If you don't like potential energy go to "Change Parameters" and set g = 0.0 !
It will not make much difference.

There really is no problem in the MMF case.
All forces between the different parts are derived from nothing else than masses being accelerated or decelerated (if we ignore gravity for simplicity reasons). And vice versa these accelerations and decelerations are the result of nothing else than these forces between the parts. So you can be absolutely sure that all resulting speeds of the moving parts and the whole bike will be in perfect compliance with the laws of nature.

Maybe you have different laws, but they don't matter to anybody.

A perpetual motion machine doing no external work is a perfectly normal thing.

??? So, increasing the amount of oxygen consumed by adding another muscle into the mix, so one can do more work, increases muscle efficiency?

The study had about a simple a set up as is possible looking at a single muscle across a single joint as I understood it. Seems to me that would be looking at the best case for observing muscle efficiency. Your explanation makes no sense to me. I would have explained the difference using another mechanism.

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Frank,
An original Ironman and the Inventor of PowerCranks

Actually, the losses you keep talking about simply don't exist, so no, they aren't "in addition to".


Like a fox. ;-) (Hi Jim!)


He is - and if you were really interested in understanding the issues at hand you'd listen to him, as he knows this stuff as well anyone in the world.

No. Increasing the amount of contracting muscle (thus increasing power output more than leg VO2) makes the overall average efficiency more reflective of that of contracting muscle, i.e., the "dilutional effect" of resting muscle (and other tissues, e.g., skin, bone) is reduced. The result is that muscle (really, limb) efficiency tends to be higher when measured during cycling vs. single-legged knee extension (during which the hamstrings are not contracting).


Then you don't understand the experimental model very well. The measured O2 consumption, metabolite exchange, etc., reflects that of all tissues that 1) drain into the femoral vein, but 2) are proximal to the pneumatic cuff placed just above the knee. IOW, you measure not only the metabolism of the knee extensors, but also of the (passive!) knee flexors, etc. Thus, while it is about as close as you can get in a human to an isolated, perfused muscle preparation, it is not the same as one.

'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.'

Frank, I've done my pedagogical best. I've led the horse to water -- no, I've brought the water to your parched, truth-starved lips, but you spat the live-giving liquid aside contemptuously as you breathed out your desperate appeal to the Patent Office.

If you don't want to do the work to enlighten yourself, that's up to you; but you forfeit any credibility on the subject until you do.

My friend, when the velocity of the thigh is tied to the rotational velocity of the wheel and the speed of the bicycle as it is in the MMF model it is simply not possible to have the kinetic energy contained in the varying speed of the thigh be compensated for by the variation in speed of the wheels and bicycle since kinetic energy relates to the square of the velocity. If the pedals were 90º apart where one thigh is accelerating while the other is decelerating it might be possible to have your theoretical "perpetual motion machine" then (as it might be possible to come up with a configuration where the total energy variation is zero) but it isn't with the pedals at 180º.

If it were you would be able to easily do the math to show me how I am wrong. Instead, all we get is "trust me" you are wrong.

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Frank,
An original Ironman and the Inventor of PowerCranks

Well, one should be able to compensate for the resting metabolism of the rest of the leg. Doesn't seem to hard to do. Did this author not do that?

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Frank,
An original Ironman and the Inventor of PowerCranks

Another try:
Do we agree, that it is possible, to calculate the exact state of motion of every part of the MMF and the speed of the whole system for any given angle of the crank and a given speed of rotation of the crank?

If yes, we can calculate the energy of every part and of the whole system. Ok?
It might be a bit complex, but we do not really need to do it. It is enough, that we know we could. Then we would have an equation which of course also works backwards:
Then we can calculate for any other crank angle the corresponding crank rotation speed, which determines the motions of all parts that add up to the same total energy.

There is no problem at all.
You seem to think too complicated.

yes
yes
Ugh, it is very complex because the speed of the various parts of the system are tied together because of the pedals, chain, and wheels. Change the speed of the thigh x amount changes the speed of the rotating wheel and the bicycle overall a fixed amount depending upon where the thigh is and on the gearing/wheel size of the bike. Further, the change in the energy of the thigh is not the same for every angle of the pedaling arc. So we have a fixed relationship between these various speeds yet a varying energy change in the thigh that you hope to come up with some relationship between the parts that can exactly compensate. You have to find a solution to the problem that satisfies every condition. If it can be done you should be able to set it up and everything will cancel out and you end up with a constant. To tell me that you can do it so you don't have to doesn't work with me. I can't do it so I am going to have to rely on you. I look forward to seeing your solution.
No problem at all! LOL. As I said, I look forward to seeing your solution. Lots of mathematicians hang out here. I would accept their solution to this problem also.

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Frank,
An original Ironman and the Inventor of PowerCranks

 

Everybody else agrees that the two camps of the disagreement are as stated above by Frank?

