Parabolic drag polar ($C_D=C_{D0}+k{C}_L^2$)

A parabolic drag polar models drag coefficient as
The zero-lift drag coefficient supplies the intercept. In a simple finite-wing approximation, $k$ can represent lift-induced drag; fitted whole-aircraft values can include additional lift-dependent losses. The model is conditioned on a configuration and operating regime, rather than universally valid through aerodynamic stall.

Its dimensional form explains a useful speed tradeoff. In steady level flight let lift equal gravitational weight $W$, with fixed mass density $\rho$ and wing area $S$. Substitution of $C_L=W/(qS)$ and $q=\rho V^2/2$ yields
The first contribution grows with speed; the second decreases as the required lift coefficient falls. Differentiation gives the minimum-drag speed $V_{\mathrm{MD}}=(b/a{)}^{1/4}$.

Minimizing required propulsive power instead gives
The minimum-power speed is therefore different from the minimum-drag speed in this model. For stipulated $\rho =1.2kg/{\mathrm{m}}^3$, $S=20{\mathrm{m}}^2$, $W=10000\mathrm{N}$, $C_{D0}=0.020$ and $k=0.050$, the speeds are about $36.30$ and $27.58m/s$, respectively.

These are idealized level-flight comparisons of required propulsive power. Engine efficiency, allowable lift coefficients and a speed-dependent polar can move an operational optimum. Relating the result to minimum sink speed in unpowered flight requires the additional shallow-glide approximation; it does not identify minimum fuel flow for an arbitrary engine.

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.

The minimum-drag speed $V_{\mathrm{MD}}$ minimizes the dimensional drag required for steady level flight at fixed gravitational weight, configuration and atmospheric conditions. It is not the speed at the smallest drag coefficient: changing speed changes the dynamic pressure and the lift coefficient needed to support the load.

For the parabolic drag polar $C_D=C_{D0}+k{C}_L^2$, assume $C_{D0}>0$, $k>0$ and constant mass density $\rho$. Substituting $L=W=qS{C}_L$ givesThe ideal finite-wing induced term motivates the quadratic dependence on lift coefficient; a measured whole-aircraft fit needs its own regime assessment.^[ref]

The arithmetic-geometric mean inequality proves a global bound for every $V>0$:Equality holds when $a{V}^2=b/V^2$, so the two modeled drag contributions are equal. Equivalently, the optimum lift coefficient is $C_L^{\mathrm{MD}}=\sqrt{C_{D0}/k}$.

An original model uses $a=0.20$ and $b=512,000$ when $V$ is in m/s and $D$ in newtons. It gives $V_{\mathrm{MD}}=40$ m/s and $D_{\min }=640$ N, with 320 N from each term. At 30 m/s, the terms are 180 and about 568.89 N, giving a larger total of 748.89 N. The slower state nevertheless requires less mechanical power, illustrating the distinct minimum-power speed objective.

At fixed weight, minimizing drag maximizes the lift-to-drag ratio. Under a shallow, steady-glide approximation this relates to best glide through the air, whereas minimum sink speed concerns altitude loss per time. The FAA polar explicitly distinguishes the corresponding operating points.^[ref] Wind, variable coefficients or an unattainable lift coefficient can change the applicable optimum. These model results are not flight instructions or substitutes for measured aircraft performance.