A drag that needs a discontinuity

D’Alembert’s paradox says a closed body in a steady, inviscid, irrotational flow has no drag, and the result survives being argued four separate ways, every time. Above Mach one it is false, and the reason given there is that the flow is no longer irrotational: a shock leaves entropy behind it, entropy gradients make vorticity, and the theorem’s hypotheses are broken before its conclusion is reached.

That account is correct and it is about supersonic flight. The interesting case is the one below Mach one, where the free stream is subsonic everywhere, the body is thin and closed, the flow is inviscid by assumption — and there is a pocket of supersonic flow on top of the section with a shock at the end of it.

Every hypothesis of the theorem is satisfied by the free stream and one of them is violated in a region a third of a chord long that the body made for itself. What that costs is the subject here, and the arithmetic is unusually clean: the drag is a single integral over the section’s own surface, it has to vanish when the theorem’s conditions hold, and it can be computed from a solution that was obtained without any reference to it.

The integral that ought to be zero #

For a thin symmetric section at zero lift the streamwise pressure force is

$$C_d = \tau\int_0^1 C_p(x)\,f'(x)\,\mathrm{d}x,$$

where $f$ is the thickness distribution. The integral has no obvious sign. The front half of the section slopes one way and the rear half the other, so a pressure distribution that is symmetric about mid-chord contributes nothing: the suction over the rear pulls forward exactly as much as the pressure over the front pushes back.

Nothing in that integral is arranged to cancel. It is taken over a numerical solution computed without any symmetry imposed, on a grid that does not know where mid-chord is, and the cancellation has to happen because the flow is right or not at all. That makes it the strongest available check on the solver, and it is the reason to compute it before computing anything interesting.

On the shock-free members of the family it comes out at $2.4\times10^{-6}$, against $5.1\times10^{-2}$ on a member with a shock. That is a ratio of $4.7\times10^{-5}$: five decades between a flow that obeys d’Alembert’s theorem and one that does not, on the same section, with the same equation, and with the free stream subsonic in both.

The residual is worth one more sentence, because it is a measurement rather than a rounding. A trapezoidal integral over forty points of a curve whose own values are known to about $10^{-6}$ should produce about $10^{-6}$ of noise, which is what it produces. So the shock-free drag is not “small”; it is the size of the arithmetic, and the calculation has no room left to hide a physical drag in. Had it come out at $10^{-4}$ the interesting question would have been what was wrong with the solver rather than what was interesting about the flow.

The same statement in the other direction is what makes the supercritical number trustworthy. A drag of $5\times10^{-2}$ is four decades above the noise, so it is not sensitive to the grid, to the convergence tolerance, or to where the integration starts and stops. Two numbers on either side of a gap that wide do not need error bars to be told apart.

Where the cancellation breaks, chord by chord #

The subcritical distribution is not exactly symmetric — compressibility is not a symmetric effect — but it is symmetric to the order the small-disturbance equation keeps, and the residual is the $2.4\times10^{-6}$ above.

A shock destroys the symmetry in a particular way. The supersonic pocket runs from somewhere near a third of the chord to the shock, and through it the pressure keeps falling while the surface is already sloping downwards. Then the shock returns the pressure to something near its subcritical value in the width of a cell. The recovery that should have been spread over the rear half of the chord, pulling the section forward, happens all at once at the end of it, and the forward pull is never collected.

That is the mechanism, stated without entropy. The entropy statement is the same thing seen from outside: the total pressure behind a shock is lower than in front of it, so the flow leaving the section cannot recover the free-stream pressure, and the deficit is a momentum deficit. Both readings give the same number and the first one is the one visible in the figure.

Where the energy goes, if not into viscosity #

A drag is a rate of doing work, and it is worth asking where the work ends up in a calculation that contains nothing capable of turning work into heat.

In the model, nowhere. The small-disturbance equation is written for a velocity potential, a potential flow has no mechanism for dissipation, and the “shock” is a place where the numerical scheme permits a jump rather than a physical structure with a thickness. The drag appears as a momentum deficit in the pressure field and the calculation stops there.

In the world, the energy goes into heat inside the shock, over a distance of a few mean free paths, by exactly the viscosity and conduction the calculation omits. And here is the part that is genuinely strange: the amount of energy destroyed does not depend on how much viscosity there is. The jump conditions across a shock follow from conservation of mass, momentum and energy alone; the entropy rise they force is a function of the upstream Mach number and nothing else. Halving the viscosity halves the thickness of the shock and leaves the loss exactly where it was.

That is the same structure a hydraulic jump has, and the same one a sudden expansion in a pipe has. It is the third appearance of one structure: a loss whose size is fixed by conservation and whose mechanism is viscous. The reason an inviscid calculation can compute it at all is that the size is all it is being asked for.

