This post again just summarizes what others have said elsewhere (see Anu Dudhia on “Basic Physics of Rowing”).
The drag of a rowing shell with a velocity is given by:
The drag factor is proportional to the wetted surface of the hull A and is be calculated with the following equation (from Sliasas and Tullis):
with = 997 kg/m³ the density of water, the wetted surface area of the shell and c a non-dimensional skin friction drag coefficient with a value of 0.00225. Slasias and Tullis claim (from personal communication with Kleshnev) that a heavy men’s 4 with a total crew mass of 376 kg and a shell mass of 50kg has a wetted surface area of 5 m². Thus, they obtain =6.0 N s²/m².
This surface area is a function of the water displacement which is equal to the total mass of the hull plus the crew: , assuming N rowers with an average rower mass . Thus, we can construct a scaling relation to calculate the relative drag for different boats and crews from a known value. Taking the single scull as the starting point, this equation is:
For I take the value for a single scull quoted by Marinus van Holst to be 3.5 N s²/m² for a 14kg single scull with a rower weighing 70kg (=84 kg). Using these values, I obtain, for the heave men’s four a value of =9.5 N s²/m². This number is 50% higher than the value of Kleshnev / Slasias / Tullis.
For now, I am using the value of Van Holst in my numerical model. Van Holst himself wrote me (same kind of private conversation that Kleshnev had with Slasias and Tullis) that my power numbers look large. Perhaps this is the reason, Marinus?