How many balls can you place on a snooker table?

[Wizard]

All I can say is you have way too much spare time !!!
Just fetching my abacus to work that little problem out

[Wizard]

Ok I have a total of 78 columns therefore half are 34 and half are 33
Half of 78 is 39
Therefore (39 * 34) + (39 *33) = 2,613

[SquaredCue]

Thank you very much, Wizard!

Although it might look like such calculations take a long time, I can assure you that working out problems like this is negligible to the work put even in a small feature of SnookerQ. Speaking about balls, for example, modelling and calculating the collision of two balls require much, much more effort and hours, not to mention the countless fine tuning touches we added to make the game playable and enjoyable. So, don't worry about my time. Instead, I'd like every visitor to enjoy the time spent on the game as much as possible!

Now we know that we can place at least 2613 balls on the bed of the table! That's a lot! But can we improve that? No more balls fit anywhere on the table with this layout, so it's not easy to guess if we can. However, we know that we have unutilized space at the top (blue area, 7.7 mm high) and at the right side of the table (red area, 31.977 mm wide):

Unutilized area at the top and at the right

sntmnobcols2.png (25.7 KiB) Viewed 25513 times

How can we use those areas if no balls fit there? Well, we can use the blue area to squeeze the ball columns even closer to each other! The honeycomb style packing is definitely the most dense method so how come? It's easier then one would think: Let's place ball 34 to the top and let's distribute this extra 7.7 millimetres equally over the 33 gaps between the balls of the first column. This way the balls of the second column can move a little bit to the left until they are touching the balls of the first column again! And of course, balls in the 3rd column can move to the left twice as much, 4th column balls can move to the left three times as much, and so on. We can have a considerable gain in the last column, making the width of the red area larger.

If x denotes the gap between balls in a column, the following simple equation can be written to describe the full height of the rectangle, which is the width of the snooker table:

(34*52.45) + (33*x) = 1791

Thus, x=0.2333 mm

The new, smaller distance between the columns is denoted by y and it can be calculated easily if we consider the following diagram:

Calculating y, the distance between ball columns

sntmnobcols3.png (11.11 KiB) Viewed 25513 times

The result of the calculation is the following: y = 45.356 mm.
The distance between columns was 45.423 mm in the honeycomb style placement so this is not too much but we have 78 columns, thus the new width of the red area increases to 3582 - 52.45 - (77*45.356) = 37.178 mm! We gained more than 5 millimetres width in the red area!

So, what do we do with it? Extra balls still don't fit there! The trick is to use the full width of the red area to place the balls of the last column as close to the top as possible! Instead of long boring calculations, let's solve this placement graphically. The following picture shows how dense the balls of the last column can be placed next to the balls of the previous column:

how-many-balls-on-a-snooker-table-method-red.png (8.47 KiB) Viewed 25513 times

So, how many balls do fit in the "last column"? Let's count it on the full picture:

how-many-balls-on-a-snooker-table-full-red.png (29.97 KiB) Viewed 25513 times

That's 42 balls in the last column instead of the 33 balls we had before, so we could place 9 extra balls on the table!

Thus, the final answer is 2613 + 9 = 2622 balls.

Is that really the final answer? No! As we will see, we can place a dozen more balls on the table! Can you guess how?

[Captain-Fantastic]

Thank @#%& for that!!!

[Wizard]

Okay so there's 2622 balls and 6 of them are colours, 1 is a cue ball and the rest are reds
Assuming the ridiculous (like you got no room to move but let's just say if you did) what is the maximum break?

@Captain-Fantastic just when you thought it was over

[Juggernaut]

Hopefully I haven't fallen for an obvious trap but according to your fantasy rules, the possible highest break will be 20,955 with a free ball. 20,947 without a free ball. Please tell me I haven't fallen into a cheeky trap

I did misread it and fell for a trap! The above was edited.

[Wizard]

Hopefully I haven't fallen for an obvious trap but according to your fantasy rules, the possible highest break will be 20,955 with a free ball. 20,947 without a free ball. Please tell me I haven't fallen into a cheeky trap

I did misread it and fell for a trap! The above was edited.

I could be evil and say it was a trap or I could say ask Ronnie what the maximum break is but well done on the maths mate.

So...tick...correct calculation but in fact the maximum break is 147 unless you get extreme luck off the break

[BIG-G]

Hopefully I haven't fallen for an obvious trap but according to your fantasy rules, the possible highest break will be 20,955 with a free ball. 20,947 without a free ball. Please tell me I haven't fallen into a cheeky trap

I did misread it and fell for a trap! The above was edited.

