Minimal watts per rating

[jcross88]

I wasn't sure where to post this, but I have a somewhat abstract question to ask about the relationship between stroke rate and wattage on the erg.

Two different stroke rates can have the same wattage if the higher rate has less force per stroke. Two different wattages can have the same stroke rate if the higher wattage has more force per stroke. There is no upper limit on pace for any given rating, up until the erg itself can't withstand the force (chain breaks, metal bends, etc). Even pulling a 1:15 pace (830 Watt) at 20 spm is possible, but would take a very high force per stroke (roughly 386 lbs).

What is the lower limit of wattage you can achieve for a given stroke rate? A non-zero force is necessary simply to accelerate the drive chain along a reasonable stroke length the required speed to achieve, say, a 26 rating. Is there a way to calculate it?

Your total stroke duration from catch to catch, in seconds, is 60/rating. Force is energy divided by distance, and energy is work (wattage) times duration. So you get:

duration = 60/rating
energy = watts*duration
force = energy/drive_length

force = (60*watts)/(rating*drive_length), which can be divided by acceleration due to gravity to give force in kilos, and converted to pounds if desired.

Implied in this question is a general assumption that your drive and recovery ratio is reasonable. Presumably you could achieve a very low wattage for a given rating if you took a very long time on your drive, carefully pulling as gently as possible, and recovering as fast as possible. They say the ideal ratio is in the ballpark of 1:2.

So lets say a 26 rating, with a 1.5m drive length, and a strict 1:2 drive to recovery ratio; What's the lowest I can possible get my wattage under those circumstances?

Also, if a rating strictly implies a minimal wattage, can it also imply a reasonable maximum wattage? For instance if I did somehow pack 386 pounds of average force into my drive and pull 830 watts at 20 spm, my drive speed would be very very fast. Given a stroke duration of 60/20 = 3 seconds, my drive speed would be 1.5 m/s, which can't possibly carry 386 pounds of average force. At what point does a given wattage and rating combo drift so far from good form that a higher rating (or lower wattage) is implied?

[jcross88]

Your total stroke duration from catch to catch, in seconds, is 60/rating. Force is energy divided by distance, and energy is work (wattage) times duration.

Whoops, typo there. Wattage is power, not work, obviously.

[Carl Watts]

The wattage displayed on the monitor is average so if you look at the Wattage applied during the DRIVE time only it is much much higher. Once you drop to below 20spm the Wattage goes through the roof.

For the sake of simplicity, 1:15 pace at 830W at 20spm on a 0.5sec drive time is nearly 5000 Watts during the drive. As its not possible to put that energy in as a perfect flat 5000W for the whole duration of the drive then the peak power in the middle of the drive is much higher than that. Concept 2 would be integrating the area under the curve and then working this out as an average per second, its not per minute.

The only thing that links your different Watts, pace and spm and drag factor is your heartrate. If you want to try and start optimizing anything you would be looking for a minimum HR to achieve any given pace.

The pace is a cubic law so the power required rapidly decreases at 3 minute pace and there would be a crossover at which point your using more energy just moving your body weight up and down the slide than is going into the flywheel.

[jcross88]

As I understand it, the wattage that appears on the performance monitor is average watts per stroke. Wattage is work divided by time, and work is force times distance. The force you apply over the distance of your stroke length gives you joules of work done, and if you divide those joules by the time it takes to complete a whole stroke (which includes the drive and recovery), you'll get watts per stroke.

You're right though, I misspoke in the last part of my original post. The way I worded that did seem to imply that the average force of the stroke is somehow equivalent to the peak force of the drive, which is false. Like you said, the drive isn't a flat, constantly applied force. Instead it's a curve on a graph where the y-axis is force and the x-axis is the distance of your stroke length. The area under that force curve is force times distance, equal to work done.

So then the wattage is the average watts per stroke, averaged not only over the drive, but also the recovery, during which no force is being generated.

Please correct me if any of that is wrong.

But this is somewhat beside the point, and my question remains; It would certainly take tremendous force in the drive to get a high average watt per stroke at a low stroke rating. But to achieve a certain stroke rating a minimal force is necessary. What is it?

I'm not trying to optimize anything, really. It's a sort of purely theoretical question. If I have a 20 stroke rating, then each complete stroke is 3 seconds long. If I am adhering to a 1:2 ratio for drive and recovery, then the drive portion is 1 second, and the recovery is 2 seconds. If my drive length is 1.5m, then my average drive speed must be 1.5 m/s. Attempting to accelerate the handle will meet resistance and require force to overcome it. What is the least force that will achieve the necessary drive speed to meet the above conditions?

