A ship in a canal really is a venturi, and it chokes below the wave speed

Not half a venturi took apart one of the most repeated explanations in the subject. The air over a wing is said to be squeezed as in a venturi, so it speeds up and its pressure falls. Every step is true of a stream tube, and the whole is still not an explanation, because nothing forces the air into the narrow channel: it is a result of the circulation rather than a cause, and the same constriction appears on the wing’s underside and chords above it. The venturi needs walls.

It is fair to ask where the story does apply, then, and the most striking case has been in front of everyone for two centuries. A ship in a canal has walls. Water moving past it, in the ship’s frame, has to go through a section reduced by the ship’s own area, and there is nowhere else for it to go. The continuity equation is doing what the venturi diagram says it does. And unlike a venturi on a bench, the fluid here has a free surface, which can fall. That brings in something a pipe of water could never have.

So this is the misconception’s mirror image. The story fails on a wing and it is correct here, and being correct here it goes on to predict things that nobody would have reached by looking at a wing.

Two equations and a limit #

Work in the ship’s frame, where the canal’s water comes towards it at speed $U$ through a section $A=Bh$, of width $B$ and depth $h$. The ship blocks an area $A_s$, a fraction $m$ of $A$. Past the ship the speed is $V$, and the water’s surface has fallen by $z$ over the whole width. Water is incompressible, and the surface is at atmospheric pressure, so

$$U\,Bh = V\,(Bh - A_s - Bz), \qquad gz = \tfrac12(V^2-U^2).$$

With $F=U/\sqrt{gh}$, the Froude number on the depth, and $\zeta=z/h$, these become one equation in one unknown. It is what Schijf wrote in 1949, and it contains no empirical constant. Nothing here is more elaborate than a first course in fluid mechanics.

The result at small speeds is what a venturi says. The ship’s area is taken out of the channel, the return flow is about $mU/(1-m)$, and the surface falls by Bernoulli’s amount. A ship sinks in step with the surface around it, and that is squat: the extra draught of a ship under way in shallow water, which has put a good many ships on the bottom of channels whose charted depth exceeded their static draught by a metre. It is quadratic in speed, so it’s easy to ignore until it isn’t.

The interest is at the other end, and it comes from a feedback. As the surface falls, the section gets smaller and the speed must rise. As the speed rises, Bernoulli wants the surface to fall further. For a small $F$ the feedback is mild. At a certain $F$ it runs away, which means that there’s no surface position that satisfies both equations. It’s not that the numerical root is hard to find. The two curves simply do not meet.

And the limit has a physical meaning, which is the whole point. At that value, the stream beside the ship is moving at exactly its own wave speed $\sqrt{gd}$ for the local depth $d$. A disturbance can no longer travel upstream past the ship. The section has choked. It’s the same event as a venturi’s throat reaching Mach 1, in which the downstream pressure stops being able to tell the upstream flow anything, and a choked throat that buys time has this in it too. Water in a canal is effectively incompressible as far as its density goes. It doesn’t matter. A free surface is its own kind of compressibility, in which the depth does what density does, and $\sqrt{gh}$ does what the speed of sound does.

The closed form, which is a hump #

Solving the two equations at the limit, where the feedback’s derivative vanishes, gives

$$F_L = \left(2\sin\frac{\arcsin(1-m)}{3}\right)^{3/2}, \qquad F_U = \left(2\sin\frac{\pi-\arcsin(1-m)}{3}\right)^{3/2}.$$

The first is the lower limit, and a tenth’s blockage gives 0.623. The second, the upper limit, gives 1.39. Between those two there is no steady flow past that ship at all.

That expression for $F_L$ doesn’t look like it should have a simple origin, and yet the source is transparent. Taking a fixed area $A_s$ out of a section whose surface can drop is exactly the algebra of raising the canal’s bed by a fixed height $mh$. The equations are identical, since depth over the hump is $h-mh-z$, which is the same as the section past the ship per unit width. The hump’s problem is what the depth that costs least solves with its specific energy: a flow choking over a hump of height $\Delta$ has

$$\frac{\Delta}{h}=1+\frac{F^2}{2}-\frac32F^{2/3}.$$

A trigonometric formula from a cubic’s roots then gives the expression above. So a ship is a moving hump, and the two cases differ only in which frame the observer stands. A rock in a river at a Froude number of 0.6, with a height of a tenth of the depth, is just on the point of making the river go critical over it, which is what a smooth tongue of water does before it tumbles into a hydraulic jump just downstream. A ship of a tenth’s blockage at a Froude number of 0.62 is on the point of the same thing, and it shows the same drop in the surface alongside.

