mechanical vs hydraulic clutch physics
Pedal force * pedal travel = throwout bearing force * throwout bearing travel
should apply to either type of clutch (work in = work out). So assuming pedal travel and the pressure plates are the same, the pedal effort should be the same. Am I missing something? Or are cable clutches limited to a shorter pedal travel for some mechanical reason? Something doesn't add up.
Last edited by Gary Z; May 25, 2006 at 03:35 PM.
pedal force = throwout bearing force/slave cylinder piston area * master cylinder piston area * master cylinder moment arm / pedal moment arm
and
pedal travel = throwout bearing travel * slave cylinder piston area/master cylinder piston area * pedal moment arm/ master cylinder moment arm
multiply the two equations together and you get
pedal force * distance = throwout force * distance
How is it not as simple as that? The fact that you're using hydraulic advantage (is that a term?) instead of leverage doesn't change the "work in = work out" relationship.
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