[Carl Watts]

Are you using ErgData ?

Thats really fun for those that like numbers but personally most of it is meaningless to me. I use it because it gives you the best graphs in the C2 LogBook and it has the total stroke count and the drag factor.

Pays not to over analyze things unless your really at that Elite level, just get on the Erg and row.

[MartinSH4321]

...
I'm not trying to optimize anything, really. It's a sort of purely theoretical question. If I have a 20 stroke rating, then each complete stroke is 3 seconds long. If I am adhering to a 1:2 ratio for drive and recovery, then the drive portion is 1 second, and the recovery is 2 seconds. If my drive length is 1.5m, then my average drive speed must be 1.5 m/s. Attempting to accelerate the handle will meet resistance and require force to overcome it. What is the least force that will achieve the necessary drive speed to meet the above conditions?

It depends on drag factor. At a DF of 0 you would produce 0 watts at any stroke rate (after you accelerated the flywheel), at infinite DF you would produce 0 watts at 0 stroke rate as the flywheel won't move at all.
Everything inbetween this theoretical extremes should be able to calculate, but I can't

[jamesg]

Length 1.5 and rating 26 will never be a low power exercise, due to the force needed to accelerate the flywheel. Even on drag 47 I had to deliver around 170W. Suggest you try it.

If you use ergdata you can see length and average force, as well as rating and power, and choose the combination that suits you best.

[jcross88]

It depends on drag factor. At a DF of 0 you would produce 0 watts at any stroke rate (after you accelerated the flywheel), at infinite DF you would produce 0 watts at 0 stroke rate as the flywheel won't move at all.
Everything inbetween this theoretical extremes should be able to calculate, but I can't

I suspected that drag factor was the missing component. The drive encounters resistance because of the angular momentum of the flywheel, the air resistance slowing the flywheel down, and mechanical friction of the machine components. The first strokes accelerating a stationary flywheel will require more work to get it up to speed, and every subsequent stroke only maintains that speed. If there were no resistance components after the initial strokes, then no work would need to be done by the drive to accelerate the flywheel rotation. It would be like pedaling on a moving bicycle, but not fast enough to engage the gearing and accelerate; the pedals spin with very little resistance. Conversely, if the drag were high enough to halt the flywheel entirely in the time span of a single stroke recovery, then every drive would require the work of starting a stationary flywheel.

I know the PM does some proprietary math to determine how much power your drive imparts, and so higher and lower drag factor won't allow you to "go faster" or "go slower" on the erg. In a low drag setting, the flywheel decelerates more slowly, requiring faster, more explosive force to maintain speed. In a high drag setting, the flywheel decelerates quickly, requiring a persistent, more constant force to maintain speed. A low drag factor seems more appropriate for a high rating, and a high drag factor for a low rating.

What are the dimensions of the drag factor? Is a 130 drag factor measured in torque per second or something like that? Or is it a dimensionless constant? In either case, how does the PM use the number mathematically?

Length 1.5 and rating 26 will never be a low power exercise, due to the force needed to accelerate the flywheel. Even on drag 47 I had to deliver around 170W. Suggest you try it.

If you use ergdata you can see length and average force, as well as rating and power, and choose the combination that suits you best.

I was wondering about this question during my warm up and cool down. Lately my warm up and cool down has been 2:00 intervals practicing with a metronome to improve my cadence. For a warmup I start at a 20 rating, and step up to a slightly higher rating than what I plan to row for the session, usually over 3 or 4 intervals. For the cool down I step back down to a 20 rating. I focus entirely on form, cadence, and heart rate, paying no attention to wattage. It's a somewhat tedious process and while in lala land I stumbled on this question: How low can I get my wattage for each rating?

[jcross88]

Per the Concept2 website https://www.concept2.com/indoor-rowers/ ... etting-101:

Between each stroke, the PM measures how much your flywheel is slowing down to determine how sleek or slow your “boat” is. This rate of deceleration is called the drag factor. On your next stroke, the PM uses the drag factor to determine from the speed of the flywheel how much work you are doing.

So the drag factor is the rate of angular deceleration, which is measured in radians per second squared. That deceleration must be caused by a net external torque, measured in newton-metres or joules per radian.

One interpretation could be that a drag factor of 130 = 20.7*2*pi and therefore equals a loss in angular velocity of 20.7 revolutions per second, every second. That seems unrealistic however, since 20.7 revolutions per second is 1242 rpm, and I very much doubt the flywheel is losing nearly 2500 rpm in the two seconds of recovery for a 20 stroke rate row.

