Square Wheels on the Right Road, math that surprises you lesson

A square wheel on a flat road lifts its axle every time it tips over a corner, so a ride on it would bump up and down. Rounding the corners would cure that, but there is another way, which keeps the square exactly as it is and changes the road instead. This page finds the road on which a square rolls without slipping while its center travels along a perfectly level line, and shows why each bump of that road must be a piece of one particular curve.

Key terms

Square wheel

Definition A square wheel is a rigid square of side $2a$ with its axle at its center. On a flat road it pivots about one corner at a time, so its center rises from height $a$ , when a side lies flat, to $a\sqrt2$ , when a corner points straight down, and falls again, a bounce of $a(\sqrt2 - 1) \approx 0.414a$ in every quarter turn. The square wheel problem asks for a road on which the square rolls without slipping, always touching the road and never cutting into it, while its center moves along a horizontal line at height $H$ . Such a road exists. It is a row of equal arches. With $L = 2a\ln(1 + \sqrt2) \approx 1.763a$ , the arch centered at $x = kL$ , for each integer $k$ , is

y = H - a\cosh\frac{x - kL}{a}

for $kL - L/2 \le x \le kL + L/2$ , where $\cosh$ is the hyperbolic cosine. Each arch is an upside-down catenary whose length is $2a$ , the side of the square. When the center is above the top of an arch, the bottom side lies flat on it and touches it at the middle of the side. Neighboring arches meet in valleys with a right angle, where a corner of the square sits while the next side takes over. The point of contact is always directly below the center. The axle stays level, but the square does not turn at a constant rate. If the center moves forward at a steady speed $v$ , then while it is above the point $x$ of the arch centered at $kL$ , the square turns at the angular speed

\omega = \frac{v}{a\cosh\bigl((x - kL)/a\bigr)},

which is $v/a$ over the top of an arch and $v/(a\sqrt2)$ in a valley. All of this is ideal geometry, a rigid square on a rigid road with no slipping. It says nothing about the vibration, impacts or comfort of a real vehicle.

In plain words The road bulges up exactly where the square would sink and dips exactly where its corners would lift it, so the two effects cancel and the center glides along a straight line. The square still turns faster over the tops of the arches than in the valleys.

Example Take a square of side $2$ , so $a = 1$ , with its axle at height $H = 1$ . The tops of the arches are then at height $0$ , the arches are about $1.763$ wide, and the valleys are $\sqrt2 - 1 \approx 0.414$ lower than the tops, the same amount as the bounce on a flat road. At the ends of an arch the road has slopes $1$ and $-1$ , so two neighbors meet at a right angle and the corner of the square fits the valley exactly. If the axle moves at $1$ unit per second, the square turns at $1$ radian per second over each top and at $1/\sqrt2 \approx 0.707$ radians per second in each valley. Each arch takes about $1.763$ seconds and turns the square a quarter turn, $\pi/2$ radians, an average of about $0.891$ radians per second.

Hyperbolic cosine

Definition The hyperbolic cosine and the hyperbolic sine are the functions

\begin{aligned} \cosh x &= \frac{e^{x} + e^{-x}}{2}, \\[0.4em] \sinh x &= \frac{e^{x} - e^{-x}}{2}, \end{aligned}

defined for every real number $x$ . The hyperbolic cosine is even, its smallest value is $\cosh 0 = 1$ , and it grows without bound on both sides. The two functions satisfy $\cosh^2 x - \sinh^2 x = 1$ , and their derivatives are $(\cosh x)' = \sinh x$ and $(\sinh x)' = \cosh x$ . For each real number $t$ the equation $\sinh x = t$ has exactly one solution,

x = \ln\left(t + \sqrt{t^2 + 1}\right),

written $\operatorname{arsinh} t$ . The name comes from the hyperbola $X^2 - Y^2 = 1$ , on which the point $(\cosh x, \sinh x)$ lies, just as $(\cos x, \sin x)$ lies on the circle $X^2 + Y^2 = 1$ .

