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dr_dave
08-28-2007, 03:32 PM
Iíve gotten several e-mail questions concerning Diagram 3 in my August '07 article (http://billiards.colostate.edu/bd_articles/2007/aug07.pdf), so I thought I would post the info here to see if people have feedback, input, and/or other questions. Here's the diagram:

http://billiards.colostate.edu/images/English_effects.jpg

Several people have suggested the throw direction is wrong due to collision- or cut-induced throw (CIT). Before giving you my answer, think about it yourself and decide if you think the diagram is correct or not. Also, many people seem to be confused by the real effects of squirt and swerve. I think Diagram 4 from the article (see below) helps clarify things.

http://billiards.colostate.edu/images/effective_squirt.jpg

I use the phrase "effective squirt" for the net effect of squirt and swerve. As Jal pointed out in the squirt measurement thread (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&Board=ccb&Number=256666&page =0&view=collapsed&sb=5&o=&vc=1), some people like to call this "squerve." My series of articles on squirt (http://billiards.colostate.edu/bd_articles/index.html) is looking at all of the details of squirt over the next few months. I will be posting them online after they appear in print.

Now back to Diagram 3. Throw direction depends on the direction of the relative motion of the surface of the cue ball in contact with the object ball. This direction is affected by both cut angle and spin. I think my January '07 (http://billiards.colostate.edu/bd_articles/2007/jan07.pdf) and February '07 (http://billiards.colostate.edu/bd_articles/2007/feb07.pdf) articles illustrate the different possibilities quite well. Please refer to them. I think the throw direction shown in Diagram 3 of my August article is appropriate given the amount of English.

Object ball throw depends on cut angle, shot speed, type and amount of English, and the amount of vertical plane spin (draw, follow, stun). My series of twelve articles on throw elaborate on all of these factors. Collision-induced throw (CIT) and spin-induced throw (SIT) are just different names for throw, depending upon the primary cause of the throw, but the effects don't really combine as separate factors.

Please let me know if you disagree with anything, or if you have further questions.

Regards,
Dave

cushioncrawler
08-28-2007, 06:58 PM
Dr Dave. I think that the diagram iz perfiktly correct, without going into whether there would be enuff english on that qball to throw the OB "back of the line". But, it might be a good idea to show the/a OB line for the/a case of CIT. But this might be diffikult to do (or too confuzing) in the one diagram.

I suppoze that a diagram (of throw and CIT) true to scale would not be very clear, but perhaps a dia true to comparative sizes of max-throw and max-CIT would be good, even if both are exaggerated. madMac.

okinawa77
08-28-2007, 07:59 PM
Dr Dave,
In my opinion/experience, your diagram is correct.
I have found that many pool players confuse and mis-use the terms. I have seen this with various skill levels. And I think when the highly skilled and experienced pool players use the terms incorrectly, it spreads through out the pool playing community like fire.

I replied to a posting earlier about using your thumb on your grip, and was slammed about suggesting experimentation as a positive concept. The response was to trust the more experienced players, rather than re-invent the wheel.
Well, I listen to every person's advice whether experienced or not, but I take it with a grain of salt. What may not work for a 20 year tour pro, may work for someone else.
There are many legendary players that used unorthodoxed strokes, and were very successful.

Cornerman
08-29-2007, 05:54 AM
I like Diagram 4. I think in Diagram 3, the arrows for "squirt angle" aren't correct from a dimensional drawing point of view. I think you just meant to point to it, but the two arrows suggest an enclosed angle. So, the bottom arrow should be pointing to the aiming line. But that's just a minor nitpick.

I have to agree that the throw line could use some modification. From a magnitude point of view, it looks nearly as big as the squirt angle. I think you have it on the correct side, but I don't think it ends up that big. I'm sure I could be proven wrong, but observation tells me differently.

Fred

ken_r
08-29-2007, 06:54 AM
Maybe a better diagram would be showing a straight on shot removing CIT from the equation.

dr_dave
08-29-2007, 09:33 AM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> Dr Dave. I think that the diagram iz perfiktly correct, without going into whether there would be enuff english on that qball to throw the OB "back of the line".<hr /></blockquote>The cut angle in the diagram is just over 40 degrees. Diagram 2 in my January '07 article (http://billiards.colostate.edu/bd_articles/2007/jan07.pdf) shows that "gearing outside English" for that cut angle is just over 50% (see my July '06 article (http://billiards.colostate.edu/bd_articles/2006/july06.pdf) to see how this relates to "tips of English"). In the diagram below, lots of English is indicated (&gt; 80%), so there is "excessive outside English" and the OB will definitely throw to the left of the impact line as shown.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>But, it might be a good idea to show the/a OB line for the/a case of CIT. But this might be diffikult to do (or too confuzing) in the one diagram.<hr /></blockquote>I think Diagram 4 in my January '07 article (http://billiards.colostate.edu/bd_articles/2007/jan07.pdf) and Diagram 5 in my February '07 article (http://billiards.colostate.edu/bd_articles/2007/feb07.pdf) show all of the possible cases fairly well. If you had something else in mind, please let me know.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>I suppoze that a diagram (of throw and CIT) true to scale would not be very clear, but perhaps a dia true to comparative sizes of max-throw and max-CIT would be good, even if both are exaggerated. madMac.<hr /></blockquote>I will try to make the diagram closer to actual scale, but as you point out, it probably won't be as clear. FYI, I have many example shots with various amounts SIT and CIT (including typical maximum values), all drawn to scale, in my series of 12 articles dealing with throw (http://billiards.colostate.edu/bd_articles/index.html).

Regards,
Dave

http://billiards.colostate.edu/images/English_effects.jpg

dr_dave
08-29-2007, 09:42 AM
<blockquote><font class="small">Quote okinawa77:</font><hr> Dr Dave,
In my opinion/experience, your diagram is correct.
I have found that many pool players confuse and mis-use the terms. I have seen this with various skill levels. And I think when the highly skilled and experienced pool players use the terms incorrectly, it spreads through out the pool playing community like fire.<hr /></blockquote>Thank you for your message. There has been a lot of misinformation and misconceptions about squirt and throw in the pool world (books, articles, instructors, pros, etc.) over the years (and even now). That's why Bob Jewett and I have written so many articles on these topics over the years.

Regards,
Dave

dr_dave
08-29-2007, 10:08 AM
<blockquote><font class="small">Quote Cornerman:</font><hr> I like Diagram 4. I think in Diagram 3, the arrows for "squirt angle" aren't correct from a dimensional drawing point of view. I think you just meant to point to it, but the two arrows suggest an enclosed angle. So, the bottom arrow should be pointing to the aiming line. But that's just a minor nitpick.<hr /></blockquote>Fred, I'm not sure what you mean by "enclosed angle." The actual CB path is enclosed by the aiming line and the initial squirt direction, but that's the way it should be. I guess I don't understand what you mean. I've cleaned up the diagram a little (see below). If you still think it misrepresents something, please let me know.

<blockquote><font class="small">Quote Cornerman:</font><hr>I have to agree that the throw line could use some modification. From a magnitude point of view, it looks nearly as big as the squirt angle. I think you have it on the correct side, but I don't think it ends up that big. I'm sure I could be proven wrong, but observation tells me differently.<hr /></blockquote>Good point. Both angles were purposely exaggerated in the diagram to illustrate the effects clearly, but I've reduced the throw angle a little in the revised diagram so it doesn't look so unrealistic. Although, I think it is common for maximum throw angles (3-6 degrees) to be much larger than typical squirt angles (1-2 degrees); although, this obviously varies with equipment and conditions.

Regards,
Dave
http://billiards.colostate.edu/images/English_effects.jpg

dr_dave
08-29-2007, 10:09 AM
<blockquote><font class="small">Quote ken_r:</font><hr> Maybe a better diagram would be showing a straight on shot removing CIT from the equation. <hr /></blockquote>Good idea.

See my November '06 article (http://billiards.colostate.edu/bd_articles/2006/nov06.pdf).

Regards,
Dave

KellyStick
08-29-2007, 11:23 AM
I've never really studied this stuff that close (Squirt mostly). I've heard of these things (Squerve) and had a sort of concept of the meaning but was never fully convinced that this stuff really exists. (Masse yes but swerve not so much, but with slightly elevated butt it makes sense) It seems most people agree that it does in fact exist so...

I seem to be able to account for this stuff most of the time. Though I don't do anything that I am conscious of, most of the time. I do occasional catch myself changing the aim point as I change my stroke or english. It's very subtle and most of the time I don't even notice me doing it... I think... It seems that my brain knows about this but it never informed me of it. Apparently I adjust automatically??

I just wish my brain had explained this to me so I wouldn't feel so ignorant right now... Also, I'm a little afraid of even thinking about this cuz that might really mess me up... like maybe I adjust subconsciously and then adjust again consciously and accidently poke my eye out when I shoot.

Jager85
08-29-2007, 11:36 AM
Dr. Dave,

I have read all of you articles in dealing with throw, squirt/swerve, and deflection. The only thing I am confused about is what gives the max amount of throw. In older articles it is explained that more english=more throw. In the newer articles you have seemed to override this with 50% english with a soft stun shot will maximize throw. In my experience playing since reading the articles it appears to me that 50% english with soft stun shots do, in fact, give more throw.

Is this correct?

Jager

Jal
08-29-2007, 12:42 PM
<blockquote><font class="small">Quote KellyStick:</font><hr> I've never really studied this stuff that close (Squirt mostly). I've heard of these things (Squerve) and had a sort of concept of the meaning but was never fully convinced that this stuff really exists. (Masse yes but swerve not so much, but with slightly elevated butt it makes sense) It seems most people agree that it does in fact exist so...

I seem to be able to account for this stuff most of the time. Though I don't do anything that I am conscious of, most of the time. I do occasional catch myself changing the aim point as I change my stroke or english. It's very subtle and most of the time I don't even notice me doing it... I think... It seems that my brain knows about this but it never informed me of it. Apparently I adjust automatically??

I just wish my brain had explained this to me so I wouldn't feel so ignorant right now... Also, I'm a little afraid of even thinking about this cuz that might really mess me up... like maybe I adjust subconsciously and then adjust again consciously and accidently poke my eye out when I shoot. <hr /></blockquote>It would be exceedingly diffcult to try to calculate this stuff out at the table, unless maybe you happened to be a mathematical savant. You've learned to make adjusments through experience, like everyone else, and that is the best way to continue to go about it.

But for someone just learning the game, an awareness of the variables involved should help speed the learning process. If you already have experience, it could help, but it could also hurt, in my opinion, for the reasons you mentioned.

Jim

Cornerman
08-29-2007, 12:57 PM
<blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote Cornerman:</font><hr> I like Diagram 4. I think in Diagram 3, the arrows for "squirt angle" aren't correct from a dimensional drawing point of view. I think you just meant to point to it, but the two arrows suggest an enclosed angle. So, the bottom arrow should be pointing to the aiming line. But that's just a minor nitpick.<hr /></blockquote>Fred, I'm not sure what you mean by "enclosed angle." <hr /></blockquote>Think "Mechanical Drawing 101." If you drew that angle and dimensioned it per Mechanical Drawing standards, and then compared it to what you have, you'll see what I'm talking about. It's a minor nitpick, nothing more.

Fred

Jal
08-29-2007, 04:01 PM
<blockquote><font class="small">Quote Cornerman:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote Cornerman:</font><hr> I like Diagram 4. I think in Diagram 3, the arrows for "squirt angle" aren't correct from a dimensional drawing point of view. I think you just meant to point to it, but the two arrows suggest an enclosed angle. So, the bottom arrow should be pointing to the aiming line. But that's just a minor nitpick.<hr /></blockquote>Fred, I'm not sure what you mean by "enclosed angle." <hr /></blockquote>Think "Mechanical Drawing 101." If you drew that angle and dimensioned it per Mechanical Drawing standards, and then compared it to what you have, you'll see what I'm talking about. It's a minor nitpick, nothing more.

