### Which Way did the Bicycle Go?

“A bicycle, certainly, but not the bicycle,” said Holmes.

In Conan-Doyle’s short story The Adventure of the Priory School  Sherlock Holmes solved a mystery by deducing the direction of travel of a bicycle. His logic has been minutely examined in many studies, and it seems that in this case his reasoning fell below its normal level of brilliance. As front wheel moves along the positive x-axis the back wheel, initially at (0,a), follows a tractrix curve (see below).

When the path of the back tyre is known

We easily derive equations for the path of the front tyre when the path of the back tyre is known. Denote the position of the point of contact of the front tyre with the ground by ${(x_\mathrm{F},y_\mathrm{F})}$ and that of the back tyre by ${(x_\mathrm{B},y_\mathrm{B})}$. We define $\displaystyle \Delta x = x_\mathrm{F} - x_\mathrm{B} \qquad\mbox{and}\qquad \Delta y = y_\mathrm{F} - y_\mathrm{B} \,.$

We assume that the distance between the two points of contact — the wheel-base ${a}$ — is constant so that $\displaystyle \Delta x^2 + \Delta y^2 = a^2 \,.$

The point ${(x_\mathrm{F},y_\mathrm{F})}$ is always in the plane of the back wheel, so it lies on the tangent to the path of the back wheel, and it is always at a constant distance ${a}$ from ${(x_\mathrm{B},y_\mathrm{B})}$. The slope of the path of ${(x_\mathrm{B},y_\mathrm{B})}$ is ${m=\mathrm{d}y_\mathrm{B}/\mathrm{d}x_\mathrm{B}}$. Using these simple geometric facts, we can write equations for ${(x_\mathrm{F},y_\mathrm{F})}$: $\displaystyle x_\mathrm{F} = x_\mathrm{B} + \frac{a}{\sqrt{1+m^2}} \,,\qquad y_\mathrm{F} = y_\mathrm {B} + \frac{ma}{\sqrt{1+m^2}} \,. \ \ \ \ \ \ \ \ \ (1)$

Knowing the path of the back wheel in the form ${y_\mathrm{B}=y_\mathrm{B}(x_\mathrm{B})}$, it is a simple matter to compute ${(x_\mathrm{F},y_\mathrm{F})}$.

For small ${m}$ we can approximate equations (1) by the following: $\displaystyle x_\mathrm{F} \approx x_\mathrm{B} + {a} \,,\qquad y_\mathrm{F} \approx y_\mathrm{B} + {ma} \,.$

Now suppose the back wheel follows a sinusoidal track, ${y_\mathrm{B}(x_\mathrm{B})=A \sin k x_\mathrm{B}}$.
Then the front point of contact is $\displaystyle x_\mathrm{F} \approx x_\mathrm{B} + {a} \,,\qquad y_\mathrm{F} \approx C \sin(k x_\mathrm{B}+\gamma)$

where ${C=A\sqrt{1+(ka)^2}A}$ and ${\gamma=\arctan{ka}}$. Thus, the front wheel also follows a sinusoidal track, but one with a larger amplitude.

When the path of the front tyre is known

If the path ${(x_\mathrm{F},y_\mathrm{F})}$ is given, the path of the back wheel can be computed, but the procedure is much more complex, involving the solution of coupled nonlinear differential equations. For details see Dunbar, et al. (2001). Fig. 1. Tracks of the front wheel (solid red) and back wheel (dashed blue) of bicycle. The front wheel follows a simple sinusoidal path. The back track soon becomes sinusoidal with smaller amplitude.

Stan Wagon (2010) gives some Mathematica code to compute the back track from the front one. Using this, the solution when the front point of contact follows a sinusoidal curve (solid red line) is shown in Fig.1. The back wheel (dashed blue line) quickly adjusts to this and asymptotes to a sinusoid with slightly smaller amplitude.  Fig. 2. Tracks of the front wheel (solid red) and back wheel (dashed blue) of bicycle. The direction of travel may be deduced from the requirement that the tangent from the back track must intersect the front track at a fixed distance a. One bike goes east, one goes west. Which is which?

More general cases are shown in Fig. 2. The front wheel now follows a more sinuous curve (solid red line). The path of the back wheel (dashed blue line) is also more complex. Note that the direction of travel can be determined by inspection. The tangent from the blue curve must intersect the red curve at a fixed distance ${a}$. For Fig. 2, one panel shows a track going left, the other shows one going to the right. Which is which? Can you out-smart the great Sherlock Holmes? (Answer at the bottom of this post).

Linear Track and Closed Loop Tracks

Suppose the front wheel is initially at ${90^\circ}$ to the back wheel and that it rolls in a straight line along the positive ${x}$-axis. Then the back wheel will be pulled along, always pointing towards the front contact point and at fixed distance ${a}$ from it. Its movement will be described by the differential equation $\displaystyle \frac{\mathrm{d}y}{dx} = -\frac{y}{\sqrt{a^2-y^2}} \,.$

But this is the equation for a tractrix (see Figure at the head of this post), a special curve with equation $\displaystyle x = a\,\mathrm{sech}^{-1}\left(\frac{y}{a}\right)-\sqrt{a^2-y^2} \,.$

or, in parametric form, $\displaystyle x(t) = a(t-\tanh t) \,,\qquad y(t) = a\,\mathrm{sech}\,t.$

Now consider the front wheel at a fixed angle less than ${90^\circ}$. It will describe a circle, and the back wheel will follow on a circle of smaller radius (Fig. 3, left panel). The area between the two circles is easily calculated at ${\pi a^2}$. It is independent of the radii of the two circles, so a pair of smaller circular tracks will enclose the same area (Fig. 3, centre panel).

A Challenge

Do the tracks have to be circular for the area between them to remain fixed? Suppose both wheels trace out simple closed curves with winding number 1 (rotating through one anticlockwise turn). An example is shown in Fig. 3, right panel. In general, when the tracks cross, we must count as positive areas where the back wheel is within the front track, and as negative those where it is outside.

The hypothesis is that the nett area is ${\pi a^2}$, independent of the details of the track. Can you prove this? Fig. 3. Left: Tracks of front wheel (red) and back wheel (blue) for bicycle on a circular track. Centre: circular tracks with smaller radii. The area between the circles is still pi a^2. Right: Schematic diagram of general case, where both wheels execute one full closed loop.

Meanwhile, back at the Ranch

In The Adventure of the Priory School, Holmes reasoned as follows:

No, no, my dear Watson. The more deeply sunk impression is, of course, the hind wheel, upon which the weight rests. You perceive several places where it has passed across and obliterated the more shallow mark of the front one. It was undoubtedly heading away from the school.

Of course, the back wheel will obliterate the front track irrespective of the direction of travel, so this says nothing about which way the bicycle was going. Even Holmes nods. ${\bullet}$ Dunbar, Steven R., Reinier JC Bosman, and Sander EM Nooij, 2001: The track of a bicycle back tire. Math. Mag., 74(4), 273-287. ${\bullet}$ J. Konhauser, D. Velleman and Stan Wagon, 1996: Which Way Did the Bicycle Go? Math. Assoc. Amer., 235pp. ${\bullet}$ Wagon, Stan, 2010: Mathematica in Action, 3rd Edn. Springer, 578pp. ISBN: 9-780-387-75366-9.