Trying to understand drag factor better

[JaapvanE]

To be pedantic, drag is not proportional (i.e. constant x) velocity. It is proportional to some power of velocity, and I think C2 uses square as the power, but I'd have to check before swearing to that.

Lets keep each other sharp to keep information correct . You are right that for pure air based rowers, the drag is proportional to the square of angular velocity. For magnetic based rowers it is proportional to the angular velocity directly. Here, the models C and D differ very subtly, as the new magnet construction actually acts as a small magnet brake. It is about 1 DF difference, but it is sufficient to make a subtle difference in results (about 0.47%). The PM5 corrects this from a results perspective, but "hides" the DF effects for the user. It is extremely subtle, but from my experience with the raw data, we typically see an almost straight quadratic curve on a D where we get a nice straight line on a C.

I have no problem with you saying that it only feels very different in the first few strokes. I also agree with you that C2 dramatically underestimates power in the first few strokes, but that's a somewhat different issue.

To sum it up, I think my position is that they are different; yours is that they are almost the same (i.e. different).

I agree with you that the force distribution on a RowErg will be different on a boat, in fact my point is that it will never be similar to a boat to begin with.

When people ponder about the effect of drag on an RowErg, wether the analogy is a heavier boat or a longer/bigger oar, I think the correct analogy is the bigger oar (or bicycle gear). The heavier boat analogy doesn't work as the power to linear velocity ratio can't be changed on a PM5. So, with more drag, each individual stroke will feel heavier (aside the first stroke), but you will get more linear speed out of it. In that sense, it is just another way to deliver power to a system, similar to a higher bicycle gear. As a bigger oar feels heavier during a stroke, that aspect behaves similar as well.

Is it completely similar to boat behaviour with real oars? Obviously not as it wasn't close to begin with. But let's agree to disagree.

[gvcormac]

I totally agree that "longer oar" ("taller gear" on a bikeERG) is a better analogy than "heavier boat."

I didn't know about the Model D magnet. That's the magneto that charges the battery? I guess I what I didn't know is that Model C didn't have one. How does it sense angular velocity?

But I have a 1st gen Model D with PM3. Later Model C's were supplied with PM3. They also know the difference?

[Tsnor]

Here, the models C and D differ very subtly, as the new magnet construction actually acts as a small magnet brake. It is about 1 DF difference, but it is sufficient to make a subtle difference in results (about 0.47%). The PM5 corrects this from a results perspective, but "hides" the DF effects for the user. It is extremely subtle, but from my experience with the raw data, we typically see an almost straight quadratic curve on a D where we get a nice straight line on a C.

Neat. How does this work ? The mag generator unit on the D powers the PM5. It supplies 5 watts or less given a USB connector can power the PM5, and is likely regulated to give the same voltage/current at any flywheel speed.

Given the wattage based on flywheel speed wind resistance is in the hundreds of watts why would you "typically see an almost straight quadratic curve on a D where we get a nice straight line on a C" vs the same curve shape offset by a few watts used by the generator. Quadratic (at least in high school) looked very different than a straight line.

Aside: if you plug a USB cable into the PM5 on a model D then you are not using the generator power, the power comes from the USB connection. Does doing that change the shape of the curve ?

[JaapvanE]

Neat. How does this work ?

In the development of OpenRowingMonitor, I added a small optocoupler circuit to my RowErg (nondestructive) so a Raspberry Pi can pick it up. When you plot the length between the impulses against time, a perfect air-rower should result in a straight line during the recovery phase. The slope of that line equal to the dragfactor times a constant (more specifically, 2π / (Flywheel Inertia * Impulses Per Rotation)).

The mag generator unit on the D powers the PM5. It supplies 5 watts or less given a USB connector can power the PM5, and is likely regulated to give the same voltage/current at any flywheel speed.

That is my understanding as well. The incoming signal is 15V in a sinuslike wave varying between 0V and 15V. It could be a rectified sine as well, which would make sense as they probably could easily add a small buffer circuit to create a stable voltage to power the PM5.

Given the wattage based on flywheel speed wind resistance is in the hundreds of watts why would you "typically see an almost straight quadratic curve on a D where we get a nice straight line on a C" vs the same curve shape offset by a few watts used by the generator. Quadratic (at least in high school) looked very different than a straight line.

It is a subtle quadratic function, so for example instead of the straight line 10 x + 7 you get 0,001 x^2 + 10 x + 7. In OpenRowingMonitor we approximate it with Linear Regression regardless. We do see some degredation in the R^2, indicating that the line isn't a perfect match for the data. Typically we have a R^2 of 0.93 on the D, where on the C we get to 0,97. We are looking at Quadratic regression, but it comes at quite some cost in terms of calculationtime. So that is something we might add later.

Aside: if you plug a USB cable into the PM5 on a model D then you are not using the generator power, the power comes from the USB connection. Does doing that change the shape of the curve ?

Nope, regarless if the electricity is usefully consumed by a PM5 or is transformed into heat, the magnetic behaviour remains the same, so does the resistance. The electricity just encounters wires with infinite resistance I guess.

[Tsnor]

Nope, regarless if the electricity is usefully consumed by a PM5 or is transformed into heat, the magnetic behaviour remains the same, so does the resistance. The electricity just encounters wires with infinite resistance I guess.

I think there is feedback here, If there is no current to the PM5 there is a lower magnetic field, therefore less load on the flywheel. But this is well outside my expertise, and might not be right.

For larger generators "...Loading: If the load on a power system is increased, the load current on the system increases, which results in increased armature current in the generator. This increased armature current results in an increased magnet field in the armature..."

and

"..For any electromechanical device, the electromagnetic torque or force produced is directly proportional to the current flowing through them, if the excitation field is unchanged. A generator has its torque opposing that produced by the engine (or whatever is turning the generator). The engine requires energy to turn the shaft against the torque of the generator, much like you need energy to lift a weight vertically against earth's gravity.

When the load draws more current, more generator torque is induced due to the higher currents, the engine faces more resistance to its motion, just like how you would feel when you are lifting a 10 kg mass and someone adds another 5 kg mass to what you already have. The speed (of the generator and you) would drop as a result (which is not good), so the control system sends additional fuel to bring the speed back..."

[JaapvanE]

When the load draws more current, more generator torque is induced due to the higher currents, the engine faces more resistance to its motion, just like how you would feel when you are lifting a 10 kg mass and someone adds another 5 kg mass to what you already have. The speed (of the generator and you) would drop as a result (which is not good), so the control system sends additional fuel to bring the speed back..."

I agree with you that it is an issue with larger generators, but then you are talking about huge currents.

But on such tiny current, with a secondary circuit using that same 15 Volt signal to communicate with the PM5, the load wouldn't differ that much. But there is only one way to find out: testing it. I'll put that on my calendar after migrating to a low-latency kernel with ORM/Raspberry (as it is suspected to be our biggest source of noise). First I need to baseline the R^2 for a plugged in machine (we use CrewNerd to extract data from the PM5 for comparison) and then we can compare the R^2 for a stand-alone machine. As the R^2 is an indication how well the straight line fits the data, that say enough.