Field

Circulation and lift

Where lift actually comes from. The Kutta condition, the Joukowski aerofoil, and lift derived rather than assumed.

A Joukowski aerofoil at 6°. A cambered aerofoil in a uniform stream. The circulation is not chosen: it is whatever value makes the flow leave the sharp trailing edge smoothly, and that single condition fixes the lift.

What actually holds a wing up

Not the shape, and not the story about air meeting up again behind. A wing lifts because there is circulation round it, and the sharp trailing edge is what decides how much.

A Joukowski aerofoil at 8°. A cambered aerofoil in a uniform stream. The circulation is not chosen: it is whatever value makes the flow leave the sharp trailing edge smoothly, and that single condition fixes the lift.

Ideal flow round a wing admits infinitely many solutions, each with a different lift, and all of them exact. One extra requirement — that the air leaves the trailing edge instead of whipping round it — picks a single one.

The lift curve, and why it is a straight line

Lift against angle of attack is a straight line, it does not pass through the origin, and its slope is very close to a number that has no business being there. All three facts fall out of the theory.

Lift with no wing at all

A spinning cylinder has no camber, no aerofoil section and no trailing edge, and it lifts exactly as hard as its circulation says it should. Which settles what lift is caused by.

The starting vortex, and the circulation it pays for. A wing that has just begun to lift, and the vortex it shed as it started. The circulation round the wing and the circulation round the shed vortex are equal and opposite, so a circuit large enough to contain both has no circulation at all — which is what Kelvin's theorem requires of a circuit that began at rest.

The vortex a wing leaves behind

A wing at rest has no circulation. A wing in flight has a great deal. Circulation round a circuit of fluid particles cannot change, so the difference had to come from somewhere — and it did, as an equal and opposite vortex dropped on the runway.

Finding the aerodynamic centre by sweeping the chord. How fast the pitching moment changes with incidence, plotted against where along the chord the moment is taken. The curve crosses zero once, and that crossing is the aerodynamic centre — the single point about which pitching the wing does not change the moment.

Where the lift acts

Lift gets drawn as an arrow, and an arrow has to start somewhere. The pressure is spread over the whole surface, so the choice of where to put the arrow is free — except that exactly one point on the chord has the property of not changing its answer when the wing is pitched.

Circulation across the span, for three planforms. How much circulation each part of the wing carries, plotted across the span. It has to reach zero at both tips, because a wing cannot carry circulation off its end, and the rate at which it falls is what determines the vorticity shed into the wake.

The price of having ends

An infinitely long wing in a fluid with no viscosity has no drag at all. Cut two ends into it and the drag appears — not from friction, which is still absent, but from the fact that circulation cannot be carried off the end of anything.

Where the flow stops, at Γ = -9. A cylinder with circulation, with the points where the flow is at rest marked. As the circulation grows the two points slide round the surface towards each other, meet at the bottom, and then leave the body — after which there is nowhere on the surface where the air is at rest at all.

How much circulation is too much

Spin a cylinder faster and it lifts harder, with no limit in the equations. What does have a limit is the flow's willingness to stop anywhere on the surface — the two points where the air is at rest slide round towards each other, meet at the bottom, and leave the body altogether.

The wall that pushes back

A wing near the ground carries more lift than the same wing in free air, and the usual explanation — a cushion of compressed air underneath — is not what the equations say. The ground is a mirror, and what changes is not the pressure under the wing but the circulation the wing is forced to carry.

The flap moves the curve and leaves its slope alone. Lift coefficient against incidence for several flap deflections. The lines are parallel: deflecting the flap gives the section lift at an incidence where it had none, and does not change how much lift each further degree of incidence buys.

What a flap does, and what it does not

Lowering a flap gives a wing lift at an incidence where it had none. It does not make the wing more responsive to being pitched — the lift curve moves sideways and its slope does not change, and those are two quite different things to buy.

$Induced drag against aspect ratio, all at C_L = 0.6. Induced drag for wings of different aspect ratio, each flown at whatever incidence makes it carry the same lift coefficient. The drag falls as one over the aspect ratio, so the longest wing here pays a fraction of what the shortest does for exactly the same load.$

The span is the whole story

Induced drag is the price of having ends, and the only thing that lowers it at a given lift is putting those ends further apart. Not more area, not a better section, not a cleverer planform — span, and everything else is a correction of a few per cent to it.

The speed depends on the core; the slope does not. A ring's self-induced speed against the logarithm of the cutoff used to compute it. The line is a quadrature of the reduced integral; the dots are an independent sum of cross products over a hundred and twenty thousand straight segments. Both diverge as the cutoff is thinned — a filament of zero thickness would move infinitely fast — so no answer here is a ring's speed until a core model is chosen. The slope, Γ/4πR, is the same whatever is chosen, and is what the check measures.

A ring moves because it is bent

A straight vortex filament induces exactly no velocity on itself — every element's direction is parallel to the line joining it to the point being evaluated, and the cross product is zero. Bend it into a ring and it drives itself forward, at a speed that diverges logarithmically as the core is thinned.

The lift does not arrive when the incidence does. Bound circulation against distance travelled, in units of the settled value. The wing starts far short of its final lift and takes tens of chords to collect the rest, because every scrap of circulation it takes has to be paid for with an opposite vortex shed behind it, and that vortex's own downwash holds the wing back until it is far away. Wagner's exact 1925 answer is drawn beside the model in the colour this site keeps for a borrowed claim: the shapes agree and the model is slower, by about a fifth at its worst.

The lift that arrives late

A wing set into motion does not have its lift. Every scrap of circulation it takes has to be paid for by shedding an equal and opposite vortex behind it, and until that debt is far downstream its own downwash holds the wing back — for tens of chords, not for an instant.

A lift curve that keeps climbing to 49 degrees. The lift of a slender delta of aspect ratio 1, split into the potential term that an attached flow would give and the vortex term the separation adds, with a conventional wing's curve behind them. The delta's is nonlinear from the start and reaches 1.68 at 49 degrees, where a conventional wing stalled at fifteen. That is the whole design case for the shape: not that it is efficient — it is not — but that it still has lift at incidences where an ordinary wing has none, which is what a delta-winged aircraft needs on approach and in a turn.

Lift out of a failure

Separation is what ends an ordinary wing's lift curve. A slender delta with sharp leading edges separates on purpose, rolls the shed sheet into a pair of vortices above its upper surface, and takes most of its lift from the suction they induce — with a curve that climbs to forty-nine degrees.

