The great rotating-weight debate

[Wayne Pein]

http://www.velonews.com/tech/report/articles/9662.0.html


Wayne

[mtb Dad]

It would be interesting to know how many Hummers total you tow up to 10
mph on a three hour mountain ride, to know if one is that big a deal.
I know it feels like I've been towing about hundreds of hummers when I
climb a lot...

[Tom_A]

Oh my...it looks like somebody needs to brush up on their kinetic
energy calculations. Looks to me to be off by a factor of almost 20X
on the relative contribution of rotational KE change to total KE change.

[Michael Press]

In article
. com>,
"Tom_A" > wrote:

> Oh my...it looks like somebody needs to brush up on their kinetic
> energy calculations. Looks to me to be off by a factor of almost 20X
> on the relative contribution of rotational KE change to total KE change.

Let us only consider the mass of the rim, tube, and tire.
Denote by v
v the speed of the bicycle,
m the mass of the wheel
w the angular speed of the wheel
r the radius of the wheel
I the moment of inertia

v = w * r
I = m * r^2

The the kinetic energy of translation is 1/2 * mv^2.
The the kinetic energy of rotation is 1/2 * mIw^2 = 1/2 * mv^2.
So the rotational energy of the wheel equals
the translational energy.

--
Michael Press

[Tom_A]

>The the kinetic energy of translation is 1/2 * mv^2.
>The the kinetic energy of rotation is 1/2 * mIw^2 = 1/2 * mv^2.
>So the rotational energy of the wheel equals
>the translational energy.

Ummm...only if ALL the mass is at the outer perimeter of the wheel.
Real wheels aren't like that.

Oh yeah...what about the remaining mass (i.e. the rider and bike) that
has KE? You're only considering the wheels.

Try again.

[Michael Press]

In article
. com>,
"Tom_A" > wrote:

> >The the kinetic energy of translation is 1/2 * mv^2.
> >The the kinetic energy of rotation is 1/2 * mIw^2 = 1/2 * mv^2.
> >So the rotational energy of the wheel equals
> >the translational energy.
>
> Ummm...only if ALL the mass is at the outer perimeter of the wheel.
> Real wheels aren't like that.
>
> Oh yeah...what about the remaining mass (i.e. the rider and bike) that
> has KE? You're only considering the wheels.
>
> Try again.

Try what? I quantify the kinetic energy in a rotating
wheel; that is all. The residuum for hub, freehub, and
cogwheels are a much smaller fraction of their
translational kinetic energy.

--
Michael Press

[tiborg]

Ron Ruff wrote:
> Great! Leave it to Zinn to further confuse the issue...
>
> "As an example, you may be moving up the hill at an average rate of 10
> MPH but in reality you could be varying from 9-11 MPH through each
> revolution.
>
> So what?" you might ask. Well Peter, the bottom line is that you are
> accelerating your wheels twice per revolution and that can add up to
> some serious effort."
>
> It should be obvious to anyone with a tiny bit of tech savy that the
> variation in speed is *less* with the *heavier* wheels! You only have
> to accelerate at each stroke because you just got through decelerating.
> Since the variation in speed is actually slightly detrimental, the
> intertia of the heavier wheels is a *bonus*... not a problem.
>
> Of course you still have to carry the extra weight up the hill, but
> there is no added inertial issue.

So we should really be more concerned about how much rolling resistance
our tires have and our overall wind profile rather than weight.

[G.T.]

"Ron Ruff" > wrote in message
oups.com...
> Great! Leave it to Zinn to further confuse the issue...
>
> "As an example, you may be moving up the hill at an average rate of 10
> MPH but in reality you could be varying from 9-11 MPH through each
> revolution.
>
> So what?" you might ask. Well Peter, the bottom line is that you are
> accelerating your wheels twice per revolution and that can add up to
> some serious effort."
>
> It should be obvious to anyone with a tiny bit of tech savy that the
> variation in speed is *less* with the *heavier* wheels! You only have
> to accelerate at each stroke because you just got through decelerating.
> Since the variation in speed is actually slightly detrimental, the
> intertia of the heavier wheels is a *bonus*... not a problem.
>
> Of course you still have to carry the extra weight up the hill, but
> there is no added inertial issue.
>

Also, "you could be varying from 9-11 MPH ". Isn't it more like 9.9 to 10.1
for a typical pedal stroke?