Fred <hr /></blockquote>I don't know what you mean by enclosed angle either Fred. In diagram 4, the squirt angle is even more exaggerated (perhaps 20 degrees?) than in diagram 3. I would recommend a different cue. (Mac's hollow steel one should do much better than that. In fact, I think around 16 degrees is the worst you could ever see.) But I don't think too many would assume either diagram is to scale. Their "logic" seems to be perfect to me?

Jim

dr_dave
08-29-2007, 04:40 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote KellyStick:</font><hr> I've never really studied this stuff that close (Squirt mostly). I've heard of these things (Squerve) and had a sort of concept of the meaning but was never fully convinced that this stuff really exists. (Masse yes but swerve not so much, but with slightly elevated butt it makes sense) It seems most people agree that it does in fact exist so...

I seem to be able to account for this stuff most of the time. Though I don't do anything that I am conscious of, most of the time. I do occasional catch myself changing the aim point as I change my stroke or english. It's very subtle and most of the time I don't even notice me doing it... I think... It seems that my brain knows about this but it never informed me of it. Apparently I adjust automatically??

I just wish my brain had explained this to me so I wouldn't feel so ignorant right now... Also, I'm a little afraid of even thinking about this cuz that might really mess me up... like maybe I adjust subconsciously and then adjust again consciously and accidently poke my eye out when I shoot. <hr /></blockquote>It would be exceedingly diffcult to try to calculate this stuff out at the table, unless maybe you happened to be a mathematical savant. You've learned to make adjusments through experience, like everyone else, and that is the best way to continue to go about it.

But for someone just learning the game, an awareness of the variables involved should help speed the learning process. If you already have experience, it could help, but it could also hurt, in my opinion, for the reasons you mentioned.<hr /></blockquote>Jim, excellent reply. I couldn't agree more.

Dave

dr_dave
08-29-2007, 04:45 PM
<blockquote><font class="small">Quote Jager85:</font><hr>I have read all of you articles in dealing with throw, squirt/swerve, and deflection. The only thing I am confused about is what gives the max amount of throw. In older articles it is explained that more english=more throw.<hr /></blockquote>This is true, but only up to a point, and the point depends on the type of shot. I'm sorry if I was misleading with this.

<blockquote><font class="small">Quote Jager85:</font><hr>In the newer articles you have seemed to override this with 50% english with a soft stun shot will maximize throw. In my experience playing since reading the articles it appears to me that 50% english with soft stun shots do, in fact, give more throw.
Is this correct?<hr /></blockquote>That's true. For a direct hit (i.e., no cut angle), about a 50% English stun shot gives the maximum SIT. But different amounts of English are required for different cut angles to maximum throw.

Regards,
Dave

dr_dave
08-29-2007, 04:46 PM
<blockquote><font class="small">Quote Cornerman:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote Cornerman:</font><hr> I like Diagram 4. I think in Diagram 3, the arrows for "squirt angle" aren't correct from a dimensional drawing point of view. I think you just meant to point to it, but the two arrows suggest an enclosed angle. So, the bottom arrow should be pointing to the aiming line. But that's just a minor nitpick.<hr /></blockquote>Fred, I'm not sure what you mean by "enclosed angle." <hr /></blockquote>Think "Mechanical Drawing 101." If you drew that angle and dimensioned it per Mechanical Drawing standards, and then compared it to what you have, you'll see what I'm talking about. It's a minor nitpick, nothing more.<hr /></blockquote>Thank you for clarifying.

Dave

wolfdancer
08-29-2007, 06:18 PM
I think I saw this diagram once before in a 1980's Computer Programming book (Basic)explaining a Markov Chain. A drunk is walking him from the bar...and the street has 6 intersections...at each intersection his probability of getting home is 2/3 and going back to the bar 1/3...which are the same odds of my ball finding a pocket.
Realizing that a pool table has 6 pockets, I have also used the Markov Chain for my runout patterns...
talk about confusing....

dr_dave
08-30-2007, 08:23 AM
<blockquote><font class="small">Quote wolfdancer:</font><hr> I think I saw this diagram once before in a 1980's Computer Programming book (Basic)explaining a Markov Chain. A drunk is walking him from the bar...and the street has 6 intersections...at each intersection his probability of getting home is 2/3 and going back to the bar 1/3...which are the same odds of my ball finding a pocket.
Realizing that a pool table has 6 pockets, I have also used the Markov Chain for my runout patterns...
talk about confusing....<hr /></blockquote>Wolfie,

I think it is time for you to self-medicate again.

Dave

Shaft
09-01-2007, 05:31 AM
I think the diagram can be correct, and it can be incorrect, depending on then amount of english used.

A moving CB always has only one axis of rotation. For a center shot, that axis is horizontal, like the axis of a car wheel. For extreme spin, that axis is near vertical, like the axis of a spinning top with little forward motion.

For slight english, the axis tips down slightly on the right for right english. For extreme english, the axis tips way down, approaching vertical. You can get a sense of where the axis is when you use a measles CB and look at the spinning dots.

The diagram is INCORRECT for SLIGHT english where the CB forward motion is greater than the RH spin. I think will be true for most long shots.

The diagram is correct if, at the instant of contact, the spin is great enough so that skin of the cue ball is moving backwards despite the CB forward motion.

(The transition from "incorrect" to "correct" happens when the axis tips more than 45degrees, but the axis is invisible, so that doesn't help much.)

The diagram is CORRECT for EXTREME english. The CB axis must be tipped more than 45 degrees.

Jal
09-01-2007, 05:19 PM
<blockquote><font class="small">Quote Shaft:</font><hr> I think the diagram can be correct, and it can be incorrect, depending on then amount of english used.

A moving CB always has only one axis of rotation. For a center shot, that axis is horizontal, like the axis of a car wheel. For extreme spin, that axis is near vertical, like the axis of a spinning top with little forward motion.

For slight english, the axis tips down slightly on the right for right english. For extreme english, the axis tips way down, approaching vertical. You can get a sense of where the axis is when you use a measles CB and look at the spinning dots.

The diagram is INCORRECT for SLIGHT english where the CB forward motion is greater than the RH spin. I think will be true for most long shots.

The diagram is correct if, at the instant of contact, the spin is great enough so that skin of the cue ball is moving backwards despite the CB forward motion.

(The transition from "incorrect" to "correct" happens when the axis tips more than 45degrees, but the axis is invisible, so that doesn't help much.)

The diagram is CORRECT for EXTREME english. The CB axis must be tipped more than 45 degrees. <hr /></blockquote>Interesting remarks, but perhaps a couple of nitpicks..

When you say the "the spin is great enough so that skin of the cue ball is moving backwards despite the CB forward motion", it's not clear if you mean moving backwards relative to the table or the point of contact on the object ball. Only the latter is necessary to produce the throw direction indicated.

As far as the amount of english required, I'm not sure I agree with the "extreme" characterization. Consider two boundary cases, the first where a stun shot is used, and the second where the cueball reaches natural roll before impact.

Since the cut angle is approximately 45 degrees (let's go with 45), a stun shot that produces zero throw (using gearing english) would require a tip offset of 0.28R, where R is the cueball's radius. This is just a little more than half of maximum english, about 0.5R. So to get the throw direction indicated, you would need a little more than 0.28R.

In the diagram though, the cueball has swerved quite a bit, indicating that it is well on its way to natural roll. If it reached natural roll, it will have lost about 2/7'ths of its initial speed, but very little sidespin. In this case, the offset necessary for zero throw would be about 0.2R.

So an offset a bit greater than 0.2R or 0.28R would be all that's necessary to throw the object ball in the indicated direction. To me, 0.4R could be considered "extreme", but not something substantially less than this.

The orientaion of the spin axis will depend on how much topspin the cueball has acquired. For stun, it will be vertical. The sidepin/final-speed ratio will be about 0.7 for the no throw (transition) condition with either stun or natural roll. If the ball's rolling, the sidespin/roll-spin ratio is also 0.7. This corresponds to a axis orientation of 35 degrees from the horizontal. So for natural roll, this would be the "transition angle" of the spin axis. (It would be 45 degrees at some intermediate state between stun and roll.)

But you might want to check my math.

Jim

dr_dave
09-01-2007, 05:29 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote Shaft:</font><hr> I think the diagram can be correct, and it can be incorrect, depending on then amount of english used.

A moving CB always has only one axis of rotation. For a center shot, that axis is horizontal, like the axis of a car wheel. For extreme spin, that axis is near vertical, like the axis of a spinning top with little forward motion.

For slight english, the axis tips down slightly on the right for right english. For extreme english, the axis tips way down, approaching vertical. You can get a sense of where the axis is when you use a measles CB and look at the spinning dots.

The diagram is INCORRECT for SLIGHT english where the CB forward motion is greater than the RH spin. I think will be true for most long shots.

The diagram is correct if, at the instant of contact, the spin is great enough so that skin of the cue ball is moving backwards despite the CB forward motion.

(The transition from "incorrect" to "correct" happens when the axis tips more than 45degrees, but the axis is invisible, so that doesn't help much.)

The diagram is CORRECT for EXTREME english. The CB axis must be tipped more than 45 degrees. <hr /></blockquote>Interesting remarks, but perhaps a couple of nitpicks..

When you say the "the spin is great enough so that skin of the cue ball is moving backwards despite the CB forward motion", it's not clear if you mean moving backwards relative to the table or the point of contact on the object ball. Only the latter is necessary to produce the throw direction indicated.

As far as the amount of english required, I'm not sure I agree with the "extreme" characterization. Consider two boundary cases, the first where a stun shot is used, and the second where the cueball reaches natural roll before impact.

Since the cut angle is approximately 45 degrees (let's go with 45), a stun shot that produces zero throw (using gearing english) would require a tip offset of 0.28R, where R is the cueball's radius. This is just a little more than half of maximum english, about 0.5R. So to get the throw direction indicated, you would need a little more than 0.28R.

In the diagram though, the cueball has swerved quite a bit, indicating that it is well on its way to natural roll. If it reached natural roll, it will have lost about 2/7'ths of its initial speed, but very little sidespin. In this case, the offset necessary for zero throw would be about 0.2R.

So an offset a bit greater than 0.2R or 0.28R would be all that's necessary to throw the object ball in the indicated direction. To me, 0.4R could be considered "extreme", but not something substantially less than this.

The orientaion of the spin axis will depend on how much topspin the cueball has acquired. For stun, it will be vertical. If natural roll, the sidepin/speed ratio will be about 0.5 for the no throw (transition) condition. Since the ball's rolling, the sidespin/roll-spin ratio is also 0.5. This corresponds to a axis orientation of about 27 degrees from the horizontal. So for natural roll, this would be the "transition angle" of the spin axis. (It would be 45 degrees at some intermediate state between stun and roll.)

But you might want to check my math.

Jim <hr /></blockquote>Wow! Very thorough observations and responses. Sounds good to me.

Dave

Jal
09-01-2007, 05:44 PM
<blockquote><font class="small">Quote dr_dave:</font><hr>Wow! Very thorough observations and responses. Sounds good to me.

Dave <hr /></blockquote>Thanks Dr. Dave. I did check my math (logic) and made some changes (last paragraph), so please ignore the version you read. /ccboard/images/graemlins/smile.gif

(I'm finding that lately I have to think this stuff through maybe ten times before having any hope of getting it right.)

Jim

wolfdancer
09-01-2007, 07:34 PM
Doc, for an academic...you have a good sense of humor...
Now as soon as I get Vista loaded on my computer, I can begin uploading diagrams to the CCB. Having some trouble getting the cds to load.
Hope everybody is having a safe driving holiday weekend
This state of the art computer sold for $1795 new... web page (http://img340.imageshack.us/my.php?image=dscn0008xq3.jpg)

Shaft
09-02-2007, 07:57 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote Shaft:</font><hr> I think the diagram can be correct, and it can be incorrect, depending on then amount of english used.

