People have claimed that lowering the weight of the wheel/tire by say 10 lbs per wheel will have an 8-fold effect at each wheel, effectively lightening the vehicle by 320 lbs. I smelled a rat here and decided to perform a little physical analysis.

The answer?

A 10 lb reduction of wheel/tire weight will result in a total overall apparent effect of reducing vehicle weight by about 65 lbs and at absolute maximum, 80 lbs.

The analysis follows:


Question: What is the effective additional mass of wheels/tires due to their rotation?

For the vehicle in motion, the kinetic energy is given by:

E_kinetic = ½ * (M_total*V² + 4*I_wheel*w²)

Where
M_total = Total vehicle mass including wheels
I_wheel = moment of inertia of a single wheel/tire combination
V = Velocity of vehicle
w = rotational velocity of the wheel (in radians/sec)

Now, V = R_tread * w where R_tread = effective radius of the tire's tread

Combining, we find:

E_kinetic = ½ * (M_total + 4*I_wheel / R_tread²) * V²

So the effective additional mass added to the vehicle due to the wheels rotation = 4*I_wheel / R_tread²

Now, I_wheel = M_wheel * R_effective²

Where
M_wheel = mass of the wheel/tire combination
R_effective = radius of gyration (which is always less than R_tread )

What is this radius of gyration (otherwise known as the radius of inertia)? It is the radius at which an infinitely thin hoop of material of identical mass would have the same moment of intertia as the body in question. It is a mathematical abstraction, but can be calculated for any object.

For example, a disk of uniform density would have an R_effective = 0.707 x R_disk

Now, finally, the effective additional mass added to the vehicle due to the wheels rotation is:

M_{due to rotation} = 4 * M_wheel * (R_effective / R_tread)²

On a per wheel basis, the EFFECTIVE TOTAL wheel mass is given by:

M_{wheel, effective} = M_wheel * (1 + (R_effective / R_tread)² )

R_effective / R_tread is always less than 1 and probably somwhere around 80% by my guess. This ratio is a function of wheel and tire weight disribution.

So in this case, M_{wheel, effective} = M_wheel * 1.64

The absolute maximum (impossible) case would be M_{wheel, effective} = M_wheel * 2

Example: New wheels and tires are fitted, lowering the weight of each wheel / tire combo by 10 lbs. Assuming the radius of inertia of the wheel and tire combination are 80% of the outer radius of the tire, the apparent effect is to lower the weight of the car by 65.6 lbs (16.4 lbs per wheel).

One further note:
I used an energy analysis because it was more convenient for me. The exact same conclusion results if the analysis is performed using Newton's Second Law (F=M*a). This is not a steady-state analysis: it applies to all linear acceleration and deceleration conditions as long as traction is maintained.

edited 11/16/02