Powercranks

[Frank Day]

"Phil Holman" <[email protected]> wrote in message news:
> > The last time I looked, the thigh had to get back up to the top and needed to accelerate from a
> > speed of zero at the bottom in order to do
> > it. The energy to do must come from somewhere. One must account for the energy of the entire
> > circle, not just the downstroke.
>
> The thigh gets back to the top through work done by the leg muscles. Look at this as an investment
> which is returned on the downstroke with no losses "just for accelerating a mass"

Ignore the PE changes. The legs balance each other pretty much. There are no substantial PE changes
when both legs are considered together. Now deal with the kinetic energy changes. How do you deal
with them. They can't be moved into PE, because it is taken care of.

>
> >
> > >
> > > > In running, if you assume that all leg-deceleration is via eccentric muscle work, the method
> > > > stands a chance, although assuming that a given amount of negative work has an equal
> > > > metabolic cost is a fudge.
>
> > > > Then let us assume it is all by eccentric muscle work in cycling also. Foot movement in
> > > > running is very close to foot movement in cycling. After all you learned to pedal unattached
> > > > to the pedals so you had to control your leg speed and foot position to maintain contact
> > > > with the pedal. This would require eccentric muscle contraction, at least at constant
> > > > cadence. How does that sound?
>
> These are the stability losses which were already included in my estimate.

What estimate? If you mean that the power loss of pedaling varies with the cube of the cadence then
we are talking about the same thing. Something makes me think you are referring to something else.

Why do you refer to them as stability losses. It is a matter of making the foot follow a prescribed
path at a prescribed speed. The foot can follow this path using external energy from a pedal or from
energy generated internal to the leg. In the analysis of the energy losses it shouldn't matter where
the energy comes from to perform this necessary movement, either way it is a loss isn't it?
>
> > >
> > > To be honest, it sounds like you are making this up as you go along. There is no reason for
> > > this kind of loss to magically fall in with the definition of "internal work". The funny
> > > chainrings demonstrate that internal work can be set to zero, but the mechanisms for internal
> > > energy loss remain.
> >
> > I am not making anything up. it is possible for people to make their legs move in a circle at a
> > "constant" velocity without being attached to any pedals. Runners do a pretty good job without
> > any special training. To do this means the muscles must do eccentric work. Why on earth do you
> > believe that just because someone puts their legs on bicycle pedals that the intrinsic wiring of
> > our neuromuscular control system suddenly changes. Where is the evidence for that? In this
> > months Scientific American there is an interesting tidbit that shows that the swing phase (hip
> > flexors) in bipedal runners (birds) takes 25% of the blood flow to the legs. If only we could
> > teach them to pedal.
> > >
> > > Experiments show the method doesn't even work that well for running. Why do you insist on
> > > fudging it to fit?
> >
> > Huh, I am not fudging anything. I am only reporting what my analysis (flawed as it is) showed
> > and how it seems to fit the experimental evidence. It certainly is a better fit than the
> > model the "right thinkers" propose, a model that says pedaling a bicycle per se requires no
> > energy cost.
>
> What they are saying is the cost is zero for the explanation you give and not pedaling costs zero.
>
> > >
> > > After a very long thread, we haven't got very far with the theoretical justifications though,
> > > have we?
> >
> > No, but at least one of the respected engineers here now agrees with me that pedaling power
> > losses are present and they vary with the cube of the cadence.
>
> I wasn't under the impression that I or anyone else has stated otherwise.

Gee, I thought AC had stated that pedaling losses were "nada" with substantial support in that view.
Perhaps I misread those posts.

Frank

 

[Phil Holman]

"Frank Day" <[email protected]> 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]...
> > > "Phil Holman" <[email protected]> wrote in message
> > >
> > > No, in a constant speed system, it requires energy to maintain the constant speed.
> >
> > The constant speed is the result of there being a resistance to work against. The energy
> > required is used to do useful work. If the resistance gradually reduced to zero, the system
> > would gradually change to non-constant speed.
>
> Yes, the constant speed is the result of there being resistance to work against. That is why it is
> so illogical to put forth a model that requires non-constant speed to show that the system is
> conservative.

I certainly don't need it to satisfy myself that it is conservative.

