Overhanging dominoes

[legend]

Some of you might know this one already, but I haven't seen it posted here before. So here it goes;
You are given an infinite amount of (idealized) dominoes with dimensions 2*4*0.5 [cm] and have to stack them them on a table. You can only stack them on top of each other.
What's the maximum overhang (horizontal distance from the edge of a domino and the edge of the table) you can crate that way?

If it's not clear what's going on, here's en example:
___________
______dddd_
_____dddd__
____dddd___
___dddd____
__dddd_____
XXXXX______

If you try to put one more domino on the top of that, the whole thing will probably tilt over.

EDIT: fixed some typos. thx jaap. and no it's the same word as in german

[jaap]

Instead of "staple", you mean "stack".
(I bet you're Dutch - the English word "staple" means Dutch "nietje")

[Qaanol]

Well, I’ve seen this before, so I won’t give a full solution, just the answer.

Spoiler:

Harmonic series diverges, so there’s no limit.

[skeptical scientist]

Spoiler:

There's no limit.

Take N dominos, and place the first so that it's edge is at the table edge, the second 1/N cm out, the third 1/N+1/(N-1), the fourth 1/N+1/(N-1)+1/(N-2), and so on. Then the center of mass of the top n dominos will be

1/n(1+2*1/2+3*1/3+...+n*1/n+n*1/(n+1)+...+n*(1/N))=1+1/(n+1)+...1/N

cm out from the table edge, while the outer edge of the n+1st domino from the top will be 1/(n+1)+...1/N+4 cm out from the table edge. That means each domino will support the dominos above it, and keep them from toppling. So to get k cm from the table edge, it suffices to take at least e^k dominos.

[lightvector]

Also,

Spoiler:

If you're allowed to stack more than one domino on top of another, rather than having to put them in just a single tower, you can do drastically better than the harmonic series by adding counterweights.

[Qaanol]

Which brings us to our next problem: for each positive integer N, using exactly N tiles, what is the greatest amount of overhang that can be realized?

That is, the single tile that overhangs the most, its amount of overhang is to be maximized among all stable arrangements of N tiles.

[Proginoskes]

Which brings us to our next problem: for each positive integer N, using exactly N tiles, what is the greatest amount of overhang that can be realized?

That is, the single tile that overhangs the most, its amount of overhang is to be maximized among all stable arrangements of N tiles.

A couple of years ago, an article about this topic showed up in the American Mathematical Montly:

November 2009
Maximum Overhang
By: Mike Paterson, Yuval Peres, Mikkel Thorup, Peter Winkler, and Uri Zwick
[email addresses removed to slow down spammers]
A 150-year-old problem asks how many identical rectangular blocks of length 1, balanced at the edge of a table, are needed to achieve some given overhang D beyond the table edge. A well-known construction based on the harmonic series requires a number of blocks which grows exponentially with D. Many people had assumed that this was optimal until a construction by Paterson and Zwick in this MONTHLY (January 2009) showed that a quantity just cubic in D was sufficient. The current paper settles this classic problem within a constant factor, proving cubic growth to be not only sufficient, but also necessary.


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

[WarDaft]

A related, but entirely different in practice, problem is how far over the edge of a desk you can extend a deck of cards. Based on some of the better domino strategies, it should be possible to exceed 3 card lengths. But it's not quite that easy.

[Goplat]

A related, but entirely different in practice, problem is how far over the edge of a desk you can extend a deck of cards. Based on some of the better domino strategies, it should be possible to exceed 3 card lengths. But it's not quite that easy.

I doubt one could create much of an overhang at all out of playing cards. A card will easily bend and allow the cards on top of it to slide off. You need something that will stay flat and has decent friction.

[WarDaft]

I managed slightly over 3 card lengths, but that was more than just one deck. Actually I'm not sure exactly how many it was because it was a Fluxx deck and not regular playing cards, and I had some leftover that couldn't have improved the structure. The cards do bend, but you can build in accordance of that. And the low friction, well, that just makes it more technically challenging. The paper actually assumes frictionless blocks, it's the bending that makes things different.