The side that cannot keep up

A hovering rotor is the most symmetric machine in this collection. Every blade meets the same wind at the same radius whatever azimuth it is at, so one distribution of circulation along the blade describes the whole disc, the thrust is an integral over one variable, and the actuator disc that represents it is a legitimate simplification of a genuinely axisymmetric flow.

Move the rotor forward at an advance ratio $\mu = V/\Omega R$ and the blade’s own speed becomes

$$U_T = \Omega R\left(\frac{r}{R} + \mu\sin\psi\right),$$

so at μ = 0.35 the advancing tip meets 1.35 ΩR and the retreating tip 0.65. Lift goes as the square of that, so the advancing side would carry four times what the retreating side does, and the machine would roll over within a revolution.

It does not, and what stops it is a control hover does not need.

The loading over the disc #

Every quantity a hover calculation reports as a function of radius is here a function of two variables, and the picture is the reason: the loading is not axisymmetric, cannot be made axisymmetric, and is concentrated on the front and rear of the disc rather than on its sides.

That last observation is the whole of the trim problem. A rotor trimmed to zero rolling moment does not have a symmetric loading — it has a loading whose first harmonic in the roll sense is zero, which is a much weaker statement. The advancing side still carries more at most radii; what has been removed is the integrated moment.

The two sides of one blade #

In hover the two are the same number by symmetry, and the solve returns them equal to fifteen digits — which is a check on the quadrature rather than on the physics, and is worth having for that reason.

By μ = 0.5 the advancing section carries fifty-eight times what the retreating one does, and the retreating section is close to carrying nothing at all: 0.09 degrees of incidence at a dynamic pressure a quarter of the advancing side’s.

The ratio grows monotonically — 1.4 at μ = 0.1, 3.1 at 0.3, 6.4 at 0.4 — and it is the single number that says what forward flight has done to a machine designed for hover.

The control the rotor needs #

The trim is a three-control problem — collective, lateral cyclic, longitudinal cyclic — solved here by a Newton iteration on the three residuals to $4\times10^{-16}$. What comes out has two features worth naming.

The collective falls with speed, from 7.48° in hover to 3.53° at μ = 0.5. That is not the rotor working less hard; it is Glauert’s inflow falling as the machine acquires a mass flow it no longer has to accelerate from rest. The induced velocity at μ = 0.5 is a ninth of its hover value, so the blade sees a much smaller induced angle and needs much less pitch for the same lift.

And the cyclic grows from nothing, because in hover there is no asymmetry to remove. It reaches −0.69° at μ = 0.5 in this rigid-bladed model.

That number is small compared with a real helicopter’s, and the reason is the model rather than the physics: a real blade flaps. Hinged at the root, it rises on the advancing side and falls on the retreating side, and the flapping changes the local incidence in exactly the sense that relieves the asymmetry — a mechanical version of the same relief a gust-alleviating wing gets from its own motion. Flapping and cyclic pitch are very nearly equivalent, which is Lock’s observation and is why a rotor can be trimmed with a swashplate at all. A rigid-bladed calculation asks the cyclic to do the whole job and still finds it a small angle, because the loading it is cancelling is an integral rather than a peak.

A circle of reversed flow #

There is a second thing forward flight introduces that hover has no room for, and it is exactly geometric.

Where $r/R < -\mu\sin\psi$ the blade’s own section is moving backwards through the air. In the disc plane that region is exactly a circle of diameter μR sitting on the hub on the retreating side, so its area fraction is $\mu^2/4$ — with no π in it, because both areas carry one.

Measured against computed: 0.002499 against 0.0025 at μ = 0.1, 0.02246 against 0.0225 at 0.3, 0.06251 against 0.0625 at 0.5. Four decimal places, from a count.

That is a kinematic statement with no aerodynamics in it whatever — it needs only the definition of $U_T$ — and it is the cleanest available demonstration that the disc’s two-dimensionality is unavoidable. At μ = 0.5 a sixteenth of the rotor is working backwards, and no radial function can say so.

Where the machine runs out #

The stalled fraction is zero in hover by symmetry and grows to 2.07 per cent at μ = 0.5. It includes the reversed-flow region, which is stalled in the trivial sense of working backwards, and it grows because the retreating side has to make up in incidence what it has lost in dynamic pressure.

This is what limits a helicopter’s speed, and it is worth separating from the limit everybody names. The advancing tip does approach the speed of sound — at μ = 0.5 with a tip speed of 220 m/s it is at about Mach 0.97 — and that is a real constraint. But the retreating side’s problem arrives at the same time and is worse: a blade that has stalled produces no lift, a large pitching moment and a great deal of vibration at once per revolution, and the vibration is usually what stops the pilot rather than the lift.

