No, you haven't shown it is actually possible to move the KE around with the speed constraints placed on the various parts because they are connected by a fixed gear. Here is an example of what I mean. We know that the energy varies as the thigh moves up and down and that this energy will increase as the cadence goes up, with the square of the cadence. So, let me throw out a couple of scenarios. Let's say the thigh weighs 14 kg and the max speed variation is 1 m/s. Since there are two thighs and they accelerate and decelerate at the same time we have 28 kg showing a speed variation of 1 m/s. This energy variation must be transferred to the bike to keep the total energy constant. Let us put gearing on the bike that has the bike move 1 meter per crank revolution. If the mass of the rider/bike is 70 kg wouldn't we expect to see a speed variation of 28/70 m/s in the bicycle to account for the decrease in energy as the thighs move from max to minimum speed (1/4 of a second at a cadence of 60). If the bike was moving at 1 m/s at the start it now has to be moving at 1.4 m/s 0.25 seconds later and then at 0.5 seconds be back to 1 m/s as it transfers kinetic energy back into the thighs as they speed up. This is a 40% speed variation every 0.25 seconds. If we double the cadence, we increase the speed variation necessary to account for the energy difference but decrease the amount of time allowed to change the speed. Under this principle, high cadences, rather than "smoothing" the speed variation of the bicycle would increase the speed variation. Go to any track and see if how much speed variation in those bikes there are and then report back.

If we change the gearing, everything changes. If we change the masses, everything changes. If we change the cadence everything changes. Yet, you (and others) are telling me the MMF model will work under any and all mass, gearing, and cadence conditions. Give it a push and it will go forever. I don't see how it can work under any condition.

While you may think that if you could suddenly develop that perfect material that these energies could magically be transferred you need to show me that it is possible mathematically. If it is not possible mathematically then it is not possible, regardless of the materials you use. If the energy cannot be transferred to keep the total energy constant then there must be energy losses and the model will eventually come to a stop.

So, until you can come here and tell us how the MMF model works under any and all conditions to exactly transfer the right amount of energy back and forth between the thighs and the bicycle I will accept your silence as your quiet surrender. :-)

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Frank,
An original Ironman and the Inventor of PowerCranks

Prove it.

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Frank,
An original Ironman and the Inventor of PowerCranks

Let me alter my example to make this "problem" of the energy transfer more obvious. Let's gear the bicycle such that one revolution of the cranks moves the bike 1 cm. According to the MMF model this should not be a problem, give the bike a push and those legs are going to go around very fast (afterall, there is no friction and the materials are perfect) but there are going to be very big "thigh" energy variations and how the hell do these get transferred to the bike with that gearing.

Go for it. I look forward to seeing your solution.

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Frank,
An original Ironman and the Inventor of PowerCranks

Dr. Coggan, Could you provide a reference for the above highlighted statement? Thank you in advance.

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Frank,
An original Ironman and the Inventor of PowerCranks

I'll make you a deal...I'll lay out the math of the whole thing for you as soon as you identify by what mechanism energy is being removed from the system if the links are rigid and the joints are frictionless.

Until we can get you to understand both the first law of thermodynamics (energy can be neither created or destroyed, it can ony change form) and Newton's first law of motion (an object in motion stays in motion, and an object at rest stays at rest, unless acted upon by an outside force) it's really useless to go any further...

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http://bikeblather.blogspot.com/

Energy cannot be removed from the system if the links are perfectly rigid and the joints are frictionless. That would not be a problem if the energy could be transferred around from one element to another to keep the total energy constant. From the example I gave above, it is clear that is not possible under all circumstances and, if your model fails in one circumstance it, most likely, fails in all circumstances. So, I don't have to explain how the energy is removed if you have perfectly rigid components because it can be shown it can't work while conforming to the laws of thermodynamics. Energy must be either stored and released in the system (through components that are perfect springs, not perfectly rigid) to maintain constant internal energy or energy must be lost from the system through material hysterisis, what actually occurs in the real world in addition to frictional losses.

Regarding your "thermodynamic principles", the thigh is constantly changing its velocity during the pedal motion. Therefore, we know that it is being acted upon by an outside force (outside of the thigh but internal to the MMF). For the MMF to work as you describe the source of that force must perfectly balance the energy change internally to the model (i.e., the total energy of the system constant). If it cannot, which it is clear to me that it cannot, then it is your model that is violating the principles of thermodynamics, not my understanding of it.

