# Which factors can help a car flip on a turn?

I am not sure if this is the right place to ask it, but this is a question that I thought today, and it gave me some curiosity to understand. Imagine that a car will curve, we can say the turn is a bit tight , what are the factors that can help it to flip? I was wondering about some aspects:

1. if the car has mass, it has inertia, so while it is curving it tends to keep the motion in the same direction that it was instants before the turn. right? So, if the car has more mass, it has more inertia, and since there is friction, one heavier car would flip easier then one lighter, considering that all other possible variables were equal.

2. Center of gravity, a car with an higher center of gravity would flip easier. The whole inertia of the car distributed to higher heights would be further of the tires(where friction acts), creating angular momentum.

3. The car being thin because it has less surface contact with the ground;

4. The car being lighter. This opposes what I've said in "a)", but a heavier car is more difficult to get off the road. A lighter car has more instability.

Am I wrong? In what I'm wrong? What do you think?

Thanks for helping. :)

## 4 Answers

Let me take the reverse of your question, because I was in a car that flipped and it's not nice. So here are the ways I can think to not flip, but if you do want to flip, do the opposite :)

The faster it travels, the more chance an instability of any kind will be amplified and the less time the driver will have to get it back under control.

The lowest c. of g. you can achieve the better. F1 cars are the obvious example.

Mass works both ways. If an instability occurs below a certain speed, it can dampen oscillations, but above that speed it can make things worse.

The addition/modification of aerofoil wings and the venturi effect of a properly designed underbody can make a huge difference. As I am sure you know, F1 cars can theoretically easily drive on the roof of a tunnel, as they have so much downforce.

From BowlofRed's comments.

Sticky tires/high friction increases the chance of flipping (on flat pavement). If the tires start slipping, then there's reduced forces available to tip the car. For most sedans, the tires will slip long before forces build to the point that tipping is possible.

A driver with fast reactions.

Draw a force diagram for the car, including downward gravity at the center of mass (call that "1", and all other forces can be relative to that), normal forces upward on each tire, and a static force of friction sideways on the tires. The forces have to add up to something that is given ahead of time, the sideways force the car needs to make the turn. You can just treat that latter thing as "x", the variable that you crank up until the car rolls over. Solve for all the other forces in units of "1". They all have to then add up to x pointing sideways. The key constraint for the car to not turn over is that the total torque has to be zero. But when x gets so large that the normal forces on the inside tires goes to zero, that's the place where the car is at the edge of rolling over. The angle from the center of mass to the outside tires is the key, so that's where the factors you mentioned come into play-- the width of the car, and the height of its center of mass. (I'm not including forces from the air, though others have mentioned that might matter. It's also a good point that there can be a limit to the size of the frictional force, which can protect against rollover but at the cost of maybe crashing into a tree!)

By the way, if you write down the forces, you will see that the mass of the car plays no role at all-- given the speed and the turn, the scale of x relative to 1 is all that matters, and that will not depend on mass, it depends only on v*v/rg, where v is the car speed, r is the radius of the turn, and g is the acceleration of gravity. But this does show that you have more of a rollover problem on the Moon-- all else being equal, the speed has to be down by the square root of the Moon's g over the Earth's g.

There are several factors which can increase the chance of flipping a car on a turn. These include:

• The centripetal acceleration on the car, which is $v^2/r$. The higher the speed of the car, and the smaller the radius at a given speed, the more likely it is that a car will flip
• The center of gravity of the car. The higher the center of gravity, the more likely the car is to flip
• Whether or not the road is "crowned". For drainage purposes, roads are not flat, but are instead modestly curved, with the high point in the center of the road. This leads to cars having a slight "outward tilt" when going around a curve, which increases the chance of flipping
• Whether or not the road is "banked". Banked curves give a car an "inside tilt" when rounding a curve, and decrease the chance of flipping
• A strong cross wind that is blowing towards the outside of the curve will increase the chance of flipping
• A rut on the shoulder will tend to "grab" the outside tires when rounding the curve, again increasing the chance of flipping
• Any substantial object that you run over with the "inside" tires will lift up the side of the car that is on the inside of the turn, again, increasing the chance of flipping
• Losing your nerve in the middle of a turn and stomping on your brakes will definitely cause the car to "rock" towards the outside of the turn, increasing your chance of flipping

Any one of the above causes increases the risk of flipping, and if two or more are encountered at the same time, the risks will compound each other. Hopefully, you will use the above info to avoid a flip, rather than cause one, because (as another poster alluded to) the inside of a car is not that soft when you are being bounced back and forth off the walls of the car.

When $h/b>gR/v^2$ where $h$ is the height of the center of mass above the ground and $b$ is half the distance between the wheels, he car flips.