Retreating-blade stall is also what makes the limit an aeroelastic problem rather than an aerodynamic one. The stall is dynamic — the blade pitches up rapidly as it goes round — so it sheds a leading-edge vortex, produces a lift overshoot and then a violent nose-down moment, and the moment twists the blade, which changes the incidence, which changes the stall. Nothing in the quasi-steady model above represents any of that.

What the machine does about it #

The trim above is the aerodynamic answer. The mechanical answers are worth listing, because between them they define what a rotorcraft is.

The flapping hinge, de la Cierva’s, which lets the blade rise where the lift is high and fall where it is low. A hinged blade carries no root bending moment, so the rolling moment it can transmit to the airframe is zero by construction — the asymmetry is absorbed as motion rather than as force.

The lag hinge and the damper, which are there because a flapping blade’s Coriolis force makes it lead and lag in plane once per revolution, and an undamped lag degree of freedom couples with the airframe into ground resonance.

The swashplate, which converts a stationary control input into a once-per-revolution pitch variation and is the device the cyclic angles above are settings on.

And the compound and coaxial configurations, which attack the problem rather than accommodate it. A coaxial rotor has two counter-rotating discs, so the advancing sides are on opposite sides and the rolling moments cancel between them — which permits the retreating side to be unloaded entirely, and is the basis of the advancing-blade-concept rotors that have reached advance ratios near 0.9. A compound helicopter offloads the lift onto a wing at speed, for the same reason.

Each of those is a response to the same fact: the loading over the disc is a function of two variables, and the only symmetric thing available is the pair of a rotor and its mirror image.

What the model is #

It is worth being explicit about the machine that has been computed, because it is a simple one.

Rigid blades. No flapping, no lagging, no torsion. The cyclic angles above are therefore not a real helicopter’s, and the loading is more asymmetric than a real one’s because nothing relieves it.

Uniform inflow. Glauert’s momentum-theory inflow, solved from $\lambda = C_T/2\sqrt{\mu^2+\lambda^2}$ and applied uniformly over the disc. A real rotor’s inflow is strongly non-uniform in forward flight — higher at the back than the front, because the wake is swept downstream — and that non-uniformity is a first-order effect on the longitudinal trim.

A capped lift coefficient, at 1.2, standing in for stall. It is what keeps the integral finite in the reversed-flow region and it is a caricature: a real section’s lift falls after stall rather than plateauing, and a dynamically stalled one overshoots first.

And quasi-steady sections. Each blade element is treated as though it were in a steady flow at its instantaneous velocity and incidence, which is the assumption the reduced frequency measures and which a rotor blade at once per revolution violates by more than most things do.

The lift the cap is standing in for #

The lift-coefficient cap deserves a paragraph, because the choice of 1.2 is doing more work than it looks.

Without it the integral is not merely inaccurate — it is dominated by the strip where $U_T$ passes through zero, where the induced angle $\lambda/U_T$ is unbounded and the section is at an infinite incidence. An uncapped model reports a rotor needing half a degree of cyclic at μ = 0.3 where the answer is nearer five, with every residual at 10⁻¹⁶ and nothing anywhere to indicate a problem.

That is a numerical artefact presenting as a physical result, and the site’s own habit is the reason it was caught: the cyclic came out an order of magnitude below what a helicopter uses, which is a check against the world rather than against the code.

Capping the lift coefficient is what a real section does and it makes the model well-posed. What it does not do is make the reversed-flow region right — a section working backwards has a rounded trailing edge facing the flow, no Kutta condition in the usual place, and a lift curve nothing like a capped straight line.

Why hover is the wrong intuition #

The general point is worth drawing out because it applies beyond rotors.

Hover is an axisymmetric problem, and axisymmetry is a symmetry of the boundary conditions that makes the solution a function of one variable. Forward flight breaks it, and what replaces it is not a small perturbation: the reversed-flow region has no counterpart in hover at all, and neither does a control that varies once per revolution.

The same relationship holds between a wing in steady flight and a wing in a gust, and between a hovering rotor and one in the vortex-ring state. In each case the simpler problem is not the limit of the harder one in the sense of being a first term: it is a different problem with a symmetry, and the symmetry is what has been lost.

And a collapse inherits the symmetry of the thing it was derived for. The actuator disc is axisymmetric by construction, so a disc theory for forward flight — and there are several — has to introduce the azimuthal variation by hand rather than derive it.

The vibration, which is the real design driver #

One more consequence of the two-variable loading is worth naming, because it is what a rotorcraft engineer spends most of the time on.

