Powercranks

[Phil Holman]

"Frank Day" <[email protected]> wrote in message
news:[email protected]...
> [email protected] (Phil Holman) wrote in message

> > The KE of the thigh is a maximum on the downstroke as the crankarm passes horizontal. From then
> > on the angular rotation of the thigh slows down. If we freeze frame at a position below
> > horizontal, a freebody diagram will reveal the force slowing down the thigh as an axial thrust
> > of the lower leg at the knee. This force is reacted by an equal and opposite force at the ankle
> > which in turn pushes down of the pedal. The KE energy is therefore transferred to useful work in
> > turning the pedal.
>
> Your description of what is happening makes no sense to me when it is applied to the real world.
> First, your model requires the pedal speed to increase on the entire downstroke and decrease on
> the entire upstroke. This is not constant velocity pedaling.

How come, if the rider was starting into a climb and decelerating the mechanism would be the same.
This is just a simple model of the transfer of KE at one part of the pedal stroke.

>
> Second, it assumes the weight of gravity is the only force to consider here.

It doesn't matter if the input force was just gravity and/or muscle input, the mechanism is the
same. KE is 1/2mv^2, all it requires is mass and velocity. When the thigh is slowed, the energy goes
into turning the pedal.

>How does your model handle pedaling when the cadence is so high that the thigh is forced to fall
>faster than gravity would make it go passively. (after all, it is not a bowling ball but hinged
>so will fall slower than a free falling object.) In that instance, there would be a retarding
>force, not on the downward movement of the thigh but on the downward movement of the pedal) on
>the majority of the downward portion of the stroke which would require energy from the flywheel
>to maintain pedal speed and then a further retarding force on the upstroke. How does that
>conserve energy?

I have no idea what you are talking about. The mechanism is the same at any cadence.

>
> Third, I assume it presumes that the transferred kinetic energy you describe is stored and given
> back by the flywheel (without any loss from the acceleration and deceleration of the system) to
> get the thigh back up when there is no evidence that this is the case.

Flywheel???? When riding along normally, a combination of muscle action and or gravity causes the
thigh to move, the KE in the moving thigh is transferred to the pedal/crank on the lower part of the
downstroke which does work against drag. End of story.

>
> Further, your model doesn't describe how the pedaling power losses come to vary with the cube of
> the cadence.

The $1million question. Cube function power *losses* are a characteristic of fluid dynamics. A
square function energy input to accelerate a mass with respect to velocity *doesn't* explain
the losses.

>While at slow speeds there may be some transfer of thigh potential energy to the pedal, this
>certainly is not evidence that this then makes pedaling at constant cadence energy conservative.

The mechanism you propose incurrs no losses. Considering just the anatomy of the rider, pedaling a
bicycle involves work against non-conservative forces and while you continue on this tack, the real
culprits are being ignored. Post your hypothesis to sci.physics or sci.engr.mech.

Phil Holman

 

[Frank Day]

"Phil Holman" <[email protected]> wrote in message news:<[email protected]>...
> "Frank Day" <[email protected]> wrote in message
> news:[email protected]...
> > [email protected] (Phil Holman) wrote in message
>
> > > The KE of the thigh is a maximum on the downstroke as the crankarm passes horizontal. From
> > > then on the angular rotation of the thigh slows down. If we freeze frame at a position below
> > > horizontal, a freebody diagram will reveal the force slowing down the thigh as an axial thrust
> > > of the lower leg at the knee. This force is reacted by an equal and opposite force at the
> > > ankle which in turn pushes down of the pedal. The KE energy is therefore transferred to useful
> > > work in turning the pedal.
> >
> > Your description of what is happening makes no sense to me when it is applied to the real world.
> > First, your model requires the pedal speed to increase on the entire downstroke and decrease on
> > the entire upstroke. This is not constant velocity pedaling.
>
> How come, if the rider was starting into a climb and decelerating the mechanism would be the same.
> This is just a simple model of the transfer of KE at one part of the pedal stroke.

I guess I am only looking at the steady state. You could also ask the question as to what happens at
the top of the hill when cadence increases. Let's just look at the steady state.

How about this scenario. An exercycle on the international space station. In that situation, there
would be no potential energy in the thigh (or anywhere else) to give to the pedal. How would you
analyze the energy requirements there?
>
> >
> > Second, it assumes the weight of gravity is the only force to consider here.
>
> It doesn't matter if the input force was just gravity and/or muscle input, the mechanism is the
> same. KE is 1/2mv^2, all it requires is mass and velocity. When the thigh is slowed, the energy
> goes into turning the pedal.

But, the masses are different so to conserve energy the velocities must be substantially different.
They are not. So, less energy is transferred to the pedal than was contained in the thigh. How do
you explain that?
>
> >How does your model handle pedaling when the cadence is so high that the thigh is forced to fall
> >faster than gravity would make it go passively. (after all, it is not a bowling ball but hinged
> >so will fall slower than a free falling object.) In that instance, there would be a retarding
> >force, not on the downward movement of the thigh but on the downward movement of the pedal) on
> >the majority of the downward portion of the stroke which would require energy from the flywheel
> >to maintain pedal speed and then a further retarding force on the upstroke. How does that
> >conserve energy?
>
> I have no idea what you are talking about. The mechanism is the same at any cadence.

I agree, the mechanism of loss is the same at any cadence or any gravity condition (see above). If
less kinetic energy is transferred to the pedals than is present and this repeats twice every 360
degree circle then energy is lost (if it is not stored as potential energy - which is not possible
in the zero gravity situation). Whether it is lost through eccentric contraction of the muscles or
flexing of the componentes with friction loss or some other mechanism, it is lost. to convince me
otherwise you will need to show me the data that shows the kinetic energy of the system remains
constant through the entire circle when pedaling at constant cadence.
>
> >
> > Third, I assume it presumes that the transferred kinetic energy you describe is stored and given
> > back by the flywheel (without any loss from the acceleration and deceleration of the system) to
> > get the thigh back up when there is no evidence that this is the case.
>
> Flywheel???? When riding along normally, a combination of muscle action and or gravity causes the
> thigh to move, the KE in the moving thigh is transferred to the pedal/crank on the lower part of
> the downstroke which does work against drag. End of story.

No, the transfer of kinetic energy is not complete. It can't be because of the constant velocity
restriction and the differing masses of these components. Show me a caclulation that shows this
transfer can be complete regardless of the mass of the rider or the gear combination chosen or the
cadence. I don't think it is possible.
>
> >
> > Further, your model doesn't describe how the pedaling power losses come to vary with the cube of
> > the cadence.
>
> The $1million question. Cube function power *losses* are a characteristic of fluid dynamics. A
> square function energy input to accelerate a mass with respect to velocity *doesn't* explain
> the losses.

Yes it does, just as a square function energy loss with regards to aerodynamic losses results in a
cube function aerodynamic power loss with speed .
>
> >While at slow speeds there may be some transfer of thigh potential energy to the pedal, this
> >certainly is not evidence that this then makes pedaling at constant cadence energy conservative.
>
> The mechanism you propose incurrs no losses. Considering just the anatomy of the rider, pedaling a
> bicycle involves work against non-conservative forces and while you continue on this tack, the
> real culprits are being ignored.

what real culprits? Give me another square function energy loss involved in this motion.

> Post your hypothesis to sci.physics or sci.engr.mech.

I have enough trouble finding the time to just posting here. If I can I will.

Frank

 

[andrewbradley]

Originally posted by Frank Day
FD:"As far as I can tell it is the need to constantly accelerate and decelerate the separate parts in a
manner that doesn't conserve energy. Under conservation of momentum, once in motion an object will
remain in motion at a constant velocity unless acted upon. Acting upon it requires energy."

That's just not correct. It all depends on whether speed increases or decrease and via what mechanism. Let's get this one straight (again): increasing KE requires energy input, decreasing it requires energy output and this can be put to use given the right mechanism - whatever the acceleration involved in aquiring or reducing speed.

FD:"Now, if one wants toget from point a to point b for the least amount of energy expenditure it seems the best
plan is to maintain a constant speed since the object in motion that is not being acted upon
requires zero energy and an object that is moving at the same average speed but is continually being
acted upon to be accelerated and decelerated requires an energy expenditure that is not zero to
cover the same distance. Now, where are the losses there? I am not sure as I am sufficiently far
away from this stuff academically and professionally that it is not on the tip of my tongue. I know
it exists though."

