how many 2.25 pool balls fit on a 9 ft 50 x 100 billiard table?
- Thread starter OTLB
- Start date
OTLB
Banned
brain
Brain must have farted , only 50 rows will fit , NOT 51 and I think the math is incorrect,I am stickin with my figurin
I going with this guy...Just look at his name
Brain must have farted , only 50 rows will fit , NOT 51 and I think the math is incorrect,I am stickin with my figurin
OTLB
Banned
You mean Brains Brain? who's on first
Trig 101
The nice thing about math is that if I'm wrong it can be easily proven. Here are the specifics, the calculation of the number of rows is predicated on several mathematical equations:
In the picture above the distance between rows is the height of the triangle (the calculation at the bottom is for packing efficiency, or volume fraction). The triangle has sides that are 2.25" and angles of 60 degrees. Therefore, the distance between rows = Sin(60) x 2.25 = 1.9485571585149869552183771341941
Next for n-rows, there are [n-1] 1.95" spaces between centers. So, the total distance will be (n-1) x 1.95 plus half a ball on either side (i.e. a full ball) - the 1.95 distance is center to center, so you have to account for the distance from the center to the rail on either side. Without doing the reverse algebra, if there are 51 rows this works out to (51 - 1) x 1.95 + 2.25 = 99.75
This is less than 100, so 51 rows will fit.
There are several reasons why the picture illustrated on the previous page might have given the wrong answer: perhaps the size of the ball is slightly out of proportion to the table; perhaps the balls are not perfectly touching or are touching on the outside of the line and not the inside, etc...
Then again, maybe I am wrong - but now it should be easy to show me where I went wrong (just be sure to use math).
The nice thing about math is that if I'm wrong it can be easily proven. Here are the specifics, the calculation of the number of rows is predicated on several mathematical equations:
In the picture above the distance between rows is the height of the triangle (the calculation at the bottom is for packing efficiency, or volume fraction). The triangle has sides that are 2.25" and angles of 60 degrees. Therefore, the distance between rows = Sin(60) x 2.25 = 1.9485571585149869552183771341941
Next for n-rows, there are [n-1] 1.95" spaces between centers. So, the total distance will be (n-1) x 1.95 plus half a ball on either side (i.e. a full ball) - the 1.95 distance is center to center, so you have to account for the distance from the center to the rail on either side. Without doing the reverse algebra, if there are 51 rows this works out to (51 - 1) x 1.95 + 2.25 = 99.75
This is less than 100, so 51 rows will fit.
There are several reasons why the picture illustrated on the previous page might have given the wrong answer: perhaps the size of the ball is slightly out of proportion to the table; perhaps the balls are not perfectly touching or are touching on the outside of the line and not the inside, etc...
Then again, maybe I am wrong - but now it should be easy to show me where I went wrong (just be sure to use math).
SidePocketMike
Registered
A very good attempt... but 1097 is incorrect! Since this is my first post, i'll forgive any flames, but I'll put up $100 that I can show a proof that you can get more than this... any takers?
Better pull out the magic eraser again....
area of a 2.25" ball is 3.974 sq. inches, not 4.5 or 5.06.
Keep in mind, a ball is round.
Geoff
count
That is a real cute pic with the triangle and hex as well as circles, not to mention all the mumbo jumbo pies and squares to the 2nd and 3rd powers and what have you. BUT all I wanted to do was put 2.25" balls in a 50 x 100" box which what I did with CAD .
Break it all down bruther and put the circles in the box and let's count em up . hell, I could be wrong, I was jus playin the game. I didn't see any mention of the Nobel prize in the beginning but I might have missed that as well..
I think about 50 per cent of this crap is mental and the rest is in your head.
BTW, what did OTLB count? He could be wrong as well. Surely not...
That is a real cute pic with the triangle and hex as well as circles, not to mention all the mumbo jumbo pies and squares to the 2nd and 3rd powers and what have you. BUT all I wanted to do was put 2.25" balls in a 50 x 100" box which what I did with CAD .
Break it all down bruther and put the circles in the box and let's count em up . hell, I could be wrong, I was jus playin the game. I didn't see any mention of the Nobel prize in the beginning but I might have missed that as well..
I think about 50 per cent of this crap is mental and the rest is in your head.
BTW, what did OTLB count? He could be wrong as well. Surely not...
OTLB
Banned
Well I figured it out using 3 balls and come up with 1097. 22 rows at 26 and 21 rows at 25 all nested. My father told me when I was young that all I ever had to do was to learn to count by 1's.
Cutting balls in half doesn't count either.
$1000, $2000 etc
Would like to see another grand total larger
There is only one outstanding wager on the mech forum
Cutting balls in half doesn't count either.
