In his most recent Rowing Biomechanics Newsletter (October, 2018), Dr. Kleshnev comments on two studies that were recently published in the Journal of Sports Sciences (see Hofmijster et al. ,2018*; *Lintmeijer et al., 2018). In these studies, we explain why the common method to determine rowing power is incorrect, we determined how large the error is and to what extent it depends on stroke rate and rowing style. We also proposed an improved proxy for determination of the true average power output of a rower, based on measurement of pin forces and oar rotation alone.

As illustrated by our recent papers, determining power output in rowing is far from trivial, and we are pleased to see that others, like us, take this matter seriously as well. In his recent newsletter, Dr. Kleshnev criticizes our approach and suggests that we make a “common mistake”. While we are as fallible as any human being, we are quite confident that there is no mistake in our analysis in this case. The arguments for this can be summarized as follows:

- Application of Newton’s Laws starts with choosing the “free body”, i.e. the system to be analyzed; here the free body “rower”. Newton’s second law states that the sum of all forces on
*any*system determines mass times the acceleration of the system’s center of mass, relative to an inertial frame of reference; for the applications at hand, any frame of reference attached to the earth surface or moving with a constant velocity relative to the earth surface can be considered inertial; see in depth analysis below for the consequences of using a non-inertial frame of reference. - For any passive mechanical system (e.g. a rigid body), the instantaneous power equation (again using an inertial frame of reference) can be derived from Newton’s second law. In case no pure moments are applied to this rigid body, the instantaneous power equation simply states that the sum of the power of the forces equals the time derivative of the total kinetic energy. Here, the power of each force acting on the rigid body is defined as the dot product of the force vector and the velocity vector (relative to the inertial frame of reference used!) of the point of application of this force. The outcome of this dot product is identical to the outcome of Kleshnev’s equation 1 in the Rowing Biomechanics Newsletter.
- For an active mechanical system (e.g. a rower), all that has to be added to the passive-system power equation is a term on the left-hand side representing the net mechanical power generated/dissipated
*within*the active system; in case of the rower, it makes sense to label this term P_{rower}, representing the net mechanical power generated by the rower. - When all terms in the power equation for an active mechanical system are averaged over a full cycle of a periodic motion, the term describing the time derivative of the total kinetic energy vanishes. This is because the total kinetic energy at the start of periodic motion cycle i+1 is then identical to that at the start of periodic motion cycle i; consequently, the average change in kinetic energy per unit time is zero during periodic motion.

Points 1-4 above are all that is required to define the average power equation for the active system “rower” during steady state periodic motion, relative to an inertial frame of reference. When we focus on the two forces acting on the rower that have a high associated power (force of oar on hands, force of stretcher on feet), equation 2 in Kleshnev’s newsletter (taken from Lintmeijer et al., 2018) is obtained. Thus, this equation follows directly from the basic theory of classical mechanics, and there is nothing incorrect about it.

In his newsletter, Kleshnev has unfortunately failed to notice that (in line with the theory summarized in 1-4 above), F_{o,r} and F_{f,r} in equation 2 are forces acting from oar and from stretcher on rower, and not the other way around, as Dr. Kleshnev suggests (“… where For/Ffr are the forces which the … rower applies to the … oar handle / stretcher”). Perhaps this misreading of our paper lies at the heart of his misinterpretation of our claims. **In any case, his conclusion that the average mechanical power generated by a rower is correctly captured by systems considering pin force and oar rotation only, is incorrect. Fortunately, as shown in ****Lintmeijer et al. (2018)****, the measurements of pin force and oar rotation can still result in a useful estimate of the true value of P _{rower} if a simple correction is made.**

**A more in depth-analysis**

*For the reader interested in a more in-depth analysis, we will explain below why the analysis of forces and power terms is complicated when the kinematics are described relative to a non-inertial frame of reference. Here, we will also show that elaboration of the power equation for a rower, analyzed using a frame of reference that moves with the boat, yields exactly the same final result as that in equation 9 of **Hofmijster et al. (2018)*

##### Formulating the power equation of a rower

Dr. Kleshnev states that “the basic equation of mechanical power P” is P = F v cos(α). In vector notation, we can simplify this as:

eq.1: P_{X= }**F _{X}**∙

**v**

_{X}

(in words: the dot product of the force vector and the velocity vector at point X)

In this equation, P_{X} is the power exchanged at point X. Note that this is something else than power produced by say the rower.