There are no losses implied if the kinetic energy loss from the decelerations is recuperated. This has lead people to invent braking systems for bikes via spring wind-up.
You can accelerate as hard and long as you like as long as your braking spring has the capacity. (Wind resistance neglected here of course).

FD:"The athlete only has so much energy to give. It has to be divided up amongst the competing
interests. There is a certain amount of "overhead" energy to keep the athlete alive. If pedaling
does require energy, then it is also an "overhead" that must be subtracted from the total energy the
athlete can give to the wheel which will make him slower than he has energy to give. If pedaling
requires no energy then all the additional energy, including this work, can be given to make the
bike go faster. We have seen experimental evidence here that pedaling is not free and various
cadences can cause substantial changes in HR without changing power to the wheel one iota. That is
what I mean. One scenario is faster or slower than the other. Does that make sense?"

This is all quite true, however these effects aren't related to the underlying system of mass movements which is conservative unless you factor in some specific resistance.

> >Or, the analysis is flawed and there really is no difference between fixed wheel pedaling and
> >free wheel pedaling, or PowerCranks pedaling and the pedaling losses are the same for all. Take
> >your pick. Me thinks the pedaling losses are probably the same for all cyclists whether they have
> >a freewheel, fixed wheel, or PowerCranks, but I can't prove it.
>
> Conservation of energy is your one line proof (at this level).

FD:"If that proves my contention it sure isn't being accepted very well."

It proves that whatever the pedalling system there is no energy loss from mass movements alone. I thought your original contention was the opposite.

How the leg-muscles get on with these systems is another matter. The fixed wheel allows the muscles maximum freedom to contract whenever and however hard they see fit. In theory, "when" could be a factor since being fully fired at times when contraction speed is very low (ie early and late in the range of joint movement) could be costly (on the assumption that muscle fibre has an optimal contraction speed) and it could also involve a bit of co-contraction due to the fire-up delay.

The round chainring and freewheel probably allows the muscles enough "freedom of operation" both on how hard and when to contract for most cadences. Clearly, powercranks don't allow the same freedom, making them difficult to pedal at speed.

Attempting to apply maximum pressure in all sectors may not be such a good idea, if you don't have to.

FD:"The experimental evidence is my
proof that the phenomenon exists. It seems to me we are arguing about mechanism. People may not like
the mechanism I believe is responsible but I have yet to see an argument that convinces me I am
wrong or that proposes a better mechanism."

But you don't have (didn't have?) a mechanism that's the point. Something is required to turn mechanical energy into heat, and your model doesn't (didn't?) include that. If you now want to run with the real world loss mechanisms then you have to establish a causal relationship between the mechanical energy variation and the losses. As I said, with a heavy chainset and pedals, you only need a slightly "unround" chainring to set "internal work" to zero - what does that say about the predictive ability of these calculations? Strikes me they would massively underestimate the losses (which is the opposite of what you wanted to show).

Riding uphill, or with a Biopace chainring, internal work calulations would predict substantially more "energy loss" than on the flat with a round or traditional elliptical - somebody would have noticed these effects, surely.

Andrew Bradley (hoping this post is readable, newsfeed playing up, google playing up...)

 

[Phil Holman]

"Frank Day" <[email protected]> wrote in message
news:[email protected]...
> "Phil Holman" <[email protected]> wrote in message
>
> I guess I am only looking at the steady state. You could also ask the question as to what happens
> at the top of the hill when cadence increases. Let's just look at the steady state.

The state has no affect on the transfer.

>
> How about this scenario. An exercycle on the international space station. In that situation, there
> would be no potential energy in the thigh (or anywhere else) to give to the pedal. How would you
> analyze the energy requirements there?

For PE, you lose a little on the down stroke but you did less work on the upstroke not having to
raise the leg against gravity so overall there is no difference. KE is no different in zero gravity.

> >
> > >
> > > Second, it assumes the weight of gravity is the only force to consider here.
> >
> > It doesn't matter if the input force was just gravity and/or muscle input, the mechanism is the
> > same. KE is 1/2mv^2, all it requires is mass and velocity. When the thigh is slowed, the energy
> > goes into turning the pedal.
>
> But, the masses are different so to conserve energy the velocities must be substantially
> different. They are not. So, less energy is transferred to the pedal than was contained in the
> thigh. How do you explain that?

Your perception is flawed. Energy is transferred from the thigh (say 10kg) to the pedal which is
providing propulsion to a bike plus rider (say 80kg) not to a 100g pedal. The equations used to
figure out the velocity relationship are 1/2*I*w^2 and 1/2*m*v^2. One for rotational KE of the thigh
and the other for linear KE of the bike plus rider. If the scenario is a thigh moving at 2
radians/sec and a cyclist at 30kph the increase in speed of the cyclist when the thigh transfers
it's KE is given by:

1/2*m*v2^2 = 1/2*m*v1^2 + 1/2*I*w^2 40*v2^2 = 40*30^2 + .3125*2^2 v2^2 = 900.0313 v2 = 30.0005kph

> >
> > >How does your model handle pedaling when the cadence is so high that the thigh is forced to
> > >fall faster than gravity would make it go passively. (after all, it is not a bowling ball but
> > >hinged so will fall slower than a free falling object.) In that instance,
there
> > > would be a retarding force, not on the downward movement of the thigh but on the downward
> > > movement of the pedal) on the majority of the downward portion of the stroke which would
> > > require energy from the flywheel to maintain pedal speed and then a further retarding force on
> > > the upstroke. How does that conserve energy?
> >
> > I have no idea what you are talking about. The mechanism is the same at any cadence.
>
> I agree, the mechanism of loss is the same at any cadence or any gravity condition (see above). If
> less kinetic energy is transferred to the pedals than is present and this repeats twice every 360
> degree circle then energy is lost (if it is not stored as potential energy - which is not possible
> in the zero gravity situation). Whether it is lost through eccentric contraction of the muscles or
> flexing of the componentes with friction loss or some other mechanism, it is lost. to convince me
> otherwise you will need to show me the data that shows the kinetic energy of the system remains
> constant through the entire circle when pedaling at constant cadence.

That's a flawed requirement of a flawed model that ignores conservation of energy.

> >
> > >
> > > Third, I assume it presumes that the transferred kinetic energy you describe is stored and
> > > given back by the flywheel (without any loss from the acceleration and deceleration of the
> > > system) to get the thigh back up when there is no evidence that this is
the
> > > case.
> >
> > Flywheel???? When riding along normally, a combination of muscle action and or gravity causes
> > the thigh to move, the KE in the moving thigh is transferred to the pedal/crank on the lower
> > part of the downstroke which does work against drag. End of story.
>
> No, the transfer of kinetic energy is not complete. It can't be because of the constant velocity
> restriction and the differing masses of these components. Show me a caclulation that shows this
> transfer can be complete regardless of the mass of the rider or the gear combination chosen or the
> cadence. I don't think it is possible.

See above.

> >
> > >
> > > Further, your model doesn't describe how the pedaling power losses come to vary with the cube
> > > of the cadence.
> >
> > The $1million question. Cube function power *losses* are a characteristic of fluid dynamics. A
> > square function energy input to accelerate a mass with respect to velocity *doesn't* explain the
> > losses.
>
> Yes it does, just as a square function energy loss with regards to aerodynamic losses results in a
> cube function aerodynamic power loss with speed.

Being a cube function alone doesn't explain the losses especially one that ignores conservation
of energy.

>
> > >While at slow speeds there may be some transfer of thigh potential energy to the pedal, this
> > >certainly is not evidence that this then makes pedaling at constant cadence energy
> > >conservative.
> >
> > The mechanism you propose incurrs no losses. Considering just the anatomy of the rider, pedaling
> > a bicycle involves work against non-conservative forces and while you continue on this tack, the
real
> > culprits are being ignored.
>
> what real culprits? Give me another square function energy loss involved in this motion.

Viscous damping force can be a function of v^2

Phil Holman

 

[Frank Day]

andrewbradley <[email protected]> wrote in message news:<[email protected]>...
> Frank Day wrote:
> > FD:"As far as I can tell it is the need to constantly accelerate and decelerate the separate
> > parts in a manner that doesn't conserve energy. Under conservation of momentum, once in motion
> > an object will remain in motion at a constant velocity unless acted upon. Acting upon it
> > requires energy."
> That's just not correct. It all depends on whether speed increases or decrease and via what
> mechanism. Let's get this one straight (again): increasing KE requires energy input, decreasing it
> requires energy output and this can be put to use given the right mechanism - whatever the
> acceleration involved in aquiring or reducing speed.

