Inelastic ends of the stroke ?

[Nomath]

....
About one year I did a lot of numerical simulations of a sequence of rowing strokes based on the equation P = C * ω³ + ω * J * dω/dt, in discrete time steps of 1 millisecond.



Now a comparison how much each of the two terms contributes. Let's look at the 6th stroke where the ω-wave is fairly stabilized.
Work in the stroke determined by numerical integration over 1 ms time steps (DF=125) : 631 J
Work estimated for the first term using ω1 = (2*ωmax + ωmin)/3 : 441 J
Work estimated for the second term using the second approximation : 188 J
Estimated combined work of the two terms using the approximation formulas : 626 J
You can see that the approximation deviates by only 1% from the true value.
...

I did these simulations more than one year ago and I took the numbers from my notes from that time.
This week I started the subroutine again. It is always to challenge to figure out the code that I wrote a long time ago, because I am not a very disciplined programmer. But after some trial and error I had it running properly and with the same parabolic force input for a simulation of 6 consecutive strokes at DF=125 as displayed in the above figure.

The results for the 5th stroke differ on two important points from the numbers given above.

• work in the 5th stroke from numerical integration over 1 ms time steps : 640.9 J

• estimated work for overcoming the drag resistance during the drive, using ω1 = (2*ωmax + ωmin)/3 : 196.5 J

• estimated work for accelerating the flywheel, using the approximation Wacc = ½*J*(ωmax² - ωmin²) : 441.4 J

Sorry that I mixed up the two contributions. In contrast to the numbers in the quoted text, the work for the acceleration is more than twice the work for the drag resistance.
But as anyone can see, the sum of the two estimated terms is 637.9 J, which is only 3.0 J (i.e. less than 0.5%) short of the accurate number.

[HornetMaX]

Hmm seems right. Even in my rough simulation the torque necessary to accelerate the flywheel (due to its inertia only) is bigger than the drag torque (by more than a factor 2).

[JaapvanE]

Hmm seems right. Even in my rough simulation the torque necessary to accelerate the flywheel (due to its inertia only) is bigger than the drag torque (by more than a factor 2).

This is unexpected to me, what other force is there to overcome here? Or are you aiming at a specific acceleration?

[jamesg]

torque necessary to accelerate the flywheel (due to its inertia only) is bigger than the drag torque

Drag + acceleration torque can equal 4 x drag torque, at low df and spm. One only pulls the handle when strictly necessary to avoid the boat stopping completely, but then we pull it hard.

[HornetMaX]

Hmm seems right. Even in my rough simulation the torque necessary to accelerate the flywheel (due to its inertia only) is bigger than the drag torque (by more than a factor 2).

This is unexpected to me, what other force is there to overcome here? Or are you aiming at a specific acceleration?

Independently of anything else, in my sim during the drive the one-way bearing is engaged for 0.4815 seconds. During this time:

• the flywheel speed goes from w1 ~= 106.752 rad/s to w2 ~= 132.540 rad/s (these corresponds to handle speeds v1 ~= 1.51042 m/s and v2 ~= 1.87530 m/s). Note that these values are comparable to what Nomath has in his plots.
• the kinetic energy of the wheel changes hence by ~309 J, meaning the average power used to accelerate the wheel (only due to its inertia) is 309 J/0.4815 s ~= 642 W.
• if you assume the acceleration constant (it is not, but it's not too far either) you can say that the average power used against drag is the average of the powers at w1 and w2 (Pdrag = k * w^3, where k = 120E-6), leading to ~ 213 W. If you want to be more precise, you integrate instead to take averages and assume constant acceleration, and that gives me ~218 W, close enough.

So yeah, the power used to fight the drag is roughly "only" one third of the power used to win the flywheel inertia (pretty in line with what jamesg is saying, looking simply at the torques).

Notice that in the above I've computed the average power over the time span where the one-way bearing is engaged (this is less than the drive duration).
If you want to compute the same powers but over the whole stroke duration (drive + recovery), then you have (I was at stroke rate 30):

• Pwheelinertia = 642 W * 0.4815 s / 2 s ~= 154 W
• Pdrag = 213 W * 0.4815 s / 2 s ~= 51 W

The sum of the two is a tad over 200 W: this is not too far from my 2K average power (235W for a 7'36.8 at SR29, from my logbook).
So now I'm a bit lost: is it really true that the PM5 counts only the power used to fight the drag ?

[Nomath]

...
Independently of anything else, in my sim during the drive the one-way bearing is engaged for 0.4815 seconds. During this time:

• the flywheel speed goes from w1 ~= 106.752 rad/s to w2 ~= 132.540 rad/s (these corresponds to handle speeds v1 ~= 1.51042 m/s and v2 ~= 1.87530 m/s). Note that these values are comparable to what Nomath has in his plots.

