Water climbs a falling wedge, and the load comes from the climbing

The pressure that depends on the past made the time derivative of the velocity potential into a pressure one can feel: the term in the unsteady Bernoulli equation that steady flow drops, which makes a blow’s pressure impulse and a pipe’s start-up take seconds. The borrowed mass the boundary decides then gave the oldest practical use of it. A seaplane’s float or a flat-bottomed hull striking water sets the water beneath it moving; the water’s added mass is half what the same plate would carry submerged, because a free surface is half of an infinite fluid; and the impact force is the rate at which that added mass is being given momentum. That is von Kármán’s account of 1929, and the essay named Wagner’s refinement of 1932 and left it.

This essay computes it, because the refinement is not a correction. It changes the wetted width by a factor that depends only on the shape of the bottom, and it changes the load by that factor’s square.

Where the hull is wet #

A body with its keel at the origin and its bottom at height $f(x)$ above the keel enters still water at speed $V$, so after a time $t$ its keel is $h = Vt$ below the original surface. While the penetration is small compared with the body, the water sees a flat plate of half-width $c$ pushed into it, with an added mass per unit length of $\tfrac{\pi}{2}\rho c^2$. The question the whole calculation turns on is what $c$ is.

Von Kármán’s answer is the obvious one: the bottom is wet where it is below the undisturbed water level, so $f(c) = h$. For a wedge of deadrise $\beta$ — the angle each side of the bottom makes with the horizontal — that gives $c = h/\tan\beta$. Wagner noticed that the water the plate pushes down has to go somewhere, and it goes sideways and up. Outside the plate the free surface rises, with a vertical velocity that grows without bound at the plate’s edge, and it rises to meet the bottom before the bottom has come down to it.

Why pushing down makes the water go up #

The rise is mass conservation, and it is worth seeing without the integral. The flow under a flat plate pushed into a half-space at speed $V$ is the flow of a plate moving broadside through an infinite fluid, cut in half along the plate’s line. Under the plate the water moves down with it. Outside the plate, on the free surface, the same flow moves the water up, with a vertical velocity $V\big(|x|/\sqrt{x^2 - c^2} - 1\big)$ that is large near the edge and dies away far from it. The volume the plate pushes down per unit time, $2cV$, comes back up through the free surface outside — mass has nowhere else to go — and because the upward velocity is concentrated near the edge, so is the rise.

The surface at a given point has been rising since the plate first started pushing, and the edge was nearer to it at later times, so its rise is a weighted history of where the edge has been. That is what makes the condition an integral over the body’s shape rather than a local statement, and it is the same structure as the pressure that depends on the past: the state now is an accumulation of a flow that has been changing. A quasi-steady estimate, which takes the surface as it was, misses all of it — the general warning of slow enough to be steady in its most expensive form.

How far ahead the water gets #

Wagner’s condition makes that precise. The rise of the surface at a point outside the plate is the time integral of the plate flow’s vertical velocity there, which depends on how the edge got to where it is; requiring the risen surface to meet the body exactly at the edge, at every instant, gives one equation for $c(h)$:

$$\int_0^{\pi/2} f(c\sin\theta)\,d\theta = \frac{\pi}{2}\,h.$$

For a wedge, $f = x\tan\beta$, the integral is $c\tan\beta$ and the wetted half-width is $c = \tfrac{\pi}{2}h/\tan\beta$: a factor $\pi/2 = 1.571$ wider than von Kármán’s, at every depth. For a circular bottom of radius $R$, locally $f \approx x^2/2R$, the integral is $\pi c^2/8R$, and $c = 2\sqrt{Rh}$ against the geometric $\sqrt{2Rh}$: a factor $\sqrt{2}$. The figure plots both bodies both ways. The water is ahead of the drawing by more than half as much again for a wedge, and by four-tenths for a round bottom.

The surface, risen #

The integral condition is compact enough to be suspicious, so the figure computes the thing it stands for. The free surface outside the plate rises at $V\big(|x|/\sqrt{x^2 - c^2} - 1\big)$, and integrating that in time at each point — with the edge moving outwards as Wagner’s condition says — gives the surface’s shape at any instant. The integrand has an inverse-square-root singularity as the edge approaches a point, which a change of time variable removes. The water outside the wedge has risen everywhere, and steeply near the edge: it meets the bottom at 1.571 times the half-width at which the bottom crosses the undisturbed level, where the bottom stands 0.571 penetrations above that level. The marched surface and the body agree at the wetted edge to seven parts in ten thousand of the penetration, which is the condition checked without using it.

