In Reply To:
Frank,
"But, once in motion the energy variation of the thighs must be absorbed somewhere in the system or it must be lost to the system because the total energy of the system can never go up once external forces are removed. It cannot be transferred to the speed of the bike because the different masses and speeds cannot be made equal because the gearing is fixed."
No one said the energy of the system in free motion had to go up. Why do you think the different masses and speeds need to be made equal? There will be a proportionality based on crank position, but so what? The rate of change of bike+rider speed is influenced by its inertia when viewed as a single mass combined with the inertia of the thighs. The rate of change of the thighs is influenced by their own inertia combined with the linear inertia of bike+rider. That inertia of bike+rider is a relatively large reservoir of kinetic energy which is added to or taken from, as appropriate. I think you must be thinking of the thighs and the bike+rider mass partly as independent systems, but they aren't. (Maybe you've spent too long thinking about independent cranks?)
Would it help to conceive of the MMF as a single-legged rider? Then gravitational potential energy enters the issue, but surely you can see that the rate of fall of the leg will be limited by the need to accelerate the bike+rider, and the leg will rise again by taking kinetic energy out of the system. Actually, in the first half of the descent, the rate of fall will not only be limited by the bike+rider acceleration, but also by the acceleration of the thigh; the rate in the second half of the descent will be limited by the bike+rider acceleration, but assisted by the deceleration of the thigh; and so on. Does that help?
I think it also might help you to make a simple Excel model and for small delta-angle of the crank work out pedal force, effect on bike speed and thigh speed over the related time interval, and work your way around the crank circle. Keep track of the sum of kinetic energy of the thighs and kinetic energy of the bike+rider. At some point the light will go on, surely. The 'work done' (component of pedal force tangential to the circle times the distance covered over the iteration cycle) is what adds to the kinetic energy of bike+rider and subtracts from the kinetic energy of the thighs. The same number is used as a basis for both.
"A perfectly rigid material cannot absorb the energy because it would fracture."
Well, if you are talking about mechanical deformation, then it would be undefined. That is obviously what was being excluded. There is still the possibility of thermal, gravitational potential, kinetic, and chemical potential energy (maybe I've missed some other types as well).
"As far as I know, that material does not exist."
Obviously. The point is to isolate different things to enhance discussion and understanding.
"So, where is the energy lost to? Energy must be lost through materials distortion and hysterisis losses (heat), the only place it can go in a frictionless system."
So the implication is that an ideal MMF could not move? The pieces would just shatter?
What your ideal MMF proposes is
a perpetual motion machine. At least I know I have one supporter on my side in this debate,
the US Patent office.
Simply show me a mechanism by which the energy of the different parts of the bicycle, when totaled up, remain constant while this bicycle is "coasting" along such that it would continue to coast forever. I will allow you to have frictionless bearings, joints, and chains. But, everything else must be real material with mass.
The difference between this problem and the "double pendulum" problem someone else presented is the double pendulum was simply converting potential energy into kinetic energy. In the MMF case there is a need to convert kinetic energy into kinetic energy through a fixed mechanism to keep the total energy constant. Good luck.
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Frank,
An original Ironman and the Inventor of PowerCranks