Greg

[Leo Lichtman]

"Ron Ruff" wrote: (clip) It should be obvious to anyone with a tiny bit of
tech savy that the variation in speed is *less* with the *heavier* wheels!
(clip)
^^^^^^^^^^^^^^^^
I think I have a tiny bit of tech savy, and it is obvious to me that the
wheels cannot have an appreciable effect. Any reduction in speed occurs
during the part of the pedaling stroke when torque goes through a minimum.
The entire bike/rider slows down a little. If a bike and rider total, say
200 lb, how much of this could possibly be in the wheels? Then, remember,
we're not ELIMINATING wheel weight, we're reducing it a little, so the
difference could amount to, maybe 0.1 to 0.2% of the total. Finally, the
effort to reaccelerate the system to its maximum speed is regained on the
next half-cycle when it decelerates again.

[[email protected]]

Leo Lichtman wrote:
> Then, remember, we're not ELIMINATING wheel weight, we're reducing it a little,
> so the difference could amount to, maybe 0.1 to 0.2% of the total.

The second article seemed to miss that point. I read through it
quickly but his analysis of pulling those Humvees used the entire
weight of the wheel, not the differences in weight between a light
wheel and a heavy wheel.

Tom

[jb]

It's easily proven with a thought experiment.

Add x pounds to each pedal.

[Ron Ruff]

wrote:
> The second article seemed to miss that point. I read through it
> quickly but his analysis of pulling those Humvees used the entire
> weight of the wheel, not the differences in weight between a light
> wheel and a heavy wheel.

The point isn't that he over estimated or otherwise erred in his
calculation... he was conceptually wrong... completely. What he failed
to notice was that the oscillation in speed moves down as well as up,
and the only effect of increased rotational inertia is to *decrease*
the oscillations... it doesn't effect the average speed. Well actually
it does *very* slightly... but in the opposite direction...ie the heavy
wheels are faster.

[David Damerell]

Quoting Michael Press >:
>So the rotational energy of the wheel equals
>the translational energy.

Quite correct... if all the mass is at the very outside. Which it isn't.
Even the rim and tyre are nontrivially inboard of that point.
--
David Damerell > flcl?
Today is First Oneiros, April.

[David Damerell]

Quoting jb >:
>It's easily proven with a thought experiment.
>Add x pounds to each pedal.

Do it as a real experiment. Some years ago I went from cycling (normally)
in trainer shoes to large steel toecapped boots. Suddenly become
mysteriously slower? Nope.
--
David Damerell > flcl?
Today is First Oneiros, April.

[[email protected]]

Ron Ruff writes:

>> The second article seemed to miss that point. I read through it
>> quickly but his analysis of pulling those Humvees used the entire
>> weight of the wheel, not the differences in weight between a light
>> wheel and a heavy wheel.

> The point isn't that he over estimated or otherwise erred in his
> calculation... he was conceptually wrong... completely. What he
> failed to notice was that the oscillation in speed moves down as
> well as up, and the only effect of increased rotational inertia is
> to *decrease* the oscillations... it doesn't effect the average
> speed. Well actually it does *very* slightly... but in the
> opposite direction...ie the heavy wheels are faster.

The problem is that acceleration on bicycles is so slight that all
this is irrelevant. If someone graphed speed variations while
pedaling on an absolute scale while riding at steady state on the flat
or for that matter during acceleration from stopped to that speed, the
whole subject would blow away.

Jobst Brandt

[Tom_A]

>The problem is that acceleration on bicycles is so slight that all
>this is irrelevant. If someone graphed speed variations while
>pedaling on an absolute scale while riding at steady state on the flat
>or for that matter during acceleration from stopped to that speed, the
>whole subject would blow away.

>Jobst Brandt

That has already been done...and then some (make sure you look at the
linked appendix D which has an instantaneous acceleration plot for a
crit):

http://www.biketechreview.com/archive/wheel_theory.htm

Yet...oddly...the subject just won't "blow away".

[[email protected]]

Tom Anhalts writes:

>> The problem is that acceleration on bicycles is so slight that all
>> this is irrelevant. If someone graphed speed variations while
>> pedaling on an absolute scale while riding at steady state on the
>> flat or for that matter during acceleration from stopped to that
>> speed, the whole subject would blow away.

> That has already been done...and then some (make sure you look at
> the linked appendix D which has an instantaneous acceleration plot
> for a crit):

http://www.biketechreview.com/archive/wheel_theory.htm

> Yet...oddly...the subject just won't "blow away".

I see no velocity curves with resolution that show speed variations
for several pedal strokes, or for that matter any acceleration data.

Jobst Brandt

[Tom_A]

The links for both Appendix B and D have a plot of instantaneous
acceleration shown in their respective second graphs. It's shown in
the same chart as HR data. Peak accels rarely are larger than .05g
even in the crit. He did peak it out at .1gs, but that was only at the
start.

