Thickness carries no load of its own and still moves the load forward

The one thing that does not add up made a precise statement out of a vague one. Laplace’s equation is linear, so flows can be laid on top of one another; the pressure is quadratic in the velocity, so pressures cannot. Expanded about a free stream, though, the pressure coefficient is $-2u'/U$ plus terms quadratic in the disturbance, and to first order in the disturbance pressures do add. Adding them is exactly as legitimate as linearising and no more. That essay then named the place this licence is used most and estimated the cost: thin-aerofoil theory computes a section’s thickness, camber and incidence separately and adds their pressures, and for a twelve-per-cent section at four degrees the neglected cross terms should be “a few per cent” — largest at the leading edge, where the disturbance is largest.

The estimate can be replaced by a measurement, because Joukowski’s map makes every piece exact. A circle through $z = c$, mapped by $\zeta = z + c^2/z$, becomes a section with a sharp trailing edge; the circle’s centre decides the shape — moved left it thickens the section, moved up it cambers it — and the flow round the circle with the Kutta condition is an exact flow round the section. So the twelve-per-cent cambered section, the symmetric section of the same thickness, the circular-arc camber line and the flat plate are four exact flows, and the question “what does adding three of them give for the fourth” has an exact answer.

Four exact flows #

The section is fixed by asking for a thickness of 12 per cent of its chord and a camber of 4 per cent, measured on its own outline, and solving for the circle that gives it: a centre at $(-0.1015, 0.0881)$ with $c = 1$. Its three parts are the same circle with one coordinate removed at a time and the incidence assigned where thin-aerofoil theory assigns it: the symmetric section, from the centre $(-0.1015, 0)$, at no incidence; the circular arc, from $(0, 0.0881)$, at no incidence; and the flat plate, from the origin, at the section’s 4°. Each is computed on its own surface, and each surface is read at the same fraction of its own chord.

On the circle the surface speed is the classical $2U|\sin(\theta - \alpha) + \sin(\alpha + \beta)|$, with $\beta$ the angle at which the centre sees the trailing edge, and the map divides it by $|1 - c^2/z^2|$. Thin-aerofoil theory’s sum is then the free stream plus the three parts’ disturbances — each part’s surface speed less the free stream — at each station on each surface. The pressure can be taken from that sum linearly, $C_p = -2\sum u'$, which is what the theory licenses, or through Bernoulli, $C_p = 1 - (1 + \sum u')^2$, which keeps the quadratic terms but still adds the velocities.

What each part is, in the linear theory #

The three parts are not arbitrary pieces; each is what thin-aerofoil theory solves for with a different singularity on the chord line, and the exact Joukowski parts are those singularity problems solved without linearising. The thickness is a distribution of sources and sinks along the chord, with strength proportional to the slope of the thickness — the construction of bodies made out of nothing, a source ahead and a sink behind closing into a symmetric shape — and it speeds the flow over both surfaces alike. The camber is a vortex sheet along the camber line whose strength is chosen so the line is a streamline; the incidence is another vortex sheet on the flat chord, singular at the nose. In the linear theory each sheet’s velocity is computed on the chord, as if the section had no thickness, and the three are added — flows add up, so the velocity fields really do.

What the linearisation gives up is twofold, and the exact parts separate the two. Each part’s own disturbance enters the pressure squared, and that is the error a Bernoulli sum repairs: it is a property of each piece alone. And each piece’s singularities sit on the chord line rather than on the surface the flow actually follows; the camber’s vortices and the incidence’s should be laid on the thick section’s surface, where the thickness has already sped the flow up. That second error is what no sum of parts can repair, because it is the parts acting on one another, and it is what remains after Bernoulli.

A quarter of the suction, lost in the adding #

The linear sum is good in the middle of the chord and poor near the nose, as the cross-term argument predicted, and worse than its estimate. At a tenth of the chord on the upper surface the exact suction is $-C_p = 1.39$ and the linear sum gives 1.06, a quarter too little; at two per cent of the chord the shortfall is 0.27 of a dynamic pressure. At half chord the sum is 0.10 too little, and towards the trailing edge it turns slightly the other way.