A moving CB always has only one axis of rotation. For a center shot, that axis is horizontal, like the axis of a car wheel. For extreme spin, that axis is near vertical, like the axis of a spinning top with little forward motion.

For slight english, the axis tips down slightly on the right for right english. For extreme english, the axis tips way down, approaching vertical. You can get a sense of where the axis is when you use a measles CB and look at the spinning dots.

The diagram is INCORRECT for SLIGHT english where the CB forward motion is greater than the RH spin. I think will be true for most long shots.

The diagram is correct if, at the instant of contact, the spin is great enough so that skin of the cue ball is moving backwards despite the CB forward motion.

(The transition from "incorrect" to "correct" happens when the axis tips more than 45degrees, but the axis is invisible, so that doesn't help much.)

The diagram is CORRECT for EXTREME english. The CB axis must be tipped more than 45 degrees. <hr /></blockquote>Interesting remarks, but perhaps a couple of nitpicks..

When you say the "the spin is great enough so that skin of the cue ball is moving backwards despite the CB forward motion", it's not clear if you mean moving backwards relative to the table or the point of contact on the object ball. Only the latter is necessary to produce the throw direction indicated.

As far as the amount of english required, I'm not sure I agree with the "extreme" characterization. Consider two boundary cases, the first where a stun shot is used, and the second where the cueball reaches natural roll before impact.

Since the cut angle is approximately 45 degrees (let's go with 45), a stun shot that produces zero throw (using gearing english) would require a tip offset of 0.28R, where R is the cueball's radius. This is just a little more than half of maximum english, about 0.5R. So to get the throw direction indicated, you would need a little more than 0.28R.

In the diagram though, the cueball has swerved quite a bit, indicating that it is well on its way to natural roll. If it reached natural roll, it will have lost about 2/7'ths of its initial speed, but very little sidespin. In this case, the offset necessary for zero throw would be about 0.2R.

So an offset a bit greater than 0.2R or 0.28R would be all that's necessary to throw the object ball in the indicated direction. To me, 0.4R could be considered "extreme", but not something substantially less than this.

The orientaion of the spin axis will depend on how much topspin the cueball has acquired. For stun, it will be vertical. If natural roll, the sidepin/speed ratio will be about 0.5 for the no throw (transition) condition. Since the ball's rolling, the sidespin/roll-spin ratio is also 0.5. This corresponds to a axis orientation of about 27 degrees from the horizontal. So for natural roll, this would be the "transition angle" of the spin axis. (It would be 45 degrees at some intermediate state between stun and roll.)

But you might want to check my math.

Jim <hr /></blockquote>Wow! Very thorough observations and responses. Sounds good to me.

Dave <hr /></blockquote>

Thanks for writing Jim. I owe you some clarification.

Your first question: My writing was unclear. Since the OB is stationary and the table is stationary, any motion relative to either of these reference points is the same.

But what I was trying to describe is the fact that the resultant direction of the CB contact point with the OB is a combination the CB spin and the CB linear motion. The CB spin might override CIT, as Dave's diagram suggests, or it might merely lessen CIT, with CIT still dominating.

(On further reflection, I take back what I said about 45 degrees being the axis that makes the diagram go from incorrect to correct. That number is correct for only very thin hits. The correct angle also depends on the contact angle with the OB.)

Jim: I am not challenging your math, but I find it difficult to follow it without diagrams. Too bad we can't draw this out on a cocktail napkin at the Corner Pocket....

The point I am making is there are many cases (sorry guys, here comes the tech-speak) where the tangential component of the FORWARD MOTION of the CB contact point is greater in the clockwise direction (using Dave's diagram) than the tangential component of SPIN of the CB contact point in the counterclockwise direction.

The answer is a function of the spin axis and the contact angle.

Paul (Shaft)

Jal
09-02-2007, 03:01 PM
<blockquote><font class="small">Quote Shaft:</font><hr>Thanks for writing Jim. I owe you some clarification.<hr /></blockquote>You're welcome Paul and you definitely do owe me some further explanation. /ccboard/images/graemlins/smile.gif And I appreciate you not getting ticked at the fine-toothed comb treatment, really.

<blockquote><font class="small">Quote Shaft:</font><hr>Your first question: My writing was unclear. Since the OB is stationary and the table is stationary, any motion relative to either of these reference points is the same.

But what I was trying to describe is the fact that the resultant direction of the CB contact point with the OB is a combination the CB spin and the CB linear motion. The CB spin might override CIT, as Dave's diagram suggests, or it might merely lessen CIT, with CIT still dominating.<hr /></blockquote>Yes, I completely agree and my statement with regard to this was poorly worded. I think we both know what's going on. I was just trying to make the point that the contact point on the cueball doesn't have to be moving in a truly retrograde (as Bob Jewett would put it) manner with respect to the table. This is because of where the contact point on the object ball is located (ie, the line from the center of the object ball to it is not at 90 degrees with respect to the cueball's velocity vector). But I'm sure we understand each other.

<blockquote><font class="small">Quote Shaft:</font><hr>(On further reflection, I take back what I said about 45 degrees being the axis that makes the diagram go from incorrect to correct. That number is correct for only very thin hits. ...<hr /></blockquote>I believe we still have some disagreement here. I don't know if you like math so I'll spare you the gruesome details, but if you'd like to pursue it further.... (I could always be wrong too.)

Edit: Okay, I see what you mean Paul. You're assuming a cueball at or near natural roll. Yes, for a thin cut the axis transistion angle would be about 45 degrees. (I kind of glossed over this on the first read.)

<blockquote><font class="small">Quote Shaft:</font><hr>Jim: I am not challenging your math, but I find it difficult to follow it without diagrams. Too bad we can't draw this out on a cocktail napkin at the Corner Pocket.... <hr /></blockquote>Please challenge, here or via PM's, if you like. My math is very challangeable and it's better than being ignored (its usual fate). It's pretty simple in this case, so I have some confidence...but you never know.

<blockquote><font class="small">Quote Shaft:</font><hr>The point I am making is there are many cases (sorry guys, here comes the tech-speak) where the tangential component of the FORWARD MOTION of the CB contact point is greater in the clockwise direction (using Dave's diagram) than the tangential component of SPIN of the CB contact point in the counterclockwise direction.

The answer is a function of the spin axis and the contact angle.<hr /></blockquote>Couldn't agree more. My nitpick was that I don't think it requires what would be considered extreme english for the spin to overcome the tangential component of the cueball's velocity at this cut angle. But "extreme" is a bit subjective and your point is well taken.

Jim

cushioncrawler
09-02-2007, 06:12 PM
Dr Dave -- Just a quick mention of FlatSpot Sqeez again. The diagram and some postings mention "impakt line" and "line of centers" and "contact point" and "instant of contact". So i shood remind everyone that the ball'to'ball impakt takes time, and that the initial contact point iznt uzually the final contact point, and the same can be said about the initial impakt line. Hencely in a friktionless world the OB will allwayz head-off at a wider angle than the initial line of centers. This "extra angle" can be over 1dg for a hi-speed cut. Hencely, to be perfiktly correct, SIT and CIT shood be shown and measured from the OB's friktionless line of travel.

There are other implikations of flatspot sqeez. It means that for perfikt gearing the qball must have a little bit of excess outside english. But, if u/me/everyone thinks about this really really hard, it bekums clear that "perfikt gearing" iz actually an impossibility (splitting atoms here). madMac.

Jal
09-02-2007, 10:05 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>...There are other implikations of flatspot sqeez. It means that for perfikt gearing the qball must have a little bit of excess outside english. But, if u/me/everyone thinks about this really really hard, it bekums clear that "perfikt gearing" iz actually an impossibility (splitting atoms here). madMac. <hr /></blockquote>Mac, the same thought occured to me (although I believe you may have mentioned this before and I didn't appreciate the point). But I'm not so sure.

True, the cut angle continuously grows during impact. But if the cueball traverses, say, a straight chord across the object ball, wouldn't a fixed spin/speed ratio be sufficient to maintain gearing?

How do you see it?

Jim

dr_dave
09-02-2007, 10:34 PM
<blockquote><font class="small">Quote wolfdancer:</font><hr> Doc, for an academic...you have a good sense of humor...<hr /></blockquote>
Thanks. For a non-academic, yours ain't too bad either.

<blockquote><font class="small">Quote wolfdancer:</font><hr>Now as soon as I get Vista loaded on my computer, I can begin uploading diagrams to the CCB. Having some trouble getting the cds to load.
This state of the art computer sold for $1795 new... web page (http://img340.imageshack.us/my.php?image=dscn0008xq3.jpg)<hr /></blockquote>Good one.

Catch you later,
Dave

dr_dave
09-02-2007, 10:40 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> Dr Dave -- Just a quick mention of FlatSpot Sqeez again. The diagram and some postings mention "impakt line" and "line of centers" and "contact point" and "instant of contact". So i shood remind everyone that the ball'to'ball impakt takes time, and that the initial contact point iznt uzually the final contact point, and the same can be said about the initial impakt line. Hencely in a friktionless world the OB will allwayz head-off at a wider angle than the initial line of centers. This "extra angle" can be over 1dg for a hi-speed cut. Hencely, to be perfiktly correct, SIT and CIT shood be shown and measured from the OB's friktionless line of travel.<hr /></blockquote>Mac, I appreciate you trying to keep us honest, but I think this effect is too subtle for most people. The effects of squirt, swerve, and throw are much more significant than ball deformation effects for most pool shots.

Regards,
Dave

cushioncrawler
09-03-2007, 01:22 AM
<blockquote><font class="small">Quote Jal:</font><hr><blockquote><font class="small">Quote cushioncrawler:</font><hr>...There are other implikations of flatspot sqeez. It means that for perfikt gearing the qball must have a little bit of excess outside english. But, if u/me/everyone thinks about this really really hard, it bekums clear that "perfikt gearing" iz actually an impossibility (splitting atoms here). madMac. <hr /></blockquote>Mac, the same thought occured to me (although I believe you may have mentioned this before and I didn't appreciate the point). But I'm not so sure. True, the cut angle continuously grows during impact. But if the cueball traverses, say, a straight chord across the object ball, wouldn't a fixed spin/speed ratio be sufficient to maintain gearing? How do you see it? Jim<hr /></blockquote>Hi Jim -- Hmmmm -- I think that u are (mostly) correct here/there -- if the qball haz gearing at "first contact", then it stands to reezon that it will continue to have gearing despite the effektiv radius of both balls reducing (and then unreducing) during impakt. But i aint (completely) wrong neither, koz it woznt really the changing effektiv radius that woz the reezon for my statement (straining to remember what in hell i woz thinking months ago when i thort all of this throo). To save credibility i will havta fall back on the fakt (only a/my theory actually) that the OB describes an "S" during impakt, before both balls part company and then go "straightish". Yes -- I think that it woz this "S" that led me to say that perfikt gearing iz an impossibility -- but i admit that this iz splitting hairs.

The excess of outside english stuff (for gearing) iz simple enuff and correct -- what i mean iz -- if the initial contact iz at say 45dg, and if the OB's (friktionless) trajectory iz at say 46dg, then the qball's (gearing) rotation needs to be based (calculated) on 46dg, thus the "excess rotation", nothing very special to it really.

