Road Shape for Square Wheels

Question

Let's say you have a bike with square wheels of side a. In order for a smooth ride, there must be these bumps in the road. Is there a formula for the area of each bump using a?

Answer 1

Let

• $a = 2b$.
• $P : (x_P, y_P)$ be the contact point between the wheel and the bump below.
• $Q : (x_Q, y_Q)$ be the center of the wheel.

Choose a coordinate system so that $Q$ is constrained to move on the line $y=0$.
This means $y_Q = 0$ identically.

Parameterize $P$ by arc-length $s$. We will assume the wheel is rolling from left to right without sliding. Furthermore, at $s = 0$, $Q$ is located at origin and lies directly above $P$. More precisely,

$$(x_Q(0), y_Q(0)) = 0,\quad\text{ and }\quad \begin{cases} ( x_P(0), y_P(0) ) = (0,-b)\\ ( x'_P(0),y'_P(0)) = (1,0) \end{cases}$$

After we roll the wheel for a distance $s$, $P$ moved to $(x_P(s), y_P(s))$. The tangent vector and upward normal vector of the bump at that point equals to $t = (x'_P(s),y'_P(s))$ and $n = (-y'_P(s), x'_P(s))$.
Since the wheel is rolling without sliding with respect to the bump. One can reach $Q$ from $P$ by a move along direction $n$ for a distance $b$ followed by a move along direction $-t$ for distance $s$.

This leads to following ODE

$$0 = y_Q(s) = y_P(s) - s y'_P(s) + b x'_P(s)$$

Together with the constraint $x'_P(s)^2 + y_P'(s)^2 = 1$ and given initial conditions, one find:

$$ \begin{cases} x_P(s) = b\sinh^{-1}\frac{s}{b}\\ y_P(s) = -\sqrt{s^2+b^2} \end{cases} \quad \iff \quad \begin{cases} x_P(s) = bt\\ y_P(s) = -b\cosh t \end{cases} \quad\text{ where }\quad t = \sinh^{-1}\frac{s}{b} $$ Let $t_0 = \sinh^{-1}(1) = \log(1+\sqrt{2})$.

The two endpoints of the bump corresponds to $s = \pm b \iff t = \pm t_0$.
At those points, $y_P(s)$ reaches its lowest value $-\sqrt{2}b$.

The area under the bump (but above this $y$) is given by the formula:

$$\begin{align}\int_{-b}^b (y_P(s)+\sqrt{2}b)x'_P(s) ds &= b^2 \int_{-t_0}^{t_0} (\sqrt{2} - \cosh t) dt = 2b^2 \left(\sqrt{2}t_0 - \sinh t_0\right)\\ &= \frac{a^2}{2} \left(\sqrt{2}\log(1+\sqrt{2}) - 1\right) \end{align} $$ Update

About how to solve the ODE, differentiate both equations by $s$, one get $$ \begin{cases} y - s y' + b x' &= 0\\ y'^2 + x'^2 &= 1 \end{cases} \quad\stackrel{\frac{d}{ds}}{\Longrightarrow}\quad \begin{cases} -sy'' + bx'' &= 0\\ y'y'' + x'x'' &= 0 \end{cases} $$ Comparing coefficients for $x''$ and $y''$, we find there is a function $\lambda(s)$ such that $$ (x',y') = ( b\lambda, -s\lambda ) $$

Since we want $x' > 0$, the normalization condition $x'^2 + y'^2 = 1$ fixes $\lambda$ to $\frac{1}{\sqrt{s^2+b^2}}$. Integrate above equation by $s$ will give us the desired solution.