Hi frlh112,
In the August post on this subject, Korina mentioned that the solar power numbers were based on a flat detector at the latitude where the projection is made. So, to project your expected energy production, you will need to take into account both the angle of your panels and the length of the day. Assuming you are in Switzerland (my guess) and some of your panels are tipped towards the south, you will actually absorb more power than the projections from Windy. The straightforward calculation is based on the cos of the angle of incidence. This is simply because the projected area of the panel varies by the cos of the angle of incidence.
In order to predict the total energy over a day, you will need to have the solar power numbers for each hour of the day, corrected for the angle of incidence. Sum the total for each hour and you have total energy in kWh. This can be converted to Joules.
If your panels also have an East/West angle component you will need to calculate the compound angle of incidence. And every day the solar power, the angles of incidence and the length of the day will vary.
Hope this helps, greggy74

@daniel-castrom Yes, and every day the Wp, angle of incidence and length of day will vary for each panel. The only thing that cannot be predicted is the effect of weather.

Hi frlh112,
In the August post on this subject, Korina mentioned that the solar power numbers were based on a flat detector at the latitude where the projection is made. So, to project your expected energy production, you will need to take into account both the angle of your panels and the length of the day. Assuming you are in Switzerland (my guess) and some of your panels are tipped towards the south, you will actually absorb more power than the projections from Windy. The straightforward calculation is based on the cos of the angle of incidence. This is simply because the projected area of the panel varies by the cos of the angle of incidence.
In order to predict the total energy over a day, you will need to have the solar power numbers for each hour of the day, corrected for the angle of incidence. Sum the total for each hour and you have total energy in kWh. This can be converted to Joules.
If your panels also have an East/West angle component you will need to calculate the compound angle of incidence. And every day the solar power, the angles of incidence and the length of the day will vary.
Hope this helps, greggy74