20050907, 22:46  #12  
"William"
May 2003
New Haven
2^{2}·3^{2}·5·13 Posts 
Quote:
And Ken already provided links to MathWorld that get the same result. What part do you think is wrong? I'll say it again with more words. We start with three balls on the bottom layer that are nestled together in an equilateral triangle, and a fourth ball that is nestled into the hole. The upper ball is centered over the centroid of the equilateral triangle formed by the centers of the lower three balls. The centroid is 2/3 of the way from any vertex to the opposite side's midpoint. The length of that segment is 5*sqrt(3) = 8.66 cm The distance from any vertex to the centriod is 2/3 of this distance 10* sqrt(3) / 3 = 5.77 dm We can form a right triangle from the center of any ball on the base to the center of the upper ball by going from the lower ball's center to the centroid then straight up to the upper ball's center. The hypotenuese of this triangle is two radii, 10 cm. The leg from the lower ball to the centroid is 10 * sqrt(3) / 3 = 5.77 cm. The other leg is the interlayer distance that is, the vertical distance from centers on one layer to centers on the next layer. From Pythagorus, it must be 10 * sqrt(6) / 3 = 8.16497 cm That's a complete derivation, twice, using nothing more than basic geometry, and supported with a link to Mathworld's derivation of the same value. If it's wrong, I deserve more than "that doesn't fly." You can point out errors in my derivation, or you can provide an alternate derivation, but unsubstantiated disagreement is not an acceptable response at this point. 

20050907, 23:15  #13 
Mar 2004
ARIZONA, USA
23 Posts 
The upper ball is always centered over the centroid of the equilateral triangle formed by the centers of the lower three balls THAT ARE TOUCHING?
How would one compensate for such a problem, mathamatically? Last fiddled with by THILLIAR on 20050907 at 23:22 
20050908, 02:42  #14 
"William"
May 2003
New Haven
2^{2}×3^{2}×5×13 Posts 
I'm sorry. I should have guessed from your first assertation that geometry was beyond your ken. Here are the centers of the four balls in a coordinate system centered on the center of the corner ball. You can confirm that every pair of balls is touching by confirming that
(x1x2)^{2} + (y1y2)^{2} + (z1z2)^{2} is 100 for each of the six pairs. The first three centers are on the plane z=0, the fourth is above that plane by 8.164996. (0, 0, 0) (0, 10, 0) (8.660254, 5, 0) (2.886751, 5, 8.164996) If you still don't understand, someone else will need to respond. This as elementary and obvious as I know how to make it. 
20050908, 11:21  #15  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×599 Posts 
Quote:
Quote:
Perhaps what you need is a diagram. I did a brief search and found http://www.chem.ox.ac.uk/icl/heyes/s...1.html#anchor4 Look under "CLOSE PACKING OF SPHERES". And this: http://www.math.binghamton.edu/zasla....20040406.html Last fiddled with by cheesehead on 20050908 at 11:26 

20050908, 18:08  #16 
Mar 2004
ARIZONA, USA
23 Posts 
What about the additional space found within the 1meter cube? Each layer has some additional space that the balls will, like a liquid, "fill". The first layer has 3.4 cm extra. All of the calculations you have given assume that all of the balls are touching; in the real world they do not stack in a given space, they fill it.
Not to be idiomatic but sometimes less is more, but more often it is just more, reality is always more, even when viewed from a lesser point of view. Last fiddled with by THILLIAR on 20050908 at 18:15 
20050908, 18:35  #17  
Jun 2003
The Texas Hill Country
2×541 Posts 
Quote:
I assure you that it is impossible to fill all of the voids in a box with spheres of a uniform size. I see nothing in the description of the problem to support your assertion that the "balls" would deform to fill a compact volume which is equal to the sum of their original volumes. Please describe exactly what "facts" form the basis for your assertion. 

20050908, 19:22  #18 
Mar 2004
ARIZONA, USA
23 Posts 
Who said anything about the balls deforming, I did not. I am simply saying that the balls do not tightly fit into the cubic space in the horizontal dimensions; and any horizontal movement of "stacked" spheres translates into vertical movement in all the layers above, this we must account for when establishing an interlayer distance. Think as if the balls are water molecules and you may get my point.

20050908, 19:47  #19  
Jun 2003
The Texas Hill Country
2·541 Posts 
Sorry. I misunderstood what you were trying to say.
However, Quote:
Perhaps you are confusing this situation with either the one where the spheres and small in comparison to the total volume or molecular effects where the molecules are mostly "empty" at the subatomic level and two can occupy the same space at the same time. In the former case, things could shift only it there is a defect in the lattice. In the second case, Billards balls do not interact with each other in the same way that molecules interact. Perhaps, I still don't understand. I don't see the analogy to water molecules. Last fiddled with by Wacky on 20050908 at 19:49 

20050908, 20:15  #20 
Jun 2005
Near Beetlegeuse
184_{16} Posts 
Wacky, what I "think" he is trying to say is that because the balls do not fit tightly into the base of the box, when the second layer is put on top the balls on the bottom layer move apart a little bit. This would mean that the tetrahedrons (if that's the right shape) are not as tall as Wblipp's maths would indicate. This process is repeated with each subsequent layer.
I'm not saying he's right, I am only interpreting what he said. 
20050908, 20:25  #21  
"William"
May 2003
New Haven
2^{2}×3^{2}×5×13 Posts 
Quote:
This is what I've called the wiggle room. I've already offered my opinion about how many additional balls can be fit into the box because of this. The space you mention on the bottom layer is mostly filled by the balls on the second layer, but that last row could roll 5 mm further and drop slightly, opening a tiny gap between the last two rows on the second layer. The the last row on the third layer could drop into this gap. Which would make it possible for the last two rows of the fourth layer to drop slightly. This process could be continued until the top layer, letting gravity packing slide the ends down slightly. There are other possibilities that you haven't mentioned. The edge row of the top layer doesn't have to nestle down into the tetrahedral packing. It could be rolled up slightly to hit top of the box. The the next rows over could be rolled up slightly to touch the displaced row. This process could be continued across the top of the box. Continuing all the way across the top, this extra space could join the extra space from the gravity packing. It's possible that these moves in combination would create enough space to squeeze in one more row of balls. This possibility is why I said "at least 1254 balls. With only 5 mm spare on each layer and less than 2 mm spare at the top, I think it's unlikely that these manuevers will gain enough room to fit more 50mm balls. I judge the likelyhood to be low enough that I'm not going to do the calculations to see if they fit. But if you think more can fit, go for it. If you're right, then we will all applaud your superior solution. Of course, be sure to show all the math so we can confirm the answer. 

20050908, 20:26  #22 
Jun 2003
The Texas Hill Country
2·541 Posts 
If that is what he is saying, he is ignoring the quantum requirement. Even if there is a slight amount of slack which lets the layers above "settle", the amount of such settling is so small that it cannot accumulate enough to add an additional layer.

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