### Brain Teasers

# Overhanging Bricks

What is the maximum overhang that you can create with an infinite supply of bricks?

You cannot use any mortar or attach the bricks to each other or anything else in any way. The bricks are simply dry stacked on level ground and jut out over a ledge.

Assume that you have an infinite amount of space to work with and that a brick is 22x10x6 cm in dimension and weighs 5kg. Also assume that we are on earth.

You cannot use any mortar or attach the bricks to each other or anything else in any way. The bricks are simply dry stacked on level ground and jut out over a ledge.

Assume that you have an infinite amount of space to work with and that a brick is 22x10x6 cm in dimension and weighs 5kg. Also assume that we are on earth.

### Hint

How far can brick 1 be from brick 0?### Answer

You can create an infinite overhang.Let us assume that the brick is of length 1.

To determine the place of the center of mass a(n):

a(1)=1/2

a(n)=1/n[(n-1)*a(n-1)+[a(n-1)+1/2]]=a(n-1)+1/(2n)

The limit for n->oo of half the Harmonic series is oo.

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## Comments

Actually, you can only create an overhang that approaches 1 full brick length. You can theoretically use an infinite number of bricks but the overhang, according to your formula, will never total greater than or equal to 1 brick length.

I thought that the answer was infinite because by using a pyramidal shape as long as the base is increased the top may go higher. The math i didn,t catch.

Feb 24, 2004

I sooo did not get this one...

HEY JAKE this is the first teaser of yours I've seen. Good one I hope we shall see more

if the tower or whatever was equal on both sides it could be an infinite amount of overhang but where are you going to find all these builders to build it???

no i don't get the formula either

The center of mass a(n) of the nth brick approaches 1 brick length. However, the brick overhangs its centre of mass by half of its length. Therefore the answer is one and a half brick lengths is the maximum overhang. In other words 33 cm (if your centre of mass formula is correct).

I think 1 1/2 is wrong. You can't get more than 1 brick length overhang. The first brick can go out 1/2, the second, 1/4, the third 1/8, etc... Add all of these up and you approach 1, not 1 1/2.

Never mind my stupid comments earlier. Thought I was soooooo smart but I was way off. Original answer is correct.

Can get more info at...

http://mathforum.org/advanced/robertd/harmonic.html

Can get more info at...

http://mathforum.org/advanced/robertd/harmonic.html

It turns out that the overhangs are related to the harmonic numbers Hn, which are defined as 1 + 1/2 + 1/3 + ... + 1/n: the maximum overhang possible for n bricks is Hn/2.

May 05, 2004

I said 6...

whoa! TOO COMPLICATED!

whoa, that went WHOOSH rite over my head. i believe your answer.

you would end up with an overhang goping up at an angle. If yopu had infinate bricks you would be able to keep going until the bottom ones broke from the weight not infinate. There would be probably 10-20 bricks in place to hold eahc one that goes out every time just to hold that 1 brick in place.

Of course there is no real answer to this teaser, because as vast as the earth is, when it comes to plains that are infinite planes, the earth is a half-vast place.

I wish Jake had spelled out the size of the overhang for the first four or five layers of brick.

I picture an infinitely high brick wall with all staggers at 50%. The working edge is unfinished, so every second brick overhangs the one below it. Repeatedly fill the gaps starting just above the bottom most overhang. Each 'column' added increases the overhang by one-half brick. and is adequately counterbalanced until infinity is reached (which it can't be.) And the bottom rows are eventually crushed by those above.

I wish Jake had spelled out the size of the overhang for the first four or five layers of brick.

I picture an infinitely high brick wall with all staggers at 50%. The working edge is unfinished, so every second brick overhangs the one below it. Repeatedly fill the gaps starting just above the bottom most overhang. Each 'column' added increases the overhang by one-half brick. and is adequately counterbalanced until infinity is reached (which it can't be.) And the bottom rows are eventually crushed by those above.

That is true stil, but by the way the problem is worded, take it abstractly. With an infinite plane, it this problem is valid, but in the real world, there is limited possibilty here.

Wait. If I have the center of mass past the first brick, won't everything fall? Brick 1 is supporting ALL OTHER BRICKS, not just brick 2. So isn't it the same from the perspective of brick 1 if the other bricks are a giant, strange brick? And doesn't that brick fall?

The infinite analysis works if and ONLY if each brick is FIRMLY supported. The overall center of mass above a brick cannot be past it. So, the center of mass for all bricks above the first is over the first. So, the overhang cannot be infinite (which leads to all collections being unsupported).

BTW, an overhang is the length in the air that is not over the first brick, so a right-side-up pyramid has no overhang.

The infinite analysis works if and ONLY if each brick is FIRMLY supported. The overall center of mass above a brick cannot be past it. So, the center of mass for all bricks above the first is over the first. So, the overhang cannot be infinite (which leads to all collections being unsupported).

BTW, an overhang is the length in the air that is not over the first brick, so a right-side-up pyramid has no overhang.

http://math.dartmouth.edu/

~pw/papers/sodaproc2.pdf

Everything you wanted to know about overhanging bricks. If you are going to post a science teaser, please do your research first. A simple google search pulled this up as the first link.

~pw/papers/sodaproc2.pdf

Everything you wanted to know about overhanging bricks. If you are going to post a science teaser, please do your research first. A simple google search pulled this up as the first link.

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