Quote:
Originally Posted by shabarivas
ok if we assume the wheel is a uniform disc... of weight M then the moment of inertia of that wheel is 1/2 * M * R^2 ... so... with a 19'' wheel - lets say it weighs 25lbs the moment of inertia is about - 1/2 * 25lbs * 19 * 19 = 4512 lb*inch^2
now lets say the new wheel weighs about - 24 lbs so a 1lb saving... moment of inertia = 1/2 * 19 * 19 * 24 = 4332 lb*inch^2
ratio of difference = 4% less moment of I w/ 1lb diff in wheel
so ... what that tells us is... roughly - shaving 1 lb off per wheel - makes it 4% easier to spin that wheel ... hope that makes sense 
EDIT: for more info: Moment of Inertia, Thin Disc |
Nice. I have some followup questions though:
1) Does it make any difference whether the weight is being dropped from the driven (rear) or rolling (front) wheels? I'm inclined to think it doesn't matter, as the engine is pushing both rotationally, assuming no appreciable tire slip being factored in for a simple case.
2) How do we apply this to the car as a whole in order to determine how much effective % difference in acceleration a weight drop at the wheels makes? Let's say I run the numbers for the whole wheel/tire package at all 4 corners, using the combined weights of the tire+wheel at each corner and the radius of the tire+wheel. Let's say I shaved off 4% per corner. That doesn't really mean the same in terms of overall acceleration numbers as, say, a 4% drop in car body weight does it?
I would imagine one would have to calculate moment of inertia stuff for all rotating masses (so the hubs, rotors, wheels, tires, etc, plus the driveshaft, the rear axle, the rear diff, the transmission, the flywheel, the engine itself, etc??? it's a long list), and then somehow combine that and the static weight of the car into some final number for how much force has to be applied to move the car a foot, and then see what % that number drops from removing weight at just the wheel as a percentage of the whole, right?