Your contention is that cycling efficiency is <<100% because of major energy losses "downstream" of when the legs are set in motion, and that this has a physical basis that allows you accurately quantify the effects of cadence. In reality, however, such losses (which are due to friction in joints, hysteresis in limbs, etc.) are very small (i.e., >90% of the energy "invested" in the legs is recovered at the pedal), and do not vary in such a predictable manner with cadence as you envision. In fact, if you were pedaling in a complete vacuum with infinitely-stiff limbs and completely friction-free joints (and a locked ankle), there would no "downstream" energy losses at all, regardless of the cadence. If there were, it would represent a violation of the 1st Law of Thermodynamics (which is why Tom A. is simply the latest of many who for years have been asking you "where does the energy go, Frank?").

Frank,

I wrote: "Why wouldn't we deny a variation?"
You replied: "So, while I would expect everyone to agree that there was a variation, I could not guarantee it."

I wrote: "... even for 100% hysteresis loss."
You replied: "Well, you are describing a perfect spring."

I wrote: "I only mention this to help disabuse you of the notion that you have found something that several who have training and practical experience in this kind of analysis are missing..."
You replied: "Seems to me you have come around to this point of view."

I wrote: "Why must one expect this?"
You replied: "Huh? Expect what?"
In this case my text did not make sense, although you could have deduced from the context that I left out the word 'not'.

I wrote: "No. Tom want you to demonstrate an understanding of basic physics before going on."
You wrote: "No, Tom is simply being recalcitrant."

Overall, I would say that your reading comprehension has become as poor as your grasp of physics. When communication itself is compromised, there is no point in continuing.

As I said before, three strikes and you're out.

You have repeatedly made groundless (and ridiculous) assertions and ducked the question when pressed, whether by not even understanding the objection or by speaking about something else or positing some new outlandish distraction. You have contributed no analytical or experimental work to this thread, except to check my math (which I've welcomed). What others have provided, you have distorted. From the height of your erroneous arrogance it is easy to dismiss the legitimate words and work of others, and pretend that the onus is on them to persuade you because you haven't changed your mind yet. Suggestions for 'easy' experiments have poured forth (some flawed), but you have not done the work to test your novel theories. In short, others must tirelessly defend the axioms of physics, whereas you have no accountability for your errors.

My advice is to take a college freshman level course in statics and dynamics, and keep re-taking it until you score over 95% (just to be sure you mastery is broad enough to rid you of your misconceptions). Then come back and discuss with the other intelligent people on this thread. The only way you can regain the respect lost during the course of this thread is to admit how wrong and warped you have been. I am not as concerned that you are wrong with respect to the facts of physics, as that in your wrongness you are oblivious to the possibility that you could be so wrong. I have no problem with a blind man bumping into things; it is something else when the blind man claims that everyone else can't see.

from your statement above let me clarify your position for everyone here. You think it is possible to violate the 2nd law of thermodynamics when riding a bicycle if we could just make some physically impossible components.

you think it is possible to transfer energy between two elements on the bike (by definition, transferring energy means they are not in equilibrium) without energy loss.

Does your "no loss" model also require infinitely stiff chain links and infinitely stiff wheels, and frame, and any other component that might store this energy variation?

you criticize me for simply stating that, in the real world, such losses actually exist and are part of the whole when trying to explain the efficiency drop between the muscles and the wheel.

Could you recommend a good physics or dynamics course...so I can come to the same understanding as the rest of you. I feel so alone and stupid.

I recommend ESS 3093, as I think it more at your level than ESS 3096:

http://www.acs.utah.edu/...log/crsdesc/ess.html

No, because if such components (including body parts) existed, the 2nd Law of Thermodynamics need not be violated.


Not in reality, no. Rather, the point of the pedagological construct is to help you recognize that you are barking up the wrong tree in claiming that there is an obligatory energy loss during pedaling that varies with the square of the cadence.


Obviously the frame, crank, b.b. spindle, and pedals would have to be infinitely stiff for the system to be "lossless", but the rest of the drivetrain is really irrelevant as the basic physics are the same regardless of whether there is a chain or not.


No. I criticize you for not realizing that such losses are 1) quite minor, 2) are not related to the cadence in any simple manner, and 3) do not exist at all in the frictionless, infinitely stiff, pedaling-in-a-vacuum hypothetical model. The latter is most important, as it illustrates your lack of understanding of simple physics.

Frank, I totally agree with you on this, and I think all others who have participated in this debate: isometric contractions, (and contractions at very low speed), waste energy. However, I thought that the arguing was whether a mechanical system of weights similar to a human leg connected to a bicycle (MMF), could turn the up and down of the thigh in speed variations of the bike with or without losses. According to your previous posts, this system would gradually slow down at a predictable rate, even if there was no friction. You were supporting this point with the fact that force was not parallel to motion. But now you say that a crankshaft is perfectly efficient even if the forces are not entirely tangential. So I guess we are all in agreement?

