Average Power vs. RMS Power

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Coming from an engineering background, this whole discussion regarding power on and indoor being different to power outdoors puzzles me, there must be a good explanation, we just haven't found it yet.

So, I started thinking about analogies in my world - electronic engineering, and looking for possible explanations.

Electrical power for a DC signal is defined as P=VI (voltage x Current). However, for an AC signal it get's a bit more complex - the average voltage for a sinusoidal wave is zero, as is the average current.

Now in order to calculate average power of an AC signal, the concept of RMS or root mean square is used. To get the rms voltage of an AC signal we divide the peak voltage by the square root of 2. Hence power for a varying signal is calculated as the rms voltage multiplied by the rms current. My apologies if I've gone too deep here!!

Now, as a gross approximation, the torque pattern on the crank approximates a sinusoid. My understanding of the behaviour of the power meter is that it samples once every second the average torque over the last second, and not the RMS. I stand to correction here!! The same applies to the cadence - which is sample once per rotation.

Given the lack of inertia on an indoor trainer, one would assume that the torque pattern on an indoor trainer is different to the torque pattern outdoors. Is it possible that the RMS power value is actually different because the wave shape is different, and our bodies are responding to the RMS values and not the average values that our power meters are displaying?

For people that are able to produce similar power indoors to outdoors, their natural torque patterns are similar in both environments.

Just a way out thought - feel free to shoot it down.

For a good overview of rms power see http://www.meyersound.com/support/papers/amp_power.htm

My understanding of the behaviour of the power meter is that it samples once every second the average torque over the last second, and not the RMS. I stand to correction here!! The same applies to the cadence - which is sample once per rotation.PMs vary in their torque sampling frequency and the relationship to crank rpm. The PT samples torque 60x per sec and every 1.26secs it averages the (~75) torque samples and sends the observation to the computer, regardless of crank or hub cadence. This sampling frequency (and the fact that it includes a variable number of crank revolutions) accounts for the variability of power with the PT. If I understand the technology correctly, the SRM samples torque once per crank revolution, so that it always includes the same number of downstrokes and observed torque (and power) is less variable.

Given the lack of inertia on an indoor trainer, one would assume that the torque pattern on an indoor trainer is different to the torque pattern outdoors.Not sure this is true.

Good work.:D

I know the PT samples the torque regularly, it's how it calculates the average of all of those samples that interests me - is it a pure average, or an RMS value - they will be different.

Now, if the wave shape is always the same, the error is a constant and you would be none the wiser. But, changing from one environment (road) to another (indoors) you've possibly changed the shape of the wave, and hence the rms value will change, therefore changing the error, and therefore changing the true power for a given displayed power.

Ok, I got a chuckle out of your subject line, so I guess it's worth a response. :)

I don't have an answer, just some thoughts:
1) The torque pattern on a bike crank is always positive, as opposed to alternating across the origin like AC voltage.
2) The link didn't explain where the square root of 2 comes from, but isn't it derived from the integral of the sine wave divided by its period? FWIW, a PM really integrates the strain gage current over the sampling period to measure torque, and then divides by the time period to determine the 'average' torque. IOW, I think they are effectively the same concept.

I could be wrong too. It's fun to think about, though.

Ok, I got a chuckle out of your subject line, so I guess it's worth a response. :)

I don't have an answer, just some thoughts:
1) The torque pattern on a bike crank is always positive, as opposed to alternating across the origin like AC voltage.
2) The link didn't explain where the square root of 2 comes from, but isn't it derived from the integral of the sine wave divided by its period? FWIW, a PM really integrates the strain gage current over the sampling period to measure torque, and then divides by the time period to determine the 'average' torque. IOW, I think they are effectively the same concept.

I could be wrong too. It's fun to think about, though.1. You can also get an "always positive" AC signal - just means there is a DC component to it - doesn't really change the maths.
2. No, root mean square means "square all the values, then take the square root of the sum of the squares" - basically the way normalised power is calculated except NP is raised to the power of 4. Note that in the case of NP - power is always a positive value as well.

see http://en.wikipedia.org/wiki/Root_mean_square

If I understand the technology correctly, the SRM samples torque once per crank revolution, so that it always includes the same number of downstrokes and observed torque (and power) is less variable.

