# Muscle power vs contraction speed

Figure 1 Muscle power vs velocity. Idealized graph, from Wikipedia

Figure 1 shows the muscle force and power vs contraction velocity. It’s a very important graph because it marks where classical mechanics stops and exercise physiology starts. Note that the graph is idealized, with no absolute values on the axis. Each athlete will have a slightly different graph, which will probably also change as a result of training. Even the position of the peak vs maximum contraction speed depends on the person.

So what’s the impact on the gearing or rigging parameters of your boat? Suppose we want the rower to achieve his maximum power at each stroke at a given stroke rate, let’s say at mid-drive. We can then calculate the rower’s velocity  (relative to the hull) at mid-drive and estimate the duration of the stroke, for a given length on the slide. Then we have to adjust the gearing such that leg force needed to achieve this speed equals the maximum force from the rowers force-contraction velocity graph. Finally, we could put into the calculation our ideal catch angle and voilà, we can determine span, inboard, scull length, and through-the-work setting. I believe this is what Dr. Kleshnev is using in his rowing speed and rigging chart.

So how do you know how your power-contraction velocity graph looks? I guess there are three ways. The first one would be to design some direct measurements. A simple home setup could be on the rowing ergometer using different ratings and drag factors, trying to achieve maximum speed at a given time interval and recording stroke/recovery ratio and displayed average power. The second method would be to rely on intuition and set the rigging parameters such that the rower feels comfortable to achieve his ideal stroke rate rowing at “race power”.  The last method would be to do on-the-water boat speed measurements at different rigging parameters and ratings.

Figure 2. Average race stroke rate vs average boat speed for Lucerne World Cup 2010 Sunday Finals

Figure 2 shows the stroke rate vs boat speed for my favourite data set, the Lucerne World Cup 2010 Sunday Finals races. Clearly, there is a trend of increasing rating with increasing speed, where the Men’s eight rows at 39 spm and the women single winner rows at 34spm. There are, however, also a few outliers, most notably the women 2- at 38.54 spm and 4.7 m/s. Either they used a lighter gearing or they rowed a shorter stroke. I presume most coaches use quite conventional rigging parameters, which means there will not be a lot of variation between them. As these experienced crews know how fast they go and at which power they need to row to not burn out before reaching the finish line, they intuitively select an appropriate rating corresponding to their “ideal race power”.

Final question: Is it really necessary that the crew rows close to its optimum muscle contraction speed? If not, then rigging optimization really boils down to setting it to what “feels good for the crew”, or engage in time-consuming on-the-water trials, and hoping we can ignore other factors such as varying weather conditions, rowing technique, psychological factors, nutrition, cheating and what else?

# From ergometer to boat – validation 2

The total energy spent in a race of duration t can be fitted to a linear curve

$U = A + B t$

From my personal ergometer trials, I calculated the energy by reading the wattage from the monitor, adding the power needed for moving the rower up and down the slide and multiplying by the race duration. Indeed, the linear relation holds.

The parameter $A$ corresponds to some fixed energy storage in the body, while the parameter $B$ represents the power level at which the rower “can go on forever”.

Now I can calculate from my recent 1km race time in the single scull (slight headwind) my maximum power over that distance. Plugging this power, my weight and rigging parameters into the model, I indeed arrive at this time. So, for at least one rower, the relation between erg times and boat times is validated.

I will not publish the values of the data right now, as I have a race next week and hope to go faster than my theoretical limits. (Also, the values are not very impressive.)

# Rower Power per kilogram – model validation

Table 1: Average crew power (per rower) for Lucerne 2010 A finals (Sunday 11 July 2010)

Table 1 shows the average velocities of winning crews in the Lucerne Regatta finals, as rowed on Sunday 11 July 2010, a hot and sunny day with no wind. I took the average crew weight from the athletes’ weight on worldrowing.com and used the average stroke rate from the race result GPS data. I had to guess the rigging data so I based my values on recommended values from maxrigging.com, as given in table 2:

Table 2: Rigging numbers used to calculate results of table 1

For boat weights I used the FISA minimum weight requirements and I assumed the coxswain to have a weight of 55kg. Some of the lightweight athletes have not published their weight on the FISA pages, so for them I have taken the maximum average crew weight (70kg for men and 57kg for women). I do not have data for all disciplines, as some of the finals were run on Saturday. I have no clue about the weather on the Saturday, but it seemed there was some wind, as all times were on the slow side.

The model was used to estimate the average power (per average crew member) assuming a flat race. In reality, the average power may have been up to 10% higher if the crew rowed at different speeds in different parts of the race.

The calculated crew power amounted to 6.9±0.2 Watts/kg for the men and 5.4±0.2 Watts/kg for the women. Comparing this with data I found for cyclists, this looks very realistic. It places the world’s best rowing crews of 2010 in the same league as ‘world class’ cyclists in 2005. It is also encouraging to see that the watt/kg index seems to be independent of weight for these data.

