cII watts/splits conversion?
Marc Gwadz
to splits/500m on the concept II erg?
thanks for any help.
marc gwadz
Alex Ozonoff
mg...@merhaba.cc.columbia.edu (Marc Gwadz) wrote:
I don't think there is a direct linear correlation between the two,
although they are definitely closely related. Looking at team results for
erg tests, average splits may be very close or identical, but watts tend to
vary much more (but not a lot). Can anyone else back me up on this one?
--
Al Ozonoff, Boston University, aozo...@bu.edu
John D. Blouch
The important thing to remember is that watts are a measure of power.
There is therefore no direct correlation between splits and watts.
Basically, if someone rows inefficiently, it will take them more power
to row the same distance. The monitor may account for this slightly when
computing the splits, but it does not account for it completely.
JB
Y150-95
Hugh Ainsley
conversions -it ++appears++ (and I need a magnifying glass for this) to
come from the "AMERICAN <Stride-Chart><Slide-Chart>(?) Corp., Wheeton,
IL?0187" the first number of the zip is a blob - 0 or 8 or 3 or something!
It might be worth phoning Concept - I guess they would have suppplied the
figures - if you find anything could you pass on phone or fax numbers - I
could do with another (better) copy!
hugh
Daniel Martin
A few times I have done two pieces on the erg and my meters rowed are close
to identical but the total watts were off by a considerable amount. Actually,
sometimes the piece where I pulled more meters (greater splits/500) had less
total watts. My guess is that there is a problem in the conversion algorithm, although I don't the internals of how the performance monitor does this.
Does anyone have any ideas as to why this happens?
--
---------------------------------------------------------------
| email: dma...@bnr.ca | The opinions expressed are mine, |
| Bell Northern Research | all mine, and are not necessarily |
| Ottawa, Ontario | those of my employer. |
---------------------------------------------------------------
() () 1996 Canadian LWT 2X
______________,__{]--,__{]--,_____________ - The race begins
`------------------/------/--------------'
~~~~~~~~~~~~~~~~~~~~~~/~~~~~~/~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~[~~~~~~[~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Christopher J. Russo
>In article <2iaa3o$g...@apakabar.cc.columbia.edu>,
>mg...@merhaba.cc.columbia.edu (Marc Gwadz) wrote:
>
>> does anyone have a conversion formula to convert watts
>> to splits/500m on the concept II erg?
>>
>> thanks for any help.
>>
>> marc gwadz
meter to convert between calories, erg splits, and watts, and some other
form of power. I got mine at the registration for the Head of the
Charles last year, so you might try contacting CII themselves. If
you're looking for a formula, I would try just fitting a regression on a
spreadsheet that you get from the gauge. I can't imagine that it would
be anything very complicated.
Christopher Russo
Fluid Mechanics Lab
cru...@jade.tufts.edu
ke...@u.washington.edu
The linear fit won't come close. For boats moving through fluids
Power required varies as v^3.
Try this formula:
Power = 1522/(time)^3
where Power is in watts
time is in minutes.
eg.
time for 500m is 1:40 = 1.67 minutes
Power = 1522/(1.67)^3 = 326 watts
This comes pretty close to the CII calculator...and the results I've
measured with my CII b ergometer.
Remember: It takes 8 times more power to row 2 times faster!!
Michael Reed
John D. Blouch <blo...@minerva.cis.yale.edu> wrote:
>In article <2iaa3o$g...@apakabar.cc.columbia.edu> mg...@merhaba.cc.columbia.edu (Marc Gwadz) writes:
>>does anyone have a conversion formula to convert watts
>>to splits/500m on the concept II erg?
>>
>>thanks for any help.
>>
>>marc gwadz
>
>The important thing to remember is that watts are a measure of power.
>There is therefore no direct correlation between splits and watts.