Between the risen water and the body, along the bottom near the edge, the water does not simply stop. It turns and runs up the body as a thin sheet — the spray every hull throws at impact — leaving from a narrow root at the wetted edge. The outer solution cannot describe that root; it is a separate, local flow, and it is where the largest pressure is.

The load, twice multiplied #

The force per unit length is the rate of change of the added mass’s momentum. At constant speed,

$$F = \frac{d}{dt}\Big(\tfrac{\pi}{2}\rho c^2\,V\Big) = \pi\rho V\,c\,\dot c.$$

Both $c$ and $\dot c$ carry the pile-up factor, so the force at a given depth is the factor squared times von Kármán’s: $\pi^2/4 = 2.467$ for a wedge, at every depth, as the figure’s ratio shows. A 10° wedge entering at a metre a second meets 5.05 kilonewtons a metre on von Kármán’s estimate at five centimetres’ penetration and 12.5 on Wagner’s.

This is the force of getting going in a form where the mass being set moving grows with time, and it is worth seeing why that makes the force larger than a fixed added mass would. The momentum is $m_a(t)V$, and with $V$ fixed the force is entirely $\dot m_a V$ — the rate at which new water is being recruited. The faster the wetted edge runs out, the faster water is recruited, and Wagner’s edge runs out $\pi/2$ times faster than the geometric one. A slam is not a body accelerating water; it is a body acquiring water. That is also why the slam is over so quickly: once the chines are wet the recruiting stops, the added mass stops growing, and the force falls to whatever the deceleration of the water already moving requires, which at constant speed is nothing.

Where the pressure is #

The outer solution’s pressure on the bottom is

$$p = \frac{\rho V \dot c\,c}{\sqrt{c^2 - x^2}} - \frac{\tfrac12\rho V^2 x^2}{c^2 - x^2}.$$

The first term is the unsteady one — the time derivative of the potential, the pressure that depends on the past — and it integrates exactly to the momentum force above. The second is the steady Bernoulli term of the flow along the bottom. At the keel the pressure is $\rho V\dot c$, 17.8 times the dynamic pressure of the entry for a 10° wedge; it rises towards the edge, and at the edge both terms are singular, which is the outer solution saying it has failed.

What replaces it there is the jet root, a small region in which the flow turns from moving outwards along the bottom with the edge’s speed to running up the body as a sheet. In a frame moving with the edge that flow is steady, the edge’s speed $\dot c$ is the stream’s, and the stagnation pressure at its root is $\tfrac12\rho\dot c^2$: 79.4 times the dynamic pressure of the entry at 10° of deadrise. The peak is not a force so much as a moving hammer, a strip a fraction of a per cent of the width across, travelling outwards at 8.9 metres a second for a one-metre-a-second entry, and it is what cracks hull plating and what a pressure gauge flush with a bottom records as a sharp spike as the edge passes over it.

A flat bottom is a hammer #

Because the edge runs out at $\dot c = \tfrac{\pi}{2}V\cot\beta$, the peak is $\tfrac{\pi^2}{4}\cot^2\beta$ times the dynamic pressure. At 10° it is 79.4; at 20°, 18.6; at 30°, 7.4. Halving the deadrise roughly quadruples the peak. As the bottom flattens, $\cot\beta$ and with it the theory’s peak grow without bound, and the theory has reached the end of its validity in a definite way: a flat plate meeting a flat surface would recruit the whole wetted area at once, at infinite edge speed, which incompressible water cannot do. What happens instead is that the water’s compressibility enters — the edge cannot outrun the speed of sound, and the peak becomes the acoustic pressure $\rho c_w V$, the same pressure a closing valve meets — and that a layer of air trapped under the flat bottom cushions the impact. Planing hulls are given deadrise for exactly this reason, and fifteen to twenty-five degrees at the transom is the usual compromise between the slam it buys off and the planing lift it costs.

A bow in a head sea #

Put a ship’s bow section on it: a deadrise of fifteen degrees, re-entering the water at five metres a second after the bow has lifted clear of a wave. The wetted edge runs out at $\tfrac{\pi}{2}V\cot 15° = 29.3$ metres a second, and the jet-root pressure is $\tfrac12\rho\dot c^2$ — 440 kilopascals in seawater, four atmospheres, sweeping out across the flare in a few hundredths of a second. When the wetted half-width has reached half a metre the force is $\pi\rho Vc\dot c$, 236 kilonewtons per metre of hull. At that instant the geometric estimate would have given 96 kilonewtons and a peak of 179 kilopascals: two and a half times too little on both counts.