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Giovanni Ciriani
http://www.GlobusSHT.com

I have no idea, how Frank comes to the conclusion, that the patent office is on his side.
But even if it were, that would not impress me much. Sometimes I have the impression, that you can patent almost every nonsense, if it's not already patented. If it actually works does not matter so much.

It would be easier and more useful not to think about springs but about perfectly stiff materials, since this is much closer to the real thing. You don't have big forces und so you don't have big deformations in a real bike, that is only rolling. There are no crash-like events as in the swinging metal balls.

But it seems, you don't understand what a simplified model is good for anyway.

You are some kind of an engineer? Unbelievable!

Hey, if you want to "simplify" things and pretend these forces and losses don't exist that is fine. As long as everyone understands that is what you have to do to make your solution work.

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Frank,
An original Ironman and the Inventor of PowerCranks

The simplifications "work" because the properties of the "links" in the system we are discussing behave more like perfectly elastic members than not...

If you think that there are some sort of massive losses due to flexing and stretching/compressing of the "links" in the system, you are sadly mistaken...and simulations such as the one in this paper (which modeled each leg as a 3-rigid-body linkage) wouldn't so closely match actual data.

http://www.me.utexas.edu/~neptune/Papers/jbme120%283%29.pdf

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http://bikeblather.blogspot.com/

BTW, part of the reason you haven't seen a "mathematical solution" demonstrated for you is that the mathematics involved aren't exactly basic and involve an Euler-Lagrange framework and a numerical solver, as outlined in this paper:

http://www.itk.ntnu.no/...ications/Ids2002.pdf

Take a close look at equation 21 and the paragraph just after it...

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http://bikeblather.blogspot.com/

In any case, I'm glad that you FINALLY have agreed that with either perfectly rigid or perfectly elastic links, and no friction in the joints, that there are no other external losses in the system. Great. Baby steps...baby steps.[/reply]
I made that assertion many posts ago. Don't you read what I write? And, I don't agree that perfectly rigid links would work. A perfectly rigid material would have no capacity to absorb any energy so would shatter at the first attempt to push against it. This assumption is even more unrealistic than the perfect spring. At least the perfect spring would work if it could be made. Unfortunately, it cannot.

At least you FINALLY seem to agree that there is substantial energy variation in the thighs that needs to be either stored some way and then recovered during the pedaling action or it must be lost as heat. Since you haven't identified what part of the bike is acting as that "perfect spring" to do this I presume you will accept a great deal of this energy is lost as heat in the real world?

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Frank,
An original Ironman and the Inventor of PowerCranks

"Here we have also included the inertia of the trike, Mcycle. This term requires further explanation:
Mcycle should include the summed inertia of the crank, the chain and the rotating wheels. If the
cycle is stationary and mounted on a trainer, then it should additionally include the effective inertia
of the trainers. If the cycle is moving, on the other hand, then it should instead include the inertia
resulting from the total mass of the rider/trike-system. The individual inertias must of course be
transformed to the crank."

Does he not believe the inertia of the leg components count for anything?

Anyhow, I have already surrendered on my ability to solve this problem mathematically but I see one serious deficiency in his analysis. He looks at the kinetic energy of the thigh and kinetic energy of the cycle without regard to how they are connected and assumes that magically the kinetic energy can be transferred without loss. For this to occur without minimal loss the natural forces on the pedal from a flacid leg would have to be close to tangential to the pedal circle. There is zero evidence that this is the case that I know of. Therefore, any such assumptions are pure conjecture.

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Frank,
An original Ironman and the Inventor of PowerCranks

1) Take a look at http://www.youtube.com/watch?v=bPtlRf6dg8c and try to count the cycles. I'd say that system actually has a fair amount of damping, and even that isn't attenuating very rapidly.
2) Tuning forks go on for a LONG time. Sure not forever, but a long time. The sound just drops below your audible hearing level yet the fork is still vibrating.
3) Regarding vibrations, from a wave propagation perspective you should consider the structural impedance. Once you ground a system, portions of the wave energy are propagated through or reflected by the interface.

Well...before I take your money, let me warn you that I used to work on these bad boys ;-)

http://www.es.northropgrumman.com/...tions/hrg/index.html

The resonator of that gyro is housed in an evacuated region for a very good reason...and we even made some with metal resonators as well. Quartz is used because it has EXTREMELY low material damping and doesn't have a fatigue limit...but metal is no "slouch" either when it comes to damping.

Hey, since stuff I designed is in space, does that officially make me a "rocket scientist"? :-)



Perfectly rigid does not mean brittle. It only means that it doesn't stretch. Rigid body mechanics and modeling are mainstays of engineering...why do you think that is?



No...it HAS been identified to you, you just refuse to believe it. Think of the total system. The moving mass of the total bike+rider system is a very large energy "capacitor".

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http://bikeblather.blogspot.com/

I guess you missed the term "M(q) + M(q+pi)"...



It's not conjecture, Frank...it's just math.

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http://bikeblather.blogspot.com/