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An aircraft is a vehicle capable of flight through the atmosphere. Its support can involve aerodynamic lift or buoyancy. A fixed-wing aircraft obtains aerodynamic lift from its aircraft wings as it moves relative to the surrounding air. An engine can supply thrust, but propulsion and the aerodynamic support of gravitational weight are distinct functions.^[ref]

For straight, level flight at constant velocity, neglect vertical thrust and align the remaining thrust with the flight direction. The force balance isHere $L$ is lift, $D$ is drag, $T$ is thrust and $W$ is gravitational weight. For a stipulated $1000\mathrm{kg}$ vehicle with $g=9.81m/{\mathrm{s}}^2$ and lift-to-drag ratio $L/D=12$, this requires $L=9810\mathrm{N}$ and $T=D=817.5\mathrm{N}$. At an air-relative speed of $50m/s$, $TV=40.875\mathrm{kW}$ is the useful propulsive mechanical power; an engine's input power can be larger because of losses.

An unpowered glider can instead maintain a steady descending velocity. If its glide angle below the horizontal is $\gamma$, resolving gravitational weight perpendicular and parallel to the flight path givesConsequently an assumed constant $L/D=12$ corresponds to $\gamma \simeq 4.76{}^{\circ }$ and 12 units of horizontal travel per unit of height lost in still air.^[ref] The loss of gravitational potential energy supplies the work against drag; the kinetic energy need not be falling during that steady descent.

These balances concern translation of the vehicle. Maintaining its orientation also requires a balance of torques about its centre of mass. An aircraft can satisfy an instantaneous vertical force balance while still having an unwanted pitching moment.

The glide angle is the inclination of a descending flight path below the horizontal. A positive $\gamma$ will denote descent here. For aerodynamic reasoning it is important to specify motion relative to the air: a steady wind can change the ground track without changing the air-relative glide angle at the same operating condition.

For a steady unpowered descent, resolve gravitational weight perpendicular and parallel to the air-relative velocity. With lift normal to that velocity and drag opposing it,The still-air horizontal distance per unit height lost is therefore $L/D$. NASA's description connects this geometry with a historical observation: the Wright brothers flew parallel to a dune slope, letting the slope provide an estimate of the flight-path angle.^[ref]

For an original constant-velocity model, set $L/D=20$ and airspeed $V=25m/s$. Then $\gamma =\arctan (1/20)\approx 2.8624{}^{\circ }$, horizontal air-relative speed is $u=V\cos \gamma \approx 24.9688m/s$, and downward speed is $w=V\sin \gamma \approx 1.24844m/s$. In still air, a stipulated $1000\mathrm{m}$ height loss corresponds to $20\mathrm{km}$ horizontally.

If a constant horizontal wind component $U_w$ is positive along the direction of travel, the ground-frame horizontal speed is $u+U_w$. With no vertical wind and $u+U_w>0$,A $10m/s$ headwind makes that ratio about $11.9900$; a tailwind of the same magnitude gives $28.0100$. All three cases retain the same air-relative angle and sink rate. Their groundspeeds and ground-frame path angles differ.

This distinction separates a measured ground trajectory from an aerodynamic lift-to-drag ratio. Estimating the latter from height and horizontal travel requires accounting for air motion and identifying the steady portion of the trajectory.

A glider is an aircraft designed for sustained flight without continuous engine thrust. Its aircraft wings obtain aerodynamic lift from motion relative to the air. It can descend steadily while its speed remains constant: loss of gravitational potential energy, rather than a continuous reduction of kinetic energy, can supply the work against drag.^[ref]

In still air and a steady unpowered glide, the mechanical power balance iswhere $V$ is airspeed and $w>0$ is downward speed. This follows from $D=W\sin \gamma$ and $w=V\sin \gamma$ in glide angle. Motion of the surrounding air changes the ground-relative energy account; rising air can offset a glider's descent relative to that air.

Two performance objectives should be distinguished. The minimum sink speed minimizes the downward rate, while the best glide speed maximizes horizontal distance per unit height lost in still air. They need not be the same speed.

For an original comparison, stipulate two steady operating points: $V_1=20m/s$ with $w_1=0.6m/s$, and $V_2=30m/s$ with $w_2=0.7m/s$. Their horizontal-distance ratios areAmong these two stipulated points, the slower one stays aloft longer for a given height loss, but the faster one covers more horizontal distance. From $1000\mathrm{m}$, the corresponding times are about $1666.7$ and $1428.6\mathrm{s}$. These invented points illustrate the competing objectives; they are not a measured performance curve.

A simple motivating experiment tracks a small unpowered model's position and height through a nearly steady portion of flight, then compares the vertical and horizontal motion. NASA identifies paper and lightweight model gliders as accessible examples.^[ref] Repeating with controlled geometry and launch conditions distinguishes the resulting motion from a claim based solely on a large wing aspect ratio.

The lift-to-drag ratio $L/D$ compares the lift supporting an aircraft with the drag opposing its motion. For $L>0$ and $D>0$, a larger ratio means less drag for a given lift. If both coefficients use the same dynamic pressure $q$ and reference wing area $S$, then $L/D=C_L/C_D$. Ratios using mismatched reference areas do not have this interpretation.

A parabolic drag polar provides a useful worked model:Here $C_{D0}$ is the zero-lift drag coefficient. In a simple finite-wing model, $k=1/(\pi eAR)$ relates lift-induced drag to wing aspect ratio and a span efficiency factor.^[ref] A fitted whole-aircraft $k$ can also incorporate additional effects; it need not be determined by geometry alone.

The maximum lift-to-drag ratio follows by differentiation:At this point the two modeled contributions to drag coefficient are equal. With the stipulated $C_{D0}=0.020$, $e=0.8$ and $AR=8$, $k\simeq 0.049736$, $C_L^{\ast }\simeq 0.6341$ and $(L/D{)}_{\max }\simeq 15.85$. The optimum is usable only if the actual wing can attain that lift coefficient in the regime where the model is valid; aerodynamic stall can intervene.

For steady unpowered descent in still air, the force balance gives $\tan \gamma =D/L$, where $\gamma$ is the glide angle. Thus the stipulated optimum corresponds to $\gamma \simeq 3.61{}^{\circ }$ and about $15.85$ units of horizontal travel per unit of height lost.^[ref] Wind and changing flight conditions alter travel over the ground. An airfoil section's ratio also should not be substituted for that of an entire aircraft, which includes other sources of drag.