How accurate are power meters?

Joe Riel writes:

Joe Riel writes:

James writes:

On 18/12/13 15:45, Joe Riel wrote:
James writes:

On 18/12/13 13:45, Phil W Lee wrote:
James considered Wed, 18 Dec 2013 08:29:23
+1100 the perfect time to write:

On 18/12/13 04:00, Phil W Lee wrote:
Frank Krygowski considered Wed, 11 Dec 2013
19:41:02 -0800 (PST) the perfect time to write:

On Wednesday, December 11, 2013 4:22:05 PM UTC-5, James wrote:
Or should I ask how inaccurate?

According to http://www.youtube.com/watch?v=5DRQwKREgvI it appears a 3%
increase in bike+rider mass results in a 14% increase in power to
maintain the same speed.

Unless I missed something, the rider+bike mass for the pro rider was
78.4kg, and at 16km/h on an 8% gradient the power meter recorded 279W
average.

When they added 2.6kg, to achieve 81kg total and
approximately a 3% increase in mass, the power required to
maintain the same speed of 16km/h was 40W higher, at 319W. A
14% increase!

I thought the power increase would be about proportional to
the mass increase, i.e. 3%.

I agree. A 3% mass increase should cause pretty close to 3%
increase in power required for a given speed on a given grade.

If 3% mass difference yielded 14% power difference, then it
seems
that reducing one's bike+rider mass by 21% (not impossible with my
bikes, especially if you change the rider!) should reduce one's power
requirement to almost zero. That's assuming things are proportional,
which is what the laws of physics claim.

I don't know the explanation for their findings, but they
don't seem to make sense.

I think you need to distinguish between the different sources of drag
before you can even start to analyse it.
Aerodynamic drag isn't going to change at all if the speed remains the
same, unless the gradient requires you to stand up to maintain speed,
in which case it could increase quite a lot, but on an 8% gradient at
10mph the main source of drag will not be aerodynamic. Frictional
losses would increase pretty much in proportion with power use.
Climbing power is what is going to be affected by mass.

I wonder if the accuracy of the power meter could be affected by
standing on the pedals to climb, as that would have a tendency to put
more weight on the pedals throughout their rotation.
That is, of course, only relevant if they are the pedal type (and may
indicate a superiority of the hub type).

Did you watch the video, Phil? The power meter was a "power tap" rear
wheel hub. It appears to have a wireless connection to a display unit
that measures average power over some time interval.

No, I didn't - I was watching something on TV at the time.
Doesn't mean I can't speculate as to why displayed readings don't
match theory though.

"don't match"? They're not even close.

Theory predicts a 3% increase, or a little over 8W. They measured a 14%
increase, or 40W.

Another possibility has occurred to me though.
On a gradient, is there more of a tendency for the speed to be less
consistent - constantly accelerating and slowing (maybe even in time
with cadence) would increase average power use, and speed would bleed
off far more easily up a gradient, meaning it would be more difficult
to keep to a consistent speed.
The extreme case (obviously not this one) has a rider almost stalling
as the cranks are vertical, then reaching peak acceleration as the
cranks are horizontal. It would take much of that component to have a
serious effect on efficiency.

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.

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.

One possibility is that during the first test, without the added weight,
the rider was supporting some of his weight with the safety harness. It
looks pretty easy to lighten up a bit on the pedals and have it reduce
the workload.


Possibly, though I think it amounts to about a 7.6kg weight reduction.
That's a fair amount of support. I would have thought that was quite
noticeable.

Maybe you can get a bit of horizontal component with a non-vertical
strap. For an 8% slope you'd only need 7.6kg*8/100 = 0.6*kgf of forward
thrust. Regardless, the existence of the harness seems a real weakness
in the experimental design. Not that it looked much like a real
experiment, to me.

In some of the video, the angle of the harness from vertical is about 30
degrees. So 1.2kgf of tension in the cable would explain the
difference.

Seems like the harness should be connected to a sliding piece to
prevent developing any horizontal force. My first thought was
whether the harness was really necessary. It didn't take long
to locate this

http://www.youtube.com/watch?v=MWCE02HrMV4

With a real bike treadmill that wouldn't be such an issue, but ...

--
Joe Riel

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.

--
JS

Phil W Lee writes:

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.

