# Mechanics (statica) of a car (centre of gravity/grip)

1. Aug 13, 2010

### Hupke

I am new here and english is only my third language so forgive me small errors.
I'm not a 12 year old so you can speak science to me. But bear in mind I'm not a theorethical physcisist with 23 phd's, just an engineering student.

The problem goes as:

You've got a car and you need to give it as much mechanical (cornering) grip as possible.

The width, lenght, massa and tire-compound are constants. (let's call them W, L, M and since it is for the friction µ(static))
For this approach the "engine" is in the centre of the car.
I mean that approximately all the weight is located in one point the (centre of gravity) that is located on W/2 and L/2)

The parameter are thus the height of the centre of gravity and the centre of the wheels.

So either the centre of gravity is below the centre of the wheels. It is at the same height or it is lower. The wheels are offcourse conected to the car by their centre.

I'm not speaking about a car accelerating or braking (there is no weight-shift and stuff). There's no mentioning of a sway-bar (although if someone could explain that aswell would be nice, since that interrests me aswell)

The car is just taking a turn. So there is a force on the centre of gravity pulling the car (let us say the car is turning right, (it wasn't specified)) to the left.
And there's good old gravity. And there is the tires conecting to the road. (!!! there are 4 tires, since the car is turning they don't have the same forces acting upon them).

The question is then how can one achieve the best possible grip on the road. What should be the height of the centre of gravity and what should be the diameter of the tires (and thus also what should be the height of their centre).

The prof also said think in car dimensions. Let us say W = L *2/5 and the car is at least L/50 high so both the wheels and the COG should be at least so high i guess.

I've then simplyfied it some more.

The centre of gravity I'll call P. (and T1 is the left front tire, T2 is right front T3 is left rear,...)

If the car turn right there needs to be a centripetal force on P pushing it right. And there will be a centrifugal force to the left on P. They are equall (the car doesn't skid).
and let's call them F (and also constant, I mean we look at the car at only one single moment)

I think everybody get's the problem now (the question not my problem).

I can't figure this one out.
The relation between F and M for instance, they just said take them as constant's.
And then I was thinking does the height of the wheel really matter (yeah and you can see it when looking at the moment but still)

Or who know's a thing or 2 about car's to get me going.

2. Aug 13, 2010

### Hupke

Forgot to mention it.
In this stadium of the question you could see it as a 2D problem.

1 Wheel left and 1 Wheel right,...

But if this was solved then I could think about adding wheight to the rods and the wheels and think about a gravitational centre that wasn't in the middle. But that would still be out of my league for this year.