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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?

Brian M. Scott
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  • @RahulNarain ok! –  Jan 10 '13 at 01:18
  • 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 01:38
  • Since you're only interested in $\gamma([0,1])$, it doesn't matter whether the arguments are $t$ or $1-t$, so the $1-t$ seems an unnecessary complication. (By the way, the closing parentheses are missing in both instances of $\tau([0,1])$.) – joriki Jan 10 '13 at 01:55
  • @joriki You are right. For question no matter if you take $\alpha-\alpha'$ or like there. But I wanted to give emphasis that the bike is going the other way. The title come from this. Therefore I took the reverse curve. Thank you I will edit the parentheses. –  Jan 10 '13 at 02:07
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    See Which way did the bicycle go? – Michael E2 Jan 10 '13 at 02:58
  • @MichaelE2 nice!! Where I find the book?Do you know? The book has the answer? –  Jan 10 '13 at 03:41
  • @EdgarMatias The book is in print. If you don't like the link I gave, try this one: MAA I believe they do solve it. It's got 190 other problems, too. – Michael E2 Jan 10 '13 at 04:31
  • @MichaelE2 ok! But I think the solution of the question above is not there! I will see. –  Jan 10 '13 at 04:37
  • Well, a straight line ... – Neal Jan 10 '13 at 04:40
  • @EdgarMatias Sorry if the solution is not there. It's been a long time since I looked at it, and my memory may be mistaken. I don't seem to have a copy. – Michael E2 Jan 10 '13 at 14:13
  • About the edits from "parameterize" to "parametrize" and back: Both are in dictionaries; both have hundreds of thousands of Google hits; both could be left to stand. – joriki Jan 26 '13 at 09:03
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    See Exercise 27 on pp. 22-23 of my differential geometry notes http://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 06 '13 at 18:43
  • @TedShifrin If the back track curve is closed, as the present problem assumes, than so is the resulting front track curve. In the exercise of your book a closed {\em front} track curve is given. In this case it indeed most associated back track curves (there are many) will not be closed. – Gil Bor Jan 08 '14 at 08:22

5 Answers5

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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.

Conifold
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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.

Daniel R
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Narasimham
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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

zyx
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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.

Gil Bor
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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 '14 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 '14 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.

CiaPan
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