 Best glide speed

The best glide speed is the airspeed giving the greatest horizontal distance per unit height lost in a specified steady glide condition. In still air this is the speed for maximum lift-to-drag ratio. It differs from the minimum sink speed, whose objective is longest time aloft for a given height loss.^[ref]

Write a glider sink polar as downward rate $w(V)>0$. The horizontal air-relative speed is $u(V)=\sqrt{V^2-w(V{)}^2}$, so the exact still-air range per unit height is $u/w$. For a shallow glide angle, $u\simeq V$ and maximizing range is approximately minimizing $w/V$. At an interior optimum,Geometrically, a line from the origin is tangent to the positive-downward-rate polar. This is not the horizontal tangent defining minimum sink speed.

For an original calculation, take a parabolic drag polar $C_D=C_{D0}+k{C}_L^2$ and approximate $L\simeq W$. At fixed air mass density $\rho$, wing area $S$ and gravitational weight $W$,Minimizing $w/V\simeq D/W$ givesStipulate $W=3000\mathrm{N}$, $S=10{\mathrm{m}}^2$, $\rho =1.2kg/{\mathrm{m}}^3$, $C_{D0}=0.020$ and $k=0.045$. Then $A=0.12$, $B=67500$ in the corresponding SI units, $V_{\mathrm{BG}}\simeq 27.39m/s$, and the model gives $w\simeq 1.643m/s$. Its shallow-glide range ratio is $V/w\simeq 16.67$. These parameters are invented; an attainable lift coefficient and a valid drag polar are prerequisites, not consequences of the optimization.

Wind changes the ground-range objective to $(u+U_w)/w$ for an along-track wind component $U_w$ and no vertical air motion. Therefore the still-air optimum need not maximize ground distance. The FAA constructs the corresponding shifted tangents on its performance polars.^[ref] A controlled performance measurement records steady airspeed and height loss at several speeds; wind, vertical air motion and changes of loading must be distinguished from the polar itself.