I didn’t find any data for women, but women I presume the difference is (at least partly) explained by the difference in body fat percentage between men and women. I tried searching the internet for ‘men women power relationship’ but that yielded results from an entirely different field.

I’d be happy to find any better data. I have no easy access to academic literature and no background in sports physiology, so this is the best I can come up with.

In conclusion, the predictive power of my model tested on rowing world cup finalists seemed quite good.

 Discipline Weight Winner Crew weight (ave) Cox weight Boat weight (ave) average speed Average Stroke Rate average power Power/kg M8+ Open GER 87.4 55.0 96 6.05 38.94 610.97 6.99 M4x Open CRO 92.0 52 5.76 37.84 644.83 7.01 M4- Open GBR 96.0 50 5.63 39.40 643.92 6.71 M2x Open FRA 90.5 27 5.28 36.25 617.86 6.83 M2- Open NZL 91.0 27 5.19 38.48 613.78 6.74 M1x Open CZE 98.0 14 4.83 35.45 644.05 6.57 W8+ Open USA 82.0 55.0 96 5.37 36.67 408.93 4.99 W4x Open GBR 78.8 52 5.22 35.93 424.69 5.39 W2x Open GBR 79.5 27 4.87 35.52 447.10 5.62 W2- Open NZL 72.0 27 4.66 38.54 379.99 5.28 W1x Open BLR 82.0 14 4.43 34.20 437.92 5.34 M4- Light GBR 70.0 50 5.54 37.86 492.07 7.03 M2x Light NZL 70.0 27 5.24 38.16 511.34 7.30 W2x Light AUS 57.0 27 4.67 34.66 315.63 5.54

# Luzern

FISA has included a fantastic source of data on their world cup results website. For example for the recent Lucerne World Cup event, results can be viewed on this page. What’s even better, clicking on “GPS” reveals data on stroke frequency and average velocity on 50m intervals for each crew. For example, here. Finally, we can also find out the weight of each participant through his/her athlete’s page. For example, Ondrej Synek, world cup winner single sculls.

I watched all finals races as they were live broadcasted on Czech TV, so I know the weather was windstill and very very hot and sunny. The absence of wind makes these data ideal for comparison with a rowing model that does not take into account wind (although perhaps 5 m/s tailwind would have been even better), and the stroke frequency and crew weights are of course important input parameters.

The high temperature and strong sun will have had a negative influence on the athlete’s performance.

# Energy balance and efficiency

I thought I knew classical mechanics pretty well, but I must say I was pretty confused after reading the write-ups on rower energy dissipation by Atkinson and Van Holst. Well, this is how I calculate the energy balance. There are three energy sinks in the system:

1. Drag. Energy dissipated by fluid friction along the boat. ${\rm d} E_{\rm D} = F_D {\rm d} s = \alpha (v_t + v_b)^3 {\rm d} t$.
2. Energy dissipated by the slip of the blade through the water. ${\rm d} E_{\rm blade} = F_{\rm blade} v_{\rm blade} {\rm d} t$.
3. Some energy may be dissipated in the rower’s body when he or she slows down at the end of the stroke or the end of the recovery

To calculate the magnitude of energy dissipated the third sink, I follow the approach taken by Van Holst, as described here and here.

Figure 1

Figure 1 shows the result of a typical single stroke simulation. The system is in equilibrium, i.e. the hull speed at the start of the stroke is constant from stroke to stroke. The simulation was done for a single scull heavy rower rowing at an average velocity of 4.74 m/s, which corresponds to a 500m time of 1:45.5.

Following Van Holst, the force moving the mass of the rower with respect to the boat is given by

$F_q = F_{\rm foot} - F{\rm handle} = m_c (\ddot{x}+\ddot{y})$

The power supplied by this force is equal to $P_q = F_q \dot{y}$ and is plotted in figure 2 (cf figure 4.6 here):

Figure 2

Positive values mean power supplied to the system by the rower, while negative values mean power absorbed from the system by the rower. This power cannot be reused and is thus lost for propulsion. Atkinson argues that the absorption of kinetic energy is done by different muscle groups than the generation, and this is reflected here by assuming it takes 1W to absorb 1W of kinetic energy, as depicted in the right hand side of the figure. (See also the discussion by Van Holst about energy consumed while walking down the stairs). In the simplified example of Van Holst, indeed some dissipation in the rower occurs. In my simulation results in figure 2, I see hardly any power dissipation, which is in agreement with the simulation results of Van Holst in the realistic system but in contrast with Atkinson’s findings.  I guess a discussion on this is needed.

Figure 3

Figure 3 shows the energy invested into the system by the rower as a function of the time during the stroke. The blue line is the energy provided by $P_q$ (the area under the graph in figure 2). The black line, “puddle energy” is energy used to drag the blade through the water creating the puddle. The red line is the propulsive energy accelerating the boat at the oarlock, and the green line is the total energy.

