Between hover and twice the hover inflow

Each actuator disc worked through before this one had a stream flowing through it in one direction. The most a disc can take let the wind blow through a turbine; a big slow push let a propeller push air backwards while it flew forwards; the wake that has to spin added the torque. In each the air arrived from one side, left from the other, and the control volume round it was a tube with an inlet and an outlet.

A helicopter rotor can climb, hover and descend along its own axis, and when it descends the air meets it from below while the rotor is still pushing air downwards. Following that one change through the same momentum balance produces something none of those problems met: a band of flight conditions in which the balance has no answer at all, bounded by two numbers that come out exactly.

One disc, two momentum equations #

Keep the thrust $T$ fixed and let the rotor climb at speed $V$, counted positive in the direction the thrust pushes air — downwards through a lifting rotor — so that descent is negative. Relative to the rotor, the stream arrives at $V$, passes the disc at $V+v$ with $v$ the induced velocity, and leaves in a wake at $V+2v$: the same half-at-the-disc, all-in-the-wake split the turbine calculation derived.

The thrust is the mass flow through the disc times the change in velocity along the tube. Climbing, the mass flow is $\rho A(V+v)$ and

$$T = 2\rho A\,(V+v)\,v.$$

Measure every speed in units of the hover induced velocity $v_h = \sqrt{T/2\rho A}$ — the value at $V = 0$ — and this is $v(V+v) = 1$, with the positive root

$$v = -\frac{V}{2} + \sqrt{\frac{V^2}{4} + 1}.$$

In hover $v = 1$. Climbing, $v$ falls towards $1/V$, because a rotor moving into fresh air meets more mass every second and needs to accelerate each kilogram less.

Descend fast enough and the whole stream through the disc moves upwards: the rotor is falling through air that rises through it, and the thrust now opposes the flow, as a windmill’s does. The mass flow is $-\rho A(V+v)$, the balance becomes $v(V+v) = -1$, and its root

$$v = -\frac{V}{2} - \sqrt{\frac{V^2}{4} - 1}$$

is a real number only when $V^2 \ge 4$. On the descending side that means $V \le -2$: the windmill-brake state begins at a descent of exactly twice the hover induced velocity, where $v = 1$ again.

The hover unit is doing more than tidying the algebra. Written in it, the thrust, the disc area and the air density have all left the problem, so the two branches and whatever lies between them are the same for every rotor that has been built: a model in a wind tunnel, a small multirotor, a helicopter and a heavily loaded tiltrotor all sit on the first figure’s curves at the same places. What differs between them is only the conversion back to metres a second, $\sqrt{T/2\rho A}$, which is the one dimensional speed a disc carrying a thrust possesses.

That is the collapse the turbine calculation found, whose ceiling of 16/27 contained no machine, and it has the same consequence here: nothing a rotor designer puts inside the disc can move the edges on this axis. Blade shape, twist, number of blades and aerofoil section all live inside the box. The designer’s only lever on the band is the disc loading, through the conversion — a larger disc for the same weight lowers the hover inflow and narrows the band in metres a second — which is the same lever a big slow push found deciding a propeller’s efficiency.

A root is not a streamtube #

The climb formula has a root at every speed, including every descent, and the dashed curve in the first figure is that root. What it does not have in descent is a flow.

A streamtube is a region air enters at one end and leaves at the other, and the whole momentum argument is an account of what crosses its two ends. Climbing at one hover inflow, the air arrives from above at 1, passes the disc at 1.618 and leaves below at 2.236: in at the top, out at the bottom, faster. Descending at three hover inflows on the windmill branch, the air arrives from below at 3, passes the disc at 2.618 and leaves above at 2.236: in at the bottom, out at the top, slower, because the rotor is taking energy from it.

Descending at one hover inflow, the climb root gives an arriving velocity of −1 and a wake velocity of +2.236. The air far above the rotor is moving up, away from it, and the air far below is moving down, away from it. Air is leaving through both ends and entering through neither, which a steady tube cannot do, and the thrust that formula assigned to it is an answer about a flow that does not exist.

Two different failures that share one interval #

The band is bounded by two separate facts, and it is worth seeing that they are separate, because each edge fails for its own reason.

The climb root’s wake velocity is $V + 2v = 2\sqrt{V^2/4+1}$, which is positive at every speed. So the climb root describes a consistent tube exactly when the arriving velocity $V$ has the same sign — for every climb, for hover, and for no descent at all. Its failure begins at the upper edge of the band and never ends.

The windmill root’s failure is of a different kind. Its equation simply has no real solution while $V^2 < 4$, and where it does have one, $V \le -2$, its wake velocity $-2\sqrt{V^2/4-1}$ is upward like its arriving velocity, so the tube is consistent. Its failure ends at the lower edge of the band.