Ha, numbers I can use to test my approxomation formulas for calculating the work for acceleration, Wacc, and the work against drag, Wdrag:
- Wacc = ½ * MoI * (w2² - w1²) = 308 J ; moment of inertia MoI=0.100 kg m²
- Wdrag = C * ((2*w2+w1)/3)³ * drivetime = 110 J ; drag coefficient C=0.000120 kg m²
At 29 spm, this yields an average stroke power of 418/2.07 = 202 W

So now I'm a bit lost: is it really true that the PM5 counts only the power used to fight the drag ?

No, the PM5 only counts the work during the drive, which includes both drag and acceleration. However, if you are in a stable stroke rhythm, the work done for accelerating the flywheel in the drive is used to compensate for the drag force during the recovery so that at the catch you start at the same flywheel speed as in the previous stroke. All our power input is for fighting the drag, sooner or later!

[JaapvanE]

No, the PM5 only counts the work during the drive, which includes both drag and acceleration. However, if you are in a stable stroke rhythm, the work done for accelerating the flywheel in the drive is used to compensate for the drag during the recovery so that shortly after the catch you start at the same flywheel speed as in the previous stroke. All our power input is for fighting the drag, sooner or later!

Indeed. That is why another approach is to determine the average speed over the entire stroke (i.e. angular distance covered divided by stroke time) and use s = (k/C)^1/3 * θ and thus v = (k/C)^1/3 * ω. This is close enough for most purposes, although in practice I don't know if this is the approach taken. I say this because when testing, I notice subtle differences: in theory given a specific θ, one would expect similar results, but the numbers differ subtly. I'm not certain whether it is due to issues with measurement/signal processing, or that the formulae contain extra parameters.

The key operative part is "so that shortly after the catch you start at the same flywheel speed as in the previous stroke.". That is what I meant with across-stroke acceleration. As soon as people start accelerating (start, sprints, etc.) this assumption will not hold, and this approximation is less usefull (as concluded by the Ulm team). For most rowers, this is quite acceptable. Only when people have a very unstable rowing style, this becomes a real issue.

[HornetMaX]

No, the PM5 only counts the work during the drive, which includes both drag and acceleration.

OK.

Indeed. That is why another approach is to determine the average speed over the entire stroke (i.e. angular distance covered divided by stroke time) and use s = (k/C)^1/3 * θ and thus v = (k/C)^1/3 * ω. This is close enough for most purposes, although in practice I don't know if this is the approach taken. I say this because when testing, I notice subtle differences: in theory given a specific θ, one would expect similar results, but the numbers differ subtly. I'm not certain whether it is due to issues with measurement/signal processing, or that the formulae contain extra parameters.

I think what you/Nomath are saying is that, with Nomath notation and under steady rowing assumption, Wacc is equal to the work done by the drag while the one-way is not engaged (note that this does not correspond to recovery, it's bigger). So if we compute the work done by the drag over the entire cycle (one-way engaged and one-way disengaged), then this work must be equal to the work put in via the handle. I'm fine with that, clear.

What we want to compute is the integral over the whole stroke of (drag_torque * dθ) = (k * ω^2 * dθ) = (k * ω^3 * dt), then divide by the stroke period to get the average power. You can't compute the average power via the average ω, in general.
This only works if you take a 1st order approximation of ω^3 at the average ωa. In that case it is true that:

(1/T) * integral(0,T, k * ω^3 * dt) = k * [(1/T) * integral(0,T, ω * dt)]^3 = k * ωa^3

But yeah, 2 assumptions are baked in: steady rowing *and* 1st order approximation validity.

[JaapvanE]

What we want to compute is the integral over the whole stroke of (drag_torque * dθ) = (k * ω^2 * dθ) = (k * ω^3 * dt), then divide by the stroke period to get the average power. You can't compute the average power via the average ω, in general.
This only works if you take a 1st order approximation of ω^3 at the average ωa. In that case it is true that:

(1/T) * integral(0,T, k * ω^3 * dt) = k * [(1/T) * integral(0,T, ω * dt)]^3 = k * ωa^3

But yeah, 2 assumptions are baked in: steady rowing *and* 1st order approximation validity.

I sent some raw data via PM, which you can plug into the MathLab model. Please note I have an issue with my flywheel as it seems to wobble, so the time between pulses fluctuates because of this (issue under investigation by C2), but as I process the data with Theil-Senn Quadratic Regression it shouldn't impact the derived Angular Velocity and angular Acceleration.