I want to create the "perfect road" for the following shaped polygon wheel.
I know it is something with $-\cosh(x)$
Another conclusion you can take when thinking about the problem is that the length of one catenary graph (between the starting point and the next catenary) is equivalent to the length of one side of the wheel.
Another thing is that the inner angles are as big as the angle that is created in the cusp when drawing the tangents through the point the catenaries intersect.
My first thought was just trying to use the formula
$L=∫_a^b\sqrt{1+(f'(x))^2}$
that is used to calculate the length and solve it for $b$ and then I would have the $x$-coordinate. But it doesn't work like that because I am not being aware of the slope that it has to have, due to the tangents. But I don't know how to apply this information into a formula.