Effect of Weight in Rowing
Recent Modifications
 04JAN08 Updated layout. New stats for recent World Championships.
1. Summary
The following table summarises the theoretical dependence of various
types of test on the oarsman's
weight W (specifically the appropriate exponential of weight).
These are listed in order of decreasing advantage for the heavier
athlete.
Test  Scaling  Notes


Erg Power (Anaerobic):
 W^{1}
 Anaerobic Power proportional to muscle volume

Erg Power (Aerobic):
 W^{2/3}
 Aerobic Power proportional to surface areas

Erg Speed (Anaerobic):
 W^{1/3}
 Erg speed varies as cube root of Power

Erg Speed (Aerobic):
 W^{2/9}
 The famous 2/9, or 0.222, formula

Sculling Speed (Anaerobic):
 W^{1/9}
 Small weight advantage (ignore boat weight)

Sculling Speed (Aerobic):
 W^{0}
 No weight advantage (ignoring boat weight)

Erg Power/Weight (Anaerobic):
 W^{0}
 No weight advantage

Erg Power/Weight (Aerobic):
 W^{1/3}
 Lightweights have an advantage

The following sections on this page explain
how these are derived.
2. Relationship between Power and Weight
In the following, the phrase 'similar physique' means having the same
build and physiology but not necessarily same weight or height, i.e. the ratio
of weights of two people with 'similar physique' will be the cube of the ratio
of their heights.
The body can produce Anaerobic Power P_{A} and
Aerobic Power P_{O}.
Anaerobic power depends on muscle bulk, i.e. for a given physique,
P_{A} will be simply proportional to total Weight W:
where c_{1} is a constant.
Aerobic power is governed by the flow of
oxygen across various membranes, so is proportional to surface areas.
Two people
of similar physique will have surface areas proportional to the square of
their heights, or the 2/3 power of their weights:
(2.2)
 P_{O} = c_{2}W^{2/3}

where c_{2} is a different constant.
3. Power/Weight ratios
From the previous section the Anaerobic Power/Weight ratio
should be a constant, i.e. over short distances on an erg, say 1000m or less,
two athletes of similar physique should have similar power/weight ratios.
On the other hand, the Aerobic Power/Weight ratio
(3.2)

P_{O}/W = c_{2}W^{1/3}

is inversely proportional to the cube root of the weight, i.e. over
long distances on an erg, say 5000m or more, the lighter athlete should have
the larger power/weight ratio.
4. Relationship between Weight and Erg Speed
An ergometer, such as a Concept, fundamentally measures power P which
is related to speed V (=distance/time) via:
where c_{3}
is a constant (nominally the same for all Concept machines,
whatever vent/gear setting is used). So, the time T to cover a given
distance D (using T=D/V) is related to power by:
where c_{4} is a constant proportional to the distance D.
Using the expressions for aerobic power P_{A}
(Eq. 2.1)
and anaerobic power P_{O} (Eq. 2.2),
the timeweight
relationship for anaerobic work (i.e. a few minutes or less) is
given by
(4.3)
 T_{A} = c_{5}W^{1/3}

where c_{5}
is some constant proportional to the distance, and for
aerobic work (i.e. 20 minutes or longer)
(4.4)

T_{O} = c_{6}W^{2/9}

where c_{6} is some constant proportional to the distance.
This second
expression is the origin of the (2/9) or 0.222 power that often appears in
formulae relating erg times to body weight. Note that it is only valid for
aerobic work.
5. Relationship between Weight and Boat Speed
Most of the resistance R (a Force) to a moving boat comes from surface
drag which is proportional
to the wetted surface area A and the square of the velocity V.
(see Basics: Section 2)
where c_{7} is some constant dependent upon hull shape.
For a given submerged hull shape, the surface area A varies as the
(2/3) power of the volume enclosed, which is proportional to the displacement,
which is almost entirely the mass of the crew (ignoring boat weight, cox,
oars).
(5.2)
 R = c_{8}W^{2/3}V^{2}

