# Balance of Propulsive Forces

** Balance of Propulsive Forces**

#### In this newsletter I continue analysis of forces
in rowing, and focus here on the balance of propulsive forces. The force *Fb*_{x}
acting on the boat was previously defined (RBN 2019/01) as the difference
between the pin *Fp*_{x} and stretcher forces* Fs*_{x}:

*Fb*_{x} = Fp_{x} – Fs_{x} (1)

The propulsive force applied
to the rower’s centre of mass CM *Fr*_{x} was defined as the
sum of reaction forces at the stretcher *Fs*_{x} and the handle *Fh*_{x}
(Fig.1), where the latter *Fh*_{x} is the sum of X
components of normal *Fh*_{n} and axial *Fh*_{a}
forces at the handle. The latter *Fh*_{a} was assumed to be
equal to the measured axial gate force *Fg*_{a}, but in fact they could
be slightly different because of the axial blade force *Fb*_{a} and
centripetal force of oar rotation *Fo*_{c} acting only on the
gate axial force* Fg*_{a}.

*Fr*_{x} = Fs_{x} - Fh_{x}
= Fs_{x} – (Fh_{n} cos(θ) + Fg_{a} sin(θ)) (2)

where *θ* is the oar angle. Fig.2 shows
these forces in M1x at 35.5spm, the same sample as in RBN 2019/01.

The propulsive force applied to the rower-boat
system *Fpsys*_{x} was defined as the sum of the forces on the
boat *Fb*_{x}
and rower’s CM *Fr*_{x} (Eq.1,2). The stretcher force *Fs*_{x}
is cancelled, so *Fpsys*_{x} is the difference between X forces* *at
the pin *Fp*_{x} and handle *Fh*_{x}:

*Fpsys*_{x} =* Fr*_{x} + Fb_{x} = (Fp_{x} –
Fs_{x})+ (Fs_{x} - Fh_{x}) = *Fp*_{x} - Fh_{x}* (3)*

Therefore, **the
stretcher force is not accounted in the system propulsive force at all, being
an internal force between the boat and the rower’s mass. The propulsive force
acting on the rower-boat system is the difference between X components of the
pin and handle forces. However, this doesn’t mean a rower should minimise the
stretcher force, as it is firmly transferred through the boat-rigger to the
gate-pin force, which reaction counterpart is the main positive component of
the system propulsive force.**

*Fb*

_{x}*Fp*

_{x}*Fs*

_{x}*Fb*

_{x}= Fp_{x}– Fs_{x}(1)*Fr*

_{x}*Fs*

_{x}*Fh*

_{x}*Fh*

_{x}*Fh*

_{n}*Fh*

_{a}*Fh*

_{a}*Fg*

_{a}*Fb*

_{a}*Fo*

_{c}*Fg*

_{a}*Fr*

_{x}= Fs_{x}- Fh_{x}= Fs_{x}– (Fh_{n}cos(θ) + Fg_{a}sin(θ)) (2)*θ*

*Fpsys*

_{x}*Fb*

_{x}*Fr*

_{x}*Fs*

_{x}*Fpsys*

_{x}

*Fp*

_{x}*Fh*

_{x}*Fpsys*

_{x}=*Fr*

_{x}+ Fb_{x}= (Fp_{x}– Fs_{x})+ (Fs_{x}- Fh_{x}) =*Fp*

_{x}- Fh_{x}*(3)*

#### During the middle of the recovery, *Fpsys*_{x}
is close to zero as expected (Fig.3,1), but **the system propulsive force becomes positive even before the catch**
(2), which could be an effect of the oar centripetal *Fo*_{c} and rotational
inertia force *F*_{o.in} (RBN 2015/05). During the recovery, the oar CM
(located on the outboard) moves forward, so it’s kinetic energy is transferred
to the system at oar deceleration before the catch (2). After the finish, the
oar inertia force is negative, which decreases the system propulsive force (3).

The X component of the oar
inertial force was defined with the angular oar acceleration *α *derived
from the oar angle, the oar moment of inertia *Io* at the pin (1.6 kgm^{2})
and distance between CM and the rotation axis *Lo*_{CM} (0.42m
from pin):

*F*_{o.in} = Io α / Lo_{CM} cos(θ) (5)

Oar inertia force explains the
increasing of the total propulsive force before the catch (Fig.3,2) and its
drop after the finish (3).

The total propulsive force
measured from the pin/gate and handle *Fpsys*_{x} was compared with
one used traditionally as the forward component of the blade force *Fb*_{x}
defined through the handle force* Fh*, gearing ratio (inboard *L*_{in}
to outboard *L*_{out}) and oar angle* θ*:

*Fb*_{x} = Fh (L_{in}/L_{out})
cos(θ) (6)

These two forces were quite similar during the
first half of the drive (Fig.4,1), but then, *Fpsys*_{x}
appeared to be much higher than *Fb*_{x} (2). To eliminate sensor
error, the propulsive force at the blade *Fob*_{x} was derived from the
oar sensors, which were completely independent from the pin sensor:

*Fob*_{x} = (Fg_{n} – Fh_{n})
cos(θ) (6)

where *Fg*_{n} is the normal gate
force and *Fh*_{n} is normal handle force. Two forces *Fpsys*_{x}
and *Fob*_{x}
matched quite well together - a small difference between them (3) could be
related to the axial force at the blade* Fba*_{x} and oar centripetal
force *Fo*_{c}, which were not measured from the oar sensors in
*Fob*_{x},
but affected the total propulsive force* Fpsys*_{x}.