I could be evil and say it was a trap or I could say ask Ronnie what the maximum break is but well done on the maths mate.

So...tick...correct calculation but in fact the maximum break is 147 unless you get extreme luck off the break

Luck you say...........

How would you know about that sir??

[Captain-Fantastic]

Okay so there's 2622 balls and 6 of them are colours, 1 is a cue ball and the rest are reds
Assuming the ridiculous (like you got no room to move but let's just say if you did) what is the maximum break?

@Captain-Fantastic just when you thought it was over

Same set up, 2622 balls, 6 of them are the colours, 1 is the cue ball, what would be the lowest scoring total clearance possible?

[SquaredCue]

Same set up, 2622 balls, 6 of them are the colours, 1 is the cue ball, what would be the lowest scoring total clearance possible?

I think the lowest possible break score after clearing the table is 2635 points.

[Captain-Fantastic]

Same set up, 2622 balls, 6 of them are the colours, 1 is the cue ball, what would be the lowest scoring total clearance possible?

I think the lowest possible break score after clearing the table is 2635 points.

How did you arrive at 2635?

[SquaredCue]

All the reds in one shot, then yellow, green, brown, blue and pink, finally pot the black and go in-off for a true clearance.

[SquaredCue]

Oops, yellow would be respotted at first... My bad... Then I modify my answer to 2637.

[Wizard]

Ok so there are 2622 balls on the table
1 is white
6 are colours
Hence 2615 are reds
therefore...

Highest break = 2615*8(red with black for total 20920) + 27 for the colours = 20947
Lowest break = 2615 (all reds off the break like to see that shot without a colour going in too) + 2 for yellow (which is then re-spotted) + 27 for the colours = 2644

Please deliver the prize via email to wizardisasmartass@wiseoldguy.com

[Captain-Fantastic]

Correct.....2644.

Your prize of £500.....will be donated on your behalf to a charity of my choice.

[Wizard]

Correct.....2644.

Your prize of £500.....will be donated on your behalf to a charity of my choice.

Thank you for the confirmation sir. Your generosity is astounding!!

[SquaredCue]

Thus, the final answer is 2613 + 9 = 2622 balls.

Is that really the final answer? No! As we will see, we can place a dozen more balls on the table! Can you guess how?

As nobody has come with any ideas, I'll show you how I could improve that result. No conceptual change really, all we'll do is we'll put rows of balls initially instead of putting columns of balls and we'll apply the same tricks as before:

Rows of balls

sntmnobrows1.png (10.32 KiB) Viewed 24285 times

Wait a minute! That's 20*68 + 19*67 = 2633 balls, which is already more than the optimized layout with the columns of balls! Can we have even more?

To find it out, let's utilize the blue area as before, by creating a small gap between the balls of a row so that the rightmost ball touches the right hand side cushion. The width of the blue area is 15.4 mm and we have 68 balls in the row, so the size of the 67 gaps, denoted by x, is the following:

x = 15.4 / 67 = 0.22985 mm

Having these horizontal gaps, the rows can be squeezed a little down to make the red area bigger. The vertical difference of ball centres between rows, denoted by y as before, is the following:

y = square_root_of( 2R squared - (R + x/2) squared ) = 45.3565 mm

Thus, the red area becomes this high:

1791 - 52.45 - 38*45.3565 = 15.0035 mm

To utilize the red area, we'll do the same trick as before: instead of putting the balls of the topmost row in the grooves of the previous row, we will put them as close to the left side as possible, hoping that we can place more than 68 balls this way. Again, let's solve this graphically:

69 balls in the topmost row!

how-many-balls-on-a-snooker-table-long-red-indexed.png (42.83 KiB) Viewed 24285 times

That's 69 balls in the topmost row, one more than before! All these efforts were not in vain!

Let's count the balls: 19 rows with 68 balls, 19 rows with 67 balls and 69 balls in the top row:

19*68 + 19*67 + 69 = 2634 balls

We can place 2634 standard snooker balls on a standard snooker table so that all the balls are resting on the table bed.

There are 22 balls in a snooker ball set so we need 120 boxes of snooker balls to do this in real.
The industry standard weight of a snooker ball is 142 grams so the 2634 balls weigh a little more than 374 kg!

2634 balls

2634.png (8.85 KiB) Viewed 24285 times

Is 2634 really the final answer? (;