Car Dynamics – Sliding and Overturning

In this post we will take a look at car performance in curves. Of central importance for our considerations is the centrifugal force. Whenever a body is moving in a curved path, this force comes into play. You probably felt it many times in your car. It is the force that tries to push you out of a curve as you go through it.

The centrifugal force C (in N) depends on three factors: the velocity v (in m/s) of the car, its mass m (in kg) and the radius r (in m) of the curve. Given these quantities, we can easily compute the centrifugal force using this formula:

C = m · v² / r

Note the quadratic dependence on speed. If you double the car’s speed, the centrifugal force quadruples. With this force acting, there must be a counter-force to cancel it for the car not to slide. This force is provided by the sideways friction of the tires. The frictional force F (in N) can be calculated from the so called coefficient of friction μ (dimensionless), the car mass m and the gravitational acceleration g (in m/s²).

F = μ · m · g

The coefficient of friction depends mainly on the road type and condition. On dry asphalt we can set μ ≈ 0.8, on wet asphalt μ ≈ 0.6, on snow μ ≈ 0.2 and on ice μ ≈ 0.1. At low speeds the frictional force exceeds the centrifugal force and the car will be able to go through the curve without any problems. However, as we increase the velocity, so does the centrifugal force and at a certain critical velocity the forces cancel each other out. Any increase in speed from this point on will result in the car sliding.

We can compute the critical speed s (in m/s) by equating the expressions for the forces:

m · s² / r = μ · m · g

s = sqrt (μ · r · g)

This is the speed at which the car begins to slide. Note that there’s no dependence on mass anymore. Since both the centrifugal as well as the frictional force grow proportionally to the car’s mass, it doesn’t play a role in determining the critical speed for sliding. All that’s left in terms of variables is the coefficient of friction (lower friction, lower critical speed) and the radius of the curve (smaller radius, more narrow curve, smaller critical speed).

However, sliding is not the only problem that can occur in curves. Under certain circumstances a car can also overturn. Again the centrifugal force is the culprit. Assuming the center of gravity (in short: CG) of the car is at a height of h (in m), the centrifugal force will produce a torque T acting to overturn the car:

T = h · C = m · v² · h / r

On the other hand, there’s the weight of the car giving rise to an opposing torque T’ that grows with the width w (in m) and mass m of the car:

T’ = 0.5 · m · g · w

At low speeds, the torque caused by the centrifugal force will be lower than the one caused by the gravitational pull. But at a certain critical speed o (in m/s), the torques will cancel each other and any further increase in speed will result in the car overturning. Equating the above expressions, we get:

m · o² · h / r = 0.5 · m · g · w

o = sqrt (0.5 · r · g · w / h)

Aside from the coefficient of friction, the determining factor here is the ratio of width to height. The larger it is, the harder it will be for the centrifugal force to overturn the car. This is why lowering a car when intending to go fast makes sense. If you lower the CG while keeping the width the same, the ratio w / h, and thus the critical speed for overturning, will increase.

Let’s look at some examples before drawing a final conclusion from these truly great formulas.

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According to caranddriver.com the center of gravity of a 2014 BMW 435i is h = 0.5 m above the ground. The width of the car is about w = 1.8 m. Calculate the critical speed for sliding and overturning in a curve of radius r = 300 m on a dry asphalt road (μ ≈ 0.8).

Nothing to do but to apply the formulas:

s = sqrt (0.8 · 300 m · 9.81 m/s²)

s ≈ 49 m/s ≈ 175 km/h ≈ 108 mph

So with normal driving behavior you certainly won’t get anywhere near sliding. But note that sudden steering in a curve can cause the radius of the your car’s path to be considerably lower than the actual curve radius.

Onto the critical overturning speed:

o = sqrt (0.5 · 300 m · 9.81 m/s² · 3.6)

o ≈ 73 m/s ≈ 262 km/h ≈ 162 mph

Not even Michael Schumacher could bring this car to overturn.

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How would the critical speeds change if we drove the 2014 BMW 435i through the same curve on an icy road? In this case the coefficient is considerably lower (μ ≈ 0.1). For the critical sliding speed we get:

s = sqrt (0.1 · 300 m · 9.81 m/s²)

s ≈ 17 m/s ≈ 62 km/h ≈ 38 mph

So even this sweet sport car is in danger of sliding relatively quickly under these conditions. What about the overturning speed? Well, it has nothing to do with the friction of the tires, so it will still be at 73 m/s.

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This was an excerpt from More Great Formulas Explained. Interested in more car dynamics? Take a look at my post on How to Compute Maximum Car Speed. For other interesting physics articles, check out my BEST OF. I hope you enjoyed and drive safe!

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