Admittedly, nothing is there that shows the data over a shorter time
span, but if you email Kraig, I'm sure he still has the power meter
files. I believe he was using an SRM and so the data should be saved
in ~1 sec intervals.

[[email protected]]

Tom Anhalts writes:

> The links for both Appendix B and D have a plot of instantaneous
> acceleration shown in their respective second graphs. It's shown in
> the same chart as HR data. Peak accels rarely are larger than .05g
> even in the crit. He did peak it out at .1gs, but that was only at
> the start.

That (0.05g) is my point, and applying that to the mass of rider and
bicycle gives infinitesimal speed variations.

> Admittedly, nothing is there that shows the data over a shorter time
> span, but if you email Kraig, I'm sure he still has the power meter
> files. I believe he was using an SRM and so the data should be
> saved in ~1 sec intervals.

Considering that one pedals at a cadence of >60, a one second interval
shows no fluctuation trend. I don't see an acceleration or velocity
plot that would reveal what occurs within a couple of crank rotations.
This would have to be done with a high resolution speedometer and
accelerometer, neither of which were involved in this test as far as
I can see.

Jobst Brandt

[Michael Press]

In article >,
David Damerell > wrote:

> Quoting Michael Press >:
> >So the rotational energy of the wheel equals
> >the translational energy.
>
> Quite correct... if all the mass is at the very outside. Which it isn't.
> Even the rim and tyre are nontrivially inboard of that point.

Yes, an upper bound on the rotational energy to
translational energy is one.

--
Michael Press

[Tom_A]

>That (0.05g) is my point, and applying that to the mass of rider and
>bicycle gives infinitesimal speed variations.

Exactly...and those peaks were sprints out of the corners of a crit.
It doesn't take much reasoning ability to realize that the
accelerations around a pedal cycle will be orders of magnitude smaller
than even these miniscule amounts, especially at steady-state.

I pointed this out since your original post not only referred to
steady-state pedal cycle variations, but also acceleration from a stop.
The crit file shows this (the start) and also all the corner
accelerations.

>Considering that one pedals at a cadence of >60, a one second interval
>shows no fluctuation trend. I don't see an acceleration or velocity
>plot that would reveal what occurs within a couple of crank rotations.
>This would have to be done with a high resolution speedometer and
>accelerometer, neither of which were involved in this test as far as
>I can see.

You are correct. However, rather than performing a test, it would be
easy to plug in a pedal force input to the equation of motion of a
cyclist and look at the resulting predicted speed variation. Sounds
like a nice "rainy day" project for me.

Actually, I think Ron Ruff may have done this already? Ron?

[Ron Ruff]

Tom_A wrote:
> Actually, I think Ron Ruff may have done this already? Ron?

Yes... 10% grade, avg power 250W (varies from 0 to 500W), 60rpm, 80kg.
Avg speed = 3.093 m/s (6.919 mph)
Variation = .149 m/s (.333 mph, +/- .167 mph)

So the total speed fluctuation is about 5% with these conditions. *Way*
less on the flat of course.

[David Damerell]

Quoting Michael Press >:
> David Damerell > wrote:
>>Quoting Michael Press >:
>>>So the rotational energy of the wheel equals
>>>the translational energy.
>>Quite correct... if all the mass is at the very outside. Which it isn't.
>>Even the rim and tyre are nontrivially inboard of that point.
>Yes, an upper bound on the rotational energy to
>translational energy is one.

Nowhere in the post I replied to do you mention that you are stating an
upper bound. You got it wrong; deal.
--
David Damerell > Distortion Field!
Today is First Aponoia, April.

[David Damerell]

Quoting >:
>The problem is that acceleration on bicycles is so slight that all
>this is irrelevant. If someone graphed speed variations while
>pedaling on an absolute scale while riding at steady state on the flat
>or for that matter during acceleration from stopped to that speed, the
>whole subject would blow away.

For one thing, we can see these effects are not significant from the fact
that tandems with cranks mounted out of phase are no faster than tandems
with conventionally mounted cranks.
--
David Damerell > Distortion Field!
Today is First Aponoia, April.

[Michael Press]

In article >,
David Damerell > wrote:

> Quoting Michael Press >:
> > David Damerell > wrote:
> >>Quoting Michael Press >:
> >>>So the rotational energy of the wheel equals
> >>>the translational energy.
> >>Quite correct... if all the mass is at the very outside. Which it isn't.
> >>Even the rim and tyre are nontrivially inboard of that point.
> >Yes, an upper bound on the rotational energy to
> >translational energy is one.
>
> Nowhere in the post I replied to do you mention that you are stating an
> upper bound. You got it wrong; deal.

And?

Will you provide an accurate quantitative analysis?

--
Michael Press