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 = 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
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.