# Why is the Weibull distribution used to describe wind speeds?

Frequently, it is stated that the distribution of wind speed at a certain location can be described by the so-called Weibull distribution: $$W(x;\lambda,k):=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{-(\frac{x}{\lambda})^{k}}$$ for $$x\geq 0$$ and $$0$$ otherwise. $$k$$ is the so-called shape parameter and $$\lambda$$ the scale parameter. If both parameters are known for a certain location, the wind speeds are said to follow the resulting distributionn.

When reading the literature, this statement is ubiquous and stated as if it were an obvious fact. Empirically, this is apparently a well tested statement, since it seems to be used in the siting of wind turbines. However, I have not yet found any satisfying answer why this specific distribution should be used instead of any other.

I am only aware of one particular case that has a bit of justification: the Rayleigh distribution. $$R(x;\lambda):=\frac{x}{\lambda^{2}}e^{-(\frac{x}{2\lambda^{2}})}$$ The Rayleigh distribution is sometimes, but rarely used for describing wind speeds. On a theoretical level the Rayleigh distribution describes the magnitude of a vector where each component is uncorrelated, normally distributed with equal variance and zero mean. This may be a very idealized picture, but it may describe a situation where each wind direction behaves completely independet. It has to be noted, that setting the parameter $$k=2$$ in the Weibull distribution retrieves the Rayleigh distribution as a special case. It could be conjectured that the general Weibull parameter may somehow describe the behavior of correlated wind directions. However, I am not aware of any derivation or proof of this conjencture.

Arguably, a one paramter distribution as the Rayleigh may not allow enough freedom to describe wind-speeds. Still, I wouldnt see apriori why not any other 2-parameter distribution is as suited to describe wind speed as the Weibull distribution.

Is anyone aware of a rigorous argument why the Weibull distribution in particular should be the best distribution to describe wind speeds?

• If it works, people use it. What sort of a 'rigorous argument' are you expecting? – Jon Custer Dec 13 '19 at 18:51
• For instance a derivation, from first principles, although this sounds hard. As an example i have posted how the related Rayleigh distribution can be motivated as modelling the magnitude of a higher dimensional vector with easy assumptions on the variation of the components. A similar argument for the general Weibull case would be interesting. – ckrk Dec 13 '19 at 18:55
• "Still, I wouldnt see apriori why not any other 2-parameter distribution is as suited to describe wind speed as the Weibull distribution." - Likely because the Weibull distribution is similar in structure to the Rayleigh distribution, so it's straightforward to see how the data deviates from "ideal" behavior. – probably_someone Dec 13 '19 at 18:58
• See en.wikipedia.org/wiki/Extreme_value_theory and en.wikipedia.org/wiki/Generalized_extreme_value_distribution for some theoretical justification. The more practical justification is that Wiebull fits many real-world phenomena be changing the value of $k$, and that there is a simple way to estimate the "best" $k$ value. – alephzero Dec 13 '19 at 19:20
• I’m with @alrphzero here. A Weibull curve is what you use when you have only a vague and weakly limited idea of what the real situation is. It’s a general purpose flexible family of curves, and is rarely chosen on strong theoretical grounds. – dmckee --- ex-moderator kitten Dec 13 '19 at 20:46