The least drag a volume can have

The first essay in this anchor established that a closed body in a steady inviscid supersonic stream has drag, which d’Alembert’s theorem forbids below Mach one. This one asks how little of it a body can get away with.

Von Kármán’s slender-body result is that a body’s wave drag depends on nothing about it except the distribution of its cross-sectional area along its length:

$$D = -\frac{\rho U^2}{4\pi}\iint S''(x_1)S''(x_2)\ln|x_1-x_2|\,\mathrm dx_1\,\mathrm dx_2.$$

Not the shape of the cross-sections. Not whether the area is in a fuselage or a wing. Not the Mach number, to this order. Two aeroplanes with the same area distribution have the same wave drag, which is the area rule read backwards and is the largest single design result in the transonic and supersonic regime.

Three routes to one number #

The double integral has a logarithmic singularity on its diagonal, which is integrable and awkward. Writing $x = \tfrac L2(1-\cos\theta)$ and expanding $\mathrm dS/\mathrm dx = \sum A_n\sin n\theta$ turns it into

$$D = \frac{\pi\rho U^2}{8}\sum n A_n^2,$$

a sum of squares — so the drag is positive for every shape, which it had better be, and the minimisation is now a question about coefficients.

Both routes are computed here and they agree to seven parts in a hundred thousand, and both agree with the closed form $64V^2/\pi L^4$ to seven decimal places.

That agreement is worth more than a check. The singular integral is the definition and the series is a convenience, and a two per cent disagreement between them would have been indistinguishable from a physical claim about the shape.

Which harmonic carries the answer #

The series makes the minimisation immediate, and the structure of the answer is worth having.

Closure forces $A_1 = 0$. The condition $S(0) = S(L) = 0$ means $\int S'\,\mathrm dx = 0$, and in the $\theta$ variable that integral picks out exactly the first sine. So a closed body has no first harmonic at all — measured here at $9\times10^{-16}$.

The volume fixes $A_2$. Integrating by parts gives $V = (\pi L^2/16)A_2$, so $A_2 = 16V/\pi L^2$ whatever the shape — measured at $0.1018592$ against a predicted $0.1018592$.

So a body of given length and volume has $A_2$ fixed for it and every other coefficient free, and the least $\sum nA_n^2$ is the one that sets all of them to zero.

The Sears–Haack body is the shape whose area distribution is a pure second harmonic, which in physical variables is

$$S(x) = S_{\max}\left[\frac{4x(L-x)}{L^2}\right]^{3/2},$$

a spindle with its maximum area at mid-length and a radius going as $[x(L-x)]^{3/4}$.

Why the area distribution is all that matters #

The claim at the top of this essay is strong enough to deserve an explanation, because it is not obvious that a fuselage and a wing should be interchangeable.

Slender-body theory represents a body by a line of sources along its axis, with the strength at each station proportional to the rate at which the cross-sectional area is growing there. That is the same construction superposition uses to make bodies out of nothing, run in reverse: a given area distribution needs a given source strength, whatever shape the cross-section is.

At supersonic speed each source radiates on a Mach cone rather than in all directions, and the drag is the work done against the pressure field those cones create — which is why the answer is a double integral over pairs of stations with a logarithm between them. The logarithm is the two-dimensional Green’s function of the wave operator, and it is where the interaction between one part of the body and another lives.

Two consequences follow immediately. Nothing about the cross-section’s shape appears, because the source strength does not depend on it. And the drag is not local: it is a double integral, so moving area from one station to another changes the contribution of every pair, which is why the optimisation is a global one and why the answer is a single smooth shape rather than a local rule.

What makes a distribution expensive #

The integral is built from $S''$, and reading that is the practical skill.

A kink in $S$ makes $S''$ a delta function, and a delta function against a logarithm is a large contribution. The cone–cylinder’s two kinks are the whole of its 2.47.

A discontinuity in $S$ — a base — makes $S''$ a derivative of a delta, which is why the theory refuses one outright rather than merely reporting a large number.

And smoothness is not enough by itself. The ellipsoid has a perfectly smooth area distribution and pays 2.75, more than the cone–cylinder, because its area distribution has too much curvature near the ends: $S\propto x$ near the nose gives $S''$ finite but the body has a blunt nose in the sense that matters, with $r\propto\sqrt x$. Sears–Haack has $S\propto x^{3/2}$ and $r\propto x^{3/4}$, which is pointed enough to keep $S''$ integrable and blunt enough to enclose volume.

That last is the trade the three-halves power resolves, and it is why the answer is a fractional power rather than something tidier.

