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Above is the definition of the probability factor from Eurocode 1-4. It is used in context of calculating the extreme wind load on a building. Ignoring some other factors, it is used as follows:

$$v_b = v_{b,0} * c_{prob}$$

where $v_{b,0}$ is the extreme wind velocity with less than 0.02 probability of being exceeded. The factor $c_{prob}$ is used to reduce the extreme wind velocity in cases when considering time periods shorter than the 50 year-return period (0.02 probability).

I assume the formula comes from Gumbel distribution:

$$p = e^{-e^{-\frac{v-\mu}{\beta}}}$$

which can be solved for $v$:

$$v = \mu-\beta \ln(-\ln(p))$$

where $p$ is the probability of encountering extreme wind $v$.

Taking the ratio $\frac{v}{v_{b,0}}$ and considering that $v_{b,0}$ is defined to have to probability of being exceeded of 0.02, we get:

$$\frac{v}{v_{b,0}}=\frac{\mu-\beta \ln(-\ln(1-p))}{\mu-\beta \ln(-\ln(0.98))}$$

Defining $K = \frac{\beta}{\mu}$, we get:

$$\frac{v}{v_{b,0}} = c_{prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))}$$

Finally to my question: Where does the exponent n come from in the Eurocode formula? As I have derived to formula in the picture, there is no need for the exponent. Is my derivation flawed? Is there something I don't know?

EDIT: Only reason I assumed Gumbel distribution was that I was able to get a similar formula using it. There seems to be nothing in Eurocodes that goes into detail as to what distribution is actually used.

  • $\begingroup$ Is your assumption about the Gumbel based on something your read in Eurocodes, or is it something you assumed because Gumbel can be used in extreme values distribution? $\endgroup$
    – NMech
    Apr 18, 2021 at 15:31
  • $\begingroup$ @NMech There seems to be nothing in Eurocodes that goes into detail as to what distribution is used, I only assumed Gumbel because that I was able to derive a similar formula using Gumbel. Using some other distributions like Weibull the resulting formula doesn't seem to be anything similar. $\endgroup$
    – S. Rotos
    Apr 18, 2021 at 15:35
  • $\begingroup$ This answer may help: engineering.stackexchange.com/a/42652/10902 $\endgroup$
    – Solar Mike
    Apr 18, 2021 at 15:48

1 Answer 1


Your derivation seems absolutely reasonable, as does your assumption about the Gumbel Distribution, with one minor detail.

IMHO, the reason why the exponent $n$ exists, is that the Eurocode does not really care about the velocity but about wind pressures (ultimately about loads on the structure). So the idea, is that you are trying to capture the extreme value of the wind pressure (though the wind speed).

So all the analysis you carried out is not about $\frac{v}{v_{b,0}}$ but about wind pressures $\frac{q}{q_{b,0}}$. So, instead its:

$$\frac{q}{q_{b,0}}= c_{\color{red}{q},prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} $$

However, because the wind pressure $q$ is proportional to the square of the wind velocity so:

$$\frac{q}{q_{b,0}} = \frac{v^2}{v_{b,0}^2}=\left(\frac{v}{v_{b,0}}\right)^2 $$

$$\left(\frac{q}{q_{b,0}} \right)^{0.5} = \frac{v}{v_{b,0}}$$

$$\left(c_{q,prob} \right)^{0.5} = \frac{v}{v_{b,0}}$$ $$\left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5} = \frac{v}{v_{b,0}}$$

However, because $c_{prob}$ is a factor that is applied to velocity:

$$ \frac{v}{v_{b,0}} = \left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5}$$

  • $\begingroup$ Ah, seems logical. The Code wording just puzzles me; they mention that $n$ maybe be given in the national annex and that the recommended value is 0.5. So I wonder how could it be then anything else than 0.5.. $\endgroup$
    – S. Rotos
    Apr 18, 2021 at 17:57
  • $\begingroup$ To the annex I had access to it was 0.5. My opinion is that it offers a degree of flexibility. $\endgroup$
    – NMech
    Apr 18, 2021 at 18:36
  • $\begingroup$ Flexibility in what way? $\endgroup$
    – S. Rotos
    Apr 18, 2021 at 18:43
  • $\begingroup$ By using a higher coefficient closer to one, then the base wind pressure increases quadratically (so you have a more stringent constraint), while reducing it will lead to lower values (thus more relaxed). $\endgroup$
    – NMech
    Apr 18, 2021 at 18:50

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