Packing SpheresDate: 08/16/99 at 11:08:39 From: Elizabeth Subject: Spheres and spaces How many 1.68-inch-diameter spheres fit into a 96.3-foot space? We have looked at your definition of spheres and several books (_Spherical Models_, _Go Figure_, _Circles: A Mathematical View_). We are not sure which formula applies. We would be happy to have a formula to do the application. Thank you very much. Elizabeth Date: 08/16/99 at 12:25:20 From: Doctor Peterson Subject: Re: Spheres and spaces Hi, Elizabeth. The answer depends on what you mean by "fit into." Often this just means "how many times more volume does the larger sphere have?", as if we melted small spheres and poured them in to entirely fill the space. If this is the meaning, then you just take the ratio of diameters and cube it: 96.3 ---- = 57.32 57.32^3 = 188,343.66 1.68 But if you actually want to fill a sphere with smaller spheres, you have to take into account the space between the spheres, and this depends on how they pack together. Mathematically, it has been determined that the densest packing of uniform spheres takes up about 75% of the space [pi/(3sqrt(2)) = 74.048%, to be precise]; other ways of packing spheres may give as little as 50% filling, with random packing filling about 64%. Here's a reference if you care for details: Sphere Packing - Eric Weisstein's World of Mathematics http://mathworld.wolfram.com/SpherePacking.html If we assume random packing, we can just multiply the volume of the larger sphere by 64% and divide by the volume of the smaller sphere. More simply, we can multiply the ratio of volumes determined above by 0.64 to get: 188,344 * 0.64 = 120,540 You'll have to decide which formula applies to the situation. Let me know if you have more details that may affect the answer. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 08/16/99 at 17:08:58 From: Elizabeth Subject: Re: Spheres and spaces Dear Dr. Peterson, The patron actually said a 96.3 CUBIC foot space. Does that affect the answer? I found conversions for inches to feet and cubic inches to cubic feet. Thanks. Elizabeth Date: 08/17/99 at 09:13:16 From: Doctor Peterson Subject: Re: Spheres and spaces Hi, Elizabeth. Yes, I'm afraid much of what I said is wrong, since I misinterpreted the problem. I assumed that the 96.3 feet was the diameter of a larger sphere. Let's try it again: >How many 1.68 inch diameter spheres fit into a 96.3 CUBIC FOOT space? The volume of a sphere is V = (4/3)*pi*r^3 so with D = 1.68 inch and r = 0.84 inch, the radius is 1 foot r = 0.84 in * ------- = 0.07 ft 12 inch the volume is V = 4/3 * 3.14159 * 0.07^3 = 0.001437 cubic feet If we divide 96.3 by this, we get 96.3 / 0.001437 = 67,026 spheres This is the number of spheres that will fit if you "melt them down" or "squash them together" to fill the space completely. Taking into account the packing density if you are actually fitting spheres into this space, we can multiply this by 64% assuming random packing to get: 67026 * 0.64 = 42,896 spheres Since the 64% is only a guess, I would round this and say about 43,000 of them would fit. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 12/16/2004 at 10:08:28 From: Martin Subject: Sphere packing - shaken or stirred? This is more of a comment. When considering packing densities it must be essential to determine if the real world question relates to a "pour them in and count them" situation, in which case the 65% answer will be about right, assuming a large enough sample. Or is it a "pour them in and shake and top-up" situation? A little judicious shaking will cause what the packing industry calls "settlement". Depending on the extent of shaking the settlement will become larger and larger until the theoretical maximum of 74% will be achieved, but only if the container is infinitely large, ie. the walls do not interfere with the settlement. So the answer should really be based on 65% rising to 74% depending upon the amount of shake-down. Date: 12/16/2004 at 12:43:51 From: Doctor Peterson Subject: Re: Sphere packing - shaken or stirred ? Hi, Martin. You're certainly right, though I would add that there may well be situations in which you would get less than the 64%, particularly in small containers with no shaking. Since we were just looking for an approximation, I didn't worry about those details, but the extra 10% will make a big difference, and is very much worth considering. By the way, if I recall correctly, this question was one of several that arose from some contest that involved filling a Volkswagen Beetle with golf balls, though I didn't know it at the time. It may be hard to shake down a whole car! If you have any further suggestions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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