Figure 4

Figure 4 shows the energy going into all the sinks (red line). Also depicted is the energy “temporarily stored” as kinetic energy (in excess of the system kinetic energy at the catch at t=0). The sum of these sinks plus storage should be equal to the energy delivered by the rower at each point in time. This is shown in figure 5:

Figure 5

The efficiency of the stroke is the ratio of energy  dissipated in overcoming the hull friction vs. the total energy dissipated in the system.

${\rm efficiency} = E_D/(E_D + E_{\rm blade} + E_{\rm rower})$

The average power exerted by the rower is given by the total energy divided by the duration of the stroke $T$ (in seconds):

$P = (E_D + E_{\rm blade} + E_{\rm rower})/T$

The typical efficiency values that I find are around 77-80%. This is significantly higher than Atkinson, probably due to the fact that I find negligible values of $E_{\rm rower}$.

It is interesting to see that a significant portion of the rower energy is delivered through the kinetic energy of the rower (“through the footboard”). However, in real life, it is very hard for a rower to decouple these two, except for slight differences due to rowing style. One could wonder for example about the efficiency of different types of catch (“trunk swing” vs “leg push”), and perhaps I will try to model it in the future. In my model, this would be implemented as a different function of rower body of mass vs handle position.

# On levers and balance of forces

Here’s a picture of the oar at two points in time. On the left-hand side of the picture, the frame of reference is the boat. This is the usual perspective of the rower moving down the slide and pulling the oar. On the right-hand side, the same oar, at the same points in time, in the reference frame of the public standing on the bank. What’s not clear from the perceptive of the rower but should be noticed by an accurate observer on the bank, is that the blade slips through the water. The oar is a lever, but the effective outboard length of the lever is given by $l_2$ rather than by $l_{\rm out}$.

To calculate the forces on the boat, let’s however first consider an idealized case without blade slip. Imagine the rower to catch poles standing in the canal, as depicted in the following figure:

The rower pulls the handle with a force $F_{\rm handle}$. Through the lever principle, this results in forces $F_{\rm oarlock}$ and $F_{\rm blade}$ at the oarlock and the blade, respectively. As the blade doesn’t slip, no work is done at the blade. Thus, the work at the handle being equal to the work at the oarlock, the oarlock force is given by:

$F_{\rm oarlock} = \frac{L}{l_{\rm out}} F_{\rm handle}$

where $L$ is the total length of the oar (measured from the handle to the center of the blade). On the right hand side of the figure, the forces on the hull are depicted. The oarlock force accelerates the boat, while the drag force $W$ and the rowers footboard force $F_{\rm foot}$ act in the opposite direction. Using Marinus van Holst’s notation, $\dot{x}$ is the velocity of the hull, $\dot{z}$ is the velocity of the center of mass, and $\dot{y}$ is the velocity of the rower with respect to the hull. The equations of motion are thus:

$F_{\rm prop} = (m_c + m_b) \ddot{z} = m_b \ddot{x} + m_c (\ddot{x}+\ddot{y})$

$m_b \ddot{x} = \frac{L}{l_{\rm out}} F_{\rm handle} - F_{\rm foot} - W$

$m_c (\ddot{x}+\ddot{y}) = F_{\rm foot} - F_{\rm handle}$

Inserting the last two equations into the one for the net propulsive force on the system, $F_{\rm prop}$, we obtain:

$F_{\rm prop} = \frac{L}{l_{\rm out}} F_{\rm handle} - F_{\rm foot} - W + F_{\rm foot} - F_{\rm handle}$

Thus, finally we obtain:

$F_{\rm prop}= \frac{l_{\rm in}}{l_{\rm out}} F_{\rm handle} - W = F_{\rm blade} - W$

Voilà, the net force on the system equals the blade force minus the drag force. This is logical, as the blade and the hull area are the only points of contact between the boat and the rest of the world.

Now, for the slipping blade situation, the balance of forces is slightly different. From work considerations, we have

$F_{\rm handle} (\dot{x}+\dot{y}) - F_{\rm oarlock} \dot{x} - F_{\rm blade} (\dot{\phi} l_{\rm out} - \dot{x}) = 0$

Where the blade force is still* given by:

$F_{\rm blade} = \frac{l_{\rm in}}{l_{\rm out}} F_{\rm handle}$

The system net propulsive force is given by $F_{\rm blade} - W$.

Still nothing new here, dear readers. In the next post, I will finally address the energy balance question. This will be an interesting one, as there are differences between the findings of the two giants (on whose shoulders I stand) about the amount of energy dissipated in the body of the rower.

I have removed the references to a virtual outboard $l_2$ from this text. The use of this virtual outboard is not necessary and leads to erroneous results. I have updated my model accordingly.