The drag a wake keeps however it rolls up

Weighing what is missing took a control volume behind a two-dimensional section and read its drag from the momentum deficit in the wake. It ended on the case the method cannot do simply. A plane behind a finite wing contains the trailing vortex system as well as the section’s wake, and a momentum survey picks up part of the vortices’ contribution in a way that depends on where the plane is and how far the vortices have rolled up. Separating the two, it said, is standard, not unique, and a genuinely interesting state for a measurement to be in.

This essay separates one of them exactly. The induced drag has a plane of its own — the far-wake plane Trefftz used — and in that plane it is an energy: the kinetic energy of the crossflow the trailing vortices induce. What happens to that energy as the wake develops is the whole question of whether the induced drag “depends on where the plane is”. In an inviscid wake the answer is that it does not, however violently the sheet rolls up. In a real wake it does, for a reason that has nothing to do with roll-up, and the amount that moves goes somewhere the same plane can still see.

The induced drag is an energy left in a plane #

Behind a lifting wing, a plane across the wake sees the trailing vortices as a two-dimensional flow: velocities $v$ and $w$ in the plane, swirling round the vortices and sinking between them. Each second the wing lays down one more length $U$ of that flow, so the power it has to supply is $U$ times the crossflow kinetic energy per unit length, and the force is

$$D_i = \tfrac12\rho\iint (v^2 + w^2)\,dA.$$

The conventions are these. Lengths are scaled on the span $b$, circulation on the root circulation $\Gamma_0$, and time of flight — which is distance behind the wing divided by $U$ — on $b^2/\Gamma_0$. The loading is written as a lifting-line Fourier series $\Gamma(\theta) = \sum a_n\sin n\theta$ on $y = -\tfrac b2\cos\theta$, and the induced drag it costs is $\rho(\pi/8)\sum n\,a_n^2$. For an elliptic load that is $\rho\pi\Gamma_0^2/8$. The same number is the kinetic energy of the flow round a flat plate of width $b$ moving through still fluid at the far-wake downwash $\Gamma_0/b$ — an added mass of $\rho\pi b^2/4$ times half the square of that speed — which is what the crossflow behind an elliptic wing is.

For scale: a wing carrying 600 kN at 70 m/s with a 60 m span has a root circulation of 152 m²/s, and its elliptic induced drag is 10.8 kN, a lift-to-induced-drag ratio of 55.

The elliptic loading is the cheapest. A parabolic load of the same lift costs 1.125 times as much crossflow energy, a span efficiency of 0.889, and a triangular one 1.386 times, an efficiency of 0.721. The elliptic load is the one whose far-wake downwash is uniform across the span, and a uniform downwash is the least kinetic energy that can carry a given momentum — which is the whole of Munk’s result, and the reason span is the whole story: at a given lift the only way to spend less energy is to spread the same downward momentum over a wider plate. For the 600 kN aeroplane the difference is not small: a parabolic load carrying the same lift on the same span would cost 12.2 kN of induced drag instead of 10.8, and a triangular one 15.0.

A smoothed sheet of vortices holds exactly that energy #

The Fourier series is one route to the number. A second route builds the crossflow itself. Represent the trailing sheet by a row of point vortices whose strengths are the drops in circulation between neighbouring stations, and add up their mutual energy. A bare point vortex has infinite energy of its own, so each is smoothed over a small length $\delta$ in the way Krasny introduced for sheet roll-up, which makes the energy finite and exactly conserved by the smoothed dynamics.

With 400 vortices the energy is 0.34971 of $\rho\Gamma_0^2$ at a smoothing of 0.02 of the span, 0.36774 at 0.01 and 0.37849 at 0.005, approaching π/8 = 0.39270 from below. The number of vortices does not matter at these resolutions — 200, 400 and 800 agree to five figures — so the approach is set by the smoothing alone, and because smoothing replaces a logarithmic kernel the difference goes as $\delta\ln\delta$ and $\delta$. Fitting those two terms at three smoothing lengths and extrapolating to none gives 0.392694, which is π/8 to 1.2 × 10⁻⁵. The energy of the vortex sheet and the induced drag of the loading are the same number, reached by routes that share nothing.

Roll-up moves the energy and keeps all of it #

Left to itself, the sheet does not stay flat. Each vortex moves in the field of all the others, the tips curl round first, and within a short time of flight the sheet has wound into two concentrated vortices.