In plain words The hyperbolic cosine is the average of a growing exponential and a shrinking one. Its graph is a U shape with a rounded bottom at height $1$ that climbs faster and faster away from $0$ .

Example For $a = 1$ , the arch of the square wheel road with its top at $x = 0$ ends at $x = \pm\ln(1 + \sqrt2) \approx \pm 0.881$ . At the right end $e^{x} = 1 + \sqrt2$ and $e^{-x} = 1/(1 + \sqrt2) = \sqrt2 - 1$ , so

\begin{aligned} \cosh x &= \sqrt2, \\[0.4em] \sinh x &= 1, \end{aligned}

and indeed $\cosh^2 x - \sinh^2 x = 2 - 1 = 1$ . The value $\cosh x = \sqrt2$ puts the end of the arch $\sqrt2$ below the axle, where the corner reaches, and $\sinh x = 1$ gives the road its slope of $1$ in size there.

Catenary

Definition A catenary is the shape of an ideal chain, perfectly flexible, unstretchable and of the same weight per unit length all along, hanging at rest in uniform gravity from two supports. With its lowest point on the $y$ -axis, it is the graph of

y = c\cosh\frac{x}{c} + d,

where the constant $c > 0$ equals the horizontal pull in the chain divided by its weight per unit length, and $d$ shifts the curve up or down. Turned upside down, the same curve is an inverted catenary, the shape of the arches in the square wheel road. That road needs $c = a$ , half the side of the square, and no other value of $c$ fits that square. The equation of the hanging chain was found in 1691 by Leibniz, Huygens and Johann Bernoulli.

In plain words Hold a necklace by its two ends and let it hang. The curve it makes is a catenary. It looks rather like a parabola or an arc of a circle, but it is neither, and the square wheel needs this exact curve.

Example Here is why the inverted catenary fits the square. Take a road point on the arch centered at $kL$ , at horizontal position $x$ , write $u = (x - kL)/a$ , and put the center of the square at height $H$ directly above that point. The vertical gap from the center down to the road is $a\cosh u$ . The road there has slope $-\sinh u$ , so its tangent line makes an angle $\alpha$ with the horizontal for which $\tan^2\alpha = \sinh^2 u$ . Then $1 + \tan^2\alpha = \cosh^2 u$ and $\cos\alpha = 1/\cosh u$ . The perpendicular distance from the center to the tangent line of the road is the vertical gap times $\cos\alpha$ ,

a\cosh u \cdot \frac{1}{\cosh u} = a.

That is exactly the distance from the center of the square to each of its sides, so a side lies along the tangent line and touches the road directly below the center. The arch curves downward, so it stays below that tangent line and the square never cuts into it. A check of the neighboring arches shows that the square clears them too. Rolling is what puts the center straight above the contact point. Without slipping, the point of contact is momentarily at rest, so the square is turning about it, and the center can move straight forward only if it is straight above that point.

Arc length

Definition The arc length of a curve is the limit of the total length of a chain of straight segments whose ends and corners lie on the curve, in order, as the longest segment shrinks to zero, when that limit exists. For the graph of $y = f(x)$ with $p \le x \le q$ , where $f$ has a continuous derivative, the limit exists and equals

s = \int_{p}^{q} \sqrt{1 + f'(x)^2}\, dx .

A wheel rolls without slipping on a road when the point of the wheel touching the road is momentarily at rest. Then the length of road passed over always equals the length of the wheel’s edge that has touched it.

In plain words Arc length is the distance an ant would walk along a curve, not the distance straight across beneath it. For a wheel that does not slip, each bit of its edge meets a bit of road of the same length.

