Road Shape for Square Wheels 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?
 A: 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.
