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BigRigTom
02-25-2008, 12:51 PM
When two pool balls meet we have two spheres that are 2.25" in diameter coming together at a very small point.

Do any of you physics types know how to mathematically determine the size of that contact point?

I read an article on the Meucci Pool Cue site once that claimed that pool is the most accurate game ever and it talked about how we make these contact points come together with a degree of accuracy that is mind boggling.

Maybe I just need more work to do so I am not wasting my brain power on this type of minutiae.

Bob_Jewett
02-25-2008, 01:44 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body">When two pool balls meet we have two spheres that are 2.25" in diameter coming together at a very small point.

Do any of you physics types know how to mathematically determine the size of that contact point?.. </div></div>
For speeds of collision about like the break shot, it is a patch about 1/4-inch in diameter. If you know the contact time and the speed of the cue ball, it's possible to calculate the diameter of the patch. That information is of more or less no use to the typical player.

As far as the size of the contact patch being related to the accuracy of the shot, that may seem intuitively obvious, but it is wrong. If you think there is a direct connection, think on it some more. The location of the center of the contact patch, a single point, along with throw and some other stuff, determines which way the ball goes.

At tennis, the player contacts about a third of the ball with his racket. That's 120 degrees. That has nothing to do with with the accuracy of ball placement at tennis.

Similarly, some people claim that you only have to worry about the location of the tip on the cue ball for accuracy of spin down to the diameter of the tip's contact patch on the cue ball. That's also wrong. For some shots, you have to be a lot more accurate than that in your tip placement.

Jal
02-25-2008, 03:26 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body">...Maybe I just need more work to do so I am not wasting my brain power on this type of minutiae. </div></div>
It's much better to have a deep understanding of why the balls never drop than to mindlessly pot them all day long. Trust me. /forums/images/%%GRAEMLIN_URL%%/smile.gif

To get an approximate but reasonably accurate estimate of the width of the array of candidate contact points that will result in a successful shot, do the following:

- Estimate the width of the target area, say, 5" for a 4-1/2" pocket maybe. We can get into great detail here, but let's keep it simple.

- Subtract the diameter of the ball from this (2.25" say).

- Multiply this by the radius of the ball (1.125" say).

- Divide this by the distance to the target pocket in inches.

As a formula:

Wc = (Wt - 2R)(R/D)

So, with the object ball a distance of 36" from a 5" target:

Wc = (5" - 2.25")(1.125"/36") = 0.086"

Throw doesn't change things much because the difference in throw across this small "area" (arc length) will be very, very small. It mainly affects the position of this arc on the object ball. An exception, to some extent, occurs with a straight on shot.

Dr. Dave has done a great deal of work in figuring the effective target (pocket) sizes at various approach angles, shot speeds and actual pocket sizes. You can look up his articles for more info on this, but for a quick and dirty estimate, just use the width of the pocket, or thereabouts.

Jim

dr_dave
02-25-2008, 03:35 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Jal</div><div class="ubbcode-body">Dr. Dave has done a great deal of work in figuring the effective target (pocket) sizes at various approach angles, shot speeds and actual pocket sizes. You can look up his articles for more info on this</div></div>FYI to BigRigTom, I have lots of illustrations and plots on this in my Nov'04-Jan'05 articles here (http://billiards.colostate.edu/bd_articles/index.html).

Regards,
Dave

BigRigTom
02-25-2008, 04:13 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Bob_Jewett</div><div class="ubbcode-body">it is a patch about 1/4-inch in diameter </div></div>

Wow Bob!
That is a lot bigger than I imagined!
It is what I was looking for though.

Using deductive reasoning:
If I invision the contact point being about the same as that little red dot on the Aramith red dot cue ball, that would be close...right.

This is strickly for my mind's eye so that I have and idea how big of target I am aiming for when I aim at the contact point on the object ball.

Thanks

BigRigTom
02-25-2008, 04:14 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: dr_dave</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Jal</div><div class="ubbcode-body">Dr. Dave has done a great deal of work in figuring the effective target (pocket) sizes at various approach angles, shot speeds and actual pocket sizes. You can look up his articles for more info on this</div></div>FYI to BigRigTom, I have lots of illustrations and plots on this in my Nov'04-Jan'05 articles here (http://billiards.colostate.edu/bd_articles/index.html).

