Let's go with slightly more reasonable assumptions. Here's a 1965 Ford Thunderbird (it was a 1966 model in the movie, close enough).

https://upload.wikimedia.org/wikipedia/commons/7/78/1965_Ford_Thunderbird_Convertible.jpg

And here are its specs. The wheelbase is 2.9m, while the total length is 5.2m. Eyeballing the photo, I'd put the wheel diameter at about 0.7m and the clearance no more than 0.2m. The front axle itself seems to be around 0.8m behind the front bumper, placing the rear axle 3.7m behind the front bumper or 1.5m in front of the rear bumper. I'll generously place two thirds of the mass in the front third of the car (engine and all), and the rest evenly distributed in the rear two thirds:

$$ \frac{dm}{dx} = \frac{m_0/3}{5.2 \cdot 2/3}:x<\frac{2}{3}5.2$$

and

$$ \frac{dm}{dx} = \frac{2 m_0/3}{5.2 \cdot 1/3}:x>\frac{2}{3}5.2$$

Now I find the moment of inertia relative to the real axle, neglecting the fact, that the mass may be distributed slightly above the height of the axle:

$$ I=\int_0^{5.2} \frac{dm}{dx}(x-1.5)^2dx=5.87m_0$$

and the torque exerted by gravity

$$ \tau =\int_0^{5.2} \frac{dm}{dx}(x-1.5)g\,dx=1.97m_0 g$$

Plugging in SI values one should get `kg*m^2`

and `kg*m^2/s^2`

units for the first and second quantity, respectively.

So as soon as there's no ground below the front wheels, we may expect the car to undergo an angular acceleration of $\ddot{\phi} = \tau/I = 3.29 rad/s^2$.

As the clearance is somewhat smaller than even the wheel's radius, we're good to deal with "small" angles. Take at point of the vehicle $L$ meters ahead of the rear axle. After time $t$ it will drop by $L\phi = L \ddot\phi t^2/2$. If the car is moving at velocity $v$, the length $L$ is given by $L_0-v t$ where $L_0$ is the wheelbase. So we need the condition that

$$(L_0-vt)\ddot\phi t^2/2<h $$

where $h$ is clearance. This should hold until $vt=L_0$. Essentially, the expression above (to the left of the inequality relation) is how much the point of the car directly above the ledge has dropped. It's a cubic expression, but it clearly has a maximum somewhere. Differentiate and find the maximum by solving, it's at $t=2L_0/(3v)$. Plugging this value in we get the constraint:

$$ \frac{2 \ddot \phi L_0^3}{27 v^2} < h $$

and solving the inequality we find $v>5.44 m/s$ (or about 20km/h).

I leave it to the curious reader to consider, that the Ford Thunderbird is a rear-wheel drive car, which can get about 6000 Newton-meters of torque on it rear wheels and has a mass of around 2200kg. This is enough data to check if the engine can keep it from flipping down off the ledge too fast and make it possible to drive off the cliff at an initially lower speed.

You're always going to need the radius, one way or another. If you are calculating from the angle of steering, you will also need the vehicle's wheelbase (the distance between the front and back tires). The radius of turning is going to be (approximately) the wheelbase length divided by the angle of steering in radians. Given the radius and the speed of the vehicle, the force acting on the tires is just the centripetal force $mv^2/r$, divided by the number of tires.

In practice I expect in a tight turn, the force is unevenly distributed, with the outer tires applying more force.

## Best Answer

I think the only time the speed of the car should matter is if it is going so fast that it skids. Otherwise, static friction between the road and the tires will be maintained, which means it will follow the same path regardless of speed.

The front tires will only turn on one axis (per tire). As such, the only way the car can move without skidding is for the tire to move in exactly the direction it is pointed, regardless of speed.

The angle of each of the front tires is carefully calculated so that all four tires are tangent to a circle about the same center point at all times. From that information, you can calculate the turn radius of the car with geometry. Which appears to be the approach you've taken.

It would probably be more useful to work with the radius of the circle in which the point directly between (and equidistant from) the two rear wheels moves rather than the point directly between (and equidistant from) the two front wheels. This allows you to work with the dimensions of the car (which will be given) instead of the angle from a point on the car to the center of the turn (which will not be given). This also has the advantage of making your triangles right triangles.