Another interpretation is that the 130 somehow relates to the torque generated by air resistance which is decelerating the flywheel?

If I could make sense of the units of the drag factor, I may be able to use them to approximate the minimal wattage.

[Nomath]

Interesting question!
You can find a general explanation of the physics in an internet paper The Physics of Ergometers.
You can find several answers to specific issues in my topic Fan Blade Physics and a Peek inside C2's Black Box .
What C2 calls the drag factor is the same variable as the drag coefficient in the above references multiplied by 10^6 (e.g. DF 120 corresponds to a drag coefficient of 0.000120 kg m². The units are the same as for the moment of inertia of the flywheel).
You will find an equation for the deceleration of the flywheel in my topic. The angular velocity does not decrease linear with time.

At a first look, I don't believe there is a quick and simple answer. When I find time, I will run some simulations.

[jcross88]

Interesting question!
You can find a general explanation of the physics in an internet paper The Physics of Ergometers.
You can find several answers to specific issues in my topic Fan Blade Physics and a Peek inside C2's Black Box .
What C2 calls the drag factor is the same variable as the drag coefficient in the above references multiplied by 10^6 (e.g. DF 120 corresponds to a drag coefficient of 0.000120 kg m². The units are the same as for the moment of inertia of the flywheel).
You will find an equation for the deceleration of the flywheel in my topic. The angular velocity does not decrease linear with time.

At a first look, I don't believe there is a quick and simple answer. When I find time, I will run some simulations.

Dude, you are speaking my language!!

Give me a bit to digest your formidable post and I'll get back to ya.

[Nomath]

I wasn't sure where to post this, but I have a somewhat abstract question to ask about the relationship between stroke rate and wattage on the erg.

Two different stroke rates can have the same wattage if the higher rate has less force per stroke. Two different wattages can have the same stroke rate if the higher wattage has more force per stroke. ......

After giving your question some more thought, I think that it contains a basic flaw. The flaw is that you cannot increase the stroke rate without increasing the handle force. The chain drive mechanism implies that handle speed and handle force are not independent. Handle speed and rotation speed of the flywheel are coupled and a higher rotation speed implies a (much!) higher handle force.

The rowing erg is a mechanism that is described physically by only one differential equation. I have written a small computer program that simulates this equation numerically by taking very small time steps, typically 1 millisecond duration. It calculates the changes in the angular velocity from applying a certain force curve (handle force as a function of time). From the speed of the flywheel the distance travelled on the slide during the drive can be calculated. For a fixed distance, the higher the rate the higher the work in the drive (work = force x distance). The higher the rate, the shorter the duration of the drive, hence the higher the average power in the drive. And keeping the time ratio drive:recovery constant means that the total average power will be higher. A shorter recovery also implies that the flywheel has decelerated less at the catch. These conditions are applied in a continuous cycle of drive and recovery until a stable cycling pattern is obtained.
The figure below shows the result of such a simulation (parameters : drag factor 120 ; drive length 1.50 m ; drive:recovery = 1:1.5 ; acceleration at the catch 8 m/s²).



I was initially surprised and puzzled by how strong the power increases with the stroke rate. Doubling the stroke rate by a factor of about 2, from 16 to 30 spm, increases the power by a factor of 10! The factor 10 appears to be caused by a factor 5 increase in average handle force and a reduction of the duration of a stroke by a factor of 2.
This result seems counter-intuitive, because stroke rates of 30 are not superhuman, but rowing at 460W for an extended time is not for ordinary athletes. Note, however, that the figure is the result of a simulation of increasing the stroke rate AND keeping the drive distance constant AND keeping the time ratio drive:recovery constant. These conditions are rarely used in practise. My guess is that most rowers who increase the stroke rate do this by decreasing the recovery time much stronger than the drive time. Also the drive length tends to become shorter with higher stroke rates.

[jcross88]

Very interesting! And thank you for your time helping to investigate. Much appreciated.

I suppose the minimal watts question occurred to me because of how I conduct my sessions. I've been rowing with a metronome which I set to three times my desired rating. If I do steady state at 24 spm, I'll set it to 72 bpm. I set the accent bell to three, so the bell rings on every catch, the first tick signals 1/3 of the stroke is complete and so my drive should be complete, and the second tick tends to be just after my knees bend. This ensures that I maintain a 1:2 ratio while also giving a very satisfying auditory correlate to my stroke positions. Because of this I hardly look at my rating during a session, and focus intently on maintaining a consistent drive length with proper form. I'm very new to rowing and I'm a bit of a perfectionist, so I've been obsessing about exactly the conditions that you describe: Fixed stroke rate, fixed drive length, fixed drive-recovery ratio.

