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Re: cadence [gciriani] [ In reply to ]
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Tigermik:
A whole industry has been established based on the above. You can purchase quite complex codes for performing multibody dynamics, be it flex, rigid, or flex/rigid. Look up a code like ADAMS as an example. But I guess companies like MSC are completely clueless and should have consulted with you first.
Are you an Adams user? If so, or if you know somebody who is, perhaps you could implement a simple model of thigh-leg-crank-and-wheel on the road and convince Frank.
Unfortunately no. My forte is flex body. I personally hate mechanisms. Too many ferris wheel homework problems getting my BSE left a bad taste in my mouth.
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Re: cadence [Tom A.] [ In reply to ]
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You insist that even if all of the elements were perfectly rigid, there was no friction at the joints, and that there was no air friction, if the system was set in motion and then "let go" it would STILL slowly contain less and less average KE over time...so again, WHERE IS THAT ENERGY GOING? Energy can neither be created nor destroyed, it can only change form. To what form is that KE being converted??
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.

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.
To say that the pedaler system will lose no energy simply violates the 2nd law. Draw the lines wherever you want. Believe whomever you want. Just don't expect me to believe this preposterous notion. The second law states quite explicitly that perpetual motion machines are impossible. The question is simply how fast is the energy lost, not whether it is lost. If you cannot accept that energy can be lost from the system (because you have become to believe in an impossible model) how on earth do you ever expect to understand how much is lost, let alone state with some certainty that the losses will be small, as Dr. Coggan did?

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Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [Frank Day] [ In reply to ]
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The question here is: Stupidity or dementia?

Maybe we could have a poll!
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Re: cadence [tigermilk] [ In reply to ]
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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.
I must say this entire thread has been quite amusing. I've been left scratching my head at numerous times and muttering "huh?" I particularly love the quote "The simple pendulum is not a machine." Sure it is. Look up machine in the dictionary and tell me what you find. I'll give you a hint - lots of references to mechanisms. The pendulum is a mechanism. A one-bar simple mechanism.

The thigh, lower leg, foot, crank system is nothing more than a 4-bar linkage. The point at the hip is constrained against translation and rotates in a plane parallel to the plane of the bike. Likewise the bottom bracket. In a frictionless system, give either the "hip" or the "bottom bracket" point an initial rotation via some external torque and you set the wheels in motion (pun intended). That system will continue to move per it's kinematic constraints ad infinitum unless acted on by some other external force/moment.

Now you can argue all you want about the kinematic constraints, but if the problem is properly constrained (i.e., it's not an unstable mechanism), you can fully define the kinematic motion of every point in the system, whether it's rigid, flexible, or a combination of flex and rigid. And you can "transfer energy" from one rigid member to another.

A whole industry has been established based on the above. You can purchase quite complex codes for performing multibody dynamics, be it flex, rigid, or flex/rigid. Look up a code like ADAMS as an example. But I guess companies like MSC are completely clueless and should have consulted with you first. Their codes must be all wrong....
ma·chine (m-shn)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.

Exactly which one of these definitions of "machine" does the simple pendulum satisfy? I simply don't see the simple pendulum as a machine. I guess you could claim that a spinning disk is a machine also. A spinning disk is something that should continue forever unless acted upon by an outside force. Both the simple pendulum and the spinning disk is in equilibrium with itself. The MMF system converts an up and down motion to a rotational motion satisfying both definitions 1a and 1b. If loaded, as when riding a bicycle, it also satisfies definition 2.

Yes, it should be possible to work this entire problem out mathematically. The only issue would be the assumptions used for calculating the losses in the materials used. What would one use for the soft tissue of the leg for instance?) If there is an adequate system for analyzing this why doesn't some one do it and prove me wrong. All we have is hyperbole coming from the believers in the preposterous notion that this machine has no losses.

--------------
Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [bjohn34] [ In reply to ]
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The question here is: Stupidity or dementia?

The correct answer is "yes".

http://bikeblather.blogspot.com/
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Re: cadence [Frank Day] [ In reply to ]
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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.
I must say this entire thread has been quite amusing. I've been left scratching my head at numerous times and muttering "huh?" I particularly love the quote "The simple pendulum is not a machine." Sure it is. Look up machine in the dictionary and tell me what you find. I'll give you a hint - lots of references to mechanisms. The pendulum is a mechanism. A one-bar simple mechanism.