That ratio is the reason Wagner’s paper is still cited in ship structural rules. A plate panel designed for the geometric load fails under the real one, and it fails locally, under the moving spike, rather than as a whole. The spike’s speed also matters to the structure: at 29 metres a second it crosses a 0.6-metre panel in twenty milliseconds, which is comparable to the natural period of such a panel, so the plate’s response is dynamic and can exceed what the peak pressure applied slowly would produce. The impulse the water delivers, the force integrated over the impact, is the quantity a hull’s frames feel; the peak is what its plating feels.

Sharper bottoms are wetted further ahead #

For a bottom shaped as $f = kx^n$ the condition gives a pile-up factor that depends on $n$ alone: $(\pi/2I_n)^{1/n}$, where $I_n$ is the integral of $\sin^n\theta$ over a quarter turn. It is $\pi/2$ for a wedge and $\sqrt 2$ for a round bottom, 1.278 for a bottom flatter still at the keel, $n = 4$, and it falls towards one as $n$ grows. Below a wedge, for a bottom that is sharper than straight at the keel, it rises: 1.72 at $n = \tfrac12$.

The trend has a reading. A sharp keel enters with little width and a steep surface; the water it displaces has a long way to rise before meeting the steep sides, and it rises a great deal relative to the width. A bottom that is flat near the keel and curls up sharply at the chines is wetted almost geometrically, until the chines arrive — after which the flow separates from them and a different calculation is needed. The load the factor feeds is the square of it times the geometric load, so the shape of the bottom changes the slam by between nothing and a factor of three, before any change of deadrise.

How the entry was checked #

Wagner’s condition is solved numerically for any body, by quadrature in the angle and bisection on the width, and it returns the closed forms: 1.57079632 for a wedge against $\pi/2$, and 1.41421356 for a parabola against $\sqrt 2$. The free surface, marched in time from the plate’s kinematics with the singularity removed by the change of variable, meets the body at the wetted edge to $7\cdot10^{-4}$ of the penetration — the integral condition confirmed by computing the thing it encodes. And the first pressure term, integrated across the bottom in the angle variable that removes its edge singularity, equals $\pi\rho Vc\dot c$ to $3\cdot10^{-14}$: the pressure and the momentum account of the force are the same force.

The convention: deadrise, penetration, and the dynamic pressure of the entry #

Deadrise is the angle each half of the bottom makes with the horizontal, so a flat bottom has zero. Penetration is the keel’s depth below the undisturbed water level. Pressures are given as multiples of $\tfrac12\rho V^2$ with $V$ the entry speed, so that “79 dynamic pressures” at a metre a second is 40 kilopascals and at ten metres a second four megapascals; the theory is linear in that scaling. The body is two-dimensional, long compared with its width, and enters at constant speed.

What the picture cannot show #

The theory is the small-penetration, small-deadrise limit, and it overpredicts for steeper wedges: the exact similarity solution for wedge entry, computed by Dobrovol’skaya in 1969 and later numerically, gives a wetted width and force below Wagner’s for deadrise angles past about twenty degrees, because the flat-plate model’s plate is too flat. It neglects gravity, which matters only once the entry has slowed to a Froude number of order one; air, which cushions flat impacts; and compressibility, which caps the peak at the acoustic pressure. A real hull also decelerates during the slam — its own inertia against the force — so the constant-speed results here are the upper limit for a given impact speed. And the jet-root peak is a property of the inner solution, which the figures place by value but do not resolve. Everything is two-dimensional, too: a bow section is a slice of a three-dimensional body, and a body that can push water fore and aft as well as sideways raises the surface less, so its pile-up factor is smaller than the slice’s — the reason slamming estimates for short, bluff bows are made with three-dimensional corrections.

Who found it, and when #

Von Kármán’s 1929 paper on seaplane floats gave the added-mass account with the geometric width. Herbert Wagner’s 1932 paper added the rise of the free surface, derived the π/2 for a wedge, and the jet-root pressure; it remains the basis of slamming estimates for ship bows, seaplane hulls and offshore structures. The matched-asymptotic treatment that joins the outer solution to the jet root was given by Armand and Cointe and by Howison, Ockendon and Wilson in the 1980s and early 1990s, and Zhao and Faltinsen’s computations of 1993 established where Wagner’s estimate stops being conservative.

Still open: a hull that slows down #

Every result here holds the entry speed fixed, and a real hull’s speed is not fixed: the slamming force decelerates it, and the deceleration depends on the hull’s mass per unit length against the water it is recruiting. The next calculation solves the entry with the body’s own equation of motion, $m\dot V = -\tfrac{d}{dt}(m_a V)$ plus gravity, so that the recruitment and the slowing are computed together, and asks at what ratio of the hull’s mass to the added mass of its full width the peak force falls below the constant-speed value by half — which would say when a light hull’s slam is limited by its own inertia rather than by its shape.