2
$\begingroup$

This is a variation on the stacking problem.
A block is a 1D object of length L and uniformly distributed mass. (with some negligible thickness).
A stack of size n is a series of n blocks placed flat one over the other(i.e. their lengths are parallel).
A stable stack is one that doesn't topple under constant gravity.
A monolayer stack is where at most $1$ block is immediately above another block.

We are interested in finding the probability that a monolayer stack constructed ground-up from $n$ blocks is stable.
The stack is constructed probabilistically as follows:

  1. Place the first block.(say with its left edge at $0$)
  2. The next block is placed randomly such that it touches the block immediately before, at least somewhere, equally probably i.e. its left edge is in $(-L,L)$,equally probably. The probability of the block landing outside is taken $0$. (say its left edge is at $-L<x_2<L$ and nowhere else)
  3. Place the third block immediately above the second with the same probability distribution. (i.e its left edge $x_3$ can land anywhere in $(x_2-L,x_2+L)$ and nowhere else)
  4. and so on till the n-th block.

Of course most such random stacks topple. Question is , what is the probability $p(n)$ that the stack stays stable during construction?

example stacks


So far...

I have figured out the following constriants
1. $\forall m \ge2$, $x_{m-1}-L<x_m<x_{m-1}+L$ (placement region) and
2. $\forall m \ge1$,$x_m-L/2<Average[x_{m+1},x_{m+2},...x_n]<x_m+L/2$

(2) is motivated by the fact that placing a block shouldn't destabilise the stack below. For this we calculate the COM of the $block_n$, then $block_n+block_{n-1}$ and so on till all such sub-stacks are found to be stable
3. $p(n)=\frac{1}{2^{n-1}} \frac ab$ for some positive rational $\frac ab \lt 1$ (from numerical analysis)

Calulating phase space volume subject to above constraints as a function of $n$ seems difficult.
Am I missing some physics which would make the calculation straightforward?
So far I have found(with $L=1$) (mc denotes via monte-carlo), $$ \begin{array} {|r|r|}\hline n & p(n) \\ \hline 1 & 1 \\ \hline 2 & 1/2 \\ \hline 3 & 7/32 (mc:0.218) \\ \hline 4 & 107/1152 (mc:0.0928) \\ \hline 5 & 2849/73728 (mc: 0.0385\pm 0.0004) \\ \hline 6 & mc: 0.0156\pm0.003\\ \hline \end{array} $$

$\endgroup$
  • $\begingroup$ plz move to math se if needed $\endgroup$ – lineage Dec 14 '19 at 2:56
  • $\begingroup$ Is your high value for $p(4)$ a typo? What does frommc mean? $\endgroup$ – G. Smith Dec 14 '19 at 6:01
  • $\begingroup$ @G.Smith had a typo before in the mc value(was 0.926)..its correct now $\endgroup$ – lineage Dec 14 '19 at 6:03
  • $\begingroup$ Where did the exact fractions come from? $\endgroup$ – G. Smith Dec 14 '19 at 6:06
  • $\begingroup$ till p(3) by hand, rest mathematica $\endgroup$ – lineage Dec 14 '19 at 6:07
0
$\begingroup$

A block is sits on top of the block below if its center is lower the bottom block. The odds of that are 1/2.

Each block in the stack, except the bottom has a 1/2 chance of spoiling the stable stack. The odds are $2^{-(n-1)}$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ center is lower its a vertical stacking please clarify $\endgroup$ – lineage Dec 14 '19 at 5:34
  • $\begingroup$ has a 1/2 chance of spoiling... not necessarily as even if the block maybe stable wrt the one immediately beneth, it may not be wrt to the one further below $\endgroup$ – lineage Dec 14 '19 at 5:38
  • $\begingroup$ at this point, the $2^{-(n-1)}$ doesn't seem to match the monte carlo or $p(3)$ $\endgroup$ – lineage Dec 14 '19 at 5:40
  • 1
    $\begingroup$ "if its center is lower the bottom block" Do you mean "over the bottom block? $\endgroup$ – Acccumulation Dec 14 '19 at 7:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.