Hmmm...from what I've heard about the CII performance meter, the
way it works is:
a) Compute the watts output. This can be done directly from the
mass of the wheel, and the speed at which it is spinning. Also,
the rate of decelleration (on the recovery) is used to allow the
monitor to self-calibrate. This allows the machine to give
consistent results at different altitudes, and different states
of bearing wear.
NOTE: Although it may be hard for many of you to believe, this
also means that no matter if the vents are open or closed, whether
you use the small or large sprocket, IT TAKES THE SAME AMOUNT OF
ENERGY TO PRODUCE THE SAME SPLITS!!! Maybe one setting is easier
for you _personally_, due to the mechanics of you body, but that
is a seperate issue!
b) From the watts. the Kcal are computed directly.
c) From the watts, a (nonlinear) function is used to compute the
/500m. There is, therefore, a direct correlation between splits
and watts. It may not be obvious, it almost certainly isn't linear,
and it pobably does not correlate to f(watts) = x/500m on the
water, but it _is_ a direct correlation...
>Basically, if someone rows inefficiently, it will take them more power
>to row the same distance.
Yes...and their splits will be higher...
> The monitor may account for this slightly when
>computing the splits, but it does not account for it completely.
>
So if watts and splits are _not_ correlated, how are they measured? As
far as I can tell, the machine only senses 1 thing: wheel speed. From
changes in that, stroke rating and watts are calculated.
>JB
>Y150-95
>
>
Michael Reed
N.Y.A.C. Rowing
ke...@u.washington.edu
>In article <1994Jan29.0...@news.yale.edu>,
>John D. Blouch <blo...@minerva.cis.yale.edu> wrote:
>>In article <2iaa3o$g...@apakabar.cc.columbia.edu> mg...@merhaba.cc.columbia.edu (Marc Gwadz) writes:
>>>does anyone have a conversion formula to convert watts
>>>to splits/500m on the concept II erg?
>>>
>>>thanks for any help.
>>>
>>>marc gwadz
For a moving boat, the power dissipation varies as v^3. This is also
true of a wheel spinning in air. Both are dissipating energy by
generating turbulence and fluid motion.
Since v = dist/time, Power varies as 1/time^3. All we need is the
constant to make the conversion. For 500 m pieces measured in decimal
minutes,
Power (watts)= 1522/time(minutes)^3
eg. time =1:40 = 1.67 minutes
Power = 1522/[1.67 x 1.67 x 1.67] = 327 watts.
eg. 2. time = 2:00 = 2.00 minutes
Power = 1522/[2 x 2 x 2] = 190 watts.
This pretty nearly agrees with the CII slide rule calculator and agrees
with my own measurements with the CII B ergometer.
Note that a 20 second change in split acounts for a huge difference in power.
the power of 3 of velocity is the reason for this. As I had mentioned
earlier, going 2 times faster requires 8 times the power. In the more
usual need for evaluation....1% greater speed requires 3% more power.
The below statement is clearly incorrect.
>>
>>The important thing to remember is that watts are a measure of power.
>>There is therefore no direct correlation between splits and watts.
>Hmmm...from what I've heard about the CII performance meter, the
>way it works is:
> a) Compute the watts output. This can be done directly from the
> mass of the wheel, and the speed at which it is spinning. Also,
> the rate of decelleration (on the recovery) is used to allow the
> monitor to self-calibrate. This allows the machine to give
> consistent results at different altitudes, and different states
> of bearing wear.
> NOTE: Although it may be hard for many of you to believe, this
> also means that no matter if the vents are open or closed, whether
> you use the small or large sprocket, IT TAKES THE SAME AMOUNT OF
> ENERGY TO PRODUCE THE SAME SPLITS!!! Maybe one setting is easier
> for you _personally_, due to the mechanics of you body, but that
> is a seperate issue!
If the above is true, CII can clearly design the constants in their
computer to calculate the power dissipation of the wheel by how quickly
the wheel slows down. They can then use the v^3 formula to calculate
the power for any other speed. In fact, the power dissipation is the
first thing for the computer to come out, the v is derivative and, of
course, so is the splits for 500.
