A turbine’s average power is not its power at the average wind speed, because power depends on the cube of the speed and the wind spends time in both calms and gusts. For winds that follow a Rayleigh distribution, the average of v³ is 1.91 times the cube of the mean speed; a 3 m rotor at 35% efficiency in a 5 m/s mean wind therefore makes about 338 W, not the 189 W obtained by cubing the mean.
Part of Advanced Wind Turbine
A power curve gives a turbine’s output at each wind speed; combined with how often each speed occurs, it gives the energy a site delivers.
Cubing The Mean Gets The Wrong Answer
Take the power law P = ½ × ρ × A × Cp × v³, where ρ = 1.225 kg/m³ is the air density, A the swept area, Cp the fraction of the wind’s power the rotor captures, and v the wind speed. Cp can never exceed the Betz limit of 16/27, about 59% (Wind Power). That page notes that home-built rotors reach 25 to 35% of the available power; this page takes 35%.
For the 3 m rotor in the carrier’s table, A = π × 1.5² = 7.07 m². Then P = 0.35 × 0.5 × 1.225 × 7.07 × v³ = 1.515 × v³ watts. At the mean speed of 5 m/s that is 1.515 × 125 = 189 W. But the wind does not sit at 5 m/s. More than half the time it is below the mean, spending hours in a calm that yields nothing, and a few hours well above, where each gust is worth far more than a calm costs, because of the cube.
The Rayleigh Distribution Fixes The Ratio
A common model of wind at a site (an assumption, to be checked against your own record) is the Rayleigh distribution, in which the chance that the speed is below v is 1 − exp(−v² ÷ 2σ²) with σ = U ÷ √(π/2) for mean speed U. Integrating v³ against it gives an average of v³ equal to (6 ÷ π) × U³ = 1.91 × U³. For U = 5 m/s, the mean of v³ is 1.91 × 125 = 239 m³/s³.
Therefore with no cut-in and no limit on output, the average power is 1.515 × 239 = 362 W. That is 1.91 times the 189 W of the naive calculation. The ratio is the same at every U, because the distribution simply rescales: raising the mean from 5 to 6 m/s raises the unlimited average by exactly (6 ÷ 5)³ = 1.73.
Real Turbines Clip And Wait
A real power curve has four regions: nothing below the cut-in speed, a cubic rise, a flat rated region where the controller or furling holds the output down, and a fall in storms. Assume cut-in at 3 m/s and a rated output of 2 kW, which this rotor reaches at v = (2,000 ÷ 1.515)^(1/3) = 10.97 m/s. The wind exceeds that only 2.3% of the time at U = 5 m/s.
Numerical integration of this curve against a Rayleigh wind gives:
| Mean speed U | P at the mean (1.515 × U³) | True average power | Capacity factor | Ratio |
|---|---|---|---|---|
| 3 m/s | 41 W | 71 W | 3.5% | 1.73 |
| 4 m/s | 97 W | 178 W | 8.9% | 1.84 |
| 5 m/s | 189 W | 338 W | 16.9% | 1.78 |
| 6 m/s | 327 W | 526 W | 26.3% | 1.61 |
| 7 m/s | 520 W | 715 W | 35.8% | 1.38 |
The capacity factor is the true average divided by the 2 kW rating. At 5 m/s it is 17%, which lies inside the carrier’s 15-25% and its 300-500 W average for a 2 kW turbine. The energy is 338 × 24 = 8.1 kWh a day. The ratio column falls at high speeds because the rating clips the top: from 5 to 6 m/s the average rises 526 ÷ 338 = 1.56 times, less than 1.73.
These are rotor-shaft figures. The alternator, rectifier and cabling take a further share, so the electrical output is lower.
Cut-In Is Set By The Generator
Cut-in is not a property of the blades alone; it is the wind at which the alternator first reaches battery voltage. The tip speed ratio λ ties the rotor speed to the wind: wind speed = rpm × π × D ÷ (60 × λ). The Coil Winding page quotes cut-in generator speeds of 180 to 280 RPM; with D = 3 m and the carrier’s tip speed ratio of 6:
- 114.6 RPM: 3.0 m/s.
- 180 RPM: 180 × π × 3 ÷ (60 × 6) = 4.7 m/s, with the wind below that for 50% of the time at U = 5.
- 250 RPM: 6.5 m/s, with the wind below that for 73% of the time.
A generator with a 250 RPM cut-in on a 3 m rotor in a 5 m/s site averages only 252 W (12.6%) against 338 W for the 3 m/s cut-in. Winding the stator for a lower cut-in is worth more than a small gain in blade efficiency (Stator Winding (Coil Design)).
Method Of Bins: Doing It With Your Own Data
Step 1: log the wind speed at hub height for a year (Wind Resource Assessment). Step 2: count the hours in each 1 m/s bin. Step 3: multiply each bin’s hours by the power at its middle speed. Step 4: add the products.
At U = 5 m/s, the Rayleigh bins give about 1,303 h at 3-4 m/s, 1,305 h at 4-5, 1,167 h at 5-6 and 948 h at 6-7; the sum over all bins is about 2,970 kWh a year, matching 338 W × 8.76 = 2,960 kWh. Real sites differ from Rayleigh in the long calms and the storms, so your own bins replace the model.
The wind law is not the water law
The cube-and-clip behaviour belongs to wind. A stream’s power is proportional to its flow and has no cube (Seasonal Variation). A 44% faster wind gives 1.44³ = 2.99 times the power; a 44% larger stream flow gives 1.44 times.
Common Mistakes
- Cubing the average speed — it halves the energy estimate.
- Assuming the turbine reaches its rating — the wind is above 10.97 m/s only 2.3% of the time in this example.
- Ignoring cut-in — a generator that needs 6.5 m/s sits idle 73% of the time at a 5 m/s site.
What’s Next
- Learn next: Site Selection & Tower Height to raise the mean speed
- Combine with: Battery Bank Sizing to cover the calm hours