Regards,
Dave </div></div>

Thank you Dr. Dave.
You can be sure I will read through all that stuff.
As always you have a wealth of information there.

Bob_Jewett
02-25-2008, 05:14 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body"> ... Wow Bob!
That is a lot bigger than I imagined!
It is what I was looking for though.

Using deductive reasoning:
If I envision the contact point being about the same as that little red dot on the Aramith red dot cue ball, that would be close...right.

This is strictly for my mind's eye so that I have and idea how big of target I am aiming for when I aim at the contact point on the object ball.

Thanks</div></div>
But as I tried to point out, on some shots you need to be much more accurate than that or you will miss. A 1/4-inch contact "span" corresponds to about +- 6 degrees of error. Many shots -- as shown in Dr. Dave's articles -- require +- 1 degree of accuracy or even better.

BigRigTom
02-25-2008, 05:59 PM
Thxs again Bob.

I am pretty sure I understand.
I wish I was accurate enough for that 6 degrees to really be a problem but if I get that close to my target (so I hit part of the 1/4 inch that you are referring too), I am usually smiling afterward....worst case scenario, I miss but...I was still very close.

Bob_Jewett
02-25-2008, 06:28 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body"> ... I wish I was accurate enough for that 6 degrees to really be a problem but if I get that close to my target (so I hit part of the 1/4 inch that you are referring too), I am usually smiling afterward....worst case scenario, I miss but...I was still very close.
</div></div>
But I think you already shoot more than accurately enough to get much closer than +- 6 degrees. Suppose you shoot a ball off the spot into a far corner pocket. To make that shot, you have to be within +- 1 degree even for a fairly generous pocket. If in fact you can make such a shot, it means that you are landing on the object ball within 1/6th of the width of the contact patch between the balls.

BigRigTom
02-26-2008, 09:27 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Bob_Jewett</div><div class="ubbcode-body">
But I think you already shoot more than accurately enough to get much closer than +- 6 degrees. Suppose you shoot a ball off the spot into a far corner pocket. To make that shot, you have to be within +- 1 degree even for a fairly generous pocket. If in fact you can make such a shot, it means that you are landing on the object ball within 1/6th of the width of the contact patch between the balls.</div></div>

Wow Again!
I do make that shot pretty consistently.
Now I have more pondering to do. This is why I love this game too. The revelations never end.

Scott Lee
02-26-2008, 01:46 PM
The size of the contact point between two pool balls is a pinpoint. The size of the contact point between the cue tip and the CB is 3mm or about 1/8".

Scott Lee

Bob_Jewett
02-26-2008, 02:24 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Scott Lee</div><div class="ubbcode-body">The size of the contact point between two pool balls is a pinpoint. ... </div></div>
Well, no, it's about 5mm in diameter for hard hits. You may want to check this out in Marlow's book. It will help to know that the ball-ball contact time is about 200 microseconds. After that, it's a fairly simple application of Hertz's Law. This was also discussed extensively in RSB, where Mike Page provided the simplified formulas that related contact sizes with speed of the shot.

If the balls were in fact hard enough to keep the contact patch to the size of a pinhead, the collisions between the balls would be very different.

dr_dave
02-26-2008, 02:25 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Scott Lee</div><div class="ubbcode-body">The size of the contact point between two pool balls is a pinpoint. The size of the contact point between the cue tip and the CB is 3mm or about 1/8".</div></div>Those are good "example" values for a slow shot, but the sizes are much larger at higher speeds. The tip contact-patch size also varies quite a bit with tip hardness and shape.

Regards,
Dave

av84fun
02-28-2008, 02:28 AM
Tom...I think you're right about the minutiae. You exceeded the appropriate amount of time wondering about the width of the contact patch by writing out your post!! (-:

There is no living human who can
A. Determine the correct patch
B. Freeze it in his mind's eye while moving around setting up for the shot and then
C. Somehow aim for some fraction of a few millimeters on an imaginary contact patch.