Close: the SRM samples at 200 Hz, but averages all the values for a single crank revolution, from which power is calculated. This means that more samples are taken when cadence is low vs. when it is high, but since the sampling frequency exceeds the Nyquist frequency by a large margin (maximum pedaling rates being <4 Hz), this isn't an issue.

Coming from an engineering background, this whole discussion regarding power on and indoor being different to power outdoors puzzles me, there must be a good explanation, we just haven't found it yet.

So, I started thinking about analogies in my world - electronic engineering, and looking for possible explanations.

Electrical power for a DC signal is defined as P=VI (voltage x Current). However, for an AC signal it get's a bit more complex - the average voltage for a sinusoidal wave is zero, as is the average current.

Now in order to calculate average power of an AC signal, the concept of RMS or root mean square is used. To get the rms voltage of an AC signal we divide the peak voltage by the square root of 2. Hence power for a varying signal is calculated as the rms voltage multiplied by the rms current. My apologies if I've gone too deep here!!

Now, as a gross approximation, the torque pattern on the crank approximates a sinusoid. My understanding of the behaviour of the power meter is that it samples once every second the average torque over the last second, and not the RMS. I stand to correction here!! The same applies to the cadence - which is sample once per rotation.

Given the lack of inertia on an indoor trainer, one would assume that the torque pattern on an indoor trainer is different to the torque pattern outdoors. Is it possible that the RMS power value is actually different because the wave shape is different, and our bodies are responding to the RMS values and not the average values that our power meters are displaying?

For people that are able to produce similar power indoors to outdoors, their natural torque patterns are similar in both environments.

Just a way out thought - feel free to shoot it down.

For a good overview of rms power see http://www.meyersound.com/support/papers/amp_power.htm

Power equals (force x distance)/time, or, in angular terms, (torque times angular displacement)/time. Torque may vary in a quasi-sinusoidal manner, but the average value of an analog signal would be precisely the same as if it were perfectly constant. Thus, the real issue is whether the digital sampling occurs at a sufficiently high frequency to obtain a representative average (essentially = area under the torque-time curve). The answer is that it does, as even the PowerTap samples at a frequency many-fold higher than (2 x) the pedaling rate.

1. You can also get an "always positive" AC signal - just means there is a DC component to it - doesn't really change the maths.
No, but it makes an arithmetic mean useful again (by eliminating the negatives). Since power doesn't include a powered function of either measured variable, there'd be no reason to use a root mean.

see http://en.wikipedia.org/wiki/Root_mean_square
Aha. Thanks. :)

Power equals (force x distance)/time, or, in angular terms, (torque times angular displacement)/time. Torque may vary in a quasi-sinusoidal manner, but the average value of an analog signal would be precisely the same as if it were perfectly constant. Thus, the real issue is whether the digital sampling occurs at a sufficiently high frequency to obtain a representative average (essentially = area under the torque-time curve). The answer is that it does, as even the PowerTap samples at a frequency many-fold higher than (2 x) the pedaling rate.
I have no problem with the sampling rate which to be strictly correct should be a minimum of 4x rpm since there are two peaks per revolution (left leg and right leg) assuming a perfect sinusoid, and needs to be higher since the torque curve is not a perfect sinusoid and therefore contains higher frequency components (harmonics).

I am questioning the use of an arithmetic mean, instead of a root mean square (if that is in fact the case).

So, for a particular rider when riding outdoors, the torque curve will be some function of time:
Torque = f(t)

When riding indoors, the torque curve is altered due to changes in inertia etc
Torque' = f'(t)

The rms values for Torque and Torque' are (possibly) different.

