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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 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.

A fixed-wing aircraft obtains most of its aerodynamic support from aircraft wings that do not continuously rotate around a shaft to generate lift. The word fixed distinguishes the lifting arrangement; it does not require every surface to be immovable or perfectly rigid. Both powered airplanes and an unpowered glider can use fixed wings.

Thrust and aerodynamic support are distinct. Relative motion through air produces lift and drag; an engine can supply the thrust needed to balance drag in steady level flight. The components must be resolved as vectors, since their directions change with the motion and attitude.^[ref]

For an original ideal turning model, suppose a point-mass aircraft maintains horizontal circular flight at constant airspeed $V$, with negligible vertical thrust and no net vertical acceleration. If its lift vector is inclined through a bank angle $\varphi$, thenThe horizontal component supplies centripetal acceleration. Eliminating $L$ gives $L/W=1/\cos \varphi$ and $R=V^2/(g\tan \varphi )$. With $V=40m/s$, $\varphi =30{}^{\circ }$ and $g=9.81m/{\mathrm{s}}^2$, the radius is about $282.5\mathrm{m}$ and the required lift is about $1.1547W$. The wing remains fixed to the body while its force direction changes in space.

This is a Newtonian force model, not a prediction that an arbitrary airplane can produce the required lift. The attainable lift coefficient, wing loading, propulsion and aerodynamic moments determine whether a particular steady condition can exist. In a steady unpowered descent, the balances instead lead to the glide angle relation.

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.

A glider sink polar relates a glider's steady downward speed relative to the air to its true airspeed, for specified loading, configuration and atmospheric conditions. Write it as $w(V)>0$ with downward motion positive. Some plots instead put descent below the horizontal axis; the sign convention must accompany the curve. The minimum sink speed and best glide speed select different points on this same performance relation.^[ref]

The curve is a measurement or model of performance, rather than a drawing of the flight path. At steady speed, its horizontal air-relative component is $u=\sqrt{V^2-w^2}$. Thus time per height lost is $1/w$, while horizontal distance per height lost is $u/w$. The best glide speed article derives the tangent construction and an original parabolic drag polar example; those results need not be recomputed to interpret a measured polar.

A useful extension is glider ballast scaling. Suppose the gravitational weight changes from $W$ to $\alpha W$ at unchanged air mass density and geometry, and suppose the same lift coefficient and drag coefficient remain applicable. At corresponding steady glide conditions, the lift-to-drag ratio and glide angle are unchanged. Since $L=\rho V^{2}S{C}_L/2=W\cos \gamma$,For an original stipulated point $V_1=25m/s$ and $w_1=0.80m/s$, increasing gravitational weight by $25\%$ gives $V_2\simeq 27.95m/s$ and $w_2\simeq 0.8944m/s$. The corresponding distance-to-height ratio is unchanged, but the same height is lost sooner. This scaling assumes the aerodynamic coefficients transfer; changes of Reynolds number, surface condition or trim can invalidate it. The FAA illustrates the loading-dependent displacement of glider polars.^[ref]

To measure a polar, record steady true airspeed and height change at several operating points while documenting configuration and loading. Ground-relative descent $s$ includes vertical motion of the air: with upward air velocity $u_z$, $s=w-u_z$. For example, an air-relative sink of $1.5m/s$ in air rising at $0.5m/s$ appears as $1.0m/s$ ground-relative descent. Repetition can reveal variability, but one trajectory cannot separate these two contributions without additional information. A measured groundspeed is likewise not automatically the horizontal component of the air-relative polar.

Groundspeed is the magnitude of an aircraft's horizontal velocity relative to the ground. It determines travel time along a ground track, whereas aerodynamic lift and drag depend on motion relative to the air. Let ${\mathbf{v}}_{g,h}$ and ${\mathbf{v}}_{a,h}$ be the horizontal ground-relative and air-relative velocity vectors, and let ${\mathbf{w}}_h$ be the horizontal wind velocity:For climbing or descending flight this horizontal convention distinguishes groundspeed from the magnitude of the full three-dimensional ground-relative velocity.^[ref]

In an original horizontal-flight example, an airspeed of $50m/s$ directed due north and a wind blowing east at $15m/s$ give ${\mathbf{v}}_{g,h}=(15,50)m/s$ in east/north coordinates. Thus $V_g=\sqrt{2725}\approx 52.20m/s$, with the track about $16.70{}^{\circ }$ east of north. Adding scalar speeds would wrongly give $65m/s$.

To keep a due-north track in the same wind, the aircraft must instead supply an air-relative westward component of $15m/s$. At the same airspeed, its northward component becomes $\sqrt{50^2-15^2}\approx 47.70m/s$, so the on-track groundspeed is smaller than the airspeed. The required wind correction angle is $\arcsin (15/50)\approx 17.46{}^{\circ }$ west of north. Heading and track are distinct directions.

Timed positions from a ground-based tracker estimate groundspeed, not airspeed. Inferring the latter also needs the wind velocity. Conversely, a Pitot-static tube does not measure travel over the ground. The same distinction explains why a headwind changes the ground range of a steady glider without changing its air-relative glide angle.

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.