PDA

View Full Version : Ball Collisions/Dominant

slim
03-28-2007, 12:36 PM
When your cutting an object ball into the pocket and using 12 o'clock high/center/follow english on the cue ball, at what cut angle (of the object ball) will the object ball become the "DOMINANT" weight force of the colliding objects, therefore..not allow the cue ball to go forward of the cut angle/tangent line? I would think 45 degrees would be the telling angle.

ceebee
03-28-2007, 01:01 PM
Bert Kinister spoke about these things in one of his videos.

He said this. If you hit a half ball shot on the object ball, with no spin to assist the Cue Ball's roll, both balls should roll equidistant.

I have experimented with this &amp; I have to agree with him. Using a percentage formula, for determining the roll of both balls, i.e. 10%/90% to 90%/10% is difficult because of unintended spin on the Cue Ball. JMHO

Good Luck...

Bob_Jewett
03-28-2007, 01:13 PM
<blockquote><font class="small">Quote slim:</font><hr> When your cutting an object ball into the pocket and using 12 o'clock high/center/follow english on the cue ball, at what cut angle (of the object ball) will the object ball become the "DOMINANT" weight force of the colliding objects, therefore..not allow the cue ball to go forward of the cut angle/tangent line? I would think 45 degrees would be the telling angle. <hr /></blockquote>
The cue ball starts out along the tangent line as it leaves the object ball regardless of what spin it has. If it has follow coming into the collision, it always will curve ahead of the tangent line. In the December 1998 article at http://www.sfbilliards.com/articles/BD_articles.html (that's the one marked 1998-12), there is a very simple, accurate system for finding the angle that the cue ball will take for any follow shot in which the cue ball is rolling smoothly on the cloth.

slim
03-28-2007, 04:31 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <hr /></blockquote>
BOB....SEE INSERTS IN CAP'S.
The cue ball starts out along the tangent line as it leaves the object ball regardless of what spin it has.
MOST OF THE TIME IT DOESN'T, WHY, BECAUSE WHEN TWO OBJECTS CONTACT EACH OTHER THEIR FIRST ACTION IS TO REBOUND AWAY, AND AS YOU KNOW, THE FULLER (NOT THE BEST CHOICE OF WORDS) THE OBJECT BALL IS CONTACTED THE MORE THE "ACTION/REACTION" PRINCIPLE APPLIES.
If it has follow coming into the collision, it always will curve ahead of the tangent line.

NOT IF THE CUE BALL IS STRUCK WITH A FULL SWING WITH THE BELOW SITUATION.

SO YOUR SAYING THAT IF I CUT AND OBJECT BALL AT A 70 DEGREE ANGLE WITH 12 O'CLOCK HIGH CENTER FOLLOW AT A HIGH SPEED, THE CUE BALL WILL CURVE AHEAD OF THE TANGENT LINE?
In the December 1998 article at http://www.sfbilliards.com/articles/BD_articles.html (that's the one marked 1998-12), there is a very simple, accurate system for finding the angle that the cue ball will take for any follow shot in which the cue ball is rolling smoothly on the cloth. <hr /></blockquote>

dr_dave
03-28-2007, 04:53 PM
<blockquote><font class="small">Quote slim:</font><hr> When your cutting an object ball into the pocket and using 12 o'clock high/center/follow english on the cue ball, at what cut angle (of the object ball) will the object ball become the "DOMINANT" weight force of the colliding objects, therefore..not allow the cue ball to go forward of the cut angle/tangent line? I would think 45 degrees would be the telling angle.<hr /></blockquote>With follow, the CB always goes forward of the tangent line. The amount and when depends on cut angle and speed. My April-July '04 and March '05 instructional articles (http://www.engr.colostate.edu/~dga/pool/bd_articles/index.html) for lots of illustrations, explanations, and examples (with video demos).

Also, you might be interested in: where the CB goes for different cases (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=234071&amp;page =0&amp;view=&amp;sb=&amp;o=&amp;fpart=&amp;vc=).

Regards,
Dave

dr_dave
03-28-2007, 05:05 PM
<blockquote><font class="small">Quote ceebee:</font><hr> Bert Kinister spoke about these things in one of his videos.

He said this. If you hit a half ball shot on the object ball, with no spin to assist the Cue Ball's roll, both balls should roll equidistant.