It looks like the destruction of the flat-sheet picture, and in one sense it is: the crossflow is now concentrated round two cores instead of spread across the span. But a two-dimensional inviscid flow conserves its kinetic energy and its linear impulse exactly, and the smoothed sheet inherits both. There is no vortex stretching in a plane flow to feed energy anywhere, which is the same absence that gives a plane flow its second conserved quantity and sends its energy up the scales rather than down.

Through six hundred steps of roll-up the energy holds to a few parts in 10⁹, and the separation of the two halves’ centroids stays at 0.7854 of the span throughout. That spacing is fixed from the first instant by the impulse, and for an elliptic load it is $\pi b/4$ — the distance at which the two concentrated vortices end up, known before any of the rolling up has happened. The pair sinks at 0.20 of $\Gamma_0/b$, against the 0.2026 that two point vortices of strength $\Gamma_0$ at $\pi b/4$ would descend at. The consequence for a survey is direct: in an inviscid wake, a plane one span behind the wing and a plane a hundred spans behind it read the same induced drag, and the visible change in the flow between them is a rearrangement of an unchanged total.

The impulse that fixes the spacing is worth one more sentence, because it is the lift. Two vortices of strength $\pm\Gamma_0$ a distance $\pi b/4$ apart carry a downward impulse of $\rho\Gamma_0\pi b/4$ per unit length of wake, and the wing lays down a length $U$ of wake each second, so the downward momentum it hands the air per second is $\rho U\Gamma_0\pi b/4$ — which is exactly the lift of an elliptic wing, the circulation account of what actually holds a wing up arriving from the far wake. The sinking pair is the air-pushed-down argument made precise: its impulse is the weight, its energy is the induced drag, and both are conserved by the roll-up that makes it look so different from the sheet that began it.

Why a plane flow cannot change its own energy has a short answer worth keeping. In three dimensions a vortex that is stretched along its axis spins faster and gains energy from the strain doing the stretching — the mechanism of the spin that feeds itself. A crossflow in a plane has its vorticity pointing out of the plane and every velocity gradient lying in it, so there is no stretching term at all. The vortices can wrap, merge and sink, and each of those moves energy from one part of the plane to another; none of them can make any or destroy any, and only viscosity, which the smoothed sheet does not have, can take it away.

Where in the plane the drag sits #

The rearrangement is worth seeing in the quantity the drag is an integral of.

Through the flat sheet the downwash is nearly uniform across the span — −0.94 of $\Gamma_0/b$ on the centreline, where lifting-line theory’s far-wake value is exactly −1 and the smoothing takes off the rest — with upwash outboard of the tips. Through the rolled-up pair the same energy is concentrated in the swirl round two cores. The integrand has moved from a broad band to two peaks; the integral has not changed. A rake that sampled only the region between the tips would see its reading fall as the sheet rolled up, and would be measuring where the energy went rather than how much of it there is.

Viscosity is what changes the split #

Roll-up cannot change the induced drag a plane reads. Viscosity can. Once the sheet has rolled up, the wake is two vortices with finite cores, and those cores grow: a Gaussian (Lamb–Oseen) vortex spreads with its squared radius increasing as $4\nu t$. The energy of a pair of them, strength $\pm\Gamma$ at separation $b_0$, has a closed form,

$$E = \frac{\rho\Gamma^2}{2\pi}\Big[\ln\frac{b_0}{\sigma} - \frac{\ln 2 - \gamma_E}{2} + \tfrac12 E_1\!\Big(\frac{b_0^2}{2\sigma^2}\Big)\Big],$$

which falls as the core radius σ grows. Choosing the core at which the pair holds exactly the flat sheet’s energy — 0.0629 of the span, 3.8 m for the 60 m wing — and letting it grow gives the fate of the induced drag against the viscous age $4\nu t/b_0^2$.

The crossflow keeps 0.971 of the induced drag at a viscous age of 10⁻³, 0.809 at 10⁻², 0.431 at 10⁻¹ and 0.089 at 1. The energy that leaves the crossflow is dissipated into heat in the vortex cores, and that heat is still in the plane: it shows as a loss of total pressure, exactly the quantity a total-pressure rake integrates to find a section’s profile drag. A survey that separates crossflow energy from total-pressure defect — which is what the modern breakdowns of wake-plane drag do — therefore reports the same wing as having more induced drag close behind it and more profile drag far behind it, with the sum unchanged. The split is not a property of the wing. It is a property of the wing and the plane together.

