Turbine power: P = ½·ρ·A·Cp·v³, with ρ = 1.225 kg/m³, A = 50.3 m² (8 m rotor), Cp = 0.42 (constant; the Betz limit is 16/27 = 0.593). Power is zero below 3 m/s and above 25 m/s, and capped at 10 kW from 9.18 m/s.

Wind: Weibull f(v) = (k/c)(v/c)^(k−1) e^−(v/c)^k, k = 2, c = 6.5 m/s, mean 5.76 m/s. Capacity factor CF = E[P] / P_rated ≈ 0.35 in the long run.

Plough: draft F = k_s·b·d·n = 40 kPa × 0.35 m × 0.25 m × 3 = 10.5 kN. Power at the wheels F·v, electric F·v / 0.85. Energy per hectare F·A / (w·η) = 32.7 kWh, the same at any speed: faster work only needs more power at the same time.

Battery: 60 kWh, charge and discharge efficiency 0.95 each (round trip 0.90). Energy balance: wind in = load served + curtailed + battery losses + change in storage (checked every game second, residual below 10⁻⁹ kWh).

Simplifications, stated honestly: constant Cp; hourly wind with no day-night cycle; perfect forecast; no rolling resistance or slip; the turbines are 8 m micro-turbines (the reference picture shows much bigger ones); the wind speed is used as hub height speed; the k_s and wind-profile values still need a check against a source.

Checks (see measure.md): Monte Carlo and integral CF agree within 0.0003; analytic and numerical P(v ≥ 3 m/s) agree; Betz maximum equals 16/27; energy per hectare is independent of speed; the energy balance closes.