bonanova Posted March 18, 2013 Report Share Posted March 18, 2013 You are stacking children's building blocks on a level table. They are perfect, identical cubes, evenly weighted, and you're not fastening them together in any way. The stack is not very straight. In fact at one point you notice the topmost block (viewed from overhead) is completely outside the footprint of the bottom block. What is the minimum number of blocks in your tower? Or is that even possible? Quote Link to comment Share on other sites More sharing options...
0 austinm Posted March 18, 2013 Report Share Posted March 18, 2013 5 Top block overhangs that under it by 1/2 side-length (For now let's do it in one dimension, aligned along a footprint edge. We'll relax that restriction at the end.) Block 2 (counting from top) overhangs block 3 by 1/4 side-length. Now we're out 3/4 of a block. At this point most people assume it's a geometric progression, that the sum converges to one, and that it will take an infinite number of blocks. That assumption is wrong. Block 3 overhangs block 4 by 1/6 side-length. (CoM of top-three blocks is 5/6 side-length back from leading edge, not 7/8. Verification is left as exercise for the reader.) Now we're out 11/12 of a block. Block 4 overhangs block 5 by 1/8 side-length. Now the leading edge is out 25/24 of a block, and top block is completely outside the footprint of the base-block. Obviously, we could offset along the second direction simultaneously without any effect; it won't reduce the minimum, though. This question gets much more interesting if we go, say, 1.5 block-lengths outside of the original footprint--shape of tower differs dramatically. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 18, 2013 Author Report Share Posted March 18, 2013 5 Top block overhangs that under it by 1/2 side-length (For now let's do it in one dimension, aligned along a footprint edge. We'll relax that restriction at the end.) Block 2 (counting from top) overhangs block 3 by 1/4 side-length. Now we're out 3/4 of a block. At this point most people assume it's a geometric progression, that the sum converges to one, and that it will take an infinite number of blocks. That assumption is wrong. Block 3 overhangs block 4 by 1/6 side-length. (CoM of top-three blocks is 5/6 side-length back from leading edge, not 7/8. Verification is left as exercise for the reader.) Now we're out 11/12 of a block. Block 4 overhangs block 5 by 1/8 side-length. Now the leading edge is out 25/24 of a block, and top block is completely outside the footprint of the base-block. Obviously, we could offset along the second direction simultaneously without any effect; it won't reduce the minimum, though. This question gets much more interesting if we go, say, 1.5 block-lengths outside of the original footprint--shape of tower differs dramatically. The balance is precarious at the 1-2 edge and the 4-5 edge, but some of the 1/24 can be spent to make them stable. If the middle blocks are rotated, hmmm... might not matter. Nice job. Quote Link to comment Share on other sites More sharing options...
0 austinm Posted March 18, 2013 Report Share Posted March 18, 2013 I can make the separation-from-original-footprint greater than 1/24 (and spend an arbitrarily-small portion of that gain to ensure it's not precarious)--can you? (7-4*sqrt(2))/8 Quote Link to comment Share on other sites More sharing options...
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bonanova
You are stacking children's building blocks on a level table.
They are perfect, identical cubes, evenly weighted, and you're
not fastening them together in any way.
The stack is not very straight.
In fact at one point you notice the topmost block (viewed from
overhead) is completely outside the footprint of the bottom block.
What is the minimum number of blocks in your tower?
Or is that even possible?
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