Now also take note that “v” in this equation is not the velocity between “object” and “subject”. Instead, it is the velocity of the point of application of **F _{X}**

*with respect to the chosen frame of reference*. This can be found in virtually every classical mechanics reference. Obviously, the choice of the (velocity of the) frame of reference determines the size of P. For this reason, it is important to consider the complete power equation (e.g., our equation 1 in Lintmeijer et al., 2018), as the separate parts of the equation are often quite meaningless. Consider for example our equation 1, which is based on the equation of motion of the rower (i.e. applying Newton’s second law on the free body “rower”):

eq. 2: ** F _{feet}** +

**F**+

_{handle}**F**= m

_{seat}_{rower}∙

**a**

_{rower }_{ }

For the sake of simplicity, let us only consider forces in the direction of boat travel (referred to by x in superscript) and assume that the horizontal seat force is zero (which is not unrealistic considering the limited mass of the seat and the fact that friction forces on the seat are very low compared to F^{x}_{handle} and F^{x}_{feet}). The equation of motion than becomes:

eq.3: F^{x}_{feet}+F^{x}_{handle}=m_{rower}∙a^{x}_{rower}

Note that the prove below still holds in 3D, only the equations become more extensive. From the equation of motion we can now go to the equation of power by multiplying each element in the equation by the velocity of the point of application of the respective forces. The right-hand side of the equation can be considered as an “inertial force”, we should multiply this part by the velocity of the center of mass of the rower. Note that the power equation in this case also contains a power source (the rower!). This power source goes on the left-hand side of the equation and is written as P_{rower} (a good reference as to how to derive the power equation in cyclic endurance sports is Van Ingen Schenau, 1990). The standard power equation of the rower thus becomes:

eq.4: P_{rower}+F^{x}_{feet}∙v^{x}_{feet}+F^{x}_{hands}∙v^{x}_{hands}=m_{rower}∙a^{x}_{rower}∙v^{x}_{rower }

Note that all the velocities are with respect to the same inertial (i.e. standing still or moving at a constant speed) frame of reference.

##### Effect of the velocity of the (inertial) frame of reference on P_{rower}

Now it can be readily appreciated that most of the terms in the equation above depend on the choice of frame of reference. In fact, all terms; *except P _{rower}*! Here’s the proof:

Say we change the velocity of the frame of reference by the velocity “v^{x}_{extra}” (v^{x}_{extra} could be the extra velocity because of a tailwind or caused by rowing in a larger boat). We now have:

eq.5: P_{rower}+F^{x}_{feet}∙(v^{x}_{feet}+v^{x}_{extra})+F_{hands}∙(v^{x}_{hands}+v^{x}_{extra})=m_{rower}∙a^{x}_{rower}∙(v^{x}_{rower}+v^{x}_{extra})

Factoring out parts of the equation yields:

eq.6: P_{rower} + F^{x}_{feet}∙(v^{x}_{feet}) + F^{x}_{hands}∙(v^{x}_{hands}) + (F^{x}_{feet}+F^{x}_{hands}) ∙v^{x}_{extra} =

m_{rower}∙a^{x}_{rower}∙(v^{x}_{rower})+m_{rower}∙a^{x}_{rower}∙v^{x}_{extra}

If you look closely at the equation above, you see that from equation 3 it follows that the terms (F^{x}_{feet}+F^{x}_{hands})∙v^{x}_{extra} and m_{rower}∙a^{x}_{rower}∙v^{x}_{extra} cancel each other out. What is left is the original power equation, equation 4.

Conclusion: *P _{rower} is independent of the choice of frame of reference, but the other terms are not.* For rowing, it does not make much sense to discuss handle power and foot-stretcher power separately.

