Has anyone bothered to work out the force required on one's pedal

to maintain motion up gradient X

in gear Y

(as a multiple of bike+rider weight)

?!?

D = distance moved on ground for 1 pedal stroke (meters)

P = distance moved by pedal for 1 pedal stroke = 2*pi*crank length (meters)

S = bike speed on the road (meters/second)

M = mass of rider and bike combined (kilograms)

G = gradient of hill = height gained / height travelled horizontally = tan(gradient as an angle)

g = acceleration due to gravity at earth's surface = 9.8

Rate of vertical ascent R = SG

Force required to lift mass M at vertical speed R, F_lift = M x acceleration = M(R+g) = M(9.8 + SG)

Model the cyclist as applying equal force to both pedals at a uniform rate over the full pedal rotation.

Ratio of vertical distance travelled to pedal distance travelled = DG/P

DG/P is also the ratio of pedal force to lifting force F_lift, except we need to divide by 2 to account for both pedals.

Therefore the force applied to the pedals in order to climb a hill of gradient R at speed S, assuming you push just as hard on each pedal for the entire pedal stroke

F_pedal = DGM(9.8+SG)/2P

D/P = CN_f/2*pi*LN_r where N_f = no. teeth on your chainring, N_r = no. teeth on your rear sprocket, C = wheel circumference.

Therefore F_p = (GMCN_f(9.8+SG)) / (4*pi*PLN_r)

I'm not sure whether there should be a gradient squared term in there, can anybody comment?

You could also model the rider as pushing on a single pedal at 90 degrees, using torque calculations. However, I'm off to get a cup of tea!