Thread: Easy questions View Single Post
 2005-09-08, 20:30 #23 THILLIAR     Mar 2004 ARIZONA, USA 278 Posts 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