Moment of inertia and friction

[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?

 

[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.

 

[jadatis]

If you want to know how tire- and car-manufacturers calculate the tire-pressure for on the road, I can help you with the spreadsheets I made with use of the formula of the ETRTO ( european).
http://cid-a526e0eee092e6dc.skydrive.live.com/browse.aspx/.Public
in this public map on my skydrive of hotmail you see mostly tire-pressure maps and loose spreadsheets. 3 are in English , the rest is in my native speek Dutch.
recalculating tire -pressure map is for recalc when non oem tires are used
calculateload1 is to compare the diferent powers that are used bij the diferent tire-organisations.
also an English pdf about a new sugestion for the formula, with comparison. It shows that the ETRTO formula I use is the savest one.
If you want to see the formula sheet of the ETRTO I can sent it to you in a personal message. For copyright reasons I did not place it on my public map.
If you use the actual weights, you can use this formula mayby for races too.