Aircraft

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 aircraft yaw axis is the body-fixed axis about which the nose rotates from side to side. In a right-handed body coordinate system with $x$ forward and $y$ toward the right wing, it is the $z$ axis directed downward through the centre of mass. It completes the orthogonal set with the aircraft roll axis and aircraft pitch axis. The rudder can generate a side force aft of the centre of mass, producing a yawing torque.^[ref]

For a force $\mathbf{F}$ acting at displacement $\mathbf{r}$ from the reference point, the yaw component of $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$ isIn an original example, a rightward $80\mathrm{N}$ tail force at $\mathbf{r}=(-3,0,0)\mathrm{m}$ gives ${\tau }_z=-240\mathrm{N}\mathrm{m}$. With positive yaw defined nose-right, this force turns the nose left. If the axis is a principal axis with moment of inertia $I_z=1200\mathrm{kg}{\mathrm{m}}^2$ and the initial motion is pure yaw, the instantaneous angular acceleration is $\dot{r}={\tau }_z/I_z=-0.20rad/{\mathrm{s}}^2$. A tail's displacement matters as well as the force magnitude.

The aircraft yaw rate $r$ is a component of body angular velocity; it is not always the rate of change of aircraft heading. A banked vehicle projects its rotation onto several body axes, and its ground track also depends on wind. General coupled motion uses Euler's equations, rather than three independent scalar torque laws. Measuring a heading change alone therefore cannot establish the applied yawing moment.

The aircraft roll axis is the longitudinal body axis through the centre of mass, conventionally directed toward the nose. Positive rotation around that forward axis lowers the right wing in the right-handed system with $y$ right and $z$ down. An aileron changes the aerodynamic loading of a wing; unequal loads on opposite sides can generate a rolling torque.^[ref]

The roll component of $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$ isAn original pure-couple example places an upward incremental force of $50\mathrm{N}$ at $y=+3\mathrm{m}$ and a downward incremental force of the same magnitude at $y=-3\mathrm{m}$. With both application points at $z=0$, their total incremental force vanishes butThe negative sign means right-wing-up rotation. For an initially pure roll about a principal axis with moment of inertia $I_x=600\mathrm{kg}{\mathrm{m}}^2$, the instantaneous angular acceleration is $\dot{p}=-0.50rad/{\mathrm{s}}^2$. This is a stipulated force couple, not an aileron-performance prediction.

The aircraft roll rate $p$ describes rotation, while bank angle describes attitude relative to a local horizontal reference. A vehicle held at a constant bank angle can have zero change of that angle while its flight direction turns. Conversely, a rolling aircraft need not already have the steady centripetal force required by the fixed-wing aircraft turning model. Attitude, angular velocity and trajectory are separate measured quantities.

The aircraft pitch axis is the transverse body axis about which the nose rotates upward or downward. In the convention used by NASA Glenn Research Center, it passes through the centre of mass and points toward the right wing tip; the aircraft roll axis points toward the nose and the aircraft yaw axis toward the underside. These are body-fixed directions, so their orientation relative to the ground changes with the aircraft.^[ref]

Choose right-handed body coordinates $x$ forward, $y$ right and $z$ down. A force $\mathbf{F}$ acting at displacement $\mathbf{r}$ from the centre of mass produces torque $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$. Its pitching moment isFor an original tail-load example, $\mathbf{r}=(-4,0,0)\mathrm{m}$ and $\mathbf{F}=(0,0,100)\mathrm{N}$ give ${\tau }_y=+400\mathrm{N}\mathrm{m}$. The downward tail load lifts the nose in this convention. If the aircraft pitch axis is a principal axis, the motion is initially pure pitch and the corresponding moment of inertia is $I_y=2000\mathrm{kg}{\mathrm{m}}^2$, then the instantaneous angular acceleration is $\dot{q}={\tau }_y/I_y=0.20rad/{\mathrm{s}}^2$. Coupled three-axis rotation requires the full Euler's equations.

The axis is a geometrical reference, not necessarily a physical hinge. Changing a control-surface load changes both forces and torques; a zero net translational force does not establish zero pitching moment. A balance measurement therefore needs the reference point and sign convention alongside its reported moment.^[ref]

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 wind correction angle is the difference between the aircraft heading and the desired ground track needed to compensate for a crosswind. A heading points the vehicle; a track describes horizontal travel over the ground. In the horizontal model here, the air-relative velocity is aligned with the heading. Wind-direction statements below specify where the air moves, not the meteorological direction it comes from.^[ref]

Let $V$ be true airspeed, let $w_{\parallel }$ be wind along the desired track, and let $w_{\perp }$ be wind to its right. Define a positive correction $\delta$ as heading right of the track. Cancelling cross-track velocity and projecting along the track givesFor the branch with a nonnegative along-track air-relative component,The condition $|w_{\perp }|\le V$ permits crosswind cancellation, but forward progress also requires $V_g>0$.

