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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 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

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Is the width of the road on the straight parts equal to the difference between the outer radius and the inner radius on the curve? – Henry Jan 29 '13 at 7:21
Yes, it is - at least it is supposed to be :) - and thanks for embedding the picture, I don't have the privileges to do so :) – LeZerp Jan 29 '13 at 7:33
If you ever add any more realism, my vote would be to include acceleration vs. deceleration. If a car can break quicker than accelerate (as most cars do) then the apex taken earlier gives more overall speed (more length for regaining speed from the turn gives better top speed in the straightaway.) – adam W Jan 29 '13 at 15:56
Sounds very interesting, Adam - if it is simple to implement, I'd love to see something on the subject - but for now it would be a lot more detailed than the model for the game :) – LeZerp Jan 29 '13 at 19:42
Yeah, it would be about as complex as the curves with changing radius...BTW You need the '@' symbol for me to know you addressed me, as in '@Adam'. I only just happened to look here again and saw your comment. – adam W Feb 1 '13 at 19:27
up vote 5 down vote accepted

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

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In retrospect, the solution can be found without any algebra by observing that adding or subtracting the same value to all radii leaves the tangencies unchanged. So subtract $r_i$ from everything and the inner arc shrinks to a point, and then we can practically read off the solution. – Rahul Jan 29 '13 at 11:14
Thanks for your answer, Rn However I have some problems putting it to use; I can't fault your diagram, that seems good to me. I got some strange results. I expected 'smooth' Rr, increasing as the curve of the track decreased, but the results show several 'bumps' at discrete angles, and the wrong trend. see: [1]: What am I missing here? – LeZerp Jan 29 '13 at 12:28
I am an idiot! Thanks again! :) – LeZerp Jan 29 '13 at 13:09

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!

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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

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So for 180º Turns you will notice that the turn radius is no more than 2•X+2•ri. – practical Jan 16 '15 at 14:01
Oh the turn in velocity is easy to calcuate so let say the track is two lane road width 7m, and Ri is 7m Ro is 14. 1/tan22.5•7+7 So the radius of the turn is 30.9 therefore the root of 2•9.8•30.9 =V^2 and so V is 24.62 m/s or 88.6 km/h and since the best line is a constant radius the speed will also remain constant. – practical Jan 17 '15 at 9:17

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