> b) From the watts. the Kcal are computed directly.
> c) From the watts, a (nonlinear) function is used to compute the
> /500m. There is, therefore, a direct correlation between splits
> and watts. It may not be obvious, it almost certainly isn't linear,
> and it pobably does not correlate to f(watts) = x/500m on the
> water, but it _is_ a direct correlation...
>>Basically, if someone rows inefficiently, it will take them more power
>>to row the same distance.
>Yes...and their splits will be higher...
>> The monitor may account for this slightly when
>>computing the splits, but it does not account for it completely.
>>
>So if watts and splits are _not_ correlated, how are they measured? As
>far as I can tell, the machine only senses 1 thing: wheel speed. From
>changes in that, stroke rating and watts are calculated.
CII has approximated the power that a 1X racing shell dissipates at
racing speed to calculate the splits for 500m. The splits for team boats
will not be close to the CII splits. However, the ergometer only
measures the power output of the rower going into mechanical motion of
the external device [forget your body for moment]. When you scull a 1X,
the mechanical output of your sculling motion goes into many things:
a) propelling the shell [if it were to proceed smoothly and straight
through the water].
b) stirring up the water with the sculls.
c) stirring up the water with the shell with extraneous motion.
Hopefully, most of it goes into (a). [Of course you get a shell where
this is minimized.]
Because we must get force on the oars, the reaction force stirs the water
up. The best rowers will have small and quiet puddles so that the energy
lost here is minimized.
The extra motion of the shell fore and aft, sideways and up and down will
dissipate a lot of energy. The best rowers will be very smooth and minimize
this as well.
Clearly the art in rowing is to minimize b and c.
I haven't yet addressed the question of the efficiency of the body
motion. You can fire your antagonists a lot and use up a lot of power
that doesn't even come out in an ergometer or a boat's motion!!
Enough for now!!
Sincerely,
Ken Young
U. of Washington
Some of it must go into (b) but small quiet puddles will minimize this.
That the shell varies in speed by +/- 30% in a rowing cycle and also
bounces up and down dissipates quite a lot of energy. T
ke...@u.washington.edu
> stuff deleted
>>A few times I have done two pieces on the erg and my meters rowed are close
>>to identical but the total watts were off by a considerable amount. Actually,
>>sometimes the piece where I pulled more meters (greater splits/500) had less
>>total watts. My guess is that there is a problem in the conversion algorithm, although I don't the internals of how the performance monitor does this.
>>
>>Does anyone have any ideas as to why this happens?
>>
See below
Power = 1522/(T)^3 = d(Energy)/d(t)
where T is the split time for 500m in minutes.
I think that you are addressing the energy for a given piece.
To do this you have to integrate the power over the time interval for the
piece. This is complicated if the power varies with time ie if you pull
a piece with some fast some slow, etc.
Energy = integral(power(t) dt). You would have to record the power
level as a function of time!! and then integrate this.
Very little energy is required for slow pieces. For a given piece at a
fixed time, the energy is minimized if the power is constant or the split
times are constant. This is true for boats and the ergometer.
Believe it or not, the CII computer is keeping track of this and is
integrating the power to give the energy used at the end.
Beware:
In the CII ergometer, the power in watts is the mechanical output of the
rower. The power in (calories/hour) is the total power of the rower
which includes the heat produced and the mechanical output. Typically,
mechanical power/(total power) = 0.2
[CII refines this relationship in their side rule calculator.]
So when you are rowing at 300 watts mechanical output, you are producing
total power of 1500 watts. You are eating up your stored energy at 1500
watts rate. The 1200 watts comes off as heat which is why you can row in
cold weather with only a tank top!!
Remember: Power is the time rate use of Energy.