The linear sum’s error is the dropped quadratic term, $-|u'|^2$, and a twelve-per-cent section at four degrees is not a small disturbance near its nose. The surface speed there is half as much again as the free stream, $u' \approx 0.55$, so the dropped term is about $0.3$ of a dynamic pressure — the same size as the error measured. Carrying the summed speeds through Bernoulli restores it, and the error near the nose falls by more than half and changes sign; along most of the chord it falls to two hundredths. Most of what “pressures do not add” costs on a section is the square of each part’s own disturbance, which no theory that adds velocities and then squares them has to give up.

What Bernoulli cannot restore is the part of the error that is not in any single piece. The remaining 0.12 at the nose is the interaction of the pieces: the flat plate’s leading-edge velocity, $\sin\alpha\sqrt{(1-x)/x}$ in thin theory, is computed on a plate with a sharp nose, and on the thick section the same circulation flows round a rounded nose of radius proportional to the thickness squared. The thickness changes where and how sharply the incidence’s flow turns, and the change is not a sum. The singular edge of a flat plate carries a finite suction force in an infinite speed; the thick section carries the same force spread over a finite peak, and a superposition of the plate’s flow with a thickness flow places that peak wrongly.

Thickness moves the load #

Thin-aerofoil theory’s most useful claim is about the loading — the difference between the lower and upper surfaces’ pressures — and the claim is that a symmetric thickness distribution carries none: it raises the speed equally on both surfaces, and their difference is the camber’s and the incidence’s alone. To first order that is true. On the exact section it is not.

At a tenth of the chord the exact loading is 1.59 and the parts’ sum 1.26; at nine-tenths, 0.40 against 0.51. The section carries more of its load forward and less aft than its camber line and flat plate would together, and more of it in total. Thickness carries no load of its own and still moves the load forward and adds to it. The mechanism is the same cross term seen through the loading: thickness speeds the flow over both surfaces, and the incidence’s and camber’s disturbances ride on that faster flow, so their contribution to the pressure difference is multiplied by the local speed the thickness sets — largest near the nose, where the thickness’s own disturbance is largest. The loading, linear in each piece, is bilinear in the pieces together.

That has a practical consequence the lift alone would hide. The pitching moment about the quarter chord, which the camber decides in thin theory, moves when load moves forward, and so does the centre of pressure; a section designed for a moment with thin-aerofoil theory and then thickened has a different moment, by an amount that does not appear until the cross term is kept. It is one reason a designer who starts from three numbers out of a camber line finishes with a panel method.

Nine per cent of the lift #

The integral of the loading is the lift, and the shift adds up. The exact section lifts $C_L = 1.026$; its camber line and flat plate, each computed exactly, lift 0.992 between them, and thin-aerofoil theory’s $2\pi(\alpha + 2h)$ gives 0.941. The section lifts nine per cent more than thin theory and three and a half per cent more than its exact parts. Swept over thickness the excess grows in proportion: 4.4 per cent at six per cent thickness, 9.0 at twelve, 13.7 at eighteen.

Joukowski’s own formula says why. The section’s lift is $\rho U\Gamma$ with $\Gamma = 4\pi a U\sin(\alpha + \beta)$ fixed by the Kutta condition, and the circle’s radius $a$ exceeds $c$ by the thickness parameter, while the chord grows by less; per unit chord the lift slope is $2\pi(1 + 0.77\,t/c)$ to first order in the thickness, with $0.77 = 4/(3\sqrt3)$. The computed factor, $1 + 0.75\,t/c$ at this camber and incidence, agrees with it to the second-order terms. So the lift curve’s slope is not $2\pi$ for a thick section and the lift curve’s textbook slope is the slope of a plate.

The factor hardly depends on anything but the thickness. For a twelve-per-cent section it is 1.093 with no camber at no incidence, 1.092 with two per cent camber, 1.090 with four at four degrees and 1.085 with six at eight degrees: the camber and the incidence change it only in the third figure, as their own second-order terms do. So the cross term is simple in the one quantity it changes most — the lift is the thin theory’s lift times a function of the thickness alone — and complicated only in where along the chord it puts that lift. A thin-aerofoil calculation corrected by multiplying its lift by $1 + 0.77\,t/c$ gets the lift of a Joukowski section to a few parts in a thousand and its loading wrong by a fifth at the nose.

Measured sections do not show the nine per cent, and the reason is the boundary layer: its displacement thickness, largest over the rear upper surface, decambers the section and takes back roughly what the thickness added, so real twelve-per-cent sections lift at very nearly $2\pi$. Thin-aerofoil theory’s agreement with measurement is in part a cancellation between the cross term it drops and the viscosity it omits — which is the most reassuring and least defensible kind of agreement there is.