The thigh, lower leg, foot, crank system is nothing more than a 4-bar linkage. The point at the hip is constrained against translation and rotates in a plane parallel to the plane of the bike. Likewise the bottom bracket. In a frictionless system, give either the "hip" or the "bottom bracket" point an initial rotation via some external torque and you set the wheels in motion (pun intended). That system will continue to move per it's kinematic constraints ad infinitum unless acted on by some other external force/moment.

Now you can argue all you want about the kinematic constraints, but if the problem is properly constrained (i.e., it's not an unstable mechanism), you can fully define the kinematic motion of every point in the system, whether it's rigid, flexible, or a combination of flex and rigid. And you can "transfer energy" from one rigid member to another.

A whole industry has been established based on the above. You can purchase quite complex codes for performing multibody dynamics, be it flex, rigid, or flex/rigid. Look up a code like ADAMS as an example. But I guess companies like MSC are completely clueless and should have consulted with you first. Their codes must be all wrong....
ma·chine (m-shn)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.

Exactly which one of these definitions of "machine" does the simple pendulum satisfy? I simply don't see the simple pendulum as a machine. I guess you could claim that a spinning disk is a machine also. A spinning disk is something that should continue forever unless acted upon by an outside force. Both the simple pendulum and the spinning disk is in equilibrium with itself. The MMF system converts an up and down motion to a rotational motion satisfying both definitions 1a and 1b. If loaded, as when riding a bicycle, it also satisfies definition 2.

Yes, it should be possible to work this entire problem out mathematically. The only issue would be the assumptions used for calculating the losses in the materials used. What would one use for the soft tissue of the leg for instance?) If there is an adequate system for analyzing this why doesn't some one do it and prove me wrong. All we have is hyperbole coming from the believers in the preposterous notion that this machine has no losses.
If you were an engineer or a pragmatist you'd recognize it as a machine.

Regarding the "MMF", you do realize that if someone were to model this, the FIRST step would be to assume a rigid body system don't you? That's engineering 101 at work there. Same thing with a mass-spring system. You start with the "unrealistic" model of no losses which actually tells you a WHOLE lot. You can bump up the fidelity and add damping and such as you get smarter or more adventurous, neither of which I'd suggest in your case for obvious reasons :) But as you add those "losses" you better have some good data. For this particular problem you could merely parametrically investigate the effects of various system damping, friction losses, and such.

Any "machine" is a construct of your own devise. As demonstrated above, you can set the analytical problem up to have no losses. If you throw this mechanism in a multibody dynamics program and specify no losses, then you will get a perpetual motion machine. You absolutely have to as you've specified no losses due to damping, friction, heat, hysteresis, etc. If you're simulation comes back otherwise, the program isn't implementing the equations of motion properly.

Your obvious mistake is trying to interject what you see in nature (the lack of perpetual motion machines) and pigeonhole it into idealized models that we engineers utilize every day in design and analysis. You see that as a flaw in the methodology. We see that as an idealization that gets us 99% of the info we want.
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Re: cadence [tigermilk] [ In reply to ]
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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.
I must say this entire thread has been quite amusing. I've been left scratching my head at numerous times and muttering "huh?" I particularly love the quote "The simple pendulum is not a machine." Sure it is. Look up machine in the dictionary and tell me what you find. I'll give you a hint - lots of references to mechanisms. The pendulum is a mechanism. A one-bar simple mechanism.

The thigh, lower leg, foot, crank system is nothing more than a 4-bar linkage. The point at the hip is constrained against translation and rotates in a plane parallel to the plane of the bike. Likewise the bottom bracket. In a frictionless system, give either the "hip" or the "bottom bracket" point an initial rotation via some external torque and you set the wheels in motion (pun intended). That system will continue to move per it's kinematic constraints ad infinitum unless acted on by some other external force/moment.

Now you can argue all you want about the kinematic constraints, but if the problem is properly constrained (i.e., it's not an unstable mechanism), you can fully define the kinematic motion of every point in the system, whether it's rigid, flexible, or a combination of flex and rigid. And you can "transfer energy" from one rigid member to another.