There is a limit to how far that reasoning carries, and it is worth marking. It works because the shock is thin compared with everything else in the problem, so the region where the model is wrong is too small to matter to the region where it is right. A weak shock is not thin in that sense — as the jump goes to zero its thickness goes to infinity — so the very cases near the drag-rise threshold, where the shock is barely a shock, are the ones where the separation of scales is worst and the computed drag is least reliable.

The drag rise is a threshold, not a slope #

The curve at the head of this essay has a shape that a drag curve normally does not: it is flat and then it is not.

Between $K = 2.4$ and $K = 1.4$ the scaled drag sits at $2\times10^{-4}$, which is the solver’s noise. The supersonic pocket opens between $K = 1.4$ and $K = 1.3$, and at $K = 1.2$ — one small step further — the drag is already ten times its noise floor. By $K = 0.9$ it is 0.83, four thousand times the flat value, and by $K = 0.5$ it is 4.36.

In Mach numbers, for an eight per cent section, that whole transition is from 0.80 to 0.87. A section that is entirely well behaved at Mach 0.80 has a drag coefficient of five hundredths at Mach 0.92, out of nothing but the arrangement of its own pressure field. This is the drag-divergence phenomenon, and the reason it is a cliff rather than a gradient is that the shock is a threshold rather than a quantity: there is no such thing as a weak fraction of a discontinuity.

That last sentence needs a qualification that the figure supplies. The drag does not jump; it turns. There is such a thing as a weak shock, and immediately past the threshold the pocket is a quarter of the chord and the jump in pressure coefficient across it is a fifth. What makes the curve a cliff is not a discontinuity in the drag but the rate at which a weak shock strengthens once it exists: the pocket lengthens, the flow at its end is faster, the jump needed to return it to subsonic is larger, and the drag goes as something like the cube of that jump. Three compounding effects in one direction produce a curve that is flat and then vertical.

This is why the number an operator cares about is not the drag but the drag-divergence Mach number — the speed at which the slope of the drag curve reaches some stated value, usually 0.1 per unit Mach number. A threshold defined on a derivative is a strange thing to design an aircraft around, and it is used because the alternative — the Mach number at which the drag first becomes non-zero — is not measurable: the first shock to appear contributes a drag below anything a balance can read. The quantity that matters physically is undetectable and the quantity that is quoted is a convention, and the curve above shows why the two are not close together.

And the drag obeys the same rule as the pressure #

The similarity scaling that collapsed the pressure distributions collapses this too, and it has to, because the drag is an integral of those distributions against a fixed shape.

$$\frac{C_d}{\tau\varepsilon} = \frac{C_d\,[(\gamma+1)M_\infty^2]^{1/3}}{\tau^{5/3}} = D(K).$$

The five-thirds power is worth reading. Wave drag goes as the five-thirds power of thickness at fixed similarity parameter, which is a steeper penalty than the linear theories anywhere else in the subject impose and a shallower one than the squared dependence a reader might guess from the fact that drag is usually quadratic in a disturbance. It comes out of the same three-term balance that produced the two-thirds in the critical Mach number: one power of $\tau$ from the section’s own slope in the integrand, and two-thirds of a power from the pressure.

Its practical form is a statement about design that does not need a solver. A wing section taken from ten per cent to twelve per cent thick, at the same similarity parameter, pays $(1.2)^{5/3} = 1.36$ times the wave drag — thirty-six per cent more for twenty per cent more thickness — and that is before the fact that a thicker section reaches a given similarity parameter at a lower Mach number.

Compounding those two effects is what makes transonic wing design feel like a trap. Holding the Mach number fixed and thickening the section lowers $K$, which moves the section down its own drag curve into a steeper part of it, and multiplies whatever it finds there by $\tau^{5/3}$. The two act together, and near the threshold the first is the larger of them by a wide margin: at $K$ just below 1.3 a twenty per cent increase in thickness moves the section from a drag of $2\times10^{-3}$ scaled to something of order one, which is three orders of magnitude, against the 1.36 the scaling contributes.

So the similarity rule’s practical message is the opposite of what a scaling law usually says. The exponent is not the thing to design around. What to design around is which side of the threshold the section is on, and the exponent only matters once the answer to that is “the wrong one”.

What it is not, which is friction #

It is worth being explicit about what has and has not been added, because “drag at high subsonic speed” is a phrase that usually means several things at once.

There is no viscosity in this calculation. The section has no boundary layer, no skin friction, no separation and no wake. If the section were run subcritically and its drag measured in an imaginary inviscid tunnel, the answer would be zero — and the answer here is, to five decimal places.

