Geometry – How to Calculate the Radius of the Racing Line Through a Turn

geometry

I am in the process of designing a board game involving car chases, and I am stumped by the following problem:

A car will have a maximum speed through a constant radius speed turn, giving a maximum safe cornering speed for a given turn radius. But, if the car follows the racing line (an apex turn?), the radius of the turn is greater, and the car can take the turn faster

A simplified situation would be sufficient to cater to my needs (no need for curves of changing radius, etc), see i.stack.imgur.com/odIyF.jpg for an illustration

enter image description here

The radius of the racing line R(race) must be dependent on 3 factors: Outer radius: R(outer) Inner radius: R(inner) And the length of the turn in degrees.

From this I should get R(race)

R(inner) and R(outer) have the same center, and the center of R(race) must be somewhere on the line that bisects the turn (midpoint of the curve)

I would love to have a formula for the 90 degree turn, but preferably I would like a general solution, where the turn can be of any angle of turn (up to 180 degrees or more). Looking at my sketches, at 180 degrees of turn, the radius of the racing line will equal R(outer), while the radius will approach infinity, as the angle of turn becomes smaller and smaller

I have tried searching online for answers, but the formula I have dug up have given results I haven't been able to reproduce when mocking up on graph paper

Best Answer

Let the outer, inner, and race radii be $r_o$, $r_i$, and $r_r$, and the angle of the turn be $\theta$.

$\hskip2in$enter image description here

The arc is tangent to the outside and inside of the road at the points marked $P_1$ and $P_2$, so $$\begin{align} x+r_o&=r_r,\\ x+r_i\cos\frac\theta2&=r_r\cos\frac\theta2. \end{align}$$ The solution is $$r_r=r_i+\frac{r_o-r_i}{1-\cos\frac\theta2}.$$