I have experimented with this &amp; I have to agree with him.<hr /></blockquote>If you are interested, TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf) shows the details for this. It turns out that the CB and OB will roll about the same distance for about a 40 degree cut angle (about a 30% ball-hit), and the CB and OB will leave at the same angle for about a 33 degree cut shot (about a 40% ball-hit). Both cut angles are less than a 1/2-ball hit (50%).

Regards,
Dave

cushioncrawler
03-28-2007, 05:35 PM
<blockquote><font class="small">Quote ceebee:</font><hr> Bert Kinister spoke about these things in one of his videos. He said this. If you hit a half ball shot on the object ball, with no spin to assist the Cue Ball's roll, both balls should roll equidistant. I have experimented with this &amp; I have to agree with him. Using a percentage formula, for determining the roll of both balls, i.e. 10%/90% to 90%/10% is difficult because of unintended spin on the Cue Ball. JMHO Good Luck... <hr /></blockquote>Charley -- Thats an interesting idea, ie the notion of a certain contact that rezults in the qball and OB rolling equidistance. But i doubt that halfball is the answer, i reckon that it would be a bit thinner than halfball. But i think that slim woz thinking of some other sort of problem, not sure.

Anyhow, Bob and Dr Dave could have added that, koz the balltoball collision "e" is allways less than 1.00, the qball allwayz follows throo a little at first (ie during impact), ie past the tangent line, before then doing its thing (ie before drawing back outside the tangent line, or accelerating further inside the tangent line, after impact).

And, even if e were 1.00, a qball would nonetheless "follow-throo" an additional perhaps 1mm during the impact event, due partly to the large strain(s) in the immediate area of the contact (ie in the "flatspots"), and due partly to overall compression in the ball(s). madMac.

Jal
03-28-2007, 06:22 PM
<blockquote><font class="small">Quote slim:</font><hr> <blockquote><font class="small">Quote Bob_Jewett:</font><hr> <hr /></blockquote>
The cue ball starts out along the tangent line as it leaves the object ball regardless of what spin it has.<hr /></blockquote>
MOST OF THE TIME IT DOESN'T, WHY, BECAUSE WHEN TWO OBJECTS CONTACT EACH OTHER THEIR FIRST ACTION IS TO REBOUND AWAY, AND AS YOU KNOW, THE FULLER (NOT THE BEST CHOICE OF WORDS) THE OBJECT BALL IS CONTACTED THE MORE THE "ACTION/REACTION" PRINCIPLE APPLIES.<hr /></blockquote>
You've gotten good explanations but I thought a diagram might help.

http://ww2.netnitco.net/users/gtech/BasicCollision.jpg

Just before the collision, the cueball is traveling in the direction of the black arrow, labeled "V" in the picture. Its velocity IS the arrow "V", a vector, and consists of two components perpendicular to each other, shown as the red and green arrows. The cut angle is "A".

If it weren't for friction between the balls, the object ball would be completely unaware of the green component, which besides being perpendicular to the red one, is in the direction of the tangent line. It would only respond to the red component, just as if the cueball was traveling in the direction of the red arrow and making a "full" hit from this direction.

After impact, the cueball would stop all movement in the direction of the red arrow (also as if it had hit the object ball full in that direction), but continue on in the direction of the green arrow unscathed (it would retain its green component in full). In most cases though, some friction will exist between the balls. This will slightly slow the cueball down in the green direction, and also give the object ball a small velocity component in the same direction (throw). And as Mac pointed out, the cueball doesn't come to an absolute complete stop in the red direction for the reasons he stated (but close).

If the cueball has topspin, the friction force from the cloth caused by this spin will point in the direction of the black arrow - the cueball's original pre-impact velocity direction. This will propel the cueball across and forward of the tangent line, and it will happen at all cut angles except 90 degrees.

Maybe more to the point is that you're right that an equal but opposite force acts on each ball. But what governs the final result is the product of this force with the time of contact (Ft), and distances the balls move during contact (Fx), where F is the averaged value of the force in any particular direction. Since t and x are common to both balls during impact, both of these products are the same for each ball (but opposite in sign), and yield the conservation of momentum and energy respectively. These dictate that the cueball stops in the red direction for instance, ideally.

In reality, the F multiplied by t and the F multiplied by x are not exactly one and the same. That is, F averaged over t and F averaged over x are slightly different (in a perfectly elastic collision they would be identical). This indicates a violation of the conservation of energy (mechanical), but this lost energy can be traced to other things such as heat and surface deformation (potential energy). Equal and opposite forces do not mean that the balls rebound from some fixed reference point on the table, such as the point of contact, equally.

