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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.

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Do you also consider driving to outside as much as possible before taking the turn? I mean the classical "outside-inside-outside" maneuver which tries to minimize the curvature. –  user13838 Nov 15 '11 at 10:56

2 Answers 2

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 $$\int_a^b (\|\gamma'(s)\|-\|\psi'(s)\|)\ ds = \epsilon \int_a^b \kappa(s).$$

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The should be no need for teleporting from side to side. As you exit a curve you aim for best entry to the next one. If the curvature is the same you hug the edge if it's the opposite you slowly change lanes to enter the next curve. Would that change the calculation. –  Nelson Nov 16 '11 at 1:30
    
Yes. The problem becomes much harder. You will essentially need to solve the variational problem $\min_{\psi} \int \|\psi'\|$ subject to $-\epsilon < \|\psi-\gamma\| < \epsilon$ and appropriate boundary conditions. I don't have a solution right now. –  user7530 Nov 16 '11 at 6:42

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.

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