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(asked by JST on the Q&A board at JMM)

Can two bicycles of different lengths leave the same set of tracks (aside from a straight line)?

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  • $\begingroup$ What assumptions are we making on the paths of the wheels? Continuous? Smooth? $\endgroup$ Commented Jan 16, 2010 at 6:03
  • $\begingroup$ <a href="mathdl.maa.org/images/upload_library/22/Polya/… a Bicycle Create a Unicycle Track</a> says it's possible if you let one length be 0. $\endgroup$ Commented Jan 16, 2010 at 6:27

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The answer, I believe, is no in the case of finite length tracks and yes in the case of infinite length tracks. I will show this in the (probably not very) simplified case of assuming that we can idenitify a back wheel track and a front wheel track from the tracks left.

Finite length: We will show that if they have the same back wheel path then the lengths of their front wheel paths are different. Let $\gamma$ be the arclength parameterization of the back wheel path. Then, for a bicycle of length $l$, the front wheel path is given by $\gamma + l\cdot\gamma'$ and so we can calculate its length by:

$\int |\gamma' + l\cdot\gamma''|$

Since we chose the arclength parameterization, $\gamma'$ and $\gamma''$ are orthogonal so this becomes

$\int \sqrt{|\gamma'| + l^2\cdot|\gamma''|^2}$

And clearly this quantity is increasing with $l$ unless $\gamma''$ is identically zero, which only occurs for a straight line. Thus, two different lengths of bicycle will leave front wheel tracks of different lengths, which therefore cannot be the same (note - I just realized this only says that the distance traveled by the two front wheels is different, but due to overlap perhaps the tracks could still have the same length. I think we should be able to discount this possibility, but it is not immediately obvious to me exactly how).

Infinite length: We will provide a general construction. Start by fixing two bikes of different lengths, call them bike 1 and bike 2 of lengths $l_1$ and $l_2$. Pick some finite length curve, call it $\gamma_1$ as the first back wheel curve, and let $\Gamma_1$ be the corresponding front wheel curve for bike 1. Then we will take $\gamma_2$ to be $\gamma_1$ followed by a backwheel curve starting at the end of $\gamma_1$ that if ridden by bike 2 will end up with the front wheel of bike 2 tracing over all of $\Gamma_1$ (of course it will also trace over much more). Then $\Gamma_2$ will be the front wheel curve associated to bike 2 with back wheel curve $\gamma_2$. Then we will let $\gamma_3$ be $\gamma_2$ followed by a backwheel curve starting at the end of $\gamma_2$ that if ridden by bike 1 will end up with the front wheel of bike 1 tracing over all of $\Gamma_2$ and $\Gamma_3$ will be the front wheel curve associated to bike 1 with back wheel curve $\gamma_3$. We repeat this process, alternating back and forth between bikes each time adding to the backwheel path what we need to cover what the previous bike covered with its front wheel tracks. This algorithm can be used to produce an infinite length backwheel path that will give the same frontwheel paths when ridden by either bike (we can consider the front wheel path abstractly as just the extension along the tangent lines of the appropriate bicycle length)

For an interesting account of bicycle curves (including some exposition on the model I am implicitly using), I recommend the paper by Levi and Tabachnikov, "On bicycle tire tracks geometry, hatchet planimeter, Menzin's conjecture and oscillation of unicycle tracks" available at http://arxiv.org/abs/0801.4396

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Perhaps circular tracks should also be excluded?

Gerhard "Ask Me About System Design" Paseman, 2010.01.15

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    $\begingroup$ I don't see that concentric circular tracks would make the bicycle's length ambiguous. The length would be half of the length of a chord tangent to the inner circle. $\endgroup$ Commented Jan 16, 2010 at 8:06
  • $\begingroup$ Why would there be an inner circle? $\endgroup$ Commented Jan 16, 2010 at 17:19
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    $\begingroup$ See the animation here to get a feel for it: en.wikipedia.org/wiki/File:Bike_weaving.gif $\endgroup$
    – TonyK
    Commented Jan 16, 2010 at 19:23
  • $\begingroup$ I am a bit unsure. Do we have the assumption that two wheels must be of same size? $\endgroup$
    – abcdxyz
    Commented Jan 17, 2010 at 15:23
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I would answer no, based on a very simple argument. (But in contrast to the Sean Howe's answer I restrict my argument to finite curves.) Consider a bicycle ridden on a circle. Its radius $r_1$ is defined by the angle of the frontwheel with respect to the bicycle. The rearwheel will draw a smaller circle with radius $r_2$. The difference of radii will depend on the distance $d$ of the axes such that $d^2 = r_1^2 - r_2^2$.

Since every curve can be considered as an approximation of a circle, this implies that different distances of the axes will supply different curves for every given $r_1$, except when the curve is a straight line, both radii being infinite.

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  • $\begingroup$ Does the fact that for two different bicycles different points on the front track match the same point on the rear track lead to a problem with this argument? $\endgroup$ Commented May 5, 2013 at 17:31
  • $\begingroup$ @Yoav: I don't think so. You are right, it is possible, for instance, to have a circle with radius $r_2 = 0$ But as soon as $r_1$ is changed also $r_2$ will follow by the given Pythagorean equation such that never limit- or start-effects can take pace. $\endgroup$
    – user112109
    Commented May 5, 2013 at 17:40
  • $\begingroup$ The relationship between inner and outer circles was mentioned earlier. However, the main question is what happens if the curves are not circles. I don't understand the sense in which "every curve can be considered as an approximation of a circle." $\endgroup$ Commented May 6, 2013 at 0:13
  • $\begingroup$ @Douglas: Every piece of the tracks can be considered to be a piece of a circle (depending only on the momentary angle of the frontwheel). $\endgroup$
    – user112109
    Commented May 6, 2013 at 10:52
  • $\begingroup$ As the front angle oscillates between $-\pi/2$ to $\pi/2$ as you trace out something which looks like a figure-$8$, which circle is this approximating? $\endgroup$ Commented May 6, 2013 at 17:46

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