Example For the arch $y = H - a\cosh(x/a)$ , the slope is $f'(x) = -\sinh(x/a)$ and $\sqrt{1 + \sinh^2(x/a)} = \cosh(x/a)$ . With $x_e = a\ln(1 + \sqrt2)$ for the end of the arch,

\begin{aligned} s &= \int_{-x_e}^{x_e} \cosh\frac{x}{a}\, dx \\[0.4em] &= 2a\sinh\frac{x_e}{a} = 2a, \end{aligned}

since $\sinh(\ln(1 + \sqrt2)) = 1$ . The arch is $2a$ long, the length of a side, although it is only about $1.763a$ wide. The match holds at every moment, not only for the whole arch. From the top of the arch to the point at horizontal distance $x$ the road has length $a\sinh(x/a)$ . At that moment the contact point is $a\cosh(x/a)$ below the center and the side is $a$ from the center, so by the Pythagorean theorem the contact is $a\sinh(x/a)$ from the middle of the side, the same length. The same no-slip condition is at the heart of Aristotle’s Wheel.

Check yourself

Change the size. Take a square of side $4$ . How much would its center bounce on a flat road, and how wide, how long and how deep are the arches of its matching road?

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Now $a = 2$ . On a flat road the center moves between heights $2$ and $2\sqrt2 \approx 2.828$ , a bounce of $2(\sqrt2 - 1) \approx 0.828$ . The matching road has arches

y = H - 2\cosh\frac{x - kL}{2}

with $L = 4\ln(1 + \sqrt2) \approx 3.525$ . Each arch is $2 \cdot 2 = 4$ long, the side of the square, and its valleys are $0.828$ below its tops. Every length has doubled. The road built for the square of side $2$ will not do, because its arches are only $2$ long and fit only sides of length $2$ .

Try a near miss. For the square of side $2$ , replace each arch by an arc of a circle through the same top and the same two valley points, $0.881$ to each side of the top and $0.414$ lower. Does the circular road work?

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It does not, although it looks almost the same. Write $x_e = \ln(1 + \sqrt2) \approx 0.881$ and $h = \sqrt2 - 1 \approx 0.414$ . A circle of radius $R$ through the top and a valley point satisfies $x_e^2 + (R - h)^2 = R^2$ , so

R = \frac{x_e^2 + h^2}{2h} \approx 1.145.

The arc is $2R\arcsin(x_e/R) \approx 2.012$ long instead of $2$ , so a side of the square cannot roll over it exactly. Its slope at a valley point is $x_e/\sqrt{R^2 - x_e^2} \approx 1.206$ in size, about $50.3^\circ$ instead of $45^\circ$ , so neighboring arcs meet at about $79.3^\circ$ and the valley is too narrow for the $90^\circ$ corner. Directly above the end of the arc, a center at height $1$ is only about $0.903$ from the arc’s tangent line there, instead of $1$ . A level axle would push the side into the road, so the square would have to rise, slip or cut in. Only the catenary passes every test.

Extend the pattern. Build the road for a regular hexagon of side $2$ . How wide and how long is each arch, how deep are its valleys, and what angle do neighboring arches make?

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The distance from the center of the hexagon to each side is $\sqrt3$ , so it plays the role of $a$ and the arches are

y = H - \sqrt3\cosh\frac{x - kL}{\sqrt3}.

The hexagon turns $60^\circ$ from one side to the next, so the side in contact tilts by up to $30^\circ$ on each side of the top. That side lies along the road, so the road’s slope, $\sinh u$ in size with $u = (x - kL)/\sqrt3$ , is the tangent of the tilt. The arch therefore ends where $\sinh u = \tan 30^\circ = 1/\sqrt3$ , at $u = \ln\sqrt3$ , where $\cosh u = 2/\sqrt3$ . Each arch is $2\sqrt3\ln\sqrt3 = \sqrt3\ln 3 \approx 1.903$ wide, and its length is $2\sqrt3\sinh u = 2$ , the side of the hexagon. The valleys are $\sqrt3(2/\sqrt3 - 1) = 2 - \sqrt3 \approx 0.268$ below the tops, which is again the bounce of the same hexagon on a flat road. The road leaves each valley at $30^\circ$ above the horizontal on each side, so neighboring arches meet at $120^\circ$ , the angle of the hexagon’s corner. The same recipe works for regular polygons with more sides. It fails for an equilateral triangle. Its arches have the right lengths and its valleys the right $60^\circ$ angle, but as the triangle rolls, its leading corner swings below the surface of the next arch, so a rigid triangle would cut into the road.