bonanova 76 Report post 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? Share this post Link to post Share on other sites

0 austinm 0 Report post 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. Share this post Link to post Share on other sites

0 bonanova 76 Report post 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. Share this post Link to post Share on other sites

0 austinm 0 Report post 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 Share this post Link to post Share on other sites

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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