Polya's orchard problem asks for what radius $r$ of trees at each lattice point within a distance $R$ of the origin block all lines of sight to the exterior of the orchard. The answer is known; $r$ should be about $1/R$. See this earlier questionthis earlier question for details and references.
Suppose instead of straight rays, visibility may follow a circular arc (of any radius) from the origin. Let $B=B(r)$ be the furthest distance visible in the plane along any circular arc when all lattice points (excluding the origin) center disks of radius $r$. It is easy to see that $B > R$, where $R=\frac{\sqrt{1-r^2}}{r}$ is the Polya distance determined by $r$:
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$r=\frac{1}{4}$, $R \approx 3.9$.
Two circular arc rays (of different radii) from the origin to beyond $R$ are shown.
$B$ is finite for any $r>0$: For otherwise the radius $a$ of the arc that shoots to $\infty$ must itself be infinite. But then it is a straight-line ray, and couldn't reach further than Polya's $R$.
Q. Can anyone see an estimate of, or a bound on $B$ as a function of the disk radius $r$?
Perhaps the furthest reaching arcs straddle the disks on the $x$- and $y$-axes?