Wow, that is news to me. I didn't realize that when the space shuttle fired its rockets that it made
a difference as to how the scientists calculate what will happen depending upon whether they are
trying to speed it up or slow it down. Now, I admit that slowing an object down can be done using a
mechanism that stores the energy and then transfers it back, but it seems to me that we have pretty
much established that that can't be the case for the cyclist with a freewheel. I am not sure what
you mean by energy output. If two equal of opposite moving masses collide the result is both objects
will "output" energy by your definition. How is that possible? Where did the energy go?
> > FD:"Now, if one wants toget from point a to point b for the least amount of energy expenditure
> > it seems the best plan is to maintain a constant speed since the object in motion that is not
> > being acted upon requires zero energy and an object that is moving at the same average speed
> > but is continually being acted upon to be accelerated and decelerated requires an energy
> > expenditure that is not zero to cover the same distance. Now, where are the losses there? I am
> > not sure as I am sufficiently far away from this stuff academically and professionally that it
> > is not on the tip of my tongue. I know it exists though."
> There are no losses implied if the kinetic energy loss from the decelerations is recuperated. This
> has lead people to invent braking systems for bikes via spring wind-up. You can accelerate as hard
> and long as you like as long as your braking spring has the capacity. (Wind resistance neglected
> here of course).
We are not talking about wind up braking systems here (which simply change the kinetic energy in the
form of potential energy I presume). This is the basis of the Toyata Prius hybrid automobile where
braking is done by generators to convert kinetic energy to the potential energy of a battery. There
is no energy storage system nor any way of giving energy back to the legs in a bicycle with a
freewheel. Stay on topic please.
> > FD:"The athlete only has so much energy to give. It has to be divided up amongst the competing
> > interests. There is a certain amount of "overhead" energy to keep the athlete alive. If
> > pedaling does require energy, then it is also an "overhead" that must be subtracted from the
> > total energy the athlete can give to the wheel which will make him slower than he has energy
> > to give. If pedaling requires no energy then all the additional energy, including this work,
> > can be given to make the bike go faster. We have seen experimental evidence here that pedaling
> > is not free and various cadences can cause substantial changes in HR without changing power to
> > the wheel one iota. That is what I mean. One scenario is faster or slower than the other. Does
> > that make sense?"
> This is all quite true, however these effects aren't related to the underlying system of mass
> movements which is conservative unless you factor in some specific resistance.
You and others have proclaimed it conservative. I see no evidence of that. You simply don't like my
explanation of where the energy goes. Yet, rather than explain where it goes, you simply declare it
to be conservative, which is impossible in a bicycle with a freewheel pedaling at constant cadence.
To be conservative you must get the stored energy lost when transmitted to the pedal in your
scenario back into the legs at some point. Simply explain how this is done. I don't understand.
> > > >Or, the analysis is flawed and there really is no difference between fixed wheel pedaling
> > > >and free wheel pedaling, or PowerCranks pedaling and the pedaling losses are the same for
> > > >all. Take your pick. Me thinks the pedaling losses are probably the same for all cyclists
> > > >whether they have a freewheel, fixed wheel, or PowerCranks, but I can't prove it.
> > >
> > > Conservation of energy is your one line proof (at this level).
> > FD:"If that proves my contention it sure isn't being accepted very well."
> It proves that whatever the pedalling system there is no energy loss from mass movements alone. I
> thought your original contention was the opposite.
I believe the energy loss is from the mass movements, more accurately, the continual mass
accelerations required by the pedaling motion with no mechanism to conserve energy.

> How the leg-muscles get on with these systems is another matter. The fixed wheel allows the
> muscles maximum freedom to contract whenever and however hard they see fit. In theory, "when"
> could be a factor since being fully fired at times when contraction speed is very low (ie early
> and late in the range of joint movement) could be costly (on the assumption that muscle fibre has
> an optimal contraction speed) and it could also involve a bit of co-contraction due to the fire-up
> delay. The round chainring and freewheel probably allows the muscles enough "freedom of operation"
> both on how hard and when to contract for most cadences. Clearly, powercranks don't allow the same
> freedom, making them difficult to pedal at speed.

Phooey, at power a powercranker couldn't tell the difference between PowerCranks and regular cranks
on a fixed wheel. When one is pedaling properly it doesn't matter which cranks are on the bike
because there is forward pressure on the pedal the entire 360 degrees. When pedaling in this fashion
it is impossible to transmit any energy to the leg from the pedal, whether coming from the bicycle
or the other pedal. Only when one got lazy or tried to relax would one be able to tell the
difference. The muscles have the same "freedom of operation" to pedal regardless of the type of
bicycle they are using or cranks on that bicycle
> Attempting to apply maximum pressure in all sectors may not be such a good idea, if you don't
> have to.

Perhaps, but this has nothing to do with what this disucssion is all about.
> > FD:"The experimental evidence is my proof that the phenomenon exists. It seems to me we are
> > arguing about mechanism. People may not like the mechanism I believe is responsible but I have
> > yet to see an argument that convinces me I am wrong or that proposes a better mechanism."
>
>
>
> But you don't have (didn't have?) a mechanism that's the point. Something is required to turn
> mechanical energy into heat, and your model doesn't (didn't?) include that. If you now want to run
> with the real world loss mechanisms then you have to establish a causal relationship between the
> mechanical energy variation and the losses. As I said, with a heavy chainset and pedals, you only
> need a slightly "unround" chainring to set "internal work" to zero - what does that say about the
> predictive ability of these calculations? Strikes me they would massively underestimate the losses
> (which is the opposite of what you wanted to show).

It really shouldn't matter if we don't understand what the exact mechanism is if the math indicates
the losses should be there. Unless the math is wrong, the losses are there. I believe there are
several possible mechanisms from eccentric muscle contractions to flexing of the non rigid
structures (all of the structures of the leg are non rigid). And I don't have to establish the
causal relationship you suggest. Mechanical energy variations are inherently "lossy". the principle
is well established. Unless one can show that these energy variations are conservative with energy
storage and retrieval mechanisms (pendulum, weight on a spring) then energy is not conserved. If
pedaling energy were really conserved as you contend, the rider would have to actually consciously
apply energy to stop pedaling. I don't know anyone who has ever noticed this phenomenon.

>
> Riding uphill, or with a Biopace chainring, internal work calulations would predict substantially
> more "energy loss" than on the flat with a round or traditional elliptical - somebody would have
> noticed these effects, surely.

Why? riding uphill the cadence is usually lower. Why would you predict pedaling losses to be greater
riding uphill? Or, with BioPace? And, people have noticed this energy robbing effect. At least the
effect that pedaling requires power and the power required increases with the cube of the cadence
has been shown in the literature. And every new PowerCranker has noticed.

Frank "this thread may slow down a lot as I seemed to have reached some posting limit" Day

 

[Phil Holman]

"Frank Day" <[email protected]> wrote in message
news:[email protected]...
> andrewbradley <[email protected]> wrote in message
news:<[email protected]>...
> > Frank Day wrote:
> > > FD:"As far as I can tell it is the need to constantly accelerate and decelerate the separate
> > > parts in a manner that doesn't conserve energy. Under conservation of momentum, once in motion
> > > an object will remain in motion at a constant velocity unless acted upon. Acting upon it
> > > requires energy."
> > That's just not correct. It all depends on whether speed increases or decrease and via what
> > mechanism. Let's get this one straight (again): increasing KE requires energy input, decreasing
> > it requires energy output and this can be put to use given the right mechanism - whatever the
> > acceleration involved in aquiring or reducing speed.

> Wow, that is news to me. I didn't realize that when the space shuttle fired its rockets that it
> made a difference as to how the scientists calculate what will happen depending upon whether they
> are trying to speed it up or slow it down. Now, I admit that slowing an object down can be done
> using a mechanism that stores the energy and then transfers it back, but it seems to me that we
> have pretty much established that that can't be the case for the cyclist with a freewheel. I am
> not sure what you mean by energy output. If two equal of opposite moving masses collide the result
> is both objects will "output" energy by your definition. How is that possible? Where did the
> energy go?