Since the rider rides at the same power (using an arithmetic mean) when on the trainer (because that is what his power meter is telling him) possibly he is riding at a different rms power - which is what he is feeling. In this case a 5% differential is not insignificant from an RPE perspective.

This argument is very similar to the usage of NP instead of AP when assessing the load being placed in the body.

As the body adapts to riding indoors the values of Torque and Torque' tend towards each other.

Electrical power for a DC signal is defined as P=VI (voltage x Current). However, for an AC signal it get's a bit more complex - the average voltage for a sinusoidal wave is zero, as is the average current.

Now in order to calculate average power of an AC signal, the concept of RMS or root mean square is used. To get the rms voltage of an AC signal we divide the peak voltage by the square root of 2. Actually, RMS is applied to AC Voltage or Current (and that's PEAK voltage and current), not AC Power. Average Power, whether AC or DC, is just that; average of power over some time period.

And if you calculate average power from peak power (which is peak voltage times peak current), average power would be half the peak power, not divided by the square root of 2 (Pavg = Vrms * Irms = [Vpeak / sqrt(2)] * [Ipeak / sqrt(2)] = Vpeak * Ipeak / [ sqrt(2) * sqrt(2) ] = Ppeak / 2).

Bicycle PMs calculate average power from sampled power values (not peak torque or peak cadence) so you don't have to worry about RMS.

Well, for some of us geeky programmer types RMS means entirely something else, but that's soooooooooooooooooooooooooooooo off topic here in this forum so I'll just shut up now.

Ken

edit: added a paragraph to show Pavg is not Ppeak / sqrt(2)

Actually, RMS is applied to AC Voltage or Current (and that's PEAK voltage and current), not AC Power. Average Power, whether AC or DC, is just that; average of power over some time period.

Bicycle PMs calculate average power from sampled power values (not peak torque or peak cadence) so you don't have to worry about RMS.

Well, for some of us geeky programmer types RMS means entirely something else, but that's soooooooooooooooooooooooooooooo off topic here in this forum so I'll just shut up now.

KenAgreed, AC power is calculated using rms voltage and current. I was incorrect in using the term average power. I was trying to refer to power generated over a period of time and not instantaneous power.

On a bicycle, power is calculated by multiplying torque and angular velocity. Using the electrical analogy, surely if torque and angular velocity is varying (in the same way that voltage varies in AC) then the correct way to calculate the torque is the rms value of torque and angular velocity and not the arithmatic mean (average).

Using the electrical analogy, surely if torque and angular velocity is varying (in the same way that voltage varies in AC) then the correct way to calculate the torque is the rms value of torque and angular velocity and not the arithmatic mean (average).Only if PMs calculated average power from PEAK torque and PEAK angular velocity values (which, BTW, would be accurate only if torque and angular velocity curves are reasonably in phase and sinusoidal just like voltage and current are assumed to be, which I doubt they are).

In bicycle PMs, sampled power is calculated from sampled torque and sampled angular velocity, not from PEAK torque and PEAK angular velocity, so there's no room for RMS to be used to calculate average power.

Ken

I have no problem with the sampling rate which to be strictly correct should be a minimum of 4x rpm since there are two peaks per revolution (left leg and right leg) assuming a perfect sinusoid, and needs to be higher since the torque curve is not a perfect sinusoid and therefore contains higher frequency components (harmonics).

I am questioning the use of an arithmetic mean, instead of a root mean square (if that is in fact the case).

So, for a particular rider when riding outdoors, the torque curve will be some function of time:
Torque = f(t)

When riding indoors, the torque curve is altered due to changes in inertia etc
Torque' = f'(t)

The rms values for Torque and Torque' are (possibly) different.

Since the rider rides at the same power (using an arithmetic mean) when on the trainer (because that is what his power meter is telling him) possibly he is riding at a different rms power - which is what he is feeling. In this case a 5% differential is not insignificant from an RPE perspective.

This argument is very similar to the usage of NP instead of AP when assessing the load being placed in the body.

As the body adapts to riding indoors the values of Torque and Torque' tend towards each other.

