What is the maximum achievable driving force (in units of Newton) of a bicycle when one pedal is loaded with the full weight of the driver? The gear shift is set to the lowest gear ratio.
Details and Assumptions:
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The cogs have each a circumference C i = 2 π r i = n i d ⇒ r i = 2 π d n i , i = 1 , 2 with a radius r i of the the cog and distance d between two teeths, that is equal the length of one chain link. The torque acting on the pedals and chainring results T 1 = r 1 F 1 = m g l The torque acting on the rear wheel and on the pinion results T 2 = r 2 F 2 = R F wheel With F 1 = F 2 the equations can solved for the driven force: F wheel = R l r 1 r 2 m g = R l n 1 n 2 m g The force is maximal for the teeth numbers n 1 = 2 4 and n 2 = 3 6 : F wheel = 3 5 cm 1 7 . 5 cm ⋅ 2 4 3 6 ⋅ 7 0 0 N = 4 3 ⋅ 7 0 0 N = 5 2 5 N