At steady speed V, the resistive power (RV) equals the
motive power P.
Using the derived weightdependence of Power from
section 2,
the speed V_{A} for anaerobic distances,
setting RV_{A} = P_{A}
(Eq. 2.1) is given by:
(5.3)
 V_{A} = c_{9}W^{1/9}

where c_{9} is some constant. Thus, over short distances,
heavier scullers
have a slight advantage over light scullers.
(only the ninth root of weight).
The speed V_{O} over aerobic distances,
setting RV_{O}=P_{O}
(Eq. 2.2) is given by:
where c_{10} is a constant.
This suggests that over longer distances, e.g.
'Head' races, light crews are as fast as heavy crews. This isn't quite true,
since the mass of the boat has been ignored, but lightweight scullers
certainly feature more regularly at the top of the Tideway Scullers Head than
in the open 1x finals of Nat.Champs.
6. Speeds of different Boat Classes
Most racing boats have essentially the same hull shape e.g. the submerged
volume of an eight has approximately twice the dimensions of length, width and
depth as that of a scull since it has to displace 8 times (=2x2x2) as much
water.
We can then use the same resistance formula
Eq.(5.2) for different boats, replacing the weight
of the individual W with the total weight of the crew
of N members NW
(6.1)

R = c_{8}(NW)^{2/3}V^{2}

To simplify things, assume each rower weighs the same and absorb
W^{2/3}
into the constant c_{8}:
(6.2)

R = c_{11}N^{2/3}V^{2}

where c_{11} is a constant proportional to the
(2/3) power of the average weight.
The total power generated will also be NxP,
where P is the
power generated by an individual, so equating resistive power
RxV with
motive power NxP, the basic boat speed is given by
where c_{12}
is a constant proportional to the cube root of the average
power of an individual. Since 2^{1/9} = 1.08,
this implies an 8% difference in
speed between different boat classes
(singles  doubles  quads, or pairs  fours  eights).
The following table shows the winning times in the Men's events in recent
World and Olympic Championships,
together with the % differences between different size
boats in the sculling and sweep categories separately.
 Sculling events
  Sweep Events

Year  1x  <Diff>  2x  <Diff>  4x
  2  <Diff>  4  <Diff>  8+

2004  6:49  5%  6:29  8%  5:57
  6:31  6%  6:07  7%  5:42

2005  7:16  9%  6:38  16%  5:35
  6:53  10%  6:12  13%  5:23

2006  6:35  7%  6:08  8%  5:39
  6:18  9%  5:44  6%  5:22

2007  6:46  7%  6:17  7%  5:49
  6:25  8%  5:54  5%  5:35

Table 6.1: difference in winning times of men's (heavyweight)
boats at recent Olympic/World
Championships
Most results fall within about 8±1%.
The 4x and 8+ in 2005 went significantly faster than predicted,
although since these two events are run on the following day to the other four
events, this might be explained by a change in wind conditions.
For other years the 8+ event is 6±1% faster than the 4, rather
than 8%, which could
be explained by the extra weight of the cox  see next section.
The following table shows a comparison of the 2/2x and the 4/4x results
for the same years.
Table 6.2: As Table 6.1, but comparing sweep/sculling events
 2man boats
  4man boats

