There is no doubt that the maximum rotational crank/pedal deflection (as opposed to any lateral deflection) will be when the crank is in the forward horizontal position close to where the pedal force is maximum. All the rotational deflection doesn’t disappear until the pedal force producing rotation of the crank is zero. This should be near bottom dead center (unless you are an appostle of Frank Day and his Powercranks). So, it is over this quadrant that the stored energy in the crank is dissipated somewhere. I believe most of the energy is lost in diminishing the rotational speed of the lower leg while only a small amount provides forward motion of the cyclist. Why do I think this? I imagine this situation being somewhat similar to a small cart being driven and pushing a much larger cart. There is a spring between the two which is compressed and transfers a pushing force from the smaller cart to the larger cart. So what happens to the energy in the spring when the smaller cart reduces its pushing force to zero over a short time interval. This is somewhat like an elastic collison where momentum is conserved. IOW, the momentum gained from the push of the spring by forward motion of the larger cart must be the same as the momentum lost by the smaller cart (the force in the spring pushes the larger cart forward and the smaller cart backwards). Now if the ratio of masses is 10:1, then the decrease in velocity of the smaller cart will be 10 times the increase in velocity of the larger cart. The change in energy (increase) of the two carts will be equal to the energy returned by the spring. I’ve taken a few liberties in ignoring some things to better describe the first order effect. For a bicycle crank, the ratio of crank motion to forward motion of the bicycle will have a significant effect on the distribution of the energy in the crank. I say this because the force on the crank is reduced considerably by gear ratio and crank arm/wheel radius ratio. Assuming this ratio is 25mph for the bike versus 5mph for the crank, a ratio of 5:1, the distribution of energy will be a combination of the equivalent mass and speed ratios. Assuming a rider mass/lower leg mass ratio of 10:1 and a speed ratio of 5:1 then the distribution of energy will be close to 50:1. That is, 50 parts lost to the lower leg versus 1 part producing forward motion. IOW, this is so small it is easier to assume all the energy is lost. My 2 cents.