AFAIK, my understanding is the only explanation that actually makes any thermodynamic sense unless you can prove to me that this energy can be conserved. As I said, I am waiting.

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Frank,
An original Ironman and the Inventor of PowerCranks

Frank,
I prepared a crude simulation of the physics of the rider-bike-thigh system. It shows quite well that the speed of the crank varies by about 0.5% between max and min, to keep KE of the bike + KE of the thigh constant.

Albeit, as I wrote, it's a crude approximation: one thigh linked to the pedal through a weightless lower leg. But the principle is the same, just the calculations get more complicated. The distorted sinusoid (rotated by 90 degrees) simulates 1/4 of a turn of the crank, therefore keeps repeating along the vertical axis. The data I used are: man+bike = 85 kg; gain between crank and wheel = 5 (2.5 gear ratio x 2 ratio between wheel 720 mm and crank 180 mm); thigh 14 kg concentrated at 1/3 of the distance hip-knee (to respect the physics of the momentum of inertia) + 5 kg of lower leg concentrated at the knee, for a total equivalent mass of 19.3 kg adding both legs; crank 0.18 m; speed before linking the legs to the pedals, corresponding to a cadence 80 rpm, V = 17 mph. In this model you bring the speed of the bike to regime, and reach a KE of the system of 2400 J. Then let the system go: you either pedal just hard enough to generate the wattage to compensate for friction and drag losses, thus maintaining KE, or let the system slow down because of friction and drag. In either case in the first few revolution the speed of the bike and the speed of the crank will keep going up and down by 0.5% to compensate for the accumulation and restitution of KE in/from the leg.

Changing the gear drastically as you suggest, to obtain 1 cm of travel for each crank revolution, will change the percentage of variation of the speed of the crank drastically.

The crude-approximation equation is: total KE = constant = 1/2 * ManBikeMass * (gain*omega*r)^2 + 1/2 * leg * (omega * crank * cos(omega*t))^2

Edit:
I had mistyped a parenthesis in my Excel spreadsheet. The corrected calculation shows a variation of speed of only 0.5%, and not 7.6% as originally written.

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Giovanni Ciriani
http://www.GlobusSHT.com

You do understand don't you that if the speed of the crank varies 7.6% in a quarter turn of the crank that the bike speed must also vary 7.6% in a quarter turn of the crank. At 20 mph and 60 rpm that means the bike has to accelerate from 20 mph to 21.52 mph in 0.25 seconds and then get back to 20 mph 0.25 seconds later and do this continually down the road, since we are talking about a fixie. While I haven't done the math here, I suspect the total energy change of the bike rider system is substantially higher than the energy change of the thighs alone, plus I submit such speed changes are impossible to see in real life, so I think you have missed something.

Help me out here. What have I missed?

Edit: I think I see what you did. You found one solution where you could get things to balance. But, that is not the MMF problem. The MMF, by definition, goes on forever regardless of the gearing and regardless of the speed, and regardless of the different masses. Change one aspect of your intial conditions and it doesn't work. What you have solved is not the MMF problem.

edit 2. In your solution with the bike at 17 mph and the cadence of 80, with a 7.6% variation in crankspeed we will find the bike needs to change speed only 1.2 mph in less than 0.2 seconds to satisfy your solution. While your solution may work for you mathematically, it doesn't work for me real world. I go back to thinking you have missed something.

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Frank,
An original Ironman and the Inventor of PowerCranks

What you are missing (and keep missing) is that you need to understand the basic, simplified solution before you start going and adding in the more "real world" conditions. Once you start adding in the aerodynamic loads (which vary with the square of the velocity), the frictional losses, and any small elastic member effects, THEN you'll start having a model that better matches your empirical observations and you'll better be able to understand the magnitudes of those effects. However, the math starts getting pretty messy fairly quickly...and based on your demonstrated lack of understanding some fundamental principles of physics, who's going to bother laying out the more complex case when you can't even grasp the simple, idealized case?

So, again, in the simplified model, how would energy be removed from the system if it can't be removed as heat? Have you discovered a new form of energy? Shall we call it "Day Virtual Energy"? Until you understand that the energy of the system is conserved there's no point in going any further and adding in other non-idealized energy storage or transfer mechanisms. I'm not giving you any math until you come to that basic understanding.