Year  2  <Diff>  2x 
 4  <Diff>  4x

2004  6:31  0.5%  6:29 
 6:07  2.7%  5:57 

2005  6:53  3.6%  6:38 
 6:12  9.9%  5:35 

2006  6:18  2.6%  6:08 
 5:44  1.5%  5:39 

2007  6:25  2.1%  6:17 
 5:54  1.4%  5:49 

We can probably discount the 2005 4x/4 difference (9.9%)
due to the change in
conditions mentioned above. The small 2x/2 difference in the 2004
Olympics may be due to an exceptionally fast 2 (Australians Tomkins & Ginn
 not many would argue with that)
since they were also
fast compared to the 4.
Those apart, results suggest that
sculling
boats are generally 2±1% faster than the equivalent
sweep boats.
7. Effect of Deadweight on Boat Speed
Now the really tricky one: how much difference does that extra kilo of
deadweight make to your speed? First of all, to point out the obvious, it
doesn't really matter where the extra kilo is: on the cox, on the boat or on
one of the rowers, it slows you down just the same.
To take a simple model of a hull, imagine a submerged portion with a
constant semicircular cross section, radius X and length Y.
The displaced mass of water, equal to the mass of crew, cox, boat and oars,
W, is given by:
where pi=3.14 and D is the density of water (about 1000 kg/m^{3}),
and the total surface area A is given by:
If a small mass dW is added, the boat sinks a by an extra depth
dZ (assume the sides of the hull are vertical at the water line) until
the equivalent additional mass of water is displaced:
and the extra wetted surface area dA is given by:
Putting these equations together, we get:
Note that a simpler assumption, that surface area increases as the (2/3)
power of mass, is really only true for comparing different boat classes, or
perhaps single scullers. However, this would lead to a factor (2/3) rather than
(1/2) in the above equation, which would not significantly affect the
answer.
From section 5, a boat's speed V is determined
by the balance between
the motive power P and the resistive power
R V, where the resistance R
itself depends on wetted surface area
A and the square of the speed (Eq. 5.1), so
Rearranging, and assuming constant power (i.e. same crew rowing the boat and
that the extra mass dW is deadweight),
where c_{13}
is a constant proportional to the cube root of the total power.
Using some basic calculus, the change of speed with surface area is given by:
and using Eq.(7.5), the relationship between speed and
deadweight is given by:
Which tells you that the percentage loss of speed is one sixth the percentage
increase in mass.
An example: assume an VIII, total mass 800 kg (=8x80kg rowers + 50kg cox +
100kg boat + 10kg oars). An extra 10 kg (=22 lbs) represents 1/80=1.25% increase
in mass. So the boat moves 1.25/6=0.2% slower. Over a 6 minute race (eg
2000m) this corresponds to 0.6 sec, or 4m (about 1/5th of a boatlength )
How fast would a coxless VIII be? Minus 50kg represents a 6.25% decrease in
mass, so the boat would be 6.25/6 ~ 1% faster. This would why the difference
between the 4 and 8+ times in
Table 6.1 is about 1% less than the theoretical 8% when
comparing 'similar' coxless boats.
Comments from Marinus van Holst (24 Oct 2000)
 Eq.(7.9) invites for integration. This yields V/Vo = (Wo/W)^(1/6).
Using the figures of your numerical example yields as expected the same
results.

[assuming dP/P = 3.dV/V
from differentiating Eq.(7.6)] I derived the equation
dP/P = (1/2).dW/W
and after integration P/Po = (W/Wo)^(1/2) at constant speed. The practical
situation is a coach confronted with the problem to replace a crew member
by a heavier one. He wants to maintain the speed of the boat and requires
that the new crew member supplies the extra power needed. Again I use
your figures and assume that the mass of the new crew member is 90 kg.
Further I assume that the total energy output was
Po = 8 * 400 = 3200 Watts then dP = 3200 . (1/2 . 10/1000) = 16 Watts, the
extra power to be delivered by the new crew member.
 The waterline, that is
the closed curve formed by the intersection of the hull and the
watersurface, is a rectangle in your model (Eq.(7.1)).
The naval architect uses the
waterline coefficient
Cwl = (area enclosed by the waterline)/(B.L)
where B is the maximum width of the waterline (=2.X),
and L the length of the waterline (=Y).
Then
dZ = dW / (Cwl.B.L.d)
For an eight Cwl is only slightly smaller than 1 but for a single
scull I estimate Cwl = about 0.6 and the influence is important. Using Cwl
there is no longer the need to make the assumption on the cylindrical
form of the hull.
8. Relationship between Erg Score and Boat Speed
Section 4 discussed the relationship between Weight
and Erg Speed, and Section 5 discussed the relationship
between Weight and Boat Speed, so it is possible to use the Weight to derive
some relationship between Erg Speed and Boat Speed.
The heavier rower exerts more power (hence a larger erg score) but also
displaces more water in a boat (hence creating more drag), so how much more
power would they have to generate to overcome the extra drag?
This is, of course, what many coaches want to know: all other aspects (such
as technique) being equal, how do you compare two rowers of different weights
and different erg scores in terms of deciding who would move a boat fastest?
If V_{E} is the erg speed then, reproducing
Eq.(4.1),
(8.1)