Ken Young
>>--
>> ---------------------------------------------------------------
>>| email: dma...@bnr.ca | The opinions expressed are mine, |
>>| Bell Northern Research | all mine, and are not necessarily |
>>| Ottawa, Ontario | those of my employer. |
>> ---------------------------------------------------------------
>>
>> () () 1996 Canadian LWT 2X
>> ______________,__{]--,__{]--,_____________ - The race begins
>> `------------------/------/--------------'
>> ~~~~~~~~~~~~~~~~~~~~~~/~~~~~~/~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>> ~~~~~~~~~~~~~~~~~~~~[~~~~~~[~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>if one were to row a set distance and maintained even splits,he
>or she could then look directly at the average watts number
>to determine the watts that have arbitrarily (i assume)
>been calculated to coorelated with a certain pace/500m.
>if you were to repeat the same piece, go the same distance
>for the same amount of time but did so in an inconsistent
>fashion (go out hard and die, or start easy and sprint),
>you would likely find that the watts are higher.
>i.e. it ismore efficient to maintain even splits.
>this is a direct result of thge fact that resistence due to
>air increase in a non-linear fashion (force increases with
>the square of the speed). one good reason not
>to do power tens in an erg test.
[Note Power = F x v ~ v^2 x v = v^3]
and Power(watts) = 1522/(split time)^3.
>so, my original post was not ask whether or not watts
>and splits are related, but to determine the actual
>formula used to make the conversion.
>marc gwadz.
Marc Gwadz
>A few times I have done two pieces on the erg and my meters rowed are close
>to identical but the total watts were off by a considerable amount. Actually,
>sometimes the piece where I pulled more meters (greater splits/500) had less
>total watts. My guess is that there is a problem in the conversion algorithm, although I don't the internals of how the performance monitor does this.
>
>Does anyone have any ideas as to why this happens?
>
>--
> ---------------------------------------------------------------
>| email: dma...@bnr.ca | The opinions expressed are mine, |
>| Bell Northern Research | all mine, and are not necessarily |
>| Ottawa, Ontario | those of my employer. |
> ---------------------------------------------------------------
>
> () () 1996 Canadian LWT 2X
> ______________,__{]--,__{]--,_____________ - The race begins
> `------------------/------/--------------'
> ~~~~~~~~~~~~~~~~~~~~~~/~~~~~~/~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> ~~~~~~~~~~~~~~~~~~~~[~~~~~~[~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
if one were to row a set distance and maintained even splits,he
or she could then look directly at the average watts number
to determine the watts that have arbitrarily (i assume)
been calculated to coorelated with a certain pace/500m.
if you were to repeat the same piece, go the same distance
for the same amount of time but did so in an inconsistent
fashion (go out hard and die, or start easy and sprint),
you would likely find that the watts are higher.
i.e. it ismore efficient to maintain even splits.
this is a direct result of thge fact that resistence due to
air increase in a non-linear fashion (force increases with
the square of the speed). one good reason not
to do power tens in an erg test.
so, my original post was not ask whether or not watts
Joris Trooster
MG> does anyone have a conversion formula to convert watts
MG> to splits/500m on the concept II erg?
Let's try a more scientific approach to this subject:
One can measure the time interval between the pulses from
the wheel (three in one round), so the wheel speed is
calculated by:
v = dx/dt
The acceleration of the wheel can be calculated by:
a = dv/dt = d^2x/dt^2
The force can be calculated by multiplying the acceleration
with a constant value, c. This value depends on the
wheel mass. The force calculated now is composed of the
force from the rower and the force from the friction
(resistance). So:
Fmeasure = Frower + Fresistance = c * a
The force from the friction is:
Fresistance = rho * v^2
Where rho is the resistance-factor. During the recover Frower = 0:
Fresistance = Fmeasure
rho * v^2 = c * a
The resistance-factor can thus be calculated during the recover:
rho = (c * a) / v^2
During the stroke, one can calculate the actual force from the rower:
Frower = Fmeasure - Fresistance = (c * a) - (rho * v^2)
Once the force is known, one can integrate the force over the
stroke length to get the work:
A = Integr(Frower) over str.length
The power the rower gives in one stroke is:
P = A/t
Where t is the total stroke length.