For an elastic collision, both energy and momentum are conserved and the masses you describe would
move off in opposite directions. For an inelastic collision, the energy is dissipated as heat but
the momentum is again, as always, conserved. Don't take my word for it, look up the relevant physics
equations. Any college freshman physics text will do.

> > > FD:"Now, if one wants toget from point a to point b for the least amount of energy expenditure
> > > it seems the best plan is to maintain a constant speed since the object in motion that is not
> > > being acted upon requires zero energy and an object that is moving at the same average speed
> > > but is continually being acted upon to be accelerated and decelerated requires an energy
> > > expenditure that is not zero to cover the same distance. Now, where are the losses there? I am
> > > not sure as I am sufficiently far away from this stuff academically and professionally that it
> > > is not on the tip of my tongue. I know it exists though."

> > There are no losses implied if the kinetic energy loss from the decelerations is recuperated.
> > This has lead people to invent braking systems for bikes via spring wind-up. You can accelerate
> > as hard and long as you like as long as your braking spring has the capacity. (Wind resistance
> > neglected here of course).

> We are not talking about wind up braking systems here (which simply change the kinetic energy in
> the form of potential energy I presume). This is the basis of the Toyata Prius hybrid automobile
> where braking is done by generators to convert kinetic energy to the potential energy of a
> battery. There is no energy storage system nor any way of giving energy back to the legs in a
> bicycle with a freewheel. Stay on topic please.

Why do you think the energy needs to be given back to the legs? The energy is used to propel the
cyclist forward and that's where it would stay if it wasn't eventually dissipated due to drag.

> > > FD:"The athlete only has so much energy to give. It has to be divided up amongst the competing
> > > interests. There is a certain amount of "overhead" energy to keep the athlete alive. If
> > > pedaling does require energy, then it is also an "overhead" that must be subtracted from the
> > > total energy the athlete can give to the wheel which will make him slower than he has energy
> > > to give. If pedaling requires no energy then all the additional energy, including this work,
> > > can be given to make the bike go faster. We have seen experimental evidence here that pedaling
> > > is not free and various cadences can cause substantial changes in HR without changing power to
> > > the wheel one iota. That is what I mean. One scenario is faster or slower than the other. Does
> > > that make sense?"

> > This is all quite true, however these effects aren't related to the underlying system of mass
> > movements which is conservative unless you factor in some specific resistance.
> You and others have proclaimed it conservative. I see no evidence of that. You simply don't like
> my explanation of where the energy goes. Yet, rather than explain where it goes, you simply
> declare it to be conservative, which is impossible in a bicycle with a freewheel pedaling at
> constant cadence. To be conservative you must get the stored energy lost when transmitted to the
> pedal in your scenario back into the legs at some point. Simply explain how this is done. I don't
> understand.

You don't understand because you have your own definition of conservative. The reason it is
conservative is because it isn't lost to the legs but is transferred and does useful work in
propelling the cyclist. Give it a rest Frank. Your lack of understanding of physics is the only
problem here.

Phil Holman

 

[N Crowley]

I don't know what all that discussion was supposed to prove but regarding the difficulty that some
riders have in trying to adjust to PC's, it can be easily explained. While PC's teach the round
pedaling style, they use an extreme version which could almost be compared to a form of ankling and
ankling is most effective at a lower cadence because time is needed for pedal power application
adjustments. In addition with PC's the range of drawing back, pulling up and forcing forward has to
be extended and this explains why some riders have difficulty when getting down in an aero TT
position, the clearance is not always there for the knee when pedal is at TDC. PC's use the basic
mental idea of round pedaling but where the difficulty arises for round pedalers is that no corners
can be cut when using them. For that very reason I see them as good training equipment for round
pedaling, the stepping stone to Anquetil's pedaling perfection, but not equipment for racing or time
trialing. I cannot understand how they could produce extra power, if they can do that, it should be
possible for round pedaling to generate that same extra power with normal cranks. Anquetil's
technique can produce substantial additional pedal power but it is done by extending the main pedal
stroke from 11 o'clock to 5 o'clock and by the assistance of maximum arm resistance whenever it is
required. Trying to get additional pedal power by attempting to work both legs at the same time
won't change anything.

 

[Andrew Bradley]

Frank Day <[email protected]> wrote in message
news:[email protected]...
> andrewbradley <[email protected]> wrote in message
news:<[email protected]>...
> > Frank Day wrote:
> > > FD:"As far as I can tell it is the need to constantly accelerate and decelerate the separate
> > > parts in a manner that doesn't conserve
energy.
> > > Under conservation of momentum, once in motion an object will remain
in
> > > motion at a constant velocity unless acted upon. Acting upon it requires energy."
> > That's just not correct. It all depends on whether speed increases or decrease and via what
> > mechanism. Let's get this one straight (again): increasing KE requires energy input, decreasing
> > it requires energy output and this can be put to use given the right mechanism - whatever the
> > acceleration involved in aquiring or reducing speed.
>
> Wow, that is news to me. I didn't realize that when the space shuttle fired its rockets that it
> made a difference as to how the scientists calculate what will happen depending upon whether they
> are trying to speed it up or slow it down.

I suggest keeping away fom rocket science, but consider the KE of the gasses in both cases
after firing.

> > How the leg-muscles get on with these systems is another matter. The fixed wheel allows the
> > muscles maximum freedom to contract whenever and however hard they see fit. In theory, "when"
> > could be a factor since being fully fired at times when contraction speed is very low (ie early
> > and late in the range of joint movement) could be costly (on the assumption that muscle fibre
> > has an optimal contraction speed) and it could also involve a bit of co-contraction due to the
> > fire-up delay. The round chainring and freewheel probably allows the muscles enough "freedom of
> > operation" both on how hard and when to contract for most cadences. Clearly, powercranks don't
> > allow the same freedom, making them difficult to pedal at speed.
>
> Phooey, at power a powercranker couldn't tell the difference between PowerCranks and regular
> cranks on a fixed wheel.

You said yourself they were difficult to pedal at "normal" cadence and wanted to know why. PCs do
limit the freedom of the muscles to "do what they want" - isn't that how they are supposed to work
their magic? Different chainring shapes can influence the ease with which you can pedal at higher
cadence on a freewheel. No big deal.

>When one is pedaling properly it doesn't matter which cranks are on the bike because there is
>forward pressure on the pedal the entire 360 degrees.

Pedalling "properly"?

Andrew Bradley

 

[Frank Day]

"Phil Holman" <[email protected]> wrote in message news:<[email protected]>...
> "Frank Day" <[email protected]> wrote in message
> news:[email protected]...
> > andrewbradley <[email protected]> wrote in message
> news:<[email protected]>...
> > > Frank Day wrote:
> > > > FD:"As far as I can tell it is the need to constantly accelerate and decelerate the separate
> > > > parts in a manner that doesn't conserve energy. Under conservation of momentum, once in
> > > > motion an object will remain in motion at a constant velocity unless acted upon. Acting upon
> > > > it requires energy."
> > > That's just not correct. It all depends on whether speed increases or decrease and via what
> > > mechanism. Let's get this one straight (again): increasing KE requires energy input,
> > > decreasing it requires energy output and this can be put to use given the right mechanism -
> > > whatever the acceleration involved in aquiring or reducing speed.
>
> > Wow, that is news to me. I didn't realize that when the space shuttle fired its rockets that it
> > made a difference as to how the scientists calculate what will happen depending upon whether
> > they are trying to speed it up or slow it down. Now, I admit that slowing an object down can be
> > done using a mechanism that stores the energy and then transfers it back, but it seems to me
> > that we have pretty much established that that can't be the case for the cyclist with a
> > freewheel. I am not sure what you mean by energy output. If two equal of opposite moving masses
> > collide the result is both objects will "output" energy by your definition. How is that
> > possible? Where did the energy go?
>
> For an elastic collision, both energy and momentum are conserved and the masses you describe would
> move off in opposite directions. For an inelastic collision, the energy is dissipated as heat but
> the momentum is again, as always, conserved. Don't take my word for it, look up the relevant
> physics equations. Any college freshman physics text will do.

