 Minimum-power speed

The minimum-power speed $V_{\mathrm{MP}}$ minimizes the mechanical propulsive power required to sustain the specified flight state. For steady level flight with drag $D(V)$, the required power is $P(V)=D(V)V$, using speed through the air. Minimizing this product is a different problem from finding the minimum-drag speed.

With the fixed-coefficient parabolic drag polar, writeDifferentiation givesThus the stationary point is the unique global minimum for positive speed:At this point, $b/V^2=3a{V}^2$: the decreasing drag contribution is three times the increasing contribution, rather than equal to it. In lift-coefficient form, $C_L^{\mathrm{MP}}=\sqrt{3C_{D0}/k}$.

For an original numerical model with $a=0.20$ and $b=512,000$ in consistent metre-second-newton units, $V_{\mathrm{MD}}=40$ m/s and $V_{\mathrm{MP}}\simeq 30.393$ m/s. The power minimum is about 22.461 kW, with drag about 739.01 N. At the drag minimum, drag is only 640 N but power is 25.600 kW. The lower power comes from the combination of force and speed, despite the greater force.

In shallow steady unpowered descent, the gravitational energy-loss rate approximately balances $DV$, so minimizing required power relates to minimum sink speed when the same assumptions hold. This is consistent with the FAA's distinct minimum-sink and best-glide points.^[ref] A changing propulsive efficiency changes the input-power objective: if $P_{\mathrm{input}}=P/{\eta }_p(V)$, a stationary point satisfies $P^{\prime }/P={\eta }_p^{\prime }/{\eta }_p$, not generally $P^{\prime }=0$.

Attainability must also be checked. The model's optimal lift coefficient can exceed the allowed range before aerodynamic stall. The formula concerns the stated required mechanical power, not a universal minimum of fuel consumption or an operational speed recommendation.