How accurate are power meters?

On Sunday, December 22, 2013 9:06:23 PM UTC-5, Dan wrote:
Frank Krygowski writes:

What you are thinking applies only to isolated systems, systems in which no energy crosses the control boundary.

If no energy is otherwise gained or lost in a system (an isolated system, by definition) then adding work will increase potential energy - or some other energy - within that system. But with a motorized (or braked) treadmill, energy flow occurs in the treadmill mechanism. So the rider puts in work (or energy) but it leaves through the mechanism, and his potential energy is unchanged.

I'd think it would be reduced. It has to go somewhere.

No. Potential energy doesn't have to go anywhere. It can be unchanged.

As someone (Joe?) said, it can also be explained via a free body diagram. The force at the rear tire's contact point, parallel to the surface, is needed to counteract the surface-oriented component of the weight. That surface-oriented weight component increases as tilt increases,

Doesn't the surface-oriented weight component decrease as the
slope increases and more of the total weight is being pulled
back down the hill by gravity? (As the incline reaches vertical,
doesn't the weight on the surface approach zero?)

Joe's just explained this well.

(Don't get mad at me - maybe I'm just not understanding your
bike tech talk.)

Oh... is the weight component *parallel* to the surface? (IOW,
not bearing on the surface at all but relative to it.) If so, okay
- it just seemed counter intuitive to "weight" and "surface". So
what happens, then, is that the weight doesn't change, but the
amount of it that can be exerted against is reduced. Right? Hmm...
still sounds to me like a *non*-surface oriented weight component.

Again, Joe's explained this. It may help to draw a sketch, draw the vertical Weight force, show it resolved into orthogonal components, and label all angles carefully. Pick a tilt angle of ten degrees or so, so you won't mix up the complementary angles in your sketch.

... so more tire force is required from the rider. That force, times his velocity (relative to that moving surface) is his bike's power output. It, too, must increase with tilt.

There are some quirks that make this a bit tricky to understand, for instance the link between metabolic power and rider "foot power"; and the action-reaction thing at the tire contact patch. I can explain those if needed.

Yes, please explain the quirks.

Briefly: The "metabolic power" exerted by the cyclist isn't necessarily the same as the mechanical power he puts into the pedals; there can be great inefficiencies. As an extreme example, if a person were on a stationary bike held upright with its wheels & cranks prevented from moving, and if that person pushed on the forward pedal as hard as he possibly could for as long as he possibly could, he'd burn a fair amount of metabolic energy. But his mechanical work output would be practically zero. A human can get very tired while doing no "work" in the sense of physics. But that's parenthetical to the problem at hand. The power meter at least purports to measure mechanical power, not metabolic power.

And the situation at the rear tire's contact patch confuses many who first try to analyze it. The bike tire pushes backwards on the road, but that force is pointed backwards; so that's not (directly) what causes the bike to move forward, nor what fights against gravity on a hill. Instead, it's the reaction force of the ground _on the bike_ that has a forward direction. The reaction force is the one that does the cyclist some good. Of course, by Newton #1 the reaction force equals the backwards force; but drawing a proper force diagram is difficult until one gets that concept.

Don't let me dissuade you guys from the analytics, but when I
approach a hill on my bike, I don't process the data too mathematic-
ally. ...

We're not riding at the moment. We're discussing engineering principles. Believe it or not, someone has to understand this stuff. It's part of what engineers do.

- Frank Krygowski

Joe Riel writes:

Dan writes:

Frank Krygowski writes:

On Saturday, December 21, 2013 11:13:59 PM UTC-5, Phil W Lee wrote:
James considered Fri, 20 Dec 2013 10:19:39

The bicycle rider has to produce the same amount of force, and hence the
same power, to hold the their position on the treadmill moving
underneath, as riding at the same (treadmill) speed up a hill of the
same gradient, ignoring wind resistance and road roughness.

No they don't - they are adding nothing whatever to the potential
energy of increased altitude.

