[Silentbob343]

LOL at all the anti rice comments

No nothing is wrong with wheels untill they get so big they cause the car to ride so high up it looks like it's about to go off road.

Or when you get wheels that are to big and heavy they actually slow your car down. Even if you get bigger wheels that are lighter than smaller stock wheels they can still slow you down due to the increase in rotational mass.

"Actually, there is a relationship between rotating and static weight. For the purposes of a car,

a=T(2/md+d/2I)

a = acceleration
T = torque driving the wheel
m = mass the wheel must "tow" from the center of rotation (COR)
d = diameter of tire
I = polar moment of inertia of wheel/tire combination
This was derived from first principles. The first term comes from F=ma where F = T/(d/2) --> d/2 is the moment arm at which the force from the ground on the tire acts. The 2nd term comes from a = alpha*(d/2) where alpha is angular acceleration. T=alpha*I.

(I) is not easy to calculate for complicated shapes like wheels... it is usually measured. The general form is (I) = sum (mi*di^2). (mi) is the "lump" of mass at a distance (di) from the COR. Break up your wheel into a zillion parts, measure the distance from the COR to each part, sum them all up using the above equation, and you get (I) :-). As an example, let's take a simple shape: for a uniformly distributed disk, (I) = md^2/8. Plugging into the above equation for (a), we get:

a=T(2/md+4/m_wd) where m_w is the weight of the wheel. Thus, we can see that for a uniformly distributed disk (like a hockey puck), the importance of rotational weight is 2x that of static weight (if the weight was static, it would contribute to the first term with the "2" factor, if it was rotational, it would contribute to the 2nd term with the "4" factor). However, a typical wheel/tire combination has most of it's weight at the outer edges, which increases (I). Thus, rotational weight is more than 2x the equivalent static weight.... probably near 3x or more (for really heavy tires with light wheels).

Wheels with a larger moment of inertia (I) require more torque to _accelerate_ at the same rate as a wheel with a smaller I. For a given horsepower output, both wheels will eventually reach the same terminal velocity, but the wheel with the larger I will take longer to reach it. It is possible for a heavier wheel to accelerate at a faster rate than a lighter wheel for a given torque. As an example, let's say we have 2 identical 15" wheels. Now let's add 1 lb to wheel #1 at the very edge of the wheel, and 2 lbs to wheel #2 near the center of the wheel. Even though wheel #2 is heavier, it has a smaller polar moment of inertia (I) and thus will accelerate at a faster rate for a given torque than wheel #1.

You can, of course, calculate the acceleration, then work your way backwards and simply define an "effective mass" as the proportionality constant between F and a. In this case you'd arrive at

M_effective = (M + 4I/d^2)
However, such a definition is neither physical, nor possibly even useful. (I) is not easy to calculate for complicated shapes like wheels so that, eventually, you'll have to take a guess at (I), like..3x.

These are not things worth memorizing. However, you do need to know that and object with its mass far from the center has a greater moment of inertia than another object (of the same mass) with its mass near the center."