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littleCajun
03-11-2005, 06:48 AM
If the amount of throw on an object ball depends on the amount of dwell time that the cue ball and object ball spend together. Together with the fact That energy in equals energy out it would seem unless the object ball is pinched against another ball, rail or something else, That the object ball and cue ball would have very little dwell time together. Think of Newtons craddle (all the little balls hanging on strings, drop one one comes out, two in two out). So I would think that very little spin if any would be transfered from the cue ball to the object ball. So instead of sping throwing a ball, It would seem to me that the squirt or deflection of the cue ball would change the contect point to throw the ball.

So my Question is if any spin is transfered is it really worth it?

Again this is for a straight on hit. I know about dirty balls and cloth conditions for cut shots and that. This is just to talk about spin transfer.

Now Open for discussion, Talk to yall later.

SpiderMan
03-11-2005, 07:33 AM
<blockquote><font class="small">Quote littleCajun:</font><hr> If the amount of throw on an object ball depends on the amount of dwell time that the cue ball and object ball spend together. Together with the fact That energy in equals energy out it would seem unless the object ball is pinched against another ball, rail or something else, That the object ball and cue ball would have very little dwell time together. Think of Newtons craddle (all the little balls hanging on strings, drop one one comes out, two in two out). So I would think that very little spin if any would be transfered from the cue ball to the object ball. So instead of sping throwing a ball, It would seem to me that the squirt or deflection of the cue ball would change the contect point to throw the ball.

So my Question is if any spin is transfered is it really worth it?

Again this is for a straight on hit. I know about dirty balls and cloth conditions for cut shots and that. This is just to talk about spin transfer.

Now Open for discussion, Talk to yall later.
<hr /></blockquote>
OK, I was a little bit misled by your title and intro, ie you are not talking about throw (a change in second-ball path from the line of common centers), but rather transfer of english (spin).

I'd say that definitely the transfer of spin is significant in many common shots. The best example of this is the one-rail bank shot. Transfer of spin is commonly used to influence the object ball's rebound angle off the rail. For example, a bank angle that is not "dead" can be widened or narrowed by transferred spin, potting the ball while still hitting straight-on to stop, draw, or follow the cueball.

SpiderMan

Cane
03-11-2005, 07:53 AM
OK, story here. I shared a hotel room with a Master Instructor, not RandyG, one who is very analytical and can't stand to not know exactly why something does or does not happen on a pool table, and he said, and convinced me later on a pool table, that you CAN throw a ball that isn't frozen to another ball, rail, etc, but that you can only throw a non frozen ball one inch in 80-something feet (great if you're playing on a 90 foot table, but not of much significance on a 9' table as that would only translate to 1/8" of contact induced throw on a 9 foot shot). I think what many perceive as throw may be the cue ball moving off of the original line of travel as a result of squirt, swerve, etc. I shoot many shots where I'll hear someone say "he threw that one it", when actually, all I did was intentionally curve or slightly masse the cue ball to get a bit of a different approach to the OB with the CB... a good example of that would be a 90° or nearly 90° shot in the side pocket that's practically unmakeable, and you simply shoot the cue ball offline (outside your optimal line of aim) with a little masse and effectively make an 85° shot out of it. The object ball, in this case, does NOT throw into the pocket, rather the cue ball is travelling on a different line than it would with no non-axial spin, when it approaches the OB. The exceptions to this are frozen ball shots, where since the ball is frozen to another, you can impart an appreciable amount of spin on an OB with a CB because of the gear effect. The target ball is frozen and dwell time is drastically increased, therefore the spin will transfer much more efficiently.

Now, with that said, a rail will impart spin on an object ball, and the ball may "throw" off of it's oringal line of travel because of that induced spin. An example, and I wish I could use that damned WEI thing, but can't, the old "trick" shot where you set the cb and ob in line on the center of the table lined up to the side pocket. You shoot the OB above the side pocket and three rail it back into the pocket closest to you, where it banks up table then when it hits the second rail it comes back down table, then when it hits the final rail it travels even more back down table to the target pocket. I have "shown off" with this shot for years, and until about a year ago, I always used extreme low inside spin to make the shot. It was proven to me, by a pair of Master Instructors, that the spin on the CB has nothing to do with the shot, it's the rail crush at high speed that imparts the spin on the OB. So, I still show off with this shot from time to time to someone who but NOT to demonstrate throw, as I used to, but to demonstrate the effects of rail crush on an OB or CB, and I can shoot it with no spin, inside, outside, high, low, just whatever spin I want to put on the ball. I was never actually imparting any appreciable amount of spin on the object ball with the cue ball, rather the CUSHION was imparting the spin. Stated simply, what I always thought was a throw shot, was simply a speed shot. Spin had no importance nor effect on the path of the object ball, rather the speed at which it hit the cushion is what put the necessary rotation on the OB to make it spin back down table and travel to the target.

