PDA

View Full Version : The 30 degree rule - Dr. Dave

Vapros
09-02-2007, 01:03 AM
I understand the 30 degree rule. What will the deflection be when the cue ball hits two balls frozen together?

cushioncrawler
09-02-2007, 02:09 AM
It would be a good excercize to set up some frozen sets/plants at varyus angles and to compare the qball deflexion angles. The angle would inkreec by i reckon allmost 10dg, ie 35dg (my verzion of 30dg) would bekum allmost 45dg. I have tryd this sort of shot at english billiards but have never seen it aktually kum up in a game.

And, the 90dg (frozen) stun angle would bekum allmost 100dg (i think). madMac.

dr_dave
09-02-2007, 10:27 PM
<blockquote><font class="small">Quote Vapros:</font><hr> I understand the 30 degree rule. What will the deflection be when the cue ball hits two balls frozen together? <hr /></blockquote>I have not looked at this specifically, but it might be difficult to come up with a simple principle like the 30 degree rule, because the deflection angle will vary with the alignment of the frozen balls.

Regards,
Dave

ken_r
09-02-2007, 11:10 PM
Why would it be any different?

Vapros
09-02-2007, 11:23 PM
Yes, I can see that it would vary in that manner, but it would be helpful to know the maximum deflection, which I would assume to be when all three balls are in a straight line at contact, and cueball rolling. Is this something you might look into? Should we expect 60 degrees, or twice the deflection for twice the stationary mass? Thanks.

Vapros
09-02-2007, 11:34 PM
See above. The 30 degree rule tells us how much deflection to expect when a rolling cue ball hits another ball of the same weight in approximately a half-ball hit. If two balls are frozen together, I'm quite certain the deflection would be more, due to the increased weight (or resistance) of the object ball.

ken_r
09-03-2007, 12:17 AM
Yes, I understand what the 30 degree rule is.

However, the number of balls doesn't change anything. Try it yourself, you can line up all 15 balls in a straight line and it won't make a difference. Assuming all the balls are perfectly straight only the 15th ball will move and the cue ball will still carom off exactly the same as if there was only one ball.

Maybe this will help

dr_dave
09-03-2007, 09:07 AM
<blockquote><font class="small">Quote Vapros:</font><hr> Yes, I can see that it would vary in that manner, but it would be helpful to know the maximum deflection, which I would assume to be when all three balls are in a straight line at contact, and cueball rolling. Is this something you might look into? Should we expect 60 degrees, or twice the deflection for twice the stationary mass? Thanks.<hr /></blockquote>Try it out. I just did. The angle increases a little, but not by much. The reason is that the two balls don't react the same way as a single ball with twice the mass would react. Even if they did, I think the deflected angle would still be closer to 30 degrees than 60 degrees. It wouldn't be too difficult to figure this out from TP A.4 (http://billiards.colostate.edu/technical_proofs/new/TP_A-4.pdf) using a modified TP 3.1 (http://billiards.colostate.edu/technical_proofs/TP_3-1.pdf); but again, the assumptions required would not be very realistic. Try it out at a table.

Regards,
Dave

Vapros
09-03-2007, 11:58 AM
Thanks. I don't have a table at home, but I'll set it up and try it at the pool room sometime this week. I'm surprised to hear that the increased deflection is so slight. Seems like something I should have already discovered, but maybe not. The shot does not come up often.

Here's another thought. I play one-pocket, and many shots are played gently, either for controlling the cue ball or to get pocket-speed on the object ball. Of course, on a spot shot from behind the line, getting to the short rail is not a problem. However, if the object ball is a bit out in front on the spot, but still on the center line of the table, and I have cue ball in hand, it frequently looks to me like a scratch, no matter where I put the cue ball down, leaving no option but to draw the rock a little to avoid the scratch. I suppose this is due to the slight variations in the 30 degree rule. Or is it just my imagination? Who else has noticed this?

cushioncrawler
09-03-2007, 05:12 PM
I seem to recall that, when i woz praktising sets/plants, i got a bigger deflexion (for rolling and for stun) even when contacting the 1st ball only 1/4 ball, and even when the qball did not form a line with the other ballz. madMac.

Jal
09-04-2007, 01:09 AM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> I seem to recall that, when i woz praktising sets/plants, i got a bigger deflexion (for rolling and for stun) even when contacting the 1st ball only 1/4 ball, and even when the qball did not form a line with the other ballz. madMac. <hr /></blockquote>Mac, I think you should see a larger deflection, but the question is how much?

Bob Jewett's "ten times fuller" system for predicting the path of the first object ball tells us something about the ratio of the implulse between the cueball and first object ball, and the impulse between the two object balls. Based on that, I have a program which attempts to calculate the cueball's final direction.

