Tech Article Title

Author

Date

Effect of Wheel Mass on Acceleration

Eduardo
Martinez 
2001

What is the effect of a change in wheel/tire mass? Several people have
quoted factors of x6 or x8 to determine the equivalent nonrotating mass variation. That is, an
increase of 10lbs/wheel could be as detrimental to performance as carrying Arnold
Schwarzenegger and Scooby Doo as passengers.
I decided to refresh my Physics and came up with the following conclusion: from the point of view of acceleration, an increase of X in wheel or tire mass is no
worse than an increase of 2X in passenger mass. 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 acceleration 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 writing 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 "nonrotating equivalent mass" E,
E = I/r^2 + M
Now suppose that we increase the mass 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 mass will be located closer that [r] and will
contribute less to the total momentum (it is not too pessimistic though: most of the mass 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 = E + 2*X
In other words, from the acceleration point of view, the equivalent nonrotating mass increment corresponding to an increment X in rotating mass is  at worst  2X.
NOTE: After doing some approximations and assumptions about mass distribution in a typical wheel+tire combo, I believe that 1.7X is a better approximation. A
10lb/wheel mass increase would not hurt acceleration worse than carrying
RinTinTin.
