Timeline for How to compute the number of regular spheres needed to fill a rectangular space
Current License: CC BY-SA 2.5
9 events
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| Nov 21, 2014 at 12:36 | history | edited | user9072 |
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| Mar 10, 2010 at 2:10 | comment | added | S. Carnahan♦ | 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". | |
| Mar 9, 2010 at 6:58 | comment | added | Kim Morrison | Reopened. My apologies for acting hastily, I agree closing was a mistake. Sorry for the inconvenience. | |
| Mar 9, 2010 at 6:57 | history | reopened |
David E Speyer Kim Morrison |
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| Mar 9, 2010 at 2:48 | comment | added | David E Speyer | 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. | |
| Mar 8, 2010 at 22:26 | comment | added | Kim Morrison | Closed as "ask wikipedia": see qwerty's link below to the sphere packing article. | |
| Mar 8, 2010 at 22:25 | history | closed | Kim Morrison | off topic | |
| Mar 8, 2010 at 21:14 | answer | added | Quimey | timeline score: 3 | |
| Mar 8, 2010 at 21:01 | history | asked | Chris Ballance | CC BY-SA 2.5 |