Four camber lines, and the angle at which each stops lifting. Four mean lines on the same chord: symmetric, a circular arc, a four-digit line with its crest at forty per cent, and a reflexed line whose tail turns up. The zero-lift angle beside each is computed from that line's own slope by quadrature and is a property of the shape alone — no incidence, no speed, no thickness enters it. The symmetric line's is exactly zero, the arc's is −2m to ten decimal places, and the reflexed line's is positive: it needs to be pointed up before it stops lifting.

Where lift starts

A wing at zero incidence is not a wing making no lift. The angle at which a section stops lifting is a property of its camber line and of nothing else — not of its thickness, not of its speed, not of the air — and it is an integral anybody can take.

Every one of these is a solution, and they lift different amounts. Lift coefficient against the circulation the aerofoil was told to carry, with the Kutta condition's own answer marked. Each point is a complete solve: the sources were found for that circulation and the surface is a wall to within 1.3e-4 at every collocation point. Every member satisfies the equations of motion and the boundary condition, and the lift runs through them at exactly 2Γ/Uc — Kutta–Joukowski appearing as a property of the family rather than as a result about any member of it. Ideal flow round a closed body does not have a unique answer, and the Kutta condition is the extra sentence that picks one.

Nothing but the edge

Cut an aerofoil into panels, put a singularity on each, and require the surface to be a wall. The system that comes out has one more unknown than it has equations, and the row that is missing is not a bookkeeping slip — it is the fact that ideal flow round a closed body has no unique answer at all.

A slotted flap at 30°, in a flow with no viscosity anywhere. Streamlines through a main element and a flap, computed by a two-body panel solve. Each element carries its own circulation and its own Kutta condition, and the two interfere through their velocity fields and through nothing else — there is no boundary layer here, no wake, no mixing region and no high-energy air. The system's lift coefficient is 2.757 against 0.698 for the main element alone at the same incidence, and the main element itself is carrying 3.98 times the circulation it carries by itself.

A slot is not a nozzle

The gap between a wing and its flap is supposed to blow fast air into a tired boundary layer. A solver with no boundary layer in it at all — no viscosity, no wake, no mixing — produces most of the lift increment anyway, and produces it on the element nobody moved.

$Two profiles, and they are the same profile. The chordwise and spanwise velocity profiles in the boundary layer of a yawed flat plate, each as a fraction of its own edge velocity. They are computed by different code — the chordwise one by shooting a third-order nonlinear equation, the spanwise one by a single pass through a linear second-order one — and they agree to 2.1e-8 over the whole layer. They are the same function of η, because the two equations reduce to the same equation. A swept flat plate has no crossflow at any sweep angle, and that is the independence principle in the only form that has no wriggle room in it.$

The wind a swept wing feels

Sweeping a wing back is usually justified by saying it meets a slower wind. It does not meet a slower wind. The equations split exactly in two, and the flow along the span is a passenger that exerts no force and changes nothing — until a pressure gradient breaks the split, and then it becomes the reason a swept wing is a different problem rather than a harder one.

The rectangular wing stalls at the root; the tapered one stalls at the tip. Section lift coefficient across the half span for three planforms, each drawn at the wing incidence where its own worst section first reaches 1.5. A rectangular wing's peak is at the root, which is where a designer wants it: the stall starts inboard, ahead of the ailerons, and the pilot feels it. A tapered wing's peak has moved out to 0.62 of the semi-span — over the ailerons — because taper takes chord away from the tip faster than it takes circulation. The elliptic wing is the degenerate case: every section reaches the limit at once, which is elegant and is the worst possible stall behaviour.

Which part of a wing stalls first

A wing has one lift coefficient and its sections have a hundred, and no section is at the wing's. Which of them runs out first is decided by the planform, it decides whether the pilot keeps the ailerons, and on any wing that was close to elliptic the standard fix costs span efficiency for every degree of it applied.

$Five quarters of the span, the same bending moment, 64/75 of the drag. The bell-loaded wing's induced drag and root bending moment against its span, both as fractions of an elliptic wing of unit span carrying the same lift. At equal span the bell is worse: its span efficiency is exactly three quarters. But its bending moment is exactly four fifths, and bending moment grows in proportion to span while drag falls as its square — so there is a span at which the bell has bought back the structure and is ahead on drag. It is at exactly five quarters, where the two curves are at 1 and 0.8533. Both numbers are rational and neither was put in by hand.$

The loading nobody used

Elliptic loading is the least-drag answer to a question no aeroplane asks. Constrain the moment the lift makes about the wing root instead of the span, and a different curve comes out — five quarters of the span for sixty-four seventy-fifths of the drag, both exact — with an upwash over the outer wing and a yaw that turns the right way.

The wake, and the velocity it gives itself. The cross-section of the wake far behind a wing with turned-up tips, with the velocity the wake induces on itself drawn as arrows normal to the trace. Induced drag is the integral of the circulation against that velocity and nothing else — this picture contains the entire quantity. It also contains no information whatever about where the surfaces were: two wings a chord apart and two wings ten chords apart produce the same picture and therefore the same drag, which is Munk's stagger theorem stated as a fact about what the arithmetic can see.

A wing that leaves the plane

A winglet is not a fence and it does not block anything escaping round the tip. The induced drag of any system of lifting surfaces depends on one cross-section of its wake and on nothing else whatever, and a wake that reaches upwards is cheaper for the same reason a wake that reaches sideways is.

Four aeroplanes, one number. Four biplanes with the same gap and the same loadings, staggered by nothing, by four tenths, by nine tenths and by one and six tenths of a chord. Their induced drags agree to the last bit of the arithmetic — the calculation cannot even express the stagger, because the Trefftz plane is a cross-section and everything drawn here projects onto the same one. That is Munk's stagger theorem, and stating it as the drag is unchanged understates it: there is no place in the computation where the stagger could be entered.

Two wings and it does not matter where

Move one wing of a biplane a chord forward and the induced drag does not change. Not approximately, not to a good approximation — the calculation that gives the induced drag has nowhere to put the stagger, because everything projects onto the same cross-section of the wake.

