How far can you overhang blocks? | Hacker News Reader

How far can you overhang blocks?

datagenetics.com

288 pointsdeletes40 comments

It works! http://www.pasteall.org/pic/show.php?id=52197

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.

Reading an article, and immediately setting out to recreate the results based on the paper.

Science needs people like you.

That's why everyone needs to own a set of well made wooden blocks at some point in their life! Blocks let you naturally pick up some really great physics and maths through play, without having to know any algebra.

It's pretty cool to do it with yard sticks. The only problem was that the ones I was using were pretty warped, so they were hard to balance. I'm pretty sure I still got the top one to be all the way over "the void" once.

The problem of finding the maximum overhang has been recently solved. The resulting construction is more complex than the one described in this article.

http://www.math.dartmouth.edu/~pw/papers/maxover.pdf

The article uses the constraint of only allowing one block per level.

It is also less interesting as they balance out the overhang by extending bricks in other direction.

Why does it make it less interesting? Overall they achieve more overhang for the same number of bricks than the other construction, n^(1/3) instead of log(n).

This is a great example of an ancient architectural design element: corbelling. Structurally, as long as you follow the 1/3 rule these creations are safe, IE you shouldn't extend the centre of mass beyond 1/3 of the width of the tower/wall. The principle shown here is similar to what's been long used, where a stretcher would be corbelled 1/3 of its width, and then a header would be corbelled 1/3 of its width, then a second header would be corbelled at 1/3 of its width, this would get an overhang of approximately 2/3 the width of the initial wall (assuming the width of the wall is thickened to provide a cantilever effect.

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 had this as a question on a freshman physics problem set. A few of us were skeptical, but we tried it and it works:

http://ugcs.caltech.edu/~ohazi/cd_tower.jpg

Aren't CD cases slightly heavier on one side? There's a pivot mechanism on one side and not the other; the CD isn't placed at the exact center of the case; and the paper insert (as well as any staples it contains) can also affect the balance. So you might be able to extend the tower a few millimeters more than what mathematical algorithms allow, by aligning the heavier side toward the desk.

Sure, but this is physics, so we can stick with our spherical cows and other first order approximations without losing too much sleep... :-)

I wonder if this pringles loop I saw on the weekend is related? https://farm9.staticflickr.com/8256/8624389266_10d1156614_z....

> Whatever the make-up of the tower below, if it is stable, we can always place the top brick so that it extends half way into the void.

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?

I think the reasoning is wrong here. Obviously a tower could be stable, but still applying force to an edge of the top could make it collapse.

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.

Think of it the other way around.

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

But the extension mechanism has nothing to do with placing the top brick. It seems like a mistake in the writeup.

Correct.

If you do this so that the top brick is overhanging by half it's length and then place another brick on it, the first brick is no longer the top brick so this no longer applies.

It is an old problem. I think it's in the classic "Concrete Mathematics" by Graham, Knuth, Patashnik but I doubt it's the original source.

Yeah, I had it in my undergrad intro to mechanics textbook. Still a great problem!

I think it was in Halliday and Resnick. I remember doing it in my high school physics class. We used the 3rd edition of that book, which was from 1979.

I've found it in both the second edition and the seventh, both asking for just the 4-block case.

Nice research!

Then that means my memory is wrong. Fallible is the mind. :)

I proved this in my college calc class - my proof was the shortest (one page) and it won me the textbooks for differential equations.

It's actually a really easy bit of reasoning once you've got the "hang" of it. (sorry, bad pun.)

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)

This recurrence relation from 1 generates the harmonic series, and so ledge(n) = H_n.

The site displays a list of the least number of blocks needed to achieve an integral multiple "overhangs" (ie. Floor (x) > y, where y is an integer). What the site doesn't mention is that after the first iteration, the harmonic series will never be an integer again.

Proof: consider the sum S = 1/1 + 1/2 + ... + 1/n. Suppose 2^k is the biggest power of 2 that's <= n. Then one of the terms in S is 1/2^k, and there is nothing else with a multiple of 2^k in the denominator. So when you clear denominators, there'll be exactly one term (that one) with an odd numerator, so the whole thing is (odd number) / (multiple of 2^k) and therefore not an integer.

If you don't have any real blocks, try Algodoo (www.algodoo.com). A fun (phun) physics program that is now free.

like the pun at the end How hungover can you get?

Maybe it's my pseudo-math background talking, but I find them relying heavily on a concept of center mass to be a cheating. They basically make very convenient assumptions about the CM that, while intuitive, still need a proof of correctness.

Center of mass is directly related to the centroid or first moment of area if you would like to look into it more. It is based on integration of a product which is why simple summation of the product for a discrete element is also valid.

I kind of agree with you here. The writeup is not clear about why the CM is so important.

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.

I remember doing the maths for this at uni. Then going in to the library to see if it worked in practice. I don't know what the librarians did when they saw the pile of carefully stacked books on the floor, but we had fun!

Isn't this why the arc is so important in structural engineering?

Takes me back to an interview questions two decades ago!

No rule regarding a counterbalance?

Rule #2: Only one block per level.

I m actually quite curious to see how far can you go without Rule #2

Take a look at this paper. http://www.math.dartmouth.edu/~pw/papers/maxover.pdf

It's about 6(n^(1/3)) compared to (1/2)(log n) in the article.

Reply on news.ycombinator.com