Distance saved by shortening the road A long distance drivers technique for saving time is to drive the width of the road you are on.  To do this you drive so that you are placing you car on the inside of all turns as much as possible.  I want to know how much distance is saved.   
From the point of view of a math problem it could be presented as two parralel  curves in 3space forming a ribbon and you want to know the difference in length of the edges compared to a radius minimizing path.  The problem can have a simplified presentation of two curves in a plane.  
I am not sure of the best equation form for the edges to make it possible to calculate the radius min line between.
 A: Let $\gamma(s)$ be the arc-length parameterized center of a road of width $2\epsilon$. Then the inner shoulder is given by
$$\psi(t) = \gamma(t) + \epsilon \gamma'(t)^{\perp}.$$
Here we orient $\gamma'^{\perp}$ so that it points inside the turn. Notice that $\psi$ has a discontinuous jump whenever curvature vanishes: we assume the truck can essentially "teleport" from one side of the road to the other whenever the direction of curvature changes. 
We now have $\psi'(t) = \gamma'(t) + \epsilon \gamma''(t)^{\perp}$, so
$$\begin{align*}\|\psi'(t)\|^2 &= \gamma' \cdot \gamma' + 2\epsilon \gamma''^{\perp}\cdot \gamma' + \epsilon^2 \gamma''^{\perp} \cdot \gamma''^{\perp}\\
&= 1 - 2\epsilon \kappa + \epsilon^2 \kappa^2\\
&= (1-\epsilon \kappa)^2,
\end{align*}$$
where $\kappa$ is (unsigned) curvature. So the difference in distance traveled for the segment of road between $s=a$ and $s=b$ is
$$\int_a^b (\|\gamma'(s)\|-\|\psi'(s)\|)\ ds = \epsilon \int_a^b \kappa(s).$$
A: A quick heuristic:  leave each curve on a tangent, also on a tangent to the next curve.  Where you have a choice, be on a tangent for the furthest curve you will have a chance to do so.  This certainly succeeds when you have two significant curves in a row, defined as not being able to get through one on a straight line.  I believe it also lets you ignore curves whenever possible.
