In Reply To:
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.
Well, here is a scenario that anyone can do (at least if they have a fixie). Put your bike on the trainer in such a way that there is no resistance applied to the tire. This way there is no forward motion of the bike to worry about in the problem, the only resistance to movement is the inertia of the rear wheel. This pretty much reduces the inertia of the bike to be less than the inertia of the thighs. The total energy of the system will equal the energy of the thighs plus the energy in the rotating wheel. Calculate what kind of speed change need occur under these circumstances to keep the energy of the system constant (I think you will find it to be substantial) and then get on the bike and ride it and see if it happens. If it doesn't happen it is only because the energy, rather than being transferred back and forth between the thighs and wheel is being lost as heat. This is probably because the forces slowing the thigh are not tangential to the circle.
The energy can be completely transferred to the other elements of the bike only if the forces are tangential to the pedaling circle. (this is discussed more below) There is nothing to ensure this is the case. In fact, there is nothing to even suggest this is the case.
In Reply To:
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.
Well, you have put the problem in a nutshell right there. The work done (energy transferred) is the dot product. Unless the force is tangential the energy cannot be completely transferred. Therefore, either the energy of the system must vary (violating the laws of thermodynamic) or energy must be lost as heat. For you to prove your point you must be able to prove that these naturally applied forces are always tangential to the circle. Good luck.
In Reply To:
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.
No. the work done is not equal. If we assume a massless lower leg and foot, the force retarding the thigh has to be in the direction of the lower leg (you make this point below but note that this direction is never tangential to the pedaling circle except perhaps briefly at around 3 o'clock on the downstroke). The work done is the integral of the force through the distance. However, as you put it, the work done into the bike is the dot product. These simply cannot be the same. Therefore, energy must be lost.
In Reply To:
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.)
Well, I would say it isn't pretty much irrefutable. See above.
In Reply To:
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.
The dot product of the forces at the knee (the only forces that do any work are those tangential to the circle drawn by the knee movement) also reduce the amount of work transmitted (or increase the amount of energy lost).
In Reply To:
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.
The MMF model allows no muscle force to be added to adjust the forces. Gravity is accounted for in the potential energy and the forces are close to equal and opposite on the two pedals so can be discounted although gravity gives forces that make it hard to understand what is really going on unless one subtracts them. The MMF model is best described in a zero gravity condition as there are fewer confounding factors.
In Reply To:
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.
Thanks to your analysis I can see that the real problem is not the linking between the cranks and the bike but the dot products of the various forces/direction of motion. The issue is still the same. It is simply not possible for the MMF to conserve energy. Thanks for helping me to clarify in my mind exactly why. Perhaps this will become clearer to you all now.
In Reply To:
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.
See above. I have revised my explanation as to why this is impossible thanks to you.
In Reply To:
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.
See above. It has nothing to do with the gear ratio. It has to do with the direction of the forces. I had alluded to that before but I see now it is the only explanation. Having perfectly rigid elements cannot overcome this problem.
In Reply To:
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.
Not if the dot-products do not transmit all the energy. Again, thanks for that.
In Reply To:
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.
Of course, this can be extended to any mechanical system. Conservation of energy works as long as all the energy is transferred. However, how much is transferred depends upon the dot product. Again, thanks for that clarification.
In Reply To:
(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.)
Has the revision of my position helped you to better understand my position? The MMF cannot possibly be a perpetual motion machine because it is not possible to transfer all the energy back and forth between the different elements. This is not because of the gearing but because of the direction of the natural forces. Unless you can show the dot product of all the energy transmission forces are always maximal I will continue to hold this position. Perhaps you can now understand why the speed variations we should be seeing if energy were conserved are never seen.
--------------
Frank,
An original Ironman and the Inventor of PowerCranks
-sh
n
)n.1. a. A device consisting of fixed and moving parts that modifies mechanical energy and transmits it in a more useful form.b. A simple device, such as a lever, a pulley, or an inclined plane, that alters the magnitude or direction, or both, of an applied force; a simple machine.2. A system or device for doing work, as an automobile or a jackhammer, together with its power source and auxiliary equipment. 