However, there are a couple of unknowns. His ten times fuller system works for small impact angles, but I don't know how well it holds up at large ones. And there's also a crucial question of energy efficiency, as compared to a two ball collision. This is not trivial to arrive at so the program just compares results using different fractions of the two ball collision efficiency.

The program shows as much an 8 degree difference in the final direction between the two and three ball collisions, and this happens when the energy efficiency is taken to be the same in both cases, and at moderate impact angles. A drop in efficiency in the three ball case, which would be expected, at least a little, brings the numbers closer together. There is only a slight difference in the cueball's direction when the efficiency is .85 that of a two ball collision. (But this is not true at large impact angles.)

It doesn't account for increased spin loss during impact in the three ball case either, which should happen to some extent, and widen the angle.

So while it doesn't yield anything definite, a noticeable difference is within the realm of possibility.

Jim

dr_dave
09-04-2007, 08:37 AM
If anybody wants to look at Bob's interference systems for aiming frozen ball shots, they are described and illustrated here (http://www.sfbilliards.com/articles/1996-02.pdf). I also presented some experimental data and suggested an additional system in a previous thread (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=201944&amp;page =0&amp;view=collapsed&amp;sb=5&amp;o=&amp;fpart=&amp;vc=).

Regards,
Dave
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr> I seem to recall that, when i woz praktising sets/plants, i got a bigger deflexion (for rolling and for stun) even when contacting the 1st ball only 1/4 ball, and even when the qball did not form a line with the other ballz. madMac. <hr /></blockquote>Mac, I think you should see a larger deflection, but the question is how much?

Bob Jewett's "ten times fuller" system for predicting the path of the first object ball tells us something about the ratio of the implulse between the cueball and first object ball, and the impulse between the two object balls. Based on that, I have a program which attempts to calculate the cueball's final direction.

However, there are a couple of unknowns. His ten times fuller system works for small impact angles, but I don't know how well it holds up at large ones. And there's also a crucial question of energy efficiency, as compared to a two ball collision. This is not trivial to arrive at so the program just compares results using different fractions of the two ball collision efficiency.

The program shows as much an 8 degree difference in the final direction between the two and three ball collisions, and this happens when the energy efficiency is taken to be the same in both cases, and at moderate impact angles. A drop in efficiency in the three ball case, which would be expected, at least a little, brings the numbers closer together. There is only a slight difference in the cueball's direction when the efficiency is .85 that of a two ball collision. (But this is not true at large impact angles.)

It doesn't account for increased spin loss during impact in the three ball case either, which should happen to some extent, and widen the angle.

So while it doesn't yield anything definite, a noticeable difference is within the realm of possibility.

Jim <hr /></blockquote>

dr_dave
09-04-2007, 10:19 AM
<blockquote><font class="small">Quote Vapros:</font><hr>Here's another thought. I play one-pocket, and many shots are played gently, either for controlling the cue ball or to get pocket-speed on the object ball. Of course, on a spot shot from behind the line, getting to the short rail is not a problem. However, if the object ball is a bit out in front on the spot, but still on the center line of the table, and I have cue ball in hand, it frequently looks to me like a scratch, no matter where I put the cue ball down, leaving no option but to draw the rock a little to avoid the scratch. I suppose this is due to the slight variations in the 30 degree rule. Or is it just my imagination? Who else has noticed this?<hr /></blockquote>With cue-ball-in-hand, I can visualize many angles where the 30-degree angle doesn't result in a scratch. But I can also see how you can scratch with any angle. If you vary the speed and the cut angle (ball-hit fraction), the subtleties of the 30-degree rule come into play. See the links under "30 degree rule" here (http://billiards.colostate.edu/threads.html) for more info. My one-page quick reference sheet (http://www.engr.colostate.edu/~dga/pool/resources/30_degree_rule_summary.pdf) summarizes most of the important points.

Regards,
Dave

Vapros
09-04-2007, 10:39 AM
Thanks for a lot of good information.

dr_dave
09-04-2007, 10:42 AM
<blockquote><font class="small">Quote Vapros:</font><hr> Thanks for a lot of good information.<hr /></blockquote>
You're welcome.

Dave

Bob_Jewett
09-04-2007, 04:50 PM
<blockquote><font class="small">Quote ken_r:</font><hr> Why would it be any different? <hr /></blockquote>
While there is no obvious answer that it has to be different, in fact it turns out that it is different.

When only two balls collide, the theory is fairly simple, and gives the 90-degree rule that everyone knows well, followed by the effect of any draw or follow on the cue ball. For follow (the so-called 30-degree rule, but known for a much longer time as the half-ball rule), the result is a maximum of about 35 degrees of deflection of the cue ball's path from its original line.

