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Old 2005-09-08, 20:30   #23
THILLIAR
 
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I said nothing about there being more than 105 balls on the first layer, but the 105 balls on the first layer do not fill in the horizontal space, there is an excess.

As already stated previously: any horizontal movement of "stacked" spheres will translate into vertical movement in all the layers above. I think you did not play with marbles enough as child, that or you are missing a few, in either case; you mention Billiards balls: when racking up in pool you have a little excess in the triangle; imagine if you will that you put the cue ball on top in the center of the other balls without shoving all the balls to the front of the triangle, then as you push the balls forward and remove any gaps between the balls the cue ball will move upwards vertically.

This we also face in the current problem, a horizontal excess creates some vertical movement thus adding a measure more than 10 cm between the centers of the three balls that form the lower points of our now nonregular tetrahedron and our "four sided mathematical shape" is now not as tall and the second layer is now closer to the bottom of the cube.
How would one calculate this phenomenon?

Last fiddled with by THILLIAR on 2005-09-08 at 20:33
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Old 2005-09-08, 22:50   #24
wblipp
 
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Quote:
Originally Posted by THILLIAR
How would one calculate this phenomenon?
The first thing you need to do is get very clear about exactly what alternative packing you are going to consider. I've already described an alternative where we pack all the balls tightly against the left wall, but let the last few rows roll downhill against the right wall. In my opinion, this is the approach most likely to fit additional balls, and that is not very likely at all. It it succeeds, it will fit another row of 9 or 10 balls.

You seem to have in mind spreading the all the rows uniformly apart, so that each layer is still flat with alternating rows of 10 and 9 balls, but these parallel rows no longer touch the other rows.

You will now have the bottom filled with rows that are slightly apart, and the equilateral triangles will become isosceles triangles, with the base along the row of touching balls. You know where every center is, so you need to find the point in 3 dimensions that is 10cm from each of the three points. You are helped slightly by the knowledge that it lies on the plane bisecting the line segment on the base, so you can tell the "y" value immediately.

A computationally effective approach may be to calculate the inter-plane height that would be required to get a 13th plane, and then solving for the row spacing that would create that height. Then see how many rows can be fit in each plane using that spacing.

Good luck. Let us know what you find out.
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