Blowin' in the wind: solution
In last issue's Outer space we developed a formula for the power output per unit time of a windmill:
![\[ P \approx 0.38 \times (U/10ms^{-1})^3 \; \; kilowatts, \]](/MI/plus/issue46/outerspace/solutionhtml1/images/img-0007.png)
|
where 
is the speed of the approaching wind. We deduced that an average wind speed of 
gives a power output approximately equal to 0.38. The question was why this calculation significantly underestimates the true output.
The answer lies in understanding that the average speed is not a good guide. If, for example, the wind speed was 
for half of the time unit and 0 for the other half, then we would get an output approximately equal to
![\[ \frac{1}{2}(0.38 \times 2^3) = 4 \times 0.38. \]](/MI/plus/issue46/outerspace/solutionhtml1/images/img-0005.png)
|
This is four times more than what we got in the original calculation.