Reading the fourth power #

The $V^2/L^4$ scaling repays a moment, because both exponents can be read off the structure of the answer.

The drag is quadratic in $S''$, so it is quadratic in the area and therefore in the volume: that is the $V^2$. And $S''$ has two derivatives with respect to a length, while the double integral supplies two powers of length from its measures and the logarithm supplies none — leaving $L^{-4}$.

The practical form is a fineness ratio statement. Writing $V\sim d^2L$ for a body of diameter $d$ gives $D\propto d^4/L^2$, so the drag coefficient based on frontal area goes as $(d/L)^2$: the inverse square of the fineness ratio. Doubling a body’s fineness ratio quarters its wave-drag coefficient, and that is the number a designer works with.

It is also why the result feels so much sharper than the equivalent subsonic ones. Induced drag goes as the inverse square of the span, which is already a strong dependence; this is the inverse fourth power of the length at fixed volume, which is stronger still, and it is why supersonic aeroplanes are shaped by one constraint in a way subsonic ones are not.

Four bodies, one length, one volume #

The comparison that makes it concrete: four closed bodies of identical length and identical volume, with only the area distribution differing.

The cone–cylinder pays two and a half times the minimum, and the reason is visible in the second figure: the drag integral is built from the second derivative of the area distribution, so a kink is expensive and it has two of them.

An incidental result worth recording: the ellipsoid of revolution and the parabolic-area spindle are the same body. An ellipsoid has $r\propto\sqrt{1-((2x-L)/L)^2}$, so $S = \pi r^2\propto 4x(L-x)/L^2$, which is the parabola. They appear in the literature as though they were two shapes and they have one drag.

The scaling, and where it points #

The dependence is the part with consequences:

$$D\propto \frac{V^2}{L^4},$$

with fitted exponents of $2.0000$ and $-4.0000$.

Halving the length of an aeroplane while keeping the same volume inside it multiplies its wave drag by sixteen. That is the strongest dependence on any single quantity anywhere in this collection — stronger than the eighth power of speed in jet noise, in the sense that length is a quantity a designer chooses and speed usually is not.

The other minimum, and why there are two Sears–Haack bodies #

The problem solved above fixes the length and the volume. There is a second problem, and it has a different answer with the same name attached.

Fixing the length and the maximum diameter rather than the volume gives a different variational problem, whose solution has $S\propto[x(L-x)]^{...}$ with a different exponent and is sometimes also called the Sears–Haack body. The two are close and not identical, and which one is meant depends on which constraint the designer actually has.

That ambiguity is worth naming because it is the whole content of an optimisation: the answer is a property of the constraint set, not of the physics. The flow supplies the functional to be minimised; what is being held fixed while it is minimised comes from outside, and changing it changes the shape.

The same is true one level up. Minimising drag at fixed volume gives a spindle. Minimising it at fixed volume and fixed base area gives the Von Kármán ogive, which is what a supersonic projectile is shaped like. Minimising it at fixed volume and fixed nose bluntness — a heating constraint, as the blunt-body argument requires — gives something else again.

Where the linear theory’s ordering can be trusted #

A practical question: if slender-body theory is quantitatively poor below a fineness ratio of eight, what is it for?

It is for the ordering and the scaling, both of which survive the approximation far better than the absolute numbers do. A body that this theory says is 2.5 times the minimum will be measured at something between two and three times the minimum, and a fourfold change in length will change the measured drag by close to sixteen.

That is the usual position for a linear theory in this collection. Thin-aerofoil theory predicts a lift-curve slope of $2\pi$ that measurements put at $5.7$, and it is still what a designer reaches for first — because the dependence on camber and incidence is right even where the constant is not.

The check for whether the ordering can be trusted is whether the competing shapes differ by more than the theory’s own error. Between Sears–Haack and a cone–cylinder, a factor of 2.5 against an error of tens of per cent: yes. Between Sears–Haack and the sine spindle, a factor of 1.37: marginal, and worth confirming.

What is not in the flow #

The length.

Every term in the answer is fixed by the flow except the two constraints, and one of them is a number nobody in this subject chooses. How long an aeroplane is allowed to be is decided by hangars, by runway geometry, by the structural weight of a long slender fuselage and by what a customer will buy.

That is unlike every other missing quantity in the essays around it. Elsewhere the answer needs something physical that the model has abstracted away — a gas, a boundary, an observer. Here it needs a decision made outside the discipline entirely, and the fourth power means the decision dominates the answer.

It is also why supersonic transports are long out of all proportion to their capacity, and why the length of Concorde was set by drag rather than by seats.

