# Basic Equations (2) – What’s the right value for drag?

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 $v_t + v_b$  is given by:

$F = -\alpha ( v_t + v_b )^2$

The drag factor $\alpha$ is proportional to the wetted surface of the hull A and is be calculated with the following equation (from Sliasas and Tullis):

$\alpha = 1.07 \times \frac{1}{2} c \rho A_{\rm shell}$

with $\rho$ = 997 kg/m³ the density of water, $A_{\rm shell}$ 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 $\alpha_{4-}$=6.0 N s²/m².

This surface area is a function of the water displacement $m_w$ which is equal to the total mass of the hull plus the crew: $m_w = N m_c + m_b$, assuming N rowers with an average rower mass $m_c$. 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:

$\alpha_{\rm boat} = \alpha_1 \left( \frac{N m_c + m_b}{m_1} \right)^{2/3}$

For $\alpha_1$ 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 ($m_1$=84 kg). Using these values, I obtain, for the heave men’s four a value of $\alpha_{4-}$=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?

## 6 thoughts on “Basic Equations (2) – What’s the right value for drag?”

1. Bill Atkinson

Hi Sander:

This is my first look at your excellent site.

I am confining my comment now to your “Basic Equations(2)”:

1. I imagine it to be a typographical error, but you have s^2/m^2 reversed in the units of the expression for drag factor. I think it has no effect on your results.

2. What is the source of the scaling equation for wetted surface? I believe that your 50% discrepancy may be here.
Wetted surface as a function of displacement is strongly related to wetted shape. A soda straw, with a closed bottom, floating vertically has a virtually linear (one to one) relation between displacement and surface area. A very flat object, like a large floating wooden board- in the limit- has a wetted surface that is virtually independent of displacement.
The function given is ok if it more or less takes into account the shape of the wetted volume; which I guess is basically a half-cylinder not quite as long as the shell.

3. I get drag factors of the order of 7.0 Ns^2/m^2 for a four with a wetted surface of about 6m^2.

Best regards,
Bill

1. sanderroosendaal Post author

Actually, I am just in the process of refining my drag equations, and rereading my old Fluid Mechanics books. Will post this in a few days. Basically, I will calculate the Reynolds nr based on ittc 1957 formula and I arrive at similar values as you do.

2. Bill Atkinson

Sander:

Yes, the drag coefficient is a function of Reynolds’ number as you note re the ITTC curves. In my model ROWING I use the same 1957/1963 ITTC line as you do.
I made no attempt at all to model wave, roll, yaw, and pitch drag as too complex and not a big part of the skin and form drags. Scragg and Nelson say 2% to 5% of total drag.

From data in Scragg and Nelson I was able to develop an approximation for the relation between displacement and wetted surface for volumes approximating the general shape of shells.

Regards,
Bill

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