Please consider the case where I have 'N' rods of length L (and width W) placed on a one- or two-dimensional surface with dimensions [0, A] in 1D, and [ [0, A], [0, B] ] in 2D. For the two-dimensional case, L and W are << A or B.

As a function of the number of rods N, and the relative dimensions of the rods and the surface on which they are placed, is there a reasonably easy derivation for the number of expected intersections between rods / the probability of an intersection occurring? I feel like this should have been solved somewhere in the literature, but I was unable to find anything.

Edit - When I mentioned the rods should be 'placed', I failed to clarify that my treatment has been to assume that one end of each rod is placed with uniform random probability somewhere inside the specified dimensions of the one- or two-dimensional surface, the angle of the rod should be random, and any rod sections outside the bounded surface should either be treated or ignored depending on how easy it makes treating boundary conditions.

  • $\begingroup$ How are they placed? (Of course, somehow 'randomly', but I could imagine that at least in the two dimensional case, there are different natural ways how one could think of placing them 'randomly', affecting the result. So it might be good to specifiy this or at least to state that this is not so, or you do not care about this aspect.) $\endgroup$
    – user9072
    Apr 12, 2011 at 14:09
  • $\begingroup$ Thanks, I hopefully just clarified what I meant. I basically want the easiest possible treatment for the boundary of the surface. $\endgroup$
    – Rob Grey
    Apr 12, 2011 at 14:57
  • $\begingroup$ This isn't quite what you're asking about, but percolation of randomly placed rods often goes by the name "stick percolation" in the literature. See for instance this paper of Rahul Roy for some rigorous work math.bme.hu/~balint/oktatas/perkolacio/percolation_papers/… , and this paper of Jiantong Li and Shi-Li Zhang for some numerical work link.aps.org/doi/10.1103/PhysRevE.80.040104 $\endgroup$
    – j.c.
    Apr 12, 2011 at 16:49
  • $\begingroup$ Thanks jc... I wouldn't have thought to use the word 'stick'. $\endgroup$
    – Rob Grey
    Apr 12, 2011 at 17:05

1 Answer 1


To compute the expected number of intersections, use the fact that expectation is additive. The expected number of intersections is just $\binom{N}{2}p$ where $p$ is the probability of an intersection.

To compute the probability of an intersection you can make your life easier (with essentially no cost to the accuracy) by assuming your surface/line wraps around. This means there are no special places on your surface. Now you need to compute the probability that two rods intersect.

In one dimension imagine the first rod is placed somewhere. What is the probability that the second rod intersects it? $2L/A$ (either the second rod is placed so its left end lies inside the first rod; or that its right end lies inside the first rod).

In two dimensions, fix the position of the first rod. Then suppose the second rod is inclined at angle $\theta$ to the first rod (we can assume that $0<\theta<\pi/2$). We now need the area of the set of positions of the "top left" vertex of the second rod such that there is an intersection with the first rod. This set of positions is an octagon (whose area (as long as my geometry is correct) is $2L^2\sin\theta+2W^2\sin\theta+2WL(1+\cos\theta)$). (In the case where $W\ll L$ it reduces to a rhombus for which the area is just $L^2\sin\theta$). The intersection probability is then just $(2/\pi)$ times the integral of this area over $\theta$ divided by $AB$. i.e. $(4(L^2+W^2)/\pi+LW(2+4/\pi))/AB$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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