eduardo

What is the effect of a change in wheel/tire weight? Several people have quoted factors of x6 or x8 to determine the equivalent non-rotating mass variation. That is, an increase of 10lbs/wheel could be as detrimental to performance as carrying Arnold Schwarzenegger and Scoobie Doo as passengers.

I decided to refresh my Physics and came up with the following conclussion: from the point of view of acceleration, an increase of X in wheel or tire weight is no worse than an increase of 2X in passenger weight. Not 6x, not 8x, just 2x worst case. This is why.

At any given speed/gear combination there is maximum torque T available at the wheels. The torque does two things: (1) opposes the rolling resistance and the aerodynamic drag and (2) accelerates the car. The equation that describes the equilibrium is:

T = I*u + M*r*a + D*r

*: denotes multiplication
T: torque at the wheels
I: total moment of inertia of the rotating parts
u: angular accelaration of the rotating parts
M: total mass of the vehicle
r: external radius of the tires
a: linear acceleration
D: total drag and rolling resistance

The first term, I*u, is the torque used in making the wheels rotate faster. The second, M*r*a, is the torque used to accelerate everything (wheels included) in the direction of movement. The third is the torque used to cancel the drag and rolling resistance.

Another way of writting the equation above is:

T/r = I*u/r + M*a + D

Now the terms are forces. The left side is the force available at the contact patch.

If the tires are not slipping, the angular acceleration and the linear acceleration are related in the following way:

a = u*r

Replacing in the equation and moving things around, we get

T/r - D = (I/r^2 + M)*a

^: denotes exponentiation (r^2 means "r squared")

We can say that the left side is the force available for acceleration. Such force accelerates an "non-rotating equivalent mass" E,

E = I/r^2 + M

Now suppose that we increase the weight of the wheels and tires by an amount X. Both the total mass and the moment of inertia will increase; let's call the new mass M' and the new moment of inertia I'. Obviously,

M' = M + X

What about I'? Well, the moment of inertia of a "punctual mass" [m] (a mass concentrated in a point) at a distance [r] from the axis of rotation is [m*r^2]. That is, the moment of inertia depends critically on the distance between the mass and the axis of rotation.

In a real wheel+tire combination the mass is distributed in different amounts at different distances from the center. In order to compute the total moment of inertia we would need to know the mass distribution and use integral calculus. We can do a simpler thing though. We can make the pessimistic assumption that all of the mass increment is located on the periphery of the tire, that is, at a distance [r] from the center. This assumption is pessimistic because in a real wheel some of the weight will be located closer that [r] and will contribute less to the total momentum (it is not too pesimistic though: most of the weight is located pretty far from the center, if not at the periphery). So now we can compute I',

I' = I + X*r^2

The new "equivalent mass" is,

E' = I'/r^2 + M' = I/r^2 + M + 2*X

In other words, from the acceleration point of view, the equivalent non-rotating mass increment corresponding to an increment X in rotating mass is - at worst - 2X.

NOTE: After doing some approximations and assumptions about weight distribution in a typical wheel+tire combo, I believe that 1.7X is a better approximation. A 10lb/wheel weight increase would not hurt acceleration worse than carrying RinTinTin.

Comments anyone?

ALensink

I read something similar in a European Car article from last December (I think December was the month). The article concluded that a good approximation or rule of thumb would be that the multiplier would be 2X for tires and 1.5X for wheels, for the reasons you describe. I wouldn't know myself, since I'm no physicist.

nick n

Put this in the F.A.Q.!!!

eduardo

How does one do that?

nick n

Click on "F.A.Q." above. Click on the highlighted e-maill address in the third paragraph, I think. Then copy and paste your post! *You'll need to put it in a catagory such as "Wheel and tire" From there, you will need to set a title for it.
Hope it gets in. Your post should help clear up some of the unsprung weight issues, from my point of view.
*I've never personally posted a F.A.Q. but I know Its not difficult.

Nick Norman
99.5 1.8Tqms (still has dent)

eduardo

Thank you

nick n

Your welcome.