# Minimum Time Through Curve

Consider the following conditions:

• A level road with the path $\vec{r} = \left<x(s),\ y(s)\right>$

• A max speed of $v_{max}$

• A vehicle with the mass $m$

• Tires with a static friction coefficient of $\mu_s$

• Constant gravitational acceleration magnitude of $g$

The road is traveled from $s_0$ to $s_1$ at the maximum legal speed and without the tires loosing traction. In terms of the values given, what is the minimum time, $t$, of travel through the road segment?

I approached the problem by getting velocity in terms of radius:

$$F_c=F_f=\frac{m\left|\vec{v}\right|^2}{R}=F_g\mu_s=mg\mu_s$$

$$\left|\vec{v}\right|=\sqrt{R\mu_sg}$$

Instantaneous radius, $R(s)$, can be found as the unsigned inverse of curvature:

$$R(s)=\left|{\frac{1}{\kappa(s)}}\right|=\frac{\left(x'^2+y'^2\right)^{3/2}}{x'y''-y'x''}$$

$$\left|\vec{v}(s)\right|=\frac{d\left|\vec{x}(s)\right|}{dt}=\sqrt{R(s)\mu_sg}$$

But at this point I am confused how to proceed. I tried doing a couple things that I was unsure if where true, but most only seemed to complicate the problem, such as this one.

$$\frac{d\left|\vec{x}(s)\right|}{dt}=\frac{d\sqrt{x(s)^2+y(s)^2}}{dt}$$

$$\int dt=\int\frac{d\sqrt{x(s)^2+y(s)^2}}{R(s)\mu_sg}$$

$$t(s)=?$$

So I a unsure whether I am one step away from the answer or taking an entirely wrong direction. Any help will be appreciated.

Note: I haven't completed multi-variable calculus yet, so if answers require such, please explain thoroughly. Thanks!

• I don't think this problem can be solved analytically in an easy way in the general case. The racing line is usually a curve with constant radius, but it doesn't minimise the time given the constraints that you are imposing. Perhaps you should look into numeric solutions – Phoenix87 Sep 8 '18 at 22:03