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 14645 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 14645 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 14645 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 14645 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?