$Tip to tip, two wings cost exactly half of what they cost apart. The induced-drag saving of a pair of wings against the gap between their tips, as a fraction of what the two pay flying alone. With the tips touching the pair is one wing of twice the span carrying twice the lift, and the arithmetic of that is exact: the drag halves, and the computation returns 0.499257 of the separate figure. Pull them apart and the saving falls away with the square of the distance. Birds fly in a V because the tips are the part worth overlapping, and the spacing that pays is a small fraction of a span rather than any distance a formation could hold by eye.$

The lift beside a wing

Fly two aeroplanes with their wingtips touching and the pair costs exactly half what the two cost apart. Not approximately half — the arithmetic is a closed form, because two wings tip to tip are one wing of twice the span, and induced drag goes as the square of it.

Where the wake ends up

The sheet a wing sheds rolls up into two cores within a few spans, and nobody can compute the roll-up cheaply. Nobody has to: what the cores conserve is fixed before they form, and for an elliptically loaded wing the answer contains π and comes out at 78.5 per cent of the span.

Two roots walking towards each other, and one of them crosses. The roots of the characteristic quartic in the complex plane as the airspeed is raised from nothing to 105 metres per second — growth rate across, frequency up. At rest the two sit on the imaginary axis at the uncoupled frequencies. As the speed rises, the aerodynamic coupling drags them towards each other in frequency while pushing one left and the other right, and at 80.8 metres per second the right-hand one crosses the axis. Everything about the failure is in this picture: the coalescence, the crossing, and the fact that the flutter frequency is neither of the two the structure started with.

The shake that is not resonance

A steady airstream contains no oscillation at any frequency, so nothing is driving anything. What happens instead is that the aerodynamic forces couple two structural motions that were independent, drag their frequencies together, and turn one damping negative — and the wing takes the energy out of the air itself.

The surface in the wake

A wing has no opinion about its own incidence, which is why it needs a second surface behind it. How much that surface is worth depends on how much of the wing's downwash it is sitting in, and the factor of two everybody quotes for that is the value at infinity — where no tailplane has ever been put.

A wake that closes on itself

Prandtl's best wing system is a rectangle, not a wing. Solving for the loading on a closed wake turns up a circulation that costs nothing and does nothing — a gauge freedom in the middle of an optimisation — and a drag that keeps falling with no floor under it.

The wall is a boundary condition, solved for rather than reflected. A model's trailing vortices in a closed working section, with the sheet of sources that makes the wall a wall drawn as a displacement of the wall itself. The interference is an upwash — the model looks better than it is, and the correction factor comes to 0.1242 against the exact value 0.1250 that the image at the inverse point gives for a circle. The arrows are the interference velocity alone, with the model's own downwash removed, drawn to a common scale set by the longest of them.

The walls are in the answer

A wind tunnel measures a wing in a box the aeroplane will never fly in, and the box is worth a fifteenth of the induced drag. The correction is exactly an eighth for a closed circular section and exactly minus an eighth for an open jet, and the sign is the whole argument.

The lift at the mean angle

A wing in rough air flies at every angle in turn, so what it averages is the average of its lift curve rather than the lift at its average angle. Where the curve bends over near the stall the two differ by five per cent at three degrees of gust and by fourteen at five, and the drag goes the other way.

A finite force from an infinite speed

Pressure on a flat plate can only act along the plate's normal. Kutta–Joukowski says the force is perpendicular to the free stream. Those two directions differ by the angle of attack, and the discrepancy is made up at a single point where the velocity is infinite and the area is zero.

Three answers for the lift-curve slope, with three different signs. Thin-aerofoil theory has no thickness term at all and returns 2π for every section. The exact potential solution rises: 2π(1 + 0.766 t/c), measured off the Joukowski map. Real sections do the opposite, because the boundary layer thickens towards the trailing edge and decambers the section. Two of these curves are computed here; the third is what measurement says.

The half that carries nothing

Thin-aerofoil theory splits a section into a camber line that carries all the lift and a thickness distribution that carries none — at any incidence, exactly none. That is very nearly true, and what it discards decides the peak suction, the critical Mach number and where the boundary layer gives up.

Same disc, same solidity, different number of blades, different answer. Thrust and power coefficients for six blade counts at a fixed solidity. They span a factor of 1.101 in thrust, and every one of those rotors is the same actuator disc: same area, same blade area, same tip-speed ratio. The disc theory cannot distinguish them because the blade count is not one of its variables.

A disc that knows no blades

Momentum theory replaces a rotor with a surface across which the pressure jumps, and gets the Betz limit, the induced velocity and the whole energy argument out of it. It has no chord, no section and no number of blades — and at one fixed solidity, two blades and twenty give thrust coefficients ten per cent apart.

The side that cannot keep up

A hovering rotor is axisymmetric, so one radial distribution of circulation describes the whole disc. Move it forward and the advancing blade meets one and a half times the tip speed while the retreating one meets a half — and lift goes as the square of that. The rotor does not roll over, because the pitch is made a function of azimuth.

Deviation: how far the flow leaves from the blade angle, against solidity. The angle between the outlet flow and the blade, for a row of flat plates at 30° stagger meeting a flow at 45°. An open row barely turns the flow at all and the deviation is nearly the whole of the intended turning; a tight row guides it to within a thousandth of a degree of the blade angle. Nothing about the blade changed between the two ends of this curve.

A row is not a set of aerofoils

An isolated aerofoil's incidence is measured from the free stream. A compressor blade's cannot be, because an infinite row of identical blades above and below it has a circulation that is part of its own free stream — and the velocity the theorem uses is a vector mean that exists nowhere in the machine.

The trailing-edge speed against circulation, for a sharp edge and a round one. Sampled one grid point off the trailing edge. The sharp edge is singular at every circulation but one: the speed there is about U at the Kutta value and 11.3U half a Kutta circulation away, and it grows without bound as the sample approaches the edge. The round edge has no such point. That is the whole of the Kutta condition's justification, and it needs the corner.

The condition that can be bought

Ideal flow round a closed body has one solution for every circulation, and the Kutta condition picks one. Its justification is entirely the sharp edge: every other circulation puts an infinite velocity there. Take the corner away and nothing chooses — which is not a curiosity, it is what a circulation-control aerofoil is.

How the weight is shared between the two surfaces, against the centre of gravity. The wing's load and the tail's, for an aeroplane in level flight, against the static margin. The two must sum to the weight and their moments must cancel, and those two equations decide the split. At a forward centre of gravity the tail carries down and the wing carries more than the weight; the crossing is where the tail carries nothing.