This is hard for me to explain, but for those that don't understand how rail crush works, I'm going to give it a shot. When you shoot a ball into a cushion, the cushion compresses. If you shoot that ball in at an angle, the "uphill" side of the cushion compresses more than the downhill side. So, you have more contact between the uphill side and the ball, and this imparts spin on the ball. Lets look at it this way... you have more "pressure" from the cushion on the uphill side of the ball than you do on the downhill side of the ball, so the side with more pressure will "hold", allowing the side with less pressure to spin away from the cushion. Think of it like this... if you could sit a ball in the middle of the table and hit it pure center ball with your cue tip while putting some kind of friction on the right side of the ball to keep it from moving away as fast as the left side, then the ball would leave with left spin. That's the same thing that happens with rail crush. The side with more pressure, the uphill side, is "held in place" for a very slight amount of time, so the imparted spin will be on the downhill side. If you shoot a ball at a right hand angle into a cushion, the cushion will impart left hand spin on the ball.

Of course, all this is just my opinion and my perception and I welcome any challenges to my statements. Be easy on me... it took a lot of brain power to get through a post this long!!!

Later,
Bob

Bob_Jewett
03-11-2005, 01:17 PM
<blockquote><font class="small">Quote littleCajun:</font><hr> If the amount of throw on an object ball depends on the amount of dwell time that the cue ball and object ball spend together...
<hr /></blockquote>
That assumption is completely false, so the following conclusions, which may have some truth in them, need to be shown some other way. The contact time for tip-ball contact is about 1000 microseconds. For ball-ball it is about 200 microseconds. If what you said were true, which it is not, a reasonable conclusion would be that you can get 1/5 as much spin on the ball with another ball as with the tip. That doesn't happen.

The important thing is the friction between the balls. If you have access to a beginning physics text book, check out the ideas behind impulsive forces. Frictional forces between sliding objects are proportional to the force pressing the objects together. That means that the time of contact is unimportant; what is important is how sticky the surfaces are.

dr_dave
03-11-2005, 03:20 PM
I think the answers posted already (Bob, Cane, SpiderMan) are excellent. You might also find my previous response on an earlier thread (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=182724&amp;page =0&amp;view=collapsed&amp;sb=5&amp;o=&amp;vc=1) useful. Some video clips are cited there for viewing.

Regards,
Dr. Dave

GeraldG
03-11-2005, 04:52 PM
Dr. Dave,
In the thread that you referenced, you said something very interesting to me, not that it would have any real practical value that I can think of, but it's interesting. You first stated that spin can be transferred from a cueball to an object ball, which I fully believe...I've seen enough evidence to convince me that's true. Left spin on the cueball can induce right spin on the object ball and vice-versa and bottom spin on the cueball can induce a small amount of topspin on the object ball. No problem so far, except that I've never seen any forward overspin produced this way such as you would get with a follow shot on the cueball. What I've seen is that the natural forward roll takes over much quicker when there is bottom spin on the cueball and the initial sliding of the object ball is greatly reduced (nearly eliminated). Then you went on to say that topspin on the cueball can induce bottom spin on the object ball. This is where I have a problem. I've never seen that demonstrated at all. The videos I've seen (using striped balls for clarity) show that the most that can be accomplished is a momentary sliding of the object ball followed by natural forward roll. I've never seen a case where there was enough adherance between the surfaces of the balls to break the the friction between the object ball and the cloth and get the mass of the object ball spinning backwards against the forward movement of the object ball and the increased friction and influence of that forward motion (which is trying to make the cueball roll forward). There just doesn't seem to be enough transfer of spin to cause that to happen. Do you have any video yet to show actual backspin being induced to an object ball from topspin on the cueball?

There is a huge difference in the amount of friction between the ball and the cloth that must be overcome when you compare making a ball spin in it's vertical axis versus making it spin on it's horizontal axis.