There is also no lift, so this is not induced drag, and there is no three-dimensionality, so it is not the wave drag of a body’s volume distribution that the area rule is about. It is the drag of a two-dimensional section at zero lift, in a subsonic free stream, arising entirely from a shock the section made for itself.

What a real section would add to this is substantial and it mostly makes things worse. The shock separates the boundary layer beneath it, the separation thickens the effective body, the thickened body strengthens the shock, and the pressure drag from the separated region can exceed the wave drag itself. This calculation is the floor rather than the answer, which is the useful thing about it: a section cannot do better than the inviscid number, and the inviscid number is already a cliff.

The floor is a real floor and not a figure of speech, and the reason is that the mechanism cannot be designed away by any means that leaves the section’s thickness and Mach number alone. Once there is supersonic flow on a section in a subsonic stream, there must be a compression somewhere to return it, since the flow at the trailing edge has to be at something near free-stream conditions. The compression can be spread out — that is what a supercritical section’s flat rooftop is for, and it is worth a great deal — but it cannot be removed, and a spread-out compression that does not quite shock is a delicate thing that exists at one Mach number and not the one either side of it.

So the engineering answer to the cliff is not to abolish the shock. It is to arrange the pressure distribution so that the shock is as weak as possible for the lift being carried, which means getting the suction that produces the lift spread over as much chord as possible rather than concentrated in a peak. That is why a modern transonic section looks nothing like a classical one: flat on top, with its camber gathered near the trailing edge, and a lower surface that does work a classical section leaves to the upper.

What the picture cannot show #

The shock is one cell wide and its position is known to a cell. The drag integral is dominated by the jump, so the drag is known to about the same relative precision as the jump’s size, which a grid-refinement study would put at a few per cent rather than at the five decimal places the shock-free case achieves.

The small-disturbance equation’s shock is not a Rankine–Hugoniot shock. It satisfies the jump condition of the approximate equation, which agrees with the exact one to the order the approximation keeps. At the strong-shock end of the family — where the pocket covers most of the chord — that agreement is not close, and the drags quoted there are order-of-magnitude statements.

Entropy appears nowhere in the calculation. The small-disturbance equation is written for a potential, and a potential flow cannot carry entropy gradients at all. The drag it produces is real and its mechanism, inside the model, is the asymmetry of the pressure distribution rather than a total-pressure loss. That the two agree is a property of the approximation and is not obvious.

And the section is a parabolic arc. The threshold parameter, the drag rise’s shape and the five-thirds coefficient all depend on the shape function, which the similarity rule does not collapse.

Who found it, and when #

The drag rise near Mach one was a flight observation before it was a calculation — propeller tips in the 1920s, dive-recovery accidents in the 1940s — and it was understood qualitatively as soon as shocks were known to form on subsonic aircraft. What took until the 1970s was computing it, because computing it means capturing a shock in a mixed-type equation, which is exactly what Murman and Cole’s switched relaxation made possible.

The surprising connection is with an entirely different kind of loss, and it is the same accounting. A sudden expansion in a pipe destroys mechanical energy at a rate that can be computed from momentum alone, with no viscosity in the answer, because the jet mixes and mixing is irreversible. Here a section destroys total pressure with no viscosity in the answer, because the flow shocks and a shock is irreversible. In both cases the viscosity is what actually does the destroying and in neither case does it appear, because the rate is set by how much has to be destroyed rather than by how fast the destruction proceeds. A shock is thin because the viscosity is small; it is strong for reasons that have nothing to do with viscosity at all.

Still open: what the drag does when the shock reaches the trailing edge #

The drag curve flattens at the bottom of the range, and the flattening is an artefact of a geometry rather than a result.

At $K = 0.5$ the supersonic pocket covers three-quarters of the chord and the shock stands near the trailing edge; at $K = 0.1$ the pocket has not grown, because there is no more chord for it to grow into, and the drag has stopped rising. What the calculation is describing there is a section whose shock has run off the back, and everything about that configuration is outside the small-disturbance equation’s competence: the shock interacts with the trailing edge, the flow behind the section is no longer a small perturbation of the free stream, and in a real flow the boundary layer has separated from the shock foot all the way to the trailing edge.

The calculation that would say something honest about the bottom of the range is the same relaxation run on the full plane with a wake, so that the shock is allowed to leave the section rather than to pile up against its last cell — and, separately, the same family run through an exact Euler solver so that the small-disturbance jump condition can be compared with the real one at the place where it is worst. Whether the flattening survives either of those, or whether the drag keeps climbing to the sonic free stream and beyond, is the difference between a curve with a maximum in it and a curve without one, and the two say quite different things about where an aircraft’s drag rise ends.