P_{E} = c_{3}V_{E}^{3}

is the power
generated by the rower.
Modifying Eq.(5.2) slightly to take account of
a certain amount of deadweight D (boat, oars, cox) shared amongst
each rower, the boat speed V_{B}
is given when the Resistive power
R V_{B} matches the generated power:
(8.2)

c_{3}V_{E}^{3} = c_{8}(W+D)^{q}V_{B}^{3}

where q is (2/3) for a single scull, but probably more like
(1/2) for crew boats  it depends on how you assume the additional wetted
surface area varies with increased mass
(see Section 7).
The most convenient way to use this relationship
(Ve versus Vb, taking into account W)
is to
standardise the erg scores using
some weightdependent adjustment factor F:
(8.3)
 Speed:
 V_{B} = V_{E}F

(8.4)
 Distance:
 D_{B} = D_{E}F

(8.5)
 Time:
 T_{B} = T_{E} /F

(8.6)
 where
 F = ((W_{0}+D) / (W+D))^{q/3}


D_{E},T_{E} are the distance or time obtained on the
erg,

D_{B},T_{B} are the predicted distance or time
obtained in a boat.

W_{0} is some arbitrary 'standard' weight for a rower.
Putting in some numbers: taking D=15 kg (share of deadweight),
q=(1/2), and choosing W_{0}=75 kg, the Adjustment Factor
becomes:
(8.7)
 F = (90 / (W+15))^{0.167}

where W is the rower's mass in kg. Again, one could argue that a
(2/9) power (=0.222) is more appropriate than (1/6) (=0.167).
So if a 85 kg oarsman pulls a 5 km erg in 19 minutes (=1140s),
and a 70 kg oarsman takes 19.5 minutes (=1170s),
their equivalent 'boat speeds', normalised for a 75 kg oarsman, would be:
(8.8)
 (85kg):

T_{B} = 1140 / ( (90/100)^{0.167} ) = 1160s = 19m 20s

(8.9)
 (70kg):
 T_{B} = 1170 / ( (90/85)^{0.167} ) = 1159s = 19m 19s

i.e. a bit of a nightmare for the person who has to select between the two.
(Using the 0.222 power would give times 1167s and 1155s respectively, so the
lighter rower would win).
9. Effect of Boat Weight on Boat Speed
If boats ran smoothly, then a lighter boat hull would always be a faster boat
since boat+rower would have less mass, therefore less wetted surface area,
therefore less drag
(Section 5).
However, the boat speed varies throughout the stroke, and the amplitude
of oscillation is larger for lighter boats, requiring extra power to maintain
a given average speed (Basics, Section 5).
So could it be that there is some nonzero optimum boat mass for which the net
drag is minimised?
To maintain an average speed V_{0} requires a power
P(t)
which varies with time thoughout the stroke cycle as
(9.1)
 P(t) = a (M_{R}+M_{B})^{2/3} (V_{0} + V_{B}f(t))^{3}

where a is some constant; M_{R},M_{B} =
mass of the rower, boat;
V_{B} is the amplitude of the boat speed variation;
and f(t) is some antisymmetric function varying
from 1 to +1 with a mean of 0
which describes the
change of boat speed variation with time, which we will take to be
a square wave, as in Fig 5.1 in Basics (realistically, f probably resembles something
inbetween square and sine waves, and isn't antisymmetric in that the positive
part isn't a mirror image of the negative part).
Averaging this over a complete stroke cycle (since f(t)
is antisymmetric,
odd powers of V_{B} average to zero)
(9.2)
 P = a (M_{R}+M_{B})^{2/3} (V_{0}^{3} + 3V_{0}V_{B}^{2})