The body the theory has to refuse #

There is a shape the machinery must decline, and it is worth showing because the routine does not fail on it.

Take a Sears–Haack forebody and cut it off square, scaled to the same length and volume. Slender-body theory does not apply: the integration by parts that produces the drag integral discards a boundary term that vanishes only for a body that closes.

Asked anyway, the routine returns $4.28\times10^{-3}$ — less than Sears–Haack’s $8.15\times10^{-3}$. A reader shown that would conclude that the way to reduce wave drag is to cut the tail off.

The site’s routine refuses it, and this is exactly the habit gas established for a deflection past the detachment angle: a solver that quietly returns the nearest thing draws a smooth, captioned, false picture.

The area rule, which is this result read sideways #

If the drag depends only on the area distribution, then two configurations sharing one share their drag — and a designer can move area about freely.

That is Whitcomb’s area rule. A fuselage is waisted where the wing joins it, so that the combination’s total cross-sectional area distribution is the smooth one a good body of revolution would have, and the transonic drag rise falls by a third. The visible consequence — the coke-bottle fuselages of 1950s aircraft — is this integral, applied.

At Mach one the rule is applied to the area distribution cut by planes normal to the flow. Above it, the cuts are made by Mach planes and the rule becomes the supersonic area rule, with a different cut for every roll angle and an average taken over them. Same integral, different sections.

The Mach number that is missing, and where it comes back #

The formula at the top of this essay contains no Mach number, which ought to be alarming for a supersonic result. It is worth resolving, because the resolution is what separates the easy version of the area rule from the hard one.

For a body of revolution the absence is genuine. The linearised supersonic equation is made elliptic by stretching the transverse coordinates by $\beta = \sqrt{M^2-1}$, and for a slender axisymmetric body that stretching changes the source strengths and the Green’s function in compensating ways: the drag integral comes out the same at every supersonic Mach number. A Sears–Haack body is the minimum-drag shape at Mach 1.2 and at Mach 3, and by the same margin over its rivals.

For anything with a wing it is not. What the theory actually cares about is not the area cut by a plane perpendicular to the flow, but the area cut by a Mach plane — a plane inclined at the Mach angle, since that is the surface along which a disturbance travels. On a body of revolution an oblique cut through a slender shape gives the same area distribution to leading order, which is why the distinction never surfaced above. On a wing–body it does not: a Mach plane slices a swept wing differently from a normal plane, and differently again depending on the roll angle at which it approaches.

So the supersonic area rule is a family of area distributions rather than one — a cut for every roll angle — and the quantity to be smoothed is the average of the drags computed over them. As $M \to 1$ the Mach angle goes to ninety degrees, every cut becomes the normal one, the roll dependence disappears, and the rule collapses to the transonic version: one curve, cut square, smoothed.

Which is why the transonic area rule is a shape anybody can see and the supersonic one is not. The waisted fuselages of the 1950s are a design smoothed for a single cut at Mach one. An aircraft area-ruled at Mach 1 is not area-ruled at Mach 2, and one optimised at Mach 2 carries a shape whose logic is invisible in any single cross-sectional plot.

It also sharpens the essay’s opening claim. Two aeroplanes with the same area distribution have the same wave drag is exact — provided the distribution meant is the Mach-plane one, at the Mach number in question, averaged over roll.

Where it meets the boom #

There is a second use of the same object, and the two design problems are related without being the same.

A sonic boom’s F-function is a linear functional of the same equivalent area distribution that this integral treats quadratically. So a body optimised for one is not optimised for the other: minimum drag wants the distribution smooth and symmetric, and minimum boom wants its front stretched, which costs drag.

The exchange rate between them is the central trade of a quiet supersonic aeroplane, and it is why such an aeroplane looks nothing like a minimum-drag one.

The model limit #

Four, and the first is the largest.

Slender-body theory. The whole result is a first term in an expansion in the body’s fineness ratio, and it is quantitatively poor for anything with a fineness ratio below about eight. What it gets right at any fineness ratio is the ordering of shapes and the scaling.

Linear, so no shocks. The theory has no shock waves in it, and a real supersonic body has them. The drag it computes is the wave drag in the acoustic sense, and the additional loss through the shocks is extra.

No lift. The equivalent area used by the area rule includes a contribution from the lift accumulated ahead of each station, and this essay’s bodies are at zero lift throughout. A lifting configuration’s minimum-drag shape is a different problem.

And a closed body. The refusal above is the hypothesis being enforced. A real aeroplane has a base — an engine nozzle, a blunt tail — and its base drag is a separate calculation that this one cannot supply.