Why a small ship matters so much #

The most striking thing about these curves is near the origin. For a small blockage the hump’s formula expands about $F=1$ to give $m\approx\tfrac23(1-F)^2$, so

$$F_L\approx 1-\sqrt{\tfrac32 m},$$

and the distance of the limit from the wave speed goes as the square root of the blockage. So a blockage of one part in a hundred doesn’t change the critical speed by anything like one part in a hundred. It reduces it by twelve per cent, to 0.878, and a blockage of 2 per cent reduces it to 0.827 by this formula and 0.829 by the exact one. The wave speed is a flat maximum of the throughput, and the relation between area and speed is flat there, so the smallest obstacle has a very large effect on the speed at which it chokes.

The compressible gas has the identical shape. For $\gamma=1.4$, a one-per-cent area reduction has $A/A^*=1/0.99$, and that chokes a subsonic tunnel at Mach 0.89 rather than 1. That is why a transonic wind tunnel needs a model of about half a per cent of its section, or walls with holes in them, as the number that does not depend on the tunnel comes at from another angle. A tunnel run at Mach 0.9 with a one-per-cent model is choked at its own model, and no increase in the fan’s speed can raise the Mach number upstream. The same limit also explains why so many observations of a high-speed ship in a shallow channel report the same sudden sharp increase of resistance near but well below the wave speed: the arithmetic puts it there, for a ship of modest size.

What happens inside the band #

The steady equations have nothing to say about a ship inside that band, and yet a ship with enough power can go there. What happens has been computed and seen in tanks. Water piles up ahead of the ship, since it can’t get past fast enough, and a long, low wave of elevation forms ahead of the bow. This steepens, and a train of solitary waves is released upstream, with a new one forming every so often. Behind the ship, a depression extends, and the ship is effectively going uphill on its own wave. Constantine laid out the one-dimensional picture in 1960, and Ertekin, Webster and Wehausen computed the soliton trains in 1986. The drag rises steeply in this band, and it’s where a canal’s own speed limit effectively lies.

Above the band, the algebra picks the other branch, and Bernoulli changes sides.

On that branch the flow past the ship is supercritical, and in supercritical flow a contraction makes the water slower and deeper, the opposite of the usual venturi. It’s the same reversal compressible flow makes at Mach 1, and it has been noted for centuries by people watching water go over a weir. So the ship rises. Its squat has become a lift, and its wave no longer travels ahead.

That reversal has a famous discovery. Around 1830 a horse pulling a boat on the Glasgow and Paisley canal took fright and bolted, and the man in charge of the boat, William Houston, is said to have noticed that the boat rose and went faster with apparently less effort. It became a working service, with light passenger boats pulled at eight or nine miles an hour by horses galloping along the towpath, in a canal so shallow that this was past the critical speed. John Scott Russell’s first report on waves was an attempt to understand this: on the Union Canal nearby, in 1834, he saw the wave leave a boat that had suddenly stopped, and followed it on horseback. The solitary wave is a child of this calculation.

A real ship in a real canal #

The numbers scale directly. The largest ships in sea-level canals have sections of the order of a thousand square metres, in canals of perhaps six thousand square metres and about 24 metres deep. So the blockage is about a sixth.

So a large ship at the Suez Canal’s roughly eight knots has a fall in the surface of half a metre and a return current of nearly a metre a second. That is not dramatic. The curve is steep above eleven knots, and at fifteen knots the fall is seven times as large, 3.05 metres. The canal’s speed limit therefore has a clean mechanical reason. A real limit is set well below choking, because at half the critical speed the wash and the erosion of the banks are already bad, and because a real ship is off-centre.

That last point is the one this model has no way of expressing, since it’s a single number across the section. A ship nearer to one bank has a narrower, faster channel on that side, a lower pressure there, and is pulled towards the bank: the bank effect, a venturi with a preferred direction. That force is also what makes handling in a canal hard and can make a ship yaw away from a nearby bank at its bow while its stern is sucked in. It is outside the model.

