Well, I didn't use the numbers from the article, rather I just balanced the tower top-to-bottom, which after a quick look is what's happening in the article. With some tolerance of course, these bricks aren't very straight.
Science needs people like you.
Architecture solved a lot of these by mixing corbelling and jettying. Corbelling permits you to gain ~2/3 of your wall width in overhang when done optimally. So with a 1' thick wall, you can gain 8" of overhang, which admittedly in house construction isn't much. The interesting thing here, is when you put a jetty on it (which relies on the cantilever principle) you've extended the cantilever distance you can get without compromising the structural strength, why is this interesting? Well your original 8" overhang on a masonry wall might have gained you a 1'6" total extra length on your upper level. Now if your longest available lumber is ~24' long. You safely gain a jetty of 1/6th the lumber length, meaning 4'. In combination, you just gained ~9'6" of length and width to your upper level. So with a 40' x 40' main floor, 1600sqft, and with a 49'6" x 49'6" second floor, you get 2450sqft second floor. With your initial brick corbelling you would only get an extra 120sqft.
This is the main reason tudor style housing, with masonry main floors and timber second exploded in popularity in their era. Masonry was expensive, timber was cheap. However, timber was especially prone to rot and insect damage within the height range of rising damp. Tudor style allowed the best of timber construction (jettying) with the stability and longevity of masonry.
I don't understand the meaning of this statement. If a brick extending halfway into the void is stable, and we can place a brick extending halfway into the void on any stable brick, then we can create an infinitely overhanging tower by placing an infinite series of bricks each of which extends halfway over the void relative to the tower beneath it.
Clearly this is not the case. What does 'stable' mean (precisely) in the above quote?
What's actually happening is that at each step, you already have a max-overhang tower of N bricks, then you add another brick below, to produce a max-overhang tower of N+1 bricks. Because the old N-brick tower is stable, you can place it anywhere on the new bottom brick as long as its center of gravity is not off the edge of the new brick.
The top brick can be placed so that its centre of mass is positioned over the edge of the brick below. This two block system is stable.
The two block system can be placed on a third block so that the centre of mass of the two blocks is positioned over the edge of the third block. This three block system is then stable.
From there, it's turtles all the way down
Then that means my memory is wrong. Fallible is the mind. :)
Let the blocks have length 2 and look at the problem backwards: the topmost book's furthest-from-the-ledge edge is x=0, then x increases as we go towards the ledge. For the case of one block, the center of mass is clearly now at x=1 and that is the farthest away the ledge can be. We will have a function ledge(n) detailing the farthest the ledge can be as the number of books increases.
The recursive rule is that when we place another book, the edge of the book must be where the ledge was -- any further and the books above must topple; any less far and the ledge is not the furthest out it could possibly be. So we just proved that the greedy algorithm is correct. ;-) The greedy algorithm then says that we lay out book n + 1 to have a center of mass at ledge(n) + 1, while the above n books had a center of mass at ledge(n). The combined center of mass is therefore:
ledge(n + 1) = [n * ledge(n) + 1 * (ledge(n) + 1)] / (n + 1)
= ledge(n) + 1/(n + 1)
They could use a diagram of opposed forces along the interface between the block and the table showing how, if there is no force applied to both the left and right sides of the CM, there is no way to get the torques to sum to zero. That would connect their assertions to the fundamental equations of state.
It's about 6(n^(1/3)) compared to (1/2)(log n) in the article.