Fixed-wing aircraft

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 bank angle measures an aircraft's lateral tilt relative to a local horizontal reference. In the ideal fixed-wing aircraft turning model, the lift vector tilts with the vehicle. For more general motion, vehicle attitude, flight-path direction and aerodynamic force direction must be distinguished; a banked vehicle need not follow a level circle.^[ref]

For steady horizontal circular motion with negligible vertical thrust and lift tilted by $\varphi$, the force balance developed in fixed-wing aircraft gives the aerodynamic load factorAt $\varphi =60{}^{\circ }$, the required lift is twice the gravitational weight. If the maximum usable lift coefficient is held fixed, $L=\rho V^{2}S{C}_L/2$ implies the model's limiting airspeed scales asA stipulated wings-level limit of $20m/s$ therefore becomes $28.28m/s$ at this loading. This is a comparison at fixed configuration and air mass density, not a claim that a particular aircraft can sustain the required condition. Aerodynamic stall and structural limits constrain the accessible turn.

A banked glider also needs a different loading balance from the shallow straight-glide model used to define minimum sink speed. Increasing bank angle cannot be analyzed merely by rotating a drawing of its still-air trajectory. Measured performance must specify the turn condition as well as airspeed; the FAA's performance discussion illustrates how bank and loading affect glider operation.^[ref]

The aerodynamic load factor $n=L/W$ compares lift with gravitational weight. The stated convention takes $L$ along the chosen positive lift direction, so $n$ is a signed force ratio, not simply the magnitude of total acceleration divided by $g$. A steady level condition can have $n=1$ with zero inertial acceleration.^[ref]

A focused example is a wings-level pull-up at the bottom of a vertical curved path. Neglect other vertical forces and take upward normal acceleration as positive. The force balance givesFor stipulated true airspeed $V=50m/s$, curvature radius $R=500\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the upward centripetal acceleration is $5m/{\mathrm{s}}^2$ and $n\simeq 1.510$. With $m=600\mathrm{kg}$, lift is $8886\mathrm{N}$, while the net upward force is $3000\mathrm{N}$. Neither is the other.

In the steady horizontal-turn model, the condition instead is $n=1/\cos \varphi$; bank angle derives that balance. It does not apply unchanged to the vertical pull-up. At fixed air mass density, wing area and usable maximum lift coefficient, the relation $L=\rho V^{2}S{C}_L/2$ gives limiting speed proportional to $\sqrt{n}$ for positive $n$. For this example, the factor is about $1.229$ relative to the corresponding one-load condition.

A force balance measures aerodynamic load; a trajectory measurement estimates inertial acceleration. Comparing them requires the gravitational weight term and correct directions. A reported load factor also needs its force-axis convention: thrust, side forces and coupled motion can matter to what an instrument actually records. The definition alone establishes neither structural capability nor available lift before aerodynamic stall.

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.

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.

The minimum sink speed is the airspeed at which a glider has its smallest steady downward speed relative to the air, for a specified configuration, loading and flight condition. It maximizes time aloft from a given height in still air. It does not generally maximize horizontal range, which defines best glide speed.^[ref]

For a differentiable glider sink polar $w(V)>0$, an interior minimum satisfies $w^{\prime }(V)=0$, with a positive second derivative sufficient for a strict local minimum. For the original shallow-glide model developed in best glide speed,With $A,B>0$, this givesIndeed $w^{\prime \prime }=(6AV+2B/V^3)/W>0$ for $V>0$. The speed ratio is about $0.7598$, within this stipulated parabolic drag polar approximation. A real polar need not follow that numerical ratio, especially near aerodynamic stall.

Using the same original $W=3000\mathrm{N}$, $A=0.12$ and $B=67500$ gives $V_{\mathrm{MS}}\simeq 20.81m/s$ and $w_{\min }\simeq 1.442m/s$. From a stipulated $1000\mathrm{m}$ height, the model's times are about $693.6\mathrm{s}$ at minimum sink and $608.6\mathrm{s}$ at best glide. Yet the approximate range ratios are $14.43$ and $16.67$, respectively. The slower optimum buys time rather than maximum distance.

A horizontal wind changes range without changing air-relative sink at a fixed condition. Vertical air motion changes ground-relative height loss as well: descending at $w$ through air rising at $u_z$ gives ground-relative descent $w-u_z$. Observing constant height therefore does not establish zero air-relative sink. A bank angle also changes the required lift and hence the appropriate polar; the straight-glide optimum cannot simply be reused for a tight turn.