Originally Posted by
njlonghorn
Like the OP, the question has crossed my mind, but in a different context. As I spin up a steep hill, creeping forward at 5 mph or so despite a cadence in the 90s, I often wonder how far my feet are travelling.
Unlike the OP, I had no freakin' idea how to make the calculations.
Logically, the difference between d_feet and d_bike would be much larger in this context, but how much larger?
Your intuition is correct: your feet travel a much larger fraction of the distance in this case. Here's a plot of your shoe trajectory for one pedal revolution at 95 rmp cadence and 5 mph speed (axes in meters):
Note that the cycloid shape is much more rounded and less stretched out. In this case, integrating along the cycloid to determine its length I get:
d_shoe/d_bike = 1.16
This is very close to the value of d_shoe/d_bike=1.19 obtained when treating the circular motion of the pedals and linear motion of the bike independently. So it turns out that the OP's initial thinking gives a good approximation in the limit of a slow speed and high cadence (as one might expect).
