A downwind turn costs no airspeed, unless the wind changes or the pilot does

An angle, not a speed made the first and largest correction to how most people think about a stall: a wing stalls at an angle of attack, and the speed printed on the dial is only a stand-in that happens to hold under one set of conditions. A stall that is a place took it past the peak into a deep stall. There is a neighbouring question, argued with more passion than either, and it has a clean answer that its participants seldom accept.

Here is the argument, and it can be heard at any flying club. An aircraft is flying into a strong wind, and its speed over the ground is low. It turns round and flies downwind, and its speed over the ground is high. Therefore something has accelerated it. An aircraft has inertia, and the engine cannot have done it, so the aeroplane must have lagged behind and been left short of airspeed. In a low turn, this is how the student gets into a stall. It feels right, the ground really does rush up, and accident reports really do contain turns to downwind at low altitude. And in a steady wind every step after the first is false.

The equations do not contain the wind #

Let the aircraft have velocity $\mathbf V$ through the air, and let the air itself move at a steady, uniform $\mathbf W$. Its velocity over the ground is $\mathbf V+\mathbf W$, so its acceleration over the ground is

$$\frac{d}{dt}(\mathbf V+\mathbf W) = \frac{d\mathbf V}{dt} + \frac{d\mathbf W}{dt},$$

and the last term is exactly zero. A steady, uniform wind has no acceleration; a frame moving with the air is as good an inertial frame as the ground. So Newton’s second law for the aircraft in the air’s frame has no $\mathbf W$ anywhere. The aerodynamic forces depend on airspeed and angle of attack, gravity is the same in both frames, and the thrust is whatever the engine gives. Nothing in the equations of motion knows the wind exists.

So a turn in a wind is the same turn as a turn in still air, seen by someone standing on something moving past. Langewiesche made the argument in Stick and Rudder in 1944 in a form that has not been improved on: an aeroplane in flight knows nothing of the wind. A person in a balloon drifting at 15 metres a second feels no breeze and a candle in the basket burns straight up, which is also why a boat can run short of wind only relative to the water it floats on.

That is the entire argument, and what follows is only the working. But there is an honest objection. Over the half-turn the aircraft’s speed over the ground went from 25 to 55 metres a second. Its kinetic energy over the ground rose per kilogram by $\frac12(55^2-25^2)=1200$ joules, by a factor of 4.84. That energy is real in the ground’s frame. Where did it come from?

Who pays for the ground’s kinetic energy #

The answer takes a moment, because of what’s being measured. A wing’s lift is perpendicular to its motion through the air. Over the ground, the aircraft’s velocity is not along that direction, so its lift is not perpendicular to its motion over the ground. In the ground frame, the horizontal component of lift does work on the aircraft.

During the half-turn from upwind to downwind, the horizontal lift points along the turn’s inside. The ground velocity has the wind’s component added to it along a fixed direction, and at every heading through that half-turn the lift has a component along it. Integrate the power and the answer per kilogram is

$$\int \mathbf a_{\text{lift}}\cdot(\mathbf V+\mathbf W)\,dt = \mathbf W\cdot\int \mathbf a_{\text{lift}}\,dt = \mathbf W\cdot\Delta\mathbf V ,$$

with the first form’s $\mathbf V$ part giving no work, since lift is perpendicular to $\mathbf V$. The velocity change over the half-turn is 80 metres a second, from $-40$ to $40$ along the wind, and $\mathbf W$ is 15, for an answer of 1200 joules per kilogram. The march gives 1200 from the integrated force and 1200 from the speeds.

So the wing pays it, by the same lift it was already producing, with no extra angle of attack and no extra drag. The energy is not an expense, because nothing in the aircraft’s account has changed. It is an entry in another observer’s accounts. On the upwind half of the circuit, lift does negative work in the ground frame, and the energy goes back. Kinetic energy isn’t a Galilean invariant, as a stone dropped in a moving train demonstrates, and a statement that “the energy has to come from somewhere” must always be followed by “measured by whom?” It is the same test that the rule for a leaking cart fails: a statement that changes with the steady speed of the person making it is not a statement about forces. A boat’s apparent wind is the same thing seen from the water, where the speed that matters to the sail is the one measured from the deck.

Inertia sees the same change either way #

The appeal to inertia deserves its own answer, since it is what makes the mistake persuasive. Inertia resists a change in velocity, and it does so in a stated inertial frame, so it is fair to ask how large a change the aircraft’s inertia is being asked for.