Hey, if u want to see a picture of madMac and madSherie and the guys from Great Bear Log Homes, all standing in the corner of my future billiards room, then google great bear log homes (Australia) and look at the testimonials section. madMac.

cushioncrawler
09-03-2007, 02:22 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr> Dr Dave -- Just a quick mention of FlatSpot Sqeez again. The diagram and some postings mention "impakt line" and "line of centers" and "contact point" and "instant of contact". So i shood remind everyone that the ball'to'ball impakt takes time, and that the initial contact point iznt uzually the final contact point, and the same can be said about the initial impakt line. Hencely in a friktionless world the OB will allwayz head-off at a wider angle than the initial line of centers. This "extra angle" can be over 1dg for a hi-speed cut. Hencely, to be perfiktly correct, SIT and CIT shood be shown and measured from the OB's friktionless line of travel.<hr /></blockquote>Mac, I appreciate you trying to keep us honest, but I think this effect is too subtle for most people. The effects of squirt, swerve, and throw are much more significant than ball deformation effects for most pool shots. Regards, Dave<hr /></blockquote>Dr Dave -- Yes, but we must keep above (and ahead of) the plebz at AZB. madMac.

dr_dave
09-03-2007, 09:36 AM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr> Dr Dave -- Just a quick mention of FlatSpot Sqeez again. The diagram and some postings mention "impakt line" and "line of centers" and "contact point" and "instant of contact". So i shood remind everyone that the ball'to'ball impakt takes time, and that the initial contact point iznt uzually the final contact point, and the same can be said about the initial impakt line. Hencely in a friktionless world the OB will allwayz head-off at a wider angle than the initial line of centers. This "extra angle" can be over 1dg for a hi-speed cut. Hencely, to be perfiktly correct, SIT and CIT shood be shown and measured from the OB's friktionless line of travel.<hr /></blockquote>Mac, I appreciate you trying to keep us honest, but I think this effect is too subtle for most people. The effects of squirt, swerve, and throw are much more significant than ball deformation effects for most pool shots. Regards, Dave<hr /></blockquote>Dr Dave -- Yes, but we must keep above (and ahead of) the plebz at AZB. madMac. <hr /></blockquote>Good point, and well stated.

Dave

cushioncrawler
09-03-2007, 05:05 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>....There are other implikations of flatspot sqeez. It means that for perfikt gearing the qball must have a little bit of excess outside english. But, if u/me/everyone thinks about this really really hard, it bekums clear that "perfikt gearing" iz actually an impossibility (splitting atoms here)......

.....To save credibility i will havta fall back on the fakt (only a/my theory actually) that the OB describes an "S" during impakt, before both balls part company and then go "straightish". Yes -- I think that it woz this "S" that led me to say that perfikt gearing iz an impossibility -- but i admit that this iz splitting hairs.....<hr /></blockquote>Father, bless me for i have sinned.

Sin No1..... That "S" iz actually a "J" -- dont know why i keep calling it an S when it must be an elongated J.

Sin No2..... And, i might be (must be) wrong about the impossibility of "perfikt gearing". No matter how u look at it, if 2 (identikal) balls have perfikt gearing at first contact, then perfikt gearing must exist for the whole contakt, unless some outside agency acts on one of the balls. "Conservation of (rotational) momentum" actually means little more than "action equals reaction" -- karnt touch this.

Sin No3..... And, if the initial line of centers iz at say 45dg, then the needed gearing-spin must relate to 45dg, not 46dg, even if the OB duz head off at 46dg (due to sqeez). This iz simple enuff to vizualize, depending on how much red u have drunk. madMac.

Jal
09-04-2007, 12:08 AM
You know Mac, some of this is almost excrucuiating to think about. In the end, you either drop it, go insane, or come away with a deeper understanding. The first and third options generally elude me.

I have to agree with no. 1, reluctantly, since it was one of my favorite bits of minutia.

I don't see that no. 2 is necessarily true, ie, perfect gearing must hold throughout. The throw on both balls from any friction (static or kinetic) would guarantee that angular momentum is conserved. The RXP stuff would compensate for the change in spin of both balls.

The amount of throw generated by this would be exceedingly small, since it would take only the slightest touch of friction to bring them back into gearing. Looking through a microscope at the details over some very small interval of time, I think it would be hard to tell if it was in a state of frictionless gearing, or that some minute force had developed to re-synchronize the surfaces.

No. 3 is a little hard for me to see since I am cold stone sober at the moment. I think the problem is not knowing the exact path the cueball takes between the entry and exit points. Have you come to some clear way of looking at this?

Jim

dr_dave
09-04-2007, 08:52 AM
I've been trying to follow your discussion on perfect gearing, but I'm not sure I understand what you guys are suggesting. I still think the flat-spot squeeze effect is too small to consider for most shots, as compared to all of the other effects that come into play. But for now, let's assume it is a large effect. With gearing outside English, and flatspot squeeze, do you think there could be tangential impulse that would throw and tranfer spin to the object ball?

Thanks,
Dave

<blockquote><font class="small">Quote Jal:</font><hr> You know Mac, some of this is almost excrucuiating to think about. In the end, you either drop it, go insane, or come away with a deeper understanding. The first and third options generally elude me.

I have to agree with no. 1, reluctantly, since it was one of my favorite bits of minutia.

I don't see that no. 2 is necessarily true, ie, perfect gearing must hold throughout. The throw on both balls from any friction (static or kinetic) would guarantee that angular momentum is conserved. The RXP stuff would compensate for the change in spin of both balls.

The amount of throw generated by this would be exceedingly small, since it would take only the slightest touch of friction to bring them back into gearing. Looking through a microscope at the details over some very small interval of time, I think it would be hard to tell if it was in a state of frictionless gearing, or that some minute force had developed to re-synchronize the surfaces.

No. 3 is a little hard for me to see since I am cold stone sober at the moment. I think the problem is not knowing the exact path the cueball takes between the entry and exit points. Have you come to some clear way of looking at this?

Jim <hr /></blockquote>

Jal
09-04-2007, 11:17 AM
<blockquote><font class="small">Quote dr_dave:</font><hr>...With gearing outside English, and flatspot squeeze, do you think there could be tangential impulse that would throw and tranfer spin to the object ball?<hr /></blockquote>Hi Dr. Dave,

Really, I don't know? I'd of course be interested in what you'd have to say...if you were interested enough to mull it over. But for sure, the effect of having to adjust the spin of the balls in order to maintain gearing is small (but not really exceedingly small as I stated above).

Suppose the cut angle is increased by 1 degree by the contact time on an 80 degree cut shot. The difference in surface speed, keeping spin constant, between the initial and final contact would be V[sin(82)-sin(80)], ie, 81 is the average of 80 and 82. To maintain gearing would require that 1/7'th of this surface speed be taken up as throw (as per TP A.14). The throw angle would then be approximately:

Atan[(1/7)V(sin(82)-sin(80))/Vcos(81)] = 0.29 deg

So it does border on significant and is certainly much more than zero throw. The question is, does the spin really have to undergo a change (with concomitant throw) to maintain gearing? In other words, I'm tossing it back to you? /ccboard/images/graemlins/smile.gif

Jim

dr_dave
09-04-2007, 11:26 AM
Jim,

Thanks for clarifying and for giving me some things to think about. I'll have to add this topic to "my list." You and Mac have convinced me that flat-spot squeeze is at least worthy of consideration. Unfortunately, my list keeps getting longer and longer (especially lately), so it might take me a while to get around to it.

Catch you later,
Dave

<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr>...With gearing outside English, and flatspot squeeze, do you think there could be tangential impulse that would throw and tranfer spin to the object ball?<hr /></blockquote>Hi Dr. Dave,

Really, I don't know? I'd of course be interested in what you'd have to say...if you were interested enough to mull it over. But for sure, the effect of having to adjust the spin of the balls in order to maintain gearing is small (but not really exceedingly small as I stated above).

Suppose the cut angle is increased by 1 degree by the contact time on an 80 degree cut shot. The difference in surface speed, keeping spin constant, between the initial and final contact would be V[sin(82)-sin(80)], ie, 81 is the average of 80 and 82. To maintain gearing would require that 1/7'th of this surface speed be taken up as throw (as per TP A.14). The throw angle would then be approximately:

Atan[(1/7)V(sin(82)-sin(80))/Vcos(81)] = 0.29 deg

So it does border on significant and is certainly much more than zero throw. The question is, does the spin really have to undergo a change (with concomitant throw) to maintain gearing? In other words, I'm tossing it back to you? /ccboard/images/graemlins/smile.gif

Jim






<hr /></blockquote>

cushioncrawler
09-04-2007, 04:42 PM
<blockquote><font class="small">Quote Jal:</font><hr>...I have to agree with no. 1, reluctantly, since it was one of my favorite bits of minutia....<hr /></blockquote>Yeah, with this S and J stuff, i had (have) fallen into the Ex-Wife-Syndrome. Everyone who haz been marryd 3 or 4 times haz suffered. U start out by calling the new-wife by the old-wife's name, at the worst moments. Then, for a while, the brain learns to swap to the correct name even while the (wrong) name iz starting to leev yor throat. Then, kumz the worst time -- here u have now ingrained the new wife's name properly, but havent yet shaken off the "swaparoo" -- u start to say the new-wife's name, but, horror, swap to the ex-wife's name -- and there iz no "edit" button. Actually, madMac'n'madSherie's honeymoon woz hilarious -- bothly being new ex's -- we often both blurted out the wrong name at the same (worst) moment. madMac.

cushioncrawler
09-04-2007, 05:00 PM
<blockquote><font class="small">Quote Jal:</font><hr>.... I don't see that no. 2 is necessarily true, ie, perfect gearing must hold throughout. The throw on both balls from any friction (static or kinetic) would guarantee that angular momentum is conserved. The RXP stuff would compensate for the change in spin of both balls. The amount of throw generated by this would be exceedingly small, since it would take only the slightest touch of friction to bring them back into gearing. Looking through a microscope at the details over some very small interval of time, I think it would be hard to tell if it was in a state of frictionless gearing, or that some minute force had developed to re-synchronize the surfaces.....<hr /></blockquote>Jim -- I havta go along with Newton here, ie that Action = Reaction. I often "uze" an non-moveable (non-yielding, infinitely hard'n'heavy) wall to help my thinking. But, in the prezent instance, the hard wall lies. If i follow yor thinking correctly -- a ball hitting the wall at an angle, with (initially) perfikt gearing, would suffer some tangential (YY) friktion, koz of the deformation in the flatspot and (moreso) the deformation of the (now eggshaped) ball. This might be (will be) so, but, the correct wall to uze in this thort-experiment needs to be infinitely compliant in the YY direction, koz, whatever happenz to the qball allso happenz to the OB. The OB acts like a hard wall in the XX plane, but iz compliant in the YY plane (and ZZ duznt kum into things here). Duz this fix things??? madMac.

cushioncrawler
09-04-2007, 05:52 PM
<blockquote><font class="small">Quote Jal:</font><hr>.... No. 3 is a little hard for me to see since I am cold stone sober at the moment. I think the problem is not knowing the exact path the cueball takes between the entry and exit points. Have you come to some clear way of looking at this?....<hr /></blockquote>Jim (and Dr Dave) -- Just recapping.

...... The force acting on the OB (and qball) iz allwayz acting along the line'of'centers at every instant ((pleez ignore this blatent introduction of a circular arguement)). The initial LOC (at first contact) iz say 2dg off the final LOC (at last contact), and the OB flyz off nearnuff midway (ie the sqeez here would add 1dg to the cut angle).

...... Nothing new here (above). It will be eezy for me (later in 2007) to calculate the OB's trajectory (koz i allready have graphs relating speed and force and flatspot diameter and time, allbeit for a small cheap poolball). The OB trajectory will be a sort of elongated J. The final line of the OB will "miss" the "starting point" by perhaps 0.2mm (no big deal, but its the thort that counts).

