Effect of Weight in Rowing

(Part of Physics of Rowing)

Recent Modifications

• 04-JAN-08 Updated layout. New stats for recent World Championships.

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(2.1)

PA = c1W

where c₁ 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)

PO = c2W2/3

where c₂ is a different constant.

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(3.1)

PA/W = c1

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)

PO/W = c2W-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.

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(4.1)

P = c3V3

(4.2)

T = c4P-1/3

where c₄ 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 time-weight relationship for anaerobic work (i.e. a few minutes or less) is given by

(4.3)

TA = c5W-1/3

where c₅ is some constant proportional to the distance, and for aerobic work (i.e. 20 minutes or longer)

(4.4)

TO = c6W-2/9

where c₆ 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.

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(5.1)

R = c7AV2

where c₇ 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 = c8W2/3V2

At steady speed V, the resistive power (RV) equals the motive power P.

Using the derived weight-dependence 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)

VA = c9W1/9

where c₉ 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:

(5.4)

VO = c10

where c₁₀ 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.

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(6.1)

R = c8(NW)2/3V2

To simplify things, assume each rower weighs the same and absorb W^2/3 into the constant c₈:

(6.2)

R = c11N2/3V2

where c₁₁ 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

(6.3)

V = c12N1/9

where c₁₂ 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.

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(7.1)

W = ½pi D X2Y

where pi=3.14 and D is the density of water (about 1000 kg/m³), and the total surface area A is given by:

(7.2)

A = pi X Y

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:

(7.3)

dW = 2 D X Y dZ

and the extra wetted surface area dA is given by:

(7.4)

dA = 2 Y dZ

Putting these equations together, we get:

(7.5)

dA/A = ½dW/W

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

(7.6)

P = c7AV3

Rearranging, and assuming constant power (i.e. same crew rowing the boat and that the extra mass dW is deadweight),

(7.7)

V = c13A-1/3

where c₁₃ 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:

(7.8)

dV/V = -(1/3)dA/A

and using Eq.(7.5), the relationship between speed and deadweight is given by:

(7.9)

dV/V = -(1/6)dW/W

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 boat-length )

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.

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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)

PE = c3VE3

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)

c3VE3 = c8(W+D)qVB3

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 weight-dependent adjustment factor F:

(8.3)	Speed:	VB = VEF
(8.4)	Distance:	DB = DEF
(8.5)	Time:	TB = TE /F
(8.6)	where	F = ((W0+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.
• W₀ is some arbitrary 'standard' weight for a rower.

Putting in some numbers: taking D=15 kg (share of deadweight), q=(1/2), and choosing W₀=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):	TB = 1140 / ( (90/100)0.167 ) = 1160s = 19m 20s
(8.9)	(70kg):	TB = 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).

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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 non-zero optimum boat mass for which the net drag is minimised?

To maintain an average speed V₀ requires a power P(t) which varies with time thoughout the stroke cycle as

(9.1)

P(t) = a (MR+MB)2/3 (V0 + VBf(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 in-between 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 (MR+MB)2/3 (V03 + 3V0VB2)

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)

VB = MRVR / (MR+MB)

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 M2/3(V03 + 3V0(MRVR/M)2)
(9.5)	=	a V0 ( V02M2/3 + 3(MRVR)2 M-4/3) )

This power is minimised (for a fixed V₀, 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 =	(MRVR/V0).61/2
(9.7)	MB =	MR ( 2.45 (VR/V0) - 1 )

A (fast) sculler covering 2000m in 7min moves at V₀=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 V₀⁵ + 10V₀³V_B² 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).

(9.8)

MB = MR ( 6.48 (VR/V0) - 1 )

So using the same value of (V_R/V₀)=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.

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