> Further, it does not require any resistance to pedal at constant speed. When demonstrating my
> cranks on a spincycle I typically put zero resistance on the wheel (so the only resistance is in
> the bearings and belt drive) and people never have trouble pedaling at a constant speed until the
> speed gets to high then they just fall apart. So, why do you say that when the load gets to zero
> the speed will vary?

Probably because I'm still talking about a single leg. <snip> most of remaining (cross
purpose) dialogue

>To conserve energy people need to go faster over the top and bottom and slower on the downstroke
>and upstroke while I bet they will tend to go faster on the downstroke/upstroke, where all their
>power is and where they are used to pushing. How does that pattern conserve energy? Is there any
>data looking at pedal speed pattern at various cadences under zero load to see what really happens?
>I doubt it.

Zero load is of no interest. In actuality, the speed variation when pedaling on the road is
miniscule. From the sinusoidal force input and known mass of the rider, a double integration of the
acceleration yields advancement or retardation, relative to a constant speed pedal. This is an
extremely small time interval. I actually did some previous work on calculating the pedal time
lag/travel for a variety of speeds, rider weight, power output and cadence (see below). 1 u sec =
.000001 sec. The loss from such variations would be insignificant.

This is the calculation for one test case. The table below gives values over a broader spectrum.

Power 385 Watts Cadence 90 rpm Velocity 39.846 ft/sec Av. Force 7.1236 lbs Force Amp 6.5 lbs Weight
200 lbs Peak a 6.5/200 x 32.2 = 1.05 ft/sec/sec

a (acceleration) = 1.05 Sin 18.85t V (velocity) = -.0557 Cos 18.85t d (distance) = -.0030 Sin 18.85t

Test Weight(lbs) V(ft/sec) P(watts) C(RPM) dT(u sec)

R1 150 15 250 70 2501 R2 150 30 150 90 227 R3 150 40 350 100 241 R4 200 10 400 60 11134 R5 200 35
300 90 303 R6 200 45 500 80 387 R7 250 8 450 80 13627 R8 250 20 200 95 515 R9 250 55 600 120 128

Phil Holman

 

[Phil Holman]

"Frank Day" <[email protected]> wrote in message
news:[email protected]...
> "Phil Holman" <[email protected]> wrote in message news:
>
> > I wasn't under the impression that I or anyone else has stated otherwise.
>
> Gee, I thought AC had stated that pedaling losses were "nada" with substantial support in that
> view. Perhaps I misread those posts.

What AC said was the external cost of turning over the pedals was nada, meaning that all the losses
were internal due to viscoelastic considerations. I don't think the discussion on the magnitude of
the cost was ever disputed. Now I think I understand where you are coming from, but let me
paraphrase just to make sure. PE is zeroed out because over a pedal cycle, one leg is rising and one
is falling so there is no change. What *you* are saying however is, because KE is not constant then
this requires input energy which is essentially lost. Please confirm.

Phil Holman

 

[Frank Day]

"Andrew Bradley" <[email protected]> wrote in message news:<Fsa_b.10016$Y%[email protected]>...
> Frank Day <[email protected]> wrote in message
> news:[email protected]...
>
> > I, for one, am skepticaal that higher cadence recruits less fast twitch fibers, because I
> > believe the same muscles that are putting force on the pedals are accelerating the thigh and leg
> > up to speed which allows the providing of pressure. (This is all based upon thought experiment
> > and not analysis of the literature). At the same power this should actually increase muscle
> > fiber recruitment even though force on the pedal goes down for the same power at higher
> > cadences. I am sure someone has looked at emg muscle recruitment as it varies with cadence. I
> > would be interested in knowing what these results are.
>
> You should get into the literature more. Have a look at this:
> http://www.me.utexas.edu/~neptune/Papers/msse32(7).pdf I see a couple of issues with the results
> but they are quite plausible.
>
> The "spin" approach seems to assume that it is quite difficult to attain optimal contraction speed
> for the muscle fibres, so less force means less recruitment. That may or may not be the case but
> it has to be traded of against the increased power and thus fibre recruitment that is required at
> higher cadence due to the extra power required. If the study here is right this brings optimal
> cadence down under what the spin community would expect.

