|Erg Power (Anaerobic):||W1||Anaerobic Power proportional to muscle volume|
|Erg Power (Aerobic):||W2/3||Aerobic Power proportional to surface areas|
|Erg Speed (Anaerobic):||W1/3||Erg speed varies as cube root of Power|
|Erg Speed (Aerobic):||W2/9||The famous 2/9, or 0.222, formula|
|Sculling Speed (Anaerobic):||W1/9||Small weight advantage (ignore boat weight)|
|Sculling Speed (Aerobic):||W0||No weight advantage (ignoring boat weight)|
|Erg Power/Weight (Anaerobic):||W0||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.
The body can produce Anaerobic Power PA and Aerobic Power PO.
Anaerobic power depends on muscle bulk, i.e. for a given physique, PA will be simply proportional to total Weight W:
|(2.1)||PA = c1W|
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|
|(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.
|(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 PA (Eq. 2.1) and anaerobic power PO (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.
|(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 VA for anaerobic distances, setting RVA = PA (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 VO over aerobic distances, setting RVO=PO (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.
|(6.1)||R = c8(NW)2/3V2|
To simplify things, assume each rower weighs the same and absorb W2/3 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 21/9 = 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|
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.
|2-man boats||4-man boats|
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.
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:
|(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|
|(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.
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 VE 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 VB is given when the Resistive power R VB 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|
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).
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; MR,MB = mass of the rower, boat; VB 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 VB 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 VB 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 VB to the amplitude VR 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 = MR + MB 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 MB 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, MR, VR) when the differential with respect to M (or MB - same thing) is zero. Skipping the algebra, this is when
|(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 VR=1 m/s, so MB = -49% MR, 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:
|(9.8)||MB = MR ( 6.48 (VR/V0) - 1 )|
So using the same value of (VR/V0)=1/4.76, the optimum boat weight is now MB=+28% MR, 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.