On my warm ups and cool downs I'm primary focused on my heart rate, but on the session proper I'm focused on my wattage. I was in the bad habit of rowing too hard on my cool downs so I made a little game for myself to intentionally get as low wattage as I could. I suppose the question occurred to me because empirically I found it harder to get watts lower at a higher rating.

[jamesg]

For low rate training at say 18, suggest you try 4:1 recovery - pull time ratio. 2:1 is for racing 2k.

At 18 we have a 3.33 second stroke and at ratio 4:1, pull time is 0.67s, quick enough to move the boat or flywheel fast and so require some effort. Low drag and correct rowing/erging sequences will be needed.

Rhythm is a major factor: typical ratios are Quickstep (3:1) for fast UT work at up to 23 (60/23 = 2.6 and pull time 0.65s) and Waltz (2:1) for 2k race cruise at 32 (60/32=1.9 and 1.9/3 = 0.63s). Rowing and erging without these rhythms can be extremely uncomfortable and ineffective, especially afloat. On the erg there will be conflict between the strictures of HR training and how rowing is actually done.

Ergdata shows both pull speed and length on C2 machines, so these times can be estimated directly. At 20-22 I usually see 1.2m and 1.8m/s, so pull time around 0.67s; but on drag 90 and at about 34kg average pull force.

[Nomath]

... I've been rowing with a metronome which I set to three times my desired rating. If I do steady state at 24 spm, I'll set it to 72 bpm. I set the accent bell to three, so the bell rings on every catch, the first tick signals 1/3 of the stroke is complete and so my drive should be complete, and the second tick tends to be just after my knees bend. This ensures that I maintain a 1:2 ratio while also giving a very satisfying auditory correlate to my stroke positions. ....

... Ergdata shows both pull speed and length on C2 machines, so these times can be estimated directly. ...

I want to point to a difference in timing of the drive and the recovery between the human rower and the PM/ErgData

For the athlete the drive starts at the catch, which is the point where the handle is nearest to the flywheel. By definition, at the catch the handle speed is zero. The drive ends when the handle is at the point furthest to the flywheel. Again at this point the handle speed is zero, and the recovery starts. Drive and recovery are defined by the change in moving direction of the handle.

For the flywheel/PM/ErgData the drive starts when the handle speed matches the corresponding speed of the flywheel. At this moment the clutch couples the drive axis of the sprocket wheel to the flywheel axis. There is a time lag between the catch and the coupling moment. Near the end of the drive, when the handle speed gets below the corresponding rotation speed of the flywheel, the clutch decouples and the flywheel starts an autonomeous deceleration.

Hence the drive time 'seen' by the flywheel/PM/ErgData is significantly shorter than the drive executed by the athlete. Also the drive length will be shorter ; the drive speed displayed by ErgData will be higher. Correspondingly the duration of the recovery is significantly overestimated by ErgData.

The figure below showing a computer simulation of the rotation period of the flywheel as a function of time, illustrates these points.



The drive starts at time=0 and lasts 1.0 sec. With an initial acceleration of the handle at the catch of 8 m/s², it takes 0.18 sec to match the speed of the flywheel. At 0.92 sec there is a minimum in rotation period (i.e. flywheel speed peaks). Although the proprietary PM software is undisclosed, my guess is that the PM counts this minimum in rotation speed as the end of the drive. During the recovery the rotation time increases linearly with time with a slope 2π * c/J ; J is the moment of inertia of the wheel. After 2.7 sec from the start of the stroke the rotation period equals the period at the start and the same cycle can be repeated.
Note that for the athlete the drive lasts 1.0 and the recovery lasts 1.7 sec, a recovery/drive ratio of 1.7. For the PM/ErgData the drive probably lasts 0.74 sec and the recovery lasts 1.96 sec, a recovery/drive ratio of 2.6

The figure also illustrates a point mentioned in my before-last post: a higher handle speed implies much higher handle force. The arrows in the graph show what the power supplied by the drive train does to the rotation period of the flywheel. For an input power equaling c * ω³ , c is the drag coefficient and ω the angular velocity, the period does not change. Because the handle speed during the drive is proportional to ω and power = speed x force, this implies that the handle force should at least increase proportional to ω² to accelerate the flywheel.