Stack poker chips in a staircase-like fashion

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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!
 
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disregardthat

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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.
 
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No, you cant go infinitely far to the right. At some point it is going to tip over.
 
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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.
What if we assume the stacking will have a hyperbolic shape?
 
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NO! No matter what, NO! Just do a force balance and you can find out how many chips you can stack.
 

Danger

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The diagram seems to indicate that they're being stacked on edge, so I don't think you'll even get two layers. :tongue:
 
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You can go infinitely far, if you stack them in a particular shape.
 

russ_watters

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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.
 

Gokul43201

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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 [itex]\sum _k (d/2k) [/itex] 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.
 
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DaveC426913

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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).
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.
 
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Gokul43201

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

DaveC426913

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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).
 
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Gokul43201

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Gokul, the PDF does NOT demonstrate the OP's configuration.
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 all use counterbalances. Total cheating!
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.

Cesiumfrog's is the mathematical explanation, but does not adequately demonstrate the result.
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 [itex](1/N)\sum_k (1/2k) < 1/2 [/itex]. This follows trivially that each term is bounded above by 1/2.

But, not being satisifed with theory, I had to prove it...
I have no idea what you mean by this!

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

Gokul43201

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Is it just me or does weight, surface texture (friction), and the shape of the object matter?
It is just you! :biggrin:

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.
 
It is just you! :biggrin:

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.
reference please
 

DaveC426913

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

and the picture is the background configuration on figure 1.
Yes, there it is. How delightfully obfuscatory.

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 [itex](1/N)\sum_k (1/2k) < 1/2 [/itex]. This follows trivially that each term is bounded above by 1/2.
I can doubt a claim of a proof faster than I can doubt a physical demonstration.

Why? https://www.physicsforums.com/showpost.php?p=179681&postcount=50". :wink:



But, not being satisifed with theory, I had to prove it...
I have no idea what you mean by this!
...which is why ... I had to see it done for real.



I've done it with a stack of CDs. They're much flatter and more parallelized
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).
 
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DaveC426913

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reference please
Because in any of the proofs, none of those things are variables. They are immaterial to the problem/solution. In fact, the dimesions of the blocks aren't a factor either, as I just realized in my previous post.
 
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Gokul43201

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That would be cool. How high did you have to stack it?
It wasn't nearly optimal (measurement limitations, time limitations), and I think it took me like 45 CDs to get an overhang of a little over 2L. With CDs you need to be careful about having a level and flat surface to build on.
 

DaveC426913

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Cool. And totally non-intuitive.

According to the paper I linked to, you could have done 2L with a mere 31 CDs. A mere 11 gets you an overhang of 1.5L.

This would make a great bar bet.
 

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russ_watters

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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 [itex]\sum _k (d/2k) [/itex] 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.
It doesn't look to me like that stacking configuration is the same as with the OP. I see what you do - by overloading the bottom, you can make the top of the stack go further to the right without moving the center of gravity outside the footprint of the bottom chip. Clever, but the OP shows a linear stack.
 
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Gokul43201

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Clever, but the OP shows a linear stack.
I don't believe the linearity was intended in the depiction. After all, this is a well-known puzzle (it's called the overhang problem) that's quite old and does the rounds every now and again. Moreover, the OP could only have depicted the correct configuration if he knew the solution (so naturally, any details in his depiction other than the broad idea must be ignored).
 

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