What you are thinking applies only to isolated systems, systems in which no energy crosses the control boundary.

If no energy is otherwise gained or lost in a system (an isolated system, by definition) then adding work will increase potential energy - or some other energy - within that system. But with a motorized (or braked) treadmill, energy flow occurs in the treadmill mechanism. So the rider puts in work (or energy) but it leaves through the mechanism, and his potential energy is unchanged.


I'd think it would be reduced. It has to go somewhere.

As someone (Joe?) said, it can also be explained via a free body diagram. The force at the rear tire's contact point, parallel to the surface, is needed to counteract the surface-oriented component of the weight. That surface-oriented weight component increases as tilt increases,


Doesn't the surface-oriented weight component decrease as the
slope increases and more of the total weight is being pulled
back down the hill by gravity? (As the incline reaches vertical,
doesn't the weight on the surface approach zero?)

(Don't get mad at me - maybe I'm just not understanding your
bike tech talk.)

Oh... is the weight component *parallel* to the surface? (IOW,
not bearing on the surface at all but relative to it.) If so, okay
- it just seemed counter intuitive to "weight" and "surface". So
what happens, then, is that the weight doesn't change, but the
amount of it that can be exerted against is reduced. Right? Hmm...
still sounds to me like a *non*-surface oriented weight component.

The gravitional force is directed downward.

Empirical evidence seems to bear this out.

This force can be
decomposed into two orthogonal vectors, one perpendicular to the road,
the other parallel to it.

Okay, now we're breaking it down into constructs that
nature and physics are oblivious to. But that's okay;
I see what you're trying to get at.

It is the force parallel to the road that you
have to work against.

Not counting bunny hops.

As the road gets steeper that vector gets larger
(W*sin(theta)). The perpendicular force (W*cos(theta)) gets smaller.

As the incline approaches vertical, the weight on the surface
approaches zero, right?

Its decrease is insignificant;

Until you hit loose gravel or wet leaves.

... besides, its only affect is to reduce the
rolling resistance, which is already small.


I find it easier to ride a wheelie on a slight uphill grade -
for at least a few reasons, I think. Has to do somewhat with
my usual technique (no braking involved).

I sometimes get a dreamlike vertigo feeling of falling
completely away from the surface of a hill - not while
riding, though, it just comes over me sometimes as I think
about climbing an extremely steep hill. I have the impression
that I know the feeling from experience, though it could just
be something weird in my mind / body - not altogether sure.

Dan writes:

Joe Riel writes:

Dan writes:

Frank Krygowski writes:

On Saturday, December 21, 2013 11:13:59 PM UTC-5, Phil W Lee wrote:
James considered Fri, 20 Dec 2013 10:19:39

The bicycle rider has to produce the same amount of force, and hence the
same power, to hold the their position on the treadmill moving
underneath, as riding at the same (treadmill) speed up a hill of the
same gradient, ignoring wind resistance and road roughness.

No they don't - they are adding nothing whatever to the potential
energy of increased altitude.

What you are thinking applies only to isolated systems, systems in which no energy crosses the control boundary.

If no energy is otherwise gained or lost in a system (an isolated system, by definition) then adding work will increase potential energy - or some other energy - within that system. But with a motorized (or braked) treadmill, energy flow occurs in the treadmill mechanism. So the rider puts in work (or energy) but it leaves through the mechanism, and his potential energy is unchanged.


I'd think it would be reduced. It has to go somewhere.

As someone (Joe?) said, it can also be explained via a free body diagram. The force at the rear tire's contact point, parallel to the surface, is needed to counteract the surface-oriented component of the weight. That surface-oriented weight component increases as tilt increases,


Doesn't the surface-oriented weight component decrease as the
slope increases and more of the total weight is being pulled
back down the hill by gravity? (As the incline reaches vertical,
doesn't the weight on the surface approach zero?)

(Don't get mad at me - maybe I'm just not understanding your
bike tech talk.)