More lift than weight

An aeroplane in level flight is drawn with one arrow up and one arrow down, equal and opposite. That equation collapses the whole configuration onto one number, and it is not true of any aeroplane with a tail behind it: there are two surfaces, two equations, and the second decides the split.

Two theories, one composite, and the aspect ratio between them. The lift-curve slope against aspect ratio. Prandtl's lifting line is exact as the aspect ratio goes to infinity and Jones's slender-wing theory is exact as it goes to zero, and each is generous outside its own limit. Helmbold's formula reduces to both with no free constant, which is what a composite expansion is, and runs under them where they disagree.

Where the line stops being a line

Prandtl's lifting line replaces a wing with a single bound vortex and its trailing sheet, and the formula that comes out is the most quoted in low-speed aerodynamics. Solved numerically at aspect ratio one it returns its own closed form to sixteen decimals — and the answer is forty-one per cent too high.

The local sweep of the isobars, across the span. The sweep of the half-load line at each spanwise station, for four geometric sweeps. Over the middle of the span it is the wing's own sweep, which is the simple theory being right. At the root it collapses — by twenty-one degrees at a geometric thirty-five — and at the tip it falls again. That root region is where the shock forms first on every swept wing ever built, and it is why they have waisted fuselages.

The sweep a root does not have

Simple sweep theory is one of the cleanest arguments in aerodynamics: an infinite yawed wing cannot know about the velocity along its own span, so only the normal component matters. A real wing has a root and two tips, and at the root of a thirty-five-degree wing the isobars are swept fourteen.

$Wagner's function and Küssner's, from one solver and two inputs. Lift as a fraction of its steady value, against distance travelled in semichords. The step in incidence and the sharp-edged gust are the same unsteady problem with two different right-hand sides, and Jones's exponential fits to both are drawn over the solve. From four semichords on they are nearly the same curve — which is why they get interchanged.$

Two answers to one question

Unsteady aerofoil theory collapses a wing's whole history onto one function of one variable, and every quasi-steady gust calculation convolves something with it. There are two such functions, not one: a wing that is pitched changes its boundary condition everywhere at once, and a wing flying into a gust has not met most of the gust yet.

$Rolling effectiveness against dynamic pressure. The rolling moment an aileron produces, as a fraction of what it would produce on a rigid wing. It falls from one, passes through zero at the reversal pressure, and goes negative: beyond that point deflecting the aileron down rolls the aeroplane the other way. There is no oscillation anywhere in this figure and no frequency — it is a static failure.$

The control that works backwards

Divergence is the static aeroelastic failure everybody names, and a wing with its elastic axis at its aerodynamic centre cannot diverge at any speed. It can still reverse — deflect the aileron down above a certain dynamic pressure and the aeroplane rolls the other way — because the aileron's own nose-down moment is there whatever the elastic axis is doing.

One formula, and it does not ask what the shape is

Kutta–Joukowski gives the lift of any two-dimensional body from one number. Six bodies with nothing else in common are put at that number here and come out with identical lifts — and with pitching moments, load distributions and suction peaks that are not even close.

How flat the optimum is. The same curve near its maximum, with the bands within a tenth and a half of one per cent of the best shaded. Every taper ratio from 0.31 to 0.42 is within a tenth of a per cent of the optimum, and the whole band from 0.25 to 0.51 is within half a per cent. The optimum is exact, and choosing it rather than its neighbour buys nothing a wing can measure.

The optimum that does not matter

Elliptic loading gives the least induced drag, exactly, and the proof is one line: the penalty is a sum of squares. Which is also why the optimum is flat enough that a quarter of the design space sits inside a tenth of a per cent of it.

The damping crosses zero once, and the crossing is the boundary. The least damping ratio of the four aeroelastic modes against speed. It falls through zero at 80.843 metres a second, and at that speed the crossing root's real part is four parts in 10¹⁸ — which is what an algebraic condition on a quartic with real coefficients looks like when it is solved numerically.

The speed where the damping is exactly zero

Flutter is an algebraic condition on a quartic: one root crosses the imaginary axis, at one speed, exactly. And the quantity that locates it is so nearly flat there that the standard way of finding it from flight test overshoots by a quarter.

Two camber lines, one lift and one moment. A NACA 2412 mean line and the same line with a fourth harmonic added to its slope. They are eight tenths of a per cent of the chord apart, which is forty per cent of the section's own camber, and they have the same lift and the same pitching moment at every incidence — to the last bit of double precision.

Three numbers out of a camber line

Thin-aerofoil theory takes a whole function and returns a lift and a moment. Only three coefficients of that function survive: two camber lines matched in the first three, and eight tenths of a per cent of chord apart, have the same lift and the same moment at every incidence and load distributions thirty-seven per cent apart.

The wake's memory, as a gain and a phase. Theodorsen's lift deficiency against reduced frequency. It is one at zero frequency — the quasi-steady limit, where the wake has had time to convect away — and falls to a half at high frequency, with a phase lag peaking near 15 degrees in between.

The lag that makes flutter possible

This site's own flutter model set the lift deficiency to one and recorded in its notes that doing so throws away the lag which stabilises the torsion mode. Putting the lag back moves the flutter speed from 80.8 metres a second to 131, and removes the need for the structural damping that was covering the artefact.

One incidence, two lifts. Lift coefficient against incidence for a wing pitched sinusoidally through the stall, with the static curve for comparison. The loop is traversed anticlockwise: at twelve degrees the wing carries 0.22 more lift going up than coming down.

Two lifts at one incidence

A wing pitched up and down through the stall does not retrace its own lift curve. At twelve degrees it carries 0.22 more lift going up than coming down, and the loop that opens between the two is the work the airstream does on it — which is where the energy for a stall flutter comes from.

What is left behind, and what it is made of. The two rolled-up vortices behind a large aircraft, drawn to scale against its span. They sit at 78.5 per cent of the span apart, each carries 508 square metres a second of circulation, and the pair descends at 1.72 metres a second under its own induction.

A wake that says what made it

Two vortices sitting behind an aeroplane carry its weight and its span in a form that can be read back out: circulation times separation times density times speed is the lift, exactly. The reading is exact for about a minute and worth nothing after four.

A blade passing the vortex it shed a passage ago. The lift the blade section feels as it passes a tip vortex at five per cent of the radius. The pulse is a doublet rather than a bump — upwash on one side and downwash on the other — so the blade is pushed one way and then the other in the width of a few chords.

