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I was recently walking around the track at UCSC, and I noticed that the track didn't always curve inward. Sometimes it curved the other way. Compare this (A convex track): http://www.irelandaerialphotography.com/dr_f2_3838.jpg to this (A concave track): http://www.motoyard.com/trackday/images/track-chuckwalla-l.gif . In the latter, there are portions of the track where the outer edge ( the edge bordering the infinite face, if this was a planar graph) is longer than the inner edge (the edge bordering the inner field). In a convex track, the outer edge is always longer than the inner edge. But is it possible to construct a (concave) track such that the inner edge is the same length as the outer edge? If not, how does one prove that it is impossible? The track must have a width greater than zero. (If the width is zero, the inner and outer edge will be one and the same, and the question is meaningless.) As an extension, is it possible to construct a track such that all paths along it are of the same length? A path is defined as the set of all points equidistant from the outside edge all the way around the track and equidistant from the inside edge all the way around the track, though these two distances do not have to be equal.

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This question can be formalized in several different ways. Which of the following models of a "track" did you have in mind? $$ $$ 1. Let $C_1$ and $C_2$ be two Jordan curves such that for any $x\in C_1,$ $\text{dist}(x,C_2)=d>0$ is independent of $x$ (and similarly if the roles of $C_1$ and $C_2$ are reversed). $$ $$ 2. Let $C_1$ and $C_2$ be two smooth simple closed curves that do not intersect such that for any $x\in C_1,$ $y=x+dn(x)\in C_2,$ where $n(x)$ is a unit normal vector to $C_1$ at $x$ and $d$ is independent of $x.$ (Additionally, all segments $xy$ are disjoint.) –  Victor Protsak Aug 16 '10 at 2:22
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You should use "non-convex" instead of "concave" here. –  Benoît Kloeckner Aug 16 '10 at 11:28
    
Is this the track you are talking about? maps.google.com/… –  Nate Eldredge Aug 31 '10 at 20:18
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Both links are now 404. –  Rasmus Bentmann Feb 5 at 15:42

7 Answers 7

I think what you're coming up against is the simplest version of the Gauss–Bonnet theorem.

Say we have a reference runner running around the inside track. A runner next to her will have to speed up or slow down to keep pace, depending on "how much the track is turning", a parameter known as the 'curvature' of a plane curve. You can get the total distance of the nearby lane by integrating the curvature over the entire track (this is always an integer multiple of $2\pi$ — see Total curvature) and multiplying by the width of the lane. So if your track doesn't intersect itself then you get total curvature $2\pi$. In short, you can never get away from this phenomenon, and in fact the shape of the track doesn't matter at all.

In 3 dimensions you can fix this of course, by running on the surface of a cylinder, or more interestingly, you can electrically charge your runners and shoot them in a tokamak fusion reactor - then you will finally get a fair race.

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+1 for the idea of electrically charging runners and shooting them! –  Mariano Suárez-Alvarez Aug 31 '10 at 2:49

Here's an answer that only works only on a technicality. I think the formulation of the question should rule out this answer. Draw a wiggly closed path, then draw a less wiggly closed path around it which has the same length. It should look something like this:

   __________
 /            \
|  /\/\/\/\/\_ |
|  \_/\/\/\/\/ |
 \____________/

The inside and outside boundaries have the same length. Moreover, every "path" has the same length simply because there are no paths:

A path is defined as the set of all points equidistant from the outside edge all the way around the track and equidistant from the inside edge all the way around the track, though these two distances do not have to be equal.

