Units at force plot

[PeterF]

Hello,

when I use the method or function newforceplot (int) I get a array of numbers like this:
0,0,151,151,154,154,154,156,156,160,160,160,160,159,159,162,162,163,161,161,163,163,158,154,148,148,142,133,133,127,118,118,109,100,93,93,84,75,75,63,54,49,49,....

When I put this data into a graphic it looks similar to the force plot on the erg monitor. But what kind of data are this numbers? Any ideas?

Kind regards
Peter

[Citroen]

At a guess it's giving you Kg force values, so divide by 10 and I think you'll get Newton-metres (Nm).

Strictly that should be divide by 9.80665 (standard gravity) to convert a Kg force value into a Nm.

[jamesg]

Those numbers look too big to be kg and too small to be Newtons. Maybe pounds, tho' of course they could be just arbitrary.

You can guesstimate it by knowing your Watts, stroke length and rating, then using 1 W = 1 Nm/s.

Example: if you pulled 200W at rating 20, net stroke length 1.2 m, then from:

(Average Force in N x Net Length in m x Rate)/60 = Nm/s = Watts, so Force in N = Watts * 60 / (Length x Rate),

the average force would be 200 * 60 / 1.2 * 20 = 500N or roughly 50 kg = 110 lb, which at first glance looks near the average of the numbers you show.

[Citroen]

the average force would be 200 * 60 / 1.2 * 20 = 500N or roughly 50 kg = 110 lb, which at first glance looks near the average of the numbers you show.

Why do the American still use those stupid imperial units (that we have mostly let go of), have they not learned that SI units are more sane?

[jamesg]

I read that a Mars mission was lost because the weight of a part made in Europe was thought expressed in lb, while it was in fact in kg. Ooops. Money up the spout, or rocket?

[gregsmith01748]

Hi,

I remembered that a few rowing biomechanics newsletters had data on handle force.

This one: http://www.biorow.com/RBN_en_2011_files ... News01.pdf

Shows handle force plots for both static and dynamic ergs that are in newtons. I think by comparing these numbers, along with the generated watts to your data you should be able to come up with a calibration factor.

[PeterF]

Thanks for your answers.
I'm still on collecting same data from the erg.
But Greg, I assume you could be right, the data for the y-axis are the force in Newton, and the y-axis are the stroke length (the dimension is still open). When I did some quick calculation it could possible.

Beispiel Kraftkurve C2.jpg (28.24 KiB) Viewed 19045 times

[Citroen]

If you can give us time, distance and stroke rate we can calculate avg. distance per stroke and that helps to resolve the distance * force part of the equation.

[Snail Space]

I remember collecting similar data into Excel using the PMI interface and VBA. How have you collected your data? I guess from the forum name that you hhave used the SDK and some fiendishly clever C programming.

I didn't think beyond them being arbitrary numbers but I was intrigued by the time increments between the data points. It was a few years ago, so I can't remember whether they were at fixed time intervals (I don't think so) or if the intervals were a function of stroke duration (and therefore vary with stroke rate and drive:recovery ratio).

Cheers,
Dave.

[kingpeterking]

Hi All - reopening a very old post.

I am gathering force curve data from the BT interface - this looks fine
35 53 53 65 69 74 74 85 85 90 95 99 104 110 110 113 114 117 122 124 124 127 123 126 126 130 123 123 126 127 130 129 122 119 118 108 102 95 77 63 62 52 44 31 19 17 6

I am also getting the stats from other commands like STROKEDATACHARACTERISTICS and ADDITIONAL STROKE CHARACTERISTICS
For this stroke the pertinent pieces of data are
Drive Length = 1.48m
Drive Time = 0.74 secs
Recovery Time = 1.93 secs
Peak Drive Force = 130 lbf
Average Driver Force = 92.9 lbf
Work Per Stroke = 544.1 J
Stroke rate = 21
Speed = 4.067 ms-1 (2:03 split equivalent to 186 W)

The Peak Drive Force and Average Drive Force all match close enough for what I want to do.
BUT
When I calculate the Work Done using the area under the curve method I get a completely different answer.
I take each force curve reading - covert from lbf to N using 4.48 as the conversion
I work out the distance between readings based on having got 47 readings back from the force curve in 1.48m = 3.1cm per reading
This gives me a total of 607 Joules
Quite different to the 544.1 reported in Stroke Data Characteristics.

Then linking this back to the Watts figure isn't working.
I am using Power = Work x Time where Time = 21/60 = 0.35
Which gives 212 W rather than the 186 calculated from the Speed.

Any help on any of this is gratefully received.


Pete

[Nomath]

I think that your assumption that the 47 readings correspond to a drive length of 1.48 meter is wrong. Hence the calculated distance of 3.1 cm per reading.