In fact, for every hour spent on the topic of aiming IN GENERAL, 20 hours ought to be spent learning how to play routes and control the CB so that shots are easy enough that aiming is trivialized.

Regards,
Jim

cushioncrawler
02-28-2008, 04:45 AM
Tom -- I did a couple of tests for contact. In one i colored the OB in lead pencil, and hit it horizontally off a 2nd storey balcony uzing a Qball (and a cue), at varyus speedz. The size of the mark(s) on the Qball gave me the size (dia) of the flatspot(s). The distance(s) out in the middle of the lawn where the OB fell (landed) gave me the horizontal speed(s), (knowing the vertical distance and uzing a skoolkid equation). In the other tests the Qball woz hanging on a cotton pendulum, dropped from high, swinging into the waiting OB, the ht(s) of the Qball gave me the speed(s) (uzing a skoolkid equation) and ditto the pencil mark(s) told me the dia(s) of the flatspot(s). I can send u this info if u like. madMac.

av84fun
03-07-2008, 06:20 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Tom -- I did a couple of tests for contact. In one i colored the OB in lead pencil, and hit it horizontally off a 2nd storey balcony uzing a Qball (and a cue), at varyus speedz. The size of the mark(s) on the Qball gave me the size (dia) of the flatspot(s). The distance(s) out in the middle of the lawn where the OB fell (landed) gave me the horizontal speed(s), (knowing the vertical distance and uzing a skoolkid equation). In the other tests the Qball woz hanging on a cotton pendulum, dropped from high, swinging into the waiting OB, the ht(s) of the Qball gave me the speed(s) (uzing a skoolkid equation) and ditto the pencil mark(s) told me the dia(s) of the flatspot(s). I can send u this info if u like. madMac. </div></div>

That's one of the funniest posts I've read in a long time.

THANKS!
(-:
Jim

cushioncrawler
03-08-2008, 12:04 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: av84fun</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Tom -- I did a couple of tests for contact. In one i colored the OB in lead pencil, and hit it horizontally off a 2nd storey balcony uzing a Qball (and a cue), at varyus speedz. The size of the mark(s) on the Qball gave me the size (dia) of the flatspot(s). The distance(s) out in the middle of the lawn where the OB fell (landed) gave me the horizontal speed(s), (knowing the vertical distance and uzing a skoolkid equation). In the other tests the Qball woz hanging on a cotton pendulum, dropped from high, swinging into the waiting OB, the ht(s) of the Qball gave me the speed(s) (uzing a skoolkid equation) and ditto the pencil mark(s) told me the dia(s) of the flatspot(s). I can send u this info if u like...</div></div>That's one of the funniest posts I've read in a long time. THANKS!..</div></div>Jim -- The krapamiths leev a nice big impakt mark, very vizible if u hit the dark colored part of the objekt ball (az we all know) and i think that a say halfball cut shot helps, it tends to leev more rezidue (i think). In fakt its the big size of the impakt mark, and the large amount of rezidue that makes a krapamith the krap that it iz. The rezidue acts like an oil (or like ball bearings), and koz its very varyable the balltoball friktion iz very varyable. Twoznt so (i think) with the oldendayz ivorylenes and bonzolines (certainly) and clay ballz etc. I remember lightly wiping off most of the rezidue (off one such mark), and then setting up a 2-ball frozen setshot (plant iz not the correct terminology) with the ballz touching at 2 such marks, ie a wiped mark on each ball, and i got the same throw az u would get if u uzed a chalkmark tween the ballz. Shows the sort of high-jinks u can expekt with the soft light hi-gloss modern krapamith. madMac.

Sid_Vicious
03-08-2008, 12:49 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Bob_Jewett</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body">When two pool balls meet we have two spheres that are 2.25" in diameter coming together at a very small point.

Do any of you physics types know how to mathematically determine the size of that contact point?.. </div></div>
For speeds of collision about like the break shot, it is a patch about 1/4-inch in diameter. If you know the contact time and the speed of the cue ball, it's possible to calculate the diameter of the patch. That information is of more or less no use to the typical player.

As far as the size of the contact patch being related to the accuracy of the shot, that may seem intuitively obvious, but it is wrong. If you think there is a direct connection, think on it some more. The location of the center of the contact patch, a single point, along with throw and some other stuff, determines which way the ball goes.