...... The "path the qball takes" during impakt duznt matter. If Dr Dave filmed from overhead, he could moov (and orientate) the camera at a constant speed and direction (and orientation), such that the OB and qball mooved equally (from top and bottom) towards the center of the pickture (impakt), and then exited equally, ie it would all look like a perfikt mirror image (except that the OB iznt white). Here iz that wall (mirror) -- the wall runs across the center of the picture, horizontally (and duznt moov). Hmmmm -- Actually, here i/we have defined a wall that iz fixed in the XX and the YY planes, yet "duz the trick", ie this here wall duznt needtabe YY-compliant.

Jim (Dr Dave) -- Duz any of this go to the heart of the issue??? In a nutshell, can a perfikt mirror image be created and can it be maintained??? If it can (and i think it can), then perfikt gearing can exist. This mirror image stuff would havta apply too (and i think it duz) to any and all bits of the flatspot(s), even if "blown up" to fill the whole picture.

Hmmmm -- One little issue. Perfikt gearing duznt mean that all of the atoms in one ball have the same angular velocity. It only means that any molecule on one ball (so to speak), "touching" any molecule on the other ball, haz the same linear velocity (at any/every instant). Here, once "properly touching", the velocity(s) will all be in the YY plane (so to speak), ie XX velocity(s) will be zero. madMac.

Jal
09-04-2007, 06:48 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr>.... I don't see that no. 2 is necessarily true, ie, perfect gearing must hold throughout. The throw on both balls from any friction (static or kinetic) would guarantee that angular momentum is conserved. The RXP stuff would compensate for the change in spin of both balls. The amount of throw generated by this would be exceedingly small, since it would take only the slightest touch of friction to bring them back into gearing. Looking through a microscope at the details over some very small interval of time, I think it would be hard to tell if it was in a state of frictionless gearing, or that some minute force had developed to re-synchronize the surfaces.....<hr /></blockquote>Jim -- I havta go along with Newton here, ie that Action = Reaction. I often "uze" an non-moveable (non-yielding, infinitely hard'n'heavy) wall to help my thinking. But, in the prezent instance, the hard wall lies. If i follow yor thinking correctly -- a ball hitting the wall at an angle, with (initially) perfikt gearing, would suffer some tangential (YY) friktion, koz of the deformation in the flatspot and (moreso) the deformation of the (now eggshaped) ball. This might be (will be) so, but, the correct wall to uze in this thort-experiment needs to be infinitely compliant in the YY direction, koz, whatever happenz to the qball allso happenz to the OB. The OB acts like a hard wall in the XX plane, but iz compliant in the YY plane (and ZZ duznt kum into things here). Duz this fix things??? madMac. <hr /></blockquote>I'm not so sure about the wall Mac; that's another difficult problem.

What I was getting at, is that any friction that develops between the balls, even if it started out as perfect gearing , will not change the angular momentum of the system of the two balls. As you indicated, this friction is an internal force of the two ball system, and therefore cannot change any dynamic property of the combined system, including angular momentum. That doesn't mean though, that the spins of the balls can't change, and in fact surely will if some friction develops. And since the friction acts in opposite directions, both balls will acquire the same amount of spin from this (a change in spin in the case of the cueball).

This is not a violation of the conservation of angular momentum because there is a different kind of angular momentum involved here as well. If an object is moving in a straight line through space, and you pick any arbitrary point in space such that the moving object is a vector distance R from that point, it has an angular momentum about it equal to RXP, where P is its linear momentum mV. This is the vector cross product of R and P. If the object is also spinning with angular momentum L, it also has this angular momentum about that same point (it has L about every point in space).

So what happens when friction changes the spin and linear momentum of each ball, is that the new sum of the spin angular momenta (the L's), plus the new angular momenta associated with their new linear momenta (the RXP's), remains the same. The changes in the RXP's exactly cancel the changes in the L's. This is all referenced to some arbitrary point in space. (The combined angular momentum is different for different points in space, but doesn't change for any particular point.)

So, the upshot is that we can't appeal to conservation of angular momentum to simplify things.

I did look at this some more, and it turns out that if you assume a perfectly symmetric situation, the cueball will have exact gearing spin as it departs. By symmetric, I mean that the object ball's post-impact direction exactly splits (bisects) the line connecting the initial and final contact points, and its speed is the cueball's pre-impact velocity component in this direction. This is not true in reality, but it indicates that much less friction is needed to maintain gearing (if any) than the calculation I did earlier indicates (which was a worst case scenario).

While I doubt that anything like perfect gearing obtains throughout, which was your original point (that it's not perfect), it's probably also true that very little throw results from this.

Jim

cushioncrawler
09-04-2007, 09:14 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr>.... I don't see that no. 2 is necessarily true, ie, perfect gearing must hold throughout. The throw on both balls from any friction (static or kinetic) would guarantee that angular momentum is conserved. The RXP stuff would compensate for the change in spin of both balls. The amount of throw generated by this would be exceedingly small, since it would take only the slightest touch of friction to bring them back into gearing. Looking through a microscope at the details over some very small interval of time, I think it would be hard to tell if it was in a state of frictionless gearing, or that some minute force had developed to re-synchronize the surfaces.....<hr /></blockquote>Jim -- I havta go along with Newton here, ie that Action = Reaction. I often "uze" an non-moveable (non-yielding, infinitely hard'n'heavy) wall to help my thinking. But, in the prezent instance, the hard wall lies. If i follow yor thinking correctly -- a ball hitting the wall at an angle, with (initially) perfikt gearing, would suffer some tangential (YY) friktion, koz of the deformation in the flatspot and (moreso) the deformation of the (now eggshaped) ball. This might be (will be) so, but, the correct wall to uze in this thort-experiment needs to be infinitely compliant in the YY direction, koz, whatever happenz to the qball allso happenz to the OB. The OB acts like a hard wall in the XX plane, but iz compliant in the YY plane (and ZZ duznt kum into things here). Duz this fix things??? madMac. <hr /></blockquote>I'm not so sure about the wall Mac; that's another difficult problem.

What I was getting at, is that any friction that develops between the balls, even if it started out as perfect gearing , will not change the angular momentum of the system of the two balls. As you indicated, this friction is an internal force of the two ball system, and therefore cannot change any dynamic property of the combined system, including angular momentum. That doesn't mean though, that the spins of the balls can't change, and in fact surely will if some friction develops. And since the friction acts in opposite directions, both balls will acquire the same amount of spin from this (a change in spin in the case of the cueball).

This is not a violation of the conservation of angular momentum because there is a different kind of angular momentum involved here as well. If an object is moving in a straight line through space, and you pick any arbitrary point in space such that the moving object is a vector distance R from that point, it has an angular momentum about it equal to RXP, where P is its linear momentum mV. This is the vector cross product of R and P. If the object is also spinning with angular momentum L, it also has this angular momentum about that same point (it has L about every point in space).

So what happens when friction changes the spin and linear momentum of each ball, is that the new sum of the spin angular momenta (the L's), plus the new angular momenta associated with their new linear momenta (the RXP's), remains the same. The changes in the RXP's exactly cancel the changes in the L's. This is all referenced to some arbitrary point in space. (The combined angular momentum is different for different points in space, but doesn't change for any particular point.)

So, the upshot is that we can't appeal to conservation of angular momentum to simplify things.

I did look at this some more, and it turns out that if you assume a perfectly symmetric situation, the cueball will have exact gearing spin as it departs. By symmetric, I mean that the object ball's post-impact direction exactly splits (bisects) the line connecting the initial and final contact points, and its speed is the cueball's pre-impact velocity component in this direction. This is not true in reality, but it indicates that much less friction is needed to maintain gearing (if any) than the calculation I did earlier indicates (which was a worst case scenario).

While I doubt that anything like perfect gearing obtains throughout, which was your original point (that it's not perfect), it's probably also true that very little throw results from this. Jim<hr /></blockquote>Jim -- Hmmmm, i am afraid that i am starting to agree with u.

...... If the qball goze eggshaped during impakt then the qball spin-rate will fall (koz the rotational inertia of an egg-shape iz greater).

...... If the spin-rate falls, then we are entitled to think that ball'to'ball slippage (or at least friktion) exists. But, this slippage might actually be (qball) overspin (in relation to the OB), it depends on the depth (loss of radius) of the flatspot(s).

...... In the end, whether we all agree on whether "perfikt gearing" exists (or iz possible) will depend on whether we all agree on the definition.

...... If both ballz are spherical at first contact, then they are likely to be both spherical at last contact. Thusly, if we have gearing at first contact, and if friktion (mu) iz zero, then we will have gearing at last contact. But we will allmost certainly have slippage in the meantime (koz mu = zero).

...... Above -- If friktion (mu) iznt zero, then we wont/karnt have gearing at final contact.

...... With a strict'n'tight definition of "perfikt gearing" then neither event makes the grade.

...... This unhappy situation iz not the fault of the Prezident. It iz due to (a)the flatspot, and (b) the egg-shape(s), and (c) the J trajektory of the OB (which iz due to the flatspot mainly).

...... "What I was getting at, is that any friction that develops between the balls, even if it started out as perfect gearing....". (Jim -- Why not try boxer shorts??)

Jim -- Well dunn. Any more thorts?? madMac.

Jal
09-04-2007, 09:48 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>...... The force acting on the OB (and qball) iz allwayz acting along the line'of'centers at every instant ((pleez ignore this blatent introduction of a circular arguement)). The initial LOC (at first contact) iz say 2dg off the final LOC (at last contact), and the OB flyz off nearnuff midway (ie the sqeez here would add 1dg to the cut angle).<hr /></blockquote>Yep.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>...... Nothing new here (above). It will be eezy for me (later in 2007) to calculate the OB's trajectory (koz i allready have graphs relating speed and force and flatspot diameter and time, allbeit for a small cheap poolball)....<hr /></blockquote>The forces between the balls get up there in the thousands of pounds. Do you have data even in this region?

<blockquote><font class="small">Quote cushioncrawler:</font><hr>...... The "path the qball takes" during impakt duznt matter. If Dr Dave filmed from overhead, he could moov (and orientate) the camera at a constant speed and direction (and orientation), such that the OB and qball mooved equally (from top and bottom) towards the center of the pickture (impakt), and then exited equally, ie it would all look like a perfikt mirror image (except that the OB iznt white). Here iz that wall (mirror) -- the wall runs across the center of the picture, horizontally (and duznt moov). Hmmmm -- Actually, here i/we have defined a wall that iz fixed in the XX and the YY planes, yet "duz the trick", ie this here wall duznt needtabe YY-compliant.

Jim (Dr Dave) -- Duz any of this go to the heart of the issue??? In a nutshell, can a perfikt mirror image be created and can it be maintained??? If it can (and i think it can), then perfikt gearing can exist. This mirror image stuff would havta apply too (and i think it duz) to any and all bits of the flatspot(s), even if "blown up" to fill the whole picture.<hr /></blockquote>Mac, contrary to what I may have insinuated earlier, I don't drink very often....but, I'm having a glass of vino after reading this. Very nice!

Looking at it from the center of mass frame does certainly clear away some of the fog. It's obvious from this that the center of the contact area must travel in a straight line in this frame, and therefore in the table frame too. I'm not sure yet what this means for gearing (any more thoughts?), but it's a very good thing to know regarding all this flatspot business. Thanks Mac.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>Hmmmm -- One little issue. Perfikt gearing duznt mean that all of the atoms in one ball have the same angular velocity. It only means that any molecule on one ball (so to speak), "touching" any molecule on the other ball, haz the same linear velocity (at any/every instant). Here, once "properly touching", the velocity(s) will all be in the YY plane (so to speak), ie XX velocity(s) will be zero. madMac. <hr /></blockquote>Still chewing on this.

Jim

cushioncrawler
09-05-2007, 02:14 AM
Jim -- It looks like i have converted u and u have converted me, and our conversions crossed in cyberspace. Now we are argueing on opposite sides again. This would never happen on the NPR side of the forum. However, there is the danger (perhaps) that both of us are wrong. More wine. I'll be back. madMac.