Bob_Jewett
03-11-2005, 05:11 PM
<blockquote><font class="small">Quote GeraldG:</font><hr> ... No problem so far, except that I've never seen any forward overspin produced this way such as you would get with a follow shot on the cueball. What I've seen is that the natural forward roll takes over much quicker when there is bottom spin on the cueball and the initial sliding of the object ball is greatly reduced (nearly eliminated).... <hr /></blockquote>
When the cue ball has draw on it and it hits an object ball full, the object ball is expected to move away with the speed of the cue ball and to start with a small amount of forward spin. It is never expected to have what has been called "unit follow" or smooth rolling on the cloth. I have seen one case of this that seemed to be caused by skid/cling/kick but it is never going to occur with normal collisions. Perhaps the most useful application of the small amount of follow transferred to the object ball is the frozen combination in which the front object ball is "forced through" the ball behind it, and it does not follow the normal tangent or kiss line. (That shot can also be made without drawing the cue ball, but it's harder.)

As far as transferring backspin to the object ball, I think this happens but is very hard to see. I think Byrne shows a shot that might demonstrate it in either his "Advanced" book or "Wonderful World" book.

As far as "breaking the friction between the object ball and the cloth," I think the surface that the ball is on is entirely unimportant at the instant of contact. The acceleration of the object ball due to the cue ball hitting it is around 500 g's, with corresponding very brief and very huge "pushing" and "throw" forces, and those forces are going to be much higher than the cloth-ball force from friction across the cloth during that instant.

I think the real problem is that any slight backspin the cue ball puts on the object ball will be worn off by the cloth very quickly as the object ball slides away from the cue ball.

dr_dave
03-11-2005, 05:23 PM
<blockquote><font class="small">Quote GeraldG:</font><hr> Dr. Dave,
In the thread that you referenced, you said something very interesting to me, not that it would have any real practical value that I can think of, but it's interesting. You first stated that spin can be transferred from a cueball to an object ball, which I fully believe...I've seen enough evidence to convince me that's true.<hr /></blockquote>
FYI, an interesting example of this was described in a related message (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=182735&amp;page =0&amp;view=collapsed&amp;sb=5&amp;o=&amp;vc=1).

<blockquote><font class="small">Quote GeraldG:</font><hr>I've never seen any forward overspin produced this way such as you would get with a follow shot on the cueball. What I've seen is that the natural forward roll takes over much quicker when there is bottom spin on the cueball and the initial sliding of the object ball is greatly reduced (nearly eliminated).<hr /></blockquote>
Agreed (e.g., with the example in the link above).

<blockquote><font class="small">Quote GeraldG:</font><hr>Then you went on to say that topspin on the cueball can induce bottom spin on the object ball. This is where I have a problem. I've never seen that demonstrated at all. The videos I've seen (using striped balls for clarity) show that the most that can be accomplished is a momentary sliding of the object ball followed by natural forward roll. I've never seen a case where there was enough adherance between the surfaces of the balls to break the the friction between the object ball and the cloth and get the mass of the object ball spinning backwards against the forward movement of the object ball and the increased friction and influence of that forward motion (which is trying to make the cueball roll forward). There just doesn't seem to be enough transfer of spin to cause that to happen. Do you have any video yet to show actual backspin being induced to an object ball from topspin on the cueball?<hr /></blockquote>
I plan to shoot this with the high-speed camera the next time I get my hands on it. But I think you are absolutely correct (i.e., my original statement may have been in error). Good job ... very insightful.

Even if some bottom spin were transferred, it would be a very small amount and would wear off almost immediately after impact as the object ball starts sliding and eventually rolling. Also, even if this effect did exist, I think one would be hard-pressed to think of an example of how it would be useful in a game situation. A direct hit on two frozen object balls doesn't really present much of an opportunity (as it does in the above example) because cue ball follow would move the first object ball after impact anyway.

<blockquote><font class="small">Quote GeraldG:</font><hr>There is a huge difference in the amount of friction between the ball and the cloth that must be overcome when you compare making a ball spin in it's vertical axis versus making it spin on it's horizontal axis. <hr /></blockquote>Agreed.

Regards,
Dr. Dave

dr_dave
03-11-2005, 05:30 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr>As far as "breaking the friction between the object ball and the cloth," I think the surface that the ball is on is entirely unimportant at the instant of contact. The acceleration of the object ball due to the cue ball hitting it is around 500 g's, with corresponding very brief and very huge "pushing" and "throw" forces, and those forces are going to be much higher than the cloth-ball force from friction across the cloth during that instant.<hr /></blockquote>
Excellent point, and I agree completely (although I haven't verified the "500 gs"). But I still want to see the results of some high-speed video tests. I hope to do them soon.

