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article details
Author Ryan Leecock
Categories All Honda/Acura
Create Date January 18, 2002 18:06
Last Update February 08, 2005 18:23
Theory of a Rotational Mass

by Jerome Soh

What is the physical relationship between an increase in wheel/tire = weight (rotating mass) and static weight?

It depends on how the weight of the wheel/tire is distributed.

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).

> things to rotate, you need a torque.  Simply put, heavier wheels require
> more torque to get them rotating at the same speed as lighter wheels.

No and no. You seem to have the right idea, but please let me clarify. 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. The problem is that, as you stated,

>(I) is not easy to calculate for complicated shapes like wheels

so that, eventually, you'll have to take a guess at (I), like,

>.... probably near 3x
>or more (for really heavy tires with light wheels).

I think this might be helpful in an FAQ, provided that a warning is added that it is not quantitatively accurate. It would, however, provide a good conceptual insight for why adding 10 pounds of wheel is worse than driving around with a 10 pound sandbag.

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