Oh... is the weight component *parallel* to the surface? (IOW,
not bearing on the surface at all but relative to it.) If so, okay
- it just seemed counter intuitive to "weight" and "surface". So
what happens, then, is that the weight doesn't change, but the
amount of it that can be exerted against is reduced. Right? Hmm...
still sounds to me like a *non*-surface oriented weight component.

The gravitional force is directed downward.

Empirical evidence seems to bear this out.

This force can be
decomposed into two orthogonal vectors, one perpendicular to the road,
the other parallel to it.

Okay, now we're breaking it down into constructs that
nature and physics are oblivious to. But that's okay;
I see what you're trying to get at.

It is the force parallel to the road that you
have to work against.

Not counting bunny hops.

As the road gets steeper that vector gets larger
(W*sin(theta)). The perpendicular force (W*cos(theta)) gets smaller.

As the incline approaches vertical, the weight on the surface
approaches zero, right?

Yes. cos(Pi/2)=0.

Its decrease is insignificant;

Until you hit loose gravel or wet leaves.

No, it really is insignifcant. The factor there is the decrease
in coefficient of friction. Consider that for a 20% grade (damn
steep), 1-cos(arctan(1/5)) ~ 2%. That is, the normal force only
deceased by 2%.

... besides, its only affect is to reduce the
rolling resistance, which is already small.


I find it easier to ride a wheelie on a slight uphill grade -
for at least a few reasons, I think. Has to do somewhat with
my usual technique (no braking involved).

I sometimes get a dreamlike vertigo feeling of falling
completely away from the surface of a hill - not while
riding, though, it just comes over me sometimes as I think
about climbing an extremely steep hill. I have the impression
that I know the feeling from experience, though it could just
be something weird in my mind / body - not altogether sure.

--
Joe Riel

http://news.msn.com/year-in-review-2...-year#image=22

Phil W Lee writes:

Frank Krygowski considered Sun, 22 Dec 2013
10:14:24 -0800 (PST) the perfect time to write:

On Saturday, December 21, 2013 11:13:59 PM UTC-5, Phil W Lee wrote:
James considered Fri, 20 Dec 2013 10:19:39

The bicycle rider has to produce the same amount of force, and hence the
same power, to hold the their position on the treadmill moving
underneath, as riding at the same (treadmill) speed up a hill of the
same gradient, ignoring wind resistance and road roughness.

No they don't - they are adding nothing whatever to the potential
energy of increased altitude.

What you are thinking applies only to isolated systems, systems in which no energy crosses the control boundary.

If no energy is otherwise gained or lost in a system (an isolated system, by definition) then adding work will increase potential energy - or some other energy - within that system. But with a motorized (or braked) treadmill, energy flow occurs in the treadmill mechanism. So the rider puts in work (or energy) but it leaves through the mechanism, and his potential energy is unchanged.

So it's got nothing to do with the point on the circumference of the
tyre which is contacting the drag (which is the only thing that
changes when you incline the slope).

As an example, my turbo trainer contacts the tyre at a point about 20
degrees rearwards of where the tyre would normally contact the road.
This does not mean I'm going downhill all the time on my turbo
trainer.
The load comes purely from the mechanics of the trainer, both on my
turbo trainer and on the inclined treadmill.
Now, it may be that the load on the inclined treadmill is linked in
some way to the degree of inclination, but nothing in the test shows
that to be the case.

The only interface the bike makes with the treadmill is through the
contact points on the wheels. This is different from a turbo trainer,
where the axle is rigidly connected to the trainer (the connection
prevents the bike from jumping of the trainer). So where is the
resistance coming from? Do you maintain that the rider puts out
essentially no power to remain in place on a tilted, running treadmill?
If so, try visiting any gym and putting that hypothesis to the test.

--
Joe Riel

Joe Riel writes:

Dan writes:

Joe Riel writes:


snip


As the incline approaches vertical, the weight on the surface
approaches zero, right?

Yes. cos(Pi/2)=0.

Its decrease is insignificant;

Until you hit loose gravel or wet leaves.

