Do wheels of a car always travel the same distance?

Question

Consider the left wheel and the right wheel of a car (say, rear wheels). The distances two wheels travel differ when they turn left or right. But, if the car starts traveling towards the north direction and ends traveling towards the same direction, and if the car doesn't rotate fully, then I suspect that the two distances are always the same.

Look at the picture. I'll use the angle (change of the orientation angle) instead of the curvature or the radius of curvature. The car (1) goes straight (angle = 0), (2) turns left (angle = $\pi/3$), (3) goes straight (angle = 0), (4) turns right (angle = $-\pi/3$). The two wheels travel the same distance at (1) and (3). The right one travels more in (2) and the left one travels more in (4). But in (2) or (4), the difference between two circular arcs depends only on the angle and the distance between the wheels, which is a constant. So (2) and (4) cancel out and two wheels travel exactly the same distance. That is,

$$\Delta d = d_1-d_2=r_1\theta-r_2\theta=(r_1-r_2)\theta$$

depends only on $r_1-r_2$ and $\theta$.

So the answer is yes in the trivial case shown in the image. But it's just the case when the curvature is (piecewisely) constant-valued and discontinuous, which is quite unnatural. In fact, the car moves smoothly and the curvature changes continuously and I cannot think of the blue and red curves as combinations of circular arcs and line segments.

How can I verify that it is still affirmative in the general case?

Answer 1

This is a nice question. Assuming the car makes no full rotation, you are correct in your assertion when the path is gently curved.

Let $\alpha(s)$ be the position of the center of the axle (assuming it's the rear axle and can't turn). Then $\alpha'(s)$ is orthogonal to the axle, and we can assume $\|\alpha'(s)\|=1$. Let $\alpha'(s)=T(s)$ and say $T(s),N(s)$ forms a positive-oriented orthonormal basis for each $s$. We have the equation $T'(s)=\kappa(s)N(s)$, where here $\kappa$ has a sign. The two wheels are at positions $\beta_\pm = \alpha \pm \lambda N$, where $2\lambda$ is the length of the axle. The basic curvature equations are $T'=\kappa N$, $N'=-\kappa T$, and so $$\beta_\pm' = \alpha'\pm\lambda N' = T\mp\lambda\kappa T = (1\mp\lambda\kappa)T.$$ The distances the wheels travel are given by$^*$ $$\int\|\beta_{\pm}'(s)\|ds = \int ds \mp \lambda \int\kappa\, ds.$$ The first term gives the distance traversed by $\alpha$. It is well-known (see, e.g., section 3 of chapter 1 of my differential geometry text, linked freely in my profile) that $\int \kappa\,ds$ gives the net change of the angle of the axle. Your assumption, then, is that this is $0$ (and no multiples of $2\pi$ are involved). Therefore, both $\beta_+$ and $\beta_-$ traverse the same distance.

$^*$I assumed here that $1\pm\lambda\kappa\ge 0$. However, if one quantity changes sign, then there will be a discrepancy, as one integrand will be $1+\lambda\kappa$ (say, with $\kappa$ large positive) and the other $\lambda\kappa-1$. So the difference of the integrals will amount to twice the distance traveled by the center of the axle on that interval. Of course, physically, this means that one of the wheels stops and reverses direction for a while. Is this allowed? It typically means there’s a cusp in the trajectory of that wheel.

Answer 2

A stronger and more precise statement about the relationship between the wheel motions and the rotation of the chassis is “The total rotation of the chassis is proportional to the difference between the forwards-minus-backwards motions of the two wheels”.

If we take the rolling velocity of the left and right wheels as $\dot{\alpha}_{1}$ and $\dot{\alpha}_{2}$, the wheel radius as $R$ and the length of the axle as $L$, the turning velocity of the axle is

$\dot{\theta} = (-\dot{\alpha}_{1} + \dot{\alpha}_{2})\frac{R}{L}$.

Factoring out the time portion of the derivative makes infinitesimal changes in $\theta$ depend directly on infinitesimal changes in the roll angles $\alpha$,

$d\theta = (-d\alpha_{1} +d\alpha_{2})\frac{R}{L}$,

such that we can express the derivative of the axle orientation with respect to the wheel roll angles as the one-form $A$

$\frac{\partial \theta}{\partial \alpha} = \overbrace{\begin{bmatrix} -1 & 1 \end{bmatrix}}^{A}\frac{R}{L}$.

This one-form is constant, such that its exterior derivative (i.e. curl) is zero. This means that orientation is conservative with respect to the roll angles of the wheels: once the vehicle has been placed on the ground, the total change in orientation is a function of the net angles that the wheels have rolled through, and independent of the intermediate roll angles they passed through to get there. Here, specifically, the orientation relative to the starting orientation is proportional to the difference between the net angle each wheel has rolled through,

$\Delta\theta = (-\Delta\alpha_{1} + \Delta\alpha_{2})\frac{R}{L}$

The net forward-minus-backward motion of each wheel (measured at its hub) is directly proportional to the net roll angle of the wheel around its axle,

$n = R\Delta\alpha$,

so the orientation can also be expressed in terms of the net forwards-minus-backwards motion of each wheel,

$\Delta\theta = (-n_{1} + n_{2})/L$.

This measure of rotation change of the chassis distinguishes between returning to the original orientation and making a full rotation, so we can say that for a system at $\Delta\theta=0$, the net forward-minus-backward motions of the two wheels are equal.

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As noted in https://math.stackexchange.com/a/5046155/646444, this is not quite the same as asking if the wheels have both traveled the same distance — if the curve becomes tight enough that the radius of curvature is less than $L/2$, one of the wheels will be moving backwards during that time. If this backwards motion counts as positive term for “distance traveled”, then “distance traveled” will not be the same as “net forwards minus backwards motion”.

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The underlying math here is that $A$ defines a “holonomic constraint”, i.e., a set of relationships between derivatives of two subsets of a system’s configuration variables that is “integrable” into a functional relationship.

Note that this property does not hold for translational motion: Knowing the net roll angles of the wheels does not tell us how far the car has moved, and we would need the full history of the roll angles to reconstruct this motion. This gets into the idea of “nonholonomic constraints” that relate derivatives of subsets of a system’s variables, but cannot be turned into a functional relationship. (This topic requires a bit more involved discussion than seems appropriate here; the short form is that establishing that the exterior derivative of $A$ was zero was a special case of a more general check to see if a system is holonomic or nonholonomic.)

Answer 3

Yes you are 100% correct for the simple reason that circumference is a linear multiple of radius. Therefore even if the centre of your vehicle makes a wide radius $R$ turn in one direction of $n$ radians followed by a narrow radius $r$ turn in another direction of $n$ radians, in a vehicle whose wheels are $2w$ apart, the difference in the distances travelled by the two wheels is always $((R+w)n-Rn)-((r+w)n-rn)=wn-wn=0$