ISEGORIA / MATH ENCYCLOPEDIA
Aerodynamics: how a wing makes lift
A wing works by turning the air that passes it downward, and the air pushes back. Explaining how much it turns takes three ideas. Around an aerofoil section, the flow outside the thin boundary layer is very nearly ideal, and the lift is fixed by one number, the circulation, which the sharp trailing edge selects (the Kutta condition); Joukowski found a conformal map that solves the problem exactly. A wing of finite span leaks air around its tips, so vortices trail behind it, wash the air down, and tilt the lift backwards into induced drag; Prandtl’s lifting line turns this into one integral equation. Finally, an aircraft’s drag polar decides how far it can glide and how slowly it can sink, which is the arithmetic every glider and airliner is designed around.
Before you begin: Complex numbers, vector calculus, Fourier series
Predict, manipulate, then test yourself: every experiment ends with practice questions that are generated fresh at three levels. Graphs illustrate the mathematics; they do not replace a proof.
1. The Joukowski aerofoil: lift by conformal map
Flow past a circular cylinder is easy: a uniform stream, a doublet and, optionally, a vortex of circulation \(\Gamma\). The map \(z=\zeta+1/\zeta\) is conformal everywhere except \(\zeta=\pm1\), so it carries that flow to a flow past whatever shape the circle becomes. A circle through \(\zeta=1\), centred slightly left of the origin and above it, becomes a rounded, cambered aerofoil with a sharp trailing edge at \(z=2\). Any \(\Gamma\) gives a valid ideal flow, but only one keeps the velocity finite at the sharp edge: the one that puts the rear stagnation point of the circle exactly at \(\zeta=1\). That is the Kutta condition, and with it the Kutta–Joukowski theorem gives the lift. Move the circle and the angle of attack, and read the pressure.
Worked example. Centre the circle at \(\mu=-0.08+0.08i\). Its radius is \(a=|1-\mu|=1.0830\) and the trailing edge sits at angle \(\beta=\arcsin(0.08/1.083)=4.24^\circ\) below the centre, so the aerofoil is 9.7% thick and has zero lift at \(-4.24^\circ\). At \(\alpha=5^\circ\) the Kutta condition sets \(\Gamma=4\pi\times1.083\times\sin9.24^\circ=2.184\) (with \(U=1\)). The chord is 4.022, so \(C_L=2\Gamma/c=1.086\). Thin-aerofoil theory, \(2\pi(\alpha+\beta)=1.013\), is 7% low: thickness adds lift, roughly in proportion to \(1+0.77\,t/c\). Integrating the pressure around the surface gives the same 1.086.
Watch out. This is inviscid, incompressible flow. It predicts no drag at all (d’Alembert’s paradox) and no stall: real aerofoils separate from the upper surface somewhere between 12° and 18° and lose lift, while here the lift keeps rising. Joukowski aerofoils have a cusped trailing edge with zero angle, unlike practical sections, and their pressure peak near the nose is sharper than a modern section would allow.
Why does the sharp trailing edge decide the circulation?
Ideal flow allows any circulation, and with the wrong one the air would have to whip around the sharp edge, where the speed would be infinite. Real air cannot do that: the viscous boundary layer separates and sheds a starting vortex until the flow leaves the edge smoothly. By Kelvin’s theorem the wing is left with an equal and opposite circulation, and that is exactly the Kutta value.
2. A finite wing: trailing vortices and induced drag
On a real wing the high pressure underneath spills around the tips, so the lift must fall to zero there. Helmholtz’s theorem says a vortex line cannot end in the air, so wherever the circulation \(\Gamma(y)\) changes along the span, a vortex of strength \(-d\Gamma/dy\) trails away downstream. Together these induce a downward velocity at the wing, the downwash, which lowers each section’s effective angle of attack by \(\alpha_i\) and tilts its lift backwards: induced drag, the price of making lift with a wing of finite span. Prandtl wrote \(\Gamma\) as a sine series and turned the problem into a linear system. Change the planform, the aspect ratio and the twist, and compare the loading with the ideal ellipse.
Worked example. An untwisted elliptic wing of aspect ratio 8 has a purely elliptic loading, \(A_n=0\) for \(n>1\), so \(e=1\) and its lift slope is \(2\pi/(1+2/8)=5.03\) per radian instead of the aerofoil’s \(2\pi\). At \(5^\circ\) that is \(C_L=0.439\) and \(C_{D_i}=0.439^2/(8\pi)=0.0077\). A rectangular wing of the same aspect ratio loads its tips too heavily for their chord: \(e=0.937\), \(C_L=0.422\). Tapering it to a tip chord 0.4 of the root brings \(e\) back to 0.987, almost elliptic, which is why so many wings are tapered.
Watch out. Lifting-line theory assumes an unswept wing of large aspect ratio (above about 4) with attached flow and a lift slope of \(2\pi\) per radian for every section; it uses straight trailing vortices, which in reality roll up into two tip vortices within a few spans. The drawn vortex thickness is proportional to the square root of the strength shed in each strip. Sections are symmetric, so the zero-lift angle is zero; washout twists the tip nose-down linearly.
Why do gliders have such long, thin wings?
Induced drag is \(C_L^2/(\pi e\,AR)\), so for the same lift it falls in inverse proportion to the aspect ratio. A glider flies slowly at high \(C_L\), where induced drag dominates, so an aspect ratio of 20 to 40 pays for its weight and structural difficulty. Physically, a longer span pushes a larger mass of air down more gently, and kinetic energy left in the wake is what the drag pays for.
3. The glide polar
An aircraft’s drag is roughly a constant part, skin friction and form drag, plus the induced part, which grows as the square of the lift coefficient. This parabolic drag polar decides its performance. In a steady glide the lift-to-drag ratio is the glide ratio, so the best glide is at the lift coefficient where induced drag equals parasite drag, found by drawing a tangent from the origin. A glider staying aloft in a thermal wants the least sink rate instead, which comes at a slower speed. Weight, through the wing loading \(W/S\), moves the whole speed polar to higher speeds without changing its shape.
Worked example. Take a glider with aspect ratio 18, \(e=0.9\), \(C_{D0}=0.012\) and wing loading 350 N/m² at sea level. Then \(K=1/(\pi\times0.9\times18)=0.0196\), the best glide ratio is \(1/(2\sqrt{0.0196\times0.012})=32.6\) at \(C_L=0.781\), flown at \(\sqrt{2\times350/(1.225\times0.781)}=27.0\) m/s with a sink rate of \(27.0/32.6=0.83\) m/s. Minimum sink is at \(C_L=\sqrt3\times0.781=1.354\), at 20.5 m/s, where \(C_D=4C_{D0}\) and the sink is 0.73 m/s. Adding ballast to double the wing loading leaves the glide ratio at 32.6 and raises every speed by \(\sqrt2\).
Watch out. The parabolic polar is an idealisation: \(C_{D0}\) actually varies with \(C_L\) and with speed (Reynolds number), and a real polar is measured. The stall limit \(C_{L,\max}=1.4\) is a typical value without flaps. The minimum-sink point is close to the stall here, as it is for many gliders, which is why pilots circle in thermals a little faster than the formula says. Air density is the sea-level value.
Why does weight not change the best glide ratio?
The glide ratio is \(L/D=C_L/C_D\), which depends only on the lift coefficient, not on the weight. A heavier glider simply flies the same \(C_L\) faster, since lift \(\tfrac12\rho V^2SC_L\) must equal the larger weight, so \(V\propto\sqrt{W/S}\). It covers the same distance per metre of height but more quickly, which wins races in strong conditions.