Minimal watts per rating

[jcross88]

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

I got my rower last May and, after a few weeks toying around and reading these forums, decided the beginner Pete Plan was the way to go. I'm on week 6 and have a 7500 to do today. I intend to pull 181 watts, or around a 2:04.5 pace, with a stroke rate of 23. I read the entire beginner plan before starting and noted that he recommends you keep your long days at 24 spm or lower, but not much lower. In the Full Pete Plan, he recommends long days be between 22pm and 25pm. I was doing 24, but did one day at 25 just to see what that felt like. Recently I've switched to 23, since I suspect hanging out on the lower side of the 22-25 range probably compels me to improve my stroke efficiency.

I intend to follow the full Pete Plan for at least 6 months after I finish the beginner Peter Plan, and so I can't imagine rowing below a 23 rating for anything other than warm up and cool down. I might go higher for speed training or AT intervals, but not lower. The lowest rating I go for the warm up and cool down is 20.

But you have me thinking I need to experiment with as low as 18. A 20 rating at 1:2 does indeed feel pretty silly on the erg. I'll try adding an 18 rating interval and working in the 1:4 and 1:3 ratios for the lower ratings. My warm up and cool down are only about 7 minutes long right now, 2:00 intervals with 20 second "rest" to adjust the metronome. Really they should probably be at least 8 minutes, and even more for tougher workouts.

Thursday I did 3x2000 at 2:00 pace and 25.3 rating, 1:2 ratio, 4 minute rests between. It seems a little daunting to try for a 1:3 ratio for that workout, but maybe that's just anxiety over trying something new. I'm a creature of habit.

Another benefit for the 1:2 ratio with the metronome is that I use it to time my breathing too. When I'm just starting I breathe naturally, and as it starts to get harder I switch to one breath per stroke. Once I'm really going, my heart rate is up, and I'm sweating I'll switch between one breath per stroke (or one breath per three ticks of the metronome) to one breath per two ticks of the metronome. At a 23 rating this is about 35 breaths per minute and feels pretty comfortable to me. I'm not super strict about it; It's more of a guide than anything. I have noticed that if my heart rate starts creeping over 165 I can use this breathing technique to steady it at or around 164 without sacrificing my pace. At a 1:3 ratio I could do something similar, but 1:4 starts getting awkward.

I really appreciate your guys' engagement on the topic. I'm very new and my habits are thus far fueled by naive reading, lacking the benefit of experience. I've never been in a racing shell on the water, let alone on a crew. I'm primarily erging for general fitness, but I do want to gear my training towards actually getting on the water some day, or even joining a club.

[jcross88]

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.

I've been scouring your past posts from other threads. I work in QA for software development and I must say I'm very impressed by, and appreciative of, your recent work. If only I had more people like you on my team! I also fiddle with arduino boards quite a lot. I intend to follow up on that instructables link you provided to build my own performance monitor, just for the fun of it. But also perhaps to coax out new data to help me improve my erging in real time. But alas, my time is in high demand these days, so I imagine that project will have to jockey for position on my todo list for some time - at least if my wife has anything to say about it. (She will.)

In starting out on the erg, I've really wanted to focus on my form. I suppose rowing is like any other skill or instrument, where good foundations are always the best start, and a continual meditation. Ergo, the metronome. The hope is that my smart brain will trick my dumb body like Pavlov's dog or something, so that eventually anything other than the correct cadence will feel wrong. The metronome at full winding only lasts about 26 minutes, and my longer pieces are now 31+ minutes, so the last five minutes I'm forced to maintain my form and rating without the aid of the metronome, and that will only increase as I approach 10k and 12k days.

So far, this has worked pretty well. I tend to accidentally increase my pace when I switch focus onto my rate, but eventually things even out. On my logbook graphs you can't really tell where the metronome ended. But it really is like training wheels on a bike. Of course I won't have a metronome on the water, and even if I did my cadence would be better timed by the rest of the crew, not some timepiece. Some day I'll ditch the metronome altogether, only breaking it out a few times a year to return to fundamentals.

Similarly, I've considered marking my drive chain somehow, so that I'm not leaning forward at the catch. If the marking on the chain goes past the chain guide, I've gone too far. However, considering your research, a very slight lean might not be the worst thing for your form. The argument is that your shins should be vertical, with a straight back and no hunched shoulders, because this puts you in the most powerful position for the drive. But if the real work of that drive doesn't start for 17cm into the drive, then perhaps marking the chain 17cm (or whatever distance) from an ideal position would be best. It would be sort of like a violin teacher taping positions on the neck while you learn. Another sort of training wheel.

