Shadowlife25

03-30-2007, 07:22 AM

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:

Ekinetic = � * (Mtotal*V2 + 4*Iwheel*w2)

Where

Mtotal = Total vehicle mass including wheels

Iwheel = moment of inertia of a single wheel/tire combination

V = Velocity of vehicle

w = rotational velocity of the wheel (in radians/sec)

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

Combining, we find:

Ekinetic = � * (Mtotal + 4*Iwheel / Rtread2) * V2

So the effective additional mass added to the vehicle due to the wheels rotation = 4*Iwheel / Rtread2

Now, Iwheel = Mwheel * Reffective2

Where

Mwheel = mass of the wheel/tire combination

Reffective = radius of gyration (which is always less than Rtread )

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 Reffective = 0.707 x Rdisk

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

Mdue to rotation = 4 * Mwheel * (Reffective / Rtread)2

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

Mwheel, effective = Mwheel * (1 + (Reffective / Rtread)2 )

Reffective / Rtread 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, Mwheel, effective = Mwheel * 1.64

The absolute maximum (impossible) case would be Mwheel, effective = Mwheel * 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.

from toyguymod back in '02

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:

Ekinetic = � * (Mtotal*V2 + 4*Iwheel*w2)

Where

Mtotal = Total vehicle mass including wheels

Iwheel = moment of inertia of a single wheel/tire combination

V = Velocity of vehicle

w = rotational velocity of the wheel (in radians/sec)

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

Combining, we find:

Ekinetic = � * (Mtotal + 4*Iwheel / Rtread2) * V2

So the effective additional mass added to the vehicle due to the wheels rotation = 4*Iwheel / Rtread2

Now, Iwheel = Mwheel * Reffective2

Where

Mwheel = mass of the wheel/tire combination

Reffective = radius of gyration (which is always less than Rtread )

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 Reffective = 0.707 x Rdisk

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

Mdue to rotation = 4 * Mwheel * (Reffective / Rtread)2

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

Mwheel, effective = Mwheel * (1 + (Reffective / Rtread)2 )

Reffective / Rtread 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, Mwheel, effective = Mwheel * 1.64

The absolute maximum (impossible) case would be Mwheel, effective = Mwheel * 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.

from toyguymod back in '02