imagine...?...

There may be a small effect but the usual situation is that the balls are all sliding almost immediately during the frozen ball collision. That means that even if the static force were larger, it can last for only a very short time. A simple experiment could test this idea: try combos that are frozen and then separate them by barely enough to guarantee that the first object ball will have separated from the cue ball before contacting the second ball. A small correction to the initial line of centers is needed due to the separation which allows a little cut if not struck straight along the line.

 

[Jal]

There may be a small effect but the usual situation is that the balls are all sliding almost immediately during the frozen ball collision. That means that even if the static force were larger, it can last for only a very short time.

Bob, the argument can be made that with the frozen combo, the average speed of the first ball during impact is just about half of its final speed after contact is over. Therefore, if nothing else, the kinetic COF should be accordingly larger compared to when the first ball is already up to speed before impact.

A simple experiment could test this idea: try combos that are frozen and then separate them by barely enough to guarantee that the first object ball will have separated from the cue ball before contacting the second ball....

If this test showed no significant difference, particularly at slow speeds where Dr. Dave's model, for instance, presents a sharp rise of the coefficient with diminishing speed at slow speeds, what could we conclude from that?

Jim

 

[Bob Jewett]

Bob, the argument can be made that with the frozen combo, the average speed of the first ball during impact is just about half of its final speed after contact is over. Therefore, if nothing else, the kinetic COF should be accordingly larger compared to when the first ball is already up to speed before impact.

If this test showed no significant difference, particularly at slow speeds where Dr. Dave's model, for instance, presents a sharp rise of the coefficient with diminishing speed at slow speeds, what could we conclude from that?

Jim

I like your first idea. Someone needs to do the experiment. Among other things, the distance between the balls needs to be controlled down to small fractions of a millimeter, like maybe 0.01 mm, for the results to be complete.

It would be good to have a better billiard ball collision simulator. Virtual Pool is good for some things, but at least through VP3 it did not attempt to accurately simulate what was happening during ball-ball collisions.

 

[dr_dave]

Bob, the argument can be made that with the frozen combo, the average speed of the first ball during impact is just about half of its final speed after contact is over. Therefore, if nothing else, the kinetic COF should be accordingly larger compared to when the first ball is already up to speed before impact.

That's a good point Jim. The fact that a frozen combo is a 3-ball collision instead of a 2-ball collision also adds additional potential "physics complications." I've never done a careful experiment as Bob has suggested, but we should all do this. My sense has always been that a frozen combo throws a small amount (but just a small amount) more than a non-frozen stunned ball at the same speed, and your comment on the COF would seem to support this, but only a carefully done experiment will convince me one way or the other.

If this test showed no significant difference, particularly at slow speeds where Dr. Dave's model, for instance, presents a sharp rise of the coefficient with diminishing speed at slow speeds, what could we conclude from that?

... that more study and experiments are required.

Regards,
Dave

 

[dr_dave]

I've never done a careful experiment as Bob has suggested, but we should all do this. My sense has always been that a frozen combo throws a small amount (but just a small amount) more than a non-frozen stunned ball at the same speed

FYI, I just did a careful experiment and got some results. If you just want to know the conclusion, see the bolded statement at the bottom of this message.

I used two new and clean Aramith measles balls as OBs, and wiped them after every shot, along with the CB which was another Aramith measles ball. I firmly tapped the two OBs into place (frozen, but not leaning against one another) with one OB on the head spot and the other on the head-rail side of the first with the line of centers along the table centerline. I also marked the tapped positions with little white donuts to help further ensure consistent ball placement, shot after shot. I then hit the 1st OB squarely with another CB placed between the balls and the head rail to locate and mark on the far rail the line of centers direction (which was pretty much exactly along the table centerline with every shot). I then tapped and marked the CB position about 6 inches on the head-rail side of the near OB along the center-to-edge line of the frozen-ball combo to ensure a consistent cut angle (with a square hit on the 1st ball) very close to 30 degrees. I then hit this shot (after checking that the OB were in fact frozen) about 20 times at as consistent a speed as I could and marked on the far rail where the thrown OB hit (by placing a piece of chalk on the rail with one edge pointing along the line of the thrown OB). I only checked shots where the CB stopped dead and the 2nd OB bounced off the foot rail and came back to within a 1/2-diamond of the head rail to ensure consistent speed. All of the target-speed shots threw the OB very consistently to the same position on the rail.

I then removed the donut for the lead OB, replaced and carefully tilted the lead OB back toward the CB along the CTE line a very small amount and firmly re-tapped the ball into the cloth until the OB sat about 1mm (a very small gap) away from the 2nd OB. When I was confident with the placement, I put down another donut in the new location and tapped the ball in place even more firmly. I then hit about 20 shots with this new position (resulting in a non-frozen stunned hit of the lead OB into the 2nd), with everything else the same. Again, the direction the ball headed was very consistent for the shots at the target speed.

Here are the results after measuring the distances between the head spot and rail marks carefully and doing the throw angle calculations:

throw angle for frozen balls = 5.16 degrees

throw angle for non-frozen stun shot of same speed and cut angle = 4.95 degrees​

Therefore, my experiment suggests frozen balls might throw a very small percentage more (about 4% more) than a stunned ball. To me this difference is not very significant, especially given that my experiment was not perfect (although, I was as careful as I could be).