Stay with us here on the idealized system and stop trying to compare it to empirical observations just yet...especially since you keep trying to compare a driven case (your observations) to this undriven case. If you were to place a mannequin on a fixie and push it and let it go, you would see more speed variation through the pedal cycle than you observe with the driven case you describe (i.e. a rider on a track). Additionally, if we start adding in the changing KE of the lower legs and the rotational inertias of various constituents to the model above, do you think the speed variation requirement of the system would go up or down? Like I said before; Baby steps, baby steps...

BTW, did you catch your errors in your comparison of the thigh velocity vs. the system velocity above? edit: Besides the math errors, why are you comparing the tangential velocity of the thigh relative to the hip joint to the velocity of the entire bike (i.e. the tangential velocity of the crank through the gearing)?

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http://bikeblather.blogspot.com/

I didn't add in those aerodynamic conditions to attain equilibrium. The MMF model rquires zero frictional losses for the rider to go on forever. That would include air friction. I haven't demonstrated an lack of understanding of physics or thermodynamics. The MMF model that goes on forever is the one the require making or losing energy into thin air to make it work.
As I told you, if the elements are perfectly rigid there is no way to remove the energy variations that are there (unless you can show that they are not). Therefore, you, by your advocacy for this model you are the one determined to ignore thermodynamic principles, not me.
I was only commenting on the case that was presented to me. It was clearly inadequate. I suppose you can do better. I am waiting. Make it simple on yourself, assume th rotational inertia of the wheels is zero. If you can come up with a solution that works there I presume you can come up with one when the roational inertia is greater than zero.

Oh, and, I don't see that much speed variation on the track, after the event is over and the riders are simply relaxing, trying to mimick mannequin status.
I am not specifically comparing the tangential velocity of the thigh to the bike. What is important is the KE of the thigh. The tangential velocity is simply a method of expressing the KE because they are directly related.

The issue here is not whether I have made a math error in criticizing another post but whether you (or anyone else) can come up with a mathematical solution to the MMF problem that doesn't involve the magical creation or disappearance of energy to keep the energy of the system constant. I continue to patiently wait. It is a complicated problem, it will probably take some time (like forever, LOL).

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Frank,
An original Ironman and the Inventor of PowerCranks

Actually Tom, my simplified model takes into account the changing KE of the lower legs. I hypothesized 5 kg for each. I guess it must be less than that.

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Giovanni Ciriani
http://www.GlobusSHT.com

Frank,

In response to Giovanni's comment, "the speed of the bike and the speed of the crank will keep going up and down by 7.6%", you wrote, "You do understand don't you that if the speed of the crank varies 7.6% in a quarter turn of the crank that the bike speed must also vary 7.6% in a quarter turn of the crank". I don't like your reading comprehension, but I agree that the magnitude of the number looks off. But whereas I think the variation should be less for energy to balance, you think it should be more.

You want a math solution. I gave you a way to perform the math solution (search for 'Excel' in this thread). But last night I thought of another way to try to explain it. That will come in another post.

P.S.: "It is a complicated problem, it will probably take some time (like forever, LOL)." I hope you aren't messing with us on purpose. Good, knowledgeable people are investing their time because they think you have the capability to understand and the integrity to submit to the science. Don't disappoint us.

Frank,
It is an energy-conserving system, and there is no magical creation or disappearance thereof. This is the way engineering problems are routinely solved. It's not different than the way the calculations for an internal combustion engine are done. The bike crank-leg kinematic chain is in principle identical to a piston-rod-crankshaft system. See http://en.wikipedia.org/...ton_motion_equations which has similar graphs.

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Giovanni Ciriani
http://www.GlobusSHT.com

Pedaller,
One can always make mistake, but a back of the envelope calculation should tell you if the number is off, or in the ballpark :-)

The corrected calculation gives 0.5% variation in V, which translates into an approximately 1% variation in KE. Since the ratio between speed of the pedal and speed of the bike is 5, 1 Kg on the pedal produces 1/25 of the KE due to translational speed. Because the ratio of the masses that are exchanging KE is 85/19.3 = 4.4, the ratio between total KE produced at the pedal and KE produced in translation is 25 * 4.4 = 110. The inverse ratio is 0.9% which is very close to the 1% fluctuation in speed. Therefore back-of-the-envelope check is in the ballpark.

Frank,
I don't mean with this to say that you are incorrect about optimal cadence being lower than what most bikers do.

Edit: Pedaller you were right. One calculation had a parenthesis out of order. The new back-of-the-envelope calculation is much closer to the result obtained.