I agree. I was referring to AB's description of what was going on.
>
> > > > FD:"Now, if one wants toget from point a to point b for the least amount of energy
> > > > expenditure it seems the best plan is to maintain a constant speed since the object in
> > > > motion that is not being acted upon requires zero energy and an object that is moving at the
> > > > same average speed but is continually being acted upon to be accelerated and decelerated
> > > > requires an energy expenditure that is not zero to cover the same distance. Now, where are
> > > > the losses there? I am not sure as I am sufficiently far away from this stuff academically
> > > > and professionally that it is not on the tip of my tongue. I know it exists though."
>
> > > There are no losses implied if the kinetic energy loss from the decelerations is recuperated.
> > > This has lead people to invent braking systems for bikes via spring wind-up. You can
> > > accelerate as hard and long as you like as long as your braking spring has the capacity. (Wind
> > > resistance neglected here of course).
>
> > We are not talking about wind up braking systems here (which simply change the kinetic energy in
> > the form of potential energy I presume). This is the basis of the Toyata Prius hybrid automobile
> > where braking is done by generators to convert kinetic energy to the potential energy of a
> > battery. There is no energy storage system nor any way of giving energy back to the legs in a
> > bicycle with a freewheel. Stay on topic please.
>
> Why do you think the energy needs to be given back to the legs? The energy is used to propel the
> cyclist forward and that's where it would stay if it wasn't eventually dissipated due to drag.

It would only be necessary to presume energy is given back to the legs to do the necessary
accelerations if energy is conserved. However, in the usual case, where the cyclists is putting
negative force on the pedals on the upstroke, it is not necessary to get energy back from the
bicycle to accelerate the legs because the necessary energy is diverted from the power leg to
accomplish this end. Since it is not necessary for any energy to be given to the bicycle and then
given back in order to pedal this analysis can clearly be done within the pedal system itself. I
don't think anyone is claiming that pedaling without a bicycle in a constant cadence fashion is
energy conservative so I am not sure why adding a bicycle and a little mass makes any difference.
>
> > > > FD:"The athlete only has so much energy to give. It has to be divided up amongst the
> > > > competing interests. There is a certain amount of "overhead" energy to keep the athlete
> > > > alive. If pedaling does require energy, then it is also an "overhead" that must be
> > > > subtracted from the total energy the athlete can give to the wheel which will make him
> > > > slower than he has energy to give. If pedaling requires no energy then all the additional
> > > > energy, including this work, can be given to make the bike go faster. We have seen
> > > > experimental evidence here that pedaling is not free and various cadences can cause
> > > > substantial changes in HR without changing power to the wheel one iota. That is what I mean.
> > > > One scenario is faster or slower than the other. Does that make sense?"
>
> > > This is all quite true, however these effects aren't related to the underlying system of mass
> > > movements which is conservative unless you factor in some specific resistance.
> > You and others have proclaimed it conservative. I see no evidence of that. You simply don't like
> > my explanation of where the energy goes. Yet, rather than explain where it goes, you simply
> > declare it to be conservative, which is impossible in a bicycle with a freewheel pedaling at
> > constant cadence. To be conservative you must get the stored energy lost when transmitted to the
> > pedal in your scenario back into the legs at some point. Simply explain how this is done. I
> > don't understand.
>
> You don't understand because you have your own definition of conservative. The reason it is
> conservative is because it isn't lost to the legs but is transferred and does useful work in
> propelling the cyclist. Give it a rest Frank. Your lack of understanding of physics is the only
> problem here.

Let us assume that it is possible to transfer all the kinetic energy to the bicycle. Where is the
evidence that this acutally occurs? In order to do this the body must completely relax and allow the
energy transfer to occur unimpeded. If the body tries to control the movement through eccentric
contractions, then, even though it may be possible to be completely energy conservative and transfer
all that kinetic energy completely to the bike that doesn't mean that riders will do so. Especially
if the rider was trained to pedal on bicycles not being attached to the pedals such that he must
keep contact with the pedal through muscular control. If this were the case, the person would not
necessarily transefer much leg kinetic energy to the pedal at all and this energy would be lost
through eccentric contraction. If that were the case, this could account for much, if not all, of
the observed energy loss with cadence.

Perhaps, this is one of the mechanisms of the PowerCranks power improvements. Unpublished studies
have shown the EMG timing of the various muscle contractions in relationship to pedal position
changes when one goes from ordinary cranks to PowerCranks. How would you interpret data that shows
for the same power and cadence that after training with PowerCranks HR is 10, 15 or 20 beats lower.
This explanation would get away from the objections of AC and others that it takes just as much
energy to lift as to push so such improvements are impossible.

So, even though it may be possible that one can make constant cadence pedaling conservative I don't
see any data that suggests it actually
is. In essentially every study looking at pedaling efficiency there are huge losses that are
unaccounted for (in general, the potential of the muscles is about double what is actually
delivered to the wheel). Further, no one here has put forth an alternative explanation to
account for the experimentally observed pedaling energy loss that varies with the square of
the cadence. I think it just as valid to interpret that riders are unknowingly inserting
these losses as it is to assume that people are able, once they become attached to the pedals
as adults, where slipping is not a problem, to relax and contract perfectly to maximize
energy transfer.

Frank

 

[Frank Day]

"Andrew Bradley" <[email protected]> wrote in message news:<kx7%[email protected]>...
> Frank Day <[email protected]> wrote in message
> news:[email protected]...
> > andrewbradley <[email protected]> wrote in message
> news:<[email protected]>...
> > > Frank Day wrote:
> > > > FD:"As far as I can tell it is the need to constantly accelerate and decelerate the
> > > > separate parts in a manner that doesn't conserve
> energy.
> > > > Under conservation of momentum, once in motion an object will remain
> in
> > > > motion at a constant velocity unless acted upon. Acting upon it requires energy."
> > > That's just not correct. It all depends on whether speed increases or decrease and via what
> > > mechanism. Let's get this one straight (again): increasing KE requires energy input,
> > > decreasing it requires energy output and this can be put to use given the right mechanism -
> > > whatever the acceleration involved in aquiring or reducing speed.
> >
> > Wow, that is news to me. I didn't realize that when the space shuttle fired its rockets that it
> > made a difference as to how the scientists calculate what will happen depending upon whether
> > they are trying to speed it up or slow it down.
>
> I suggest keeping away fom rocket science, but consider the KE of the gasses in both cases
> after firing.
>
> > > How the leg-muscles get on with these systems is another matter. The fixed wheel allows the
> > > muscles maximum freedom to contract whenever and however hard they see fit. In theory, "when"
> > > could be a factor since being fully fired at times when contraction speed is very low (ie
> > > early and late in the range of joint movement) could be costly (on the assumption that muscle
> > > fibre has an optimal contraction speed) and it could also involve a bit of co-contraction due
> > > to the fire-up delay. The round chainring and freewheel probably allows the muscles enough
> > > "freedom of operation" both on how hard and when to contract for most cadences. Clearly,
> > > powercranks don't allow the same freedom, making them difficult to pedal at speed.
> >
> > Phooey, at power a powercranker couldn't tell the difference between PowerCranks and regular
> > cranks on a fixed wheel.
>
> You said yourself they were difficult to pedal at "normal" cadence and wanted to know why. PCs do
> limit the freedom of the muscles to "do what they want" - isn't that how they are supposed to work
> their magic? Different chainring shapes can influence the ease with which you can pedal at higher
> cadence on a freewheel. No big deal.

Well, I guess I should have specified after adaptation. Most users, it seems, are able to ride for
30-60 minutes at a cadence of 75 to 80 in 2-3 months, 2-3 hours at a cadence of 90 in 6-9 months and
at a cadence of 100-110 after two years if they choose too. And, I guess, in a sense they do limit
the freedom of the muscles to do what they want as they prevent them from putting backward force on
the pedals.

>
> >When one is pedaling properly it doesn't matter which cranks are on the bike because there is
> >forward pressure on the pedal the entire 360 degrees.
>
> Pedalling "properly"?

Pedaling in circles is what most coaches seem to preach as the ideal. Obviously not everyone here
agrees. It was what I meant by "properly".