How can we calculate the split-time and rowed-meters from this?
Well, one problem is that the wind-resistance (v^2) isn't equal
to the water resistance (v^3). A good model for a boat in the
water is:
+ ------- -------
Frower ---- o -----| 1/m |------| 1/s |---------- v_boat
-| ------- ------- |
| |
| ------- ---------- |
-----| rho2 |-----| v_boat^3 |-----
------- ----------
The boat speed is calculated by integrating the force, and
a feedback loop for the water resistance. The boat speed
depends thus on the force from the rower, the boat mass (m)
and the water resistance (rho2). One can calculate the distance-rowed
by integrating the boat speed:
-------
v_boat ------| 1/s |----- x_boat
-------
So there is NO direct correlation between the power and boat-speed,
but it is a step function. For example, with constant force the
speed increases like a step function:
|
|
| . . . .
| .
v | .
| .
| .
----------------------
t -->
So, my conclusion is that there is no direct correlation between
power and split-time. In steady state condition, however, there is.
Besides, the split-time is correlated to the force, rather than the
power.
I implemented the first part on a computer. Calculation of the
power comes very close to the power calculated by the performance
monitor. I plot the force (Frower !) against the stroke-length
on the screen to get the force-curves (looks great ;-)
I tried to implement the last part, to calculate the split-time
AND the meters-rowed. But I didn't figure out what to use for
boat-mass (m) and boat-friction, resistance (rho2). I'm pretty
sure I use the right formulas, and that the split-time isn't
calculated by a simple formula like const/v^3. The formula I
presented here is also capable of calculating the distance
rowed _during_ the stroke (so no update _after_ a stroke).
Joris (troo...@jjt.iaf.nl)
ke...@u.washington.edu
>ke...@u.washington.edu () writes:
>>CII has approximated the power that a 1X racing shell dissipates at
>>racing speed to calculate the splits for 500m. The splits for team boats
>>will not be close to the CII splits.
> I don't pay attention to watts, except to measure watts per pound,
>since as a coxswain or very lightweight rower, I'm curious to know if
>I'm pulling my weight.
Since the resistance of a shell mostly varies as wetted area and not the
displacement, the proper measure of potential on-the-water speed is not
power/weight. The wetted area approximately varies as weight^(2/3).
So: Power = force x velocity ~ wetted area x v^2 x v ~ weight^(2/3) x v^3
and v^3 ~ Power/(weight)^(2/3)
and v ~ [power/(weight)^(2/3)]^(1/3)
This gives a correction factor for erg scores which closely corresponds
to the one used at some NE universitys.
Editorial.
I think that we should use this for ergomania and not have weight classes.
> I believe the splits on the C-II erg correspond closely to a coxed
>four on the water, not a single. For example, it may be twenty years
>before a heavyweight man rows his single to 2K victory in six minutes.
I quite agree with you in real rowed boats. In my earlier note, I was
referring to the power requirements of a 1X in a towing tank where there
are no power loss factors due to scull inefficiencies and extraneous
motions of the shell.
The story is this. Van Dusen has measured the resistance of 1X in steady
state conditions. At 4.8 m/s the measured resistance is about 15 lb.
The power dissipated is Power = force x velocity where we must use
consistent units. 15lb = 67 Newtons.
Power = 67 x 4.8 = 321 watts.
This is approximately the relation stated by the CII ergometer computer.
A 1:40 500m piece needs about 330 watts. Try it on your ergometer.
So if you could row a boat with 100% propulsive efficiency and keep the
boat at a constant velocity...then you could achieve the impossible!!
I have correlated erg times with race times for the rowers in our club.