At tennis, the player contacts about a third of the ball with his racket. That's 120 degrees. That has nothing to do with with the accuracy of ball placement at tennis.

Similarly, some people claim that you only have to worry about the location of the tip on the cue ball for accuracy of spin down to the diameter of the tip's contact patch on the cue ball. That's also wrong. For some shots, you have to be a lot more accurate than that in your tip placement.</div></div>

THAT, My Friend is as-it-is, yet ellusive. I understand but still work for the execution...sid

av84fun
03-08-2008, 01:48 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Sid_Vicious</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Bob_Jewett</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body">When two pool balls meet we have two spheres that are 2.25" in diameter coming together at a very small point.

Do any of you physics types know how to mathematically determine the size of that contact point?.. </div></div>
For speeds of collision about like the break shot, it is a patch about 1/4-inch in diameter. If you know the contact time and the speed of the cue ball, it's possible to calculate the diameter of the patch. That information is of more or less no use to the typical player.

As far as the size of the contact patch being related to the accuracy of the shot, that may seem intuitively obvious, but it is wrong. If you think there is a direct connection, think on it some more. The location of the center of the contact patch, a single point, along with throw and some other stuff, determines which way the ball goes.

At tennis, the player contacts about a third of the ball with his racket. That's 120 degrees. That has nothing to do with with the accuracy of ball placement at tennis.

Similarly, some people claim that you only have to worry about the location of the tip on the cue ball for accuracy of spin down to the diameter of the tip's contact patch on the cue ball. That's also wrong. For some shots, you have to be a lot more accurate than that in your tip placement.</div></div>

THAT, My Friend is as-it-is, yet ellusive. I understand but still work for the execution...sid </div></div>

Where were you guys when I was flunking trig??? I know where I was...in the student union robbing freshmen!!!
(-:

HALHOULE
06-15-2008, 01:55 AM
BALLS TOUCH EXACTLY AT THE EQUATOR OF ALL BALLS.
AND THAT TOUCH IS A MERE PIN PRICK. SO, NOW WE ARE TRYING
TO FIND THAT PIN PRICK IN ORDER TO POCKET BALLS. IS THAT
LOGICAL ?? THERE IS NO PIN PRICK. THERE ARE 3 HUNDRED
DEGREES AROUND THE OBJECT BALL OR CUE BALL. NOW WHAT.
THERE IS A SOLUTION THAT WILL COMPLETELY SOLVE THE PROBLEM. THERE IS A WAY TO POCKET ANY SHOT, AT ANY ANGLE
AND ALWAYS USE THE SAME AIM FOR ANY AND ALL SHOTS.

Rail Rat
06-15-2008, 10:18 AM
Hello Bob. I got the back yard engineers way of figuring that out.

Paint the QB with thinned out black paint. While its slightly wet, tap it against an OB, now measure the diameter of the black spot on the OB.



- brad

BigRigTom
06-15-2008, 11:48 AM
I copied this statement from a physics book.

<div class="ubbcode-block"><div class="ubbcode-header">Quote:</div><div class="ubbcode-body">Two Spheres are in contact if the distance between thier centers is less than or equal to the sum of their radii. If they are in contact there will be prcisely one contact point.</div></div>

There must be a geometrical formula for calculating the size of the contact point based on the size of the two spheres (both being 2.25 inches in diameter in the case of pool balls).

If we ignore the fact that pool balls compress on collision it should be fairly simple to calculate the contact point size since both spheres are the same size.


I found these NOT SO Simple Equations on Wikipedia.... (http://en.wikipedia.org/wiki/Great-circle_distance#Spherical_coordinates)

Jal
06-15-2008, 12:39 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body">
There must be a geometrical formula for calculating the size of the contact point based on the size of the two spheres (both being 2.25 inches in diameter in the case of pool balls).

If we ignore the fact that pool balls compress on collision it should be fairly simple to calculate the contact point size since both spheres are the same size.</div></div>

The answer is (drum roll please):

1/x

where x is larger than any number you can think of (infinite).