When more than two balls are involved in the collision, the physics is very, very complicated. To find the paths of the balls, you need to look at the details of the collision. Those details can be ignored (while maintaining reasonable accuracy) if only two balls are involved.

If you want to see a more extreme effect than two balls frozen, try the half-ball scratch off a triangle of three balls spotted on the head spot, like the front three balls of a full rack. I think you will find that the amount the cue ball is deflected is much larger than for either a single ball or two frozen balls.

cushioncrawler
09-04-2007, 06:57 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr> I seem to recall that, when i woz praktising sets/plants, i got a bigger deflexion (for rolling and for stun) even when contacting the 1st ball only 1/4 ball, and even when the qball did not form a line with the other ballz. madMac.<hr /></blockquote>Mac, I think you should see a larger deflection, but the question is how much? Bob Jewett's "ten times fuller" system for predicting the path of the first object ball tells us something about the ratio of the implulse between the cueball and first object ball, and the impulse between the two object balls. Based on that, I have a program which attempts to calculate the cueball's final direction. However, there are a couple of unknowns. His ten times fuller system works for small impact angles, but I don't know how well it holds up at large ones. And there's also a crucial question of energy efficiency, as compared to a two ball collision. This is not trivial to arrive at so the program just compares results using different fractions of the two ball collision efficiency. The program shows as much an 8 degree difference in the final direction between the two and three ball collisions, and this happens when the energy efficiency is taken to be the same in both cases, and at moderate impact angles. A drop in efficiency in the three ball case, which would be expected, at least a little, brings the numbers closer together. There is only a slight difference in the cueball's direction when the efficiency is .85 that of a two ball collision. (But this is not true at large impact angles.) It doesn't account for increased spin loss during impact in the three ball case either, which should happen to some extent, and widen the angle. So while it doesn't yield anything definite, a noticeable difference is within the realm of possibility. Jim<hr /></blockquote>Jim -- Dont forget that Bob's systems are for when the qball itself iz frozen to one or more OB's. But Vapros iz talking about a non-frozen qball, but where the OB iz frozen to another OB. Re the actual effekt on the deflexion angle, i know it aint much, less than 10dg, but my big table iz in little bits up in the loft of the stable, and my little table iz in the stable but iz minus its cushionz, and haz lots of boxes on it stacked up to allmost the ceiling, so i might do some tests when i set the table(s) up nearer Xmas. madMac.

cushioncrawler
09-04-2007, 07:08 PM
<blockquote><font class="small">Quote Vapros:</font><hr> Thanks. I don't have a table at home, but I'll set it up and try it at the pool room sometime this week. I'm surprised to hear that the increased deflection is so slight. Seems like something I should have already discovered, but maybe not. The shot does not come up often.

Here's another thought. I play one-pocket, and many shots are played gently, either for controlling the cue ball or to get pocket-speed on the object ball. Of course, on a spot shot from behind the line, getting to the short rail is not a problem. However, if the object ball is a bit out in front on the spot, but still on the center line of the table, and I have cue ball in hand, it frequently looks to me like a scratch, no matter where I put the cue ball down, leaving no option but to draw the rock a little to avoid the scratch. I suppose this is due to the slight variations in the 30 degree rule. Or is it just my imagination? Who else has noticed this?<hr /></blockquote>Vapros -- The trouble iz in the definition -- it all dependz on how u define "deflexion angle". The "slight variations in the 30dg rule" are totally due to definition. I define "deflexion angle" differently -- in my definition the deflexion angle duznt suffer any "variations" whatsoever (not inklooding here the short-range parabolic stuff).

Basically, its like this. Az u bring the qball (in-hand) closer and closer to the OB, u will find that eventually u will find a pozzy where a softly played rolling shot will allwayz scratch, no matter what the contact on the OB (not inklooding here a full-ball contact or contacts near full-ball). madMac.

Bob_Jewett
09-04-2007, 07:42 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> ... Jim -- Dont forget that Bob's systems are for ... But Vapros iz talking about a non-frozen qball, but where the OB iz frozen to another OB. ... <hr /></blockquote>
That is exactly the case that the "10 times fuller" system covers -- when two object balls that are frozen together are struck by the cue ball. This is a very, very important shot in the US game of one pocket, and it has been written about extensively in that context. I think you must be thinking about the "2 times fuller" system which indeed is for a cue ball frozen to the object ball. The two systems work for very different reasons, but they have similar rules of thumb.