$1000, $2000 etc
Would like to see another grand total larger
There is only one outstanding wager on the mech forum
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I'm willing to bet, but because its your first post we'd have to post the money. We can both Paypal to OTLB (if he's willing) before you post your proof. Who's going to judge?
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OTLB
Banned
Yes, I am willing to help. When ready post money at paypal to ont@snet.net
I knew this would be a fun thread,
I will judge as well as everyone else, no cheating
SidePocketMike
Registered
OK... Just to be clear... I've only been a member/lurker for 2 weeks. You won't be offended if I ask about to see if you (OTLB) are a trust-worthy "holder"... Will you?
The wager is for $100 USD and I just have to post a verifiable proof showing more than 1097 balls, correct! No tricks... Similar to brain's proof... All balls on the table, no cushion compression, no stacking, all balls on the same plane... You and the forum to judge!?
OTLB
Banned
You made the bet not me. asking me if I am trustworthy is saying I am not
Get someone else
Brains numbers were verfied by me as I did it with actual balls before I started the thread.
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SidePocketMike
Registered
Take it easy... I meant no offence... Brain offered your name up to hold... And I'm just going to ship my $100 no questions asked... Fine... I'll just post my proof tonight when I get home and brain can ship me the $100 directly. I posted the wager and he's the one that accepted the wager... If any can find a hole in my proof, I'll gladly transfer the money... Gentleman's wager!
SidePocketMike
Registered
The 1098 Ball Proof
Place 22 balls evenly spaced along the 50" rail so that the end balls are both in the corner of the table (touching both rails).
The distance between balls centers should be d = 47.75/21 =~2.2738095238095238095238095238095 inches
Place the 21 balls in the next row so each ball is in contact with 2 of the balls in row 1.
Let the distance between the parallel rows of balls be called a...
and a = sqrt( 2.25^2 - .5 * d) =~1.9416352804680586159754603442379 inches
Place 49 rows of balls in this manner, alternating 22, 21, 22, 21, etc.
49 rows of balls = 25x22 + 24x21 = 1054 balls
The rectangular space taken up by these balls is the full 50" width of the table by length L
L = 48a + 2.25 =~95.44849346246681356682209652342 inches
What is significant here is that there is now ~4.55" x 50" of felt left. More than enough to arrange 2 additional rows containing 22 balls each or 44 more balls for a grand total of 1098.
The spreading of the rows to use all 50" of the table allows the distance a here to be a tiny bit smaller than in Brains example where the balls are tight.
His value for a = 2.25 x sin 60 = 1.9485571585149869552183771341941
The difference per row is less than 7 thousands of an inch, but it gets you one more ball on the table.
I patiently await my $100.
Added pic: Gaps between balls exaggerated for clarity
Place 22 balls evenly spaced along the 50" rail so that the end balls are both in the corner of the table (touching both rails).
The distance between balls centers should be d = 47.75/21 =~2.2738095238095238095238095238095 inches
Place the 21 balls in the next row so each ball is in contact with 2 of the balls in row 1.
Let the distance between the parallel rows of balls be called a...
and a = sqrt( 2.25^2 - .5 * d) =~1.9416352804680586159754603442379 inches
Place 49 rows of balls in this manner, alternating 22, 21, 22, 21, etc.
49 rows of balls = 25x22 + 24x21 = 1054 balls
The rectangular space taken up by these balls is the full 50" width of the table by length L
L = 48a + 2.25 =~95.44849346246681356682209652342 inches
What is significant here is that there is now ~4.55" x 50" of felt left. More than enough to arrange 2 additional rows containing 22 balls each or 44 more balls for a grand total of 1098.
The spreading of the rows to use all 50" of the table allows the distance a here to be a tiny bit smaller than in Brains example where the balls are tight.
His value for a = 2.25 x sin 60 = 1.9485571585149869552183771341941
The difference per row is less than 7 thousands of an inch, but it gets you one more ball on the table.
I patiently await my $100.
Added pic: Gaps between balls exaggerated for clarity
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I fully agree with the 1098 number and your reasoning.
As the money was not posted, there was no bet (that was my condition). The issue is not that you're right, it's whether you would have paid if your were not - otherwise it's not really a bet.
That being said since OTLB posted the challenge, if he says that there was a bet I'll Paypal you the money right away (you can PM me your Paypal address).
EDIT: on second thought, PM me your Paypal address, I'll pay you anyway. Welcome to AZB forums.
As the money was not posted, there was no bet (that was my condition). The issue is not that you're right, it's whether you would have paid if your were not - otherwise it's not really a bet.
That being said since OTLB posted the challenge, if he says that there was a bet I'll Paypal you the money right away (you can PM me your Paypal address).
EDIT: on second thought, PM me your Paypal address, I'll pay you anyway. Welcome to AZB forums.
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