The significant difference in
*Fpsys*_{x}-*Fb*_{x}
during the second half of the drive (4) could be explained by a shift of the
centre of pressure at the blade inwards, which decreases actual outboard and
increases the blade force, so *Fb*_{x} defined through the
handle force and gearing appeared to be underestimated. This was my previous
idea, when I analysed changing the ratio of handle/gate forces (RBN 2014/02).

To verify that the above increase of the blade
force is true, various versions of the propulsive force were averaged over the
stroke cycle at various stroke rates (17.2-41.5 spm) and
speeds (2:16-1:31 s/500m). The drag force at the boat hull (RBN 2019/01)
was also averaged, all these four forces were used to plot relative speed, then
power trend lines were built over the data points (Fig.5).

The best match was found between the drag force
and *Fb*_{x} - the blade
propulsive force obtained with the traditional method of using the handle force
only (r=0.991). **On average, the blade propulsive
force was 8.3% higher than the boat drag force, which can be easily explained
by additional air drag resistance**. Two other versions of the propulsive
force also had very high correlation with the drag force (r=0.998), but *Fpsys*_{x} (obtained from pin-handle) was 53.2% higher,
and *Fob*_{x}
(from gate-handle) was 57.1% higher than the drag force. This placed doubts
on my previous hypothesis about there being shorter actual outboard and higher
blade force during the second half of the drive.

A possible solution to this puzzle, the mysterious
extra force at the pin-gate, is to relate it to some inertial force in the
rower-boat system. To do this, the found above difference
in propulsive forces *d**F*_{prop}* = Fpsys*_{x}-* Fb*_{x} was compared with accelerations of the
boat *a*_{boat}
and rower’s CM *a*_{rower} (Fig.6), where *a*_{boat} was measured directly and *a*_{rower} was derived from measured seat and rower’s trunk movements added to the boat
acceleration. Velocities of the handle and rower’s body segments are given as a reference of the phases
of the stroke cycle.

Here, I should refer to my
model of power conversion into kinetic energy (RBN 2018/11), where point 9 can
be appended as follows: “Two components of the system, the rower and the boat,
not only receive kinetic energy through the oar, but also exchange it between
themselves. … Energy exchange also happens during the drive: after the catch,
the rower actively accelerates his CM using the leg drive, taking kinetic
energy from the boat, which receives negative acceleration”. At the catch, this
exchange happens through the stretcher force, which is not accounted in
propulsion equation No. 3, so it is not reflected in the difference of
propulsive forces *d**F*_{prop}. A small value of this difference
at the catch (Fig.6,1) could be related to oar linear inertia force, which
pushes the pin and looks like propulsive force.

During the beginning of the
drive, the boat acceleration became similar to the rower’s CM acceleration (2),
so no exchange of kinetic energy happens between them, and *d**F*_{prop} gets
close to zero.

In the middle of the drive,
the boat acceleration became higher than the rower’s acceleration (3), then the
latter became negative (4). This means that the rower transfers kinetic energy
to the boat, which happens through the gate-pin forces, but is not related to
the blade propulsion and reflected in the highest difference of forces *d**F*_{prop} (5).

To understand these mechanics
better, the following example can be used (suggested by Dr. Volker Nolte,
although for another reason). Imagine that a rower moves back and forth in the
boat with the oar firmly fixed in the oarlock without rotation, with no blade
work in the water, and only the gate force is measured. At the “catch”, the
rower pushes the stretcher, but this force is not counted as it is an internal force.
At the end of the “drive”, the rower needs to decelerate his movement and pull
the handle, which is measured as the gate force and looks like “propulsive”
force.

Currently, all this is a
hypothesis and there is still a lot of work required to understand the
mechanics better and to develop methods of analysis for propulsive and internal
forces in the rower-boat-oar system. Say, in the example above, the rower pulls
the fixed oar producing the handle force, which subtracted from the gate force
in eq.3, so “propulsive” force should be zero. Perhaps, some other factors are
at play here: Gearing? Seat friction? Axial oar forces?

In conclusion, the **mechanics in the rower-boat-oar system are quite a complex issue, where
the main challenge is separation of propulsive and internal forces mixed in the
measurements at the handle, gate-pin and stretcher. Currently, a complete** **explanation for the phenomenon of higher gate-pin
force during the second half of the drive is still not completely clear to us.
All suggestions and hypotheses are very welcome**.

*Acknowledgements*: Thank you to WinTech
Racing Boats and Oarsport Ltd. for their support of this study.

**©****2019 Dr. Valery Kleshnev ***www.biorow.com*

*Fpsys*

_{x}*Fo*

_{c}*F*

_{o.in}*α*

*Io*

*Lo*

_{CM}*F*

_{o.in}= Io α / Lo_{CM}cos(θ) (5)*Fpsys*

_{x}*Fb*

_{x}*Fh*

*L*

_{in}*L*

_{out}*θ*

*Fb*

_{x}= Fh (L_{in}/L_{out}) cos(θ) (6)*Fpsys*

_{x}*Fb*

_{x}*Fob*

_{x}*Fob*

_{x}= (Fg_{n}– Fh_{n}) cos(θ) (6)*Fg*

_{n}*Fh*

_{n}*Fpsys*

_{x}*Fob*

_{x}*Fba*

_{x}*Fo*

_{c}*Fob*

_{x}*Fpsys*

_{x}*Fpsys*

_{x}*Fb*

_{x}*Fb*

_{x}*Fb*

_{x}*Fpsys*

_{x}*Fob*

_{x}*d*

*F*

_{prop}*= Fpsys*

_{x}-*Fb*

_{x}*a*

_{boat}*a*

_{rower}*a*

_{boat}*a*

_{rower}*d*

*F*

_{prop}*d*

*F*

_{prop}*d*

*F*

_{prop}*Acknowledgements*

*www.biorow.com*