infinite Posted August 31, 2012 Share Posted August 31, 2012 The diameter of a tennis ball is 2.63 in. The size of a particular room is 43 ft X 10.5 ft X 30 ft. How many tennis balls can fit in this room? Link to comment Share on other sites More sharing options...
Greg H. Posted August 31, 2012 Share Posted August 31, 2012 (edited) It depends on how they end up stacked. If you assume the least optimal stacking method (treating each ball like a cube with a side equal to the diameter of the ball), the minimum number is 1,286,635 (give or take). Edit to add You'd also have to assume that the tennis balls on the lowest tiers of the stack do not deflect from the weight of the ones on top of them. Edited August 31, 2012 by Greg H. Link to comment Share on other sites More sharing options...
imatfaal Posted August 31, 2012 Share Posted August 31, 2012 1. This question is a great example of why we should all use metric measurement. 2. Room volume 2.3 x10^6 cubic inches Ball volume 9.5 cubic inches Maxium possible packing density pi/sqrt(18) Max Number of possible balls 1.8x10^6 There is a fairly chunky error and is an over approximation as there will be gaps at walls and corners. This is based on the greengrocer stack (each layer is hexagonal array like honeycomb and each ball of next layer fits in dip between three) which Gauss proved in the 19th century was the most efficient regular packing of spheres. We have recently seen almost proofs in which computers have (almost) exhaustively searched irregular packing and the greengrocer stack is still the most efficient Link to comment Share on other sites More sharing options...
Phi for All Posted August 31, 2012 Share Posted August 31, 2012 Sounds like a homework question. Link to comment Share on other sites More sharing options...
imatfaal Posted August 31, 2012 Share Posted August 31, 2012 Sounds like a homework question. Zut! Should have noticed that - oh well Greg and I have answered it now. Link to comment Share on other sites More sharing options...
Greg H. Posted August 31, 2012 Share Posted August 31, 2012 1. This question is a great example of why we should all use metric measurement. 2. Room volume 2.3 x10^6 cubic inches Ball volume 9.5 cubic inches Maxium possible packing density pi/sqrt(18) Max Number of possible balls 1.8x10^6 There is a fairly chunky error and is an over approximation as there will be gaps at walls and corners. This is based on the greengrocer stack (each layer is hexagonal array like honeycomb and each ball of next layer fits in dip between three) which Gauss proved in the 19th century was the most efficient regular packing of spheres. We have recently seen almost proofs in which computers have (almost) exhaustively searched irregular packing and the greengrocer stack is still the most efficient So that's why they stack fruit that way. I never knew that. Link to comment Share on other sites More sharing options...
imatfaal Posted August 31, 2012 Share Posted August 31, 2012 You will note that Mathworld states that the Kepler Conjecture (that Hexagonal Close Packing is the most efficient of any form) has been proved, whilst Wikipedia is only almost certain. It's such a simple problem - and greengrocer's automatically do it - but it taxed the minds of Gauss and Kepler to name but two of those who have worked on it (but what a pair of minds!) http://www.maa.org/devlin/devlin_9_98.html http://en.wikipedia.org/wiki/Sphere_packing http://mathworld.wolfram.com/SpherePacking.html Link to comment Share on other sites More sharing options...
Fuzzwood Posted August 31, 2012 Share Posted August 31, 2012 So that's why they stack fruit that way. I never knew that. Body centered cubic. You even find that type of stacking between atoms. Link to comment Share on other sites More sharing options...
John Cuthber Posted August 31, 2012 Share Posted August 31, 2012 Actually it's perfectly possible to get ten times that number of tennis balls into the room. But not all at the same time Link to comment Share on other sites More sharing options...
Phi for All Posted August 31, 2012 Share Posted August 31, 2012 Actually it's perfectly possible to get ten times that number of tennis balls into the room. But not all at the same time I could do it. If they're still considered tennis balls after being shredded. Link to comment Share on other sites More sharing options...
Greg H. Posted August 31, 2012 Share Posted August 31, 2012 (edited) I could do it. If they're still considered tennis balls after being shredded. Do shredded tennis balls still have a diameter? Edited August 31, 2012 by Greg H. Link to comment Share on other sites More sharing options...
Phi for All Posted August 31, 2012 Share Posted August 31, 2012 Do shredded tennis balls still have a diameter? Depends on how finely shredded they are. At some measurable level I'm sure they would. Link to comment Share on other sites More sharing options...
dimreepr Posted August 31, 2012 Share Posted August 31, 2012 Depends on how finely shredded they are. At some measurable level I'm sure they would. Is a ball still a ball when deflated? Link to comment Share on other sites More sharing options...
Phi for All Posted August 31, 2012 Share Posted August 31, 2012 Is a ball still a ball when deflated? If you saw one under your car tire, you'd still say, "What the heck is... oh, it's a tennis ball." Link to comment Share on other sites More sharing options...
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