Here is the applicable portion of the abstract of the referred paper: "Purpose: Based on the resistance-
rpm relationship for cycling, which is not unlike the force-velocity relationship of muscle, it is
hypothesized that the cadence which requires the minimal muscle activation will be progressively
higher as power output increases. Methods: To test this hypothesis, subjects were instrumented with
surface electrodes placed over seven muscles that were considered to be important during cycling.
Measurements were made while subjects cycled at 100, 200, 300, and 400 W at each cadence: 50, 60,
80, 100, and 120 rpm. These power outputs represented effort which was up to 32% of peak power
output for these subjects. Results: When all seven muscles were averaged together, there was a
proportional increase in EMG amplitude each cadence as power increased. A second-order polynomial
equation fit the EMG:cadence results very well (r 2

 

[Andrew Bradley]

[email protected] (Frank Day) wrote in message news:<[email protected]>...
> "Andrew Bradley" <[email protected]> wrote in message news:
>
> > Despite the space shuttle diversion, Frank has (more or less) factored gravity and PE out of the
> > calculation by considering (arbitrarily) both thighs together as the system.
>
> It didn't seem to me an arbitrary decision to look at both legs because it appears "obvious" that
> one is going up when the other is going down so any losses should cancel out.

It is purely arbitrary whether you choose to analyse the mass movements in one leg, two legs or half
a leg. None repeat NONE are closed systems mechanically and therefore knowing the mechanical energy
variation of ANY of these systems tells us NOTHING about "heat losses".

People have been going into all kinds of specific mechanical detail to demonstrate to you what is a
fundamental principle that needs no such proof. The conservation of energy principle took a while to
fully establish, please don't try to revise it now.

Besides, is it not more "obvious" to consider each leg separately for a circular pedaller who will
not transfer any power from one leg to the other? You could imagine the bb cut in half and a
chainring on either side.

If you were designing these chainrings to eliminate the mechanical energy variation each ring would
speed it's foot only through BDC rather than both TDC and BDC as with the two leg ring. (The cranks
would not remain aligned at 180 degrees either.)

If that does not convince you that these calculations are arbitrary, nothing will.

Andrew Bradley

 

[Andrew Bradley]

[email protected] (Frank Day) wrote in message news:<[email protected]>...
> "Phil Holman" <[email protected]> wrote in message
> news:<[email protected]>... A couple of things. Constant
> acceleration and deceleration is very inefficient use of energy. Does your car get better gas
> mileage if you average 20 miles per hour for a 20 mile trip but have to stop every block for a
> stop sign (varying between 40 and 0 miles per hour) or if you can go 20 miles per hour constantly
> for the 20 miles. Because of the energy variation in the legs requires input to keep it going. It
> doesn't matter where it is coming from. It costs energy.

What specious rhubard Frank! Man this is a strange way to promote PC's. Wasting all this bandwidth
challenging the expertise of a PC-using engineer.

Think of a pendulum. Think of a mass oscillating on a spring. Think of the mechanically closed
system that is a rider on an ergometer with no resistance and a hefty flywheel to maintain near
uniform pedal speed. Neglecting resistances he need input no power if on a fixed wheel. If he has a
freewheel he will have to input some power, but this will cause the system to constantly accelerate
to higher energy levels. No mechanical energy can be lost unless you factor in specific loss
mechanisms. Believe it!

 

[Phil Holman]

"Phil Holman" <[email protected]> wrote in message news:<[email protected]>...
> "Frank Day" <[email protected]> wrote in message
> news:[email protected]...
> > "Phil Holman" <[email protected]> wrote in message news:
> >
> > > I wasn't under the impression that I or anyone else has stated otherwise.
> >
> > Gee, I thought AC had stated that pedaling losses were "nada" with substantial support in that
> > view. Perhaps I misread those posts.
>
> What AC said was the external cost of turning over the pedals was nada, meaning that all the
> losses were internal due to viscoelastic considerations. I don't think the discussion on the
> magnitude of the cost was ever disputed. Now I think I understand where you are coming from, but
> let me paraphrase just to make sure. PE is zeroed out because over a pedal cycle, one leg is
> rising and one is falling so there is no change. What *you* are saying however is, because KE is
> not constant then this requires input energy which is essentially lost. Please confirm.

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.

