A body with no lift, and a moment anyway

An aeroplane is mostly not wing. It is a long body of revolution with a wing through the middle of it, and the long body has an aerodynamic personality of its own that is easy to miss because ideal flow says it carries no lift — which is true, and which is not the same as saying it does nothing.

It does something considerable. At any incidence a fuselage carries a couple: nose-up, growing with the square of the speed, entirely destabilising, and present in a flow with no viscosity, no separation and no wake. It is the reason a tail is a substantial aerodynamic surface rather than a trim tab, and it can be computed in about a line — provided the boundary condition is put somewhere the surface is not.

The condition moved onto the axis #

The trick is Munk’s, from 1924, and it is the cheapest useful approximation in the subject.

If a body is slender — long compared with its width — then in each plane perpendicular to the axis the flow looks nearly two-dimensional: the cross-flow is $U\sin\alpha$, the section is a circle of the local radius, and the answer for a circle in a cross-stream has been known since Rankine. The three-dimensional problem collapses into a sequence of plane ones.

What makes it cheap is where the boundary condition goes. It is not imposed on the surface at all. A distribution of doublets is put on the axis, with strength equal to the local cross-sectional area, and the surface is left to fall where it may.

That is the same move as a wall made by reflection, where a boundary condition is replaced by a singularity somewhere it is not, and as a body made of nothing, where a shape is a level set of a flow rather than a thing imposed on it. The family resemblance is not accidental: all three are the same economy, which is to satisfy a condition at a surface by putting something inside it.

What comes out, in one line #

The apparent mass of a circle of area $A$ in a cross-flow is $\rho A$ per unit length. The cross-flow at incidence $\alpha$ is $U\sin\alpha$, and the momentum in each plane is therefore $\rho A U \sin\alpha$ per unit length. A body moving forward at $U\cos\alpha$ sweeps through those planes at that rate, so the force per unit length is the rate of change of the plane’s momentum:

$$\frac{dL}{dx} = \tfrac{1}{2}\rho U^2 \sin 2\alpha \;\frac{dA}{dx}.$$

The lift is the integral of that, which is $\tfrac{1}{2}\rho U^2\sin 2\alpha$ multiplied by $[A]_{\text{nose}}^{\text{tail}}$ — the area difference between the two ends. For a closed body that is zero, and the computation here returns $4\times10^{-18}$.

The moment is the same integrand weighted by $x$, and there is nothing to cancel:

$$M = \tfrac{1}{2}\rho U^2 \sin 2\alpha \; \mathcal{V},$$

with $\mathcal{V}$ the body’s volume. Not its length, not its frontal area, not any measure of how blunt it is — its volume, which is the one property a designer chooses on completely different grounds.

The check that this is not an accident of the algebra: the site computes the moment twice, once by quadrature over the tabulated loading and once by that volume formula, and the two agree to a part in ten thousand — the trapezoid rule’s own error over the shape at eight hundred stations.

Cut the tail off and it lifts #

The lift being an area difference has a consequence that is easier to believe once it has been drawn.

So a body with a blunt base is a lifting body and a closed one is not, in ideal flow, purely because of where the integral stops. The same arithmetic run on a wing — where the cross-sectional area jumps at the trailing edge rather than tapering to nothing — is the slender-wing lift that a delta wing’s low-speed behaviour is built on, and it is why a delta’s lift-curve slope is $\pi\mathrm{AR}/2$ regardless of what its sections are.

A couple has no address #

There is a consequence of the lift being exactly zero that is worth its own paragraph, because it is the one part of this subject where a moment needs no reference point.

A moment about a point $x_0$ is $M - x_0 L$. With $L = 0$ the second term vanishes, so the moment is the same about every point on the body, and about every point off it. It is a pure couple. There is no centre of pressure, no aerodynamic centre, nothing to plot against incidence — asking where a fuselage’s lift acts is asking where a quantity that is zero is applied.

This is also why the fuselage’s contribution to an aircraft’s stability appears in the books as a shift of the neutral point rather than as a force. A pure couple cannot be located, so it cannot be added to the wing’s lift at some station; what it does is change the rate at which the total moment varies with incidence, which is the same as moving the point about which the whole aircraft is neutral. For a conventional airliner the fuselage moves it forward by something like a tenth of the mean chord, and that tenth has to be paid for in tail area.