Velocity is a vector, and the change in it over the half-turn is the same in both frames. In the air, the velocity goes from 40 metres a second one way to 40 the other, which is a change of 80. Over the ground it goes from 25 one way to 55 the other. That is also a change of 80, since adding a constant wind to both ends subtracts away. The argument that a turning aircraft must speed up from 25 to 55, a change of 30, has been looking at the lengths of two arrows and forgotten that they point in opposite directions. Every observer agrees that 80 metres a second of velocity has been added, and the lift that does it is the same lift.

It takes 22 seconds at this bank, at a mean acceleration near 3.6 metres per second squared, and the wind appears nowhere in that sum. An aircraft in a steady wind therefore has no extra inertial demand on the downwind half of its turn, and no relief on the upwind half. A demand for more force than the wing already produces would be a demand for a different turn.

Then why do people die in it #

They do, so something is real. The clearest account is that the pilot is the feedback loop that the equations don’t contain.

Close to the ground, in a turn, a pilot is taking a large share of the sense of speed from what’s going past the window. Through the half-turn in this wind, the ground’s rush rises by 30 metres a second. The natural response is to slow down, and if the eye has been the instrument, the pilot pulls back and takes off speed. Nothing else has been going on, and nothing else is needed to stall the aeroplane.

The margin here is thin, as the figure shows. An aeroplane stalling at 30 metres a second level, flying at 40, in a 30° bank stalls at $30/\sqrt{\cos 30^\circ} = 32.2$. A pilot who gives back only a quarter of the visual rise in that half-turn is left with the airspeed 32.5, and is on the buffet. Steepen the bank to 45°, as someone trying to tighten a turn over a ground feature will, and the stall rises to 35.7. A wind of 15 metres a second is strong, but the visual effect is largest in an aircraft flying slowly at a low height, which is exactly what a low-level turn back to a field is. What makes this deadly is that the instrument that disagrees is the small dial on the panel, and the window is enormous.

There is also a companion illusion that works through the rudder. A pilot in a low turn into a crosswind sees drift across the ground, and it looks like skidding. An instinctive correction with rudder produces real skidding, with the inside wing going slower, and a stall from there is a spin. A stall of that sort is the roll that feeds itself, and at a low height it is not recoverable. Neither of the mechanisms has anything to do with inertia.

When the wind really does take airspeed #

The steady, uniform assumption is the whole result. Remove it, and $d\mathbf W/dt$ is back in the equations, following the aircraft, and it enters the airspeed directly:

$$\frac{dV}{dt} = \frac{T-D}{m} - g\sin\gamma - \frac{d\mathbf W}{dt}\cdot\hat{\mathbf V} .$$

The last term is the wind’s rate of change along the aircraft’s own path, and it can be of either sign. Near the ground the wind has a gradient, roughly logarithmic in height, as the layer over any surface does. A turn with a descent or climb in it moves the aircraft through that gradient.

The first thing to notice about that figure is the size. Over a whole turn in a strong wind, a descent through the profile changes the airspeed by a metre or two. That is real, and it is not nothing, but it does not stall an aircraft with a margin of 8. The second thing is the sign: it depends on the combination of heading and of height being gained or lost. A descending turn loses airspeed while heading into the wind, because the headwind weakens on the way down, and gains it while heading downwind. A climbing turn does the reverse. So in a steady shear a level turn has no effect at all. The argument that the wind gradient makes the downwind turn specially dangerous isn’t correct as stated, either. A pilot turning downwind and climbing is losing airspeed. A pilot turning downwind and sinking is gaining.

Those signs are precisely what soaring seabirds use. Rayleigh pointed out in 1883, in a note on the flight of birds, that an albatross gains energy from a wind gradient by climbing into it and descending away from it, which makes both limbs of its cycle favourable. It is called dynamic soaring. Radio-controlled gliders working the shear behind a ridge line have reached several hundred kilometres an hour by it. A real albatross’s energy comes from the last term of the equation above, taken with the right sign, and it is a different saving from the one a flying V takes from its neighbour’s upwash.

A wind that diverges is the dangerous one #

The case that kills airliners has nothing to do with turning. The airspeed term is

$$\frac{d\mathbf W}{dt}\cdot\hat{\mathbf V} \approx V\,\frac{\partial W_s}{\partial s},$$

for a wind whose component along the path changes along the path. It depends on the divergence of the wind along the flight path, and heading enters only through whichever component that is. Suppose the wind grows in the direction it blows, a diverging flow. Flying along that direction, the tailwind rises and the airspeed falls. Flying against it, the headwind falls away and airspeed falls. Both ways. A diverging wind takes airspeed on every crossing, whichever way the aircraft is going.