A whole industry has been established based on the above. You can purchase quite complex codes for performing multibody dynamics, be it flex, rigid, or flex/rigid. Look up a code like ADAMS as an example. But I guess companies like MSC are completely clueless and should have consulted with you first. Their codes must be all wrong....
ma·chine (m-shn)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.

Exactly which one of these definitions of "machine" does the simple pendulum satisfy? I simply don't see the simple pendulum as a machine. I guess you could claim that a spinning disk is a machine also. A spinning disk is something that should continue forever unless acted upon by an outside force. Both the simple pendulum and the spinning disk is in equilibrium with itself. The MMF system converts an up and down motion to a rotational motion satisfying both definitions 1a and 1b. If loaded, as when riding a bicycle, it also satisfies definition 2.

Yes, it should be possible to work this entire problem out mathematically. The only issue would be the assumptions used for calculating the losses in the materials used. What would one use for the soft tissue of the leg for instance?) If there is an adequate system for analyzing this why doesn't some one do it and prove me wrong. All we have is hyperbole coming from the believers in the preposterous notion that this machine has no losses.
If you were an engineer or a pragmatist you'd recognize it as a machine.

Regarding the "MMF", you do realize that if someone were to model this, the FIRST step would be to assume a rigid body system don't you? That's engineering 101 at work there. Same thing with a mass-spring system. You start with the "unrealistic" model of no losses which actually tells you a WHOLE lot. You can bump up the fidelity and add damping and such as you get smarter or more adventurous, neither of which I'd suggest in your case for obvious reasons :) But as you add those "losses" you better have some good data. For this particular problem you could merely parametrically investigate the effects of various system damping, friction losses, and such.
Yes, I do realize that. However, the drawback of that system is it is not a very good model if one is trying to model internal losses. How could it be? Why can't anyone but me here see that? The rigid body model assumes internal losses are negligible. But, experimentally we know that when pedaling internal losses are not trivial, at least when the model is unloaded, according to Papadapalous. How is it posible that a model that assumes losses are trivial is still being touted as being appropriate model to answer the question as to whether they really are?
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Any "machine" is a construct of your own devise. As demonstrated above, you can set the analytical problem up to have no losses. If you throw this mechanism in a multibody dynamics program and specify no losses, then you will get a perpetual motion machine. You absolutely have to as you've specified no losses due to damping, friction, heat, hysteresis, etc. If you're simulation comes back otherwise, the program isn't implementing the equations of motion properly.
I understand that. A rigid model is useful for looking at forces and stresses on materials but not much else. As I said above, how is such a model useful for modeling real world energy losses if that is what you are interested in studying?
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Your obvious mistake is trying to interject what you see in nature (the lack of perpetual motion machines) and pigeonhole it into idealized models that we engineers utilize every day in design and analysis. You see that as a flaw in the methodology. We see that as an idealization that gets us 99% of the info we want.
No, my "mistake", at least from your view, is not falling into line and accepting the prevaling view that an idealized model that ignores the possibility of internal losses is a good model to analyze internal losses in real life. I don't buy it. I am surprised that you are taking this side. What is your evidence that this is a good model for this purpose?

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Frank,
An original Ironman and the Inventor of PowerCranks
Last edited by: Frank Day: Oct 27, 09 6:53
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Re: cadence [Frank Day] [ In reply to ]
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Frank Day:
How could it be? Why can't anyone but me here see that? The rigid body model assumes internal losses are negligible. But, experimentally we know that when pedaling internal losses are not trivial, at least when the model is unloaded, according to Papadapalous.
Frank, I think your misunderstanding resides on the fact that the losses internal to the leg, when one is not pushing on the pedals, are caused by the lengthening and shortening of the muscle fibers (a loss due to friction of sliding myo-filaments). And we all agree with that. However, you are going further asserting that there are losses even without taking that into account: that is your statement is regarding an ideal physical system without internal friction. At that point you say that there are losses just because forces are not aligned to motion. The latter is an incorrect statement from the physics point of view. I know that your intuition tells you that when you muscles push in a way not aligned to motion you are inefficient, and we all agree with that. But there are no inefficiencies caused by the simple mechanics of the feet strapped to the pedals. Somewhere you mix the two points of view and there is where the misunderstanding starts

Giovanni Ciriani
http://www.GlobusSHT.com
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Re: cadence [Frank Day] [ In reply to ]
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Coggan (and I assume others) have mistakenly stated that the pedaling motion involves no energy loss, even in real life.

