This Forum and Equations

How big are the pockets!!!
You can squeeze in extra balls ya know!!!
Are they within spec?
I want a 3D model of the geometry before anyone can answer this!!!!!!

Also, 42.
Great number.
Thanks for all the fish.
 
The correct answer is one. Take 10 balls and drop them on a 9' table. When the balls stop rolling you will have one configuration.
 
Do you consider orientation of the balls as well as position? If you account for number of contact points from ball to rail/pocket points/ball contacts, the counting can increase based on predictable math trends.

However it might not be countably infinite.
 
Ye
vapoolplayer said:
Just for curiosity sake, wouldn’t it be more than 1258?

For example, pretend the pockets are corners and not pockets. You place the cue ball tight to a corner. Now move it a microscopic distance from the side rail, but still on the short rail. Then continue that microscopic movement all the way down the rail until you touch the other side.

Those are all different positions that ball could land in and be touching the rail.

Now you have to move it away from the short/end rail a microscopic distance. And then start working backwards.


Wouldn’t that equate to a ridiculously high number that’s not just as simple as 1258 positions?
Click to expand...
Yes I agree it would be a lot more. In practical terms though it would be a lot less of a lot more than a lot because so many patterns would be so similar and play pretty much the same. It would certainly be a lot of a lot more than a lot. Anyway, that's my lot.
 
SBC said:
Actually you are on to something.

My surface area calculation doesn't take into account the space between balls. So it's less.

22 balls wide, 44 balls long = 968

So the answer is
968x967x966x965x964x963x962x961x960x959=
6.8944225e+29
Or
689,442,250,000,000,000,000,000,000,000
Click to expand...

I agree. I could be wrong, but that’s what I come up with.

And that would be the minimum number if we wanted to get stupid.

Kinda nuts.
 
SBC said:
Actually you are on to something.

My surface area calculation doesn't take into account the space between balls. So it's less.

22 balls wide, 44 balls long = 968

So the answer is
968x967x966x965x964x963x962x961x960x959=
6.8944225e+29
Or
689,442,250,000,000,000,000,000,000,000
Click to expand...
There’s an extra inch of space the long way (1/2” the short way), so alternate rows could be shifted to fill in some of the gaps (like how balls are racked). That would allow more rows to fit in each direction - I’ll leave that calculation to those with more patience.

pj
chgo
 
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Many years ago a bunch of us at the pool hall spent the better part of an afternoon trying to calculate the maximum number of rails a banked ball could travel without falling into any pocket. After thrashing through what seemed like infinite variables we finally came up with a calculation that we all more or less agreed with. That's when another buddy who had just come it asked, "Did you account for the deflection when/if the ball hits a tit?" 😖
 
What boggles my mind is that with all the endless possibilities, you see the same shots and patterns come up time and time again.

Doing my best to recognize those end game patterns. What a great way to spend time after the mugwumps are down for the night.
 
SBC said:
Actually you are on to something.

My surface area calculation doesn't take into account the space between balls. So it's less.

22 balls wide, 44 balls long = 968

So the answer is
968x967x966x965x964x963x962x961x960x959=
6.8944225e+29
Or
689,442,250,000,000,000,000,000,000,000
Click to expand...

This finds how many balls positions are possible if you pick 10 of them but using 'permutation' takes the order they are chosen into account, which I think is not proper. Also, I think you then need to multiply this result by 10! because you are performing an additional permutation when you actually assign balls to those spots.

Anyways I also think 'choose' is the proper choice for selecting the positions and not 'permutation' which results in:

(968 choose 10)*10!

Which I would perform but my calculator needs charging....

The true answer (upper bound) is probably limited by the number of weaves/square inch of the worsted cloth since the balls tend to settle in between them and also what pat said about how you can gain space by stacking balls in their gaps.. But this answer is a good lower bound (meaning the true answer is in excess of it).
 
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SBC said:
Easy
Surface area of pool table is 50x100= 5,000 square inches

A 2.25 inch diameter ball takes up 3.976 inches of surface area at it's circumference...

Take 5,000 inches divided by the surface area of a ball....that's 1,258. That's how many balls that could fit on a 9 foot table.

To figure all the possible combinations you need to do a mathematical calculation credit a Permutation Without Repetition.

The formula would be 1,258 times the next 9 descending whole numbers for each less spot on the table taken by the next ball....so

1,258 x 1,257 x 1,256, x 1,255 x 1,254 x 1,253 x 1,252 x 1,251 x 1,250 x 1,249 equals

9,577,054,998,202,898,152,157,349,600,000
That's 9.577 quintillion
Or 9.577 million trillions

It's a shit ton, but there's the math.
Click to expand...
Looks like my bank balance back in 2005 j/k

It’s a huge number. This looks like a pretty good proof. All jokes aside.

Nice work!

Best
Fatboy😃
 
Bob Jewett said:
You need to watch Corey Deuel more.

As for the OP.... Those who think moving a ball a tenth of an inch doesn't make any difference to its position should try spotting a second ball at one pocket. Leave it a tenth of a millimeter from the first ball and your opponent will be down upon you like a rabid weasel.

I'll go for a grid that is in 100ths of an inch. That's a little too coarse to make sure two balls are frozen, but let's see what it gets us.

The first ball can be in any of 4800 locations across the table and 9800 along the length. Call it 5000 and 10000 for convenience. That first ball can go in any one of 50,000,000 locations. There are about as many locations -- within a little -- for each additional ball That gives us
50,000,000 to the 16th power for all the possible locations as a very conservative estimate. That is about
1526 00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
unless I slipped a decimal point. It would take a long time to practice each one.

I've seen someone quote Albert Einstein saying there are 6 million shots on the pool table. I don't think he said that.
Click to expand...
Only brought up to make a point ( like Earl says, this game is not easy... in my many years of play I wonder if I ever shot the same shot twice...
 
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