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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 wing aspect ratio is the dimensionless measurewhere $b$ is wingspan and $S$ is projected wing area. For a rectangular wing planform, $S=bc$ and therefore $AR=b/c$, where $c$ is airfoil chord. For other wing planforms, dividing the span by a chosen root chord need not give the wing aspect ratio.^[ref]

Its aerodynamic importance comes from lift-induced drag. The wingtip vortices of a finite aircraft wing change the local flow direction; maintaining lift then entails an additional rearward force. A standard model writesHere $e$ is a span efficiency factor, $q$ is dynamic pressure and $C_L=L/(qS)$ is lift coefficient. The model compares a specified spanwise loading in a regime where this finite-wing approximation is appropriate; $e$ need not stay unchanged when the design changes.^[ref]

For an original example, take $S=16{\mathrm{m}}^2$, $q=500\mathrm{Pa}$, $C_L=0.6$ and $e=0.8$. Both wings then supply $L=4800\mathrm{N}$. Spans of 8 and 12 metres give $AR=4$ and $9$, respectively, and lift-induced drag approximately $286.48$ and $127.32\mathrm{N}$. Under those stipulated conditions, increasing span by 50% reduces that drag component by $1-(8/12{)}^2=5/9$, about 55.6%.

The boxed expression also exposes a common comparison error. At fixed $L$, $q$, $e$ and $b$, changing only the reference wing area changes both wing aspect ratio and lift coefficient; their effects cancel in this model's dimensional lift-induced drag. A larger numerical $AR$ is therefore not, by itself, a promise of smaller total drag. Structural loads, wing loading, surface losses and aerodynamic stall impose further design constraints.

Wing loading is the gravitational weight of an aircraft divided by its reference wing area:It is a whole-aircraft loading measure. In SI its units are $N/{\mathrm{m}}^2$, not $kg/{\mathrm{m}}^2$. The latter units describe mass per wing area, $m/S$, and require multiplication by gravitational acceleration to obtain $W/S$.

For an original unit check, $m=500\mathrm{kg}$, $S=10{\mathrm{m}}^2$ and $g=9.81m/{\mathrm{s}}^2$ give $m/S=50kg/{\mathrm{m}}^2$ but $W/S=490.5N/{\mathrm{m}}^2$. Although this has the units of pressure, it is not the local pressure distribution on the wing surface.

In straight level flight with negligible vertical thrust, lift balances gravitational weight.^[ref] Combining this with the definition $L=\frac{1}{2}\rho V^{2}S{C}_L$ of the whole-wing lift coefficient givesAt fixed air mass density and required coefficient, airspeed therefore scales as the square root of wing loading. For stipulated $W/S=600N/{\mathrm{m}}^2$, $\rho =1.2kg/{\mathrm{m}}^3$ and $C_L=1$, the result is $31.62m/s$. Doubling only the loading gives $44.72m/s$, a factor of $\sqrt{2}$.

If a model assumes a maximum attainable $C_{L,\max }$, the same equation gives a corresponding minimum speed for that straight-level condition. It is not a universal speed for every attitude or configuration: a turning condition can require $L>W$, and a changed airfoil, surface condition or Reynolds number can change the attainable coefficient.

A geometric and weighing check determines $m$, the relevant loading state and projected wing area. Those measurements establish $W/S$; a separate aerodynamic experiment is needed to establish the applicable lift coefficient. Equal wing loading does not establish equal lift-to-drag ratio or identical aerodynamic stall behavior.