As I mentioned above, a correct solution requires knowing the details of the interaction of the frozen balls, and a Hertz's-Law approximation comes out a little different from a Hooke's-Law approximation. See the February, 1996 article at http://www.sfbilliards.com/articles/BD_articles.html for the details of how to calculate shots with the 10-times-fuller system.

cushioncrawler
09-04-2007, 08:16 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr>.....That is exactly the case that the "10 times fuller" system covers -- when two object balls that are frozen together are struck by the cue ball. This is a very, very important shot in the US game of one pocket, and it has been written about extensively in that context. I think you must be thinking about the "2 times fuller" system which indeed is for a cue ball frozen to the object ball. The two systems work for very different reasons, but they have similar rules of thumb...... See the February, 1996 article.....<hr /></blockquote>Bob -- Thanks for that. The 10-times-fuller system iz for potting the 1-ball, what i call a "plant" (some snooker players wrongly call it a "set"). Snooker books mention simple plants, ie where the qball "steers clear" of the action (ie where the 1-ball trajectory iz nearnuff allwayz 90dg), but none dares to look (nor peek even) into yor 10-times sort of stuff, ie where the qball steers into the action and sort of double-hits the 1-ball.

But unless i am mistaken, Vapros iz more concerned about where the qball goze (in the 10-times system), rather than where the 1-ball goze. The qball's final resting place iznt mentioned in yor feb96 article (but it might be in one of yor other articles). madMac.

Jal
09-04-2007, 09:23 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>...But unless i am mistaken, Vapros iz more concerned about where the qball goze (in the 10-times system), rather than where the 1-ball goze. The qball's final resting place iznt mentioned in yor feb96 article (but it might be in one of yor other articles). madMac. <hr /></blockquote>Mac, I brought up the 10x system because it gives you the ratio of the impulses at the two interfaces.

Jim

cushioncrawler
09-04-2007, 09:30 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr>...But unless i am mistaken, Vapros iz more concerned about where the qball goze (in the 10-times system), rather than where the 1-ball goze. The qball's final resting place iznt mentioned in yor feb96 article (but it might be in one of yor other articles)...<hr /></blockquote>Mac, I brought up the 10x system because it gives you the ratio of the impulses at the two interfaces....<hr /></blockquote>Jim -- So, the qball would "feel" that that OB weighed M plus M/10, ie 1.1M ???? madMac.

Jal
09-04-2007, 10:17 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr>...But unless i am mistaken, Vapros iz more concerned about where the qball goze (in the 10-times system), rather than where the 1-ball goze. The qball's final resting place iznt mentioned in yor feb96 article (but it might be in one of yor other articles)...<hr /></blockquote>Mac, I brought up the 10x system because it gives you the ratio of the impulses at the two interfaces....<hr /></blockquote>Jim -- So, the qball would "feel" that that OB weighed M plus M/10, ie 1.1M ???? madMac. <hr /></blockquote>I'm not sure about that Mac, it depends on the impact angle. By impact angle I mean the angle between the lines of centers of the two pairs of balls...the first object ball being in common...

So if J12 is the impulse between the cueball and first object ball (ignoring friction), and J23 the impulse between the two OB's, and phi is the impact angle, then:

J23 = (9/10)(J12)cos(phi)

for the 10x system to work.

That and the conservations laws are enough to give you the cueball's post impact speed along the line of centers with the first OB, IF you know how much energy is lost to friction, spin, hysteresis. I hoped it might be relatively insensitive to this, but it isn't. Since static friction enters the picture at the 2-3 interface, it's not easy to figure.

Jim

Vapros
09-04-2007, 10:45 PM
Are you guys still discussing my question? What if the shot clock goes off before I finish my calculations? I don't think I understand all I know about this . . .

cushioncrawler
09-05-2007, 01:59 AM
<blockquote><font class="small">Quote Jal:</font><hr>.... So if J12 is the impulse between the cueball and first object ball (ignoring friction), and J23 the impulse between the two OB's, and phi is the impact angle, then: J23 = (9/10)(J12)cos(phi).... for the 10x system to work. That and the conservations laws are enough to give you the cueball's post impact speed along the line of centers with the first OB, IF you know how much energy is lost to friction, spin, hysteresis. I hoped it might be relatively insensitive to this, but it isn't. Since static friction enters the picture at the 2-3 interface, it's not easy to figure...<hr /></blockquote>Jim -- I reckon that there iz an error here. J12 duznt depend much on J23. What we need to know iz how much of J23 adds to J12. That iz a different question. For example, what if there iz some daylite between OB1 and OB2!? -- here J23 will be az big az ever, but the contribution of J23 to J12 will be zilch. madMac.

cushioncrawler
09-05-2007, 02:26 AM
<blockquote><font class="small">Quote Vapros:</font><hr> Are you guys still discussing my question? What if the shot clock goes off before I finish my calculations? I don't think I understand all I know about this....<hr /></blockquote>Vapros -- I think we are. But, re Q2 (or iz it Q3), what iz the exact arrangement of the balls, where u reckon that a scratch looks certain for any/all contacts (or something??). Duz my description of the peculiar situation (where all contacts do indeed rezult in a scratch) relate to your arrangement?? madMac.

ken_r
09-05-2007, 06:45 AM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <blockquote><font class="small">Quote ken_r:</font><hr> Why would it be any different? <hr /></blockquote>
While there is no obvious answer that it has to be different, in fact it turns out that it is different.