Okay, I get your point now - I thought you were saying that somehow the difference in inertial loading between indoor and outdoor cycling resulted in the powermeter being "tricked", whereas really what you're hypothesizing is that there's a physiological (or at least perceptual) difference due to a difference in the pattern of force application.

Only if PMs calculated average power from PEAK torque and PEAK angular velocity values (which, BTW, would be accurate only if torque and angular velocity curves are reasonably in phase and sinusoidal just like voltage and current are assumed to be, which I doubt they are).

In bicycle PMs, sampled power is calculated from sampled torque and sampled angular velocity, not from PEAK torque and PEAK angular velocity, so there's no room for RMS to be used to calculate average power.

KenTorque and velocity waves should be in phase because there is no imaginary component in the load. I.e. to go back to the electrical analogy, they are in phase if the load is purely resistive not capacitive or inductive.

I fully understand that the torque is sampled, but torque is sampled at a rate of 70Hz (for Powertap). For the angular velocity I think that they rely on the rpm value which is only sampled once per revolution. The values of these samples are then averaged and rolled up into one sample per second. I believe the way this averaging should be performed is using the rms method and not the arithmatic mean.

Calculating the rms value for a non-sinusiod would most likely require the use of Fast Fourier transforms to break the wave down into it's constituent sinusoids and calculating the rms value of each of these - certainly dividing the torque by sqrt(2) for a non-sinusoid is not valid.

Okay, I get your point now - I thought you were saying that somehow the difference in inertial loading between indoor and outdoor cycling resulted in the powermeter being "tricked", whereas really what you're hypothesizing is that there's a physiological (or at least perceptual) difference due to a difference in the pattern of force application.Yes, just like a one hour ride at 200 watts and a one hour ride with 30 minutes at 100 watts and 30 minutes at 300 watts would yeild the same AP but different NP - to plagarise your work :rolleyes:

In fact, when I first saw NP my initial reaction was that it was essentially the equivalent of rms - the principles are the same just on a different scale.

This is what got me thinking - there seems to be significant anecdotal evidence that riding on a indoor trainer is harder than riding outdoors. What could possibly cause the same power reading to feel different. The power meter doesn't know where it is, the legs don't know - they are just trying to push the pedals. Sure the brain does, but writing that off to purely motivational doesn't sit well with me.

So, looking using your concepts of NP and AP, and the arguments that the changes in inertial load force the rider to push through dead spots when riding on a trainer - I thought that the torque pattern may well change between the two environments. And, just as two rides with the same AP can yeild different NP's could we not be sitting with the same situation when moving between indoors and outdoors? The two different torque patterns would yeild the same average torque (which is being dislayed by the power meter and used by the rider to regulate his power), but different rms torque causing the rider to experience a different load?

Torque and velocity waves should be in phase because there is no imaginary component in the load. Well, if we're really wanting to geek-out here, the system does have inertia (ie, energy storage), so it's almost a certainty that peak torque values will be out of phase from peak angular velocity values. :p

I fully understand that the torque is sampled, but torque is sampled at a rate of 70Hz (for Powertap). For the angular velocity I think that they rely on the rpm value which is only sampled once per revolution. The values of these samples are then averaged and rolled up into one sample per second. Isn't 'rolled up' just a fancy word for integrated over the sampling period? Going back to the wiki derivation, if one integrates [i(t)]^2 over time and then multiplies by R, the actual energy consumption is determined, which can then be converted to Pav by dividing by the time interval. Isn't that what the PT is doing by sampling T(t) on a known frequency, multiplying by angular velocity once (to determine energy), and then dividing by 1.26sec at the end to find Pav for that 1.26 period?

I believe the way this averaging should be performed is using the rms method and not the arithmatic mean. It still comes back to "why use RMS, when neither the torque nor speed component are squared in the calculation of power?" In my understanding, RMS is used in electrical power not because of the sinusoidal nature of I(t), but because I(t) is *squared* in the calculation of P(t).