Between $V = -2$ and $V = 0$ both have failed, and no third branch is available: the balance is a statement about a tube with one inlet and one outlet, and those are the only two ways the air can pass through a disc that pushes it one way. The band is not where the theory becomes inaccurate. It is where the object the theory is about has ceased to exist.

At each edge, one end of the tube stands still #

The two edges have a common physical reading, and it is the part of the calculation that explains what the air does inside the band instead.

At the upper edge, hover, the air that feeds the rotor from far away is not moving relative to it. The rotor still draws air in, but nothing brings it: the arriving stream has speed zero. At the lower edge the reverse holds. The windmill branch’s wake velocity $-2\sqrt{V^2/4-1}$ is exactly zero at $V=-2$, so the air that has passed through the disc and been slowed by it is left at rest relative to the rotor, going nowhere.

That is the vortex ring state seen from momentum theory. A rotor sheds vorticity at its blade tips continuously, as any lifting surface with ends must, and in any consistent tube the stream carries it away — downwards in climb, upwards in fast descent. Inside the band there is no stream to do it. The shed vorticity accumulates round the rim of the disc as a ring, and a ring moves because it is bent: its own induced velocity carries it, erratically, back through the rotor it came from. The thrust becomes unsteady, and more collective pitch, which puts more vorticity into the ring, deepens it.

The same figure says why the escape the blade-element essay describes works: fly forwards so that the rotor meets fresh air. A horizontal stream through the disc is a velocity that does not vanish at either edge, and it carries the tip vorticity off sideways whatever the vertical component is doing.

The descent that needs no power is inside the band #

A rotor falling steadily with no engine, driven by the air rising through it, is in autorotation, and the momentum balance can say where that condition would have to be.

The induced power is $T(V+v)$, and in hover units it is simply $V+v$, the velocity through the disc. On the climb branch that velocity is always positive, so the rotor always does work on the air; on the windmill branch it is always negative, so the air always does work on the rotor. The two branches end at $+1$ and $-1$, at hover and at a descent of two hover inflows respectively, and the power passes through zero somewhere between them — inside the band, where neither branch is valid.

Ideal autorotation, zero induced power, is the condition $v = -V$, the faint diagonal of the first figure. It crosses the band and meets neither curve. So momentum theory, which gives a turbine’s ceiling and a propeller’s efficiency exactly, cannot say at what rate a helicopter with a failed engine will come down. That number is measured: rotors in wind tunnels, and aircraft in flight, autorotate at descent rates inside this band, towards its lower edge, with the power needed to overcome blade drag pushing them a little further down than the ideal would be.

The word is shared with the roll a stalled wing sustains in a spin, and the two are different mechanisms with one thing in common: in both the air, not an engine, supplies the power that keeps a rotation going. A rotor in autorotation takes that power from air rising through it; a spinning wing takes it from the asymmetry of lift past the stall. Neither is described by a balance drawn for the steady, attached case.

This qualifies a sentence the blade-element essay gives as an escape from the vortex ring state, that lowering the collective into autorotation takes the rotor to the other side of the momentum curve. Lowering the collective lowers the thrust and so the hover induced velocity, which moves a given descent rate further down the axis in hover units, towards and past the lower edge. Autorotation itself is not on the other side; it is inside the band near that edge, in a state momentum theory does not describe.

A wind turbine on the same axis #

A wind turbine is a disc in an oncoming stream with its thrust opposing the flow: precisely the windmill-brake state. It can be placed on this axis exactly.

For a turbine in a wind $U$, the descending rotor’s stream is the wind, so $V = -U$ and the induced velocity is $aU$. The thrust coefficient on the turbine’s own area is $C_T = 4a(1-a)$, and the hover induced velocity of a disc carrying that thrust is $U\sqrt{C_T}/2$. In hover units the turbine therefore sits at

$$V = -\frac{2}{\sqrt{C_T}},\qquad v = \frac{2a}{\sqrt{C_T}},$$

and substituting shows the point satisfies $v(V+v) = -1$ identically: every turbine on the momentum curve is a point on the windmill-brake branch. Betz’s optimum, $a = 1/3$ and $C_T = 8/9$, is a rotor descending at $3/\sqrt2 = 2.1213$ hover inflows — just outside the band, which is why the ideal turbine’s momentum balance is safe.

The turbine’s own boundary is the band’s lower edge. $C_T$ reaches its maximum of one at $a = 1/2$, which is $V = -2$ exactly, and the blade-element essay records what happens beyond it: the wake that should run backwards goes turbulent, entrains outside air, and the measured thrust coefficient keeps rising past one where the formula turns down. Every thrust coefficient above one on a heavily loaded turbine is a point inside the band a descending helicopter enters. The wind-energy name for it is the turbulent wake state; the rotorcraft name for the part near hover is the vortex ring state; the arithmetic says they are two ends of one interval.

