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Let $\alpha:[0,1]\to \mathbb R^2$ be a smooth closed curve parameterized by the arc length. We will think of $\alpha$ like a back track of the wheel of a bicycle. If we suppose that the distance between the two wheels is $1$ then we can describe the front track by

$$\tau(t)=\alpha(t)+\alpha'(t)\;.$$

Suppose we know the two (back and front) trace of a bicycle. Can you determine the orientation of the curves? For example if $\alpha$ was a circle the answer is no.

More precisely the question is:

Is there a smooth closed curve parameterized by the arc length $\alpha$ such that

$$\tau([0,1])=\gamma([0,1])$$

where $\gamma(t)=\alpha(1-t)-\alpha'(1-t)$?

If trace of $\alpha$ is a circle we have $\tau([0,1])=\gamma([0,1])$. Is there another?

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@RahulNarain ok! –  user52188 Jan 10 '13 at 1:18
    
There's something wrong with your definition of $\gamma$, since you're applying it to $[0,1]$ but it's only defined on $[-1,0]$ (since $\alpha$ is only defined on $[0,1]$). Don't you mean simply $\gamma(t)=\alpha(t)-\alpha'(t)$? –  joriki Jan 10 '13 at 1:38
    
@joriki thank you –  user52188 Jan 10 '13 at 1:54
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See Which way did the bicycle go? –  Michael E2 Jan 10 '13 at 2:58
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See Exercise 27 on pp. 22-23 of my differential geometry notes math.uga.edu/~shifrin/ShifrinDiffGeo.pdf ... In general the differential equation cannot be solved explicitly and it is highly unlikely that $\tau$ will be a closed curve. –  Ted Shifrin May 6 '13 at 18:43
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5 Answers 5

Yes, Franz Wegner constructed pairs of smooth closed curves that are not circles and can serve as pairs of bicycle tracks traversed in either direction. They can be expressed analytically in terms of Weierstrass's $\sigma$ and $\zeta$ functions. Interestingly enough such curves also describes shapes that can float in any position, and trajectories of electrons moving in a parabolic magnetic field.

Short description and a picture are here http://www.tphys.uni-heidelberg.de/~wegner/Fl2mvs/Movies.html#animations, mathematical details and more pictures here http://arxiv.org/pdf/physics/0701241v3.pdf.

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To me it looks like this. The tracks are two concentric circles. Back wheel turns in a circle radius $b$, frame length is constant = $a$, (instead of 1) tangent to this circle. Front wheel turns on a circle radius $\sqrt{a^2 + b^2}$. You turned handle bar by angle $\alpha = \arctan \frac{a}{b}$. If $\alpha = 90\,^{\circ}$, $b=0$, an extreme special case when back wheel does not move on ground.

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After the which way did bicycle go book, there has been some systematic development of theory related to the bicycle problem. Much of that is either done or cited in papers by Tabachnikov and his coauthors, available online:

http://arxiv.org/find/all/1/all:+AND+bicycle+tracks/0/1/0/all/0/1

http://arxiv.org/abs/math/0405445

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Take a figure eight curve, made of two tangent circles, not necesarily of the same size.

Or an arbitrary chain of tangent circles.

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When you switch from one circle to another, the front wheel takes a very different path depending on the direction of travel. So this only yields a solution if the reversed front wheel's path of the closed curve does not need to match the front wheel's path of the reversed closed curve. (And even then you need to go around the circles multiple times to obscure the direction of travel.) But this is indeed an interesting observation and does make it clear that the question's answer can depend in delicate ways on its phrasing! –  Matt Jan 13 at 13:14
    
And of course if there are non-circular solutions, the same chaining operation can be applied to them as well (with the same caveats). –  Matt Jan 13 at 13:16
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Would you take an (acrobatic) exercise of riding the bike backwards with the front wheel turned back? If you succeed, and managed to keep the front wheel going along a straight line, then the rear wheel would go along a tractrix – despite which direction you chose to ride.

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