I assume that you have a model D. It has a 1/4 in chain, hence a chain pitch of 6.35 mm, that runs over a 14 teeth sprocket. A chain displacement of 8.89 cm causes 1 revolution of the flywheel. A chain displacement of 148 cm thus causes 16.6 flywheel turns. There are 6 magnets on the flywheel, so there are about 100 sensor readings of the flywheel speed in each drive.

Now the question is how C2 measures the force. Mathematically, the force is determined by the following equation
F = k*DF*ω + k*J*dω/dt
ω is the angular velocity (unit : 1/s) ; DF is the drag factor (in units kg m²) ; J is the moment of inertia of the flywheel (0.1 kg m²) ; k is the transfer constant (70.68 radians/meter)
The problem is in obtaining a reliable value for the derivative dω/dt , because the ω values are already quite noisy. There are various ways to do this, but usually several data points are used to interpolate the derivative.

Looking at the readings, I find it likely that C2 has also omitted the readings at low forces, say below 50-100N. Your force curve starts at about 200N. Real handle force measurements, see typical examples here, show that the force curve starts near zero and increases gradually, not abruptly.

[JaapvanE]

There are 6 magnets on the flywheel, so there are about 100 sensor readings of the flywheel speed in each drive.

That is what I'm seeing too. 80 to a 100 datapoints on a 6 magnet RowErg. The force curve presented has 47 datapoints, so it probably is a model C or older D (3 magnets).

Now the question is how C2 measures the force. Mathematically, the force is determined by the following equation
F = k*DF*ω + k*J*dω/dt
ω is the angular velocity (unit : 1/s) ; DF is the drag factor (in units kg m²) ; J is the moment of inertia of the flywheel (0.1 kg m²) ; k is the transfer constant (70.68 radians/meter)

The problem is in obtaining a reliable value for the derivative dω/dt , because the ω values are already quite noisy. There are various ways to do this, but usually several data points are used to interpolate the derivative.

Another approach is to use:
Torque = Df * ω^2 + α * Inertia
Where α is the angular acceleration. And as you point out: it is difficult to determine ω and α in a robust (noisefree) manner.

And then
Force = Torque / Arm

Where the arm is the radius of the driving sprocket in meters.

This is what we use in OpenRowingMonitor. I actually tried to reconcile both approaches as theoretically they should lead to the same results. Issue is that when you calculate the energy and average it across the stroke, you get a different result than the "normal" power calculation. In part because the average P = Df * average ω^3 is an approximation, but it doesn't explain all variation.

Looking at the readings, I find it likely that C2 has also omitted the readings at low forces, say below 50-100N. Your force curve starts at about 200N. Real handle force measurements, see typical examples here, show that the force curve starts near zero and increases gradually, not abruptly.

It could be noise filtering, or it could be stroke detection. C2 defines a stroke as an accelerating flyhweel, which isn't 100% correct as it excludes forces lower than the drag force.

[Nomath]

.. The force curve presented has 47 datapoints, so it probably is a model C or older D (3 magnets).

That could be it. At least it brings us closer to the reported Work per stroke of 544.1 J
On these older models successive readings correspond to a chain travel of 8.89/3 = 2.96 cm.
The integral under the force curve then becomes 571 J.
Not yet the reported 544.1 J but a lot closer.
Check the numbers!

[Nomath]

..Then linking this back to the Watts figure isn't working.
I am using Power = Work x Time where Time = 21/60 = 0.35
Which gives 212 W rather than the 186 calculated from the Speed.

Power=Work/Time ! Hence 544.1/(0.74+1.93) = 203.8 W, not 212W
If your split of 2:03 is correct, the corresponding power is 2.8 x (500/123)³ = 188 W, not 186W
Again, check your numbers.

[kingpeterking]

Thanks everyone - this has been super helpful. The differences I have now are a few percentage points rather than orders of magnitude.

On power- I can derive this in three different ways and get three different numbers
Joules / (drive + recovery ) = 203W
Joules / (60 / Stroke rate) = 190W
2.8 * ms-1 pow 3 = 188W

I don't think that drive + recovery is a reliable way to calculate this as over the whole piece there are lots of 'gaps' as the total drive time + total recover time does not equal the length of this piece. The other two numbers are close enough for what I need.

On Work - the force curve has some interesting characteristics.
1. It doesn't start at zero - there is clearly missing data on the left
2. In every force curve (from 800+) strokes in this piece there are repeated values - 53,53 74,74 etc.. My hypothesis is that the PM algorithm is retrospectively smearing the Joules across the curve and converting to Ints and hence losing some precision.
3. The total work calculated by the curve is 110% greater than the work reported in the characteristics report

I have a long flight tomorrow, so will try to reconcile this - I will try to smooth the curve using a spline, add the missing left numbers, and then see if there is a consistent way to scale the curve so the area under it matches the characteristics data.

Thanks again