A Tower of Blocks and the Harmonic Series, Page 3
We now see a pattern!
In general, for the top K blocks, we can move the blocks over precisely 1/(2K) to balance on the pivot edge.
Here is the argument once more.
If we have already moved N-1 blocks, so the top block has moved ½ + ¼ + 1/6 + 1/8 + …+ 1/( 2 (N-1) ), this system of blocks has center of mass precisely at x = 0, since it is balanced on the pivot edge. The Nth block has mass 1 and center of mass at ½ to the left. So the center of mass of the system of N blocks is at
(total moment) / (total mass) = ( (N-1)*0 + 1*(½) ) / N = 1/( 2N ).
Thus, if we start at the top and carefully tap N blocks almost to the precise balance points, then the top block has moved over almost ½ + ¼ + 1/6 + 1/8 + …+ 1/( 2 N ). We can calculate this distance for various numbers of blocks.
| Number of blocks N | Distance ½ + ¼ + 1/6 + 1/8 + …+ 1/( 2 N ) |
| 3 | 0.9167 |
| 4 | 1.0417 |
| 5 | 1.1417 |
| 10 | 1.4645 |
| 20 | 1.7989 |
| 31 | 2.0136 |
| 100 | 2.5937 |
| 227 | 3.0022 |
| 1 000 | 3.7427 |
| 1 000 000 | 7.1964 |
We see that it takes at least 31 blocks to balance the top block two lengths over the bottom, and 227 blocks to extend three lengths. If you happen to have one million blocks, and if you could balance them perfectly, and no winds or vibrations ruined your tower, then you could extend the top block a little over seven lengths beyond the bottom block. (Here are pictures of 31 blocks and 227 blocks.)
A crazy question: can you build a tower where the top block is one hundred lengths over the bottom block?! This requires us to study the "harmonic series." One can calculate that it would take more blocks than there are subatomic particles in the universe.
Have fun building with blocks!
April 2008 Return to Robert Styer Home Page