When only two balls collide, the theory is fairly simple, and gives the 90-degree rule that everyone knows well, followed by the effect of any draw or follow on the cue ball. For follow (the so-called 30-degree rule, but known for a much longer time as the half-ball rule), the result is a maximum of about 35 degrees of deflection of the cue ball's path from its original line.

When more than two balls are involved in the collision, the physics is very, very complicated. To find the paths of the balls, you need to look at the details of the collision. Those details can be ignored (while maintaining reasonable accuracy) if only two balls are involved.

If you want to see a more extreme effect than two balls frozen, try the half-ball scratch off a triangle of three balls spotted on the head spot, like the front three balls of a full rack. I think you will find that the amount the cue ball is deflected is much larger than for either a single ball or two frozen balls. <hr /></blockquote>

Yes, with 3 balls frozen in a pyramid you will get a different result. The original question was about 2 frozen balls but this is certainly relevent. In a pyramid formation you get the effect of hitting a heavier ball and some rebound would be expected.

I did some rather unscientific testing on my table and confirmed what I expected. With two frozen balls the cue ball comes off at "30 degrees" just as a single ball (I tried with the two balls frozen at many different angles). I put 9 balls in a straight line and the cue ball comes off the same. I put the same 9 balls in a S shape, zig zagged them, etc. As long as each ball only impacted or "pushed" one ball (a pyramid the head ball "pushes" two) the cue ball came off at "30 degrees".

With a 3 ball pyramid the angle dropped to about 28 degrees and a 6 ball pyramid or 9 ball rack was about 26 degrees.

Vapros
09-05-2007, 08:56 AM
My question concerned a simple shot in which the object is near the center line, but a little out in front of the spot, and the cue ball is in hand, behind the line. No other balls on the table are needed. My impression is that any rolling shot will scratch, but the truth seems to be that any such shot will be close to a scratch. That keeps it from being a comfortable situation. I've had a partial answer here - that the 30 degree line can be altered a little with a speed adjustment. As I see it, it remains a judgment call, and it is my problem to locate the natural deflection and to adjust from there. Sort of living on the edge, one might say. A little draw, to get the cue to the long rail, looks like the only safe solution. The safe area on the short rail shrinks up to almost nothing.

Deeman3
09-05-2007, 09:54 AM
Vapros,

Are you describing the scratch arc near the spot?

Vapros
09-05-2007, 10:36 AM
Maybe. I never heard of the scratch arc. Tell me about it.

Jal
09-05-2007, 12:16 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>Jim -- I reckon that there iz an error here. J12 duznt depend much on J23. What we need to know iz how much of J23 adds to J12. That iz a different question. For example, what if there iz some daylite between OB1 and OB2!? -- here J23 will be az big az ever, but the contribution of J23 to J12 will be zilch. madMac. <hr /></blockquote>Mac, the expression was derived in accordance with the direction of the first object ball as a function of impact angle, which is an empirical result, I assume. If you want to rearrange things, then:

J12 = (10/9)J23/cos(phi)

but this doesn't really tell you a heck of a lot more. It needs a correction for friction at the 2-3 interface, but then you introduce shot speed dependency. Since the 30-degree rule derivations don't generally include ball/ball friction, I didn't bother either (especially since you need some data here). This was just a quick and dirty way of getting an idea of what the range of deviations from the 30-degree rule might be.

Jim

Deeman3
09-05-2007, 12:18 PM
While I have not seen it on a piece of paper, there is a an arc that starts near the corner pocket and sweeps up through a point near the spot in a complete semi-circle that basically indicates the spots a scrtach will occur with a naturally rolling cue ball. I say I have never seen it in print as I learned it like many of the one pocket "guidelines" i.e kiss shots, when a small child so I don't actually use it other than "knowing" when the scratch is "on". So, of course, when faced with this shot, you either have to apply follow or draw to avoid the sure scratch. Of course as the object ball/cue ball contact relationship becomes more thin, the less you can do about altering the path of the cue ball.

Has anyone seen this in print or is this just another of my non-scientific intuitive jaberworsts? I assume, with all the calculations these scientists do, one would have long ago plotted that line from the 30 (or 31.66 degree rule or whatever) /ccboard/images/graemlins/confused.gif and shown exactly what I am explaining. Of course there is more than one scratch arc but those would involve the side pockets.