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Giovanni Ciriani
http://www.GlobusSHT.com

My friend, those numbers do not demonstrate to me that it is an energy conserving system. Just tell me what your calculations tell you is the total energy of the system at max V and at minimum V. I simply don't believe they are the same. Oh, and the MMF problem is not quite the same as the piston problem you linked to since there is no translation of a separate mass into a longitudinal speed (edit: and the "piston" is offset). That is the problem we are trying to solve.

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Frank,
An original Ironman and the Inventor of PowerCranks

So, to solve the problem we have to have the changes be the same. And your back of the envelope calculation shows they are close, so that is good enough for you to "prove" to everyone that they are the same. Is that what you just said?

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Frank,
An original Ironman and the Inventor of PowerCranks

Thanks for your calculations!

But I think you made some assumtions that each make the speed variations of the bike higher than in (racing) reality.
- The speed of the bike is a little low.
- The gear ratio is a bit low (about 53-21).
- The legs are a bit heavy (I think about 40% of the body weight would be closer).
- Concentrating the mass of the lower legs in the knee results in big speed variations for this mass. In reality the speed of the lower legs varies only little.
- And finally you probably have not calculated the rotating wheels, have you?

@Frank
Have you still not found any knowlageable person in your company, who can help you out of your ignorance?

LOL. When someone can actually come here and show some calculations that proves the point that I am ignorant then I will look for some help. What about you? Surely you can do the calculations to show that the energy is conserved under all circumstances, the MMF model we are talking about.

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Frank,
An original Ironman and the Inventor of PowerCranks

 
Quote

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Lidlracer
But I think you made some assumtions that each make the speed variations of the bike higher than in (racing) reality.
- The speed of the bike is a little low.
- The gear ratio is a bit low (about 53-21).
- The legs are a bit heavy (I think about 40% of the body weight would be closer).
- Concentrating the mass of the lower legs in the knee results in big speed variations for this mass. In reality the speed of the lower legs varies only little.
- And finally you probably have not calculated the rotating wheels, have you?[/reply] Lidlracer,
I had misplaced a parenthesis in my Excel spreadsheet (I have corrected the graph), and now the fluctuations in speed are only 0.45%, which is hopefully closer to what everybody expects. I think you are right that your suggested changes would make the model more realistic, and reduce the speed fluctuation even further. And no I have not added the rotating wheels, for sake of simplicity.

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Giovanni Ciriani
http://www.GlobusSHT.com

Frank,
I have the utmost respect for you, and I mean it. So let's maintain a courteous tone.

Back-of-the-envelope calculations are not demonstrations of the physics of the problem, they are merely verifications that the numerical calculation were performed correctly. Actually by redoing it in my head I found there was indeed a parenthesis out of order, as I have re-posted (0.45% fluctuation now).

Even though piston-rod-crankshaft, or the crude model I calculated in a spreadsheet, do not reflect 100% the physics of cycling, the principle remains the same. The KE of the system, before losses caused by friction or myo-filaments sliding against each other, remains the same. The thigh will be accelerated up and down, by the motion of the pedals, and the opposite deceleration up and down will transfer back to the bike the same amount of energy. The same happens to the piston and crankshaft. So that is not the place where you have to look for the losses. I agree with you that the muscles are not efficient when pushing in a direction not perpendicular to the cranks, because they cannot convert in energy the part of the push that is isostatic. But this is an important consideration that comes after the physics of the problem.

The losses have to be searched in the sliding of myo-filaments against each other. But to find these losses at the there is a more practical method, the force-velocity curve of the whole pedalling leg. To determine the optimal cadence one has to look at the fact that the muscle has different efficiencies of converting metabolic energy into energy of movement. Specifically, looking at power-velocity curves for the whole leg, one can determine the most efficient speed, and then work backwards to determine the optimal cadence. I would expect it to be different for different athletes and for different demands, efforts, climbing etc.

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Giovanni Ciriani
http://www.GlobusSHT.com

The second sentence in that statement proves that the first sentence is false.

Why am I having flashbacks to the "wheel speed variations around the turns of a track" thread? :-/

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http://bikeblather.blogspot.com/

this thread is really long. can we get a platforms of AC vs FD.

:D

The piston and crankshaft problem is simply not the same problem as the MMF problem since, in the bicycle, the "piston" (hip) is fixed and the "connecting rod" is a two piece affair with a "knee" that moves up and down as the camshaft (crank) goes around. The losses are entirely different. In analyzing this MMF problem I am not looking at the muscle losses but only at the physics of the problem. I see no way the energy can be conserved.