It's not really clear what you're looking for. If there were no compression or imperfections of the surface geometry, that would mean that the contact "point" is really a point, ie, its radius or area would be smaller than any finite value.

If you don't ignore compression, then this article on Hertz Law applies (warning, lots of math):

http://www.oxfordcroquet.com/tech/gugan/index.asp

What compression does is to make the time of contact longer than infinitesimal. If it were infinitesimal, then the force between the balls would have to be infinite, which is not practical in our world.

Jim

Bob_Jewett
06-16-2008, 03:52 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body"> ... If we ignore the fact that pool balls compress on collision it should be fairly simple to calculate the contact point size since both spheres are the same size. ...
</div></div>
Pool balls do compress on contact. If they did not compress, then they would contact at one point -- the contact "patch" would have a radius of zero.

The physics book you were quoting assumes -- incorrectly for pool balls -- incompressible spheres.

Much of using physics to study pool is in deciding what's important and what's not. The fact that balls compress on contact is almost never important in any normal shot. This means that for such shots you can assume the contact is instantaneous and occurs at a single point, and you can ignore the contact patch. You have to realize which consequences of that assumption you can ignore as well. If the contact is instantaneous, the resulting pressure is much, much larger than seen in a neutron star, and the contact of the balls will result in a thermonuclear holocaust. The whole world will be destroyed. In reality, the world is not destroyed by each pool shot, and you can ignore that bogus consequence of your somewhat incorrect simplification.

Rail Rat
06-16-2008, 04:02 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Bob_Jewett</div><div class="ubbcode-body"> Pool balls do compress on contact. If they did not compress, then they would contact at one point -- the contact "patch" would have a radius of zero.

The physics book you were quoting assumes -- incorrectly for pool balls -- incompressible spheres.

</div></div>

Are you absolutely positive Pool balls compress?

Since they are made of very hard, solid synthetic material it seems to me they would fracture before they would push in?

If they do it must be no more than a fraction of a millimeter.

Bob_Jewett
06-16-2008, 06:26 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Rail Rat</div><div class="ubbcode-body"> ... Are you absolutely positive Pool balls compress?
... </div></div>
I've done the experiment. The contact patch is as large as 1/4-inch. This is close to the contact patch size predicted by physics for the contact time. Wayland Marlow measured the contact time and described the experiment in his book.

All objects compress. The size of any contact -- even with diamonds -- is not zero.

Also, a minor point, but the material used to make pool balls is not particularly hard. It's harder than rubber, and harder than some aluminum alloys, but it is softer than copper, brass and cast iron, according to http://www.machinist-materials.com/hardness.htm All of those materials also compress.

Deeman3
06-17-2008, 08:18 AM
You can see this with a bit of old carbon paper laid over the front of the rack on a break. It will astonish you how much compression there really is.

BigRigTom
06-17-2008, 09:08 AM
Thanks Bob for the clarification on the compression thing.

I knew that the balls compress but only because I had read it several times in various places. I did not know the specifics.

The "contact patch" size was actually what I was wondering about and what prompted my original question, I just didn't know enough about the subject to know the correct term.

The "contact point" is what I try to assess when I aim, so.... no matter how small it is, it is still my initial point of aim on those shots where I visualize the ghost ball contacting the object ball.

Deeman3
06-17-2008, 10:05 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body">Thanks Bob for the clarification on the compression thing.

I knew that the balls compress but only because I had read it several times in various places. I did not know the specifics.

The "contact patch" size was actually what I was wondering about and what prompted my original question, I just didn't know enough about the subject to know the correct term.

The "contact point" is what I try to assess when I aim, so.... no matter how small it is, it is still my initial point of aim on those shots where I visualize the ghost ball contacting the object ball. </div></div>

Tom, I know you know, that the contact point is not the right point in the ghost ball aim unless the cue ball is rolling across the object ball upon collision. Otherwise, you will throw the object ball so the contact point then becomes elsewhere to make the shot, hit the target.

Jal
06-17-2008, 10:31 AM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: BigRigTom</div><div class="ubbcode-body">Thanks Bob for the clarification on the compression thing...</div></div>
Stanley Milgram offers his gratitude too.

Jim