Well, if we're really wanting to geek-out here, the system does have inertia (ie, energy storage), so it's almost a certainty that peak torque values will be out of phase from peak angular velocity values. :p

Yes, but that stored energy is not returned to the system in the way a capacitor does. Mechanical load would have an imaginary component if the system had significant flex it it i.e. the initial force flexed the crank (thereby storing the energy) then the crank un-flexes thereby releasing the energy back into the system.


Isn't 'rolled up' just a fancy word for integrated over the sampling period? Going back to the wiki derivation, if one integrates [i(t)]^2 over time and then multiplies by R, the actual energy consumption is determined, which can then be converted to Pav by dividing by the time interval. Isn't that what the PT is doing by sampling T(t) on a known frequency, multiplying by angular velocity once (to determine energy), and then dividing by 1.26sec at the end to find Pav for that 1.26 period?

I think I understand you correctly, yes, that is the way (I think) the PT is calculating power - which effectively is an arithmetic mean.

The mechanical/electrical analogy I am drawing is between the electrical power equation P(t)=V(t)I(t), and the mechanical power equation P(t) = T(t) v(t) (where v = angular velocity).


It still comes back to "why use RMS, when neither the torque nor speed component are squared in the calculation of power?" In my understanding, RMS is used in electrical power not because of the sinusoidal nature of I(t), but because I(t) is *squared* in the calculation of P(t).
RMS is not related to the fact that current is squared. RMS is "a statistical measure of the magnitude of a varying quantity"

Okay, I've decided to use an example to try to convey my thinking. Unfortunately this example yeilds a result that is exactly the opposite to what is being observed in practice (am thinking about this :confused: )

Lets say that when a rider rides on an indoor trainer, and has to push the pedals through the dead spots, the torque samples for one cycle look like this:

1 2 3 4 3 2 1 0 1 2 3 4 3 2 1 0

For this situation the average torque for the cycle is 2 (sum of samples/# of samples)

and the rms (or statistical measure of the magnitude) of torque is 2.34 (square root of the sum of the squares)
Now that rider goes out on the road, and due to the inertia of his mass moving forward he does not apply force through the dead spots, so the torque samples for one cycle look like this:

0 0 0 4 4 4 4 0 0 4 4 4 4 0 0 0

For this situation the average torque for the cycle is 2 and the rms value is 2.82

In these two situations the power meter would give the same reading because it would be using the value of 2 (i.e. the arithmetic mean). However, would the legs not be experience the rms value and hence a significant difference between the two environments.

This example would lead us to conclude that riding on the road should feel harder than riding on a trainer - however, I have assumed that this is what is happening to the torque pattern between the two environments - maybe I have it the wrong way round. Maybe the dead spot is bigger when riding indoors because the flywheel's inertia only has to rotate a very light wheel through the dead spot. So possibly the torque pattern riding on the road is smoother than the torque pattern when riding on a trainer. (Hmmn, got to dig through the memory banks on classical mechanics)

Yes, but that stored energy is not returned to the system in the way a capacitor does. Mechanical load would have an imaginary component if the system had significant flex it it i.e. the initial force flexed the crank (thereby storing the energy) then the crank un-flexes thereby releasing the energy back into the system.
That's true, there's really no 'spring', but since torque (force) directly relates to the first derivative of velocity (acceleration, rather than velocity itself), the torque curve will lead the velocity curve by 90-deg. Angular velocity continues to increase until net torque equals zero, which occurs 90-deg after the torque peak.

RMS is not related to the fact that current is squared. RMS is "a statistical measure of the magnitude of a varying quantity"
Okay, I'll buy that. I guess I'm missing the physical revelance of using RMS values for electrical power, instead of just using the arithmetic average power. Interesting thoughts, but it doesn't look like I can add any more here. :)

Maybe the dead spot is bigger when riding indoors because the flywheel's inertia only has to rotate a very light wheel through the dead spot.
Nu-uh. On the road the dead spot could be 30 seconds if you only wanted to kick once and coast. That'd never work on a trainer.