[Hatch]

Does anyone know the calculation to find out how much your wheels weigh while spinning? Hence how much actual weight you will save going with light weight rims.

Jeff

 

[chronoti]

erm i havent taken physics for about a year now and that was in high school ... but i was really sure that even if something is spinning it still whighs the same. they would have less enerita thus requiring less force needed to move them and slow them down but other than that i guess minus orginal whight by new whight.

 

[Hatch]

I thought that the more a wheel weighs the more power it takes the push those wheels, and the faster it sprins the more resistance there is, i could be wrong but i coulda sworn...

 

[SBCelicaGT]

Uhh... What? The wheels do not change weight no matter how fast you are going. Do you mean momentum? Theoretically, one could use kinetic and potential energy equations to figure out how much faster your car will be if you save 5lbs per wheel. Or you could use angular momentum equations to figure out the differences. You could even figure our the difference in horsepower transmission from the axle to the wheel if you wanted to.

In reality, with all the variables in an accelerating car, it would be pointless to do any sort of calculation at all.

 

[BenG]

The problems with wheels is that the bigger you go, the more mass on the ouside of the wheel (ie tire and barrel) and the farther the mass is from the center of rotation, the more energy it takes to get the whole thing going (something to do with intertia and momentum). So it's not that the wheel weighs more as it spins, its that the evergy needed to start and stop the wheels motion is perceived as more weight on the car. 4 or 5 pounds on the circumference on each wheel can potentially feel like you're carrying an extra 100 lbs in the car. You'd need to know your physics to figure out the exact numbers, and even then, those calculations are only good for comparing between two different wheel weights and sizes. It's a relative thing, not an absolute thing.