# Stack poker chips in a staircase-like fashion

1. Dec 26, 2006

### bigdave

k so I have a bet going with one of my friends, and here's the deal.

If you stack poker chips in a staircase-like fashion, lets say going left to right like this:
____0000
___0000_
__0000__
_0000___
0000____

I say that you cant get to the point where the top poker chips left-edge is further right than the bottom chips right edge. My friend says that the chips can go infinitely far in the right direction if you have enough of them. Are either of us right? an explanation would be great to... thanks!

Last edited: Dec 26, 2006
2. Dec 26, 2006

### disregardthat

Well, you can't go on for infinity, that's for sure, hypothetically speaking.
Not thinking of the low friction which will make it fall very soon, the weight of the chips on one side will make the lowest chip tip over. With the low friction, the other chips will glide away soon. I don't have the mathematics, but i think my explanation is correct.

Look at it as a tower that is not attached to the ground, tilting in one direction. If you build it tall enough it will eventually fall. The same principle goes for stacking pokerchips. Just that the chip over the lowest chip will glide from the lowest chip way before the "tower" will fall because of the weight concentrated on the side of the chip not being backed up by any chip.

Last edited: Dec 26, 2006
3. Dec 26, 2006

### Cyrus

No, you cant go infinitely far to the right. At some point it is going to tip over.

4. Dec 26, 2006

### MeJennifer

What if we assume the stacking will have a hyperbolic shape?

5. Dec 26, 2006

### Cyrus

NO! No matter what, NO! Just do a force balance and you can find out how many chips you can stack.

6. Dec 26, 2006

### Danger

The diagram seems to indicate that they're being stacked on edge, so I don't think you'll even get two layers. :tongue:

7. Dec 26, 2006

### cesiumfrog

You can go infinitely far, if you stack them in a particular shape.

8. Dec 26, 2006

### cesiumfrog

Explanation (I liked another site better, can't find it again tho.)

9. Dec 26, 2006

### Staff: Mentor

If you stack them at an angle in a linear fashion, they tip over when the center of the middle chip is outside the footprint of the bottom chip. The center of the middle chip is the center of gravity and it must always be above the footprint of the tower for the tower to be stable.

10. Dec 26, 2006

### Gokul43201

Staff Emeritus
The OP's friend is correct.

This is an old problem, and the answer is that with an unlimited number of chips you can make the topmost chip go as far to the right as you wish (infinite overhang is possible). With n chips, simply displace the k'th chip from the top by d/2k to the right of the chip below it (d is the chip diameter). The harmonic series $\sum _k (d/2k)$ goes to infinity as k increases indefinitely, and the CoM of any subpile (from the top) is clearly above (divide above sum by number of chips and then take limit of large n) the topmost brick below that subpile.

Last edited: Dec 26, 2006
11. Dec 26, 2006

### DaveC426913

1] Are you saying you can do this with the configuration as shown in the OP's post?
2] Judging by past experience you're presumably not insane, so presumably #1 is not true, so can you please demonstrate the configuration you ARE talking about?

It sounds to me like you're going to get a curved stack of chips that asymptotically appproaches vertical. However, you are wrong in your conclusion. It will not extend infinitely far to the right. It will not exceed d to the right.

Last edited: Dec 26, 2006
12. Dec 26, 2006

### Gokul43201

Staff Emeritus
Yes, with the configuration shown in the OP's figure, except that instead of constant spacing (as seems to be depicted), the spacings are as described in my post above.

It'll be easier if I just find a link...give me a minute.

Edit: There's one in post#8 by cesiumfrog. Here's another one: http://www.cs.tau.ac.il/~zwick/papers/overhang-SODA.pdf

To find more hits google maximum overhang bricks.

Last edited: Dec 26, 2006
13. Dec 26, 2006

### Cyrus

Hahah, wow. I stand corrected. I never knew so much work went into stacking bricks. :tongue2:

14. Dec 26, 2006

### DaveC426913

Gokul, the PDF does NOT demonstrate the OP's configuration. They all use counterbalances. Total cheating! Cesiumfrog's is the mathematical explanation, but does not adequately demonstrate the result.

I found this: http://oak.ucc.nau.edu/jws8/classes/137.2006.1/index.html
And this: http://oak.ucc.nau.edu/jws8/classes/137.2006.1/images/overhang31blocks.gif

But, not being satisifed with theory, I had to prove it...

And tie-me-up-with-electrical-tape-pass-current-through-my-body-and-call-me-wiggly it really can be done! (see attachment)

I'd like to see this done with something more precise than Lego bricks - which are flawed enough to limit the maximum amount of stacking (you can see how the top bricks are not horizontal).

Last edited: Jul 20, 2007
15. Dec 26, 2006

### Gokul43201

Staff Emeritus
Yes, it does. If you can't see it there, then all the demonstration required is in my post (two up from this one).

They do not ALL use counterbalances. The description of the classical (harmonic) solution (devoid of counterbalances) is given in the first paragraph of the linked pdf, and the picture is the background configuration on figure 1.

I didn't read everything on that link, but there's a full mathematical proof in my post too (only missing some steps). To see the proof that the CoM of any pile of chips above a given chip lies over that chip, all you need to do is show that $(1/N)\sum_k (1/2k) < 1/2$. This follows trivially that each term is bounded above by 1/2.

I have no idea what you mean by this!

I've done it with a stack of CDs. They're much flatter and more parallelized.

Last edited: Dec 26, 2006
16. Dec 27, 2006

### Chaos' lil bro Order

Is it just me or does weight, surface texture (friction), and the shape of the object matter? I wonder if there is a formula to treat all these at once.

17. Dec 27, 2006

### Gokul43201

Staff Emeritus
It is just you!

The only requirement on the shape imposed by the problem is that the top and bottom surfaces be parallel, and all chips be identical. An infinite overhang can be achieved independent of weight, surface texture or other aspects of the shape than those specified above.

18. Dec 27, 2006

### Chaos' lil bro Order

19. Dec 27, 2006

### Gokul43201

Staff Emeritus

20. Dec 27, 2006

### DaveC426913

I am saying that, while there is a description, and even a proof, there isn't a demonstration i.e. visual.

Yes, there it is. How delightfully obfuscatory.

I can doubt a claim of a proof faster than I can doubt a physical demonstration.

Why? This is why.

...which is why ... I had to see it done for real.

That would be cool. How high did you have to stack it?

Actually, it's not dependent on the object used is it? I should be able to simulate any of those configs listed in the reference I added:
4 blocks (the first with overhang > L = length of one block).
11 blocks (the first with overhang > 1.5 L).
31 blocks (the first with overhang > 2 L).
83 blocks (the first with overhang > 2.5 L).

Last edited: Dec 27, 2006