That's partially true---no power is *necessary*. That is, the rider
could soft-pedal, the bike would move backwards on the treadmill until
the harness stopped any further bicycle movement, then remain there as
the treadmill continues running. However, as James has responded, to
remain in place without such an external support requires the equivalent
amount of force and power as actually climbing a hill, minus air
resistance. Try walking in place on an escalator.

--
Joe Riel

On Thursday, December 19, 2013 6:27:06 PM UTC-5, JoeRiel wrote:

That's partially true---no power is *necessary*. That is, the rider
could soft-pedal, the bike would move backwards on the treadmill until
the harness stopped any further bicycle movement, then remain there as
the treadmill continues running. However, as James has responded, to
remain in place without such an external support requires the equivalent
amount of force and power as actually climbing a hill, minus air
resistance. Try walking in place on an escalator.

I'll "third" that. The power required should be the same as if he were actually climbing a hill.

I can explain the theory, but it might be more interesting to relate the times I've done VO2 max tests, etc. on inclined treadmills. I wasn't on a bike, but instead I was walking/jogging. But I can assure you, when that thing tilts upward, it absolutely makes a huge difference in power required.

- Frank Krygowski

Frank Krygowski writes:

On Thursday, December 19, 2013 6:27:06 PM UTC-5, JoeRiel wrote:

That's partially true---no power is *necessary*. That is, the rider
could soft-pedal, the bike would move backwards on the treadmill until
the harness stopped any further bicycle movement, then remain there as
the treadmill continues running. However, as James has responded, to
remain in place without such an external support requires the equivalent
amount of force and power as actually climbing a hill, minus air
resistance. Try walking in place on an escalator.

I'll "third" that. The power required should be the same as if he
were actually climbing a hill.

I can explain the theory, but it might be more interesting to relate
the times I've done VO2 max tests, etc. on inclined treadmills. I
wasn't on a bike, but instead I was walking/jogging. But I can assure
you, when that thing tilts upward, it absolutely makes a huge
difference in power required.

Phil's intuition that, since the bicycle is stationary no work is being
done, is well-founded if not quite right. Look at the treadmill
surface, which must apply a tangential force of weight*sin(angle of
elevation). It is moving in the same direction as the force it applies,
hence work is being done on it, and that work has to come from
somewhere.

Modern exercise treadmills have enough powered electrical equipment in
them that it is hard to be sure that physical work is being done, but
this was not always the case. For a good part of the nineteenth century
the treadmill was employed as a punitive device in some British prisons,
the prisoners being employed at hard labor. In some cases the treadmill
was used to grind grain, but in at least one case it just spun a fan in
a courtyard to no good effect, it being judged that the satisfaction of
doing useful work was a reward not merited by the wretched prisoners.

--

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.

You are looking at it in the wrong frame of reference. The proper frame
is that of the top surface of the treadmill, which is continually moving
backwards and downwards.

Suppose the cyclist uses the brakes to stop the wheels from turning. At
that point the cyclist is clearly doing no work. But he will now be
moving backwards and down. To remain in position he has to apply power
against the treadmill.

--
Joe Riel

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.

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, 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. But there's nothing fundamentally wrong with use of a tilting treadmill with a bike.

- Frank Krygowski

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.

--
Joe Riel

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.

... 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.


Sure, "climbing" a treadmill would be at least similar to climbing
a hill, I'd think (just less fun coming back the other way :-)

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.

But there's nothing fundamentally wrong with use of a tilting treadmill with a bike.


Not if all you want to do is expend energy.

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. Sure I make an estimate for the right gear, and as I climb
there is a *ton* of processing going on, but it's "quirky" to try
and explain. And when I get to the top, the only measure ment I
need is to look down at the valley below, feel my heart pumping and
my body recover with each inspiration (breath). If I was going to
race in the Tour de France or something, then measured power - comp-
arative to my competitors - might matter in preparation for the race.
But I'm not, and riding a bike is, to me, an experience that says the
hell with all that. About the most I'll do is check my watch when it
*feels* like I'm maintaining a faster average speed than usual.

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. This force can be
decomposed into two orthogonal vectors, one perpendicular to the road,
the other parallel to it. It is the force parallel to the road that you
have to work against. As the road gets steeper that vector gets larger
(W*sin(theta)). The perpendicular force (W*cos(theta)) gets smaller.
Its decrease is insignificant; besides, its only affect is to reduce the
rolling resistance, which is already small.


--
Joe Riel