A wing over the sea barely touches it

Every account of ground effect so far has treated the ground as rigid. The cushion that is not there argued that the lift a wing gains near the ground is not a cushion of trapped air but the ground’s reflection of the wing’s own vortex system; the cushion that is there after all found the pressure under a low wing that does behave like one. A hydrofoil loses most lift on the way up then turned the problem over: a foil under a free surface, whose pressure deforms the surface above it, losing lift as it rises because the surface gives way.

It ended on the other half, which is what every ground-effect craft over the sea actually is. A wing above water presses on the water. The water is not rigid — it can be pushed down, and it can carry waves away — so the reflection that gives the ground effect might be weakened, and the waves might cost a drag. The craft’s designers worry about a hump in the drag at low speed, and the popular picture of a ground-effect craft riding on a cushion that presses a trough into the sea suggests the sea gives a great deal. This essay computes how much it gives, and the answer is almost nothing, for a reason that fits in one number.

The wing’s footprint #

Model the wing as its bound vortex, of circulation $\Gamma = L/\rho_a U$, at a height $h$ above the surface, with an image vortex below to keep the air from passing through the surface. Along the surface the pair slows the air by $\Gamma h/\pi(x^2 + h^2)$, and by Bernoulli the pressure there rises by $\rho_a U$ times that:

$$p(x) = \frac{L}{\pi}\,\frac{h}{x^2 + h^2}.$$

That is the wing’s footprint on the sea, and it carries a fact worth stating: its area is the whole lift. In two dimensions the ground holds the wing up — every newton of lift is a newton pressing on the surface, spread over a width about equal to the height. For a wing of five-metre chord at a lift coefficient of 0.8 at fifty metres a second, the lift is 6.13 kilonewtons per metre of span, and at one metre up the peak pressure on the water is 1.95 kilopascals. Held still, that would push sea water down 194 millimetres.

The question is what moving water does with it.

The worst speed, and how bad it is #

Deep water moving at $U$ under a steady pressure responds wavenumber by wavenumber, and the only wavenumber that carries energy away is the one whose waves travel at $U$: $k_0 = g/U^2$. The drag that is made of waves derived the wave resistance of a body from that pole, and the same residue gives it for a pressure: $R = k_0^2\,|\hat P(k_0)|^2/\rho_w g$, with $\hat P$ the footprint’s Fourier transform. The Lorentzian’s transform is $L\,e^{-kh}$, so

$$\frac{R}{L} = \frac{\rho_a}{\rho_w}\;\frac{C_L}{2}\;(k_0 c)\,e^{-2k_0 h}.$$

Everything about the result is in that line. The wave drag rises from zero at low speed, where the waves are so short that the footprint, smooth on the scale of $h$, cannot make them; it falls again at high speed, where the waves are so long that the footprint is a point and $k_0c$ is small. Between, it peaks where $k_0 = 1/2h$, at the speed $\sqrt{2gh}$ — 4.43 metres a second for a wing one metre up, whatever its chord — and there it is $(\rho_a/\rho_w)\,C_Lc/4eh$ of the lift: for this wing $4.4\cdot10^{-4}$. At a cruise of fifty metres a second it is $9.3\cdot10^{-6}$.

For comparison, the induced drag of a wing of aspect ratio three in strong ground effect is a few per cent of its lift and the profile drag about one per cent. The wave drag of the wing is three orders of magnitude below either at its worst. The hump in a ground-effect craft’s drag at take-off is real, but it belongs to its hull or floats ploughing through the water before the wing has lifted them, not to the wing’s pressure on the sea.

Why the worst speed is set by the height #

The worst speed, $\sqrt{2gh}$, contains the height and not the chord, and the reason is the footprint’s shape. A pressure spread over a width $h$ excites waves whose length is comparable to $h$ efficiently and much longer or shorter ones poorly, and the footprint’s transform says exactly how: $e^{-kh}$ for the short ones, and a factor $k$ in the drag for the long ones, whose energy per unit drag is small. The product peaks at $k = 1/2h$, a wavelength of $4\pi h$ — 12.6 metres for a wing one metre up — and the speed whose waves have that length is $\sqrt{2gh}$. The chord enters only through the lift, as a factor on the whole curve.