I never said any such thing.

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At least Coggan now admits that is not true

There is nothing to admit, since your claim above is patently false.

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although he believes the losses are small (without any evidence to support that contention, I might add).

As I have said before, see Jim Martin's work (among others).
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Re: cadence [Tom A.] [ In reply to ]
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The question here is: Stupidity or dementia?

The correct answer is "yes".
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Re: cadence [gciriani] [ In reply to ]
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Frank Day:
How could it be? Why can't anyone but me here see that? The rigid body model assumes internal losses are negligible. But, experimentally we know that when pedaling internal losses are not trivial, at least when the model is unloaded, according to Papadapalous.

Frank, I think your misunderstanding resides on the fact that the losses internal to the leg, when one is not pushing on the pedals, are caused by the lengthening and shortening of the muscle fibers (a loss due to friction of sliding myo-filaments). And we all agree with that. However, you are going further asserting that there are losses even without taking that into account: that is your statement is regarding an ideal physical system without internal friction. At that point you say that there are losses just because forces are not aligned to motion. The latter is an incorrect statement from the physics point of view. I know that your intuition tells you that when you muscles push in a way not aligned to motion you are inefficient, and we all agree with that. But there are no inefficiencies caused by the simple mechanics of the feet strapped to the pedals. Somewhere you mix the two points of view and there is where the misunderstanding starts
The 2nd law says there must be losses when something real "pedals" a bicycle (even an MMF, just not an "ideal" MMF) but it doesn't say how large the losses will be, only that they will be there. How is it you feel comfortable using a model that doesn't allow any internal losses to analyze how large the internal losses, that must be there, really are? Why is it everyone feels just dandy ignoring this question while calling me stupid for not ignoring it?

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Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [Frank Day] [ In reply to ]
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Frank Day:
How could it be? Why can't anyone but me here see that? The rigid body model assumes internal losses are negligible. But, experimentally we know that when pedaling internal losses are not trivial, at least when the model is unloaded, according to Papadapalous.


Frank, I think your misunderstanding resides on the fact that the losses internal to the leg, when one is not pushing on the pedals, are caused by the lengthening and shortening of the muscle fibers (a loss due to friction of sliding myo-filaments). And we all agree with that. However, you are going further asserting that there are losses even without taking that into account: that is your statement is regarding an ideal physical system without internal friction. At that point you say that there are losses just because forces are not aligned to motion. The latter is an incorrect statement from the physics point of view. I know that your intuition tells you that when you muscles push in a way not aligned to motion you are inefficient, and we all agree with that. But there are no inefficiencies caused by the simple mechanics of the feet strapped to the pedals. Somewhere you mix the two points of view and there is where the misunderstanding starts
The 2nd law says there must be losses when something real "pedals" a bicycle (even an MMF, just not an "ideal" MMF) but it doesn't say how large the losses will be, only that they will be there. How is it you feel comfortable using a model that doesn't allow any internal losses to analyze how large the internal losses, that must be there, really are? Why is it everyone feels just dandy ignoring this question while calling me stupid for not ignoring it?

We don't...it's just that when modeling one starts with the simplified, idealized model FIRST so that you can get a measure of the forces acting on the various links, pivots, and boundary conditions. THEN you begin to start incorporating the less idealized, more realistic effects so that there is a better understanding of their magnitudes, etc.

YOU however, can't even get the your brain wrapped around the SIMPLEST model and keep insisting that even if all friction was assumed to be zero and the links are infinitely rigid, then there would STILL be some sort of energy losses just from the basic physics of the device. That is demonstrably NOT so since it would violate both the 1st law of thermodynamics and Newton's first law of motion. As Andy pointed out, the 2nd law of thermodynamics is NOT being violated since a priori it is assumed for the modeling purposes that entropy is just not DECREASING.

