Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
Closed as "ask wikipedia": see qwerty's link below to the sphere packing article. –  Scott Morrison Mar 8 '10 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 '10 at 2:48
    
Reopened. My apologies for acting hastily, I agree closing was a mistake. Sorry for the inconvenience. –  Scott Morrison Mar 9 '10 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 '10 at 2:10
add comment

1 Answer 1

Maybe this is useful: Sphere Packing

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.