The scatter plot of these two times shows a big dispersion but there is a
positive correlation. The rowers who have the longer split times usually
have race time which is close. Rowers with short split times have race
times which are much longer. The rowers are all dedicated competitive
master rowers. I take it that the less powerful try to be more cunning
and more artful in their rowing. I'd like to hear from others about
their experience in this. [An alternative explanation for the phenomena
is that it's easier to row cleanly if you're rowing kinda slowly!!]
Ken Young
Univ. of Washington.
>Geoffrey
>--
>Geoffrey S. Knauth, <g...@marble.com> BCS-NeXT, LPF
>Marble Associates, Inc., (617) 891-5555 CRASH-B, Cambridge BC
ke...@u.washington.edu
>Hallo Marc,
> MG> does anyone have a conversion formula to convert watts
> MG> to splits/500m on the concept II erg?
>Let's try a more scientific approach to this subject:
I thought that we were all being scientific!!
> v = dx/dt
> P = A/t
Note that this is just P = F x X/t = F x v
which is what I had earlier stated.
>Where t is the total stroke length.
t must be stroke time (I hope)
>How can we calculate the split-time and rowed-meters from this?
>Well, one problem is that the wind-resistance (v^2) isn't equal
>to the water resistance (v^3). A good model for a boat in the
>water is:
I'm sorry...Water resistance for a boat moving in the Reynolds number
range of rowing shells varies as v^2 NOT v^3. Look in any
book on hydrodynamics.
In any case with power = constant x v^n and v = distance/time
with distance = 500 m and time is the split time, then
power = constant /(split time)^n
where n = 3. and the constant is 1522.
Power = 1522/t^3. The 1522 is chosen for t measured in minutes.
See my remarks to Geoff regarding realism to real rowed boats.
> + -------
------- > Frower ---- o -----| 1/m |------| 1/s |---------- v_boat
> -| ------- ------- |
> | |
> | ------- ---------- |
> -----| rho2 |-----| v_boat^3 |-----
> ------- ----------
>The boat speed is calculated by integrating the force, and
>a feedback loop for the water resistance. The boat speed
>depends thus on the force from the rower, the boat mass (m)
>and the water resistance (rho2). One can calculate the distance-rowed
>by integrating the boat speed:
> -------
> v_boat ------| 1/s |----- x_boat
> -------
If you mean that you apply power some of the time and zero power during
recovery...I agree. The idea of power in rowing is the average power for
a stroke cycle which is a legitimate idea.
>So there is NO direct correlation between the power and boat-speed,
>but it is a step function. For example, with constant force the
>speed increases like a step function:
> |
> |
> | . . . .
> | .
> v | .
> | .
> | .
> ----------------------
> t -->
Of course, you will accelerate a boat from rest but then you would obtain
a constant average speed for a fixed average power applied.
>So, my conclusion is that there is no direct correlation between
>power and split-time. In steady state condition, however, there is.
>Besides, the split-time is correlated to the force, rather than the
>power.
Power and force are correlated!! P = F x v.
>I implemented the first part on a computer. Calculation of the
>power comes very close to the power calculated by the performance
>monitor. I plot the force (Frower !) against the stroke-length
>on the screen to get the force-curves (looks great ;-)
>I tried to implement the last part, to calculate the split-time
>AND the meters-rowed. But I didn't figure out what to use for
>boat-mass (m) and boat-friction, resistance (rho2). I'm pretty
>sure I use the right formulas, and that the split-time isn't
>calculated by a simple formula like const/v^3. The formula I
>presented here is also capable of calculating the distance
>rowed _during_ the stroke (so no update _after_ a stroke).
There is much direct evidence for
Power = const x v^3.
The best evidence for this comes from towed tank tests. For rowed boats
the best evidence comes from correlating Power, velocity and weight for
given boats. This can be done for looking at the speed of team boats of
various sizes and team boats with similar talent and the power and weight
of the crews. The nicest scaling law is the one which relates the speed
of a boat to the size of the crew. n = no. of rowers.
v ~ n^(1/9). This fits the record speeds of crews from n=1 to 8. The
derivation uses Power = const x v^3.
