# The mean direction of waves in a directional distribution

When modelling ocean waves, a directional distribution $D_f(\theta)$ is used together with a frequency spectrum $S(f)$ to describe the energy of waves at a particular frequency $f$ and angle $\theta$.

I understand that the directional distribution can be written as a Fourier series i.e. $$D_f(\theta) = \frac{1}{2\pi}\left[ 1 + 2\sum_{n=1}^{\infty}\{a_n\cos(n\theta) + b_n\sin(n\theta) \} \right]$$ where $a_n = \int_0^{2\pi} D_f(\theta)\cos(\theta)\,d\theta$ and $b_n = \int_0^{2\pi} D_f(\theta)\sin(\theta)\,d\theta$.

In Kuik (1988), the mean wave direction, $\theta_0$, is found by calculating $$\theta_0 = \arctan\left(\frac{b_1}{a_1}\right)$$ where $b_1$ and $a_1$ are the first order Fourier coefficients.

Alongside this definition, the author refers the reader to Borgman (1969) but I can't find this paper on the web.

My question is why is it only the first order Fourier coefficients $a_1$ and $b_1$ used in this calculation?

EDIT: After giving this more thought, I think that the fact they are Fourier coefficients is somewhat of a coincidence.

If the directional distribution is seen as the PDF (as the integral of it is equal to 1) then $a_1$ and $b_1$ are more like the expected values that the cosine and sine of the angle $\theta$ take. The average values can then be used in the $atan2$ function to determine the mean angle.

## 1 Answer

It is helpful to understand the meaning of Fourier coefficients in the directional wave spectrum analysis.

A pitch-and-roll buoy measures time series of water elevation $\eta$, and slopes in both Cartesian directions, $\eta_x$ and $\eta_y$. The Fourier components $a_0$, $a_1$, $b_1$, $a_2$, $b_2$, are related to cross-spectra of elevation and slope time series (introduced by Longuet-Higgins 1963):

$$a_0 = \dfrac{C_{11}}{\pi}$$

$$a_1 = \dfrac{Q_{12}}{\pi}$$

$$b_1 = \dfrac{Q_{13}}{\pi}$$

$$a_2 = \dfrac{C_{22}-C_{33}}{k^2\pi}$$

$$b_2 = \dfrac{2C_{23}}{k^2\pi}$$

where the cross-spectra are:

$$C_{11}(f) = C[\eta(f)\eta(f)] = \int_0^{2\pi} E(f,\theta)d\theta$$

$$Q_{12}(f) = -Q[\eta(f)\eta_x(f)]k^{-1} = -\int_0^{2\pi} E(f,\theta)\cos{\theta}d\theta$$

$$Q_{13}(f) = -Q[\eta(f)\eta_y(f)]k^{-1} = -\int_0^{2\pi} E(f,\theta)\sin{\theta}d\theta$$

$$C_{22}(f) = C[\eta_x(f)\eta_x(f)]k^{-2} = \int_0^{2\pi} E(f,\theta)\cos^2{\theta} d\theta$$

$$C_{23}(f) = C[\eta_x(f)\eta_y(f)]k^{-2} = \int_0^{2\pi} E(f,\theta)\sin{\theta}\cos{\theta} d\theta$$

$$C_{33}(f) = C[\eta_y(f)\eta_y(f)]k^{-2} = \int_0^{2\pi} E(f,\theta)\sin^2{\theta} d\theta$$

From here you can see that the mean direction, as defined in the paper that you cite, is:

$$\theta_0 = \arctan\left({\dfrac{b_1}{a_1}}\right) = \arctan\left({\dfrac{Q_{13}}{Q_{12}}}\right) = \arctan\left({\dfrac{\int_0^{2\pi}E(f,\theta)\sin{\theta}d\theta}{\int_0^{2\pi}E(f,\theta)\cos{\theta}d\theta}}\right)$$

Because $Q_{12}$ and $Q_{13}$ are proportional to the integrated energy in the zonal and meridional directions, respectively, the arctangent of their ratio yields mean wave direction.

The authors could use higher moments to calculate direction, but this would have different physical meaning. In this specific case, using $\theta = \arctan\left({b_2/a_2}\right)$ would yield peak (dominant) wave direction in the context of pitch-and-roll buoys.

You see that the above analysis is inherently limited by the quantities that are measured, specifically $\eta$, $\eta_x$, and $\eta_y$. If higher order quantities, say $\eta_{xx}$, $\eta_{xy}$, $\eta_{yy}$ were available, the directional spectrum could be described at a higher accuracy even using just Fourier decomposition.

Further reading:

• Thank you again - you answered another question of mine a few days ago, you're very helpful! – RH_data_maths Jan 30 '17 at 21:43