ABQ_Poolhead
09-09-2007, 05:27 PM
Dave, I would like to study the effects of swerve more, but haven't found much info on it. I'm curious to see graphs indicating amount / distance of swerve in relation to speed and tip offset? I didn't see this info on your site. Did I miss it, or am I jumping the gun?
Thank you for all the information you share with us.
ABQ

Cornerman
09-10-2007, 07:06 AM
<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> Dave, I would like to study the effects of swerve more,<hr /></blockquote>The effects of swerve? Maybe you mean the cause of swerve?

Swerve itself is an effect. And other than the fact that the cueball will not travel a straight line (and therefore hit the object ball somewhere other than the aimline/aimpoint), what other effects could you be talking about?

Fred

dr_dave
09-10-2007, 08:17 AM
<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> Dave, I would like to study the effects of swerve more, but haven't found much info on it. I'm curious to see graphs indicating amount / distance of swerve in relation to speed and tip offset? I didn't see this info on your site. Did I miss it, or am I jumping the gun?
Thank you for all the information you share with us.
ABQ<hr /></blockquote>My October '07 article (http://billiards.colostate.edu/bd_articles/index.html) will discuss swerve some more. My November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf) presents and illustrates a masse-shot aiming technique. This might also be of interest to you because masse is just swerve on a bigger scale. If you want to see the math and physics details of swerve and masse shots (with some example data and cue ball trajectories), see TP A.19 (http://billiards.colostate.edu/technical_proofs/new/TP_A-19.pdf). The difficulty with swerve is that the amount of cue ball drift depends on cue elevation, the type and amount of English, shot speed and distance, the type and condition of the cloth, and ball conditions. Swerve is one of those things I think is best dealt with intuitively (once you understand the basic factors). You can also try to minimize it by keeping the cue as level as possible and by using faster shot speed (where possible and appropriate). You can also wipe the cue ball with silicone spray. That does wonders to keep swerve at a minimum; although, your opponent might not appreciate it (unless he is a trick shot artists ... they like silicone spray ... it helps make masse shots easier and more dramatic).

Regards,
Dave

ABQ_Poolhead
09-10-2007, 08:33 AM
Fred, I'm interested in Swerve because it cost me a match. Until that match, I hadn't thought swerve was enough of a factor to affect my game. Now that I know better, I want to learn about it so that I can adjust for it.

ABQ_Poolhead
09-10-2007, 08:56 AM
Dave, thank you for your reply. I look forward to reading your article. As I repled to Cornerman, swerve recently cost me a match. I attempted a safety on my problem ball that would've really screwed my opponent and fixed my problem. The distance was probably about 6' on an 8' table with 860 Simonis. I used a soft hit with approx. 1 tip right english (probably optimum conditions for pronounced swerve), then watched in horror as swerve took place. Failed to contact my object ball, and my opponent easily ran out to take the match. Obviously, I now know not to attempt that shot again unless I want the swerve. A better understanding of swerve on my part will help me to adjust for it, and even use it to my advantage. Thanks for any help you can give me.
I really enjoyed your throw articles and learned a lot. When I learned through your research that throw is strongest on a cut shot when the cue ball is sliding, I was able to employ that immediately in my play, with positive results. Thanks for your diligent study, and for freely sharing.
ABQ

Bob_Jewett
09-10-2007, 11:48 AM
<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> ... A better understanding of swerve on my part will help me to adjust for it, and even use it to my advantage. ... <hr /></blockquote>
Briefly:

The more elevated the cue is the larger the swerve angle. The more draw you use with the side, the larger the swerve angle. The more draw you use, the longer the swerve will take, as the cue ball is curving until it is rolling smoothly which is when all the draw has worn off. The stickier the cloth, the faster the swerve will take (and the sooner the draw will wear off). The harder you shoot, the longer the swerve will be (assuming you are not starting the cue ball with smooth rolling besides the side). You elevate on almost all shots; if you did not elevate, side spin would not cause swerve.

The swerve angle can actually be calculated from the system for calculating masse paths, since they are essentially the same phenomenon. The masse calculation (which was originally published by Coriolis in 1835) is covered in Robert Byrne's "Advanced Technique" book.

Cornerman
09-10-2007, 12:09 PM
<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> Fred, I'm interested in Swerve because it cost me a match. Until that match, I hadn't thought swerve was enough of a factor to affect my game. Now that I know better, I want to learn about it so that I can adjust for it. <hr /></blockquote>Okay.

I'm not alone in saying that of the three major things that affect the outcome of a shot (throw, swerve, squirt), swerve is highest on the ladder.

Fred

Jal
09-10-2007, 01:57 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr>...The swerve angle can actually be calculated from the system for calculating masse paths, since they are essentially the same phenomenon....<hr /></blockquote>In my opinion, this system is the single most amazing fact about billiards. That it could possibly be so direct as a line drawn from the base of the cueball to the spot on the table pointed at by the shaft, is really remarkable. And while it's relatively easy to prove if you already know the answer, what clue might have led Coriolis to it?

True, it needs some correction when the real world nasties get a hold of it (at least for squirt), but still...

Jim

ABQ_Poolhead
09-10-2007, 02:32 PM
Fred, I respectfully disagree. I've never seen an object ball "swerved" into a pocket, but it's not too hard to throw the ob into a pocket. A long shot on a 9' table using 1 tip english on cue ball makes me have to adjust almost a full ball to hit the object ball where I want (med. speed). Swerve almost seems to me like the silent killer. The shot I previously mentioned deviated less then half an inch over 6 feet, and that at slow speed where the masse had optimum effect. That's my take, at least.
ABQ

ABQ_Poolhead
09-10-2007, 02:35 PM
Thanks for the info, Bob. I have that book you mentioned, and will re-read it. I will also experiment with the info you gave me until. Thanks again.
ABQ

dr_dave
09-10-2007, 02:51 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr>The more elevated the cue is the larger the swerve angle. The more draw you use with the side, the larger the swerve angle. The more draw you use, the longer the swerve will take, as the cue ball is curving until it is rolling smoothly which is when all the draw has worn off. The stickier the cloth, the faster the swerve will take (and the sooner the draw will wear off). The harder you shoot, the longer the swerve will be (assuming you are not starting the cue ball with smooth rolling besides the side). You elevate on almost all shots; if you did not elevate, side spin would not cause swerve.<hr /></blockquote>Great summary, Bob. This is the type of knowledge and understanding I think helps people with their intuition-building practice.

<blockquote><font class="small">Quote Bob_Jewett:</font><hr>The swerve angle can actually be calculated from the system for calculating masse paths, since they are essentially the same phenomenon. The masse calculation (which was originally published by Coriolis in 1835) is covered in Robert Byrne's "Advanced Technique" book.<hr /></blockquote>If people want a convenient and well-illustrated presentation of the method with examples, they can refer to my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf). The Coriolis version is particularly difficult to comprehend for a non-technical reader (especially the original French version).

Regards,
Dave

Cornerman
09-10-2007, 03:57 PM
<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> Fred, I respectfully disagree. I've never seen an object ball "swerved" into a pocket, but it's not too hard to throw the ob into a pocket. A long shot on a 9' table using 1 tip english on cue ball makes me have to adjust almost a full ball to hit the object ball where I want (med. speed). Swerve almost seems to me like the silent killer. The shot I previously mentioned deviated less then half an inch over 6 feet, and that at slow speed where the masse had optimum effect. That's my take, at least.
ABQ <hr /></blockquote>You say you disagree, then go on and say all the reasons why you would actually agree.

Fred

dr_dave
09-10-2007, 04:24 PM
ABQ,

Bob's list (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=257319&amp;page =0&amp;view=collapsed&amp;sb=5&amp;o=&amp;vc=1) gives you the basic understanding you need to start building better intuition. I also agree with Fred (per his post (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=257315&amp;page =0&amp;view=collapsed&amp;sb=5&amp;o=&amp;vc=1)) that swerve is at the top of the list of English concerns. Squirt is relatively easy to judge and compensate. Throw has many subtleties; but with good understanding, it is also fairly easy to manage. I think squirt is one of those things that, even with good understanding, just requires a lot of feel (intuition created by years of successful practice and experience).

Good luck with your intuition building,
Dave

PS: Thanks for the nice words about my stuff.

<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> Dave, thank you for your reply. I look forward to reading your article. As I repled to Cornerman, swerve recently cost me a match. I attempted a safety on my problem ball that would've really screwed my opponent and fixed my problem. The distance was probably about 6' on an 8' table with 860 Simonis. I used a soft hit with approx. 1 tip right english (probably optimum conditions for pronounced swerve), then watched in horror as swerve took place. Failed to contact my object ball, and my opponent easily ran out to take the match. Obviously, I now know not to attempt that shot again unless I want the swerve. A better understanding of swerve on my part will help me to adjust for it, and even use it to my advantage. Thanks for any help you can give me.
I really enjoyed your throw articles and learned a lot. When I learned through your research that throw is strongest on a cut shot when the cue ball is sliding, I was able to employ that immediately in my play, with positive results. Thanks for your diligent study, and for freely sharing.
ABQ <hr /></blockquote>

cushioncrawler
09-10-2007, 05:52 PM
<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> Dave, thank you for your reply. I look forward to reading your article. As I repled to Cornerman, swerve recently cost me a match. I attempted a safety on my problem ball that would've really screwed my opponent and fixed my problem. The distance was probably about 6' on an 8' table with 860 Simonis. I used a soft hit with approx. 1 tip right english (probably optimum conditions for pronounced swerve), then watched in horror as swerve took place. Failed to contact my object ball, and my opponent easily ran out to take the match. Obviously, I now know not to attempt that shot again unless I want the swerve. A better understanding of swerve on my part will help me to adjust for it, and even use it to my advantage. Thanks for any help you can give me.
I really enjoyed your throw articles and learned a lot. When I learned through your research that throw is strongest on a cut shot when the cue ball is sliding, I was able to employ that immediately in my play, with positive results. Thanks for your diligent study, and for freely sharing.ABQ<hr /></blockquote>ABQ -- I am wondering why u would want to play a soft shot with 1 tip of cue offset.

......Did u want some (masse) swerve??

......If YES, why??

......Or, didnt u realize that tip-offset givz swerve??

......I reckon that u kood set up that shot, and play it the way u tryd 10 times, and get 10 very different rezults. This iz of course koz soft long shots are fraught. But, soft long shots with 1 tip of english must be suicide, even on a "good" table.

......U say u were horryfyd to see some swerv. For a soft shot swerv would happen in allmost an instant. So, i am thinking that what u saw woz some late-swerv. Now, there iz no such thing az late masse-swerve (for a soft hit) -- once a ball iz "rolling" there can be no masse-swerv.

...... So, (perhaps) what u saw shood perhaps be called something else -- Veer -- Drift -- Kurv -- or Roll-Off -- or ?????.

......Veer etc can be caused by the table being out of level, or by a "river" of chalk under the cloth, or by the nap (if any) being slightly direktional and thusly affekting the roll direktion. Theze effekts are all due to gravity.

......There iz one cause of Veer that iznt due to gravity. A spinning ball will veer (try to veer) away from the spin az it slowz.

...... So, the above raizes a question -- which way did the qball swerv, "with" the spin (ie normal swerv), or, "away" from the spin (in which case it would havta be due to one of (any or all of) the above reezon (and others).

......Allso, many players would prefer to shoot soft shots by uzing drag, ie hardish, ie to minimize problems.

......But, (repeating) why uze soft english?????????????? madMac.