No, it really is insignifcant. The factor there is the decrease
in coefficient of friction. Consider that for a 20% grade (damn
steep), 1-cos(arctan(1/5)) ~ 2%. That is, the normal force only
deceased by 2%.


I was being a bit of a smartass (bunny hops, indeed :-)

But we all know the effect is significant - if not due to
significantly reduced total weight in that vector, then due
to proportion of weight shifted from rear to front wheel (?)
to compensate for the vector rotation relative to the
gravitational force (?)

snip

Phil W Lee writes:

Joe Riel considered Sun, 22 Dec 2013 13:13:04 -0800
the perfect time to write:

Phil W Lee writes:

James considered Fri, 20 Dec 2013 10:19:39
+1100 the perfect time to write:

On 20/12/13 08:25, Phil W Lee wrote:
James considered Thu, 19 Dec 2013 08:37:20
+1100 the perfect time to write:

On 19/12/13 05:16, Phil W Lee wrote:
James considered Wed, 18 Dec 2013 14:16:15
+1100 the perfect time to write:

In the extreme case on the road where the maximum speed causes a
significantly higher wind resistance than the wind resistance at the
average speed, yes.

No, I'm referring to the power needed for acceleration, which would be
affected by a weight increase, where wind resistance wouldn't.

It's also stored and returned. Ke = 1/2 * m * v^2 No net change if you
start and finish at the same speed.

But this test was done indoors on a treadmill for bicycles, inclined at
8% gradient. There was no significant change in speed and the rider
remained at pretty much the same location on the treadmill the whole time.

Then it's a bull**** test anyway, as it disregards most of the
relevant factors.

Well, they were _trying_ to be all scientific, but the test setup seems
to have failed them, considerably.

What's worse is they didn't realise their error by simple analysis of
the results to see that they matched the theory.

OK, I've now watched the video.
I can see one glaring and fundamental flaw.
The only "climbing" involved is that the angle of the bike changes -
it never increases it's altitude, so there's no actual climbing at
all.
If there's no change in altitude, there is no power necessary to
"climb", no matter what mass is added to or removed from the
bike/rider.

The bicycle rider has to produce the same amount of force, and hence the
same power, to hold the their position on the treadmill moving
underneath, as riding at the same (treadmill) speed up a hill of the
same gradient, ignoring wind resistance and road roughness.

No they don't - they are adding nothing whatever to the potential
energy of increased altitude.

Here's a slightly different way to think about this. Suppose you are in
an elevator that descends with a constant velocity. The elevator is an
inertial frame of reference; the laws of physics are expressed the same
way there as in a stationary elevator.

Assume there is a ladder on the inside of the elevator. Climbing the
ladder will feel *exactly* the same as climbing the ladder when the
elevator is stopped. The two conditions are not distinguishable
inside the elevator. The work you have to exert to climb a number of
rungs is identical in both situations.

Now, assume you are climbing at a rate that matches the speed of the
elevator, so that relative to the shaft you are stationary. That
situation is analogous to the treadmill.

That would be relevant if the cyclist only started pedaling after the
treadmill had reached full speed.

That makes no sense. Anyhow, he could do that if he wished. That is,
with the harness there, just coast until the treadmill is at full speed
(the cyclist will be pushed to the back of the treadmill). Then start
pedaling fast enough to move forward so the harness is longer exerting
any force.

You'd need a very long treadmill for that to work.

A treadmill is an endless loop. Have you never tried a stairmaster?


--
Joe Riel

On Tuesday, December 24, 2013 3:04:56 AM UTC-5, Dan wrote:
Joe Riel writes:



Dan writes:



Joe Riel writes:





snip





As the incline approaches vertical, the weight on the surface

approaches zero, right?



Yes. cos(Pi/2)=0.



Its decrease is insignificant;



Until you hit loose gravel or wet leaves.



No, it really is insignifcant. The factor there is the decrease

in coefficient of friction. Consider that for a 20% grade (damn

steep), 1-cos(arctan(1/5)) ~ 2%. That is, the normal force only

deceased by 2%.