Bob_Jewett
09-05-2007, 12:39 PM
<blockquote><font class="small">Quote Vapros:</font><hr> My question concerned a simple shot in which the object is near the center line, but a little out in front of the spot, and the cue ball is in hand, behind the line. No other balls on the table are needed. My impression is that any rolling shot will scratch, but the truth seems to be that any such shot will be close to a scratch. That keeps it from being a comfortable situation. I've had a partial answer here - that the 30 degree line can be altered a little with a speed adjustment. As I see it, it remains a judgment call, and it is my problem to locate the natural deflection and to adjust from there. Sort of living on the edge, one might say. A little draw, to get the cue to the long rail, looks like the only safe solution. The safe area on the short rail shrinks up to almost nothing.

<hr /></blockquote>
The deflection for a rolling cue ball changes a lot with the fullness of hit. It has a maximum for roughly a half-ball hit and that maximum is not 30 degrees. It is closer to 35 degrees. The deflection of the cue ball goes to zero for both full and very thin hits. If you place the cue ball by a side cushion with ball in hand, and you roll the cue ball with follow, the scratch will be more or less impossible for the shot you describe.

Further, the angle of the cue ball's deflection does not depend on the speed of the shot. This is illustrated in http://www.sfbilliards.com/follow_curves.pdf where the final paths of the cue ball for various speeds will end up parallel to each other. The separation of the paths is due to the sliding that goes on during the curving part of the path, and there is more sliding for faster shots, but angles are normally only measured between straight lines (or a line and a tangent) so the angles are technically equal.

There is a spot for the object ball on the center-line of the table about 10 inches towards the head spot from the foot spot that has the problem you are worried about. If the cue ball is on the head spot, and the object ball is pocketed with a slowly rolling cue ball into either corner pocket, the scratch is almost certain.

If the cue ball is moved away from the head spot, the chance for a scratch disappears.

Vapros
09-05-2007, 12:55 PM
Now we're straining my brain. On this arc, are you saying the scratch is 'on' for those shots directed at the far corner pocket? I assume this is the case. Since the spot shot, itself, is not a scratch shot, (nor are the shots on any ball nearer the head rail) then the arc must extend farther toward the foot of the table, no? More like the traditional parabola. If so, then I guess the shot about which I enquired would be out there near the peak of the arc. Maybe that's my answer, then. If my object ball is on the arc, then it looks like a scratch, no matter where I put the rock down behind the line, thus requiring an adjustment of some kind.

Bob_Jewett
09-05-2007, 01:08 PM
<blockquote><font class="small">Quote Deeman3:</font><hr> While I have not seen it on a piece of paper, there is a an arc that starts near the corner pocket and sweeps up through a point near the spot in a complete semi-circle that basically indicates the spots a scrtach will occur with a naturally rolling cue ball. ... <hr /></blockquote>
Onoda described the semi-circles for dead scratches for stun shots in an article in BD in the early 1990's. I have extended that discussion to include draw shots for half-ball cuts in my latest BD column. The former column is available on the SFBA web site on the misc files page. Those diagrams assume an object ball and a pocket and a specific kind of shot. The resulting arcs have their ends at two pockets and are mostly independent of the position of the cue ball.

For a given cue ball position and a given pocket, there is a range of object ball positions that will result in a scratch in that pocket for a rolling cue ball and a half-ball hit on the "correct" side of the object ball. The "locus" of those object ball positions is an arc of a circle. If you ignore the diameters of the balls, the ends of the arc are in the pocket and at the cue ball. The number of degrees in the arc is not 180 as it would be for a stun shot, but rather 70 degrees which is twice the angle the cue ball is deflected for a rolling half-ball shot. (I had not thought of this extension of the "semi-circles of death" and will include it in a future column.)

A related diagram is in several books about English Billiards. Those books show all the places on the table for an object ball that you can scratch from for a rolling half-ball shot with the cue ball in hand in the D. Those diagrams don't have single arcs because you can move the cue ball in the D, so the "scratchable" locations are arc-ish areas.

Vapros
09-05-2007, 01:09 PM
I understand that the angles will be technically equal at all speeds, but the higher speed, in effect, moves the angle's point of origin back down the shot line and away from the object ball (due to the little side-trip along the tangent line) and thus defeats the scratch. Am I correct in this?

Jal
09-05-2007, 01:15 PM
<blockquote><font class="small">Quote Deeman3:</font><hr> While I have not seen it on a piece of paper, there is a an arc that starts near the corner pocket and sweeps up through a point near the spot in a complete semi-circle that basically indicates the spots a scrtach will occur with a naturally rolling cue ball. I say I have never seen it in print as I learned it like many of the one pocket "guidelines" i.e kiss shots, when a small child so I don't actually use it other than "knowing" when the scratch is "on". So, of course, when faced with this shot, you either have to apply follow or draw to avoid the sure scratch. Of course as the object ball/cue ball contact relationship becomes more thin, the less you can do about altering the path of the cue ball.<hr /></blockquote>Deeman, that's the scratch circle for a stun shot you're describing. For a rolling cueball, there is no such simple curve because the 30-degree thing is measured from the cueball's original direction (unlike the 90-degree rule which is measured from the object ball's direction). For any particular location of the object ball, the cueball will head off in different directions depending on where it's coming from.