ABQ_Poolhead
09-11-2007, 08:54 AM
Mac, I played the shot that way because I had only one ball left on table to shoot other than the eight. My ball was almost frozen to the foot rail 2-3 inches to the right of the pocket. My intent was to bump the object ball off the rail and leave blocking the pocket, and for the cue ball to stay toward that same end of table (hence the slow speed) but go sharply right to partially hide behind the 8-ball and at least partially hook my opponent.
Haven't heard of "veer" before. Could you please elaborate?
ABQ

dr_dave
09-11-2007, 11:23 AM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr>The more draw you use with the side, the larger the swerve angle. The more draw you use, the longer the swerve will take, as the cue ball is curving until it is rolling smoothly which is when all the draw has worn off.<hr /></blockquote>I think this is an important topic that many people don't fully understand. My October '07 article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf) has some good illustrations for the swerve-delay effect and how it impacts "effective squirt" ("squerve"). I think the difference in swerve angle between a draw and follow shot with the same amount of English is small as compared to the effect of the swerve delay. The results of the experiments we discussed by phone are in the article. Thanks again for your helpful ideas.

I hadn't thought a lot about the increase in swerve angle with a draw shot as compared to a follow shot with the same amount of English. I know the aim point on the table (point "A" in my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf)) changes with height of the ball contact point (point "B") and with the cue elevation (which might be slightly larger for a draw shot), but are there some other effects you think cause the difference? Also, how significant do you think this difference is? I think this is tough to explain and demonstrate, because a typical draw shot will have less "effective squirt" ("squerve") than a typical follow shot with the same amount of English. It appears to some that the draw shot is squirting more and/or swerving less, but we know the squirt is the same and the swerve angle is more. What do you think is the best way to explain and demonstrate the importance of these effects? Maybe you or I can do a follow-up article to expand on the ideas and illustrations in my October '07 article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf). What do you think?

Regards,
Dave

Bob_Jewett
09-11-2007, 11:58 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> ... I hadn't thought a lot about the increase in swerve angle with a draw shot as compared to a follow shot with the same amount of English. I know the aim point on the table (point "A" in my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf)) changes with height of the ball contact point (point "B") and with the cue elevation (which might be slightly larger for a draw shot), but are there some other effects you think cause the difference? ... <hr /></blockquote>
No, I'm just going from the Rule of Coriolis. Of course you might get the obstinate contrarian who uses a level cue for draw and elevates for follow, and they would violate my rule of thumb.

I think that Diagram 1 in your October article only applies to a "semi contrarian" in the sense above. A normal player will see more angle change on a draw shot rather than the equal angle shown there.

Jal
09-11-2007, 02:27 PM
<blockquote><font class="small">Quote dr_dave:</font><hr>...I hadn't thought a lot about the increase in swerve angle with a draw shot as compared to a follow shot with the same amount of English. I know the aim point on the table (point "A" in my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf)) changes with height of the ball contact point (point "B") and with the cue elevation (which might be slightly larger for a draw shot), but are there some other effects you think cause the difference? Also, how significant do you think this difference is? ...<hr /></blockquote>Dr. Dave, in diagram 1, suppose the stick elevation is 3 degrees for the follow shot, and 4 degrees for the draw shot. It looks like the offsets in the vertical and horizontal directions are the same, with a total offset of (1/2)R. It appears the ball has reached natural roll in both cases.

Given this, and for what it's worth, calculations yield swerve angles of about 0.78 and 2.2 degrees, respectively.

Jim

cushioncrawler
09-11-2007, 04:22 PM
<blockquote><font class="small">Quote ABQ_Poolhead:</font><hr> Mac, I played the shot that way because I had only one ball left on table to shoot other than the eight. My ball was almost frozen to the foot rail 2-3 inches to the right of the pocket. My intent was to bump the object ball off the rail and leave blocking the pocket, and for the cue ball to stay toward that same end of table (hence the slow speed) but go sharply right to partially hide behind the 8-ball and at least partially hook my opponent. Haven't heard of "veer" before. Could you please elaborate? ABQ<hr /></blockquote>I sometimes say veer or kurv or drift (my terms i think) to differentiate from (masse) swerv. But i reckon that if u set up that shot again, u kood play it az many times az u like and u wouldnt get the qball to do what u wanted, ie to go sharply right (off the OB). Sidespin (outside english here) on the qball duznt much inkreec the deflexion angle off an objekt ball (inside english duz dekreec the deflexion angle a little, but outside english duz next to zero). What inkreecez the deflexion angle (talking about outside english here) iz the swerv. Now, for a soft shot, there aint much swerv (ie not worth the effort probably), and, in any case, by the time the qball gets to the objekt ball any english would have evaporated anyhow. So, i reckon that if u set up that shot and played it that way, and played it with zero english, then u wouldnt see much difference, except that zero english would give better control etc. Not talking about very short range shots here, at short range english can be effektiv and controlable. madMac.

dr_dave
09-11-2007, 05:11 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr>...I hadn't thought a lot about the increase in swerve angle with a draw shot as compared to a follow shot with the same amount of English. I know the aim point on the table (point "A" in my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf)) changes with height of the ball contact point (point "B") and with the cue elevation (which might be slightly larger for a draw shot), but are there some other effects you think cause the difference? Also, how significant do you think this difference is? ...<hr /></blockquote>Dr. Dave, in diagram 1, suppose the stick elevation is 3 degrees for the follow shot, and 4 degrees for the draw shot. It looks like the offsets in the vertical and horizontal directions are the same, with a total offset of (1/2)R. It appears the ball has reached natural roll in both cases.

Given this, and for what it's worth, calculations yield swerve angles of about 0.78 and 2.2 degrees, respectively.<hr /></blockquote>Thanks Jim. I haven't checked the numbers, but they sound reasonable to me.

Even when the final angles aren't very different (only about 0.4 degrees difference in your numbers, which is difficult to visualize in a diagram), the "effective squirt" (squerve) can be radically different. Do you agree? That was the point I am making in Diagram 1 of the October '07 article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf).

Regards,
Dave

Jal
09-11-2007, 05:45 PM
<blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr>...I hadn't thought a lot about the increase in swerve angle with a draw shot as compared to a follow shot with the same amount of English. I know the aim point on the table (point "A" in my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf)) changes with height of the ball contact point (point "B") and with the cue elevation (which might be slightly larger for a draw shot), but are there some other effects you think cause the difference? Also, how significant do you think this difference is? ...<hr /></blockquote>Dr. Dave, in diagram 1, suppose the stick elevation is 3 degrees for the follow shot, and 4 degrees for the draw shot. It looks like the offsets in the vertical and horizontal directions are the same, with a total offset of (1/2)R. It appears the ball has reached natural roll in both cases.

Given this, and for what it's worth, calculations yield swerve angles of about 0.78 and 2.2 degrees, respectively.<hr /></blockquote>Thanks Jim. I haven't checked the numbers, but they sound reasonable to me.

Even when the final angles aren't very different (only about 0.4 degrees difference in your numbers, which is difficult to visualize in a diagram), the "effective squirt" (squerve) can be radically different. Do you agree? That was the point I am making in Diagram 1 of the October '07 article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf).

Regards,
Dave <hr /></blockquote>I really hadn't thought much about the "delaying" action of the draw, and it took a little bit more thought to convince myself of it (even though it's pretty clear that the friction force points in a different direction than with follow.) So it was an interesting revelation. Thanks for asking and I think I agree, but without having looked at it further, I'm wondering. It's clear that given enough distance, the draw path will cross the follow path, since the swerve angle is larger while the squirt angle is same. But will this happen before roll has set in, after roll has set in, or can it be either?

So my reservation is regarding what state the cueball is in as it meets the object ball. But certainly, with enough travel distance, the sideways displacement (squerve offset) will be very different. (There's also a maximal difference in the paths somewhere along the way to where they cross. It's that that I'm fuzzy about.)

Jim

P.S. If the numbers have some importance, I would certainly check the math, but just to note that it was .78 versus 2.2 degrees.

dr_dave
09-11-2007, 05:59 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr>...I hadn't thought a lot about the increase in swerve angle with a draw shot as compared to a follow shot with the same amount of English. I know the aim point on the table (point "A" in my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf)) changes with height of the ball contact point (point "B") and with the cue elevation (which might be slightly larger for a draw shot), but are there some other effects you think cause the difference? Also, how significant do you think this difference is? ...<hr /></blockquote>Dr. Dave, in diagram 1, suppose the stick elevation is 3 degrees for the follow shot, and 4 degrees for the draw shot. It looks like the offsets in the vertical and horizontal directions are the same, with a total offset of (1/2)R. It appears the ball has reached natural roll in both cases.

Given this, and for what it's worth, calculations yield swerve angles of about 0.78 and 2.2 degrees, respectively.<hr /></blockquote>Thanks Jim. I haven't checked the numbers, but they sound reasonable to me.

Even when the final angles aren't very different (only about 0.4 degrees difference in your numbers, which is difficult to visualize in a diagram), the "effective squirt" (squerve) can be radically different. Do you agree? That was the point I am making in Diagram 1 of the October '07 article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf).

Regards,
Dave <hr /></blockquote>I really hadn't thought much about the "delaying" action of the draw, and it took a little bit more thought to convince myself of it (even though it's pretty clear that the friction force points in a different direction than with follow.) So it was an interesting revelation. Thanks for asking and I think I agree, but without having looked at it further, I'm wondering. It's clear that given enough distance, the draw path will cross the follow path, since the swerve angle is larger while the squirt angle is same. But will this happen before roll has set in, after roll has set in, or can it be either?<hr /></blockquote>I think it can be either, depending on shot speed, cue elevation, and ball/table conditions. I think for typical cue elevations, speeds, and conditions, a follow shot with English will have less effective squirt ("squerve") than a draw shot with the same English and similar cue elevation over typical shot distances. How's that for a lawyer statement? I should probably crunch some numbers and do more experiments, but I think my statement is true.

<blockquote><font class="small">Quote Jal:</font><hr>So my reservation is regarding what state the cueball is in as it meets the object ball. But certainly, with enough travel distance, the sideways displacement (squerve offset) will be very different.<hr /></blockquote>Good point.

<blockquote><font class="small">Quote Jal:</font><hr>P.S. If the numbers have some importance, I would certainly check the math, but just to note that it was .78 versus 2.2 degrees. <hr /></blockquote>Sorry about that. I guess the difference is 1.4 degrees, not 0.4 degrees. My bad.

Regards,
Dave

Jal
09-11-2007, 06:09 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>...Sidespin (outside english here) on the qball duznt much inkreec the deflexion angle off an objekt ball (inside english duz dekreec the deflexion angle a little, but outside english duz next to zero). What inkreecez the deflexion angle (talking about outside english here) iz the swerv. <hr /></blockquote>Mac, I don't really see much of a difference between inside and outside?

<blockquote><font class="small">Quote cushioncrawler:</font><hr>Now, for a soft shot, there aint much swerv<hr /></blockquote>As I understand it Mac, the amount of swerve (swerve angle) is essentially the same, soft or hard, if roll is reached before impact. No?

However, since it "takes" early on for a soft shot, the angle of attack on the object ball will be less affected. But, I think the purpose of the english was to change the cushion rebound.

Jim

cushioncrawler
09-11-2007, 08:37 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr>...Sidespin (outside english here) on the qball duznt much inkreec the deflexion angle off an objekt ball (inside english duz dekreec the deflexion angle a little, but outside english duz next to zero). What inkreecez the deflexion angle (talking about outside english here) iz the swerv.<hr /></blockquote>Mac, I don't really see much of a difference between inside and outside?<blockquote><font class="small">Quote cushioncrawler:</font><hr>Now, for a soft shot, there aint much swerv<hr /></blockquote>As I understand it Mac, the amount of swerve (swerve angle) is essentially the same, soft or hard, if roll is reached before impact. No?However, since it "takes" early on for a soft shot, the angle of attack on the object ball will be less affected. But, I think the purpose of the english was to change the cushion rebound. Jim<hr /></blockquote>Jim -- The way i see it iz that an excess of outside-english adds to Vxx (ie the tangential vel) while Vyy (ie the center to center vel) iz (nearnuff) unchanged (ie Vyy depends on "e" and topspin etc) -- thusly OE duz inkreec the qball deflexion angle a bit. And, inside-english dekreecez Vxx while Vyy iz unchanged -- thusly IE dekreecez the QBDA moreso than OE inkreecez the QBDA.