THAT'S where the disconnect is Frank...inside YOUR head. You say that the idealized model once set in motion would slowly come to a stop. That is NOT so until we start adding losses into the system model. But, there's no way of knowing the magnitudes of those losses until you do the force analysis of the simplified, idealized model FIRST.

We're just trying to get you to the first step...and you can't even get THERE...

I fully expect you to come back with some sort of response claiming how there HAS to be losses even in the idealized "lossless" model, so go right ahead and just keep mixing up idealized vs. non-idealized cases in your head and complaining about how everyone in the universe gets this wrong and you're the ONLY one who is getting this right.

http://bikeblather.blogspot.com/
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Re: cadence [Tom A.] [ In reply to ]
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Frank Day:
How could it be? Why can't anyone but me here see that? The rigid body model assumes internal losses are negligible. But, experimentally we know that when pedaling internal losses are not trivial, at least when the model is unloaded, according to Papadapalous.



Frank, I think your misunderstanding resides on the fact that the losses internal to the leg, when one is not pushing on the pedals, are caused by the lengthening and shortening of the muscle fibers (a loss due to friction of sliding myo-filaments). And we all agree with that. However, you are going further asserting that there are losses even without taking that into account: that is your statement is regarding an ideal physical system without internal friction. At that point you say that there are losses just because forces are not aligned to motion. The latter is an incorrect statement from the physics point of view. I know that your intuition tells you that when you muscles push in a way not aligned to motion you are inefficient, and we all agree with that. But there are no inefficiencies caused by the simple mechanics of the feet strapped to the pedals. Somewhere you mix the two points of view and there is where the misunderstanding starts
The 2nd law says there must be losses when something real "pedals" a bicycle (even an MMF, just not an "ideal" MMF) but it doesn't say how large the losses will be, only that they will be there. How is it you feel comfortable using a model that doesn't allow any internal losses to analyze how large the internal losses, that must be there, really are? Why is it everyone feels just dandy ignoring this question while calling me stupid for not ignoring it?

We don't...it's just that when modeling one starts with the simplified, idealized model FIRST so that you can get a measure of the forces acting on the various links, pivots, and boundary conditions. THEN you begin to start incorporating the less idealized, more realistic effects so that there is a better understanding of their magnitudes, etc.

YOU however, can't even get the your brain wrapped around the SIMPLEST model and keep insisting that even if all friction was assumed to be zero and the links are infinitely rigid, then there would STILL be some sort of energy losses just from the basic physics of the device. That is demonstrably NOT so since it would violate both the 1st law of thermodynamics and Newton's first law of motion. As Andy pointed out, the 2nd law of thermodynamics is NOT being violated since a priori it is assumed for the modeling purposes that entropy is just not DECREASING.

THAT'S where the disconnect is Frank...inside YOUR head. You say that the idealized model once set in motion would slowly come to a stop. That is NOT so until we start adding losses into the system model. But, there's no way of knowing the magnitudes of those losses until you do the force analysis of the simplified, idealized model FIRST.

We're just trying to get you to the first step...and you can't even get THERE...

I fully expect you to come back with some sort of response claiming how there HAS to be losses even in the idealized "lossless" model, so go right ahead and just keep mixing up idealized vs. non-idealized cases in your head and complaining about how everyone in the universe gets this wrong and you're the ONLY one who is getting this right.
Sorry my friend, I do accept that the idealized model could work, presuming we all agree that to do so it must violate the 2nd law. The 2nd law simply states that such a machine is impossible, idealized or not. That doesn't mean that the simplifid model could not be useful in this analysis because it would help one to understand the forces on the materials so one could calculate the losses, if one knew how the materials would react to such forces, but no one here has even suggested that is why the simplified model is useful. Instead, the implication has been that because the simplified model shows no losses that this means that the actual losses in real life are small or insignificant, a leap in logic that has no basis.

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Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [Frank Day] [ In reply to ]
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I do accept that the idealized model could work, presuming we all agree that to do so it must violate the 2nd law. The 2nd law simply states that such a machine is impossible, idealized or not. That doesn't mean that the simplifid model could not be useful in this analysis because it would help one to understand the forces on the materials so one could calculate the losses, if one knew how the materials would react to such forces, but no one here has even suggested that is why the simplified model is useful. Instead, the implication has been that because the simplified model shows no losses that this means that the actual losses in real life are small or insignificant, a leap in logic that has no basis.