Jim

Vapros
09-05-2007, 01:17 PM
I have clouded my own post by referring to the head of the table as the foot, and vice versa. Sorry about that.

Deeman3
09-05-2007, 01:31 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote Deeman3:</font><hr> While I have not seen it on a piece of paper, there is a an arc that starts near the corner pocket and sweeps up through a point near the spot in a complete semi-circle that basically indicates the spots a scrtach will occur with a naturally rolling cue ball. I say I have never seen it in print as I learned it like many of the one pocket "guidelines" i.e kiss shots, when a small child so I don't actually use it other than "knowing" when the scratch is "on". So, of course, when faced with this shot, you either have to apply follow or draw to avoid the sure scratch. Of course as the object ball/cue ball contact relationship becomes more thin, the less you can do about altering the path of the cue ball.<hr /></blockquote>Deeman, that's the scratch circle for a stun shot you're describing. For a rolling cueball, there is no such simple curve because the 30-degree thing is measured from the cueball's original direction (unlike the 90-degree rule which is measured from the object ball's direction). For any particular location of the object ball, the cueball will head off in different directions depending on where it's coming from.

Jim

<font color="blue"> Jim,

Yes, you are correct. I said that wrong. /ccboard/images/graemlins/blush.gif</font color> As I also said, I don't use that system as even a reference anymore. I beleive I know when the scratch is on. I get itchy. /ccboard/images/graemlins/grin.gif

<hr /></blockquote>

Cornerman
09-05-2007, 02:09 PM
<blockquote><font class="small">Quote Deeman3:</font><hr> While I have not seen it on a piece of paper, there is a an arc that starts near the corner pocket and sweeps up through a point near the spot in a complete semi-circle that basically indicates the spots a scrtach will occur with a naturally rolling cue ball.

Has anyone seen this in print or is this just another of my non-scientific intuitive jaberworsts? <hr /></blockquote>This is an absolutely essential concept, IMO, but it's for 90° tangent stun shot.

Scratch arc, critical semi-circle, 90° scratch circle, etc.

Fred

Bob_Jewett
09-05-2007, 03:01 PM
<blockquote><font class="small">Quote Jal:</font><hr> ... Deeman, that's the scratch circle for a stun shot you're describing. For a rolling cueball, there is no such simple curve because the 30-degree thing is measured from the cueball's original direction ... <hr /></blockquote>
I think the circular arc that I described above qualifies as a reasonably simple curve of "fatal" object ball positions for a given cue ball position. It's just not independent of the position of the cue ball as for the stun semi-circles.

Deeman3
09-05-2007, 03:12 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> ... Deeman, that's the scratch circle for a stun shot you're describing. For a rolling cueball, there is no such simple curve because the 30-degree thing is measured from the cueball's original direction ... <hr /></blockquote>
I think the circular arc that I described above qualifies as a reasonably simple curve of "fatal" object ball positions for a given cue ball position. It's just not independent of the position of the cue ball as for the stun semi-circles. <hr /></blockquote>

<font color="blue"> I love that term, "Fatal" object ball positions. It seems a perfect description. </font color>

Fran Crimi
09-05-2007, 03:20 PM
[ QUOTE ]
I love that term, "Fatal" object ball positions. It seems a perfect description.

DeeMan
http://Zazzle.com/Deeman*

<hr /></blockquote>

Ha! I was thinking the same thing, Dee. The other time i've heard the word used was the infamous Microsoft error message: "a fatal error has occurred." When you see that one on your computer, you know you're about to go through a pulling-your-hair-out-type experience.

But I digress...

Fran

Deeman3
09-05-2007, 04:23 PM
Fran,

I think I spotted your "Big Hair" on TV yesterday in a match from about 1996, back in the big hair days. It was one of those where they tried to show two matches on one broadcast but gave up after a couple of shows of switching back and forth from a cliffhanger shot to a ho-hum one on the next table. Drove me nuts...I was out of the country during those years so the matches were new to me. /ccboard/images/graemlins/cool.gif

Anyway, you looked enthralled with the match. LOL

dr_dave
09-05-2007, 04:28 PM
<blockquote><font class="small">Quote Vapros:</font><hr> I understand that the angles will be technically equal at all speeds, but the higher speed, in effect, moves the angle's point of origin back down the shot line and away from the object ball (due to the little side-trip along the tangent line) and thus defeats the scratch. Am I correct in this? <hr /></blockquote>You are correct. See the diagrams in my one-page summary of the 30-degree rule (http://billiards.colostate.edu/resources/30_degree_rule_summary.pdf). You can also see a plot that shows how the CB angle varies with ball-hit fraction (cut angle).

