# How to (not) tip a car over

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First, let me state that my physics was below average when I studied it and it's been years since. And not just a few. Also, I did it in a different language.

So. I have a few questions about a debate I started with a few people about a car's stability and I'm kinda stuck.

First, is a lighter car harder or easier to tip over? I know center of gravity is a big factor, but is it the only one? I mean, if a 600Kg vehicle and a 1 ton vehicle have the same wheel span and the same center of gravity, shouldn't the heavier car tip over at a higher speed? On the other hand, the lighter car generates less friction so it is more likely to skid rather than tip over. Both main forces are linear (centrifugal and part of gravity). If it tips instead of skidding, then only the force required to lift the two outer wheels is required, it gets easier as it raises off the ground.

Secondly, it seems that, under ideal conditions, one would have the same Gs (lateral acceleration, to be exact) at skid point regardless of speed. Let me make this clearer:

Let us assume that at 20 km/h, in a 5 meter radius turn, the car skids out of the curve (tires lose traction). If we have an accelerometer in the car, it would read -say- 1 m/(s*s).

Let us also assume (skipping math) that at 40 km/h, in a 10 meter radius turn, the acceleration inside would also be 1 m/(s*s). Would it still skid? Sooner? Later? Does the fact that the wheels rotate faster have anything to do with that?

If not, is it sane to have an accelerometer in the car that could issue a skid point alarm assuming surface and tires are constant?

Am I missing something?

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Hmmm... Sooo many questions all in one go.

basicly a car is meant to stay put on its most stable side (on all 4 wheels).

the lighter car will naturaly be easier to tip (or off center) that the heavier one, once both are off center and ready to "Roll" there will be no difference in rate (barring mechanics of body shape etc...).

both will settle equaly.

now if youre talking about doing this on the hoof rather than stationary, the same effect will occur, the heaviest car will still take more power to over turn, getting a 1 ton to 40mph will consume more fuel (power) that getting a Half ton car to the same speed.

Now you have inertia!

but! both will still settle at the same time given the same parameters with only weight being altered

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The centre of mass, in relation to the pivot point (front wheels/front bumper) and also the height of the centre of mass are important as to whether a car will tip forwards.

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[...]both will still settle at the same time given the same parameters with only weight being altered

Thanks for the info, I'll try to clear my thoughts. The centrifugal force and the force needed to lift the car are both similar, but they have different formulae. Looking back into my mind I remember something in the order of m*v*v/r knowing notations are mass, speed and radius.

The force keeping the car on the ground is around G (m*g), ignoring the fact that once it lifted the wheels it becomes easier (a cosine is in there somewhere).

With g constant and v and r identical this seems to point to the angle alone (the position of the center of mass and wheels).

I know you just said that, just trying to retrace steps. Am I basically close?

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Also, my second question is still unanswered. If I was to manufacture an accelerometer and skid the car (curve until it loses traction), measuring that, would it be safe (in more or less real world) to have such a device issue a warning at 90% of that acceleration?

It would be useful at the same speed, since I know it works. But if the speed would be double, would the acceleration be the same? Am I missing something obvious that would send me spinning? Like a square in v*v that would make it useless right off the drawing board?

It seems fair that the same force would be required to tip a car regardless of speed, but is there something I miss? Different tire behaviour? Car tipping? Center of mass tipping over (let's not forget it has suspensions)? This isn't something I can test 400 times before refining

Note: I'm really trying to keep it short but I can't ...

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Here. The way I see it, the car begins with no significant friction from the road, because the wheels freely turn. As the car rotates horizontally relative to the direction of motion, the friction gradually increases as the tires begin to slip in a direction off-centre from their normal rotation. At some point, the translational energy from the ground in friction with the tires is transferred to the car, forcing it's centre of mass to rise high enough to allow free rotation of the vehicle in space with minimal contact.

Now any contact with any part of the car efficiently converts the translational energy of the car into rotational energy, slowing it down, and speeding up the rotation until the car is essentially rolling sideways. Finally enough energy is converted to heat from collision with the earth to cause the car to stop rolling.

Note that there are two basic cases: skidding on a straightaway, and skidding in a curve. In the straighaway case, cars rarely tumble unless struck. The Centre of Mass is low, and the tires can be designed for high friction in the direction of travel, and low friction sideways on a smooth ashphalt surface.

In the curve case, no tire design can prevent the car flipping, since the road ends immediately and the rough terrain to follow provides maximum friction with the car, forcing it to roll.

Typically, skidding can be controlled with steering technique on a straightaway, while losing control in a curve is hopeless.

Typically, on a straightaway, the heavier car will be harder to tip, since the height of the CM will be roughly the same (except for a Jeep!) but the amount of mass (and hence energy) that requires lifting is higher.

DRIVING NOTE: Whether sliding or tipping, control of the vehicle can be restore by (slight) steering into the direction of motion. One should be careful to avoid obstacles ahead however, such as other cars.

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