TITLE: Geodesics in $\mathbb{R}^2 \times \mathbb{S}^1$ under "segment" metric
QUESTION [13 upvotes]: Represent the position of a unit-length, oriented segment $s$ in the plane
by the location $a$ of its basepoint and
an orientation $\theta$: $s = (a,\theta)$.  So $s$ can be
viewed as a point in $\mathbb{R^2} \times \mathbb{S^1}$.
Now I'll define a metric on this space.
Define the distance $d(s_1,s_2)$ between two positions of
unit-length segments as the average distance between their corresponding
points:

     


Above the distances are about 0.31, 0.61, and 0.53, left-to-right.
So if the endpoints of $s_i$ are $a_i$ and $b_i$,
then $d(s_1,s_2)$ is the average of the Euclidean distances
between $(1-t) a_1 + t b_1$ and $(1-t) a_2 + t b_2$ as $t$ varies in $[0,1]$.
This is indeed a metric, I believe, because the triangle inequality
holds between corresponding points in three positions of the segment.
This metric is intended to capture the intuitive notion of how much
work is required to move $s_1$ to $s_2$.
My question is: What are the geodesics in this space under this metric?
Certainly a pure translation of $s$ is a geodesic.
It seems that a pure rotation by at most $\pi$ of $s$ about a 
point $p \in s$ should also be a geodesic, but even this is
not so clear to me.  Certainly a rotation about a point not on $s$
is (generally) not a geodesic.  Of course the main interest would be
in geodesics that mix translation and rotation, showing (locally) optimal
repositioning paths.
I investigated this long ago when working on motion-planning algorithms
("moving a ladder"), but got quite blocked on this natural question.
This superficially seems related to the 
Kakeya needle problem, 
but the metric I propose does not measure swept area.
Perhaps it has been studied in some guise previously.
If so, a pointer would be appreciated.  Thanks!
Addenda. (26Sep11.) I just ran across this book, by V. A. Dubovit͡s︡kiĭ,
which seems relevant:
The Ulam problem of optimal motion of line segments,
Translation Series in Mathematics and Engineering, Optimization Software, 1985.
It may take some time for me to locate a copy...
(11Nov11). I finally have this book in my hands.
The Preface by Hestenes says,

Dubovitskij has succeeded in solving in closed form a generalization of a problem
  of S[.] Ulam..:
  Among all continuous motions of an oriented line segment $S$ in $\mathbb{E}^n$ from one
  position to another, which preserves its length [...], find one for which the sum
  of the lengths of the paths swept by its endpoints is minimal.

The concentration here on the motion of the endpoints—in contrast to 
the average distance metric I proposed—seems to render these results as not directly relevant, although nevertheless quite interesting.

REPLY [10 votes]: In addition to Anton Petrunin's answer, here is a trick to simplify (and in some sense solve) the geodesic equation.
Since the metric has three-dimensional group of isometries (generated by rigid motions of the plane), the corresponding Noether's integrals make the geodesic flow completely integrable. More precisely, the first derivative is uniquely determined by the position and the values of the integrals. Actually a two-dimensional group is sufficient because for the third integral one can always take the speed (= the norm of the velocity vector) of a geodesic.
So let's use invariance under parallel translations only. We have a manifold of all unit segments in the plane, and a tangent vector to that manifold can be thought of as the vector field along a segment (representing velocities of all its points). This vector field has a form
$$
 v(t) = tx + (1-t)y, \qquad t\in[0,1]
$$
where $x,y\in\mathbb R^2$ are such that the vector $x-y$ is orthogonal to the segment. The norm of this tangent vector is $\int |v(t)| dt$ (as in Anton's answer), and the Noether integral corresponding to translations (if my quick computation is correct) boils down to the following: if $v$ is a velocity vector of a geodesic represented in the above form, then
$$
 \int_0^1 \frac{v(t)}{|v(t)|} \ dt = V_0
$$
where $V_0$ is a constant (for each geodesic) vector in the plane. This vector $V_0$ (and, say, the assumption that the geodesic is unit-speed) determines the parameters $x$ and $y$ in the formula for $v(t)$ uniquely (as a function of the segment's current direction), so the geodesic equation is reduced to a 1st order ODE. Furthermore, the above integral can be found as an explicit (but weird) function of $x$ and $y$, so the equation becomes easy to solve, at least numerically.
By the way, I second Jean-Marc Schlenker's proposal to consider quadratic mean of the distances (i.e. $L^2$ norm rather than $L^1$), especially if you have any physics-related application in mind. In this case the metric is Riemannian, the energy is just the standard kinetic energy of the moving segment, so the Noether intergrals are just the standard conservation laws: the linear momentum and the angular momentum. And the geodesics are very simple: the segment rotates at a fixed angular speed while its midpoint (the barycenter) moves along a straight line with a constant speed.