Jim

dr_dave
03-28-2007, 06:28 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote ceebee:</font><hr> Bert Kinister spoke about these things in one of his videos. He said this. If you hit a half ball shot on the object ball, with no spin to assist the Cue Ball's roll, both balls should roll equidistant. I have experimented with this &amp; I have to agree with him. Using a percentage formula, for determining the roll of both balls, i.e. 10%/90% to 90%/10% is difficult because of unintended spin on the Cue Ball. JMHO Good Luck... <hr /></blockquote>Charley -- Thats an interesting idea, ie the notion of a certain contact that rezults in the qball and OB rolling equidistance. But i doubt that halfball is the answer, i reckon that it would be a bit thinner than halfball. But i think that slim woz thinking of some other sort of problem, not sure.

Anyhow, Bob and Dr Dave could have added that, koz the balltoball collision "e" is allways less than 1.00, the qball allwayz follows throo a little at first (ie during impact), ie past the tangent line, before then doing its thing (ie before drawing back outside the tangent line, or accelerating further inside the tangent line, after impact).

And, even if e were 1.00, a qball would nonetheless "follow-throo" an additional perhaps 1mm during the impact event, due partly to the large strain(s) in the immediate area of the contact (ie in the "flatspots"), and due partly to overall compression in the ball(s). madMac. <hr /></blockquote>Mac,

TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf) has all of the details (except for ball deformations, which are extremely small and neglected).

Regards,
Dave

cushioncrawler
03-29-2007, 01:43 AM
<blockquote><font class="small">Quote Jal:</font><hr> .....If it weren't for friction between the balls, the object ball would be completely unaware of the green component, which besides being perpendicular to the red one, is in the direction of the tangent line. It would only respond to the red component, just as if the cueball was traveling in the direction of the red arrow and making a "full" hit from this direction.....<hr /></blockquote>Jim -- Or, if there is gearing at (before) impact. madMac.

cushioncrawler
03-29-2007, 02:09 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> .....Mac, ...TP A.16 .... has all of the details (except for ball deformations, which are extremely small and neglected).....<hr /></blockquote>Dr Dave -- Good stuff as usual. If i read it correctly, an 0.307 contact gives the qball and OB equal speeds. Re rolling distance(s), technically, one could allso factor-in the different skidding distances (mm) travelled by the qball and OB, but this is very difficult.

If i read it correctly, the equations and graphs take into account that the OB skids and decelerates, and that the qball skids and accelerates, before rolling(s) begin, and the equal speeds are when rolling. madMac.

Bob_Jewett
03-29-2007, 11:28 AM
<blockquote><font class="small">Quote slim:</font><hr> ... SO YOUR SAYING THAT IF I CUT AND OBJECT BALL AT A 70 DEGREE ANGLE WITH 12 O'CLOCK HIGH CENTER FOLLOW AT A HIGH SPEED, THE CUE BALL WILL CURVE AHEAD OF THE TANGENT LINE? ... <hr /></blockquote>
Yes. Here is a way you can show it. Put a ball on the center spot, in the exact middle of the table, between the two side pockets. Place the cue ball to cut the ball into the side pocket with your stick over a corner pocket. Place an object ball in the center of the far end cushion. If you make the ball in the side, and the cue ball has follow when it hits the object ball, the cue ball will contact the ball on the far end rail or maybe even curve forward past it. That's my prediction. Please try the shot and let us know what you find. If the angle is not steep enough for you, just move the cue ball away from the side rail.

slim
03-29-2007, 01:41 PM
Bob, I know the cue ball archs back towards the true tangent line when a power follow shot is executed, but when you hit an object ball full at at high speed the action/reaction of the two collisions distorts the tangent line as the cue ball tries to get back too and past the tangent line but usually hits a rail or??? and never will allow the cue ball to return and get back inside the tangent line.

Bob_Jewett
03-29-2007, 02:52 PM
<blockquote><font class="small">Quote slim:</font><hr> Bob, I know the cue ball archs back towards the true tangent line when a power follow shot is executed, but when you hit an object ball full at at high speed the action/reaction of the two collisions distorts the tangent line as the cue ball tries to get back too and past the tangent line but usually hits a rail or??? and never will allow the cue ball to return and get back inside the tangent line. <hr /></blockquote>
I think that if you do the test I suggest from various angles you will see that the cue ball is never pushed back of the tangent line when it has follow unless you are using a light cue ball.

cushioncrawler
03-29-2007, 05:00 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote ceebee:</font><hr> Bert Kinister spoke about these things in one of his videos. He said this. If you hit a half ball shot on the object ball, with no spin to assist the Cue Ball's roll, both balls should roll equidistant. I have experimented with this &amp; I have to agree with him. Using a percentage formula, for determining the roll of both balls, i.e. 10%/90% to 90%/10% is difficult because of unintended spin on the Cue Ball. JMHO Good Luck... <hr /></blockquote>Charley -- Thats an interesting idea, ie the notion of a certain contact that rezults in the qball and OB rolling equidistance. But i doubt that halfball is the answer, i reckon that it would be a bit thinner than halfball. But i think that slim woz thinking of some other sort of problem, not sure.