No, the implication (fact/conclusion, actually) is that such losses are non-existent as you envision them.
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Re: cadence [Andrew Coggan] [ In reply to ]
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I do accept that the idealized model could work, presuming we all agree that to do so it must violate the 2nd law. The 2nd law simply states that such a machine is impossible, idealized or not. That doesn't mean that the simplifid model could not be useful in this analysis because it would help one to understand the forces on the materials so one could calculate the losses, if one knew how the materials would react to such forces, but no one here has even suggested that is why the simplified model is useful. Instead, the implication has been that because the simplified model shows no losses that this means that the actual losses in real life are small or insignificant, a leap in logic that has no basis.

No, the implication (fact/conclusion, actually) is that such losses are non-existent as you envision them.
Perhaps you could enlighten us as to how I envision them and how, in the alternative, these losses actually occur (and, how big you see them being). I have put on my learning cap perfessor, ready to learn.

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Frank,
An original Ironman and the Inventor of PowerCranks
Last edited by: Frank Day: Oct 27, 09 10:00
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Re: cadence [Andrew Coggan] [ In reply to ]
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As I have said before, see Jim Martin's work (among others).
Could you provide a reference to the Martin work to which you refer? And, to any of the "others" to which you refer? Thanks.

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Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [Frank Day] [ In reply to ]
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Sorry my friend, I do accept that the idealized model could work, presuming we all agree that to do so it must violate the 2nd law. The 2nd law simply states that such a machine is impossible, idealized or not.

See...you're still not getting it. The idealization DOES NOT violate the second law, it merely assumes that the entropy does not DECREASE in the system, i.e. it is in equilibrium. Energy neither flows into or out of the system. The 2nd law of thermodynamics DOES NOT state that the idealized machine is impossible.



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That doesn't mean that the simplifid model could not be useful in this analysis because it would help one to understand the forces on the materials so one could calculate the losses, if one knew how the materials would react to such forces, but no one here has even suggested that is why the simplified model is useful.

I'm not going to go through the whole thread again to find all the cases, but I know I have said so, amongst others. The issue all along has been that you have INSISTED that there is an energy loss irrespective of the internal losses in the materials, at the joints, etc....such that even assuming that all of those losses didn't occur, there STILL would be some other loss just by virtue of the configuration of the machine. That is just not so.



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Instead, the implication has been that because the simplified model shows no losses that this means that the actual losses in real life are small or insignificant, a leap in logic that has no basis.

No...the implication ALL ALONG is that there aren't any losses separate from those internal losses, of which you have been adamant in saying that there is...

http://bikeblather.blogspot.com/
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Re: cadence [Tom A.] [ In reply to ]
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Sorry my friend, I do accept that the idealized model could work, presuming we all agree that to do so it must violate the 2nd law. The 2nd law simply states that such a machine is impossible, idealized or not.

See...you're still not getting it. The idealization DOES NOT violate the second law, it merely assumes that the entropy does not DECREASE in the system, i.e. it is in equilibrium. Energy neither flows into or out of the system. The 2nd law of thermodynamics DOES NOT state that the idealized machine is impossible.
If you say so.
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That doesn't mean that the simplifid model could not be useful in this analysis because it would help one to understand the forces on the materials so one could calculate the losses, if one knew how the materials would react to such forces, but no one here has even suggested that is why the simplified model is useful.

