???? converting windspeed to power out

[(unknown)]

Hi, I found a link that shows mathematical process for converting wind speed to power out for HAWT. Will the same formulas work for VAWT's?  

 http://www.orotronics.com/pages/Calculating-Power-In-The-Wind.html

[stop4stuff]

more or less.... substitute the swept area with the height & diameter of the vawt for a ballpark figure, i.e. vawt of 2m height & 2m diameter = 4m^2 swept area & then take into account your vawt is likely to have less efficiency than a hawt (unless you're building a Lenz 2 type turbine)

[KilroyOdin]

Hi, thank you. A ball park number is all I am looking for.

[KilroyOdin]

I am starting a new project and wanted to make sure if it would have a chance of working

[dsmith1427]

Don't you have to take into account that half of the area opposes the wind?  Therefore the swept area would be the radius (half the diameter) multiplied by the height.  

[TomW]

Seems you would need to factor in a negative value for the drag of the returning bucket?

The devil is in the details.

Tom

[ghurd]

I tried to do that once.

It seems like all the long math came out to about 35~40% of the swept area for a 55 gallon drum type VAWT, as related to HAWT math.

The returning bucket sees a higher "wind speed" than the working bucket.

Very confusing for me.  My numbers were probably not very close, but the principle of subtracting power for the returning bucket is sound, IMHO.

G-