I do not understand why you both are advocating advanced coursework, when all that is needed is a basic Dynamics course like MEEN 020. I think that if Frank Day still does not understand our explanation, it is in part our fault for not making the explanation crystal clear. Proposing to look at the laws of thermodynamic doesn't do anything to clarify the subject matter. Probably we need simpler explanations of the type Dr. Richard Feynman used to propose to his students.

Explain away. How is it the pumping thigh when connected to a rotating shaft can transfer energy back and forth between the two without loss, therefore violating the 2nd law of thermodynamics.

I am sure Feynman would have had no trouble explaining this to his students, or maybe he would. Anyhow, give it a go.

I do not understand why you both are advocating advanced coursework, when all that is needed is a basic Dynamics course like MEEN 020. I think that if Frank Day still does not understand our explanation, it is in part our fault for not making the explanation crystal clear. Proposing to look at the laws of thermodynamic doesn't do anything to clarify the subject matter. Probably we need simpler explanations of the type Dr. Richard Feynman used to propose to his students.

Although MEEN 020 is a prerequisite of MEEN 128 (both of which are undergraduate courses), it doesn't get into the level of detail necessary to create the equations of motion for the system. So...I guess he should do both then :-)

I suggest you take a walk over from the exercise science part of your school to the physics (or engineering) side and make that argument about how to get around that pesky, inconvenient, 2nd law of thermodynamics when analyzing an energy conservation problem and see what they say.

I know it is a lot easier to simply come here and simply suggest you know the answer and I, being incredibly stupid, couldn't possibly understand even if I did take those course but I am not going to let you get away with that. If you can't explain how it is possible here then I submit it is not possible and I am right.

Do you mean a problem something like this one?

http://en.wikipedia.org/...um_%28mathematics%29

The simple pendulum is not a machine. All the parts are in equilibrium. The only energy conversion that takes place is between kinetic and potential energy of the individual molecules. There is no energy transfer between the parts. Swinging back and forth using an idealized model does not cause an energy variation in the system which would violate the 1st law since all the kinetic energy changes that occur can be accounted for by potential energy changes. I accept that analysis as valid. Unfortunately for your position in this discussion though, it cannot be applied to the pedaling model.

Such a system without loss is impossible. Where the loss is going is not important, the second law says it is impossible. The system requires energy transfer between the components of the system and the second law of thermodynamics says that since the parts are not in equilibrium (if the were there would be no energy transfer) entropy must increase which means energy is lost from the system. This is not the simple pendulum problem. While I would admit that a system that is thermodynamically impossible could be called a "simplified" system I would submit it is not a "good" or valid system for analysing this real life problem. Because of your "belief" in this model both you and Coggan (and I assume others) have mistakenly stated that the pedaling motion involves no energy loss, even in real life. At least Coggan now admits that is not true although he believes the losses are small (without any evidence to support that contention, I might add). Papadapalous has published that unloaded pedaling involves substantial losses but that somehow loading the bike fixes the problem. Such reasoning, in view of this thermodynamic problem, seems wishful thinking trying to make real life come in line with the "simplified" MMF model, which "everyone" knows to be true but we here at this thread now know to be impossible. This model has done nothing but lead "scientists" down the wrong garden path, IMHO.

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Frank,
An original Ironman and the Inventor of PowerCranks

And around the circle we go again...talk about calling ME recalcitrant...

How about this one then? Seems to me that there are some transferences of KE between the parts...

http://en.wikipedia.org/wiki/Double_pendulum

Ummm...I'll bet on the understanding of the underlying physics of the problem by Coggan, Martin, Papadapolous, etc. over your lack thereof any day.

You don't seem to understand the simple definitions of "system" or "equilibrium" very well either. The "system" is, basically speaking, where you decide to "draw the box around the mechanism", and "equilibrium" means that energy is neither flowing into or out of the "box". Inside the "box", it is possible for energy to be converted from one form to another, and/or back, ad infinitum without violating any laws of physics or thermodynamics since the underlying simplifying assumption is that none of the energy can be lost as heat.

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http://bikeblather.blogspot.com/

If the parts were not in equilibrium at the start then there would have to be energy losses over time due to the 2nd law. If they were in equilibrium at the start then I suspect not. It would depend upon the starting conditions I suppose.

The simple pendulum is not a machine. All the parts are in equilibrium. The only energy conversion that takes place is between kinetic and potential energy of the individual molecules. There is no energy transfer between the parts. Swinging back and forth using an idealized model does not cause an energy variation in the system which would violate the 1st law since all the kinetic energy changes that occur can be accounted for by potential energy changes. I accept that analysis as valid. Unfortunately for your position in this discussion though, it cannot be applied to the pedaling model.