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Computing the volume of a sphere is straightforward 4/3*pi*R^3

As is the volume of a rectangular space length*width*height (e.g. 10*10*6)

How might I go about determining how many spheres would fit into a rectangular space, assuming the spheres are solid and not flexible?

flag	asked Mar 8 2010 at 21:01 Chris Ballance 11●2

	Closed as "ask wikipedia": see qwerty's link below to the sphere packing article. – Scott Morrison♦ Mar 8 2010 at 22:26
	I don't think this is off topic, so I'll sketch an answer: According to the Kepler conjecture, which has probably been proved by Hales, the densest possible packing of spheres in 3-space has density pi/\sqrt{18}. So, if your region has volume V, it can contain at most $V (\pi/\sqrt{18})/(4/3 \pi) = 2 \sqrt{2} V$ unit spheres. If all the sides of your box are much longer than unit length, this bound will be close to achievable. I have the impression that there are no good exact bounds for finite regions, but I am not an expert. – David Speyer Mar 9 2010 at 2:48
	Reopened. My apologies for acting hastily, I agree closing was a mistake. Sorry for the inconvenience. – Scott Morrison♦ Mar 9 2010 at 6:58
	When you say "regular", do you mean all spheres have diameter 1? Even in this case, it is a difficult computational problem. There are obvious upper and lower bounds given by truncating/embedding as much of a hexagonal close packing as possible. I've been told that any particular case (including the asymptotic infinite one) is a finite computation "by quantifier elimination". – S. Carnahan♦ Mar 10 2010 at 2:10

Quimey Vivas · Answer 1 · 2010-03-08 21:14:51Z UTC

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Maybe this is useful: Sphere Packing

answered Mar 8 2010 at 21:14

How to compute the number of regular spheres needed to fill a rectangular space