Optimal cadence is an entirely different problem and, in my opinion, has so many contributing factors that it can only be known through trial and error (testing) of each individual athlete.

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Frank,
An original Ironman and the Inventor of PowerCranks

Look, if this is so simple all you need do is show that the energy of the MMF model remains constant under all mass, speed, and gearing conditions. As of yet, no one has come up with a single condition in which this requirement is met. Until then, as far as I am concerned, you can stop with the name calling and accept the fact you can't prove your assertion. Accept the fact there are energy losses from the pedaling motion. The only real issue is how large are they.

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Frank,
An original Ironman and the Inventor of PowerCranks

Frank,

You wrote, "If the mass of the rider/bike is 70 kg wouldn't we expect to see a speed variation of 28/70 m/s in the bicycle to account for the decrease in energy as the thighs move from max to minimum speed (1/4 of a second at a cadence of 60)."

First, assuming that the effective mass of the thighs is 28kg and the effective rate is 1 m/s (not too realistic, since the thighs are supported on the hip end), and making a bunch of simplifying assumptions, the rate ratio is way off. If you start at 1 m/s and 70 kg, Ek = 0.5mv^2 = 35J. The thighs will attain kinetic energy of (0.5)(28)(1^2) = 14 J. The theory is that this would be drawn off of the kinetic energy of the bike. So there will only be 35 - 14 = 21J of Ek. Since this is 0.5mv^2, we expect v^2 to be 21/(0.5m) = 21/35 = 0.6. So v = sqrt(0.6) = 0.775 m/s.

Alternatively, if you start with Ek at 35J when the legs are at 9:00 and 3:00, Ek will go up, and 0.5mv^2=49, v = sqrt(49/35) = 1.18 m/s.

So whereas you posit a change of 40%, you should have said 18-22%. In reality, the effective speed of the thighs is much less, the bike speed is much more (the gearing is much taller). If you found a thunderthighs whose mass together with his bike was 70 kg, and put him on a fixie with a 20 tooth chainwheel and a 40 tooth cog on the back (you could almost climb walls with that granny gearing) and tell him to ride around at 3.6 km/h, then for sure the variation in speed would be noticeable. If you can supply a 60 kg guy with thighs that achieve 14J of kinetic energy, and this wonky bike gearing, then yeah, let's put him on the track. You'll see it happen.

Even changing one of these parameters -- speed -- makes a huge difference. If v = 10 m/s, then Ek = 3500J. Then the new Ek would be 3500 +/- 14, and the new v would be sqrt( (3500+/-14)/35 ) = 9.98 to 10.02 m/s. So increase the speed 10x to realistic speeds, and the speed variation decreases from 20% to 0.2% of the starting speed, ie, by a factor 100. And we can go to the track and [fail to] observe such invisible changes in speed too, if you want, even if you find your 60kg thunderthigh guy to ride the bike.

The issue of tracking the energy (re-proving the law of Newton and of thermodynamics, as Tom A says -- which we shouldn't have to do) is made somewhat complex if the entire system is looked at simultaneously. But let's break it down. For a given point in time, let us call the force exerted on the pedal spindle by the foot F (a vector). The spindle moves along the tangent (circumference) of the crank circle. Let us call the differential motion vector dC, corresponding to the motion covered in a time interval dt. The work done by the foot is the dot product of these two vectors: dE = F dot dC, which I'm going to write as F.dC for brevity.

Regardless of whether this pedal force is due to gravity, muscular force, inertia or whatever, it does not matter. Assuming an efficient drivetrain, this will translate into kinetic energy. If we write E = 0.5mv^2 in differential form, it is dE = m*v*dv, or dv = dE/mv. (Note that the change in velocity for a small change in energy decreases with increasing velocity. This is why at higher speeds the speeding/slowing effect is much less, and at lower speeds it is quite perceptible.) So the expected change in rate is dv = dE/mv = F.dC/mv. It is trivial to extend this to angular energy as well, so we will ignore that aspect for the sake of simplicity.

Now, looking at the vectors F and dC as defined, one can see that just as the foot exerts F on the pedal spindle, so the spindle exerts -F on the foot. The work done by the spindle on the foot is -F.dC. Interesting: precisely the energy added to the bike+mass system is the energy lost by the leg system, and this applies also in the case of negative energy (if the force opposes the direction of motion): the energy obtained from the bike+mass system is the energy gained by the leg system.

That is how the energy is conserved, and it is pretty much irrefutable. (It does, after all, agree with the fundamental laws Tom A has been citing.)