Ken Young
Univ. of Washington
>Joris (troo...@jjt.iaf.nl)
fee...@nova.enet.dec.com
|>For all you physicists, would-be physicists, scientists and just plain
|>curious.
|>
|>Mike Vespoli has begun tank testing his hulls using a repeated (30x/min)
|>acceleration and deceleration, to more accurately represent the forces
|>which a racing shell experiences. Traditional tank testing (if done at
|>all by manufacurers) has relied on constant velocity towing, as far as I
|>know.
|>
|>The experiments have been done by SAIC (science Applications
|>International Corporation) in California. Those interested might want to
|>talk with Mike or Dave Trump, to find out what kinds of differences were
|>noted in steady vs. periodic motion.
|>
For what it is worth tank testing done by VanDusen has also been done in this
manner. In addition the actual hulls were used (not scaled down models)
which means actual results do not have to be extrapolated.
-Jay Feenan
Robert Eikel
>In the CII ergometer, the power in watts is the mechanical output of the
>rower. The power in (calories/hour) is the total power of the rower
>which includes the heat produced and the mechanical output. Typically,
>mechanical power/(total power) = 0.2
>[CII refines this relationship in their side rule calculator.]
I thought human muscles were approx. 38% efficient. (?)
Robert
ke...@u.washington.edu
>not be surprised if rowing were less efficient (20%) due to the time
>and effort spent on the recovery, which does not deliver power to the
>oar handle.
In cycling, one foot is recoverying while the other is driving. In any
cyclical human motion...there is always recovery so the rowing recovery
isn't so special.
There is a basic stored energy to mechanical work efficiency that doesn't
include the recovery. You can think of the efficiency of just one
contraction of one muscle. This work output/energy used = 0.25. This
is discussed in the references :
TA McMahon, Muscles, Reflexes and Locomotion. Princeton Press
Whitt and Wilson, Bicycling Science.
They both have extensive references to original sources.
Running is the sport with the higher efficiency. Some speculate that
this is so because of the use of the recovery of the energy stored in the
legs elastically.
>I suppose that an individual muscle could be 38% efficient, although
>that would imply quite a lot of internal dissipation within the joints,
>ligaments, tendons, etc.
The ligaments and tendons tend to be elastic and so don't dissipate
energy if used properly.
which couple that power to the pedals or oar
>handle. In defining efficiency, one must carefully define the
>boundaries of the system considered.
Let's separate body efficiency with the efficiency of rowing. The 25% or
so that we have been discussing is the body's efficiency in converting
stored energy to mechanical work in the limbs before other losses like
oar slippage is taken into account.
If you start at the boat house,
>and finish at the boat house, then you have heated up the river and
>atmosphere, but done no net work, and so the efficiency is zero. When
>you run down a hill, you do negative net mechanical work. :)
>Does anybody have a reference for a rigorous discussion of the
>mechanical efficiency of rowing or cycling, in particular, the
>variation of efficiency as a function of cadence and resistance?
Whitt and Wilson have a collection of power curves vs pace for cyclists
at various power levels.
At about 300 watts, the power peaks at about 95 rpm which corresponds to
a foot speed of 2.5m/sec. Using foot speed instead of rpm allows us to
compare bicycling with rowing.
My calculation of foot speed for an elite rower at 4.7m/s using 300 cm
sculls is about2.5 m/sec also. The foot speed is the speed of one's foot
relative to one's butt....same as in cycling.
In any case, this is an experimental science. We need a machine for
rowing that allows us to adjust the resistance so that we can find the
foot speed for which we develop the maximum efficiency for a given power
output. If the CII model B had more gears, I could carry this out.
Perhaps the model C will allow this. To prove the case we need to not
just approach the peak but to go over the top. Has somebody out there
tried this out?
Sincerely,