I need to simulate a realtime wind's trend from his windrose distribution (made from statistical data).

How winds change direction and speed in function of time? I cannot pick a random value from that distribution because I'll simulate some unrealistic situation (Example: at minute 1, wind blows from N at 3 kn, on minute 2 wind blows from SE at 10 kn)

Edit n.1

Actually i'm taking data from these datasets: https://catalog.data.gov/organization/735d0118-b2a7-4246-a0e3-608b0fa8d1ec?res_format=CSV&tags=wind-speed

  • $\begingroup$ The wind rose will have been sampled at a certain interval. That is the only sensible interval to pick, except for going into physical modeling. $\endgroup$ Jun 15, 2017 at 12:32
  • $\begingroup$ The only option I can see to improve realism a bit for most locations would be to require wind changes to favor gradual changing over jumps most of the time, especially in quadrants with a large percentage of the distribution. Without an actual distribution of wind transition, you can't know how often it jumps around, but you're right to recognise that typically wind, when up, will persist and gradually change quite often. No way to make it actually "real" don't believe, but could randomize angle change with a weighted function of distribution and a variability constant you choose? $\endgroup$ Jun 16, 2017 at 23:24

1 Answer 1


You can't. A windrose aggregates over time, so you lose all the information about patterns over time. You're trying to do a time-based simulation, but your data source discards all time-specific information.

You need a different set of data, to base your simulation on. One that contains the dimension of time.

The changes in direction and speed over time vary hugely depending on the local climate and topology.

  • $\begingroup$ Is it possibible $\endgroup$
    – tandrea
    Jun 15, 2017 at 20:13
  • $\begingroup$ No, it is probably not possible to create a useful model in this case. You need a different kind of data $\endgroup$
    – 410 gone
    Jun 16, 2017 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.