I was being a bit of a smartass (bunny hops, indeed :-)



But we all know the effect is significant - if not due to

significantly reduced total weight in that vector, then due

to proportion of weight shifted from rear to front wheel (?)

to compensate for the vector rotation relative to the

gravitational force (?)



snip

I remember some early MTBs that were very hard to ride up steep hills, especially off road ones, because they had high rise stems that made it very hard to get the rider's weight over the front wheel to keep the wheel in contact with the ground. Steep hills and very low gears seems to equal a lot of torque that wants to lift the front wheel.

Cheers

On Tuesday, December 24, 2013 12:57:33 AM UTC-5, Phil W Lee wrote:
Frank Krygowski wrote:

What you are thinking applies only to isolated systems, systems in which no energy crosses the control boundary.

If no energy is otherwise gained or lost in a system (an isolated system, by definition) then adding work will increase potential energy - or some other energy - within that system. But with a motorized (or braked) treadmill, energy flow occurs in the treadmill mechanism. So the rider puts in work (or energy) but it leaves through the mechanism, and his potential energy is unchanged.

So it's got nothing to do with the point on the circumference of the
tyre which is contacting the drag (which is the only thing that
changes when you incline the slope).

While your meaning isn't clear to me, it does seem you've accepted that the potential energy doesn't need to change.

(BTW, in case Joe's elevator-in-a-ladder example wasn't clear, he was talking about the elevator occupant climbing the ladder while the elevator was descending. I don't think his post specified "descending." But as he said, in that situation, the guy's potential energy w.r.t. the outside world would not change; yet he'd have to do work to climb the ladder.)

As an example, my turbo trainer contacts the tyre at a point about 20
degrees rearwards of where the tyre would normally contact the road.
This does not mean I'm going downhill all the time on my turbo
trainer.

A turbo trainer is quite a different situation. The power needed to drive the TT's roller will be the same no matter where the roller contacts the tire periphery, assuming all else is equal. Incline of the bike will also make no difference except for the fact that humans (most of us, anyway) seem to be better at producing power when we're oriented more or less vertically..

But with a bike clamped on a trainer stand, gravity has no effect. You could hang 100 pounds of lead from the bike frame and the pedaling would feel the same.

The load comes purely from the mechanics of the trainer, both on my
turbo trainer and on the inclined treadmill.

Yep.

Now, it may be that the load on the inclined treadmill is linked in
some way to the degree of inclination, but nothing in the test shows
that to be the case.

If the rider is maintaining constant position w.r.t. the outside world, he must cause his rear tire to apply enough force parallel with the surface to exactly balance the component of his bike+rider weight parallel to the surface. That's a requirement for equilibrium in that direction. If the tilt increases, that component of the weight increases, so his force output must increase.

If you want to get into energy flow in the mechanism, we can cover that too - but probably not until Thursday, at least on my end.

- Frank Krygowski

Phil W Lee writes:

Basic physics tells us we need not apply any energy to remain at the
same speed and direction,

Only if there are no external forces. There is one in this case,
gravity.

so whatever force we are having to overcome
must be applied by the mechanism, not by the attitude of the bike.

Any gravitational component of the force the mechanism applies (the
"rollback" force) could be overcome by holding onto a handrail,

Of course. But the cyclist is not allowed to hold onto a handrail.

think it's fairly obvious that the inclination would need to be pretty
extreme before that sapped very much energy.

An easier way to think of this is to consider the situation with the
treadmill tilted but not running. I hope you agree that the cyclist
has to apply a continuous torque to the wheel to remain in one
position. That is not the case when the tilt is 0. Applying that
torque, with the treadmill not running, takes zero power.

With the treadmill running, the cyclist has to continue to apply that
same torque. But now, because the treadmill is running, the wheel must
turn to remain in one position (relative to the outside). The power the
cyclist must exert is then the constant torque times the angular
velocity of the wheel.

--
Joe Riel