A blade that flies through what it shed

A rotor blade meets the tip vortex the blade in front of it left, a fifth of a second earlier, at a few per cent of the radius. What it feels is a doublet — five and a half degrees of upwash and then five and a half of downwash within a few chords — and the peak goes as one over the miss distance.

What a downstream blade sees going past. The axial velocity a blade in the second row meets, over one revolution, as it passes through the wakes of thirty upstream blades. Each dip is one wake, and the blade meets all thirty of them every time it goes round.

A row that meets the row before it

A compressor blade is loaded and unloaded thirty times a revolution by geometry it does not have. The forcing sits at the blade count of the row in front of it and at multiples of that, and how far up the harmonics it reaches is decided by how far apart the two rows are.

The induced velocity, after a step in thrust. A rotor's induced velocity following a thirty per cent increase in thrust applied at fifty milliseconds. It does not jump: the air the disc has to accelerate has an apparent mass, and the response is a first-order climb to the new momentum-theory value.

The inflow that takes time to arrive

Momentum theory gives a rotor's induced velocity from its thrust, instantly. It does not arrive instantly: the air the disc has to accelerate has a mass, and the response is a first-order climb with a time constant of 33 milliseconds — a twentieth of the time the wake itself takes to convect a radius.

$What every wing pays to bend its root less. Least induced drag against the root bending moment of the lift, both as fractions of the elliptic monoplane of the same span and lift, for the monoplane and for box wings with gaps of a tenth, a fifth and two-fifths of the span. Each curve is a parabola with its minimum at that wing's unconstrained optimum. The box wings' minima sit to the right of the monoplane's — they load their lift further out — and their parabolas are shallower: the gap-of-a-fifth box can bend its root 21 per cent less than the elliptic monoplane and still match its drag.$

A lighter spar turns a box wing into a biplane

Prandtl's best wing system has two-thirds of a monoplane's induced drag at a gap of a fifth of the span, and part of that saving is carried by circulation turning the corner into its fins. Ask the box to bend its root less and it pays about half what a monoplane pays — but a biplane with no fins pays nearly as little, and by the time the spar is a fifth lighter the fins carry almost nothing and the box has become the biplane it was built from.

At a fixed impulse every orbit is a curve of constant energy. Axial separation of two coaxial rings against the radius of the ring that started smaller, for six pairs with the same total impulse, all starting in one plane. Pairs starting nearer equal (inner curves) trace closed loops: the separation swings from one sign to the other as the rings take turns, and the radius swings with it. Pairs starting further apart run off to one side and never return. The pale curves are the separatrix, the energy of two equal free rings — it passes through the plane of the start at a ratio of 0.340 and its arms reach out to infinite separation.

Two rings leapfrog only if they start alike

Two coaxial smoke rings passing through each other in turn is the most famous thing vortices do, and it is not what two rings do in general. Set them off from one plane with different radii and below a definite ratio the smaller one draws ahead and never comes back. Which of the two happens is decided before either ring moves, by whether two separate rings could hold the pair's energy.

One coefficient, two errors

A closed tunnel's walls push a measurement one way and an open jet's push it the other, so a wall somewhere between them should push it neither. There is such a wall, its coefficient is a length, and it nulls the blockage at 1.184 tunnel half-heights — and the streamline curvature at 1.590.

A gust, the aeroplane's response, and the load it is left with. A one-minus-cosine gust twenty-five chords long, as an angle of attack, against distance flown, and the lift it produces on two aeroplanes free to rise — a light single with a mass parameter of 16 and an airliner at eleven kilometres with 149. Both loads lag the gust, because lift takes several chords to build. The light aeroplane's also peaks lower and falls away early: it has started to rise with the gust, which lowers its own angle of attack.

The aeroplane rises out of its own gust

A gust pushes a wing up, and the wing, being attached to an aeroplane that is free to move, starts rising — which lowers its angle of attack and takes away part of the load the gust was delivering. How much depends on one number, and the formula every airworthiness code used for forty years is a fit to the calculation this essay does again.

A gust spectrum at a point, across a span, and as a load. The spectrum of vertical gust velocity at a single point — von Kármán's, falling as the minus five-thirds power — and the same gust averaged across a wing whose span is a tenth of the turbulence scale and across one as wide as the scale. Averaging removes the wavenumbers shorter than the span, one power steeper. Through Sears' filter for a chord of a five-hundredth of the scale, the tenth-span load falls as the minus eleven-thirds power, steep enough to converge.

The span takes away the infinity, not the gust

A two-dimensional wing in turbulence reverses its load infinitely often, because the gust spectrum never falls fast enough. A real wing averages the gust across its span, and the average keeps ninety-five per cent of the gust's strength and removes the infinity. The load's crossing rate settles at about two per turbulence scale flown and stops depending on where anyone cut the spectrum off.

A wake ends by bending, not by fading

The two vortices behind an aeroplane do not simply weaken until they are gone. In still air each one bends in the other's strain, the bend grows by a factor of e every thirty-three seconds behind an airliner, and after a couple of minutes the pair has pinched itself into a chain of rings. The wavelength it chooses is eight and a half times the spacing, and the vortex cores hardly enter.

$A V seen from above, and what each member pays. Nine wings in a V swept 45 degrees, tips one span apart, each carrying the same lift, seen from above with the flight direction up the page. Beside each is its induced drag as a fraction of what it would pay flying alone. The leader at the point pays 0.89; every other member pays between 0.35 and 0.39. The saving has flowed backwards through the formation.$

The bird at the point pays for the V

A flock's saving in a V is fixed by where its members sit across the stream, and the angle of the V cannot change it by a single per cent. What the angle changes is who gets the saving. It flows backwards through the formation, so that in a V swept forty-five degrees the leader pays nine-tenths of what it would pay alone while every bird behind it pays under four-tenths.

A fair V is a curved V

In a straight V of birds somebody always pays more than somebody else: at the fairest angle the leader and the two birds at the tips pay half again what the birds between them pay. The V in which every bird pays exactly the same can be solved for, and it is not straight. Its arms leave the leader almost abreast and bend back ever more steeply, to thirty-two degrees at the tips of a flock of nine. The share every bird then pays is the flock's average, fixed by Munk's theorem before any position is chosen — so fairness costs nothing, and the only thing that can make the flock cheaper is flying closer together sideways.

