...interesting...thanks....I was hoping to see more equations!!!

 

magician,

if you're familiar with the formulae, what would be outcome of a 2" bigger diameter but 4 pound lighter per wheel??

i.e. 16" dia. wheel to a 18" dia wheel. and a 20 pound wheel to a 16 pound wheel.

the 16" wheel = 20 pounds, the 18" wheel = 16 pounds.

 

^ Good one! The answer should be most revealing. I suspect that under 60 mph, not much difference, but above 100?

 

I would have thought the largest impact would be on initial acceleration (going to a lighter wheel), due to the increased amount of torque getting to the ground, as opposed to simply trying to spin the wheel.

 

I'm sure that if the equation is calculated using those numbers that Nimesh provided, we'll see that it would have small consequences for getting going. With centrifugal force, weight is amplified tremendously with increasing rotational speed.

 

I'll do the straight F = ma part first.

If each wheel drops 4 lbs. (20 lbs. - 16 lbs.) then the total weight of the car drops 16 lbs. (4 wheels). If the weight of the car and driver were, say, 3,000 lbs. with the original wheels, the new weight would be 2,984 lbs. The new acceleration would be 3000/2984 = 1.0054 times the old acceleration, a gain of one-half of one percent. Road & Track lists 0-60 at 5.5 sec and

 

Say that fast five times.

Thanks for taking the time to do this. So the people that are saying they are experiencing slower acceleration (not measured, just seat of the pants) are experiencing a placebo effect?

 

What about the issue that rotational weight is (theoretically) "worth" 6 to 8 times static weight. (Ie: if you shaved 2 lbs off each wheel, 8lbs total, at 6 times what that is worth in static weight, this is like shaving off 48 lbs).

 

Mea culpa!

I erred in converting inches to feet. (Sheesh!) I've fixed it.

 

No, but your comment stemmed from my original (incorrect) calculations. Now that I've corrected my error, you can correct your post, and blame it on me.