Two locations report exactly the same average wind speed: 6 m/s.
Would you expect them to contain roughly the same wind-energy potential?
It sounds reasonable.
It can also be badly wrong.
Imagine Site A experiences a steady wind of 6 m/s.
Site B spends half the observed period at 4 m/s and the other half at 8 m/s.
The arithmetic mean at both sites is identical:
Site A average = 6 m/s
Site B average = (4 + 8) / 2 = 6 m/s
Yet wind power does not respond linearly to wind speed.
It follows:
Pwind = ½ρAv3
For our simplified Site A:
63 = 216
For Site B, average the cubic contribution of the two wind conditions:
(43 + 83) / 2 = (64 + 512) / 2 = 288
Same average wind speed.
Very different average wind-power potential.
The reason is a deceptively important mathematical fact:
Average(v3) ≠ [Average(v)]3
And that is why an innocent-looking average can conceal precisely the information that matters most when evaluating a wind-energy resource.
The Problem Isn’t the Average. It’s What We Ask It to Tell Us
Averages are useful.
If thousands of wind-speed observations have been recorded over months or years, the arithmetic mean compresses all those measurements into one understandable number.
That convenience is also the problem.
Consider these three simplified sites:
Site A: 6, 6, 6, 6 m/s
Site B: 4, 4, 8, 8 m/s
Site C: 2, 2, 10, 10 m/s
Every site has exactly the same average wind speed:
6 m/s
Yet these are clearly not the same wind regimes.
Site A is remarkably steady.
Site B fluctuates moderately.
Site C alternates between relatively weak and strong wind.
If our desired output responded linearly to wind speed, averaging first might not create such a serious problem.
Wind energy doesn’t.
Wind Speed Is Cubed
The power available in moving air can be expressed as:
Pwind = ½ρAv3
where:
ρ = air density
A = swept area
v = wind velocity
Mukund R. Patel’s Wind and Solar Power Systems emphasizes the consequence: wind power varies linearly with air density but with the cube of wind speed.
That cube gives stronger winds disproportionate importance.
If wind speed doubles:
23 = 8
There is eight times as much available wind power, assuming the same air density and swept area.
If wind speed triples:
33 = 27
There is 27 times as much.
So an hour of stronger wind can contribute far more energy than several hours of weaker wind.
This is where a simple average starts losing important information.
Let’s Return to Our Three 6 m/s Sites
Ignore the common constants ½ρA for a moment. If we only want to compare the effect of wind speed, we can compare v³.
For Site A:
(6³ + 6³ + 6³ + 6³) / 4 = 216
For Site B:
(4³ + 4³ + 8³ + 8³) / 4 = 288
For Site C:
(2³ + 2³ + 10³ + 10³) / 4 = 504
All three have:
Average wind speed = 6 m/s
But their average v³ values are:
Site A = 216
Site B = 288
Site C = 504
Site C’s value is more than twice Site A’s despite their identical arithmetic mean wind speeds.
This is a deliberately simplified example rather than a real wind-resource assessment, but it exposes the underlying problem beautifully.
The mean alone cannot tell us how the wind speeds were distributed around that mean.
What We Really Need Is the Wind-Speed Distribution
Suppose we monitor a potential wind site for a long period.
Instead of throwing away most of our information by immediately calculating one average, we can ask:
How often was wind speed between 0 and 1 m/s?
How often between 1 and 2 m/s?
How often between 2 and 3 m/s?
And so on.
Now we can construct a frequency distribution.
pornip.com
September 22, 2026 at 1:52 am
WOW,Very wonderful
Zorays
September 22, 2026 at 2:51 am
Thank you for the kind words! I am glad you enjoyed the explanation of why average wind speed can be misleading for energy calculations.
— Zorays