Frank

 

[Frank Day]

[email protected] (n crowley) wrote in message news:<[email protected]>...
> I don't know what all that discussion was supposed to prove but regarding the difficulty that some
> riders have in trying to adjust to PC's, it can be easily explained. While PC's teach the round
> pedaling style, they use an extreme version which could almost be compared to a form of ankling
> and ankling is most effective at a lower cadence because time is needed for pedal power
> application adjustments. In addition with PC's the range of drawing back, pulling up and forcing
> forward has to be extended and this explains why some riders have difficulty when getting down in
> an aero TT position, the clearance is not always there for the knee when pedal is at TDC. PC's use
> the basic mental idea of round pedaling but where the difficulty arises for round pedalers is that
> no corners can be cut when using them. For that very reason I see them as good training equipment
> for round pedaling, the stepping stone to Anquetil's pedaling perfection, but not equipment for
> racing or time trialing. I cannot understand how they could produce extra power, if they can do
> that, it should be possible for round pedaling to generate that same extra power with normal
> cranks. Anquetil's technique can produce substantial additional pedal power but it is done by
> extending the main pedal stroke from 11 o'clock to 5 o'clock and by the assistance of maximum arm
> resistance whenever it is required. Trying to get additional pedal power by attempting to work
> both legs at the same time won't change anything.

Spoken like someone who has never ridden them, let alone trained on them. Of course it is possible
to generate the same extra power on regular cranks as on PowerCranks as long as someone can pedal in
the PowerCranks fashion. The problem is in the learning how.

Frank

 

[n crowley]

Originally posted by Frank Day


Spoken like someone who has never ridden them, let alone trained on them. Of course it is possible
to generate the same extra power on regular cranks as on PowerCranks as long as someone can pedal in
the PowerCranks fashion. The problem is in the learning how.

Frank





True, but maybe you can clarify where the extra power is
produced. With this increase of power in progress, what
percentage of overall power is being produced by the leg
between 6 and 12 o'clock. Also when an experienced PC user
is pedaling at a cadence around 100, will he be able to produce
that same extra power.

 

[Andrew Bradley]

[email protected] (Frank Day) wrote in message news:<[email protected]>...
> "Phil Holman" <[email protected]> wrote in message
> news:<[email protected]>...
> > "Frank Day" <[email protected]> wrote in message
> > news:[email protected]...
> > > andrewbradley <[email protected]> wrote in message
> news:<[email protected]>...
> > > > Frank Day wrote:
> > > > > FD:"As far as I can tell it is the need to constantly accelerate and decelerate the
> > > > > separate parts in a manner that doesn't conserve energy. Under conservation of momentum,
> > > > > once in motion an object will remain in motion at a constant velocity unless acted upon.
> > > > > Acting upon it requires energy."
> > > > That's just not correct. It all depends on whether speed increases or decrease and via what
> > > > mechanism. Let's get this one straight (again): increasing KE requires energy input,
> > > > decreasing it requires energy output and this can be put to use given the right mechanism -
> > > > whatever the acceleration involved in aquiring or reducing speed.
>
> > > Wow, that is news to me. I didn't realize that when the space shuttle fired its rockets that
> > > it made a difference as to how the scientists calculate what will happen depending upon
> > > whether they are trying to speed it up or slow it down. Now, I admit that slowing an object
> > > down can be done using a mechanism that stores the energy and then transfers it back, but it
> > > seems to me that we have pretty much established that that can't be the case for the cyclist
> > > with a freewheel. I am not sure what you mean by energy output. If two equal of opposite
> > > moving masses collide the result is both objects will "output" energy by your definition. How
> > > is that possible? Where did the energy go?
> >
> > For an elastic collision, both energy and momentum are conserved and the masses you describe
> > would move off in opposite directions. For an inelastic collision, the energy is dissipated as
> > heat but the momentum is again, as always, conserved. Don't take my word for it, look up the
> > relevant physics equations. Any college freshman physics text will do.
>
> I agree. I was referring to AB's description of what was going on.

When a given mass loses mechanical energy, it's environment gets it in one form or another. Thats
what I meant by "energy output". What _ is_ the problem?

> > Why do you think the energy needs to be given back to the legs? The energy is used to propel the
> > cyclist forward and that's where it would stay if it wasn't eventually dissipated due to drag.
>
> It would only be necessary to presume energy is given back to the legs to do the necessary
> accelerations if energy is conserved.

Mechanical leg energy is not conserved as mechanical leg energy but transformed into a useful form
at times, what on earth is your point?

>
>However, in the usual case, where the cyclists is putting negative force on the pedals on the
>upstroke, it is not necessary to get energy back from the bicycle to accelerate the legs because
>the necessary energy is diverted from the power leg to accomplish this end.

Of course. It is you who keeps saying energy has to be given back for conservation. Any leg can
input the energy required to increase the mechanical leg energy, and that energy gets transformed
later into work against resisatances and work done on drivetrain.

>Since it is not necessary for any energy to be given to the bicycle and then given back in order to
>pedal this analysis can clearly be done within the pedal system itself.

Of course it can, but you have to remember the system is not closed.

>I don't think anyone is claiming that pedaling without a bicycle in a constant cadence fashion is
>energy conservative so I am not sure why adding a bicycle and a little mass makes any difference.

Because the muscles themseleves have to constrain the motion, pedalling in the air is likely to be
rather more wasteful than pedalling with cranks, but this is beside the point.

> Let us assume that it is possible to transfer all the kinetic energy to the bicycle. Where is the
> evidence that this acutally occurs?

Don't need any. In the real world, the amount of kinetic energy recuperated might be less than that
gobbled up by internal resistances. So what?

The muscle work and any recuperated mechanical leg energy goes into powering bicycle and overcoming
internal resistances. It is daft to try to divide up the energy. There is no causal relationship
established between leg energy loss and any internal loss mechanism. There's no reason to assume it
represents excentric muscle work. As i said, with a heavy chainset and a slightly elliptical ring,
internal work will be zero - but would any excentric muscle work cease?

To say internal work is "one loss among others" is like saying "There'll be a loss of
1066milliwatts, amongst other things, because that represents the year of the Norman Conquest"

I am glad to see you now realise the need to link internal work to a specific loss, though.

Andrew Bradley

 

[Dvt]

n crowley wrote:
> what percentage of overall power is being produced by the leg between 6 and 12 o'clock.

100%.

By the way, noel, I saw this really cool movie that I think you should see. I don't remember
the name of the movie, who was in it, or any of the details, but you should see it. You'd
really like it.

Dave dvt at psu dot edu

 

[Frank Day]

n crowley <[email protected]> wrote in message news:<OQi%[email protected]>...
> Originally posted by Frank Day
>
>
> Spoken like someone who has never ridden them, let alone trained on them. Of course it is possible
> to generate the same extra power on regular cranks as on PowerCranks as long as someone can pedal
> in the PowerCranks fashion. The problem is in the learning how.
>
> Frank
>
>
>
>
>
> True, but maybe you can clarify where the extra power is produced. With this increase of power in
> progress, what percentage of overall power is being produced by the leg between 6 and 12 o'clock.
> Also when an experienced PC user is pedaling at a cadence around 100, will he be able to produce
> that same extra power.

I don't know. I see there are several potential mechanisms for the power improvement and I have no
real data to let me come to any firm conclusions. The Luttrell study suggests to me that the 10%
effeciency improvement they saw is probably mostly coming from just unweighting the backstroke. How
much more can be gained with time as people actually learn to apply force on the upstroke remains to
be seen and cannot be determined without pressure plate pedals. Also, it will require analysis by
pressure plate pedals to see if some of it comes from changing the direction of the applied force to
be more tangential. Further, it will take an EMG analysis to see if, perhaps, some of this
improvement is coming from a better recovery of the kinetic energy in the legs as I proposed might
be happening in another recent post.

All I have been able to do is document that real improvements occur. The fact that I do not know
yet exactly how the improvements are parsed amongst the many various possibilities does not mean
they are not real. It is really for the researchers out there who are interested in this stuff
to do that.

Whether a PC'er will be able to produce the same improvement at high cadences is not known right
now. Right now even experienced PC'ers cannot ride at the same cadence on PC's as they do on regular
cranks, falling about 20 beats less. This may simply be that they are inadequately trained (such
training may take 10 years) or it may be that it will be impossible. It is to early to know.
However, it is really not important. The important thing is that training with PowerCranks give the
rider power improvements (whether they are riding on PowerCranks OR regular cranks) that they cannot
achieve training using past techniques. How big an advantage is possible and exactly how these
changes come about is still undetermined.