Next we'll assume that the amplitude of the boat speed oscillation
V_{B}
is entirely due to the relative motions of the rower and boat
(ignoring the component due to the centre of mass of the whole system
speeding up/slowing down during the stroke/rest parts of the cycle 
a better approximation at high rates than at low rates). In that case we
can relate V_{B} to the amplitude V_{R}
of the relative velocity variation between the rower and the boat
(which will be larger for higher rates and longer strokes, but can be regarded
as a fixed number for these purposes):
(9.3)
 V_{B} = M_{R}V_{R} / (M_{R}+M_{B})

Combining Eq.(9.2) and Eq.(9.3), and
making things slightly tidier by writing M = M_{R} + M_{B} for the total
mass of rower+boat, we get mean
power P in terms of a single variable M (variable in the
sense of containing the boat mass M_{B}
which we are interested in)
(9.4)
 P =
 a M^{2/3}(V_{0}^{3} + 3V_{0}(M_{R}V_{R}/M)^{2})

(9.5)
 =
 a V_{0} ( V_{0}^{2}M^{2/3} + 3(M_{R}V_{R})^{2} M^{4/3}) )

This power is minimised (for a fixed V_{0}, a, M_{R}, V_{R}) when the
differential with respect to M (or M_{B}  same thing) is zero.
Skipping the algebra, this is when
(9.6)
 M =
 (M_{R}V_{R}/V_{0}).6^{1/2}

(9.7)
 M_{B} =
 M_{R} ( 2.45 (V_{R}/V_{0})  1 )

A (fast) sculler covering 2000m in 7min moves at V_{0}=4.76 m/s.
Taking a stroke length (defined as motion of the rower's centre of mass
relative to the boat) of 1m, rating 30, gives V_{R}=1 m/s,
so M_{B} = 49% M_{R}, i.e.
ideally negative and half the mass of the
rower!
However, the factor 2.45 (=√(6)) is very sensitive to some of the
assumptions made, and turns out to be a minimum value. In particular:
 If the surface area increases with displacement
as M^{1/2} rather than M^{2/3}
(Section 7) this gives an extra factor (3/2)^{1/2}=1.22
 If the wave drag is also significant (Basics, section 2), the power could be argued to vary
as the 5th power of velocity rather than the cube. This gives an extra
factor √(10/3)=1.83 (NB: taking only the V_{0}^{5} + 10V_{0}^{3}V_{B}^{2}
terms of the
binomial expansion for Eq.(9.2)).
 If the motion of the rower is considered sinusoidal rather than a square
wave, this gives an extra factor (π/√(8))=1.11 (which includes
π/2 from the
modification of V_{R}).
Combining all of these gives an extra factor π.√(5/8)=2.48, so instead
of 2.45 we get 2.45x2.48=6.08, and Eq. (9.7) becomes:
(9.8)
 M_{B} = M_{R} ( 6.48 (V_{R}/V_{0})  1 )

So using the same value of
(V_{R}/V_{0})=1/4.76,
the optimum boat weight is
now M_{B}=+28% M_{R}, i.e.
28% of the sculler's weight (eg 21kg for a 75 kg
sculler  well above the FISA minimum of 15kg).
From the above, I would conclude only that the ideal boat weight weight is not
necessarily "as light as possible", and a more detailed model is required to
provide an actual value.