Anyhow, Bob and Dr Dave could have added that, koz the balltoball collision "e" is allways less than 1.00, the qball allwayz follows throo a little at first (ie during impact), ie past the tangent line, before then doing its thing (ie before drawing back outside the tangent line, or accelerating further inside the tangent line, after impact).

And, even if e were 1.00, a qball would nonetheless "follow-throo" an additional perhaps 1mm during the impact event, due partly to the large strain(s) in the immediate area of the contact (ie in the "flatspots"), and due partly to overall compression in the ball(s). madMac. <hr /></blockquote>madMac -- U forgot to mention Flatspot-Sqeez. Its like this. We talk about the tangent line, but, there are at least 2 tangent lines. The first is the common tangent to the 2 circles (balls) at initial contact. The 2nd is the common tangent at final contact. There can be over 1.0dg difference between theze 2 lines (for a hi-speed impact). In other wordz, the OB can depart on a trajectory over 0.5dg "off" the initial tangent, ie nearnuff along the "mean" of the 2 tangents. This iz all due to that fact that the impact-time iz not zero (seconds), and that the FlatSpot size iz not zero (mm). Balltoball friction (ie throw) is an entirely different animal, and duznt affect Flatspot-Sqeez.

Hmmmmm. I wonder what the size of the FlatSpot-Sqeez iz for that miracle cutback double on the other thread. Let me see, qball speed 4m/s, contact 1/128th, hmmmmm, might be over 1dg here, ie FSS might be that "missing" 3rd dose of throw there. madMac.

Jal
03-29-2007, 07:21 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>Dr Dave -- Good stuff as usual.<hr /></blockquote>Let me second that.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>If i read it correctly, an 0.307 contact gives the qball and OB equal speeds.<hr /></blockquote>I believe so Mac. As he indicates, this is a cut angle of 43.8 deg, although it will vary with shot speed some. If you figure this without taking into account loss of spin and velocity from ball/ball friction, it works out to about 33 degrees. There is a pretty big difference then between the two treatments. So big in fact that I thought some error might have been made, and proceeded to find one...only the error was on my part. Sorry about the run-around Dr. Dave.

At any rate, here is a case where seemingly small details make a fairly big difference.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>...If i read it correctly, the equations and graphs take into account that the OB skids and decelerates, and that the qball skids and accelerates, before rolling(s) begin, and the equal speeds are when rolling. madMac. <hr /></blockquote>You read it as I read it, but it might be possible that at some cut angles the cueball slows down (but still changes direction and thus accelerates by definition). It will take a graph or a fair amount of algebra to find out.

Jim

Bob_Jewett
03-29-2007, 08:38 PM
<blockquote><font class="small">Quote dr_dave:</font><hr> ... TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf) has all of the details (except for ball deformations, which are extremely small and neglected).
... <hr /></blockquote>
Another thing that seems to have been neglected and is very important is that the object ball will slow to 5/7 of its speed right after the collision as it acquires natural roll.

The theoretical answer has to come out to what is observed: the cue ball and object ball roll the same distance for a soft, rolling shot for something very close to a half-ball hit. This result is well known and absolutely standard for carom play.

dr_dave
03-30-2007, 08:58 AM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> ... TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf) has all of the details (except for ball deformations, which are extremely small and neglected).
... <hr /></blockquote>
Another thing that seems to have been neglected and is very important is that the object ball will slow to 5/7 of its speed right after the collision as it acquires natural roll.

The theoretical answer has to come out to what is observed: the cue ball and object ball roll the same distance for a soft, rolling shot for something very close to a half-ball hit. This result is well known and absolutely standard for carom play. <hr /></blockquote>Funny you should mention that. I also noticed this after I read Mac's message. I've just been too busy the last few days to check all of my work and make the change. I should have everything fixed by later today.

