# A random variation on Polya's orchard problem

Polya's orchard problem is as follows:

"How thick must the trunks of the trees in a regularly spaced circular orchard grow if they are to block completely the view from the center?"

See, e.g.,

Allen, Thomas Tracy. "Polya's orchard problem." American Mathematical Monthly (1986): 98-104. JSTOR link

or this earlier MO question. This problem has been well-studied. One crude way to phrase what is known is that, if the trees/disks centered at each lattice point have radius $r$, the furthest distance one can see from the origin is $R \approx 1/r$.

Here is my question. Suppose instead of disks centered on each lattice point, we have a randomly oriented segment (a $1$-dimensional disk) centered on each lattice point, of length $2r$, i.e., of radius $r$.

Q. Is it still the case that the furthest one can see from the origin is expected to be $R=c/r$ for some constant $c$?

Here is an example, with $r=3/8$. The "Polya radius" is about $2.47$, but in this one random instance, visibility extends about $3$ times further, $R \approx 8$: In $\mathbb{R}^3$, the same question can be asked with now $2$-dimensional randomly oriented disks centered on each lattice point. And of course the question generalizes to $\mathbb{R}^d$.

• One might even guess that the expected size is all that matters, which would give $c=\pi/2$. – Will Sawin Aug 31 '14 at 2:15
• @WillSawin: Could you expand on your remark? I don't follow how you arrive at $\pi/2$... – Joseph O'Rourke Aug 31 '14 at 2:23
• I'm imagining looking through a long field of objects. It seems to me that the distance I see is going to be governed by the average visible angle the objects take up. For a line of radius $r$, this is $2/\pi$ times the visible angle of a disc of radius $r$. (average value of $\sin \theta$ for $0<\theta<\pi$). – Will Sawin Aug 31 '14 at 2:35