Leapfrogging rings end by merging, not by parting

Two smoke rings that leapfrog in an ideal fluid do it for ever, because their energy binds them. A real fluid drains the energy, and the obvious guess is that the rings drift out of the bound state and part. They do not: diffusion drains the pair's energy and the energy of every possible pair of free rings together, and the margin that binds them never goes negative. What viscosity does instead is fatten the cores until, as one ring threads the other, the two are close enough to merge. That takes a number of passes in proportion to the Reynolds number — one to three at the few thousand of a laboratory smoke ring.

The wake a flapping bird leaves is a wave in the air. Side view, the air at rest: the path the leader's wingtip traced, where its tip vortex now lies, over two wingbeats; lengths along the flight path in wake wavelengths — the distance flown in one beat — and heights in spans, for a tip swinging a fifth of a span each way. A follower three-tenths of a wavelength behind that beats three-tenths of a beat later traces the same path and flies along the leader's vortex all the way; one that beats half a beat off that traces the mirror image and meets the vortex only twice a beat.

The follower beats in time with the wake, not the bird

A gliding bird can sit in its neighbour's upwash and stay there. A flapping bird's wake is a wave left in the air — the path its wingtip traced, rising and falling with every beat — and a bird behind gains only if its own wing is where that wave is when it arrives. The best timing is a rule with no aerodynamics in it: lag the bird ahead by the time the wake took to come, so that each wingtip retraces the path of the one before. Directly behind, the rule flips by half a beat, and it buys a smaller loss rather than a gain.

A box wing's fins earn their keep in the spar

Constrain a box wing's root bending moment and its fins stop saving drag: it becomes a biplane. But that constraint counted the lift's moment, not the spars', and the fins join the spars into one frame. Joined at the tips, the two spars hand part of the moment to a couple across the gap — tension in one, compression in the other — which costs far less material than bending. With fins as stiff as the spars that is three-tenths of the moment; with rigid fins it stops at a third, whatever the gap, because a tip joint can only guide a tip. Counted in the spar, the box has less drag and a lighter spar than the elliptic monoplane at once.

A canard pays for its stability in induced drag

The argument for a canard is that both of its surfaces lift upwards, while a tail pushes down and makes the wing carry the difference. Munk's stagger theorem turns the question into arithmetic: two surfaces' least induced drag depends only on how the lift is split between them, not on which is in front. Static margin sets the split. At a margin of a tenth of a chord a tail carries a small upload and costs under one per cent; a canard must carry a fifth of the weight on a third of the span and costs twelve, and the more stable it is made, the more it pays.

One bowl, and a line across it for every way of flying. The least induced drag of a wing, a canard and a tail together, as a function of the canard's and the tail's shares of the lift, drawn as rings of equal drag about the bowl's foot, where each carries 1.9 per cent. A static margin of a tenth of a chord and a wing pitching moment draw a straight line of trimmed splits across the bowl. The tail aircraft trims where its line crosses the axis of zero canard load; the three-surface aircraft slides along the same line to the point nearest the foot.

A third surface is worth a square

A canard and a tail each leave an aircraft one free share of its lift, and the static margin spends it. Give an aircraft both and one share stays free after trim, so the split can slide along a line to the point of least induced drag. What that slide is worth turns out to be a square: half the bowl's curvature times the square of how far the tail aircraft already trims from twice the bowl's foot. For a cruising wing the saving is under one per cent and less than the canard's own skin friction; it pays only for a wing whose pitching moment is as large as a flapped section's.

Four vortices bend twice as fast, on shorter waves. Growth rate of the fastest bending wave against its wavelength, both in the units of the equivalent single pair — one unit of growth is one e-fold in the time the wake sinks by its own spacing. Crow's pair grows at most at 0.827, at 8.54 spacings. A flap vortex of 0.3 of the tip's strength at 0.4 of its station keeps a Crow band just below that and adds a second, at 1.57 and 1.8 spacings; one of 0.5 at 0.3 merges the two into one band peaking at 1.64 and 4.4 spacings.

Four vortices bend faster, and bend the wrong one

With its flaps down a wing sheds four vortices, not two: a strong one at each tip and a weaker one at each flap's edge, and on each side the two circle one another as the wake sinks. The orbit makes the wake unstable at waves a fifth of Crow's length, growing twice as fast. But the wave that grows is the weak flap vortex being bent round the strong tip vortex, which hardly moves, while the long bend that pinches the tips together grows no faster than before. Only a counter-rotating inner vortex makes the whole wake unstable, and it does so at every wavelength.

$A follower can wander further fore and aft than sideways. A flapping follower's saving, averaged over a wander of its place with the standard deviation shown, as a fraction of its own induced drag: fore and aft, holding the phase that suits its average place, for tips swinging a tenth, a fifth and four-tenths of a span; and sideways, beating in phase. The saving in place is 0.837. Sideways it has halved at a wander of 0.185 spans; fore and aft, with a fifth-span swing, only at 0.804, a third of the wake's wavelength, and with the four-tenths a cruising bird swings, at 0.37. With a tenth-span swing it never halves.$

A flapping follower can drift fore and aft, but not sideways

A bird in a flapping V has to be in the right place and beat at the right phase for that place, and no bird holds its place exactly. Drifting fore and aft costs it phase, at a full beat for every wavelength of the wake; drifting sideways takes it off the leader's tip vortex. The first is two to four times cheaper than the second, it can be bought back by re-timing the beat within about one beat of the drift, and the second cannot be bought back at all. So the precision a follower needs is sideways, and the attention it needs is on its timing.

$Holding the neutral point with a canard ahead takes a tail two and a half times the size. The tail area, as a fraction of the wing's, that puts the neutral point where the tail aircraft with a tail of 0.2 has it, against the tail's height above the wing's plane in spans: with no canard, with the canard, and with the canard but its wake at the tail removed. At the reference height the tail aircraft needs 0.2; with the canard it needs 0.496, of which 0.22 is for the canard's own lift ahead of the centre of gravity and 0.077 for its wake at the tail. A tail 0.25 span up needs 0.354 with the canard and 0.154 without.$

The tail a canard needs costs more than the canard saves

A canard added to an aircraft as a second trimming surface saves induced drag only if the neutral point is held where it was, and holding it is not free. The canard's own lift ahead of the centre of gravity pulls the neutral point forward, its wake reaches the tail and weakens it, and the tail has to grow to put the neutral point back. For a canard of a tenth of the wing's area the tail grows two and a half times, and its skin friction is twenty to thirty counts against a saving of one to twenty. A T-tail escapes most of the wake and a third of the bill, and still only a flapped wing comes near to paying.

