TITLE: Reorienting a ladder among $\mathbb{Z}^2$ poles
QUESTION [10 upvotes]: Imagine an object, which I'll call a ladder $\cal{L}$, a "racetrack" shape
composed of a rectangle of length $L$ capped at either end by
semicircles of radius $r$; so it is $L+2r$ tip-to-tip.
View every lattice point of $\mathbb{Z}^2$ as a point obstacle
(a pole).  The ladder is forbidden to include a pole in its
interior, but $\cal{L}$ may touch poles on its boundary.
$\cal{L}$ is initially sitting in the plane in a horizontal orientation.
          


My question is:

Q. 
  Under what conditions on $L \ge 0$ and $r \le \frac{1}{2}$can $\cal{L}$ be reoriented
  via continuous motions to a vertical orientation?

Throughout the motions, no pole may be interior to $\cal{L}$.
If $L=0$, $\cal{L}$ is a disk of radius $r$;
if $r=0$, $\cal{L}$ is a segment of length $L$.
I believe that, for $r=0$, there is no upper bound on $L$: an arbitrarily
long segment can be reoriented.
(I made this an exercise (7.3) in a
textbook.)
And certainly when $r=\frac{1}{2}$, and, say $L=1$, reorientation
is not possible.
I can see that, for $r=\frac{1}{2}$,
any $L \le \sqrt{2}-1$ can be reoriented
(because then $\cal{L}$ fits inside a circle of radius $\sqrt{2}/2$).
Almost everything else is unclear to me.
                    


An analogous question may be posed for an appropriately defined "ladder"
amidst $\mathbb{Z}^d$ obstacles.
Any insights, even for specific $(L,r)$ values, or corrections
to my beliefs above, are welcomed.  Thanks!

REPLY [10 votes]: At the EuroCG conference that ended as I post this, Sándor Fekete (Braunschweig)
solved this question modulo a few details, with useful input from Günter Rote (Berlin). 
Sándor's crucial observation is that, when the ladder
is in an extreme configuration, it is touching three lattice points, which form an empty lattice
triangle, which by Pick's Theorem, has area $\frac{1}{2}$, so the $L \times r$ rectangle
has area 1, and so $r \sim 1/L$.  There are many details to convert this to a formal proof,
but it seems this insight yields the right dependency between $r$ and $L$ to allow reorientability.
Curiously, at the conference a banquet bus was forced to execute a maneuver not dissimilar
from the 15-move back-&-forth path illustrated in my first figure! :-)