What the approximation costs, exactly #

An approximation that cannot say how wrong it is has not finished. This one can, because the ellipsoid’s answer is known in closed form and has been since Lamb: the exact moment is

$$M = (k_2 - k_1)\,\tfrac{1}{2}\rho U^2\,\mathcal{V}\sin 2\alpha,$$

with $k_1$ and $k_2$ the apparent-mass coefficients along and across the body. Slender-body theory is the statement that $k_2 - k_1 = 1$.

Twelve point eight per cent at a fineness of six is a usable error and an honest one. It is also worth noting which way it goes: the slender estimate is always high, so a tail sized against it is a tail with margin, which is why the approximation survived into design practice rather than being replaced.

The sphere, which the theory gets entirely wrong and entirely right #

The factor is a difference of two coefficients, and there is one body for which they are equal.

For a sphere $k_1 = k_2 = \tfrac{1}{2}$ exactly, so $k_2 - k_1 = 0$ and the Munk moment is exactly zero. A sphere at incidence has no incidence: every direction is the same direction, and there is nothing for a couple to be about.

That is the theory’s worst case and it is instructive rather than embarrassing. The approximation does not degrade gracefully to some rough answer; it degrades to a definite wrong one, because the quantity it approximates is a difference of two things that are individually large. An approximation to a difference is a different object from an approximation to a quantity, and the sphere is the cleanest demonstration of that in the subject.

What this means for an aeroplane #

A fuselage’s moment is destabilising, which means it acts to increase whatever incidence the aircraft already has. Left alone the aircraft would diverge. The tail’s job is to supply a restoring moment larger than it, at every speed, which is possible because both go as $U^2$ and their ratio is therefore a matter of geometry.

The number a designer wants from this essay is small and useful: at six degrees, a fuselage of fineness six carries a nose-up moment of $0.208\,\tfrac{1}{2}\rho U^2\mathcal{V}$, and the exact ellipsoid says $0.181$ of the same quantity. Both are proportional to volume, so stretching a fuselage to carry more passengers makes it more unstable in exact proportion to the extra volume, and the tail has to grow with it.

One more consequence is worth extracting, because it explains a shape. The moment goes as the volume and the tail’s restoring moment goes as its area multiplied by its arm, so an aircraft that grows in bulk faster than it grows in length becomes progressively harder to stabilise. That is the arithmetic behind a familiar silhouette: wide-body airliners carry proportionally larger tails than narrow-bodies of the same length, and flying boats — whose hulls are both bulky and short — carried tails that look absurd until the volume in the numerator is noticed.

The equilibrium it drives towards, which is sideways #

The couple goes as $\sin 2\alpha$, and that single factor says more about the shape of aircraft than the constant in front of it does.

It vanishes at $\alpha = 0$ and again at $\alpha = 90^{\circ}$, and it is largest at forty-five degrees. Both zeros are equilibria and only one of them is stable. Perturb a body from zero incidence and the moment is nose-up, which increases the incidence, which increases the moment — the classic signature of an unstable equilibrium. Perturb it from broadside and the moment acts to restore it. A slender body free to rotate in a stream settles broadside on, and it does so in a fluid with no viscosity, no separation and no wake to weathercock about.

That is not a hypothetical. It is what an unfinned store dropped from an aircraft does, what a bomb without a tail does, and what an airship without fins would do — which was Munk’s actual question in 1924 and is why his report is about airship hulls rather than about aerodynamics in the abstract. The reason arrows have feathers, darts have flights and rockets have fins is not that a long body is naturally stable; a long body is naturally unstable, and increasingly so as it is made longer, because the destabilising couple is proportional to the volume it gains.

Every stabilising surface on every vehicle in this essay exists to beat a term that gets bigger as the vehicle gets more useful. The couple is proportional to volume, and volume is passengers, fuel or payload. Nothing about a better shape reduces it — the inertia factor only approaches one from below, so a more slender body gets more of the theoretical couple rather than less. There is no design that escapes it, only designs that pay for it.

And the same couple acts in yaw #

There is a symmetry in the derivation that is easy to read past. Nothing in it distinguished up from sideways: the cross-flow was $U\sin\alpha$ in some plane perpendicular to the axis, and which plane was never specified. A body of revolution at an angle of sideslip has exactly the same couple, about the yaw axis, with the same volume in it and the same sign.

