A study investigating the effective inertial weight of wheels. Warning: technical in nature.
11-15-2002, 11:38 AM
#1
Member
Thread Starter
Join Date: Sep 2000
Posts: 2,636
Likes: 0
Received 0 Likes
on
0 Posts
A study investigating the effective inertial weight of wheels. Warning: technical in nature.
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<sub>kinetic</sub> = ½ * (M<sub>total</sub>*V<sup>2</sup> + 4*I<sub>wheel</sub>*w<sup>2</sup>)
Where
M<sub>total</sub> = Total vehicle mass including wheels
I<sub>wheel</sub> = 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<sub>tread</sub> * w where R<sub>tread</sub> = effective radius of the tire's tread
Combining, we find:
E<sub>kinetic</sub> = ½ * (M<sub>total</sub> + 4*I<sub>wheel</sub> / R<sub>tread</sub><sup>2</sup>) * V<sup>2</sup>
So the effective additional mass added to the vehicle due to the wheels rotation = 4*I<sub>wheel</sub> / R<sub>tread</sub><sup>2</sup>
Now, I<sub>wheel</sub> = M<sub>wheel</sub> * R<sub>effective</sub><sup>2</sup>
Where
M<sub>wheel</sub> = mass of the wheel/tire combination
R<sub>effective</sub> = radius of gyration (which is always less than R<sub>tread</sub> )
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<sub>effective</sub> = 0.707 x R<sub>disk</sub>
Now, finally, the effective additional mass added to the vehicle due to the wheels rotation is:
M<sub>due to rotation</sub> = 4 * M<sub>wheel</sub> * (R<sub>effective</sub> / R<sub>tread</sub>)<sup>2</sup>
On a per wheel basis, the EFFECTIVE TOTAL wheel mass is given by:
<b>M<sub>wheel, effective</sub> = M<sub>wheel</sub> * (1 + (R<sub>effective</sub> / R<sub>tread</sub>)<sup>2</sup> )</b>
R<sub>effective</sub> / R<sub>tread</sub> 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<sub>wheel, effective</sub> = M<sub>wheel</sub> * 1.64
<b>The absolute maximum (impossible) case would be M<sub>wheel, effective</sub> = M<sub>wheel</sub> * 2</b>
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
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<sub>kinetic</sub> = ½ * (M<sub>total</sub>*V<sup>2</sup> + 4*I<sub>wheel</sub>*w<sup>2</sup>)
Where
M<sub>total</sub> = Total vehicle mass including wheels
I<sub>wheel</sub> = 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<sub>tread</sub> * w where R<sub>tread</sub> = effective radius of the tire's tread
Combining, we find:
E<sub>kinetic</sub> = ½ * (M<sub>total</sub> + 4*I<sub>wheel</sub> / R<sub>tread</sub><sup>2</sup>) * V<sup>2</sup>
So the effective additional mass added to the vehicle due to the wheels rotation = 4*I<sub>wheel</sub> / R<sub>tread</sub><sup>2</sup>
Now, I<sub>wheel</sub> = M<sub>wheel</sub> * R<sub>effective</sub><sup>2</sup>
Where
M<sub>wheel</sub> = mass of the wheel/tire combination
R<sub>effective</sub> = radius of gyration (which is always less than R<sub>tread</sub> )
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<sub>effective</sub> = 0.707 x R<sub>disk</sub>
Now, finally, the effective additional mass added to the vehicle due to the wheels rotation is:
M<sub>due to rotation</sub> = 4 * M<sub>wheel</sub> * (R<sub>effective</sub> / R<sub>tread</sub>)<sup>2</sup>
On a per wheel basis, the EFFECTIVE TOTAL wheel mass is given by:
<b>M<sub>wheel, effective</sub> = M<sub>wheel</sub> * (1 + (R<sub>effective</sub> / R<sub>tread</sub>)<sup>2</sup> )</b>
R<sub>effective</sub> / R<sub>tread</sub> 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<sub>wheel, effective</sub> = M<sub>wheel</sub> * 1.64
<b>The absolute maximum (impossible) case would be M<sub>wheel, effective</sub> = M<sub>wheel</sub> * 2</b>
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
11-15-2002, 11:55 AM
#5
Elder Member
Join Date: Mar 2000
Posts: 15,742
Likes: 0
Received 0 Likes
on
0 Posts
Thats close, but not quite right...
Keep in mind, wheels don't have an even weight distribution from the hub to their outer surface... most of the weight is on the very outer part of the wheel.
I didn't go through your math completely, but I think there's something else off with it -- vehicle weight shouldn't be a factor at all. The reason why reducing unsprung and rotational mass helps the car isn't because it makes the car lighter, but because that weight (both unsprung, and more importantly, the higher mass at a distance from the hub) impacts significantly the drive train loss. (Some percentage of your total engine output has to be used to both pass energy across your suspension to the body of the car, and to accelerate the mass of the wheels)
I didn't go through your math completely, but I think there's something else off with it -- vehicle weight shouldn't be a factor at all. The reason why reducing unsprung and rotational mass helps the car isn't because it makes the car lighter, but because that weight (both unsprung, and more importantly, the higher mass at a distance from the hub) impacts significantly the drive train loss. (Some percentage of your total engine output has to be used to both pass energy across your suspension to the body of the car, and to accelerate the mass of the wheels)
11-15-2002, 12:02 PM
#7
Elder Member
Join Date: Dec 2000
Posts: 12,498
Likes: 0
Received 0 Likes
on
0 Posts
george, he's calculating it as if all the weight was all the way
out at the edges of the tire...in other words max possible contribution from wheel weight....but i agree it loooks suspect...i'll take a look at it this weeekend...