Phil Holman

 

[Frank Day]

"Phil Holman" <[email protected]> wrote in message news:<[email protected]>...
> "Frank Day" <[email protected]> wrote in message
> news:[email protected]...
> > "Phil Holman" <[email protected]> wrote in message news:
> >
> > > I wasn't under the impression that I or anyone else has stated otherwise.
> >
> > Gee, I thought AC had stated that pedaling losses were "nada" with substantial support in that
> > view. Perhaps I misread those posts.
>
> What AC said was the external cost of turning over the pedals was nada, meaning that all the
> losses were internal due to viscoelastic considerations. I don't think the discussion on the
> magnitude of the cost was ever disputed.

You could have fooled me.

> Now I think I understand where you are coming from, but let me paraphrase just to make sure. PE is
> zeroed out because over a pedal cycle, one leg is rising and one is falling so there is no change.
> What *you* are saying however is, because KE is not constant then this requires input energy which
> is essentially lost. Please confirm.

Yes.

Frank

 

[Frank Day]

"Phil Holman" <[email protected]> wrote in message news:
> >To conserve energy people need to go faster over the top and bottom and slower on the downstroke
> >and upstroke while I bet they will tend to go faster on the downstroke/upstroke, where all their
> >power is and where they are used to pushing. How does that pattern conserve energy? Is there any
> >data looking at pedal speed pattern at various cadences under zero load to see what really
> >happens? I doubt it.
>
> Zero load is of no interest. In actuality, the speed variation when pedaling on the road is
> miniscule. From the sinusoidal force input and known mass of the rider, a double integration of
> the acceleration yields advancement or retardation, relative to a constant speed pedal. This is an
> extremely small time interval. I actually did some previous work on calculating the pedal time
> lag/travel for a variety of speeds, rider weight, power output and cadence (see below). 1 u sec =
> .000001 sec. The loss from such variations would be insignificant.

zero load is only of interest for demonstration purposes. If a load is required to explain where the
"losses" go (such that they can be considered not losses) or to keep the cadence constant, then the
model breaks down when someone pedals at no load at a constant cadence. It seems to me a model
should be valid under all conditions.
>
> This is the calculation for one test case. The table below gives values over a broader spectrum.
>
> Power 385 Watts Cadence 90 rpm Velocity 39.846 ft/sec Av. Force 7.1236 lbs Force Amp 6.5 lbs
> Weight 200 lbs Peak a 6.5/200 x 32.2 = 1.05 ft/sec/sec
>
> a (acceleration) = 1.05 Sin 18.85t V (velocity) = -.0557 Cos 18.85t d (distance) = -.0030
> Sin 18.85t
>
> Test Weight(lbs) V(ft/sec) P(watts) C(RPM) dT(u sec)
>
> R1 150 15 250 70 2501 R2 150 30 150 90 227 R3 150 40 350 100 241 R4 200 10 400 60 11134 R5 200 35
> 300 90 303 R6 200 45 500 80 387 R7 250 8 450 80 13627 R8 250 20 200 95 515 R9 250 55 600 120 128

I agree that such instantaneous speed changes are insignificant which is why we can consider pedal
speed to be constant.

Will you not agree that even when we were learning to pedal as a child, when we were not attached to
the pedals, that part of the learning process was learning how to keep the foot in contact with the
pedal? Didn't this require coordinated eccentric muscle contraction to do this well. During this
time wren't we also ingraining this motion into our subconscious as the "normal" way to pedal. What
has anyone done to change it (other than get on PowerCranks)? Why on earth would someone presume to
say that such losses do not occur? Where is the evidence that they don't? If pedaling requires so
little external energy, why is it so hard to pedal at high cadences on PowerCranks?

 

[Frank Day]

[email protected] (Andrew Bradley) wrote in message news:<[email protected]>...
> [email protected] (Frank Day) wrote in message
> news:<[email protected]>...
> > "Andrew Bradley" <[email protected]> wrote in message news:
> >
> > > Despite the space shuttle diversion, Frank has (more or less) factored gravity and PE out of
> > > the calculation by considering (arbitrarily) both thighs together as the system.
> >
> > It didn't seem to me an arbitrary decision to look at both legs because it appears "obvious"
> > that one is going up when the other is going down so any losses should cancel out.
>
> It is purely arbitrary whether you choose to analyse the mass movements in one leg, two legs or
> half a leg. None repeat NONE are closed systems mechanically and therefore knowing the mechanical
> energy variation of ANY of these systems tells us NOTHING about "heat losses".

