I am completely perplexed by this whole “torque” thing as described here. It’s been a few years since I took Statics/Dynamics but as I recall: Torque = Rvector “cross” Force-vector. Thus Torque is a vector with a direction perpendicular to both the Rvector (the linear vector from the force application point to the torque generation point), and the Force-vector.
A magnitude of torque is attainable by multiplying the magnitudes of the Rvector (length of crank-arm) and the Force-vector and the “sine” of the angle between them.
On a bike, a person generates torque in the crank. If one is pedalling nice round circles, the force application is always perpendicular to the crank-arms. So the angle is always 90, and sine of 90 is 1 (max).
By contrast, if one is standing on a pedal at the bottom (not rotating), the force application direction is parallel to the line of the crank-arm, the angle is 0 (or 180), and sine of the angle is 0. That’s why standing on the pedal at the bottom does not generate rotation, even though the force is maximal.
The torque generated in the crank is not used in the crank, but is converted back into a linear force in the chain: the new Rvector is the line from the center of the crank to the top where the tangent chain contacts the chain-ring, and the new Force-vector is along the top of the chain. These are nearly perpendicular; designed that way for obvious reasons.
The force in the chain is converted back into a torque in the rear wheel, driving it. Again the R-F angle is nearly 90 for maximum transmission.
Going back to the point of initial power generation (the crank), the basic assumption made above for maximal generation was “pedalling round circles”. By contrast: If the only force generated were an up and down force (similar to what happens when pedalling out of the saddle, or pistoning), then the F-vector is only perpendicular to the cranks at 3-oclock and 9-oclock positions. At 12 and 6-oclock positions, the F-vector will be parallel to the R-vector producing Zero Torque.
But… Force application and conversion is only part of the story. Energy conversions and work need to be addressed in this equation as well. Unfortunately, we’re getting WAY off the subject of the original post and for that I apologise.
I did (somewhat) follow what you were driving at Frank. Unfortunately, my knowledge of biomechanics is minimal so I don’t feel qualified to properly address your points. I do have some thoughts, but no real answers for you. Personally, I’d combine 2 & 4: these would be conversions of energy in the muscle (heat) due to less than perfect (muscularly) pedalling. I think this is your primary area of concern. Number 3 seems somewhat irrelevant to your concerns (non-muscular). Your estimate for 1 seems right to me: energy lost due to the need to recover the leg to the top. Pulling up on the pedal (round pedalling) ensures continuous power transmission to the wheel, but a portion of that “pull-up” energy is used simply to raise the leg, foot, shoe, and pedal. Some of that energy might be recovered on the downstroke, but there would be losses. Good luck, and I look forward to hearing about any conclusions you reach.
Adam