Regards,
Dave

GeraldG
03-11-2005, 05:35 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <blockquote><font class="small">Quote GeraldG:</font><hr> ... No problem so far, except that I've never seen any forward overspin produced this way such as you would get with a follow shot on the cueball. What I've seen is that the natural forward roll takes over much quicker when there is bottom spin on the cueball and the initial sliding of the object ball is greatly reduced (nearly eliminated).... <hr /></blockquote>
When the cue ball has draw on it and it hits an object ball full, the object ball is expected to move away with the speed of the cue ball and to start with a small amount of forward spin. It is never expected to have what has been called "unit follow" or smooth rolling on the cloth. I have seen one case of this that seemed to be caused by skid/cling/kick but it is never going to occur with normal collisions. Perhaps the most useful application of the small amount of follow transferred to the object ball is the frozen combination in which the front object ball is "forced through" the ball behind it, and it does not follow the normal tangent or kiss line. (That shot can also be made without drawing the cue ball, but it's harder.)

<font color="red"> Yes, I can see that as an application for induced follow. When I said I couldn't think of a practical application, I was really referring to induced draw. </font color>

As far as transferring backspin to the object ball, I think this happens but is very hard to see. I think Byrne shows a shot that might demonstrate it in either his "Advanced" book or "Wonderful World" book.

<font color="red">I have Byrne's books (IMO the best pool books in existence). I can see it in theory, but have never seen anyone demonstrate it in practice. </font color>

As far as "breaking the friction between the object ball and the cloth," I think the surface that the ball is on is entirely unimportant at the instant of contact. The acceleration of the object ball due to the cue ball hitting it is around 500 g's, with corresponding very brief and very huge "pushing" and "throw" forces, and those forces are going to be much higher than the cloth-ball force from friction across the cloth during that instant.

<font color="red">That's true, but then you have essentially slide at the moment of impact. The forward force of the ball sliding across the cloth produces friction in it's own right and is trying to induce the ball to roll forward with much more force than any amount of backspin that could be induced from cueball spin. My point is that it seems that any amount of backspin that could be produced by forward spin of the cueball would be immediately overcome by this force. </font color>

I think the real problem is that any slight backspin the cue ball puts on the object ball will be worn off by the cloth very quickly as the object ball slides away from the cue ball.

<font color="red"> Yep. So quickly, in fact that I think it could be said that for all practical intents and purposes no useable backspin can be induced. </font color>

<hr /></blockquote>

The above is simply my opinions and observations and is certainly subject to be incorrect. /ccboard/images/graemlins/smile.gif

Bob_Jewett
03-11-2005, 06:26 PM
<blockquote><font class="small">Quote dr_dave:</font><hr> ...(although I haven't verified the "500 gs"). ... <hr /></blockquote>
This is pretty easy to deduce from already available collision information. Marlow measured collision times as about 200 microseconds. Hertz' Law says that the time should decrease only slightly for higher speed shots, so it can be taken as a constant for ball-ball collisions. Measuring the flattened patch on a ball gives about the same time. A fast break shot is about 10 meters / second. A normal shot is about 1 meter / second. For the former, the average acceleration is 50,000 meters /second /second, or about 5000 g's, so the peak acceleration is about 10,000 g's. For the softer shot, the peak would be about 1000 g's for an average about 500.

To actually see the mechanics of the collision -- such as gradual flattening of the two balls and recovery -- would take a frame rate of 20 to 50 thousand per second and much better focusing than I've ever seen for such videos. The flat patch during the collision is about 1/4 inch in diameter which is a flattening of only about 1/6 of a millimeter or the thickness of two sheets of paper.

GeraldG
03-11-2005, 06:52 PM
Very, very cool information, Bob.

The question I have now is this: For a break shot, since the speed is on average, approximately 10 times the speed of a normal shot, but the mass in motion remains constant, what would be the amount of flattening of the balls? I wouldn't think that it would be linear as compared to the amount of flattening in a normal shot because the amount of time they are in contact also remains constant. In other words, it should flatten some more, but not 10 times as much. Plus there are other variables that would come into play, such as the inherent elasticity of the material the balls are made of, the increased internal pressure at the point of impact due to compression, the loss of energy due to impact/compression (absorption).

Bob_Jewett
03-11-2005, 07:48 PM
<blockquote><font class="small">Quote GeraldG:</font><hr>
The question I have now is this: For a break shot, since the speed is on average, approximately 10 times the speed of a normal shot, but the mass in motion remains constant, what would be the amount of flattening of the balls?... <hr /></blockquote>
The simple way to analyze the collision says that the distance the balls penetrate into each other (leaving a common flat spot) goes directly with the speed. The diameter of the flat spot goes as the square root of the speed. So, if a 1m/s shot gave a contact spot diameter of 2mm, at 10m/s the spot would be 6mm in diameter, roughly.