I'll accept the arbitrariness of the choices available to analyze. However, whatever one chooses
one can then learn whether and how much energy is lost from or must be put into that "system" to
keep it working. The system doesn't really care where the energy comes from to keep it working,
just that it gets it.

>
> People have been going into all kinds of specific mechanical detail to demonstrate to you what is
> a fundamental principle that needs no such proof. The conservation of energy principle took a
> while to fully establish, please don't try to revise it now.

I'll try not to. Not sure what I have violated here.
>
> Besides, is it not more "obvious" to consider each leg separately for a circular pedaller who will
> not transfer any power from one leg to the other?

That is fine although most people are not circular pedalers so there is some interplay and it is
possible on regular cranks. If one is using PowerCranks it is impossible to transfer energy from one
pedal to the other, yet, to watch them, they seem to be pedaling normally. So analyze one leg.

> You could imagine the bb cut in half and a chainring on either side.

Or PowerCranks
>
> If you were designing these chainrings to eliminate the mechanical energy variation each ring
> would speed it's foot only through BDC rather than both TDC and BDC as with the two leg ring. (The
> cranks would not remain aligned at 180 degrees either.)

At first I thought that would be impossible but you are only talking about foot speed, not chain
speed so I think it is possible. To do this would not require a broken axle, just dual chain
rings I think.
>
> If that does not convince you that these calculations are arbitrary, nothing will.

I accept they are arbitrary. I was just trying to analyze the real pedaling situation, not some
potentially possible way of doing so to show it is possible to conserve energy. The fact that one
has to modify the bicycle in such a way to conserve energy strongly suggests that normal pedaling
(without such modifications) is not conservative, wouldn't you say?

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]>... A couple of things. Constant
> > acceleration and deceleration is very inefficient use of energy. Does your car get better gas
> > mileage if you average 20 miles per hour for a 20 mile trip but have to stop every block for a
> > stop sign (varying between 40 and 0 miles per hour) or if you can go 20 miles per hour
> > constantly for the 20 miles. Because of the energy variation in the legs requires input to keep
> > it going. It doesn't matter where it is coming from. It costs energy.
>
> What specious rhubard Frank! Man this is a strange way to promote PC's. Wasting all this bandwidth
> challenging the expertise of a PC-using engineer.
>
> Think of a pendulum. Think of a mass oscillating on a spring. Think of the mechanically closed
> system that is a rider on an ergometer with no resistance and a hefty flywheel to maintain near
> uniform pedal speed. Neglecting resistances he need input no power if on a fixed wheel. If he has
> a freewheel he will have to input some power, but this will cause the system to constantly
> accelerate to higher energy levels. No mechanical energy can be lost unless you factor in specific
> loss mechanisms. Believe it!

The pendulum and mass oscillating on a string are cosntantly changing PE and KE to maintain the
system energy constant. The only losses are the viscoelastic losses in the string or spring and air
resistance losses.

While it is possible your example may hold for a fixed wheel. Something bothers me about it but I am
not sure what - I think it is the mass of the flywheel (or rider) will be such that the energy
cannot be fully transferred from the legs to the flywheel so total energy cannot be kept constant.
If total energy is not constant then there are additional energy losses and the rate the system will
slow down will be determined by this energy variation over and above the usual losses. Even if that
analogy does hold, the same is not true for the cyclist with a free wheel. It is not possible to put
energy back into the legs from the flywheel in this instance. Therefore, to keep pedaling the
pedaling energy variation must come from the legs. Either this extra energy the legs are putting is
has to make the bike faster, or it robs from the energy the athlete has to give to the pedals and
makes it slower. 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.

Frank

 

[Dvt]

Kraig Willett wrote:
> Redshoe Brand Shoes makes similar claims:
>
> http://www.biketechreview.com/redshoe/brochure.htm
>
> Redshoe Brand Shoes seem to be working for me, as I've seen ~25% power improvements in only
> 8 weeks:
>
> http://www.biketechreview.com/projectredshoe.htm
>
>
> YMMV,
>
> -kraig

LOL! Nice one, Kraig. Let me know if these sell. I might just buy some stock in the company.