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 figure also illustrates a point mentioned in my before-last post: a higher handle speed implies much higher handle force.

I'm sorry, I think I've confused myself. You're saying that it is flawed to believe that you cannot increase the stroke rate without increasing the handle force?

If the chain drive mechanism doesn't engage the clutch because the sprocket wheel never spins as fast as the flywheel, then the flywheel won't accelerate and therefore the PM won't even register the drive. It would be a 0 watt drive, so to speak. This would be possible if you pulled very hard to ramp the flywheel up to a high speed, set a low drag factor so it doesn't slow down too much, and then pulled slowly enough on the handle to keep the sprocket rotation speed low.

In this case the only resistance you'd feel on the handle is the return mechanism for the entire drive. The average force of the drive would need to be enough to slightly overcome this, but not enough to reach a speed that rotates the sprocket wheel as fast as the flywheel at any point in the flywheel deceleration.

But this scenario assumes you pull very hard and then very lightly. If you even out the pulls to be more regular for an already up to speed flywheel, then the only factor that remains is the drag factor. At a drag factor of zero (flywheel housing is a vacuum, perfect bearings, etc) your regular pulls will eventually result in every pull not registering on the PM. At a very large drag factor, the flywheel will necessarily slow to a point in the drive where you can't help but rotate the sprocket wheel faster, engaging the clutch an registering wattage.

If the rating is high enough, and the 1:2 ratio is enforced (or perhaps 1:3, what have you), then you may get squeezed so that in order to displace the handle in the allotted time of the drive, the sprocket rotation must necessarily match the ever decelerating flywheel.



Another consideration I've had is the return mechanism, which exerts a constant force opposite to the drive. It isn't much, about 1.25 kg at the storage hook for the handle, and 2 kg at the end of the slide, but it isn't zero. At least some part of the drive force has to be used to overcome the return mechanism. Is it accounted for in the work calculation, or is it like dust on the flywheel modifying it's moment of inertia - negligible?

[jcross88]

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.

May I ask what model you are using to produce these results? I've worked out my own model that shows a minimal wattage of 276.9 is necessary for DF=120, drive length=1.5, drive:recovery=1:1.5. I've found that, all other parameters held constant, minimal wattage has a cubed relationship to stroke rate.

To find the answer, I began by looking at a stable stroke cycle. I used differential equations to find the instantaneous angular velocity of the flywheel during the drive and recovery portions. By iterating several simulated strokes I can approximate ω₀, the lowest angular velocity of the flywheel in the stable cycle. After initial strokes driving a non-rotating flywheel, the flywheel achieves ω₀. After the drive portion, the angular velocity increases to ω₁. After the recovery, the angular velocity decreases back to ω₀; so on and so forth. For the parameters mentioned above, I found ω₀ to be about 119.57 radians per second. This is a rotation period of about 0.0525 seconds.

By taking the formula for instantaneous angular velocity and integrating it with respect to time, we can determine the total displacement angle of the drive portion of the stroke.

I should mention that I am neglecting your discovery of the first phase of the drive, where the handle is accelerating the drive sprocket to the same angular velocity as the flywheel. This phase ultimately has the effect of shaving a small fraction of a second from the drive time, and a few centimetres from the stroke length. I'm working to incorporate it into my model, but for the time being we may simply redefine the drive portion of the stroke as beginning when the performance monitor detects it, when the flywheel begins accelerating.

By neglecting this first phase, we may convert a drive length into drive sprocket radians by dividing by radians, which are about 0.0141 meters. The power supplied that achieves a stable cycle whose drive portion displacement angle is equal to the drive length of the rower, in radians, is the minimal power necessary for that stroke rate.

It's just as you demonstrated with your graph


Imagine these three scenarios:

In the first, our handle velocity is such that the drive sprocket displacement angle exactly matches the flywheel angular velocity as it decelerates. It would be like walking behind a car rolling forward under its own momentum and placing your hand on the bumper. You aren't providing any force, just keeping up with it as it slows down. This would result in a 0W drive. You have to hope that the car has enough momentum to cross the finish line in time.