Now in the case of a massless calf+foot and a frozen ankle joint, the force F will always be colinear with the line from the knee joint to the pedal spindle. One can then perform the same sort of analysis on the knee joint to show that the thigh will see a change in energy of dE. Thus if thigh kinetic energy Ek is 0.5Jw^2 (pretend w is omega) -- from which dw = dE/Jw -- then dw = -F.dC/Jw.

We can take this further. We can have the force deviate from the line of knee to pedal spindle, by means of mass in the feet and calves, and by means of a torque on the calf (quads or hams working). And we can add gravity overall. The basic relationships still hold, and the analysis can be chased up limb by limb to show that no energy is lost or gained. But I think you would rather deal with a particular objection than go into all that.

So now to the problem (in your mind), namely that these changes in rate are mechanically linked to each other. I have suggested you work the differentials in Excel. I have also simplified the problem to the 'lump of weight on one pedal problem' to help you see that your interpretation of the physical processes is incorrect. You have taken me up on neither analysis, and have instead insisted that others are missing something, without having done the real work to prove it yourself. I now pitch to you the third ball. It's up to you to swing. If you let this one go by, right through the strike zone... well, you know the saying about three strikes.

You are concerned that this is a fixie, ie, that is, the magnitude of dC must be proportional to v*dt. So w (speed of thigh) is some function of v (speed of bike). Since the kinetic energy of the bike+rider varies as v^2, and the thigh motion does not vary as v^2, then how can the sum of the energies possibly remain constant?

The first part of the answer is to simplify the question. Since we are talking about flow of energy (delta energy), the differential form is much more appropriate for analysis. Instead of looking at v and w, we should be looking at dv and dw. The analysis at the start of this post does essentially that, and gives a numerical basis for showing how the energy can balance without causing a problem. In explicit differential form, though, it can be phrased this way: the linear dE of the bike+rider and the angular dE of the thigh must be equal in magnitude, so J*w*dw = m*v*dv. I don't see any problem with maintaining that over time. If we look at rate of change of angular and linear velocity, dv/dt = (Jw/mv)*dw/dt. This means that the bike velocity will change in some proportion to the change in thigh velocity, and that proportion will always be in the ratio of the scalar momentums of the two.

Now your question will be, why isn't it something to do with the gear ratio? My answer is that it does depend on the gear ratio. If you double the gear ratio, the velocity of the bike would be doubled also for the same thigh angular velocity. This means that the ratio of the momentums will halve. Or, to use the example with numbers up above, if the gear ratio is increased tenfold, the ratio of the momentums will change by a factor of ten, and the change in bike velocity will become 1/10 what it was before the gear ratio change; that is, the ratio of dv to dw will change to 1/10 what it was, precisely reflecting the gearing change.

Now, depending on the physical layout, there might be any weird number of relationships describing dw/dt and dv/dt, so I am going to simplify the proof by dealing with the subject inductively. Assume a correct gear ratio for storing the energy from the bike in the thighs and back again from the thighs in the bike. If the energy 'balance' (transfer of equal quantities out of one part of the system and into the other, as kinetic energy storage) can be attained for this gear ratio, it can be attained for all gear ratios, by using the reasoning of the previous paragraph. Therefore either the energy balance always holds true, regardless of gear ratio, or it never holds true, regardless of gear ratio. Now, if the energy balance did not hold, it could be used to make a perpetual motion machine that performs work for free and violates the laws of thermodynamics. That is, if the relationship were such that the ratio of momentums were greater than that expected by the gear ratio, more energy would appear in the linear
kinetic energy of the bike than was put into it. If the relationship were such that the ratio of momentums were less than that expected by the gear ratio, the wheels could be used to drive the pedals, and more angular kinetic energy would appear in the thighs than was put into the system via the wheels. Since the patent office doesn't approve of either of these scenarios, the only option left is the one in which there is a perfect energy balance, and, as previously stated, this holds for all gear ratios, and holds for all speeds within any given gear ratio.

I don't think it is conceptually difficult to see that this whole line of argumentation can be extended to the more general case of multiple reciprocating masses, gravity effects, and muscular inputs.

So the bottom line is that is it you that has been peddling a work-for-free perpetual motion machine, not the rest of us.

(I had intended to send this earlier, but had yet to finish it when some engagements interfered. In the meantime some new posts have come in that make some of the same points as were made near the top, but to avoid the work of re-writing, I'm leaving it. Apologies to the others for the duplication of your points.)