I'll comment on a physics rationale in a follow-up message.

Anybody else do an experiment yet? If not, try to duplicate what I did and see if you get similar results.

Regards,
Dave

 

[dr_dave]

Bob, the argument can be made that with the frozen combo, the average speed of the first ball during impact is just about half of its final speed after contact is over. Therefore, if nothing else, the kinetic COF should be accordingly larger compared to when the first ball is already up to speed before impact.

Jim,

Based on the results of my experiment, this must not be a significant effect. Maybe during the part of the collision where the significant throw force develops, the deformation and sliding is very similar with the two cases. In other words, the initially stationary frozen ball might not generate much deformation, normal force, and friction until its speed approaches the speed of the stunned ball during impact. Also, maybe the friction develops differently during the contact times for each case, but the summation of these forces over time ends up with close to the same net result.

I think only a very sophisticated finite-element analysis could yield meaningful insight, assuming all of the important physics were modeled in the analysis (COR variance with speed, elastic wave effects, possible nonlinear material properties, etc.). Because the collision event is so complex physically and happens over such a ridiculously short amount of time, it is difficult to rationalize what might actually be going on, IMO.

Regardless, the experimental data seems to imply that there isn't much of a difference in throw between a frozen-combo and a stun shot. I'll be curious to see if you or others have any other physics-based insights into why this might be the case.

Regards,
Dave

 

[dr_dave]

Regardless, the experimental data seems to imply that there isn't much of a difference in throw between a frozen-combo and a stun shot. I'll be curious to see if you or others have any other physics-based insights into why this might be the case.

One thing I would be even more certain about is that for small cut angles (less than about 20 degrees), the amount of throw should not vary among frozen and stun shots even fort shots of different speeds, because the balls "gear" together during the collision. This effect is clear in experimental data, theory, and at the table. For illustrations, a physics-based explanation, and demonstrations, see:

Throw - Part II: results (BD, September, 2006)
TP A.14 - The effects of cut angle, speed, and spin on object ball throw (see pages 2 and 3)
NV B.86 - Cut-induced throw (CIT) and spin-induced throw (SIT), from VEPS IV (starting at the 7:28 point in the video)​

Most people would probably prefer the video.

Regards,
Dave

 

[dr_dave]

... the experimental data seems to imply that there isn't much of a difference in throw between a frozen-combo and a stun shot.

One reason some people might think frozen balls throw more than a stun shot is that frozen balls can be hit softly, easily creating maximum throw. With frozen balls, stun is guaranteed. However with a normal cut shot, where there is distance between the CB and OB, it is very difficult to ensure stun with a soft shot. Therefore, it is rare to get maximum throw with a cut shot. Although, if you do hit a true stun shot at soft speed, the OB will throw very close to the same amount as a frozen ball would.

Regards,
Dave

 

[Bob Jewett]

One thing I would be even more certain about is that for small cut angles (less than about 20 degrees), the amount of throw should not vary among frozen and stun shots even fort shots of different speeds, because the balls "gear" together during the collision. ...

Yes, but the first object ball is also touching the cue ball. That should tend to keep it from rotating. It's not clear to me what happens to the 3rd ball but it seems like something should.

 

[Jal]

Dr. Dave,

Many thanks for taking the time to do the test, for doing it so carefully (although we wouldn't expect anything else), and then describing it so precisely.

On the possibility that an explanation might be had short of the full FEA treatment, I'm still holding out some hope that gearing may be limiting the throw in the frozen ball case. Bob has a tough objection to that, but I have a counter argument...not a killer counter argument to be sure, but one of sorts (i.e., a "considerable'" time/phase lag between the forces acting at the two interfaces). While this would be impossible to prove in a quantitative fashion, obviously, some of your high-speed videos may directly answer the question as to how free the first object ball is to spin. I'll be looking at them very carefully tomorrow.

Jim

 

[dr_dave]

One thing I would be even more certain about is that for small cut angles (less than about 20 degrees), the amount of throw should not vary among frozen and stun shots even fort shots of different speeds, because the balls "gear" together during the collision. This effect is clear in experimental data, theory, and at the table. For illustrations, a physics-based explanation, and demonstrations, see:

Throw - Part II: results (BD, September, 2006)
TP A.14 - The effects of cut angle, speed, and spin on object ball throw (see pages 2 and 3)
NV B.86 - Cut-induced throw (CIT) and spin-induced throw (SIT), from VEPS IV (starting at the 7:28 point in the video)​

Most people would probably prefer the video.

Yes, but the first object ball is also touching the cue ball. That should tend to keep it from rotating.

Good point Bob. I agree. I should have thought of that ... but I didn't.

It's not clear to me what happens to the 3rd ball but it seems like something should.

It seems to me that the gearing resistance provided by the CB would tend to reduce the throw some. On the other hand, the 3-ball (vs. 2-ball) collision dynamics might be creating some other non-obvious effects that either increase or decrease the throw.

It sounds like I need to repeat my frozen-vs-stun experiment for a smaller cut angle. Maybe some other day.

Regards,
Dave