A blade going round in a non-axisymmetric flow produces a force that varies once per revolution. Summing over $B$ blades, most of those harmonics cancel at the hub: what survives is the harmonic at $B$ per revolution and its multiples. So a four-bladed rotor shakes its airframe at four times the rotor frequency, and a five-bladed one at five, and the amplitude of that shake is set by how non-uniform the disc loading is — which is what the figures above have been measuring.

That is why the loading over the disc, rather than its integral, is the quantity of interest. The integral is the thrust, and the thrust is easy; the harmonics are the vibration, and the vibration is what limits the fatigue life of the airframe and the comfort of everybody inside it. Higher-harmonic control — a swashplate driven at $B$ per revolution rather than once — exists entirely to reshape the part of the loading a trim calculation does not touch.

The same distinction between an integral and its distribution runs through the whole of this field: a wing’s spanwise loading integrates to one lift and decides where the stall starts, and a bell-shaped loading integrates to the same lift as an elliptic one and produces a different yaw. In each case the scalar is the easy half.

What sets the speed of a helicopter #

The three limits in the figures are usually described as one — “retreating blade stall” — and they are three, arriving in an order that matters.

The reversed region arrives first and costs nothing. Its area is exactly $\mu^2/4$ of the disc, so at μ = 0.3 it is 2.25 per cent, and the blade is at a small radius there where the dynamic pressure is small anyway. It is a nuisance in the trim rather than a limit.

Stall arrives next, and it is what the cyclic runs out against. The retreating blade must produce its share of the lift at a fraction of the advancing blade’s dynamic pressure, so its incidence must be higher, and beyond an advance ratio near 0.4 there is no cyclic setting that trims the rotor with every section below its stalling incidence. The measured stalled fraction in the figures grows steeply through that region and then flattens — not because the stall has stopped spreading but because the reversed region has taken over the part of the disc that would have stalled next.

And compressibility arrives on the other side. The advancing tip sees $\Omega R(1 + \mu)$, which at a tip Mach number of 0.65 and μ = 0.4 is Mach 0.91. Drag divergence there costs power and noise, and it is why the advancing tip is swept and thinned on every fast helicopter built.

Those three are why a conventional helicopter stops near 170 knots and why every attempt to go faster — a compound, a tilt-rotor, a stopped rotor — is an attempt to stop asking one rotor to do both jobs. The limit is not in the engine. It is in the asymmetry the figures above measure.

The same diagram with the engine removed #

Every calculation above has the shaft driving the rotor. Take the drive away and the identical blade-element picture answers what happens next, with one sign changed.

With no torque the machine descends, so air comes up through the disc, tilting the resultant velocity at each section forward. Over a band of radius that tilt is enough to swing the section’s aerodynamic force ahead of the shaft axis, so that part of the blade is driving the rotor; further out it still opposes, and is driven. The descent rate settles where the two balance and the net torque is zero.

That is autorotation, and it is a stable equilibrium rather than a trick: let the rotor slow and the driving band widens, let it speed up and the band shrinks. The rotor keeps turning on the air’s energy, and the height being given up is the fuel.

The landing draws on something else — the rotor’s own rotational kinetic energy, spent in the flare by raising collective — which is why blade inertia is a design parameter with a certification consequence, and why a rotor is deliberately heavier than the aerodynamics alone would ask. It is also why the dangerous corner is not engine failure at speed but the low, slow one, where there is neither airspeed to trade nor height to build rotor speed in.

Where the machinery came from #

Glauert’s 1926 paper on the rotor in forward flight is where the inflow model used here comes from, and it is a remarkable piece of work: he wrote the momentum theory for a disc at an angle to the stream, derived the flapping equations, and computed the trim, all before any helicopter flew.

Juan de la Cierva’s autogyro of 1923 is the machine that made the problem urgent. His first three prototypes rolled over on take-off for exactly the reason this essay opens with, and the flapping hinge — introduced on the fourth — is the fix. It is one of the few cases in this subject where a piece of aerodynamics was solved by a mechanical invention rather than by a calculation, and the calculation came afterwards.

Lock’s demonstration that flapping and cyclic pitch are equivalent, and Hohenemser’s and Sissingh’s subsequent work on the trim, are what turned that invention into a design method.

What this leaves #

The disc’s axisymmetry is a symmetry of hover and nothing else. What forward flight introduces — azimuthal variation, a reversed-flow circle, a retreating side that runs out — is not carried by any radial function, and the machine’s speed limit lives entirely in it.

The next essay takes a different kind of collapse: a blade row replaced by a turning device, and what a solidity decides that an isolated aerofoil cannot.