Regards,
Dave

cushioncrawler
09-05-2007, 05:13 PM
<blockquote><font class="small">Quote Vapros:</font><hr> My question concerned a simple shot in which the object is near the center line, but a little out in front of the spot, and the cue ball is in hand, behind the line. No other balls on the table are needed. My impression is that any rolling shot will scratch, but the truth seems to be that any such shot will be close to a scratch. That keeps it from being a comfortable situation. I've had a partial answer here - that the 30 degree line can be altered a little with a speed adjustment. As I see it, it remains a judgment call, and it is my problem to locate the natural deflection and to adjust from there. Sort of living on the edge, one might say. A little draw, to get the cue to the long rail, looks like the only safe solution. The safe area on the short rail shrinks up to almost nothing.<hr /></blockquote>Ahha -- Yes, i see that Bob's comments relate to this. My earlyr comments dont apply, but i will add to them here anyhow koz it might be interesting.

When the qball and OB are a long wayz apart, if u play a rolling shot off the right'hand'side of the OB, for say each of 16 thicknesses of contact, ie full-ball being 16/16, half-ball being 8/16, and a thin cut say 1/16, and if u somehow made a chalk-line of the qball'z trajectory for each (16 in all), u would find that the chalk-lines looked like a fan, with no lines crossing. I dont really expekt anyone to be able to accurately and consistently "divide" the OB into 16 (so to speak) and play theze sorts of shots and actually mark 16 such lines, most of this iz really only hypothetical.

Now, if u brort the qball and OB much closer together, and repeated this (drawing 16 lines), u would reech the situation where some of the lines started to cross, ie no more "nice fan shape".

When the qball iz say less than 1 ball clear of the OB, most of the 16 lines will cross each other. There will be a sort of small "zone" where most of the lines intersect. If u put a pocket at that intersection zone, then most of the 16 contacts will rezult in a scratch. Obviously the full-ball (16/16) contact wont scratch, and neither will 15/16 and 14/16 (depending on how close the qball iz to the OB), but every other contact will scratch. That iz what i woz trying to describe.

This little phenomenon would have been well known to old timers who played straight rail and baulkline (ie no pockets), for their cannon play. madMac.

Fran Crimi
09-05-2007, 05:36 PM
<blockquote><font class="small">Quote Deeman3:</font><hr> Fran,

I think I spotted your "Big Hair" on TV yesterday in a match from about 1996, back in the big hair days. It was one of those where they tried to show two matches on one broadcast but gave up after a couple of shows of switching back and forth from a cliffhanger shot to a ho-hum one on the next table. Drove me nuts...I was out of the country during those years so the matches were new to me. /ccboard/images/graemlins/cool.gif

Anyway, you looked enthralled with the match. LOL <hr /></blockquote>

It could have been me but I doubt it. I rarely if ever sat in the stands. I was usually pacing in the background if not playing. As for the hair, it's still big. I call it the Italian curse --- natural curls --- My stylist says a little prayer when she sees me coming. /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif

Fran

cushioncrawler
09-05-2007, 05:38 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr>Jim -- I reckon that there iz an error here. J12 duznt depend much on J23. What we need to know iz how much of J23 adds to J12. That iz a different question. For example, what if there iz some daylite between OB1 and OB2!? -- here J23 will be az big az ever, but the contribution of J23 to J12 will be zilch. madMac. <hr /></blockquote>Mac, the expression was derived in accordance with the direction of the first object ball as a function of impact angle, which is an empirical result, I assume. If you want to rearrange things, then: J12 = (10/9)J23/cos(phi).... but this doesn't really tell you a heck of a lot more. It needs a correction for friction at the 2-3 interface, but then you introduce shot speed dependency. Since the 30-degree rule derivations don't generally include ball/ball friction, I didn't bother either (especially since you need some data here). This was just a quick and dirty way of getting an idea of what the range of deviations from the 30-degree rule might be....<hr /></blockquote>Jim -- When i have time i will havta do some tests and see how the plot compares to yor equation. That 10/9 might proov nearnuff, and that Cos() might be nearnuff too. But it will all be empirical in the end, a theoretical attack would be a nightmare. It haz taken me years to learn (memorize) the 1/2 ball deflexion angle (and 1/4 ball and 3/4 ball), but i doubt that me nor anyone else would ever find it worthwhile to actually learn frozen deflexion angles. In English Billiards i would simply play some sort of cannon instead of an in-off. madMac