No elevon reaches past a quarter chord

A wing with no tail trims itself with the back of its own section, and the back of a section is a short lever. Thin-aerofoil theory prices it exactly: the load an elevon adds acts at most a quarter of a chord behind the quarter-chord point, however small and far aft the elevon is. So every unit of trim moment costs at least four units of lift, where a tail three chords back costs a third, and a camber change shaped as a pure couple would cost nothing.

A wing averages a gust through its own loading

A finite span averages a turbulent gust and so tames the load it causes, but it does not average uniformly. The reverse-flow theorem says exactly how it weights the span: by the loading the wing makes at a uniform incidence. An elliptic wing therefore filters the gust the way a round aperture diffracts light, passes eight per cent more of its short scales than a uniform strip, and crosses its mean load up to four per cent more often.

A lagged re-timing passes no error down the V. The variance of each bird's phase error, in radians squared, against its place in one arm of a V, for fore-and-aft wander of half a span correlated over four beats. Holding a fixed phase, every bird's error is its offset from the bird ahead, 3.16. Re-timing with a one-beat lag, every bird's error is 0.632 — the first follower's and the thirtieth's alike, to the last digit. Re-timing as smoothly but through two half-beat lags, the error grows from 0.819 at the first follower to 1.05 at the tenth and 1.08 at the thirtieth, and is still growing slowly there.

A wandering flock passes no error down the V

In a flapping V every bird re-times its beat to the wake of the bird ahead, whose own beat is imperfectly timed, so the errors ought to pile up along the arm. With the simplest way of re-timing they do not, at all: the thirtieth bird is off its phase by exactly as much as the first. A one-beat lag and its complement add to one at every frequency, and that identity telescopes the whole arm. Re-time more smoothly and the errors do accumulate — by about a third, and then they stop.

The camber lines that carry them. The camber lines that carry the four loads at their ideal angles, heights in chords at a design lift coefficient of one. The uniform load's line is symmetric about mid-chord, 5.52% high; tapering the load over the last fifth moves the peak to 0.515 of the chord and raises it to 6.79%; tapering it over the whole chord, a = 0, puts the peak at 0.323.

A load carried to the trailing edge needs a hook

Thin-aerofoil theory runs backwards: ask for a chordwise load and it returns the camber line that carries it. Ask for the simplest load of all, the same everywhere, and the line comes back with a vertical tangent at both ends. The one at the nose is the price of a clean entry; the one at the tail is the Kutta condition being broken, and the hook it makes is what a Gurney flap is. Every NACA a-series line is a way of not paying it.

A washed-out rectangular wing is at its best at one lift coefficient. The span efficiency of a rectangular wing of aspect ratio 8 against lift coefficient, untwisted and with 2°, 4° and 6° of washout. Untwisted, it is 0.9367 at every lift coefficient. Each washout has one lift coefficient where it is best — 0.254, 0.509 and 0.763 — proportional to the twist, and each reaches the same peak, 0.9911, there. Below the peak the efficiency falls fast.

Washout is right at one lift coefficient

A twisted wing's loading is the sum of two shapes in a proportion that changes with speed, so its induced drag is a parabola in lift coefficient with its least at one design point, and that point moves in proportion to the twist. On a rectangular wing four degrees of washout puts it at a lift coefficient of 0.51 and recovers almost all the span efficiency the planform lost. On a well-tapered wing the same four degrees puts it at nineteen, and the washout is a drag paid at every speed for the stall alone.

A trainer's sortie wants the washout of its cruise, not of its average. The span efficiency of a rectangular wing of aspect ratio 8 averaged over a training sortie — eight minutes' climb, thirty of cruise, ten of slow flight and twelve of circuits — against its linear washout. Untwisted it is 0.9367. The best washout for the sortie is 3.43°, reaching 0.9774: the optimum for a lift coefficient of 0.436, the distance-weighted harmonic mean of the sortie's. The cruise alone would want 2.6°; the time-averaged lift coefficient, 0.671, would want 5.28° and lose 1.2% of the efficiency.

One twist is chosen for the speed flown per metre

A twisted wing is best at one lift coefficient and an aeroplane flies at many. Average its induced drag over a flight the way fuel and height are actually spent, per metre, and the twist that wins is the one for the root-mean-square speed of the flight — so the cruise outvotes the slow flight, a glider's circling barely counts, and a trainer's usual three or four degrees of washout turn out to be right for reasons nobody fitted them for.

A bending moment never chooses a span

Hold a wing's lift and its bending moment and ask for the least induced drag, and the answer does not stop: the span grows for ever and the tips push down. Both celebrated answers — Prandtl's bell at √(3/2) of the span for 8/9 of the drag, and the root-moment wing at 4/3 for 27/32 — are the span at which the least-drag loading's tips stop lifting. The structure sets how much is saved; a condition nobody states sets where the saving ends.

One wing root feels the rolling gust its lift cannot

The reverse-flow theorem says the gust weight of any load is the loading the wing would carry at an incidence shaped like that load's arm. For the lift that is the familiar uniform-incidence loading. For the bending moment at one root it is the loading of a wing washed in towards one tip only — a weight that is not symmetric, and so passes the rolling half of a gust that the lift and the two roots together cancel. That half is a sliver of the variance and adds up to a fifth to how often the root's moment crosses its mean.

The washout a mission wants rises with the lift coefficient; a twist driven by dynamic pressure rises the other way. Washout against lift coefficient for the trainer's sortie: each lift coefficient's own best washout, 7.87° per unit of it; the best fixed washout, 3.43°; the best twist that varies with dynamic pressure, which has to wash in as the aeroplane speeds up; and a twist of the natural sign, one degree of extra washout at cruise growing with dynamic pressure, from a nose-down pitching moment. The four segments sit at Cₗ 0.82, 0.33, 1.3, 0.92.