Where the adding works #

None of this condemns the method. At zero incidence the linear sum’s error at a tenth of the chord is 0.06 rather than 0.33; at two degrees 0.19; at eight, 0.71. The cross term grows with the lift, so a section at its design lift coefficient, where the camber does the work and the incidence is small, is exactly where adding the pieces is most accurate, and that is where thin-aerofoil theory is used: to choose a camber line for a design lift and a thickness for a structure, separately. The theory fails where the section is asked for lift beyond its design point — at the incidences of take-off and landing — and there it fails at the nose, in the suction peak that decides the stall.

The cross-term argument put the rule as “carry the field, not its consequences”. The exact section adds a sharpened version: carry the velocities, square them once, and expect the residue to be the interaction of the pieces. On a section that residue is thickness times lift, it is first order in each, and it puts load forward.

What the lost suction would have predicted #

The suction peak matters out of proportion to its share of the lift because the boundary layer has to climb back from it. The adverse gradient behind the peak, from a $-C_p$ of 1.39 at a tenth of the chord down to the trailing-edge value, is what decides whether the layer stays attached, and where the straight line of the lift curve stops is where it does not. A stall estimate made from the linear sum sees a peak a quarter too weak and a recovery a quarter too gentle, and so predicts a stall later than the section will have it; at eight degrees, where the linear sum’s peak is two-thirds of the exact one, the optimism is large. Carried through Bernoulli the summed speeds get the peak to within a few per cent, which is why the practical versions of thin-aerofoil theory square the velocities before reporting a pressure.

The same cross term decides which part of a wing stalls first when the sections along a span differ in thickness. A thicker root section carries its load further forward and peaks harder at the same local lift than its camber line alone would say, and the error is in the direction that brings the root’s stall forward, not back.

The convention, and what the picture leaves out #

Every quantity is for an inviscid, incompressible flow with the Kutta condition at a sharp trailing edge. Thickness and camber are measured on the exact outline as fractions of its own chord; each part is read at the same fraction of its own chord, which is the comparison thin-aerofoil theory makes, since it applies every piece on the chord line. The parts’ disturbances are their surface speeds less the free stream, taken with their signs on each surface. Joukowski sections have cusped trailing edges and a particular thickness distribution; a NACA section’s interaction differs in detail, and only a panel method would give it. And the flow is inviscid: the boundary layer that cancels the thickness’s lift excess on a real section is the subject of the body the outer flow actually sees.

Who found the parts #

Joukowski published his sections in 1910 and Kutta the condition that fixes their circulation in 1902 and 1910. Munk built thin-aerofoil theory on superposition in 1922, and Glauert set it in its textbook form in 1926, with the thickness, camber and incidence problems solved apart and added. That thickness raises the lift slope of a Joukowski section by $1 + 0.77\,t/c$ was in Glauert’s book; Riegels and Lighthill in the 1940s and 1950s worked out how to make the linear theory’s leading-edge pressures right by a factor depending on the local nose radius, which is a correction for exactly the cross term measured here.

How it was checked #

The flat plate’s upper-surface speed matches thin-aerofoil theory’s exact plate solution, $\cos\alpha + \sin\alpha\sqrt{(1-x)/x}$, to five parts in ten million at four stations — the plate being the one part for which the linear and the exact theories coincide. The fitted circle gives a section 12.000 per cent thick with 4.000 per cent camber, to a part in ten million. The normal force from integrating the exact section’s surface pressure along the chord equals Kutta–Joukowski’s lift times $\cos\alpha$ to one part in a hundred thousand. The lift excess at six, twelve and eighteen per cent thickness lies within a hundredth of Joukowski’s $1 + 0.77\,t/c$. And the section refuses a circle centred to the right of the axis and one too high to map to a section.

Still open: the interaction on a section a panel method draws #

Joukowski sections are a convenient family because every part is exact, and an unrepresentative one because their thickness peaks at a quarter chord with a cusped tail. The next calculation does the same decomposition on a NACA four-digit section with a panel method: the section, its symmetric thickness form, its camber line and a flat plate, each solved by panels at matching stations, and asks whether the loading shift and the lift excess follow the same law in thickness, how the shift depends on where the thickness is concentrated, and how much of it a nose-radius correction of Riegels’s kind recovers.