

How far can you overhang blocks? - deletes
http://www.datagenetics.com/blog/may32013/index.html

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ambrop7
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.

~~~
femto
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.

~~~
andrewflnr
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.

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pradn
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>

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

~~~
lkozma
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).

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electromagnetic
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.

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ohazi
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>

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kijin
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.

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

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minikomi
I wonder if this pringles loop I saw on the weekend is related?
[https://farm9.staticflickr.com/8256/8624389266_10d1156614_z....](https://farm9.staticflickr.com/8256/8624389266_10d1156614_z.jpg)

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lenazegher
> 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?

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garethadams
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

~~~
Dylan16807
But the extension mechanism has nothing to do with placing the _top_ brick. It
seems like a mistake in the writeup.

~~~
turtlepower
Correct.

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praptak
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.

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

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dalke
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.

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

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dalke
Nice research!

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

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

~~~
drostie
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.

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LowKarmaAccount
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.

~~~
gjm11
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.

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jjw14
If you don't have any real blocks, try Algodoo (www.algodoo.com). A fun (phun)
physics program that is now free.

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bliker
like the pun at the end _How hungover can you get?_

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huhtenberg
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.

~~~
abduhl
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.

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niccl
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!

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

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marcosscriven
Takes me back to an interview questions two decades ago!

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jasonlingx
No rule regarding a counterbalance?

~~~
praxulus
Rule #2: Only one block per level.

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tensaix2j
I m actually quite curious to see how far can you go without Rule #2

~~~
deletes
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.

