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I read somewhere that for ever pound, times that by 4 (4 rims) then by 8 (I don't know why 8). So according to this calc. you save 128 Lbs.


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SO if my 18 inch chrome wheels weight 31 pounds each, then do I do the opposite ... 20.5 -31 = 10.5 x 4 x 8 = 336 pounds heavier than stock ???

Add in the 250+ pounds of extra stereo in the car and I'm rolling with 600 extra pounds in my car at all times. OUCH!



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WOW, anyone ever take physics? Those calculations are totally bogus! Think about it, its not just the weight of the wheel but where the weight is located. If the weight was perfectly centered as a thin rod in the center of the wheel then the difference between a 16lb and a 20 lb wheel times 4 would equal 16lb, minus the effect it has from not being suspended weight, it is the same difference as if you had removed your spare tire from the trunk (assuming it was 16lb). Now if the weight was uniform along the circumfrence of the wheel and you reduced the weight by 4 pounds for each wheel then there would be a gain in acceleration as less energy would be required to rotate the wheel. In your case if it feels like its faster and the wheels are lighter, most likely it is because there is less weight concentrated along the circumfrence then what you had before. It is possible to have a heavier set of wheels net a bigger gain in acceleration then a lighter set, if the lighter set has the weight concentrated along the circumfrence and the heavier set has the weight concentrated in the center of the wheel. Just some food for thought.


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Come to think of it, could this be the added stress the powertrain has to pull along, not actual Lbs?
"It's like 128lbs"?


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NICE Wheels Marty!!


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Originally posted by 96mercury:
WOW, anyone ever take physics?




Yes!
Now that that is out of the way....

True, rotational mass is dependent on distance from the center of axial rotation. But calculating them for a wheel is really quite a bit of work, so everybody just uses the 1 lb off the wheel is equivelant to 8 pounds of static mass.
Although I am not sure if that is an average approximation along the diamter or if does assume that the loss weight is along the circumfrence of the wheel.


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