A wing that twists with its speed washes out the wrong way

A fixed washout is right at one lift coefficient, and a flight that spends its time at four of them pays for the spread. A wing whose twist changed in flight could follow — but the twist a real wing gets for free is driven by dynamic pressure, and it adds washout as the aeroplane speeds up, exactly where less is wanted. On a trainer's sortie a degree of it at cruise raises the induced energy by a third of a point. The best twist driven by dynamic pressure washes in, and only one that follows the incidence itself removes the spread entirely.

A symmetric panel with no spring stiffness washes out exactly as each lift coefficient wants. The outer panel's washout against the wing's lift coefficient over the trainer's sortie, with a spring stiffness of 0 cruise dynamic pressures: each lift coefficient's own best, 5.31° per unit of it; the best fixed washout, 2.31°; a symmetric panel at its best hinge and preload, 1.34% above an elliptic wing; and a panel with a cambered trainer section's moment of −0.05, 9.31%. The fixed twist spends 2.65% and the floor is 1.34%.

A pivoted wingtip follows the lift only with no spring stiffness and no camber

The washout a trainer's sortie wants grows in proportion to the lift coefficient, and the twist a wing gets for free follows the dynamic pressure instead. A tip panel free to pivot on a hinge ahead of its own aerodynamic centre can do better, and on one condition it does exactly what is wanted: held by a constant torque and carrying no camber moment, its own lift coefficient stays in proportion to the wing's, and every segment of the sortie flies at its own best twist. Every unit of spring stiffness gives part of that back, a cambered section's moment turns the panel into the wrong law altogether, and the free panel sheds a quarter of a gust's lift for nothing.

$Sweep slows every load's crossings by about a tenth at 35°, and leaves one root's excess over the lift standing. Against the sweep, for an elliptic wing of aspect ratio nine whose span is eight hundredths of the turbulence's integral scale: left, the lift's and one root's crossing rates as fractions of the straight wing's; right, one root's rate over the lift's. At 35° the lift crosses its mean 0.899 times as often as on the straight wing and the root 0.887; the root's excess over the lift goes from 1.19 to 1.17, and at 45° to 1.16.$

Sweep slows a wing root's gust cycles and leaves their excess over the lift

On a straight wing every station meets a gust at the same instant, and one wing root's bending moment crosses its mean nearly a fifth more often than the lift, because it reads the rolling half of the gust. Sweep the wing back and the outer stations meet the gust later, by their distance aft. That delay averages the gust along the flight path as a second, softer chord would: at 35° every load crosses its mean about a tenth less often, the root slightly more than the lift since its weight sits further out, and the variance hardly changes at all. The rolling excess survives almost whole.

Losing the leading-edge suction doubles a slender wing's drag due to lift and multiplies a long wing's by about half its aspect ratio. The drag due to lift with no leading-edge suction, Cₗ²/Cₗα, over the induced drag a full suction leaves, Cₗ²/(πAe), against aspect ratio, from a vortex lattice for rectangular, delta and elliptic flat wings; with the closed form 1 + √(1 + A²/4) that Helmbold's lift slope gives an elliptic wing. Rectangular: 2.15 at A = 1, 3.46 at 4, 5.33 at 8, 7.11 at 12; delta: 2.22 at 0.5 and 2.83 at 2. The slender limit is 2.

A wing that cannot hold its suction gains little from its span

The leading-edge suction is a small force concentrated at one edge, and on a finite wing it decides what the aspect ratio is worth. With it, the drag due to lift is the induced drag, which falls as the span grows. Without it, the drag due to lift is the lift tilted back by the incidence, and that falls only as fast as the lift slope rises — which stops at the section's 2π. A slender delta that loses its suction pays exactly twice its induced drag; a wing of aspect ratio eight pays more than five times it. Without the suction no span can lift a wing's best lift-to-drag ratio past a ceiling set by the section alone.

Counting the spar's weight stops the span, and the best wing then sheds a little load from its tips to grow five per cent. Total drag at a fixed payload, induced drag of the lift that carries the spar too, against span, both in units of the best elliptic wing's, for a wing whose area is held so that its spar thins as it lengthens. Elliptic, the best span carries a spar of a quarter of the lift; with the third harmonic free the best ratio is ε = -0.1356 (closed form (√33 − 9)/24 = -0.1356), the span 1.05 times and the drag 0.9575 times the ellipse's, with the spar still a quarter.

A spar that must lift itself stops the span at a quarter of the weight

Held only to a lift and a bending moment, the least-drag wing never stops growing. An aeroplane's wing stops because its spar has weight, and the weight has to be lifted. Count it, and the best span is the one at which the spar is exactly a quarter of the lift when the wing's area is fixed, and a third when its chord is — whatever the loading. The best loading then sheds some of its tip load, but not all the way to Prandtl's bell; with the chord held and nothing else charged, it is the bell exactly, and the more of the drag the wing's own area makes, the nearer it comes back to the ellipse.

A stiffness floor keeps the span and gives the tips their load back

A spar sized by strength alone thins to nothing at the tips, and charged as lift it stops the best span where it weighs a quarter of everything the wing carries, with the tips a little unloaded. A real outer wing is sized by stiffness, and stiffness sets a floor under the spar's weight that no reduction of the tip load can lower. Put that floor in and the tips get their load back — the best loading slides from a sixth of the way to the bell towards the ellipse — the quarter does not survive, rising to two fifths as the floor spreads over the outer two thirds of the span, and the span itself barely moves. A weight per metre cannot stop a span.

$A strut two fifths of the way out halves the spars of a box wing; near the root or the tip it buys little. The flange volume of both spars of Prandtl's best box at a gap of a fifth of the span, with one strut on each side joined rigidly to both spars, over the volume with none, against the strut's station as a fraction of the semi-span, for struts a tenth as stiff in bending as a spar, as stiff, and ten times stiffer. For a strut as stiff as a spar the least is 0.5 at 0.4 of the semi-span; at 0.1 and 0.9 it is 0.839 and 0.929.$

A strut inside a box wing is invisible to the wake, and halves the spar

Prandtl's best box wing has, everywhere inside it, air moving straight down: no sideways flow at all. A strut joining its two wings anywhere along the span therefore meets no flow it could turn, carries no load, and leaves the induced drag exactly where it was — the wake cannot see it. The spar can. A strut two fifths of the way out, as stiff as a spar and rigidly joined, halves the flange material of both spars, though it is pushed and pulled by almost nothing: the two wings lift equally and bend together, so the strut works through its joints, and a pinned one does nothing. Its only aerodynamic price is friction, and a slim strut pays it.

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