I'm trying to forecast the peak power output for a wind + solar system. The solar output as a function of time of day and month can be forecasted easily. I also have the historical wind speed for a number of years, down to the day and hour.

The question is, in order to build the Weibull distribution, should I take the average wind speed for a given hour + month combination? Because if I ignore the time of day factor, then I can't accurately forecast the peak wind + solar output.

  • $\begingroup$ It is the 100 year events that really mess up predictions. $\endgroup$ Commented Jan 22, 2021 at 15:15

3 Answers 3


In theory the smaller the interval the better the focus.

On the other hand, some windmills manufacturers have their preferred interval to forecast power output and many forecast formats have the scale of 3 hours over 48 hours total span.

Prediction of meteorological variables Wind power generation is directly linked to weather conditions and thus the first aspect of wind power forecasting is the prediction of future values of the necessary weather variables at the level of the wind farm. This is done by using numerical weather prediction (NWP) models. Such models are based on equations governing the motions and forces affecting motion of fluids. From the knowledge of the actual state of the atmosphere, the system of equations allows to estimate what the evolution of state variables, e.g. temperature, velocity, humidity and pressure, will be at a series of grid points. The meteorological variables that are needed as input for wind power prediction obviously include wind speed and direction, but also possibly temperature, pressure and humidity. The distance between grid points is called the spatial resolution of the NWPs. The mesh typically has spacing that varies between few kilometers and up to 50 kilometers for mesoscale models. Regarding the time axis, the forecast length of most of the operational models today is between 48 and 172 hours ahead, which is in adequacy with the requirements for the wind power application. The temporal resolution is usually between 1 and 3 hours. NWP models impose their temporal resolution to short-term wind power forecasting methods since they are used as a direct input. quote from Wikipedia.

here is a useful article in Wikipedia: power forecast


TLDR: If you are planning to use the weibull curve for the power output of a wind turbine you'd better use yearly data and fit the curve.

Weibull distribution and shape factor

Weibull distribution has two parameters (in some variations):

  • Scale: which is close to the concept of average. Its units are actually the units of the average (or m/s in this case).
  • shape: which gives it a characteristic shape. Sometimes denoted with a k

Below you can see the effect of shape :

enter image description here

So for a given scale, the shape can be more skewed or symmetric.

Typically, wind velocity distributions have a skewed distribution so values are between 1.5 and 2.0.

For those parameters, usually the mean value is less than the Weibull scale factor. See example below:

enter image description here

The more skewed the distribution, the more the average will deviate from the scale factor.

Variation of wind speed velocity with time interval

As the measurement interval is decreased the wind speed will appear more jagged. As the measurement interval increases the wind speed will appear more smooth. See below for 1 min and 10 [min] comparison.

enter image description here

To capture that behaviour, in a wind data measurement even if the logger can measure (and can store in non volatile memory) every second, usually the data are stored in rows of either 10min or 1 hour intervals. And usually the following data for measurement interval is used:

  • average wind speed for interval (or 1 hour average)
  • standard deviation of wind speed for interval
  • max gust of wind speed for interval

The importance of the above, is that the power output of the wind turbine will be completely different if there is high standard deviation in the wind speed. Wind turbines will perform more efficiently if the wind speed standard deviation is small i.e. constant speed.

Seasonal variations

Another thing is that you need to take into account is the seasonal variations. Wind speed through January, is not the same as in the other months. Following is the variation for 2018 in a single location.

enter image description here

How to proceed

According to your question you have data to the day for every month in the past couple of years. (following your comment I call yearly data all the data you collect in a year).

Assuming you are doing a Monte Carlo simulation or similar to predict power output, what you can do is take all that data and select one of the following:

  • fit a Weibull distribution (e.g. with maximum likelihood method)
  • calculate a kernel density estimation.
  • Create a Markov chain

Another option (more accurate) would be to take all the data you have for each month separately, and perform one of the above. Then you will have created 12 Weibull distribution you can sample randomly.

  • $\begingroup$ Not sure what you mean by 'yearly data'. What is special about the Weibull distribution around the average annual wind speed as opposed to 12 Weibull distributions around 12 monthly average wind speeds or the Weibull distribution around the two-yearly average wind speed? $\endgroup$ Commented Dec 23, 2020 at 10:49
  • $\begingroup$ I am not sure what you mean by " two-yearly average wind speed" $\endgroup$
    – NMech
    Commented Dec 23, 2020 at 10:57
  • $\begingroup$ I'm re-reading your question and I realised that I may have missed what you are trying to accomplish. I'd be happy to update the answer, if you could update the question, clarifying what you are trying to accomplish. $\endgroup$
    – NMech
    Commented Dec 23, 2020 at 11:45
  • $\begingroup$ Rephrasing my question below: In the answer you've posted above, the last line says ' ....you will have created 12 Weibull distributions you can sample randomly'. This is so as to have a more accurate representation of the month to month variation in the mean speed. But the wind speed varies by time of day as well, and this is critical in a wind + solar hybrid system. So what's the best way of capturing wind speed variability for the design of such a system - is it the Weibull distribution for each hour of a 24 hour day for the average day in the month? $\endgroup$ Commented Dec 25, 2020 at 7:59
  • $\begingroup$ I think I understand now. You want to capture the seasonal variations as well as the variations within a day (morning afternoon). Also, you seem to agree that the variation within a month is not relevant...So if I understand correctly you are planning to create 24 hourly distributions for the 12 months? $\endgroup$
    – NMech
    Commented Dec 25, 2020 at 8:15

Keep in mind a wind time series is not iid... if making a monte carlo from a fitted Weibull and its RVs, you'll get overall averages in a similar range but you will destroy the path dependence

Path dependence should be somewhat important if optimising for power output

enter image description here

  • $\begingroup$ My current approach is to generate conditional distributions for wind speed based on the previous n hours $\endgroup$ Commented Dec 16, 2022 at 14:04
  • 1
    $\begingroup$ Are you the original poster? If so, it seems like you have two accounts. You can request that these accounts be merged. $\endgroup$
    – hazzey
    Commented Dec 16, 2022 at 15:54

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