Dave dvt at psu dot edu

 

[Dvt]

Frank Day wrote:
>>Can you come up with an experiment that will determine optimal cadence?

> I can come up with an experiment to determine the most efficient cadence. This has been done many
> times by others.

Most efficient, as has been noted before, may not be optimal for your target market (racers). Most
powerful might be better.

> A "chart" of power vs cadence is different than an experiment that looks at Power vs cadence,
> controlling for effort.

I thought that power was a pretty good indicator of effort. What did I miss?

> Well, during a ride, the typical rider is constantly changing effort. Only in a laboratory
> situation can this be adequately accounted for.

See above. Controlling for power would seem to be the most accurate way of controlling for effort.
Or is there some biological measurement we can make that is instantaneous and reflects effort
accurately?

Dave "learning a lot" dvt at psu dot edu

 

[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]>... A couple of things. Constant
> > acceleration and deceleration is very inefficient use of energy. Does your car get better gas
> > mileage if you average 20 miles per hour for a 20 mile trip but have to stop every block for a
> > stop sign (varying between 40 and 0 miles per hour) or if you can go 20 miles per hour
> > constantly for the 20 miles. Because of the energy variation in the legs requires input to keep
> > it going. It doesn't matter where it is coming from. It costs energy.
>
> What specious rhubard Frank! Man this is a strange way to promote PC's. Wasting all this bandwidth
> challenging the expertise of a PC-using engineer.

Didn't know I was promoting PC's in this thread. I am simply trying to discuss what is really going
on in the pedaling motion. Some of my knowledge in this regards comes from my experience with PC's
so it keeps coming up. I am not trying to discredit Phil as I know he likes the cranks but that
doesn't mean his explanation as to what is going on makes any sense to me. I would like to answer
the question as to why pedaling normal cadences with PowerCranks is so hard while pedaling at
"normal" cadences with regular cranks seems so easy. Is there something fundamentally different
about the energy required to pedal the two ways or what?

>
> Think of a pendulum. Think of a mass oscillating on a spring. Think of the mechanically closed
> system that is a rider on an ergometer with no resistance and a hefty flywheel to maintain near
> uniform pedal speed. Neglecting resistances he need input no power if on a fixed wheel. If he has
> a freewheel he will have to input some power, but this will cause the system to constantly
> accelerate to higher energy levels. No mechanical energy can be lost unless you factor in specific
> loss mechanisms. Believe it!

One more thing I forgot to mention in my last post. Taking your example above. Is there a
difference between transferring energy between the flywheel and the legs to "conserve" energy (as
you alledge) and the rider actively moving his legs in a manner to keep zero pressure on the pedals
the entire circle?

In both instances the flywheel should maintain speed and rotate forever but in one there is a
requirement for an external input of energy to rotate the feet while in the other, it is claimed
energy is conserved to no external energy is required. I think the two views are incompatible.

Frank

 

[Andrew Bradley]

Frank Day <[email protected]> wrote in message
news:[email protected]...
> [email protected] (Andrew Bradley) wrote in message
news:<[email protected]>...
> > Think of a pendulum. Think of a mass oscillating on a spring. Think of the mechanically closed
> > system that is a rider on an ergometer with no resistance and a hefty flywheel to maintain near
> > uniform pedal speed. Neglecting resistances he need input no power if on a fixed wheel. If he
> > has a freewheel he will have to input some power, but this will cause the system to constantly
> > accelerate to higher energy levels. No mechanical energy can be lost unless you factor in
> > specific loss mechanisms. Believe it!
>
> The pendulum and mass oscillating on a string are cosntantly changing PE and KE to maintain the
> system energy constant. The only losses are the viscoelastic losses in the string or spring and
> air resistance losses.

The spring can be horizontal if you prefer, to factor out PE as in your calcs.

>
> While it is possible your example may hold for a fixed wheel. Something bothers me about it but I
> am not sure what - I think it is the mass of the flywheel (or rider) will be such that the energy
> cannot be fully transferred from the legs to the flywheel so total energy cannot be kept constant.
> If total energy is not constant then there are additional energy losses and the rate the system
> will slow down will be determined by this energy variation over and above the usual losses.