A large craft makes the arithmetic concrete. An ekranoplan-sized wing of twenty-metre chord flying three metres up at a lift coefficient of 0.6 has its worst speed at 7.7 metres a second, where the wave drag is $4.4\cdot10^{-4}$ of whatever lift it is then carrying — the same fraction, because $C_Lc/h$ is the same — and at a cruise of a hundred metres a second it is $7\cdot10^{-6}$. Large or small, the wing’s waves are a rounding error in its drag; what one number decides here is that they are, and the number is the density ratio.

The density ratio is the whole answer #

The factor in front is the ratio of air’s density to sea water’s, $1/837$. It appears because the pressure the wing can make is set by air’s density — $\rho_aU^2$ — while the force needed to move the surface is set by water’s — $\rho_w g$ per metre of depression, $\rho_wU^2$ per unit of dynamic response. Everything else in the formula is of order one: at the worst speed, $C_Lc/4eh$ is 0.37 for this wing.

So the same calculation with two fluids of the same density — a foil running just under a free surface of its own fluid, which is the hydrofoil’s situation — gives a wave drag of a third of the lift at the worst speed, and a surface that gives way by an amount comparable to the foil’s depth. That is why the hydrofoil essay found large effects and why this one finds small ones. The physics is the same; the density ratio moves the answer by three orders of magnitude.

The prediction can be tested, though not at sea. In a towing tank a model wing of half-metre chord at a lift coefficient of 0.8, run ten centimetres above still water at its worst speed of 1.4 metres a second, should leave waves 0.29 millimetres high and 1.26 metres long behind it — within the resolution of the capacitance wave probes tanks already use, and a direct measurement of the density ratio’s effect. A model hovercraft of the same size carrying the same small load would leave waves of much the same height: what makes a real hovercraft’s waves large is that it carries its whole weight at that speed, which a wing cannot, and that is the comparison the next section makes.

A hovercraft presses as hard and pays nearly two hundred times more #

The density ratio is not quite the whole story, because a hovercraft presses on the sea with air too, and its wave drag is notoriously large: the hump a hovercraft must climb over at low speed is often a tenth of its weight. The difference is in what sets the pressure. A hovercraft of ten tonnes on a cushion of fifty square metres presses two kilopascals on the water at every speed, including at rest, because its fans hold its weight up whatever it is doing. Treated the same way — a uniform pressure over a cushion ten metres long — its wave drag peaks at 5.6 metres a second, a length Froude number of 0.56, at eight per cent of its weight. That is the hump hovercraft designers fight.

A wing presses with a pressure that is its lift spread over its height, and its lift is $\tfrac12\rho_aU^2cC_L$. At the speeds where waves are costly — a few metres a second — the wing is not yet carrying anything worth the name; by the time it carries its weight, at tens of metres a second, the waves it could make are hundreds of metres long and its narrow footprint cannot excite them. The air’s density appears in the wave drag because it appears in the lift, and it is the coupling of the two through $U^2$ that makes the wing’s worst case so mild. A hovercraft decouples them, and pays.

Slow, the sea is dented; fast, it is lifted #

The surface itself, computed by transforming the water’s response back with a small damping that puts the waves behind the wing, shows how little the sea moves. At two metres a second the water under the wing is pushed down by 0.36 millimetres, a little more than the hydrostatic value for that speed’s smaller lift, with short waves behind. At the worst speed the trailing waves are at their largest relative to the wing, a few millimetres high and twelve metres long.

At fifty metres a second something less obvious happens: the water under the wing is lifted, by 3.7 millimetres. A steady pressure of wavenumber $k$ on deep water moving at $U$ produces a surface $-\hat p/\rho_w(g - U^2k)$, which is a depression for long wavelengths, where gravity wins, and a rise for short ones, where the water’s inertia wins and the response changes sign. At high speed the footprint is short compared with the waves the speed would make, so the sea under a fast wing bulges up towards it by an amount of order $L/\rho_wU^2$ — the pressure’s force divided by the water’s dynamic pressure, which is 2.4 millimetres for this wing at fifty metres a second. The cushion of the popular picture presses no trough into the sea at cruise; it raises a ridge a few millimetres high.

The waves it leaves #

The waves that carry the drag away have a height $A = (\rho_a/\rho_w)\,cC_L\,e^{-gh/U^2}$. At high speed the exponential tends to one and the height to $(\rho_a/\rho_w)\,cC_L$, 4.78 millimetres for this wing, while their length, $2\pi U^2/g$, grows to 1.6 kilometres at fifty metres a second. A cruising ground-effect craft leaves behind it a swell a few millimetres high and a mile long — invisible on any real sea, whose own waves are a thousand times larger, and carrying a drag of a fraction of a newton per metre of span.

