Originally posted by ND SVT:
It is still possible for a wheel/tire combo to have this combination while still possessing a greater moment of inertia. The moment of inertia is a measure of the mass distribution of an object about the center of a rotational axis. Since most wheels are symmetric about their rotational axis, the moment of inertia is proportionate to the mass of the wheel/tire and the square of the radius that defines the average distribution of the mass from the wheel center. So if this radius from the lighter/smaller-diameter tire/wheel combo was greater than that of the heavier/larger-diameter wheel, there is a possibility that that factor could outweigh the mass difference between the wheels, and produce a greater inertia.
Ok, but what about the tire. We are talking about the mass of the wheel, but even further out is the tire.
Ah, I think I've answered my own question. Doesn't a lower profile tire have to have a thicker sidewall to support the same amount of weight?
With the lighter wheel 18" (or any 18" wheel for that matter), you get a bit heavier tire. Even if you keep the width the same. Or at the very least, the distribution of mass is even more skewed away from the center of the wheel.
So what we are looking for is the point on the wheel where we reach the average mass of the wheel.
So given two wheels of the same mass, the one where the average mass point is further from the center of the wheel will have a greater moment of inertia.
I've got to stop, my head hurts now.
But I think I understand where you are coming from.
TB