In the second, the flywheel velocity is exactly equal to the necessary constant drive sprocket angular velocity if you want to achieve your full drive length in the time allotted by your stroke rate and drive:recovery ratio. You would provide just enough force to counteract the drag deceleration, but no more. This would be like pushing the rolling car just hard enough that it maintains a constant velocity, crossing the finish line just in time. However, although the flywheel velocity is constant during the drive, it still decreases during the recovery. After several strokes, the flywheel will still eventually coast down to zero.

In the third scenario, we achieve a stable rowing cycle. At the beginning of the drive the flywheel angular velocity is ω₀. We accelerate it during the drive only just enough that, after the recovery, it will again be at ω₀. However, we must ensure that ω₀ is high enough that the total displacement during the drive equals our expected drive length - no more, no less. You push the car just hard enough that you can stop a couple seconds before it crosses the finish line, knowing that you gave it only just enough energy to make it in time.

It seems that ω₀ increases linearly with increased stroke rate, and power increases by the cube of increased stroke rate. I'm not totally sure why yet.

For instance, for DF=127, drive length=1.4, drive:recovery=1:2, minimal_power=0.0151526063100137*(stroke_rate)^3

Where does 0.0151526063100137 come from? Why is it 0.013407064471879 for DF=112? I'm looking into that now.

[Carl Watts]

Another consideration I've had is the return mechanism, which exerts a constant force opposite to the drive. It isn't much, about 1.25 kg at the storage hook for the handle, and 2 kg at the end of the slide, but it isn't zero. At least some part of the drive force has to be used to overcome the return mechanism. Is it accounted for in the work calculation, or is it like dust on the flywheel modifying it's moment of inertia - negligible?

Thats always been an area of contention on here. Personally I think its more important to think of that as stored energy and not wasted energy. Perhaps its better looking at what percentage of that energy then assists you during the recovery because your legs are far more efficient in the drive than they are in the recovery. I say that because you can Leg Press far higher weights than you can ever reverse curl so your body is far more efficient at putting in the power on the drive. The final calculation may actually be a positive gain and not a net loss because of the way your body works. I would say looking at bungee jump videos, at least 50% of that energy put in in the drive comes back to help your recovery.

The obvious conclusion is that there will be an optimum bungee cord strength, not to little and not to much. The net effect is that you can probably totally ignore any power loss in the cord, getting the right cord return force for your weight/strength may actually give you an advantage.

[Nomath]

May I ask what model you are using to produce these results? I've worked out my own model that shows a minimal wattage of 276.9 is necessary for DF=120, drive length=1.5, drive:recovery=1:1.5. I've found that, all other parameters held constant, minimal wattage has a cubed relationship to stroke rate.

My model is the differential equation P = C * ω³ + ω * J * dω/dt , linearized for small time steps into ∆ω = ∆t * (P/(ω * J) - C/J * ω²).
Since when handle speed matches the flywheel speed, P = Force * Vhandle and Vhandle = k * ω (k is the gearing constant and equals 70.69 radians/m; 1/k = 0.0141 m/radian) this simplifies to :
∆ω = ∆t * (F/k - C * ω²)/J ; C is the drag coefficient ; J is the moment of inertia.

To calculate the changes in ω, you have to define a force curve. I mostly used parabolic force curves (parabolic as a function of time).
My software is a bit outdated. I have to ration the number of points in the time grid to about 15,000. I chose to model 6 consecutive strokes, each divided into 2500 time steps. At 24 spm and a drive:recovery ratio of 1:1.5 this corresponds to time steps of about 1 msec.

Below is an example of the calculation for a stroke rate of 24 spm, DF=120 and initial acceleration after the catch of 8 m/s².
I also had to choose an intial angular velocity ω0. In the example below it was set to 100 rad/sec.
The maximum force was iteratively adjusted so that the drive length equals 1.50 cm. In this example Fmax = 466 N.



The figures show that after 6 strokes the cycle converges to an initial angular velocity of about 104 rad/sec and a maximum angular velocity of 128 rad/sec. The work in a stroke is 468 J and the calculated average power is 468/2.5 = 253 W.

Below is the initial angular velocity and the maximum angular velocity as a function of the stroke rate



I do not observe a linear relationship between initial angular velocity and stroke rate. It is more like ω0 proportional to SR^1.2.
Also the calculated average stroke power does not have a cubed relationship with stroke rate. It is more like power proportional to SR^3.6.

It could well be that the discrepancy with your calculations is due to the time lag between the catch and the moment that handle speed matches the flywheel speed, which is part of my model.