Re the swerv angle being the same for soft and hard, this might be correct. Az u say, my meaning woz that the "effektiv swerv angle" iz less for a long-range shot than for the exact same sort of stroke for a short-range shot -- probably 10 times more effekt for 1/10th range.

I dont think that any english would be remaining on the qball for a slow long-range shot, hence there would be zero effekt on cushion rebound. But, i suppoze that whether any english remained would depend on the ratio of the % loss of Vroll (which itself depends on the rolling rezistance of the bedcloth) compared to the % loss of english (which would depend on the bedcloth friktion etc). If u/me thinks about it, english iz lost at a fairly constant rate per second, and can reduce to zero well before rolling stops -- whilst V loss (rolling) iznt constant per second, and karnt reduce to zero before spinning stops (talking about a lone qball here, on a standard bedcloth). madMac.

Jal
09-11-2007, 11:08 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>Jim -- The way i see it iz that an excess of outside-english adds to Vxx (ie the tangential vel) while Vyy (ie the center to center vel) iz (nearnuff) unchanged (ie Vyy depends on "e" and topspin etc) -- thusly OE duz inkreec the qball deflexion angle a bit. And, inside-english dekreecez Vxx while Vyy iz unchanged -- thusly IE dekreecez the QBDA moreso than OE inkreecez the QBDA.<hr /></blockquote>Mac, yes, if there is an equal loss or gain of speed along the tangent line, I believe you're right. It's a subtle difference, generally, but of course this is your specialty.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>Re the swerv angle being the same for soft and hard, this might be correct. Az u say, my meaning woz that the "effektiv swerv angle" iz less for a long-range shot than for the exact same sort of stroke for a short-range shot -- probably 10 times more effekt for 1/10th range.<hr /></blockquote>I think we're on the same page here too. (And the math does say that the cueball's final direction at roll - if roll - is essentially independent of speed.)

<blockquote><font class="small">Quote cushioncrawler:</font><hr>I dont think that any english would be remaining on the qball for a slow long-range shot, hence there would be zero effekt on cushion rebound. But, i suppoze that whether any english remained would depend on the ratio of the % loss of Vroll (which itself depends on the rolling rezistance of the bedcloth) compared to the % loss of english (which would depend on the bedcloth friktion etc). If u/me thinks about it, english iz lost at a fairly constant rate per second, and can reduce to zero well before rolling stops -- whilst V loss (rolling) iznt constant per second, and karnt reduce to zero before spinning stops (talking about a lone qball here, on a standard bedcloth). madMac. <hr /></blockquote>A cueball hit with english at .4R and at a speed of 3 mph (1.3 m/sec) would have an angular velocity of 47 radians/sec. Using a guestimate of spin loss at 12 rad/sec, say, it certainly seems as if it could lose all (certainly most) of its spin - more so at a smaller offset. That loss rate of 12 rad/sec is based on a ball spinning in place, which beside being crudely arrived at, may not apply to a rolling ball too well.

I don't see why the ball should lose speed (linear) at an uneven rate, except of course for the transition from sliding to roll? Are you thinking perhaps that the ball will "settle in" to the cloth a little more as it slows down?

Jim

cushioncrawler
09-12-2007, 02:21 AM
<blockquote><font class="small">Quote Jal:</font><hr>.....I don't see why the ball should lose speed (linear) at an uneven rate, except of course for the transition from sliding to roll? Are you thinking perhaps that the ball will "settle in" to the cloth a little more as it slows down? Jim<hr /></blockquote>Jim -- Hmmm -- yes u might be correct. I suppoze that if the bedcloth rolling rezistance iz say 1 in 100 then this iz similar to going up a hill of 1 in 100, and that if something rolling down a hill haz an even acceleration then something (a ball) rolling up the hill might have a nearnuff even deceleration (ie per second and per anything else). And, i think that a slow ball sinks deeper, hence yes, i think that the rolling rezistance inkreeces az the ball slows. But, the air rezistance(s) (form drag, and bedcloth sqeez and suck) probably vary az per the square of V, so, the overall GOTH and the change in the GOTH depends, in theory it could dekreece (for a while) az a ball slowz.

I remember hand-rolling a ball with spin. With a little bit of spin, the spin stopped before the rolling. With lots of spin (spin axis say 60dg off horizontal) rolling stopped first. And with the spin axis at say 50dg (i think) the spin and rolling stopped together (but my memory aint much good). madMac.

Jal
09-12-2007, 11:38 AM
Mac, I know you've adjusted some of your pendulum test results for air drag. It's probably been a while, but do you remember how this compares to rolling resistance, say at high speed (or whatever).

Jim

cushioncrawler
09-12-2007, 09:56 PM
<blockquote><font class="small">Quote Jal:</font><hr>Mac, I know you've adjusted some of your pendulum test results for air drag. It's probably been a while, but do you remember how this compares to rolling resistance, say at high speed (or whatever). Jim<hr /></blockquote>Jim -- I karnt remember my exakt methodology etc, but some of my tests and calcs gave the following.
Bedcloth rolling rezistance plus air drag rezistance (ie the total equivalent "grade of the hill") iz 1.8% at 0.1m/s, and 1.4% at 0.5m/s, and 3.3% (estimate) at 7m/s.
Air drag form factor for a ball (Cd) = 0.49 in mid-air (books say 0.47). And Cd = 0.60 for a ball on a table (my calc).
Air-suck (at rear of footprint) and air-sqeez (in the front of the footprint) i estimate to be perhaps equivalent to a Cd of 0.03 and 0.03, ie a total of 0.06.
Hence for a ball rolling on a bedcloth i estimate that the total effektiv air drag Cd iz about 0.66.
I calculated that Air-Drag GOTH = Nap-Drag GOTH = 1.4% (each) at 4.5m/s, ie total = 2.8% (probably some sort of mistake here). madMac.

Deeman3
09-13-2007, 07:39 AM
You guys need to be fitted for custom pocket protectors. /ccboard/images/graemlins/grin.gif

Jal
09-13-2007, 01:44 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr>Mac, I know you've adjusted some of your pendulum test results for air drag. It's probably been a while, but do you remember how this compares to rolling resistance, say at high speed (or whatever). Jim<hr /></blockquote>Jim -- I karnt remember my exakt methodology etc, but some of my tests and calcs gave the following.
Bedcloth rolling rezistance plus air drag rezistance (ie the total equivalent "grade of the hill") iz 1.8% at 0.1m/s, and 1.4% at 0.5m/s, and 3.3% (estimate) at 7m/s.
Air drag form factor for a ball (Cd) = 0.49 in mid-air (books say 0.47). And Cd = 0.60 for a ball on a table (my calc).
Air-suck (at rear of footprint) and air-sqeez (in the front of the footprint) i estimate to be perhaps equivalent to a Cd of 0.03 and 0.03, ie a total of 0.06.
Hence for a ball rolling on a bedcloth i estimate that the total effektiv air drag Cd iz about 0.66.
I calculated that Air-Drag GOTH = Nap-Drag GOTH = 1.4% (each) at 4.5m/s, ie total = 2.8% (probably some sort of mistake here). madMac.

<hr /></blockquote>Thanks a lot Mac. I've been diddling with the numbers a little too, and apparently air drag is a significant factor in the slowing of a ball. I in fact got a number similar to your 3.3% in the high speed case. Is that drop from 1.8% to 1.4% due to the deeper footprint, would you think?

(My main interest here is in helping Deeman, for whom I don't see much progress until he's thoroughly come to terms with this stuff. One day he'll thank us...but in what manner, I'm not sure.)

Jim

cushioncrawler
09-13-2007, 04:19 PM
<blockquote><font class="small">Quote Jal:</font><hr>.....Is that drop from 1.8% to 1.4% due to the deeper footprint, would you think?....<hr /></blockquote>Yes, plus at slow speed the ball sits on the cloth longer and hencely "irons" it better -- ie the cloth at the rear dont push (foreward) az hard no more (ie more hysteresis). I wanted to know all of this sort of stuff so that i could build a program that told u what happens to the balls in a collizion, ie inklooding the swerve and roll etc etc. It will be a big complicated program, might look at it again in 2008. At least i think i will be able to do it, not like that cushion rebound program that i attempted but found to be much too difficult. madMac.

dr_dave
09-17-2007, 02:12 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr>I think that Diagram 1 in your October article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf) only applies to a "semi contrarian" in the sense above. A normal player will see more angle change on a draw shot rather than the equal angle shown there.<hr /></blockquote>Thank you for pointing this out. I just changed the diagram slightly (in the online version of the article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf)) to make it more technically correct. I also added some analysis in TP A.19 (http://billiards.colostate.edu/technical_proofs/new/TP_A-19.pdf) covering the math and physics details (see page 7).

Regards,
Dave

dr_dave
09-17-2007, 02:19 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr>...I hadn't thought a lot about the increase in swerve angle with a draw shot as compared to a follow shot with the same amount of English. I know the aim point on the table (point "A" in my November '05 article (http://billiards.colostate.edu/bd_articles/2005/nov05.pdf)) changes with height of the ball contact point (point "B") and with the cue elevation (which might be slightly larger for a draw shot), but are there some other effects you think cause the difference? Also, how significant do you think this difference is? ...<hr /></blockquote>Dr. Dave, in diagram 1, suppose the stick elevation is 3 degrees for the follow shot, and 4 degrees for the draw shot. It looks like the offsets in the vertical and horizontal directions are the same, with a total offset of (1/2)R. It appears the ball has reached natural roll in both cases.

Given this, and for what it's worth, calculations yield swerve angles of about 0.78 and 2.2 degrees, respectively.<hr /></blockquote>Jal,

FYI, I added plots and analysis to TP A.19 (http://http://billiards.colostate.edu/technical_proofs/new/TP_A-19.pdf) (see page 7) to verify and illustrate the results of your numbers above. Our numbers match. Also, depending on shot speed and table/ball conditions, the plots support Diagram 1 (the new version anyway) in my October '07 article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf).

Regards,
Dave

Jal
09-17-2007, 03:31 PM
<blockquote><font class="small">Quote dr_dave:</font><hr>Jal,

FYI, I added plots and analysis to TP A.19 (http://http://billiards.colostate.edu/technical_proofs/new/TP_A-19.pdf) (see page 7) to verify and illustrate the results of your numbers above. Our numbers match. Also, depending on shot speed and table/ball conditions, the plots support Diagram 1 (the new version anyway) in my October '07 article (http://billiards.colostate.edu/bd_articles/2007/oct07.pdf).

Regards,
Dave <hr /></blockquote>Yea! Thanks for the confirmation Dr. Dave.

I haven't gotten around to reading (carefully) your latest technical squirt article yet (too many irons in the fire lately). It will be interesting comparing it to Ron Shepard's treatment.

Jim

dr_dave
09-17-2007, 04:07 PM
<blockquote><font class="small">Quote Jal:</font><hr>I haven't gotten around to reading (carefully) your latest technical squirt article yet (too many irons in the fire lately). It will be interesting comparing it to Ron Shepard's treatment.<hr /></blockquote>As I point out in TP A.31 (squirt physics) (http://billiards.colostate.edu/technical_proofs/new/TP_A-31.pdf), I make the same assumptions as Ron Shepard (because the assumptions seem appropriate based on the high-speed evidence we have). I go about the derivation a little differently (but not much), and the results are the same (although they look a little different).

Happy reading,
Dave