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This reminded me of the "DNA inequality". –  Gjergji Zaimi Aug 16 '10 at 12:01

Let $s\mapsto(x(s), y(s))$, $0\leq s\leq L(\Gamma)$, be a simply closed curve $\Gamma$ parametrized by arc length. Then the parallel curve $\Gamma_\epsilon$ at distance $\epsilon$ from $\Gamma$ has the parametric representation $s\mapsto (u(s),v(s))$ with $$u(s)=x(s)-\epsilon\dot y(s),\quad v(s)=y(s)+\epsilon \dot x(s).$$ Here $\epsilon$ may have either sign; I omit the discussion of this point. It follows from Frenet's formula that the line element of $\Gamma_\epsilon$ is given by $d\sigma=(1-\epsilon\kappa)ds$, where $\kappa$ denotes the curvature of $\Gamma$ and we assume $\epsilon\kappa(s)<1$ for all $s$. Integrating we obtain $$L(\Gamma_\epsilon)=L(\Gamma)-\epsilon\int_0^{L(\Gamma)}\kappa(s)ds = L(\Gamma)-2\pi \epsilon,$$ the latter equation following from the fact that the total curvature of a Jordan curve in the plane is $2\pi$ (up to sign).

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One possible formulation is to consider a smooth closed curve $\Gamma$ and to define the track as the Minkowski sum of $\Gamma$ and the circle of radius $r$. If $\Gamma$ is nice enough, so that the curvature at any point doesn't exceed $1/r$, then the outer boundary has length equal to the length of the inner boundary plus $4\pi r$. (That is, the perimeter of the Minkowski sum of two regions is equal to the sum of the individual perimeters in most nice cases.)

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If I understand you correctly then $\Gamma$ is the track's "center line", while the inner and outer boundaries of $\Gamma+B(0,r)$ are its "inner lane" and "outer lane" (under the curvature assumption). In that case, the track's width is $2r$ and the difference between the circumferences is $4\pi r.$ If $n$ is the outside unit normal vector to the inner track and the outer track is $d$ away along $n$ and both have no self-intersections then I think that no restriction on the curvature is needed and the difference in the circumferences is $2\pi d$ (cf Marco Gualtieri's answer). –  Victor Protsak Aug 16 '10 at 5:52
    
Yes, you are right of course. However the remark about the perimeter of the Minkowski sum not being additive in general is true, and my curvature assumption is precisely to avoid the intersection of outer and inner boundaries. –  Gjergji Zaimi Aug 16 '10 at 6:08
    
This is just an indication that, perhaps, Minkowski sum is not the right model (note also that $\Gamma$ is only implicitly present in the problem). –  Victor Protsak Aug 16 '10 at 6:50
    
I think it's pretty much the same as the other model. For example in the other model if you want the answer to remain $2\pi d$ then you need the assumption on the curvature for the outer boundary otherwise the inner boundary will be piecewise smooth and Gauss-Bonnet introduces some extra terms. The same assumption appears in the answer of Christian Blatter. –  Gjergji Zaimi Aug 16 '10 at 11:54

This may be enough to get you started.

In building a model railroad track, to make a complete circle, I need 12 curved pieces. Assuming I want a more elaborate layout, but want to keep things on the same level and allow no branches or crossovers ( i.e. I have only straight and curved pieces and no risers to lift track off my flat building surface), I need many curved pieces. If I start at one place, go counterclockwise around the track, and count pieces that curve left and pieces that curve right, I get 12 more pieces that curve left than curve right after I complete one loop and return to my starting place.

Is your scenario similar to building railroad track?

Gerhard "Ask Me About System Design" Paseman, 2010.08.15

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Allow self-intersections. Take the symmetric figure eight. That does the trick. And it has the bonus that each half is convex. This is what Kloeckner might have been trying to say.

( As a kid, we used to drive model slot cars on these. I imagine you can find such tracks raced by real cars. )

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Hi, Richard. Not sure about cars, but I enjoy the Dick Francis horse racing mystery/thrillers. He points out that all American tracks are left-hand turn, but British tracks vary and several are figure eight. There may be a distinction between flat and steeplechase racing in this regard, not sure. –  Will Jagy Aug 31 '10 at 1:45
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That's bad news if one of the horses gets half a lap ahead... –  Nate Eldredge Aug 31 '10 at 20:21

If you afford to use a bridge or a crossing, a figure eight track can at the same time: be (mostly in case of a bridge) horizontal, and have the same length for its two edges.

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