Moment of inertia and friction

1. Nov 24, 2009

mcsinc

I'm looking at some wheel/tire combinations for auto racing, and trying to decide which combination would out-accelerate or have more grip cornering.

combo 1: 225/45r15 tire on 15"x8" wheel; 34.2 lbs total weight; diameter 23.0"
combo 2: 195/50r15 tire on 15"x7" wheel; 28.4 lbs total weight; diameter 22.7"

Because it's a wheel, I don't think an actual moment of inertia could be calculated without some serious CAD drawings, etc. -- just theory here

Other things to consider:
- combo 1 has an 8.3" tread width; combo 2 has a 7.2" treadwidth. With 1.1" greater treadwidth, you'd think combo 1 would have more contact area, and more grip.

However, the footprint on each may be different, since each "may" use different optimal tire pressure due to sidewall stiffness (225x0.45=101.25" sidewall on #1 vs. 195x0.5= 97.5" sidewall on #2). Or is the footprint really different -- ie: would 30 psi inflated tire pressure cause the same amount of tire surface to make the same contact surface area on both tires? eg: If parked on a glass plate and looking at the shape of the contact patch, will x psi cause a different contact patch shape on combo 1 vs. 2?

- better cornering is also up for question because of the sidewall deflection, again, a function of tire pressure and also of the actual sidewall height.

- overall circumference of tire is smaller on #2, so "gearing" will be lower = faster acceleration

- the wheels are 3.3 lbs difference; tires are 2.5 lbs diff. Since most of the weight is towards the outer circumference (the rim of the wheel and the tire itself), combo 1 will have a significant disadvantage to use the car's power.

What other info would be needed? Ideas?

2. Nov 24, 2009

Eximus

For the moment of inertia of a thick walled cylindrical tube with open ends, the moment does not depend on the width. It's directly proportional to the average radius and the mass. Lighter, smaller tires have less momentum, and so transfer more power from axle to road.

$$I=m\frac{(r^{2}_{1}+r^{2}_{2})}{2}$$

Tire pressure ultimately determines the contact area if both tires are composed of the same material, and deform the same.

$$P=\frac{F}{4A}=\frac{mg}{4A}$$

$$A=\frac{mg}{4P}$$

So contact area is mostly a function of tire pressure, vehicle mass, and gravity. This result can be multiplied by some dimensionless factor like tread contact area divided by total cross-sectional area. Also note that any sidewall effects on contact area will be extremely small unless you have a very weak or large sidewall and a large deformation because of that.

Sidewall deflection in turns is more of a materials science problem. A good answer would require information about the shear modulus of the tire rubber as well as measurements of the geometry.

3. Nov 24, 2009