What's the mechanism for this "additional" energy loss?

>Even if that analogy does hold, the same is not true for the cyclist with a free wheel. It is not
>possible to put energy back into the legs from the flywheel in this instance. Therefore, to keep
>pedaling the pedaling energy variation must come from the legs.

Yes...

> Either this extra energy the legs are putting is has to make the bike faster,

Yes, you've got it.

>or it robs from the energy the athlete has to give to the pedals and makes it slower.

?????

>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).

Andrew Bradley

 

[Andrew Bradley]

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

> > Besides, is it not more "obvious" to consider each leg separately for a circular pedaller who
> > will not transfer any power from one leg to the other?
>
> That is fine although most people are not circular pedalers so there is some interplay and it is
> possible on regular cranks. If one is using PowerCranks it is impossible to transfer energy from
> one pedal to the other, yet, to watch them, they seem to be pedaling normally. So analyze one leg.
>
> > You could imagine the bb cut in half and a chainring on either side.
>
> Or PowerCranks
> >
> > If you were designing these chainrings to eliminate the mechanical energy variation each ring
> > would speed it's foot only through BDC rather than both TDC and BDC as with the two leg ring.
> > (The cranks would not remain aligned at 180 degrees either.)
>
> At first I thought that would be impossible but you are only talking about foot speed, not chain
> speed so I think it is possible. To do this would not require a broken axle, just dual chain rings
> I think.

You'd need a broken axel. A Rotorcrank system rotated appropriately would go some way to achieving
the same result. That's 3 new pedalling systems waiting to happen so far this thread.

> The fact that one has to modify the bicycle in such a way to conserve energy strongly suggests
> that normal pedaling (without such modifications) is not conservative, wouldn't you say?

I'd say it would be great if the cost of pedalling could be eliminated or reduced, but I'm afraid
these modifications don't do it.

Andrew Bradley

 

[Andrew Bradley]

Frank Day <[email protected]> wrote in message
news:[email protected]...
> I would like to answer the question as to why pedaling normal cadences with PowerCranks is so hard
> while pedaling at "normal" cadences with regular cranks seems so easy. Is there something
> fundamentally different about the energy required to pedal the two ways or what?

My guess is that these cranks impose a minimum power requirement on the upstroke muscles that
increases with cadence. For a given power output the upstroke would perhaps be responsible for an
increasing proportion of the power output with increasing cadence.

> >
> > Think of a pendulum. Think of a mass oscillating on a spring. Think of the mechanically closed
> > system that is a rider on an ergometer with no resistance and a hefty flywheel to maintain near
> > uniform pedal speed. Neglecting resistances he need input no power if on a fixed wheel. If he
> > has a freewheel he will have to input some power, but this will cause the system to constantly
> > accelerate to higher energy levels. No mechanical energy can be lost unless you factor in
> > specific loss mechanisms. Believe it!
>
> One more thing I forgot to mention in my last post. Taking your example above. Is there a
> difference between transferring energy between the flywheel and the legs to "conserve" energy (as
> you alledge) and the rider actively moving his legs in a manner to keep zero pressure on the
> pedals the entire circle?

Sure, with zero pressure on the pedals the flywheel would not notice the legs. The legs can't do
this without a variable internal resistance (eg excentric contractions).

>
> In both instances the flywheel should maintain speed and rotate forever but in one there is a
> requirement for an external input of energy to rotate the feet while in the other, it is claimed
> energy is conserved to no external energy is required. I think the two views are incompatible.

Not incompatible because one requires an internal resistance and thus heat production, the other
deosn't . Either way, energy is conserved.

Andrew Bradley

 

[Frank Day]

[email protected] (Phil Holman) 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]...
> > > "Phil Holman" <[email protected]> wrote in message news:
> > >
> > > > I wasn't under the impression that I or anyone else has stated otherwise.
> > >
> > > Gee, I thought AC had stated that pedaling losses were "nada" with substantial support in that
> > > view. Perhaps I misread those posts.
> >
> > What AC said was the external cost of turning over the pedals was nada, meaning that all the
> > losses were internal due to viscoelastic considerations. I don't think the discussion on the
> > magnitude of the cost was ever disputed. Now I think I understand where you are coming from, but
> > let me paraphrase just to make sure. PE is zeroed out because over a pedal cycle, one leg is
> > rising and one is falling so there is no change. What *you* are saying however is, because KE is
> > not constant then this requires input energy which is essentially lost. Please confirm.
>
>
> 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.

