Wondering how much that extra mass might effect things - I calculated the effect on acceleration.
I assumed the weight difference to be out at 25" diameter (thinking most of the difference is in the tread and/or the structureright under the tread, not in the sidewall) and I get the result below for a 2nd gear pull. Does this look about right? Equivilant to losing 3 ft-lbs of torque at peak acceleration.
The rotational "loss" is all but the same as the "loss" due to just putting on 20 lbs of non-rotating weight, so it's equivalent to adding 40 lbs of non-rotating weight. That is about 1% of the car weight, and 3 ft-lbs is close to 1% of torque - so it seems reasonable.
For wheels ... assuming weight change would be near the barrel diameter - 5 lb heavier/lighter would result in about 1/2 of the rotational loss as the tires - a result of the weight difference moving from 25" in to 18" (barrel of my 19" wheels). Half the wheel rotational loss and same linear loss is around 2.25 ft-lbs. That's for 5 lb lighter wheels on all 4 corners.
For brake rotors, assuming 13" diameter - it's only about 0.2 ft-lb due to the rotational effect - so it's nearly all just the linear portion of the weight - totals about 1.7 ft-lbs for a set of 4 rotors that are 5 lbs lighter each.
If I haven't made any mistakes, the extra weight isn't likely to be a large enough difference in acceleration to influence my purchase, but what does 5 lbs on the tires/wheels do to handling? - no simple F=ma calculation that I know of for that...
Comments/thoughts?
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