There is a practical corollary that turns the survey question round. A total-pressure rake reads the loss of total pressure, and in inviscid flow total pressure does not change along a streamline at all. So the whole of the inviscid induced drag — every joule of the crossflow energy — is invisible to a total-pressure survey: the fluid in the swirl has the free stream’s total pressure, and its lower static pressure is exactly balanced by its extra speed. Only a measurement that resolves the crossflow velocities, or the static pressure together with the total, can see it. The part viscosity has converted is different: it is a genuine total-pressure loss, and a rake picks it up as though it were profile drag. A survey behind a finite wing therefore splits its reading into induced and profile drag partly by physics and partly by what the instrument can see, and the viscous conversion moves drag from the column the rake is blind to into the column it measures.

How far back that matters depends on what sets the viscosity. With molecular viscosity alone, the 60 m wing’s pair would take about 3.7 × 10⁵ seconds of flight — more than four days — to reach a viscous age of 0.01, and the split would never move within any tunnel or any flight test. Real trailing vortices decay far faster, through turbulence in and around the cores and through the atmosphere they sink into, so the effective viscosity is many orders of magnitude larger than the molecular one and the split does move measurably behind a real aeroplane. The calculation here fixes what the split is at a given viscous age; what viscous age a given plane corresponds to is set by turbulence this model does not contain.

Why the breakdown is not unique #

The history of wake-plane drag breakdowns makes the non-uniqueness look like a defect of the methods. The calculation says it is a fact about the flow. The total drag in the plane is conserved as the wake develops downstream in steady flight: whatever the wing put into the air is still crossing each plane per second. Its division into crossflow energy and total-pressure defect is not conserved, because a physical process — dissipation in the cores — converts one into the other continuously. Any formula that assigns “induced” to the first and “profile” to the second will give answers that depend on the plane’s distance, and two formulations that draw the line slightly differently (between crossflow energy and axial velocity defect, or between entropy and enthalpy terms) will give different splits of the same total at the same distance.

That is the same lesson A wake that keeps the drag and forgets the body drew for a two-dimensional wake: the conserved quantity survives and everything about how it is distributed is lost downstream. Here the conserved quantity is the total drag and what is lost is the label on it. And it is the same structure as the first essay on this subject, where the momentum account of lift is exact in total and the face-by-face division of it depends on where the control volume’s faces are drawn.

Each statement set against a second route #

The extrapolated sheet energy matches the Fourier induced drag to 1.2 × 10⁻⁵. Through roll-up, the crossflow energy after six hundred fourth-order Runge–Kutta steps matches its starting value to a few parts in 10⁹, and the centroid spacing matches to rounding. The decay rate of the pair’s closed-form energy matches the enstrophy dissipation $-\rho\nu\iint\omega^2\,dA$ summed on a grid to 4 × 10⁻⁷ at a core of 0.03 of the span, and better at larger cores. The Fourier route itself returns an efficiency of exactly one for the elliptic load and less for every other loading tested. And the calculation refuses an unsmoothed sheet, whose energy is infinite, an unknown loading, a vortex pair with no core and the exponential integral at zero.

What the plane cannot show #

The flow near the wing. The Trefftz plane is far behind the wing, where the flow is two-dimensional in the plane. Close behind a real wing the sheet is not flat, the pressure has not recovered, and the plane’s integral includes terms this calculation drops.

Axial flow in the cores. Real trailing vortices carry an axial velocity along their cores — a jet or a wake — which adds a term to the momentum balance that a purely two-dimensional crossflow has no room for.

The instabilities that end a vortex pair. Two counter-rotating vortices are unstable to long-wave disturbances that pinch them into rings, and in the atmosphere that — along with stratification and ground proximity — usually ends the pair long before viscous spreading would. Those are three-dimensional and not here.

A finite aspect ratio. The loadings come from lifting-line theory, which is good for slender wings and poor for stubby ones, where the trailing sheet is not the simple function of the span loading assumed throughout.

Still open: the power a wing puts into its wake #

The next calculation is the energy account this subject has named and not done. The momentum box gives a force; the same box asked about energy gives a power, and every second the wing does $D_iU$ of work on the air, which leaves through the far plane as the crossflow energy computed here. Following that power as the wake ages — how much is carried as crossflow, how much as heat, how much radiated as the vortices sink through a stratified atmosphere and excite internal waves — would close the account with terms that are individually more physical than the momentum ones and collectively harder to measure. Beside it is the three-dimensional fate the plane cannot see: the long-wave instability that makes a trailing pair link into rings, whose growth rate sets how many spans behind an aeroplane its induced drag’s crossflow stops being a pair of vortices at all.