Thanks,
Dave

dr_dave
03-30-2007, 09:50 AM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> .....Mac, ...TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf) .... has all of the details (except for ball deformations, which are extremely small and neglected).....<hr /></blockquote>Dr Dave -- Good stuff as usual.<hr /></blockquote>Thanks.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>If i read it correctly, an 0.307 contact gives the qball and OB equal speeds.<hr /></blockquote>Actually, there was a slight error in my original posting. The revised value for equal final speeds is a ball-hit fraction of 44.8% (cut angle = 33.5 degrees). Sorry about that. Thank you for making me think about this some. Originally, I was accounting for the post-impact skid and curve of the CB, but I neglected to account for the OB skid (as described by TP 4.1 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/TP_4-1.pdf)). I still neglect OB swerve effects due to spin-transfer (see TP A.24 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-24.pdf)), but these effects were shown to be very extremely small.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>Re rolling distance(s), technically, one could allso factor-in the different skidding distances (mm) travelled by the qball and OB, but this is very difficult.<hr /></blockquote>If I understand what you mean here, I think I have it covered.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>If i read it correctly, the equations and graphs take into account that the OB skids and decelerates, and that the qball skids and accelerates, before rolling(s) begin, and the equal speeds are when rolling.<hr /></blockquote>Yes (now that the OB error has been fixed).

Thanks again for helping me spot the error,
Dave

dr_dave
03-30-2007, 11:43 AM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr>Dr Dave -- Good stuff as usual.<hr /></blockquote>Let me second that.

<blockquote><font class="small">Quote cushioncrawler:</font><hr>If i read it correctly, an 0.307 contact gives the qball and OB equal speeds.<hr /></blockquote>I believe so Mac. As he indicates, this is a cut angle of 43.8 deg, although it will vary with shot speed some. If you figure this without taking into account loss of spin and velocity from ball/ball friction, it works out to about 33 degrees. There is a pretty big difference then between the two treatments. So big in fact that I thought some error might have been made, and proceeded to find one...only the error was on my part. Sorry about the run-around Dr. Dave.<hr /></blockquote>Jim,

I actually had a slight (but important) error in my work after all (not with the CB, but with the OB). Thank you for making me look at everything closer. The revised and improved document is now posted. See TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf).

<blockquote><font class="small">Quote Jal:</font><hr><blockquote><font class="small">Quote cushioncrawler:</font><hr>...If i read it correctly, the equations and graphs take into account that the OB skids and decelerates, and that the qball skids and accelerates, before rolling(s) begin, and the equal speeds are when rolling. madMac.<hr /></blockquote>You read it as I read it, but it might be possible that at some cut angles the cueball slows down (but still changes direction and thus accelerates by definition). It will take a graph or a fair amount of algebra to find out.<hr /></blockquote>Since you guys were so helpful, I decided to add the graph for you. It is on page 4 of the revised TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf). It turns out that the CB picks up speed for all cut angles (except 90 degrees). Like you, I was a little surprised by this result, but not too much.

Dave

Bob_Jewett
03-30-2007, 11:55 AM
<blockquote><font class="small">Quote dr_dave:</font><hr>... It turns out that the CB picks up speed for all cut angles (except 90 degrees). Like you, I was a little surprised by this result, but not too much.... <hr /></blockquote>
This is a direct, simple result from the geometrical construction of the follow path. If you accept the construction, the result is immediate. See "2001-06 Drawing Draw -- continued for draw and angle shots" on http://www.sfbilliards.com/articles/BD_articles.html for the details of the construction.

dr_dave
03-30-2007, 12:00 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr>Another thing that seems to have been neglected and is very important is that the object ball will slow to 5/7 of its speed right after the collision as it acquires natural roll.<hr /></blockquote>You are right. The slowing of the OB is a major factor. I had accounted for the sliding and curving effects for the CB, but my original version did not account for the OB slowing. The complete analysis, that takes everything into consideration (except extremely small effects like ball deformation and OB swerve), can be found with many additional results in TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf). FYI, I also expanded my analysis in TP 4.1 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/TP_4-1.pdf) to show, in simple terms, where the 5/7 factor comes from. The 5/7 factor is also evident in the results of TP A.4 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-4.pdf), which is not as simple (but more general).