Frank

 

[Frank Day]

[email protected] (Andrew Bradley) wrote in message news:<[email protected]>...
> [email protected] (Frank Day) wrote in message
> news:<[email protected]>...
> > "Phil Holman" <[email protected]> wrote in message
> > news:<[email protected]>...
> > > "Frank Day" <[email protected]> wrote in message
> > > news:[email protected]...
> > > > andrewbradley <[email protected]> wrote in message
> news:<[email protected]>...
> > > > > Frank Day wrote:
> > > > > > FD:"As far as I can tell it is the need to constantly accelerate and decelerate the
> > > > > > separate parts in a manner that doesn't conserve energy. Under conservation of momentum,
> > > > > > once in motion an object will remain in motion at a constant velocity unless acted upon.
> > > > > > Acting upon it requires energy."
> > > > > That's just not correct. It all depends on whether speed increases or decrease and via
> > > > > what mechanism. Let's get this one straight (again): increasing KE requires energy input,
> > > > > decreasing it requires energy output and this can be put to use given the right mechanism
> > > > > - whatever the acceleration involved in aquiring or reducing speed.
>
> > > > Wow, that is news to me. I didn't realize that when the space shuttle fired its rockets that
> > > > it made a difference as to how the scientists calculate what will happen depending upon
> > > > whether they are trying to speed it up or slow it down. Now, I admit that slowing an object
> > > > down can be done using a mechanism that stores the energy and then transfers it back, but it
> > > > seems to me that we have pretty much established that that can't be the case for the cyclist
> > > > with a freewheel. I am not sure what you mean by energy output. If two equal of opposite
> > > > moving masses collide the result is both objects will "output" energy by your definition.
> > > > How is that possible? Where did the energy go?
> > >
> > > For an elastic collision, both energy and momentum are conserved and the masses you describe
> > > would move off in opposite directions. For an inelastic collision, the energy is dissipated as
> > > heat but the momentum is again, as always, conserved. Don't take my word for it, look up the
> > > relevant physics equations. Any college freshman physics text will do.
> >
> > I agree. I was referring to AB's description of what was going on.
>
> When a given mass loses mechanical energy, it's environment gets it in one form or another. Thats
> what I meant by "energy output". What _ is_ the problem?

It is just a term that isn't ordinarily used (at least it wasn't when I was going to school). I
understand the intent.
>
> > > Why do you think the energy needs to be given back to the legs? The energy is used to propel
> > > the cyclist forward and that's where it would stay if it wasn't eventually dissipated due to
> > > drag.
> >
> > It would only be necessary to presume energy is given back to the legs to do the necessary
> > accelerations if energy is conserved.
>
> Mechanical leg energy is not conserved as mechanical leg energy but transformed into a useful form
> at times, what on earth is your point?

For a system to be "conservative" it should require no external energy input to continue. Your
system requires continual energy output so falls down then energy output is less than that required.
I see you are saying it is all used and converted into work, but I don't see it as a conservative
system. Is launching the space shuttle, retreiving it, refueling it and getting ready to launch
again a "conservative system". According to you, yes.
>
> >
> >However, in the usual case, where the cyclists is putting negative force on the pedals on the
> >upstroke, it is not necessary to get energy back from the bicycle to accelerate the legs because
> >the necessary energy is diverted from the power leg to accomplish this end.
>
> Of course. It is you who keeps saying energy has to be given back for conservation. Any leg can
> input the energy required to increase the mechanical leg energy, and that energy gets transformed
> later into work against resisatances and work done on drivetrain.

That is true, but it doesn't require any transfer of energy to the bicycle to pedal in that way. So,
how do you account for the kinetic energy variations. It is handled through eccentric muscle
contraction. There is more than one way to deal with this energy variation. How do you know your way
is correct?
>
> >Since it is not necessary for any energy to be given to the bicycle and then given back in order
> >to pedal this analysis can clearly be done within the pedal system itself.
>
> Of course it can, but you have to remember the system is not closed.

Sure it is if there is no chain.
>
> >I don't think anyone is claiming that pedaling without a bicycle in a constant cadence fashion is
> >energy conservative so I am not sure why adding a bicycle and a little mass makes any difference.
>
> Because the muscles themseleves have to constrain the motion, pedalling in the air is likely to
> be rather more wasteful than pedalling with cranks, but this is beside the point.

No it isn't, it is the point. pedaling without pedals may be more wasteful than pedaling with pedals
but there is no proof (or, even, evidence) that pedaling with pedals is suddenly transformed to be
perfectly efficient.
>
> > Let us assume that it is possible to transfer all the kinetic energy to the bicycle. Where is
> > the evidence that this acutally occurs?
>
> Don't need any. In the real world, the amount of kinetic energy recuperated might be less than
> that gobbled up by internal resistances. So what?

No, assume no internal resistance. Where is the evidence that ANY energy is tranferred in this
fashion? This is all supposition. People have looked at this and said "it is possible, therefore it
happens". I say phooey. There is no evidence to support this and, in fact, the evidence is against
it unless there is another adequate explanation for pedaling power losses varying with the cube of
the cadence.

>
> The muscle work and any recuperated mechanical leg energy goes into powering bicycle and
> overcoming internal resistances. It is daft to try to divide up the energy. There is no causal
> relationship established between leg energy loss and any internal loss mechanism. There's no
> reason to assume it represents excentric muscle work.

yes there is. It is how we learned how to pedal as children. Watch a child learn to ride. They can't
do it until they learn how to control the eccentric forces to keep the foot "attached" to the pedal.
What is really the case is there is no reason to assume that as soon as one reaches a certain age or
seriousness about cycling that they suddenly change the way they pedal.

> As i said, with a heavy chainset and a slightly elliptical ring, internal work will be zero - but
> would any excentric muscle work cease?

I don't understand. if eccentric muscle work doesn't cease then it seems to me internal work
cannot be zero.

>
> To say internal work is "one loss among others" is like saying "There'll be a loss of
> 1066milliwatts, amongst other things, because that represents the year of the Norman Conquest"

Huh?
>
> I am glad to see you now realise the need to link internal work to a specific loss, though.

All losses have a specific mechanism. It is not necessary to know what the mechanism is to know that
it is present.

Frank

 

[Phil Holman]

"Frank Day" <[email protected]> wrote in message
news:[email protected]...
> [email protected] (Andrew Bradley) wrote in message > For a system to be "conservative" it
> should require no external energy input to continue. Your system requires continual energy output
> so falls down then energy output is less than that required. I see you are saying it is all used
> and converted into work, but I don't see it as a conservative system. Is launching the space
> shuttle, retreiving it, refueling it and getting ready to launch again a "conservative system".
> According to you, yes.

The space shuttle or riding a bicycle operate with both conservative and non-conservative forces.
Your space shuttle example illustrates your inability to make a distinction. It would be accurate to
say that the work done against gravity and acceleration is 100% conserved in the form of PE and KE,
the work done against air resistance is totally lost. What I think you are saying is, KE or PE is
not conserved due to air resistance. This is incorrect.

> > > Let us assume that it is possible to transfer all the kinetic energy to the bicycle. Where is
> > > the evidence that this acutally occurs?
> >
> > Don't need any. In the real world, the amount of kinetic energy recuperated might be less than
> > that gobbled up by internal resistances. So what?
>
> No, assume no internal resistance. Where is the evidence that ANY energy is tranferred in this
> fashion? This is all supposition. People have looked at this and said "it is possible, therefore
> it happens". I say phooey. There is no evidence to support this and, in fact, the evidence is
> against it unless there is another adequate explanation for pedaling power losses varying with the
> cube of the cadence.
>

How many more times, viscous drag force can be a function of v^2 which will be a power loss varying
with the cube of cadence. In any event, not knowing what it is exactly doesn't disqualify one from
knowing what it isn't.

Phil Holman

 

[Andrew Bradley]

Frank Day <[email protected]> wrote in message
news:[email protected]...
>Is launching the space shuttle, retreiving it, refueling it and getting ready to launch again a
>"conservative system". According to you, yes.

Why continually misrepresent the case of the people who disagree with you?

> >
> > >
> > >However, in the usual case, where the cyclists is putting negative force on the pedals on the
> > >upstroke, it is not necessary to get energy back from the bicycle to accelerate the legs
> > >because the necessary energy is diverted from the power leg to accomplish this end.
> >
> > Of course. It is you who keeps saying energy has to be given back for conservation. Any leg can
> > input the energy required to increase the mechanical leg energy, and that energy gets
> > transformed later into work against resisatances and work done on drivetrain.
>
> That is true, but it doesn't require any transfer of energy to the bicycle to pedal in that way.

You require internal resistance to magically modulate in order to prevent it which is far less
plausible.