How wide the band is in metres a second #

Hover units hide the number a pilot or a drone’s controller needs, and restoring them shows what makes the band wider.

The edge is $2\sqrt{(T/A)/2\rho}$, so it grows as the square root of the disc loading and as one over the square root of the air density. A small multirotor carrying 80 newtons per square metre of disc has a band from hover to 11.4 metres a second of descent at sea level; a helicopter at 350 has one reaching 23.9; a heavily loaded rotor at 950 reaches 39.4. At 4000 metres the same three bands reach 14.0, 29.2 and 48.2 metres a second.

The practical reading is the opposite of the intuitive one. A heavily loaded rotor in thin air is the one with the widest band, because its hover inflow is fast — which is also the rotor whose ordinary approach to a landing or a hover involves a descent rate well inside it. A light rotor with a large disc descends out of its band at a gentle rate; a small, heavily loaded one must keep its descent slower still or move sideways while it descends.

Why more pitch drives a rotor further in #

The band’s edge in metres a second depends on the thrust, and a rotor descending steadily carries its weight, so in steady descent the edge is fixed by the aircraft’s weight and the air it is in. That is what turns the instinctive recovery against itself, and the size of the effect follows from the square root in the hover inflow.

Pulling more collective pitch to stop a descent raises the thrust above the weight, and the band’s lower edge moves out with the square root of the thrust: a fifth more thrust moves it 9.5 per cent further down the descent axis, and half as much again moves it 22 per cent. For the rotor loaded at 350 newtons per square metre, whose band ends at 23.9 metres a second in steady descent, a thrust one and a half times its weight pushes the edge to 29.3 metres a second. A rotor descending at 15 metres a second sits at −1.26 hover inflows; the same descent rate with that extra thrust sits at −1.02 — further from the lower edge, not closer to it. The thrust meant to arrest the descent has moved the rotor towards the middle of the band.

Lowering the pitch does the reverse, which is the arithmetic behind the advice to do it. With thrust below the weight the rotor accelerates downwards, so its descent rate rises while its hover inflow falls, and both carry its point towards the lower edge and out through it. The exit costs height, which is why the condition is most dangerous where a descent is slow and near the ground — on the approach to a landing or a hover, exactly where the band is entered in the first place.

What the momentum picture leaves out #

The inflow is uniform and axial. A real rotor’s induced velocity varies across the disc and is highest near the tips, so parts of the disc enter the band before the average does. And the analysis is along the axis only; a component of forward speed changes the picture entirely, which is its own question.

The band is empty of prediction, not of flow. Rotors in it produce thrust, often a fluctuating one, and the induced velocity there has been measured and fitted. Nothing in the balance here says what those measurements show, and a calculation that fills the band with a smooth curve is quoting a fit.

The flow is steady. The vortex ring state is an unsteady condition, with the ring forming, shedding and reforming, and the inflow takes time to respond to any change in thrust. A rotor passing quickly through the band may not develop the ring at all.

No blades, no swirl, no ground. The disc has none of the losses the blade-element essay adds, and a rotor near the ground has an image of its wake beneath it that changes both edges — the same image that makes a hovering rotor’s carried air a real cushion rather than a figure of speech.

Who worked it out #

The windmill-brake and vortex ring states were named in the British propeller and windmill work of the 1920s, when tests on airscrews run in reverse flow showed the momentum balance failing in exactly this interval, and Glauert published an empirical curve through the missing region in 1926. The helicopter made the interval a practical question, and a wind-tunnel study at NACA in 1951 measured the induced velocity of model rotors across the whole descent range and fitted the curve rotorcraft analyses still use.

The edges themselves are older than any of it. They are the Froude–Rankine momentum balance of the 1880s, applied to a stream that may come from either side of the disc, and the only arithmetic needed to find them is the discriminant of a quadratic.

Still open: what forward speed does to the band #

Forward speed adds a velocity through the disc that does not vanish at either edge, and the momentum balance generalised to it — the induced velocity multiplied by the resultant speed through the disc, set equal to the hover value squared, which is the inflow the calculation of a rotor’s two unequal sides solves in level flight — has a real root at every descent rate once any forward speed is present. That is not the same as a consistent tube: whether the wake is carried away depends on how fast the shed vorticity moves relative to the rotor, and the boundary of the vortex ring state in the plane of forward and vertical speed is drawn from criteria about that transport speed rather than from the balance itself. Computing where the band closes as forward speed grows, and how much of the published boundary is the balance and how much is the criterion, is a calculation with a real result in it.

Beside it is the turbine side’s version of the same question: a turbine in yaw has a wake skewed off its axis, and whether its thrust coefficient can pass one without entering the turbulent wake state is the same transport argument run the other way.