For a break shot, there is the additional factor that the ball the cue ball strikes cannot move away easily, so there is some additional flattening. In the case that the struck ball doesn't move at all -- which is not quite true on a break shot -- the effect is the same as doubling the speed of the cue ball into a free ball. That would take the patch from 6mm to about 8mm in diameter.

I think it's impressive that pool balls can take that for years and still work pretty well.

GeraldG
03-11-2005, 07:54 PM
That makes perfect sense.

Yep, it is amazing. Especially when you consider the fact that the contact patch where the cloth is in contact with the cueball, on a break shot, at the instant of accelleration, is heated to a temperature of over 400 degrees F. from the friction. Between the constant collisions and the momentary heating.....that's a lot of abuse.

Rod
03-11-2005, 08:38 PM
[ QUOTE ]
So my Question is if any spin is transfered is it really worth it?
<hr /></blockquote>

In the context of your question, NO. Before rocket science gets involved; why would you spin in a straight in shot? It wouldn't make sense to do so. Well, unless you just like missing more balls. Your correct, using side spin "may" change the original contact point.

Rod

randyg
03-13-2005, 06:33 AM
Thank you Rod.........SPF-randyg

Scott Lee
03-13-2005, 10:56 AM
No kidding! KISS still rules!

Scott

Bob_Jewett
03-13-2005, 10:23 PM
<blockquote><font class="small">Quote Scott Lee:</font><hr> No kidding! KISS still rules!... <hr /></blockquote>
I agree, but there are still lots of people who think you can make a ball go into the pocket better by putting "get in english" on it. That would be one "reason" why someone might want to transfer english to a nearly straight-in shot.

randyg
03-14-2005, 06:11 AM
BOB: Those are the same people who end up attending your Pool School. Ball Spinners=Frustration.

Keep up the good work, see you in Las Vegas.....SPF-randyg

Cane
03-15-2005, 12:20 PM
<blockquote><font class="small">Quote randyg:</font><hr> Ball Spinners=Frustration.
<hr /></blockquote>

Absolutely! My dream table is one where I can run out using nothing but stop shots... no left, no right, no follow, no draw... When I was young, I was notorius for spinning a ball when it wasn't necessary. Got me in the L side more than it did in the W side. It took me most of my life to figure out that the times I really HAD to spin the cue ball hard, were the result of me spinning it when I didn't have to on the previous shot...

Of course, there are times when you are in a position where you have to "tweeest" the cue ball pretty hard, but not necessarily to make a shot, rather to gain position on the next shot.

You can tell by my cue that I don't screw the cue ball much... tip is pretty flat, because most of my shots are center ball or very close to it. I don't need a dime shaped radius to do that. Mine is more the radius somewhere between a quarter and a half dollar.

Later,
Bob

BlindPlayer
04-01-2005, 07:05 PM
I just read so much on the subject of throw on this thread that my mind is a fog. However, several things were never mentioned in relation to "dwell time" to transfer spin (or for that matter throwing the object ball off-line). I see dwell time as being affected by (1) how dirty the balls are (2) the angle of the shot; i.e. as angle increases - throw decreases (3) the speed of the shot - as speed increase - throw decreases (4) are you holding your mouth right? (5) not sure about humidity (?)

cushioncrawler
10-23-2005, 08:20 PM
Being new in town i am enjoying catching up on this stuff --- i enjoyed reading the comments here from Doc &amp; Bob &amp; Gerald etc. Re tranzduced side (tranzmitted side &amp;/or induced side) &amp; tranzduced throw, i would like to add the following comments.
In MathLand, tranzduced side iz allwayz equal to 2.5 timez the throw. If the Qball sends the Objekt Ball off at 20rps (i uzually uze rps for the speed of a ball, rolling or not, think of it az circumferencez per sec), then the object ball can be thrown (sidewayz here) up to say 1rps (if ball-to-ball friktion iz 1 in 20). If the throw iz 1rps then the tranzduced spin must be 2.5rps (in MathLand).
Likewize, if a rolling Qball hits the objekt ball full-ball &amp; sendz it off at 20rps, the object ball can be thrown downwardz at 1rps, &amp; the backspin on the object ball can potentially be 2.5rps.
The objekt ball and Qball are in contact for say 0.0003sec (0.0010sec for slow impakts), the objekt ball travelz say the equivalent of 9 flatspot depths during this contact time (7 for slow impakts), which iz a distance of only say 1.5mm.
I agree that az soon az ball-to-ball contact endz, ball-to-bed friktion takes over, the friktion force magnifyd due to the ball sinking deeper into the bedcloth during its first bounce. However, i reckon that Dr Dave will hav no problem in seeing such backspin in hiz video.