Car racing: How to calculate the radius of the racing line through a turn of varying length

Question

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

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

Answer 1

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

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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 $$r_r=r_i+\frac{r_o-r_i}{1-\cos\frac\theta2}.$$

Answer 2

You may observe diametre of the turn is equal the turn in point to the track out point. The difficulty is in determining where those two points are! One simple method to identify them is that the tangent line of the inner radius at the start of the turn points back to the turn in point. What more important is that this point is always visable to the driver. Often the apex is not visable and so is not a practical reference point. Using Constant bearing decreasing range enables us to calculate the turn position.

Let X be the with of the track.

Then the turn in point is:

1/tan|22.5º| • X

The 22.5 is the modal angle for any given intercept as it is the place right or left of your forward vision that you wil find constant bearings on a right angle intercept. That being the inner radius at start of the roads turn.

Since the inner radius intercepts the tangent line at the start of the corner. The equation becomes:

(1/tan|22.5º|•X)+ Ri

As track width also extends the inner radius Ri.:

1/tan |22.5º| • X +Ri +X

This give the radius of a 90º turn

Answer 3

A user was having a little trouble understanding @Rahul's argument, so here's a slight re-imagining.

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Let $a$ and $b$ be the inner and outer radii of the track, with arcs center at $O$. Let $c$ be be the "race" radius, with arc centered at $P$. Define $\phi := \theta/2$ as the half-angle of the turn (that is, half the measure of any of the arcs involved).

From right triangle $\triangle OPQ$, we read the relation $$\begin{align} \cos\phi &= \frac{c-b}{c-a} \tag{1}\\[6pt] (c-a)\cos\phi &= c-b \tag{2}\\[6pt] b-a\cos\phi &= c(1-\cos\phi) \tag{3}\\[6pt] c &= \frac{b-a\cos\phi}{1-\cos\phi} \tag{4} \end{align}$$ We can go on to write $$c = \frac{b \color{red}{-a+a}-a\cos\phi}{1-\cos\phi} = \frac{b-a}{1-\cos\phi}+\frac{a(1-\cos\phi)}{1-\cos\phi} = a+\frac{b-a}{1-\cos\phi} \tag{5}$$ which matches @Rahul's answer.

Answer 4

The interesting part: the optimal trajectory has no fixed radius. The racer can brake, start with small radius, accellerate during the turn, end it with a larger radius and speed!