In an original example, take $V=40m/s$, wind $8m/s$ rightward and $12m/s$ opposite the intended track. Then $\delta \simeq -11.54{}^{\circ }$ and $V_g\simeq 27.19m/s$. A constant-condition $100\mathrm{km}$ ground leg takes about $61.29\min$. With the same headwind, a rightward component of $39m/s$ can still be cancelled, but $\sqrt{40^2-39^2}-12\simeq -3.11m/s$: the resulting track-line motion is backward.

The calculation assumes a uniform horizontal wind and a steady true airspeed. Ground-position measurements can test the resulting groundspeed; they cannot determine the correction from air-relative speed alone without wind information. Wind variation and a difference between heading and air-relative velocity require a fuller vector model.

Airspeed describes an aircraft's motion relative to the surrounding air. When the kinematic magnitude is intended, it is the true airspeed,It differs from groundspeed whenever wind changes the relative motion. A ground-station velocity alone cannot establish the aerodynamic airspeed.^[ref]

An airspeed instrument normally infers motion from pressure. At low Mach number, for an aligned Pitot-static tube in the regime of the Bernoulli equation,Here $p_0$ is stagnation pressure, $p$ is static pressure and $\rho$ is air mass density. For a stipulated difference of $245\mathrm{Pa}$, $\rho =1.225kg/{\mathrm{m}}^3$ gives $V=20m/s$; the same difference at $0.6125kg/{\mathrm{m}}^3$ gives $28.28m/s$. The instrument senses a pressure difference, so interpreting its reading requires a density model and an instrument calibration.^[ref]

Indicated airspeed is the displayed reading before instrument and installation corrections. Calibrated airspeed includes those corrections. True airspeed expresses actual motion through the air, rather than simply relabeling either reading. The FAA distinguishes these quantities in its instrument chapter.^[ref]

The square-root calculation above assumes an appropriate low-speed stopping relation. At appreciable Mach number, compressibility changes that relation; a blocked passage or displaced static pressure source introduces a different measurement problem. Report which airspeed convention and corrections accompany a performance curve. Comparing speeds without that information can confound a change of air mass density with a change in the aircraft's loading. Functioning ports and known corrections are part of the inference, not extra measurements of groundspeed.

Calibrated airspeed is indicated airspeed corrected for instrument and installation errors. It remains a pressure-based speed convention; correcting the indicator does not by itself convert the reading into true airspeed. The FAA uses an airspeed-calibration chart to distinguish the indicated and corrected values.^[ref]

Represent a specified correction chart byThe correction can depend on the operating condition. In an original invented chart, suppose indicated $20$ and $40m/s$ correspond to calibrated $22$ and $41m/s$. Linear interpolation over this stipulated interval givesso an indication of $30m/s$ maps to $31.5m/s$. Neither a constant $2m/s$ correction nor a universal percentage reproduces both chart endpoints. Extrapolation outside the measured interval is an additional assumption.

At low Mach number, corrected pressure difference is approximately ${\rho }_0{V}_{\mathrm{CAS}}^2/2$, with ${\rho }_0$ the standard reference air mass density. The example's corrected pressure estimate is $(31.5/30{)}^2=1.1025$ times the estimate obtained by treating the indication as calibrated. The corresponding correction to an inferred aerodynamic force at fixed coefficient and area is therefore not merely the $5\%$ speed correction.

Instrument testing checks the mapping from applied pressures to the reading. Installed-system work also addresses static-pressure position error: a port can sample a local pressure different from the intended free-stream static pressure. The correction must accompany its configuration and reference conditions. To obtain true airspeed, account for ambient conditions as well. At appreciable Mach number, compressibility also separates calibrated and equivalent airspeed; the low-speed square-root density conversion must not be silently extended to that regime.

Indicated airspeed is the airspeed read directly from an indicator before correcting its instrument and installation errors. It is a displayed measurement, not automatically the aircraft's true airspeed or groundspeed. A conventional Pitot-static tube supplies total and static pressure through separate connections; the indicator converts their difference into a speed reading.^[ref]

A simple low-Mach number instrument model uses a standard reference mass density ${\rho }_0$ and the sensed difference $q_I$:This formula describes the stipulated ideal scale; real instrument errors need separate characterization. In particular, using ${\rho }_0$ does not assert that local air has that mass density.

