dquarasr
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How much Carbon Fiber could a woodchuck chuck if a woodchuck could chuck Carbon Fiber?How much wood could a wood chuck chuck if a wood chuck could chuck wood?
A lot.
How much Carbon Fiber could a woodchuck chuck if a woodchuck could chuck Carbon Fiber?How much wood could a wood chuck chuck if a wood chuck could chuck wood?
A lot.
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.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?
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
They don't like carbon it gets stuck in their teeth.How much Carbon Fiber could a woodchuck chuck if a woodchuck could chuck Carbon Fiber?
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.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
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
IDK if somebody brought this up yet but scattering balls over the table is a very good solution to the rack and break crisis. So at the very minimum, I care.Who really cares?
Looks like my bank balance back in 2005 j/kEasy
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.
is this in Bulgarrillions?A guhgillion
Nois this in Bulgarrillions?
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...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.
And God gave us a brain...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![]()
Well there is a formula for that also.....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...