# Rowing power and kinetic energy

Here we continue the
analysis of rowing power and its conversion into kinetic energy of the
rower-boat system. The definitions and measurements data from RBN 2019/01-02 is
used here. The main equations of the rower-boat system movement relate forces
applied to its components with their masses and accelerations. The total force
applied to the boat** F_{b}** is a difference between its
propulsive

**and drag**

*F*_{bp}**forces:**

*F*_{dr}*F _{b} = F_{bp}
- F_{dr} =Fp_{x} – Fs_{x} - F_{dr} = m_{b}
a_{b} (1)*

where ** Fp_{x}**
horizontal (X) component of the pin force,

**- X stretcher force,**

*Fs*_{x}**– product of boat mass and acceleration. Contrarily to the boat, which is a passive component in this case, the rower is an active part of the system: they applies forces to rowing equipment. So, rower’s mass is driven by reaction forces, which have the same magnitude and opposite direction to action forces: a rower pushes the stretcher backwards and reaction force**

*m*_{b}a_{b}**accelerates their CM forward, pulling the handle and the reaction force**

*F’s*_{x}**moves them backwards. The resultant force applied to the rower’s CM**

*F’h*_{x}**is also a difference between their propulsive force**

*F*_{r}**and drag (air resistance)**

*F*_{rp}**, which is quite small and can be neglected:**

*F*_{rr}*F _{r}= F_{rp}
– F_{rr} ~ F’s_{x}– F’h_{x} = m_{r} a_{r} (2)*

where ** m_{r}
a_{r}** are rower’s mass and acceleration. The force applied to
the whole system is the sum of rower’s and boat forces, so using Eq.1 and 2:

*F _{s} = F_{b}
+ F_{r} *

*= (Fp*~~Fs~~ - F~~F’s~~– F’h

_{x}–_{x}

_{dr}) + (_{x}

_{x}) =*= Fp _{x} –
F’h_{x} - F_{dr } = m_{s
}a_{s} (3)*

Having forces ** F**
applied to the boat, rowers and the whole system CM (Fig.1,a) and their masses

**, we can derive their accelerations**

*m***(Fig.1,b, the total forces must be used here including the drag):**

*a**a = F / m (4)*

Velocities ** v**
can be obtained as an integral of accelerations plus a constant offset

**to make an average velocity over the stroke cycle equal to zero:**

*v*_{off}*v _{Li} =v_{Li-1}
+ a _{d}t + v_{off} (5)*

The velocities ** v_{L}**
(Fig.1,c) are represented in the local inertial reference frame, which moves
with constant velocity equal to the average rowing velocity

**over the stroke cycle. In mechanics, this frame is called the “**

*v*_{av}**centre of momentum frame**”: “

**The kinetic energy of systems depends on the choice of reference frame: the reference frame that gives the minimum value of that energy is the centre of the momentum frame**, i.e. the reference frame in which the total momentum of the system is zero. This minimum kinetic energy contributes to the invariant mass of the system as a whole.”

To derive absolute
velocities ** v_{G}** in global reference frame based on the water,
the average rowing velocity over the cycle

**should be added to the local velocities**

*v*_{av}**:**

*v*_{L}As we discussed before
(see RBN 2018/11), there are two possible scenarios here:

· Internal power transfer between components of
the system, where velocities in the local system frame must be used;

· External power exchange with environment, where
velocities in global reference frame must be used.

Internal (local) power can
be derived as products of their propulsive forces (excluding boat drag) and local
velocities (** P_{L}=F_{p}v_{L}**,
Fig.1,d), and internal kinetic energy can be found with masses of the
components (

**, Fig.1,e).**

*E*_{kL}=0.5mv_{L}^{2}We will discuss internal
energy first. During the recovery, the oar is disconnected from the water, so there
is no propulsive force, and the only external force is the drag applied to the
boat hull, and so to the system. All other forces are internal ones, so their
power should be derived with velocity in the local reference frame. Neglecting the
small rower’s drag force, power applied to their CM in this local frame ** P_{rL}**
is:

*P _{rL} = F_{r}
v_{rL} = (F’s_{x}– F’h_{x}) v_{rL} (7)*

The local boat power ** P_{bL}
**is:

*P _{bL} = F_{bp}
v_{bL} = (Fp_{x} – Fs_{x}) v_{bL } (8)*

The power of the system**
P_{sL}** can be derived in two ways: 1) The sum of the rower’s
and boat powers

**shows energy spent on internal movements of the components within the system:**

*P*_{sInt}*P _{sInt} = P_{rL}
+ P_{bL} (9)*

2) Product of the force
and velocity at the system CM defines movement of the system as a whole piece,
but it valid only in the global frame, as only external forces applied to it:

*P _{sG} = F_{sp}
v_{sG} = (Fp_{x} – F’h_{x}) v_{sG } (10)*

During recovery, both
rower’s and boat powers are positive (1), because their forces and local
velocities are in the same direction: for the rower, they are negative (they
pull themselves backwards through the stretcher, and their CM also moves
backwards in the local frame); boat force and local velocity are positive (it
is pulled forward through the stretcher, and its velocity is higher than
average).