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

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.

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

Frank

 

[Frank Day]

dvt <[email protected]> wrote in message news:<[email protected]>...
> Frank Day wrote:
> >>Can you come up with an experiment that will determine optimal cadence?
>
> > I can come up with an experiment to determine the most efficient cadence. This has been done
> > many times by others.
>
> Most efficient, as has been noted before, may not be optimal for your target market (racers). Most
> powerful might be better.

Most powerful is anaerobic efforts that can only last a short time. Most racers are looking for "how
much power can I sustain for x hours" rather than sprinting power (unless they are a sprinter).
PowerCranks claims increased power across the board and claims a lot of that comes from improved
efficiency. Whether the rider chooses to race at the experimentally "proven" cadence is their
choice. Most choose what they think will make them the fastest whether it does or not.
>
> > A "chart" of power vs cadence is different than an experiment that looks at Power vs cadence,
> > controlling for effort.
>
> I thought that power was a pretty good indicator of effort. What did I miss?

200 watts in a 38/17 at a cadence of 120 is a lot more effort than 200 watts in a 53/12 and a
cadence of 60.
>
> > Well, during a ride, the typical rider is constantly changing effort. Only in a laboratory
> > situation can this be adequately accounted for.
>
> See above. Controlling for power would seem to be the most accurate way of controlling for effort.
> Or is there some biological measurement we can make that is instantaneous and reflects effort
> accurately?

One way to assess effort is to ask the subject. This is usually done in the Conconi protocol and it
is pretty reliable. HR (especially if max HR is known) is the most easily accessable effort "number"
and most experimental protocols will allow enough time after changing power to let the HR stabilize
so it is a better marker. O2 uptake (as a percentage of VO2 max) is also but this requires special
equipment outside the reach of most people.

Frank

 

[Frank Day]

"Andrew Bradley" <[email protected]> wrote in message news:<lgr_b.11136$Y%[email protected]>...
> Frank Day <[email protected]> wrote in message
> news:[email protected]...
> > [email protected] (Andrew Bradley) wrote in message
> news:<[email protected]>...
> > > Think of a pendulum. Think of a mass oscillating on a spring. Think of the mechanically closed
> > > system that is a rider on an ergometer with no resistance and a hefty flywheel to maintain
> > > near uniform pedal speed. Neglecting resistances he need input no power if on a fixed wheel.
> > > If he has a freewheel he will have to input some power, but this will cause the system to
> > > constantly accelerate to higher energy levels. No mechanical energy can be lost unless you
> > > factor in specific loss mechanisms. Believe it!
> >
> > The pendulum and mass oscillating on a string are cosntantly changing PE and KE to maintain the
> > system energy constant. The only losses are the viscoelastic losses in the string or spring and
> > air resistance losses.
>
> The spring can be horizontal if you prefer, to factor out PE as in your calcs.

That only changes the PE due to gravity but not due to spring compression. It is the same problem.
>
> >
> > While it is possible your example may hold for a fixed wheel. Something bothers me about it but
> > I am not sure what - I think it is the mass of the flywheel (or rider) will be such that the
> > energy cannot be fully transferred from the legs to the flywheel so total energy cannot be kept
> > constant. If total energy is not constant then there are additional energy losses and the rate
> > the system will slow down will be determined by this energy variation over and above the usual
> > losses.
>
> What's the mechanism for this "additional" energy loss?

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. 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.

>
> >Even if that analogy does hold, the same is not true for the cyclist with a free wheel. It is not
> >possible to put energy back into the legs from the flywheel in this instance. Therefore, to keep
> >pedaling the pedaling energy variation must come from the legs.
>
> Yes...
>
> > Either this extra energy the legs are putting is has to make the bike faster,
>
> Yes, you've got it.
>
> >or it robs from the energy the athlete has to give to the pedals and makes it slower.
>
> ?????

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?
>
> >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).

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

Frank