[Nomath]

...
Since when handle speed matches the flywheel speed, P = Force * Vhandle and Vhandle = k * ω (k is the gearing constant and equals 70.69 radians/m; 1/k = 0.0141 m/radian) this simplifies to :
∆ω = ∆t * (F/k - C * ω²)/J ; C is the drag coefficient ; J is the moment of inertia.
...

Oops..! Please read: Vhandle = ω/k (k is the gearing constant and equals 70.69 radians/m). The last formula is therefore correct.
Also : drive length is 150 cm.

[Nomath]

....
It could well be that the discrepancy with your calculations is due to the time lag between the catch and the moment that handle speed matches the flywheel speed, which is part of my model.

I simplified my model by assuming that during the drive the handle speed always matches the flywheel speed, hence zero 'catch slack'.
The results confirm what jcross88 described above :

• The angular velocity of the flywheel increases linearly with the stroke rate. This holds both for the angular speed at the start and at the end of the drive.

• The maximum handle force increases proportional to the square of the stroke rate.

• The power in the stroke increases with the cube of the stroke rate.

The figure below shows the results of the simulation for drive length = 150 cm ; DF = 120 ; drive time:recovery time = 1:1.5.



However, I like to point out that the simplification neglects a reality. I measured the acceleration of the handle after the catch. It ranged between 5 m/s² for a take-it-easy recreational rower, to 20-25 m/s² for an elite rower in a 100 m sprint. Note: 10 m/s², i.e 1g , is the acceleration required for a vertical jump in the air.
Take the angular velocity of the flywheel at the start of the drive with a stroke rate of 20 : 80 rad/s. This corresponds to a handle speed of 1.1 m/s. At an acceleration of 5 m/s² it takes roughly 0.2 sec to catch up with the flywheel. The 'slack distance' is then about 12 cm. At a stroke rate of 30, the catch slack would increase to 0.3 sec ; the total drive time is only 0.8 sec. The slack distance would increase to about 29 cm, which is already a substantial part of the drive length.

[jcross88]

My model is the differential equation P = C * ω³ + ω * J * dω/dt , linearized for small time steps into ∆ω = ∆t * (P/(ω * J) - C/J * ω²).

I think I've fallen prey to the classic "the perfect is the enemy of the good".

The model I've been using has two major defects that yours does not: A proper force curve and the acceleration at the catch. I wanted a precise integral to describe the flywheel speed, except that becomes very difficult for arbitrarily defined force curves. So, instead I defined the force curve, took its average force over the drive, and treated drive force as constant for the sake of simplicity. I was hoping to experimentally determine what the effective moment of inertia is for the drive sprocket and model the first phase of the drive where the sprocket angular velocity is accelerating to the flywheel velocity.

Ultimately, to find the balance of ω₀ and watts that produced the sufficient drive length, I resorted to the libreoffice (linux excel) non-linear DEPS evolutionary solver algorithm. Even after the concessions I had to make to use my already imperfect integral, I was using numeric methods to find the solution anyway. Your numeric approach makes much more sense to me now, and I've adopted it wholesale.

I must say thank you again, Mathieu, for your attention to this topic. It's not only been intellectually satisfying to me, but has made a real, practical difference in how I'm rowing. Since asking the question I've begun using a 1:3 drive:recovery ratio for steady state, and I'm coming along on a DIY sensor for collecting session statistics. I've been very busy at work lately, but I hope to have some useful code within a month or so.

I've been rowing for three months now, and from the outset I was eager to understand the role of drag factor. After watching YouTube videos of rowing personalities trying to "explain" drag factor I decided the only way to really understand it is to understand the math behind it. I now feel I have a very strong grasp of the precise mathematical consequences of drag factor variation, with respect to the other adjustable components of a session.

In addition to some arduino code for collecting sessions statistics, I plan to implement your model in javascript. It could prove to be a very effective learning tool when the next rookie like me comes along asking these all-too-familiar questions.

[jamesg]

I was hoping to experimentally determine what the effective moment of inertia is for the drive sprocket and model the first phase of the drive where the sprocket angular velocity is accelerating to the flywheel velocity.

What happens before the bearing engages is governed mainly by our body mass and stretcher force.

Ergdata shows me average handle force 35-40 kg, peak 60, my balance 82 kg, so me and my handle accelerate at not less than ½g. Chain speed shows 1.8m/s, so if V²=2as, s = 1.8²/10 = 30 cm maximum slack.