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I want to make a lamp made of ping pong balls in a large glass vase and need to calculate the the required amount of ping pong balls

Bravewart
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  • The least you can do is to give the dimensions of the cylinder and the diameters of the pingpong balls... – Jean Marie Feb 23 '20 at 19:32
  • I don't know the dimensions of the cylinder yet. I was hoping for a general formula how to do this with spheres and a cylinder. Ping pong balls have a 40mm diameter – Bravewart Feb 23 '20 at 19:34
  • For an upper bound, the cylinder volume is $V_c=\pi r^2h$ and sphere volume is $V_s=4\pi r^3/3$, so $$n< \frac{\pi r^2h}{4\pi r^3/3}=\frac{3h}{4r}.$$ – pshmath0 Feb 23 '20 at 19:41
  • @Pixel thank you. I can work with that – Bravewart Feb 23 '20 at 20:03
  • @Pixel Isn't the $r$ in the numerator the radius of the cylinder and the $r$ in the denominator the radius of a ping-pong ball? I think you've used the same variable for two different things. – saulspatz Feb 23 '20 at 20:19
  • @saulspatz yes, my case example would have a much larger cylinder radius than the ping pong ball radius. Therefore $$n<\frac{pi r^2h}{4\pi r^3/3}$$ is still right though? – Bravewart Feb 23 '20 at 20:28
  • You should use a different letter in the numerator so that $$n<\frac{pi R^2h}{4\pi r^3/3}$$ where $R$ is the radius of the cylinder in mm and $r=20$. I'm about to type an answer. Give me a few minutes. – saulspatz Feb 23 '20 at 20:47
  • @saulspatz yeah that's right, my bad that's a typo which I carried through, the equation should be $$n<\frac{3 R^2h}{4 r^3},$$ since the $\pi$'s cancel. $R$ is the radius of the cylinder and $r$ the sphere. – pshmath0 Feb 23 '20 at 21:27
  • "Sphere packing" is an active research subject (try googling with these keywords). You cannot expect simple "formulas" for the maximum number of spheres you can pile up in a special volume like a cylinder. – Jean Marie Feb 23 '20 at 22:18

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There is an extensive literature on sphere packing, but I'm not familiar with it. The section on irregular packing in the Wiki article suggests that you won't get a density of more than $64\%$. That is, the total volume of the ping pong balls will not exceed $64\%$ of the volume of the cylinder. So if $R,h$ are the radius and height of the cylinder, and $r$ is the radius of a ping pong ball ($r=20$ as I understand it) then$$ .64\frac{3R^2h}{4r^3}=.48\frac{R^2h}{r^3}$$ ought to be enough ping pong balls.

saulspatz
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