# How do I interpret the standard deviation of a directional dataset?

Lets say I start with a vonMises distribution dataset of compass directions:

x <- rvonmises(n=10, mu = circular(0, units="degrees", type="angles", zero=pi/2,modulo="2pi",template="geographics"), kappa=10)


I understand the angular mean:

mean(x)
# Circular Data:
# Type = angles
# Units = degrees
# Template = geographics
# Modulo = 2pi
# Zero = 1.570796
# Rotation = clock
#  4.796034


The mean is ~4.8$^\circ$ from true N. Fine.

But when I calculate the standard deviation, defined as $\sqrt{-2\ln{r}}$ (where $r$ is the mean resultant length), I get a number between 0 and infinity that I don't know how to interpret.

I vaguely understand that the reason for not adopting the arithmetic definition of the standard deviation is to preserve statistical properties of the directional data's distribution function. But at least I know that the arithmetic standard deviation is in the units of the original data, and can interpret it as the bounds of ~%60 of the data.

How is the angular standard deviation helpful as a descriptive statistic (i.e., how does one interpret its meaning)?

So, the interpretation is as follows. If we have a (normal) random variable $X \sim N(\mu, \sigma^2)$, then $X_c = X ~~ \text{mod} ~~ 2 \pi$ has a wrapped Normal distribution $WN(\mu, \rho)$, where $\rho$ is the resultant length and given by $\rho = e^{-\sigma^2 / 2}$. This means we could write the standard deviation of our original random variable as $\sigma = \sqrt{-2 \ln {\rho}},$ which you can of course recognize from the definition of the circular standard deviation.
• So with this interpretation it is easy to relate one set to another, but I still don't see how to convert it back into the scale of the original dataset. For a normally distributed dataset of velocities, for example, I can interpret $\sigma$ in the units of the dataset (e.g. m/s) as an idea of spread around the mean. With the circular $\sigma$, or even $\rho$, I don't see how to do this conversion. Would it make sense to map $\rho$ onto radians. So a $\rho$ of $0.12$ would be equivalent of $\pm 0.75 rad$ about the mean? – svh160 Dec 7 '15 at 21:00
• Because circular densities tend to lack tractable cumulative distribution functions, it will be hard or impossible to derive an exact mapping from $\sigma$ or $\rho$ to a given percentage of the density around the mean. However, you could use the normal case as an approximation. The circular standard deviation is given in radians and as the data becomes more concentrated, the area within one $\sigma$ of the mean will tend to contain roughly $.6827$ of the data. However, for less concentrated densities, this area will contain a higher percentage of the data. – Kees Mulder Dec 8 '15 at 15:13