For an original error example, let the actual total-minus-static difference be $245\mathrm{Pa}$, but suppose the static connection reads $12\mathrm{Pa}$ too high while the total-pressure connection is correct. Then $q_I=233\mathrm{Pa}$. With ${\rho }_0=1.225kg/{\mathrm{m}}^3$, the model shows $19.50m/s$ instead of $20.00m/s$. A positive error in one input has become a negative error in the difference. At lower pressure differences, the same offset has a larger fractional effect on the indicated speed.

A pressure-bench check can apply known differences to the instrument and test its zero and scale. It cannot alone characterize static-pressure position error caused by the ports' location on the flying aircraft. Calibrated airspeed accounts for the established instrument and installation corrections. A blocked port is another failure mode: treating every discrepancy as a correctable constant offset would conceal a changing or trapped pressure signal. The FAA specifically identifies obstruction and instrument deterioration as causes of incorrect indications.^[ref]

True airspeed is the magnitude of an aircraft's velocity relative to the surrounding air,It determines the relative motion used in aerodynamic models, while groundspeed concerns horizontal travel over the ground. Indicated airspeed and calibrated airspeed describe stages in a pressure-instrument reduction, rather than another reference frame.^[ref]

For an aligned low-Mach number pressure measurement in the regime of the Bernoulli equation,An equivalent airspeed uses the same dynamic pressure with a standard reference mass density, $V_{\mathrm{EAS}}=\sqrt{2q/{\rho }_0}$. HenceOnly where the low-speed pressure relations justify $V_{\mathrm{CAS}}\simeq V_{\mathrm{EAS}}$ may calibrated speed be substituted in this last expression without a separate compressibility correction.

For an original low-speed example, use $V_{\mathrm{CAS}}\simeq 30m/s$, ${\rho }_0=1.225kg/{\mathrm{m}}^3$ and ambient $\rho =0.800kg/{\mathrm{m}}^3$. The corrected difference is approximately $551.25\mathrm{Pa}$, giving $V_{\mathrm{TAS}}\simeq 37.12m/s$. At a fixed pressure difference, halving ambient mass density increases the inferred true airspeed by $\sqrt{2}$, rather than doubling it. This follows from the square-root dependence, not a wind correction.

Obtaining the ambient mass density can require independent static pressure and thermodynamic temperature measurements, with a suitable equation of state. For fixed $q$, a small relative density error changes the inferred speed by approximately $\delta V/V=-\delta \rho /(2\rho )$. Wind information is needed separately to predict the ground track. A single ground-travel speed cannot supply all of these quantities. The pressure inference also depends on clear passages and valid stopping-flow assumptions.^[ref]

A fuselage is the main body of a conventional aircraft, accommodating payload or equipment and connecting major assemblies such as the aircraft wings. It is part of the load-carrying structure as well as the aerodynamic shape. Not every aircraft has a separate conventional fuselage, and not every fuselage is pressurized.^[ref]

The structural role is more than enclosing a volume. Forces from attached assemblies and distributed gravitational weight must be transmitted through the body, producing bending, shear and axial loading. A pressurized body also carries loads from the pressure difference between its interior and the atmosphere. These are different loading mechanisms, even if they act on the same shell.

An original ideal model illustrates the pressure contribution. Approximate a closed cylindrical section as a thin-walled pressure vessel of radius $R$, wall thickness $t$ and internal excess pressure $\Delta p$. Force balance across a longitudinal cut of length $\ell$ gives $2R\ell \Delta p=2t\ell {\sigma }_{\theta }$. Balance of an end cap gives $\pi R^2\Delta p=2\pi Rt{\sigma }_z$. ThusThe first is hoop stress and the second axial stress. For $R=1.5\mathrm{m}$, $t=2\mathrm{mm}$ and $\Delta p=50\mathrm{kPa}$, they are $37.5$ and $18.75\mathrm{MPa}$. This stipulated membrane model excludes cutouts, joints, stiffeners and aerodynamic bending; it is not a complete fuselage stress analysis.

Shape also affects the combined flow around the body and wings. Richard T. Whitcomb's area rule work examined the longitudinal area distribution of the entire combination, rather than judging the wing and fuselage as isolated streamlined parts. That connection explains why structural packaging and aerodynamic design cannot always be optimized independently.

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.