| Okay…I can play ball…let's make this Z related. I know this topic has been covered before about people using smaller diameter tires at the drag strip to improve their final gear ratio. Link My question was how to calculate the force on the road produced by the wheels given that the torque remains the same and diameter of the wheel is changed? Since I’d rather not be called a b**** a** hypocrite again I’ll take a stab at answering my own question and you can all tear it apart and tell me how retarded I am ;-) Say I’ve got a TT with 350ft lbs of torque at the rear axel and I’m running on 25” overall diameter tires. Following the formula for torque, T=F*d where T=torque, F=the force perpendicular to the axis at distance d, and d=the radius of the wheel. That yields the equation 350ft lbs = F(lbs) * 1.0416 ft (radius of 25” tire in feet) Solve the equation for F and F = 350ft lbs/1.0416ft
F = 336.02 lbs This would be the force perpendicular to the radius of the tire…which would be the force applied to the road surface. So what happens if I decide to change my wheels to an overall diameter of 23”? For this case the torque at the axel remains the same 350ft lbs and the radius is now 0.9583 ft
350ft lbs = F(lbs) * 0.9583 ft Solve for F, F = 350ft lbs/ 0.9583 ft
F = 365.23 lbs A difference in force of 365.23 – 336.02 = 29.21 lbs
So by decreasing the overall diameter of my wheels/tires from 25” to 23” I have increased the force applied to the road by 29.21lbs per wheel. Obviously this is a simplified equation that does not account for friction, drag and many other variables…I was just wondering if any of you physics gurus could tell me if I was on the right track. Rip away…
 "Anyone that gives him the link is a b**** ass fish giver." - YugoBernie (NoVA) |