The bank’s pull has a number’s worth of consequence: it adds to the mean drawdown on the near side, so that a real limit on speed comes sooner than the one-dimensional one.

Open water has the same structure #

As the blockage falls to zero, the band shrinks towards $F=1$, and that is why a shallow sea has a critical speed as well. A ship in open but shallow water, with no walls at all, can spread the flow sideways, so no section is forced to choke, and the problem is two-dimensional. Tuck solved it in 1966 with a slender-ship theory, where the sinkage goes as

$$\frac{F^2}{\sqrt{1-F^2}}$$

times a volume factor. Its singularity at $F=1$ is the open-water form of the band, and its sign reversal above $F=1$ is the same one that the canal’s algebra gave. A ship in open shallow water sinks going subcritically and lifts going supercritically, for exactly the reason a canal does. The fact that it is $\sqrt{1-F^2}$, the Prandtl–Glauert factor, says the same as the choking argument: the linear equation of a shallow sea is formally the equation of a subsonic gas, with $F$ for Mach number.

That also marks what this has to do with the Bernoulli that does and doesn’t apply. Everything in this essay is along one streamline at the surface of a steady flow in the ship’s frame, with no viscosity. Each of those conditions is met well enough. It is a correct use of Bernoulli’s equation, of the kind the wing’s story tried to be. And what makes it correct is what the wing’s air doesn’t have: a wall.

What the picture cannot show #

Anything off the centreline. The model is one-dimensional, so a single number stands for the whole section beside the ship. Real return flow is faster beside the ship than near the bank. The result is a better predictor of the choking speed than of a local drawdown near the hull, and a long way from a good one for the bow and stern, where the water must turn.

Short ships. The derivation needs a ship long compared with the channel’s width, so that the flow is fully rearranged along it. A short ship in a wide channel lets flow pass beneath and around in three dimensions, and the limit moves. It is a matter of degree, and it means the band is sharply defined only for a long ship in a narrow channel.

Friction, the bed and the propeller. There is no viscosity in the two equations, so the flow’s boundary layers, the ship’s wake and the propeller’s race are all missing. Canal-bank erosion comes from the return flow near the bank and from the propeller wash. Neither is here.

Trim. A ship doesn’t sink evenly. A real squat combines sinkage with a bow-down or stern-down trim, and whichever end goes down is usually what touches, which is where a real grounding comes from.

The convention #

The Froude number here is on the depth, $U/\sqrt{gh}$, rather than on the ship’s length, which is the usual form for a hull in deep water. The blockage $m$ is the ship’s midship area over the canal’s section. Surface fall is positive downward. The canal is treated as rectangular, with the section taken as its width times its depth, though a real canal has sloping banks, which behave as an extra width near the surface and change the numbers by a modest amount.

How it was checked #

The two closed forms were compared with a direct scan for the last Froude number at which a real root of the two equations exists, at five blockages. They agree to the three figures that scan was run to. The local Froude number at the limit comes out at one, and it is a check with content, because that equality was not assumed anywhere. The identification with a hump’s specific energy returned the blockage at each limit to all digits. The model refuses a blockage at or beyond one, and a stationary ship.

Who found it, and when #

Scott Russell’s account of the boats and the waves is from 1837 and 1844. Schijf’s formula dates from 1949, in a paper to an international navigation congress. Constantine’s analysis of transcritical ships in restricted water is from 1960, and Tuck’s slender-ship theory of shallow-water sinkage from 1966. The upstream solitons of a transcritical ship were first computed and observed in the 1980s, by Ertekin, Webster and Wehausen and by Wu.

Still open: how often the soliton train leaves #

The band’s interior was described and not computed. The next calculation would march a shallow-water model, with Boussinesq dispersion, past a moving hump of the stated blockage at each Froude number in the band, measure the period at which solitary waves leave upstream and their amplitude, and ask whether a single rule gives the period across the band: whether it scales with the time water takes to pile up the deficit that choking leaves, $mh$ per unit width at the difference between the ship’s speed and the limit, so that a one-dimensional mass balance alone gives the soliton’s frequency.