>So, how do you account for the kinetic energy variations.

I don't see anything to account for in them.

>It is handled through eccentric muscle contraction. There is more than one way to deal with this
>energy variation. How do you know your way is correct?

Basic physics can explain your energy variations without the need for losses. You want us to believe
that internal losses will automatically make themselves equal to the energy variation, or (more
recently) just the loss to excentric muscle work That sounds like magic to me. Also, remember that
excentric work is not just non-conservative, there is an extra cost, but you don't know what that
is. Also bear in mind that theres a lot more power going to the pedals than that acounted for by the
leg energy variations and that is likely to affect excentric muscle work of itself. Just what does
your calculation predict or demonstrate? We don't even have any numbers from you.

> >
> > >Since it is not necessary for any energy to be given to the bicycle and then given back in
> > >order to pedal this analysis can clearly be done within the pedal system itself.
> >
> > Of course it can, but you have to remember the system is not closed.
>
> Sure it is if there is no chain.

You cannot have a closed, resistance free system with no chain and constant pedal speed. Drop the
constant pedal speed constraint and you can have a closed system with no resistances.

Oh i remember, you've got a mechanism for resistance now...

> >
> > >I don't think anyone is claiming that pedaling without a bicycle in a constant cadence fashion
> > >is energy conservative so I am not sure why adding a bicycle and a little mass makes any
> > >difference.
> >
> > Because the muscles themseleves have to constrain the motion, pedalling in the air is likely to
> > be rather more wasteful than pedalling with cranks, but this is beside the point.
>
> No it isn't, it is the point. pedaling without pedals may be more wasteful than pedaling with
> pedals but there is no proof (or, even, evidence) that pedaling with pedals is suddenly
> transformed to be perfectly efficient.

Another misrepresentation of what this thread has been about.

> >
> > > Let us assume that it is possible to transfer all the kinetic energy to the bicycle. Where is
> > > the evidence that this acutally occurs?
> >
> > Don't need any. In the real world, the amount of kinetic energy recuperated might be less than
> > that gobbled up by internal resistances. So what?
>
> No, assume no internal resistance. Where is the evidence that ANY energy is tranferred in this
> fashion?

Elementary mechanics is all that is required. Hold on though... have you got a resistance mechanism
or not in your model? Is a resistance required or not?

>This is all supposition.

Like Newtons laws.

>People have looked at this and said "it is possible, therefore it happens". I say phooey. There is
>no evidence to support this and, in fact, the evidence is against it unless there is another
>adequate explanation for pedaling power losses varying with the cube of the cadence.

The experimental evidence mentioned earlier said cost varies with cadence^4.blah . I see a cubic as
being rather too nice. Also, any correlation would not prove cause and effect. Phils comment on
losses would be just as convincing.

> >
> > The muscle work and any recuperated mechanical leg energy goes into powering bicycle and
> > overcoming internal resistances. It is daft to try to divide up the energy. There is no causal
> > relationship established between leg energy loss and any internal loss mechanism. There's no
> > reason to assume it represents excentric muscle work.
>
> yes there is. It is how we learned how to pedal as children. Watch a child learn to ride. They
> can't do it until they learn how to control the eccentric forces to keep the foot "attached" to
> the pedal. What is really the case is there is no reason to assume that as soon as one reaches a
> certain age or seriousness about cycling that they suddenly change the way they pedal.

Are you saying that pressure on the downward pedal (and thus energy transfer) is not allowed between
mid-stroke and BDC because a foot would slip off? This is when leg energy decreases and "payback"
would occur. If that is the case then how is the "down" leg to give the "up" leg that easy ride you
seek to eliminate with your cranks? This phenomenon would be a surprise to people whith force plate
pedals and indeed anybody that rides a bike.

> > As i said, with a heavy chainset and a slightly elliptical ring, internal work will be zero -
> > but would any excentric muscle work cease?
>
> I don't understand. if eccentric muscle work doesn't cease then it seems to me internal work
> cannot be zero.

Precisely! "Internal work" is a definition - total absolute mechanical energy variation. As such the
term is misleading. Internal work can be eliminated via a chainring but internal _loss_, which is
what we are actually interested in, will continue. Do your childish pedallers suddenly learn to
pedal like grown ups when the mechanical energy variations disappear - even though they are still
pedalling in circles?

> > I am glad to see you now realise the need to link internal work to a specific loss, though.
>
> All losses have a specific mechanism. It is not necessary to know what the mechanism is to know
> that it is present.

Conservation of energy shows there are no "losses looking for a mechanism" implicit in your
calculations so you'll have to get a loss mechanism that corresponds to it. But i repeat myself...

Andrew Bradley

 

[Frank Day]

"Phil Holman" <[email protected]> wrote in message news:<dPv%[email protected]>...
> "Frank Day" <[email protected]> wrote in message
> news:[email protected]...
> > [email protected] (Andrew Bradley) wrote in message > For a system to be "conservative" it
> > should require no external energy input to continue. Your system requires continual energy
> > output so falls down then energy output is less than that required. I see you are saying it is
> > all used and converted into work, but I don't see it as a conservative system. Is launching the
> > space shuttle, retreiving it, refueling it and getting ready to launch again a "conservative
> > system". According to you, yes.
>
> The space shuttle or riding a bicycle operate with both conservative and non-conservative forces.
> Your space shuttle example illustrates your inability to make a distinction. It would be accurate
> to say that the work done against gravity and acceleration is 100% conserved in the form of PE and
> KE, the work done against air resistance is totally lost. What I think you are saying is, KE or PE
> is not conserved due to air resistance. This is incorrect.

I don't understand. When it is sitting on the launch pad waiting to lift off the KE is zero but the
PE is the total of the energy due to gravity and the chemical energy in the tanks. It gains KE and
PE during lift off from the utilization of the chemical PE which it then loses in friction when it
reenters the atmosphere. such that when it gets to the ground again, it has lost energy, which must
be made up from an outside source to be returned to its starting state. How is this conservative
like a pendulum? It may be but I don't see it.
>
> > > > Let us assume that it is possible to transfer all the kinetic energy to the bicycle. Where
> > > > is the evidence that this acutally occurs?
> > >
> > > Don't need any. In the real world, the amount of kinetic energy recuperated might be less than
> > > that gobbled up by internal resistances. So what?
> >
> > No, assume no internal resistance. Where is the evidence that ANY energy is tranferred in this
> > fashion? This is all supposition. People have looked at this and said "it is possible, therefore
> > it happens". I say phooey. There is no evidence to support this and, in fact, the evidence is
> > against it unless there is another adequate explanation for pedaling power losses varying with
> > the cube of the cadence.
> >
>
> How many more times, viscous drag force can be a function of v^2 which will be a power loss
> varying with the cube of cadence. In any event, not knowing what it is exactly doesn't disqualify
> one from knowing what it isn't.

If viscous drag force "can" be a function of v^2 why doesn't rolling resistance vary with V^2? I
guess under some circumstances viscous drag can vary with V^2 but I am under no awareness that this
is present to any great degree in the human body, rather I was under the impression that most of the
resistance to motion we are talking about here behaves as a rubber band or spring. Even if some of
these losses were present, it doesn't go to the magnitude of these losses. It might be possible that
there are two different losses, each varying as a function of v^2. Under this scenario, the relative
magnitude of each loss then would determine its significnce to the rider.

The key argument I am trying to make now is the problem of what the rider must do to make pedaling
conservative under your scenario. Riders are not passive stick figures but biological systems with
conscious control over the movement of these various parts. Unless relaxation and contraction occurs
at the exact proper time in a multitude of muscles it would not be possible for the rider to
"conserve" energy as proposed. Like I said, just because it might be possible doesn't mean it
occurs. In fact, I believe the best analysis of all the data would say that this mechanism of energy
management is rarely, if ever, present.

Frank

 

[Terry Morse]

Frank Day wrote:

>If viscous drag force "can" be a function of v^2 why doesn't rolling resistance vary with V^2? I
>guess under some circumstances viscous drag can vary with V^2

For most flow with gases, and liquids at high velocity, drag is proportional to V^^2. For high
viscosity (or very low Reynolds number) flow in a fluid, drag varies with V.

Rolling resistance is viscoelastic and is not a high velocity fluid, thus it varies with V.

--
terry morse Palo Alto, CA http://www.terrymorse.com/bike/