Generally, **positive
power means concentric muscle contraction: force and velocity work in the same
direction (both positive or both negative, depending on the reference frame)
and muscles produce mechanical energy. Negative power means eccentric muscle
contraction: force and velocity act in opposite direction, so the muscles
consume mechanical energy produced elsewhere**.

Before the catch, the boat
power gets close to zero for a short time (2), because both its force and local
velocity change direction. Then, the boat power increases again (3), because both
the force and local velocity obtain negative direction at the same time. The
rower’s internal power becomes negative before the catch (4), because the force
becomes positive, but local velocity is still negative and only starts
increasing slightly: the rower’s mass is much heavier and can’t change
direction as quickly as the boat. Here, we have a typical example of collision
of two masses.

From mechanics, there are
two extreme kinds of collisions: elastic ones, like billiard balls, which
bounce in opposite directions after collision with nearly the same speed as before,
and inelastic - like two pieces of clay, which stick to each other and moves together
as one piece after collision. In both cases, the total momentum of the system
is not changed as it is dictated by “Momentum conservation” law. However,
internal kinetic energy of the system (sum of kinetic energies of the
components) becomes very different:

·
**In a perfectly
elastic collision, the objects exchange their kinetic energies, so internal
energy of the system is preserved** (energy is a scalar quantity, not a vector).

·
**In
inelastic collision, the kinetic energy is consumed for deformation of the
objects and dissipated as heat, so the internal energy of the system becomes
zero: there is no internal movement in the system after the collision**.

Of course, there are no
perfectly elastic collisions in pure mechanics, as part of the energy is always
consumed in deformations, but in the real world, energy can be added with
muscular contraction. Biomechanics studies in top sprint runners revealed that
up to 90% of take-off energy is delivered by the elastic properties of tendons,
and only 10% is added by muscles. Of course, movements are much slower and more
complex in rowing, so the above efficiency looks unrealistic. However, **the
more elastic collision happens between rower’s and boat masses at catch, the
more internal kinetic energy is preserved, so it can be recycled during the drive
phase and rowing efficiency increases. This means an effective catch should
look like “a bouncing ball”**. This perfectly confirms our previous concepts
of effective rowing: “catch through the stretcher”, “trampoline effect” and
“hammer and nail principle”, and adds one more argument against older concepts
of “minimising boat check”, “do not upset boat at catch”, etc. Fig.1,e shows
that the boat kinetic energy after the catch (5) becomes higher than during the
recovery, which means technique of this sculler is quite efficient.

If the rower’s and boat masses
collide inelastically at catch, a rower has to consume kinetic energy during
the recovery, which fatigues their muscles anyway, then produces it again with
muscles to start the drive, so efficiency suffers. **The worst thing a rower
can do is to slow down seat movement during the second half of recovery and
“stick to the stretcher” at the catch.**

During the drive, the
picture is much more complicated, because external power from the oar is added
and mixed up with internal power, so we put question marks on the drive phase
of Fig.1,d. External powers of the system components can be derived with
equations similar to Eq.7-10, where global velocities are used instead of local
ones (Fig.2,a). However, during the recovery, rower’s and boat global
propulsive powers are not valid as they are moved by internal stretcher force
mainly, so they are crossed out. The global system power can be derived in two
ways: ** P_{sG(r+b)}** is the sum of the rower’s and boat
powers; and

**with the propulsive force**

*P*_{sG}**and velocity**

*F*_{sP}**:**

*v*_{sG}

*P _{sG} = F_{sP}
v_{sG} = (Fp_{x} – F’h_{x} ) v_{sG} (11)*

Fig.2,b compares both
system powers above with traditional handle power ** P_{hnd}**
and sum of the body segments power

**(RBN 2004/06). Average power**

*P*_{segm}**was quite similar to**

*P*_{sG(r+b)}**(±3% at 17-41spm),**

*P*_{hnd}**was similar to**

*P*_{sG}**(±5%) and the range of the system kinetic energy variation was close to the measured work per stroke (±2%).**

*P*_{segm}Fig.2,c compares the internal
system power from Fig.1.d (on the left scale) with differences between global
kinetic power and rower’s power production (on the right scale). The curves
look very similar, which suggests a hypothesis explaining the second ones with
an effect of the first ones, but the problem is that the magnitudes of global differences
were 4-5 times higher than local powers.

We will continue the analysis
in the next bulleting, all comments and ideas are welcome.

**©****2020 Dr. Valery Kleshnev ***www.biorow.com*

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