Conversation Between imaking133 and aackshun

So the weight of where the tire mounts to the rim does not factor in at all? I have an experiment planned out this weekend, should imitate what forces the vehicle (axle) sees under load.

You guys are doing calculations based on the diameter of the rim when in actuality the tire is what is on the ground. If the diameter of a wheel/tire combination was 18 and the other was 17, then yes I can see where the 18 would take more power/time to stop, given they were the same weight. Because the rim is not what is in contact with the ground, you cant use the radius and the diameter of the rim in your equation.

I don't want to get off topic in the braking thread so I'll post this here....

Lets say the given VQ makes 250 Ft/lbs (easier math).

Using I = MR^2 and tau = I*alpha

the 18" and 17" wheels will weigh 20lbs for the exp.

The 18" wheel is .15 alpha while the 17" will be accelerating at .17 alpha or to put it in a more relatable perspective...

To accelerate a 18" wheel as much as a 17" wheel (at 20lbs a wheel) it will take 35 more ft/lbs!!!