##### The power equation of a rower in a non-inertial frame of reference

Unfortunately; determining the force on the stretcher is not trivial. What many people do to circumvent this problem is to determine handle force and then multiply this force by the handle velocity with respect to the boat. However, by doing so, they (implicitly) adopt a non-inertial frame of reference; one that is moving with the boat.

When you do this, a fictitious force should be introduced. This fictitious force helps you to understand why the gauge measuring the force the weight lifter holds above his head in the rocket in Dr. Kleshnev’s example reads higher numbers when the rocket starts its journey to the moon. (Another well-known example of a fictitious force is the Coriolis force, which explains why water in a sink on earth moves down the sink in spirals – the earth basically being one large non-inertial frame of reference).

This fictitious force applies on the center of mass of the free body of interest (in our examples, the rower) and has size opposite of the mass of the free body times the acceleration of the frame of reference.

By not taking this fictitious force into account, an error is being introduced. To understand this, we start with the general equation of motion again, but this time considered in a non-inertial frame of reference with acceleration a^{x}_{f.o.r.} (again, for the sake of simplicity, we only consider movement in the forward direction)

eq.7: F^{x}_{feet}+F^{x}_{handle} -m_{rower}∙a^{x}_{f.o.r.}=m_{rower}∙a’^{x}_{rower }

Now, here’s the power equation again, but now in a frame of reference that moves with (the speed and acceleration) of the boat (so a^{x}_{f.o.r.} = a^{x}_{boat}):

eq.8: P_{rower} + F^{x}_{feet}∙(v’^{x}_{feet}) + F^{x}_{handle}∙(v’^{x}_{handle}) – m_{rower}∙a^{x}_{f.o.r}∙v’^{x}_{rower} = m_{rower}∙a’^{x}_{rower}∙v’^{x}_{rower }

All velocities in the equation, as well as a’^{x}_{rower} are with respect to the velocity of the frame of reference (the boat). The second term of the equation is zero, since the foot stretcher moves with the speed of the boat

(v’^{x}_{feet} = 0). This leaves:

eq.9: P_{rower} + F^{x}_{handle}∙(v’^{x}_{handle}) – m_{rower}∙a^{x}_{f.o.r.}∙v’^{x}_{rower} = m_{rower}∙a’^{x}_{rower}∙v’^{x}_{rower}

Here, v’^{x}_{rower} is the speed of the rower with respect to the boat. We can write this as v’^{x}_{rower} = v^{x}_{rower} – v^{x}_{boat }(note that v^{x}_{rower} and v^{x}_{boat} are velocities with respect to the world!). Same for a’^{x}_{rower}: a’^{x}_{rower} = a^{x}_{rower} – a^{x}_{boat.}We can now write the power equation as:

eq.10: P_{rower} + F^{x}_{handle}∙(v’^{x}_{handle}) – m_{rower}∙a^{x}_{f.o.r.}∙(v^{x}_{rower} – v^{x}_{boat}) =

m_{rower}∙(a^{x}_{rower} – a^{x}_{boat})∙(v^{x}_{rower} – v^{x}_{boat})

This can be simplified into:

eq.11: P_{rower} + F^{x}_{handle}∙(v’^{x}_{handle}) + m_{rower}∙a^{x}_{rower}∙v^{x}_{boat} = m_{rower}∙a^{x}_{rower}∙v^{x}_{rower}

This is exactly the same equation as the equation that we propose in both our studies, only proven in a different way.

Note that we derived this equation by strictly applying classical mechanics to the free body rower. Suggesting that this is a “common mistake” feels like stating that Newton made some common mistakes when he wrote down his three Laws of Motion!

Now, let us have a closer look at the different elements of the equation. The right-hand part covers the change of kinetic energy of the rower. In steady state rowing, there is no net change in kinetic energy; as a result, this term is zero on average and thus usually not considered when steady state rowing is investigated. F^{x}_{handle}∙(v’^{x}_{handle}) is the term that is often calculated in many rowing studies and implemented in most commercial equipment. But, as you can see, this term is not equal (and opposite) to P_{rower}; the power produced by the rower; there is a second term: m_{rower}∙a^{x}_{rower}∙v^{x}_{boat}. In both our studies, we have proven that this term is not zero on average.