I'm not going to go through the whole thread again to find all the cases, but I know I have said so, amongst others. The issue all along has been that you have INSISTED that there is an energy loss irrespective of the internal losses in the materials, at the joints, etc....such that even assuming that all of those losses didn't occur, there STILL would be some other loss just by virtue of the configuration of the machine. That is just not so.
If you or anyone else has said that I surely must have missed it. But, regarding what I have said I must say phooeey. What I have said is there must be an energy loss and the only way I can explain it is through an energy loss in the materials. The second law requires an energy loss if energy is transferred between elements not in equilibrium. If you want to argue that not allowing energy loss by using materials that only exist in the imagination somehow allows one to get around this law then so be it, it is like arguing how many angels can exist on the head of a pin. The issue all along in this thread goes back to my comment that the movement of the thigh in the pedaling motion causes an energy loss and you during a brain fart agreed with me. When Dr. Coggan asked you if you missed that you quickly corrected yourself. The perfect MMF only came up as evidence to support your side of the argument that there is no energy cost to the motion. Of course, that example has no similarity to reality but little fact seems to not bother you or Dr. Coggan or anyone else. You would prefer to argue this little BS issue rather than discuss the original issue (probably because you know you can't win).
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Instead, the implication has been that because the simplified model shows no losses that this means that the actual losses in real life are small or insignificant, a leap in logic that has no basis.

No...the implication ALL ALONG is that there aren't any losses separate from those internal losses, of which you have been adamant in saying that there is...
Huh? Just exactly what does that mean? I don't have a clue what you mean by that statement. Perhaps it would help me if you could list all of the losses that need be accounted for between the muscles and the wheel to add up to the whole. Seems to me I asked that question many pages ago and never got an answer.

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Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [Frank Day] [ In reply to ]
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Frank, just give it up.

BTW, you do know that there's absolutely no reason to invoke thermodynamics in kinematics don't you? Lots of spacecraft, aircraft, automobiles, etc have been designed without that.

Just because you can't comprehend it doesn't make it wrong. Show some humility. I show my humility by deferring to those mathematicians, engineers, and physicists who came before my time. I suggest you do the same.

If you REALLY think you are right, read up on dynamics (get a book, not wikipedia) and get back to us.
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Re: cadence [tigermilk] [ In reply to ]
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Frank, just give it up.

BTW, you do know that there's absolutely no reason to invoke thermodynamics in kinematics don't you? Lots of spacecraft, aircraft, automobiles, etc have been designed without that.

Just because you can't comprehend it doesn't make it wrong. Show some humility. I show my humility by deferring to those mathematicians, engineers, and physicists who came before my time. I suggest you do the same.

If you REALLY think you are right, read up on dynamics (get a book, not wikipedia) and get back to us.
Well, kinematics, dynamics, thermodynamics, whatever you want to call it, they are simply different ways of looking at energy and trying to describe the world. It is all part of physics and they all have to follow the same laws, as far as I know. One should get the same solution regardless of how one looks at the problem.

Anyhow, I spent the first few pages of this thread trying to make my point without invoking thermodynamics. Once I discovered the thermodynamics argument makes my point so well I have decided it is easier to stay with it. If someone wants to come here and argue that there is no energy loss inherent in the pedaling motion all I need to is throw out the 2nd law in rebuttal. While it may seem possible to many the 2nd law simply says it is not possible. No need to get into the details as to exactly where the analysis fails, it simply does.

I would prefer this thread to have evolved to our now "discussing" how large the losses are under different circumstances (different masses, different cadences, different bike and crank construction materials) but it is not possible as long as people prefer to deny that these losses can exist or continually focus on whether it is possible to built a bike out of materials that cannot exist in order to avoid this loss. It is a smoke screen to avoid admitting one has made an error.

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Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [Frank Day] [ In reply to ]
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If someone wants to come here and argue that there is no energy loss inherent in the pedaling motion

There is no energy loss inherent in the pedaling motion.
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Re: cadence [Frank Day] [ In reply to ]
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Re: cadence [Andrew Coggan] [ In reply to ]
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If someone wants to come here and argue that there is no energy loss inherent in the pedaling motion

There is no energy loss inherent in the pedaling motion.
Well, then, IMHO, you are wrong. The pedaling motion requires the transfer of a back and forth constantly changing kinetic energy in one or two element(s) to a rotational energy in another element which, according to the second law, must involve a certain amount of energy loss. I look forward to your rebuttal.

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Frank,
An original Ironman and the Inventor of PowerCranks
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Re: cadence [Frank Day] [ In reply to ]
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The entropy of the system doesn't change, dumbass...
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Re: cadence [Tom A.] [ In reply to ]
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If the Naval Academy couldn't teach him elementary physics...what makes you think that you (or I...or anyone else) can? ;-)
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