That a wave of fixed height is left at every high speed is the same density ratio at work. The footprint’s force grows as $U^2$ and the water’s resistance to a long wave grows as $U^2$ too, so the two cancel and leave a height set by the wing’s size and lift coefficient, scaled down by $837$.

The rigid ground was assumed and is earned #

The calculation started by putting an image vortex under a flat surface — that is, by assuming the sea is rigid — and then asked how far it moves. That is circular only if the answer is large, and it is not: the surface moves by millimetres under a wing a metre up, so the image’s position is in error by a few parts in a thousand, and the footprint computed from it is correct to the same order. The ground-effect results of the earlier essays, which took the ground as rigid, carry over to the sea with corrections of that size. The wing keeps the lift gain the ground gives it.

This is the opposite of the borrowed mass the boundary decides, where a free surface and a rigid one gave different answers by a factor of two because the body was in the water and the water’s own inertia was what mattered. Here the body is in the air, and what the water would have to do to matter costs eight hundred times more than the air can ask of it.

How it was checked #

The footprint’s transform was computed by quadrature over four thousand heights either side of the wing, with the Lorentzian’s tails beyond added analytically, and agrees with $L\,e^{-kh}$ to $7\cdot10^{-8}$ at four wavenumbers. The wave resistance was then computed without using the residue: the surface was built by transforming the water’s response back to physical space, with a Rayleigh damping of half a per cent of the wave term so that the waves trail behind, and the resistance taken as the horizontal force of the pressure on the sloping surface, $\int p\,\eta'\,dx$. At six metres a second it is 0.033326 newtons per metre against the closed form’s 0.033311, a difference of $4.5\cdot10^{-4}$ that halves with the damping. The surface under the wing at fifty metres a second changes by less than half a per cent as the damping is halved twice, so the 3.7-millimetre rise is the undamped answer.

The convention: two dimensions, linear water, a point vortex #

The wing is a two-dimensional lifting line at height $h$, represented by its bound vortex and an image in a flat surface, so its footprint is Lorentzian; a real wing’s chordwise loading spreads the footprint over its chord, which lowers the wave drag at the worst speed further. The water is deep, inviscid and linear, and its response is steady. Heights are to the bound vortex; the lift coefficient is 0.8 and the chord five metres throughout. Wave drag is given as a fraction of the lift the wing carries at that speed, not of a fixed weight, which is the fair comparison for a wing that only carries its weight at speed; against the craft’s full weight the low-speed fractions would be smaller still.

What the picture cannot show #

A real ground-effect craft has a finite span, and a three-dimensional pressure patch makes a Kelvin wave pattern, transverse and diverging waves, with a resistance that depends on the span as well as the chord; it is smaller than the two-dimensional value at the worst speed, because energy spreads sideways, so the conclusion survives. The calculation leaves out the craft’s hull, whose wave drag at take-off is large and is the real hump; spray; the sea’s own waves, which at any real sea state dwarf everything computed here and set the craft’s real limits, through the height it must keep; and the air’s own boundary layer over the water. And it treats the wing as steady: a craft pitching in a seaway changes its lift and so its footprint, but the density ratio keeps the sea’s response to that small too.

Who found it, and when #

Lamb gave the waves made by a travelling line of pressure in his Hydrodynamics, and Havelock developed the wave resistance of moving pressure distributions in the first decades of the twentieth century, for ships. The same formulas were applied to hovercraft in the 1960s, where the pressure is the craft’s whole weight on a cushion and the wave drag is large — because a hovercraft’s cushion pressure, unlike a wing’s, is not limited by the air’s dynamic pressure. The ground-effect craft of the 1960s, Alekseyev’s ekranoplans the largest, were designed knowing their hulls set the take-off hump. The comparison drawn here, between a wing’s footprint and a cushion’s, is the density ratio read off Havelock’s formula.

Still open: a sea that is already moving #

The wing here skims a calm sea, and no sea is calm. A ground-effect craft over waves of a few metres sees its height to the surface change as it flies, and the ground effect with it, while the water beneath is moving with the waves’ orbital velocities. The next calculation puts the wing over a regular swell of stated height and length, follows the image system’s strength as the gap changes, and asks how much the craft’s lift fluctuates and at what encounter frequency — the quantity that actually limits ground-effect flight over the sea, against which the wing’s own waves, computed here, are invisible.