So, our conclusion is opposite of that of Dr. Kleshnev: the net power transferred from a rower to the external environment (in this case the boat and oars) is not measured correctly by the aforementioned commercially available devices. The good news however is that we also show that the error in the measurements is relatively constant. While most products seem to measure power incorrectly, the measurements can still result in a useful estimate of the true power if a simple correction is made.

##### As an encore: One final prove that we did not make a common mistake:

Dr. Kleshnev states that the ‘v’ in P = F∙v should be the velocity of the subject with respect to the object. With the object being the rower (and the subjects the foot stretcher and oar handle), this statement implies that the chosen frame of reference is again non-inertial; namely moving with the center of mass of the rower. So again, we should introduce a fictitious force, this time equal to -m_{rower}∙a^{x}_{rower} (remember: a^{x}_{rower} is the acceleration of the rower’s center of mass with respect to the world!).

The power equation for the rower in a frame of reference that is located at the center of mass of the rower now becomes:

eq.12: P_{rower} + F^{x}_{feet}∙v’’^{x}_{feet} + F^{x}_{handle}∙v’’^{x}_{handle} – m_{rower}∙a^{x}_{rower}∙v’’^{x}_{rower} = m_{rower}∙a’’^{x}_{rower}∙v’’^{x}_{rower }

Note that in this frame of reference, v’’^{x}_{rower} and a’’^{x}_{rower} are both zero. This means that equation 12 can be simplified into

eq.13: P_{rower} + F^{x}_{feet}∙v’’^{x}_{feet} + F^{x}_{handle}∙v’’^{x}_{handle} = 0

Here v’’^{x}_{feet} is the speed of the feet; and thus, the boat; with respect to the rower. We can write this as v’’^{x}_{feet} = v^{x}_{feet} – v^{x}_{rower} (note that v^{x}_{feet} and v^{x}_{rower} are again velocities with respect to the world). Same for v’’^{x}_{handle}: v’’^{x}_{handle} = v^{x}_{handle} – v^{x}_{rower. }

Substituting and rearranging the equation yields:

eq.14: P_{rower} + F^{x}_{feet}∙v^{x}_{feet} + F^{x}_{handle}∙v^{x}_{handle }= m_{rower}∙(F^{x}_{feet}+F^{x}_{handle})

Which, considering the equation of motion of the rower (equation 2) of course is the same as:

eq.15: P_{rower} + F^{x}_{feet}∙v^{x}_{feet} + F^{x}_{hands}∙v^{x}_{hands} = m_{rower}∙a^{x}_{rower}∙v^{x}_{rower }

_{ }

The original power equation for rowing, in an earth-bound frame of reference!

#### References

Hofmijster, M.J., et al., Mechanical power output in rowing should not be determined from oar forces and oar motion alone. J Sports Sci, 2018. 36(18): p. 2147-2153.

Lintmeijer, L.L., et al., Improved determination of mechanical power output in rowing: Experimental results. J Sports Sci, 2018. 36(18): p. 2138-2146.

van Ingen Schenau, G.J. and P.R. Cavanagh, Power equations in endurance sports. J Biomech, 1990. 23(9): p. 865-81.

hi,

i row for 50 years, and try to think clearly.

sometimes i send mails to dr. valery; some years ago i send him “my model” to compute “rowing”, which he did not comment. i tend to believe, the “mathematics” have not been to his liking (too simple?).

now i stumble on his webpage upon his reply to your critics on his “models of rowing-computation”. not that i could follow all of the mathematics, nevertheless i find your discussion on the “power of rowing” very amusing, when i succeed putting the boat into “the glide”.

salut

a.h.

Dear Arthur,

Thank you for your reply and that you are interested in the discussion on power output calculations in rowing.

kind regards