# Drag revisited (2)

In this post, I postulated that skin drag amounts to about 80% of drag and simply divided the skin drag by 0.8 to obtain the total drag.

Bill Atkinson pointed me to this work by Tuck and Lazauskas. First of all, form drag is estimated there by an equation from a paper by Scragg:

$C_f = k C_{\rm skin}$

with

$k = 0.0097(\theta_{\rm entry} + \theta_{\rm exit})$

where $\theta_{\rm entry}$ and $\theta_{\rm exit}$ are the half angles (in degree) of the bow and stern, respectively. Tuck and Lazauskas do a complete modeling of wave drag as well, and come up with something that we could almost call a phase diagram for rowing shells. Looking at the figure 2b here, the wave drag is around 6% of total drag for shells optimized for racing speed, a Froude number around 2 in the figure. Of course, wave drag varies quite strongly with the shell’s velocity. However, the shell velocity varies during the stroke, and it seems a good approximation to use an average value for wave drag which is only a minor portion of the total drag. Doing the integrals to calculate the instantaneous value would slow down the simulation software too much, for a relatively small improvement in accuracy.

As far as the bow and stern angles are concerned, I collected data from boat manufacturers sites about the boat length, beam and draft as a function of the displacement.The results are depicted in figure 1. The solid line is a linear fit, showing that all boats are basically of (more or less) the same shape. Assuming a parabolic boat shape, the bow half angle would be 8 degrees, with an elliptic shape around 6.3 degrees. This would give a value of $k$ between 0.12 and 0.15. So, the total drag (wave drag, skin drag and form drag) would be about 1.2 to 1.22 times the skin drag. Or, in other words, skin drag accounts for about 82% of the total drag.

Figure 1 Shell length vs beam (both in m) for Vespoli and Empacher shells

Regarding the scaling equation used for different boat types, it is clear that if all boats are basically scaled versions of the same shape, then beam, draft and length should be proportional to the 1/3 power of displacement. This is clearly shown in figures 2 and 3.

Figure 2: Beam as a function of displacement for different makes and types of shell.

Figure 3: Shell length vs displacement for different makes and types of rowing shells

Finally, figure 4 shows the wetted area as quoted on various manufacturer’s websites as a function of the displacement. The solid line shows the $\propto D^{2/3}$ scaling that I use in my model.

Figure 4: Wetted area as a function of displacement for rowing shells of various manufacturers

So, in conclusion, the mystery isn’t a mystery any more. The lower value quoted by Slasias and Tullis is just the skin drag portion calculated from a 5m² wetted area. In reality, the wetted area of a four is slightly higher, and one has to add 20% for wave drag and form drag. Looking at the scatter of the data in the various graphs, I would also say that the accuracy of my model is good within about 10%, not better. This is not a disaster, as the relative results predicted by the model are still valid. Finally, boat design is found to be extremely important, as a 5% reduction of skin drag can means a mean velocity increase of 2.6%. This amounts to more than 2 seconds per 500m for an eight.

# Drag revisited

A discussion on the Concept 2 UK Forum reminded me of the unsolved mystery of the amount of drag experienced by a rowing shell. So I went back to my old textbook on Fluid Mechanics (Robert A. Granger, “Fluid Mechanics”, Dover 1995). Assuming fully turbulent boundary layer flow, the drag coefficient for skin drag is a function of the shell’s velocity or Reynold’s number. Granger recommends the following equation:

$C_{D_f} = \frac{0.455}{(\log R_l)^{2.58}}$

for Reynold’s numbers $R_l$ larger than $10^7$. The  International Towing Tank Conference (ITTC) recommends a slightly different equation:

$C_{D_f} = \frac{0.075}{(\log R_l - 2)^2}$.

For a single with a length of 8.2m and a velocity of 8.42 m/s (1:43 per 500m), the Reynold’s number amounts to $3.3\times 10^7$ for water of 10 degrees C. This gives $C_{D_f} = 2.459 \times 10^{-3}$ with the ITTC equation. Assuming the wetted area for a single with a rower weighing 80kg to be 2.25 m² as claimed by Cambridge Racing Shells to be the value for a Fluidesign single, I obtain $\alpha_{\rm skin} = 2.77$ N s²/m². Using Granger’s equation, I get 2.81 N s²/m². For a four with a wetted surface of 6 m², I get 6.7 N s²/m² using the ITTC equation.

Skin drag is not the only form of drag. There is also form drag and wave drag. It is said that skin drag accounts for about 80% of the drag (Ana Dudhia). So, dividing my number by 0.8 I get a value of $\alpha$ amounting to 3.51 for the single scull at race speed. Pretty close to the value I have used so far. The difference is a slight dependence on the shell’s velocity for the value of $\alpha$. In figure 1, the drag force using the ITTC equation is compared with the simplified method of taking constant drag coefficient, for a single scull with a rower weighting 80kg (wetted surface of 2.25 m²). The differences are small but at maximum velocity they can be of the order of 10%.

Figure 1: Drag Force vs velocity for a single scull with a rower of 80kg. Green line: Constant drag coefficient approximation. Blue line: ITTC 1957 drag coefficient

The division by 0.8 to account for other forms of drag seems a bit arbitrary. Interestingly, when I try to validate the model on the Lucerne 2010 (no wind) boat speed and ergometer scores of Ondřej Synek, the 2010 world champion in the single, I get a consistent story. Synek’s Lucerne achievement was a power of 586W on the 2k (calculated using the ITTC equation value for drag). Assuming 25W is consumed in moving a 98kg heavy athlete up and down the Concept 2 slide, the remaining ergometer power of 561W corresponds to a 2000m time of 5:42 on the ergometer. Synek is known to have rowed 5:41.8 (Czech National record).

As Bill Atkinson rightly pointed out, the scaling equation I used to calculate for different boat types, depends heavily on the wetted shape. I took the $\propto m^{2/3}$ intuitively without giving it much thought, except that Anu Dudhia uses it here. The numbers given for wetted area on the CRS site seem to confirm this. Also, my validation with the 2010 Lucerne results gives confidence, but, admittedly, I have not given it enough thought from the basic physics side. To be continued …