<blockquote><font class="small">Quote Bob_Jewett:</font><hr>The theoretical answer has to come out to what is observed: the cue ball and object ball roll the same distance for a soft, rolling shot for something very close to a half-ball hit. This result is well known and absolutely standard for carom play.<hr /></blockquote>Here is a summary of the main results of TP A.16:

for equal final CB and OB speeds:
cut angle = 33.5 degrees (44.8% ball-hit fraction)

for equal final CB and OB angles:
cut angle = 36.1 degrees (41.1% ball-hit fraction)

for best combination of both (close to equal speeds and directions):
cut angle = 35 degrees (43% ball-hit fraction)

So slightly less than a half-ball hit seems optimal for near identical post-impact CB and OB motion.

I will sleep much better tonight knowing that the theory now matches observation.

Dave

dr_dave
03-30-2007, 12:04 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr>... It turns out that the CB picks up speed for all cut angles (except 90 degrees). Like you, I was a little surprised by this result, but not too much.... <hr /></blockquote>
This is a direct, simple result from the geometrical construction of the follow path. If you accept the construction, the result is immediate. See "2001-06 Drawing Draw -- continued for draw and angle shots" on http://www.sfbilliards.com/articles/BD_articles.html for the details of the construction. <hr /></blockquote>I agree now that the result is obvious.

Thanks,
Dave

Jal
03-30-2007, 12:35 PM
<blockquote><font class="small">Quote dr_dave:</font><hr>...Since you guys were so helpful, I decided to add the graph for you. It is on page 4 of the revised TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf). It turns out that the CB picks up speed for all cut angles (except 90 degrees). Like you, I was a little surprised by this result, but not too much.<hr /></blockquote>Thanks Dr. Dave. Very thoughtful of you and appreciated. How is it that you do this overnight (please don't say it took 15 minutes), whereas I would still be in the process of sharpening my pencil?

Jim

PS: I would revise and retract any statements attributed to me regarding the loss of spin and speed from friction having a large effect, if it could be shown that I actually made them.

dr_dave
03-30-2007, 12:51 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr>...Since you guys were so helpful, I decided to add the graph for you. It is on page 4 of the revised TP A.16 (http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-16.pdf). It turns out that the CB picks up speed for all cut angles (except 90 degrees). Like you, I was a little surprised by this result, but not too much.<hr /></blockquote>Thanks Dr. Dave. Very thoughtful of you and appreciated. How is it that you do this overnight (please don't say it took 15 minutes), whereas I would still be in the process of sharpening my pencil?<hr /></blockquote>Actually, the new graph only took about 5 minutes, because I already had all of the equations defined in my MathCAD file (this is the software I use for all of my analyses). To create the graph, I just had to click on one button and copy and paste the equation labels onto the graph. Is that pencil sharp yet? /ccboard/images/graemlins/wink.gif

Now, it took a little more than 5 minutes to learn how to use MathCAD originally. I actually teach it in one of my mechanical engineering courses, so I know it quite well.

Regards,
Dave

Jal
03-30-2007, 03:25 PM
<blockquote><font class="small">Quote dr_dave:</font><hr> Actually, the new graph only took about 5 minutes...<hr /></blockquote>That's not nice! /ccboard/images/graemlins/mad.gif

Jim

dr_dave
03-30-2007, 03:30 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> Actually, the new graph only took about 5 minutes...<hr /></blockquote>That's not nice! /ccboard/images/graemlins/mad.gif<hr /></blockquote>Sorry. But as I said, originally learning the software took more than 5 minutes.

Thanks again for helping to keep me honest (and accurate),
Dave

Bob_Jewett
04-02-2007, 11:35 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> ... I think that if you do the test I suggest from various angles you will see that the cue ball is never pushed back of the tangent line when it has follow unless you are using a light cue ball.

Please do the test. <hr /></blockquote>
Well, I'm a little disappointed that no one actually bothered to do the test, especially the OP. Sometimes you can learn a lot more in 10 minutes of time on the table than in an hour of pounding your computer.

Anyway, I did the test. There was no cut angle for which I could get the cue ball to back up from the tangent line when playing with follow. In fact the cue ball seemed to follow the 1/4-penetration rule as described in the article cited above. (Theoretically, it should be 2/7 penetration, but 1/4 compensates for friction and curve and is easier to estimate.)

Conclusion: the original hypothesis that the cue ball will jump back from the tangent line when a thin cut is played at high speed is false. The cue ball seems to follow the theoretical path quite accurately.

A side note: when I was shooting the shot (see the description above) with my cue stick over the end cushion -- about a 70-degree cut -- the angle felt like the cue ball would come back more than the system predicted, but the system was correct and my feel wasn't. I think this is because I'm not used to playing cuts that thin.