Re: rim size and torque/speed
It is about torque. A torque is just a moment about an axis. An engine can deliver a certain highest torque at an rpm. A tensioned chain leaving a sprocket always leaves tangentially, which is 90deg to the radius line from the axle and is the maximum torque angle. The tangential force (tension) at both sprockets is the same, but on the front sprocket you have a smaller distance from center than on the wheel sprocket.
So the torque on a wheel is increased by the use of reductions. The wheel exerts the same torque on the asphalt. The force felt by the asphalt is only the force pushing you forward- that kick to the stomach in a fast car. A 26" wheel has a shorter distance from the center which makes the larger part of T=f*d*sin(a) the force pushing you forward (f).
So to calculate, let's say you have an engine rated to 5lb-ft torque at 2500rpm. Drive is 11t and 88t #41 sprockets. The factor between them, 8, is the same for pitch radius and tooth count (drive ratio). So
wheel torque = (5lb-ft)*(8:1 ratio) = 40lb-ft
26" wheel radius in feet: 1.085ft
29" wheel radius in feet: 1.21ft
forward force 26" = (40lb-ft)/1.085ft = 37lb
forward force 29" = (40lb-ft)/1.21ft = 33lb
the 26" wheel will pull 12% harder than the 29" at 2500 rpm.
I don't always listen to Alice in Chains..... but when I do, SO DO THE NEIGHBORS
Last edited by Tony01; 07-25-2015 at 10:11 PM.