snooker balls to cover a full size table!

[ferret]

How many snooker balls would it take to cover a full size snooker table?

[mazur]

~3103 balls

@Diameter = 2 1/16"

[ste bed]

i would say 3528 at a guess.

[jrc750]

I will say 2500

[mushka]

2380 balls.

[ste bed]

my guess was just off the top of my head but thinking abit i think mazur i pretty close

[The Statman]

It's a long time since I did maths at school but I'll have a stab!

Assume the central tolerance in terms of playing surface, 1778mm × 3569mm, and we'll ignore any effect of the pockets may have. Ball's width 52½mm.

That means we have 33 balls in the first row. Since that exact figure is more than 33½ we can be certain that the second row, which will be each ball slotted between the two on the first row, will also contain 33 and not just 32.

These two rows will be 45.47+52.5 = 97.97mm in overall width.

Total table length 3569mm ÷ 97.97mm = 36 full double-rows each containing 66 balls. 36×66 = 2,376 balls.

The remaining unfilled space at the end of the table has a width of 42.08mm and I believe we would need 45.47mm to get another row of balls to squeeze in there, so assuming the requirement is that all the balls must be in contact with the bed of the table, my answer would be:

2,376

[The Statman]

Having said that, I have recalculated it doing the first line of balls on the long edge rather than the short edge, and that brings up a higher answer:

3569mm ÷ 52.5mm = 67.98 so there is room for 67 balls on the first row and easily enough for 67 in the second row as well, the two rows being 97.97mm wide overall.

1778mm ÷ 97.97mm = 18 and a bit, so that's 18 double-rows containing 134 balls in each double-row. 18×134 = 2,412

However, in practice, that .98 of a ball's width at the end of every other row, taking into account the undercut of the cushion and the fact that only an extra 1.05mm would be needed, you can probably get a sixty-eighth ball into every other row. So let's add 18 to the total above and my final answer is:

2,430

So the answer depends on the exact meaning of the question:

What's the minimum number of balls required so that no further balls can be placed on the table = 2,376 (although I'm not fully convinced that that wording is watertight)

What's the greatest number of balls that can be placed on a standard size table such that every ball is firmly standing on the bed of the table = 2,430.

[jrc750]

Probably some trick question needing double thickness or more to "cover" all of the baize

[ferret]

Cheers Statman, basically then we need 66 full sets of balls to cover one table (22 balls a set) but we only have 22 sets which is 484 ball about 1/3rd of what we need.

Any ones want to send me a set of balls? well do 44 people want to send me a set each lol

[jrc750]

I think your maths is a bit dodgy Ferret
If statman is correct @ 2376 then you would need 108 sets of balls
(108 x 22 = 2376)

[ferret]

Forgive me I been down the pub all night that is where the questions came from to start with, my next question is how much would all the balls weigh and could a snooker table handle that weight?

[GER147]

What an exciting night at the pub that must have been!

[Terry Davidson]

It sure looks like a few people on here need to get a life!

How many pints were consumed before these questions came up? I'll bet quite a few!

Terry

[rambon]

Can I throw in a cheeky suggestion of 1 set...

Pulverise the balls into a fine powder and spread on the baize and I reckon you could cover the whole table....

I guess the true answer depends on whether you want touching balls or not, or just covered to the extent that you couldn't place another ball on the table