Wheel weight. Does it mater all that much
The main issue is if the extra traction you'll gain by going with a wider wheel/tire setup outweighs the increased weight and rolling resistance. Tire selection will play a big part in that. You could also look for lighter wheels in the size you want to keep any weight increase to a minimum.
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I'm not sure I understand this. Are you saying 4lbs of rotating mass ie wheels and tires is equal to only 1lb of extra weight in the car meaning if I added 30lbs of wheels and tires it would be the same thing as roughly 120lbs to the car in dead weight?
The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/(diameter of your rim) * mav.
Regardless, a drop in wheel weight can have significant impact on the performance of the car, and it would be relatively higher than a similar reduction in non-rotating mass.
Last edited by westtexasbuff; Jun 5, 2011 at 11:47 PM.
If my wheels weigh 25 lbs each, thats 100 lbs rotating(R). 400 total R
Lets say I replace with wheels that weigh 12.5 each (50 lbs R). 200 total R
According to the rule of thumb, I lose 2 tenths???
Pretty sweet, Thanks for posting. I tried explaining to a friend and I estimated it was x3 instead of x4. I was pretty close and glad to learn this