That is the shape of a microburst. A cold downdraught reaches the ground and spreads outward in all directions, so an aircraft on approach through it first gets a rising headwind, which raises its airspeed and tempts the crew to reduce power, and then finds it falling away and turning into a rising tailwind, with both the second parts taking airspeed. Fujita identified the phenomenon after Eastern 66 crashed at New York in 1975. Delta 191 at Dallas in 1985 is the case that led to the current detection systems. Bowles’s F-factor is the term above, divided by $g$: the specific energy lost per unit time, in the units of a climb gradient. For this figure’s wind, 10 metres a second over 150 metres, at 40 metres a second, the F-factor is $0.27$. A value of $0.1$ is generally treated as a hazard.

Reverse the sign and that loss becomes a gain. A converging wind, as along a sea-breeze front, gives airspeed on every crossing. That is a real resource, used by glider pilots along such lines, though the updraught is the larger part of what they are after.

What the picture cannot show #

The stall itself. The model continues in a straight line through the stall speed: the falling curve is not what the aircraft would do. A stall changes the forces entirely, as earlier essays say.

A real aircraft’s dynamics. A point mass has no attitude and no time lag. A real aircraft whose airspeed falls without a pilot’s input noses down as a matter of static stability, trading height for some of the loss. An entry to a shear in a real aeroplane therefore looks less abrupt than the figure’s. The aeroplane’s dynamic modes respond in a few seconds, shorter than most of these turns and longer than the crossing in the last figure.

Turbulence. The real near-ground wind is gusty and the mean profile is a fiction for one flight. Short gusts of a few metres a second change the airspeed by the gust’s own size, in either direction. Everything in this essay is about means; gusts are what a stall margin exists to absorb. The gusts’ size in a strong wind is about the same as the entire effect of the mean gradient.

Instruments and perception. The mental model of the pilot here is a one-parameter caricature. Nobody takes off an exact share of the visual rise, and the size of the illusion depends on height, on the visual scene and on experience, as human factors research has shown. The figure’s use is to show how little of it is needed.

The convention #

Airspeed is $V$; heading is measured from the direction the wind blows towards. Wind is positive along $x$. Starting at heading $\pi$ is into wind. Banks are constant, and turn rate is $g\tan\phi/V$; the stall speed in a level turn scales by $1/\sqrt{\cos\phi}$ from a wings-level 30 metres a second. In the wind-gradient figure, the wind at height $z$ is $W_{10}\ln(z/z_0)/\ln(10/z_0)$ with $z_0=0.05$ metres. Thrust is held at its trim value throughout, so that the airspeed’s changes come only from the inertial term; the model’s drag does not respond. That removes the recovery term a real aircraft has, so airspeed changes persist rather than fading, and for a turn of forty-four seconds in a clean aeroplane that is a modest overstatement.

How it was checked #

The check that carries the argument is the work one, since it compares two calculations that don’t share an operation: one integrates lift along the ground velocity through the half-turn, and the other reads two speeds at its ends. They agree to the third figure. The model refuses a stationary aircraft, a 90° bank, and a negative steady wind.

Who said what, and when #

Rayleigh’s note on soaring dates from 1883. Langewiesche’s Stick and Rudder, from 1944, contains the classic statement of the frame argument, and the downwind-turn dispute has filled pilots’ magazines ever since. Fujita named the microburst in the 1970s and published its case studies in the 1980s. Bowles’s F-factor dates from around 1990. The argument about visual illusion, as distinct from a physical loss of airspeed, is in the accident investigators’ reports from the same decades.

Still open: how much of a gust a stall margin must hold #

The calculations here treat the wind’s change as a smooth, stated line. The next calculation would drive the same aircraft with a measured gust spectrum near the ground, of a size set by the same log-law friction velocity as the mean, with the short-period and phugoid modes included so that the aeroplane’s own response counts. It would ask how often a turn at 30 degrees of bank and a stated margin meets a stalling angle of attack, how that frequency depends on bank and height, and so whether the traditional rule of adding half the gust to the approach speed covers a level turn, which is flown with a stall margin that the bank has already partly spent.