So a fuselage is destabilising in yaw as well as in pitch, by the same amount, and it needs a fin for the same reason it needs a tailplane. That the two surfaces are usually different sizes is not because the aerodynamic couple differs between the axes — for a body of revolution it does not — but because everything else about the two problems does: the wing supplies a large stabilising pitch contribution and nothing comparable in yaw, the fin has a shorter effective arm on most layouts, and the yaw case has to be sized against an engine failure rather than against stability alone.

The useful statement for a designer is the one the symmetry makes obvious. Fin volume and tail volume are answering the same term. Stretch a fuselage and both have to grow; add a bulge and both have to grow; and an aircraft whose fin looks disproportionate is usually one whose fuselage volume grew after the tail was drawn.

One asymmetry does survive, and it is worth naming because it is the exception that proves the derivation. A real fuselage is not a body of revolution: it is deeper than it is wide, or flatter, and the moment depends on the cross-sectional area rather than on the shape of the section only while the section is circular. Once it is not, the two apparent-mass coefficients across the body differ from each other as well as from the one along it, and the pitch and yaw couples separate. A wide-bodied hull is more destabilising in pitch than in yaw, and a deep narrow one the other way round — which is the first correction any real stability estimate applies to Munk’s number.

What the picture cannot show #

There is no separation anywhere in this. A real fuselage at more than a few degrees sheds a pair of body vortices from its lee side, which add a non-linear lift that ideal flow has no mechanism for and which changes the moment substantially. Allen and Perkins’ 1951 cross-flow-drag correction handles that by adding a term borrowed from measurements of circular cylinders, and it is a correlation rather than a solution — this site names it and does not compute it.

The wing is absent. A real body of revolution has a wing through it, and the interference between the two is not the sum of their separate answers.

And the theory has no way of knowing whether the body is slender. Nothing in the calculation refuses a sphere. The fineness ratio does not appear in the slender answer at all — it appears only in the exact one, which is why the figure comparing them is the whole of the error analysis.

Who found it, and the connection worth keeping #

Max Munk worked this out at Langley in 1924, in a report on airship hulls: the question of the day was whether a rigid airship would fly straight, and the answer was that it would not without fins, for the reason above. The same paper contains the stagger theorem, which is the other result this site uses of his, and the two have the same character — a statement that a complicated force depends on one simple property of the geometry and on nothing else.

The surprising connection is with sound and with ships. The apparent mass that carries this whole argument is the same object that gives a body its resistance to being accelerated, that sets the frequency of a bubble, and that decides how much water a ship’s hull drags with it. Munk’s moment is the statement that a body’s apparent mass is different along it and across it, and that any object with that property, pointed at an angle to its motion, feels a couple. Boomerangs, arrows and badminton shuttles are all designed around it, in one direction or the other.

What the moment does to a design, in numbers #

The couple is $\tfrac{1}{2}\rho U^2 \mathcal{V}\sin 2\alpha$ multiplied by the inertia factor, and putting an aircraft’s numbers into it makes the size of the thing plain.

Take a narrow-body airliner: a fuselage roughly 38 metres long and 4 metres across, so a volume near 360 cubic metres and a fineness ratio near 9.5, at 230 metres a second in air a quarter of sea-level density. At one degree of incidence the Munk moment is of order 90 kilonewton-metres, nose-up, and it grows linearly with incidence over the small range that matters.

The tail’s restoring moment is its lift multiplied by its arm. With a tail area of 30 square metres at 17 metres behind the centre of gravity, the same dynamic pressure gives roughly 250 kilonewton-metres per degree of tail incidence. So the fuselage is consuming something like a third of the tail’s authority before the wing has been considered at all, and it does so at every speed, because both terms carry the same $\tfrac{1}{2}\rho U^2$ and their ratio is pure geometry.

That ratio is the reason the destabilising contribution of a fuselage appears in every stability textbook as a fixed shift of the neutral point rather than as a force to be trimmed out. It cannot be trimmed out: it is proportional to incidence, so it changes the slope of the moment curve, and only a surface whose lift also grows with incidence can answer it.

Where the ladder goes next #

The rung above this one is the area rule: if the loading depends only on $A(x)$, then so does the wave drag of a slender body at supersonic speed, and two aircraft with the same area distribution have the same wave drag however differently they are shaped. That is Whitcomb’s 1952 result, and it follows from exactly the argument in this essay run at a different Mach number — which is what a compressible flow being an incompressible flow over a different shape makes possible.