Originally Posted by

**Shielder**
Now, lets look at wind power.

The available 'fuel' is the wind correct?

So, assuming that there are 1,000,000 wind turbines in the world, their available surface area for the wind to act on (assuming 50m blades, and neglecting the spaces between the blades) is approximately, PI*50^{2}= 7854m

That gives us a total surface area for 1,000,000 wind turbines of 7,854,000,000m^{2}.

Now, the radius of the earth is 6,378.1km. So the surface area of the earth is 4*PI*6,378,100^{2}= 5.11E+14m^{2}.

So, negating the total volume of the atmosphere and just looking at the total fuel available to these million wind turbines (i.e. the total surface area of the earth that is acted upon by the wind), we have an efficiency of 7.854E+9 divided by 5.11E+14 = **1.537e-5** or 0.0015%

And that assumes perfect power output (100% load factor), no conversion losses, etc.

What were you saying about efficiency again?