A light hull slams softly, and a round one does not

Water climbs a falling wedge, and the load comes from the climbing computed Wagner’s water entry: a V-shaped bottom pushed into still water wets a width π/2 times the geometric one, because the water it displaces rises to meet it, and the slamming force, the rate at which the added mass of that wetted width gains momentum, is $\pi^2/4$ times von Kármán’s earlier estimate. Every number there held the entry speed fixed. It ended on the obvious correction: the force that slams a hull also slows it, and how much depends on the hull’s mass against the mass of water it is setting moving.

The correction turns out not to be a correction to every hull. For a V-shaped bottom it changes the answer qualitatively — the force stops growing and peaks at a definite point fixed by the mass. For a round bottom it changes nothing at the moment that matters.

One momentum, shared #

Over the few milliseconds of a slam, gravity and buoyancy are small beside the forces involved, and the hull and the water it has set moving form a closed system in the vertical. The water’s share of the motion is its added mass, which for a wetted half-width $c$ is $m_a = (\pi/2)\rho c^2$ per unit length — the mass a body has to borrow to move through a fluid, here growing as the body goes deeper. With $m$ the hull’s mass per unit length and $V_0$ its speed at contact,

$$(m + m_a)\,V = m\,V_0, \qquad F = \frac{m^3\,V_0^2\,m_a'(h)}{(m + m_a)^3},$$

where $h$ is the penetration and the prime is its derivative. The force is the constant-speed force $V_0^2\,m_a'(h)$ multiplied by $(m/(m + m_a))^3$: one factor because the hull is slower, two more because it recruits water more slowly and each piece of it at a lower speed. An infinitely heavy hull is the constant-speed entry exactly, and the stepped entry returns it to $5\times10^{-9}$.

A wedge’s slam has a peak #

For a wedge of deadrise $\beta$, Wagner’s wetted width grows in proportion to the penetration, so the added mass grows as its square, $m_a = Kh^2$ with $K = (\pi^3/8)\,\rho/\tan^2\beta$. At constant speed the force $2KV_0^2h$ grows without limit until the chine — the edge of the bottom — wets, and then the wetted width stops growing and the load collapses. That is the shape the earlier essay drew, and for the 10° wedge here it reaches 700 kilonewtons per metre at the chine.

With the hull’s own mass in, the force is proportional to $h/(m + Kh^2)^3$, which rises, peaks and falls. It peaks where $Kh^2 = m/5$: when the water recruited weighs a fifth of the hull, at every deadrise. The momentum with no value is the warning that a fluid’s momentum can be ambiguous; here it is not, because the water’s share is its added mass, a definite number. The peak is

$$F_{\max} = \frac{250}{216}\,\frac{1}{\sqrt5}\,V_0^2\sqrt{Km},$$

and grows only as the square root of the hull’s mass. The stepped entry finds the peak where the closed form puts it, the recruited share to two parts in a hundred thousand and the force to $2\times10^{-9}$. For hulls of 0.3, 1 and 3.73 times the added mass of their full width the peaks are 99, 181 and 350 kilonewtons per metre, against the constant-speed 700.

How much the mass takes off #

The natural measure of a hull’s mass is the added mass of its whole bottom, $(\pi/2)\rho b^2$ for a half-beam $b$ — the water it would carry if the whole width were wetted. Call the ratio $\mu$. A hull lighter than five times that water peaks before its chine wets, at a force $0.2588\sqrt\mu$ times the constant-speed chine force; a heavier hull peaks at the chine, at $(\mu/(\mu+1))^3$ of it. The two expressions meet at five. The peak is half the constant-speed value at $\mu = 3.73$, and a tenth at 0.15.

That is the number the earlier essay asked for, and it is larger than intuition suggests: a hull has to weigh nearly four times the water under its whole width before its own inertia stops limiting the slam. A planing dinghy, whose hull is a few tens of kilograms per metre against more than a tonne of water under a metre’s beam, is in the regime where its slam is set almost entirely by its lightness; a laden ship’s bow section, many times heavier than the water under it, slams nearly as the constant-speed theory says.

Slowing is what softens it #

The mechanism is visible in the speeds. By the time its chine wets, a hull of $\mu$ has given $1/(\mu + 1)$ of its momentum to the water and kept $\mu/(\mu+1)$ of its speed: the lightest here is down to 23 per cent of its entry speed, a hull as heavy as the water under it to half, the heaviest to 79 per cent. The force is the hull’s mass times its deceleration, and a hull that decelerates hard early is one whose later recruitment of water happens slowly, at a speed that has already fallen — and the force goes as the speed squared.

This is also why the peak sits where it does. The constant-speed force is growing because the wetted width is accelerating outwards; the slowing is growing because the recruited mass is accumulating. At first the first effect wins, and once the recruited water is a fifth of the hull, the second does. The fifth is pure arithmetic — the cube in the momentum-sharing factor against the linear growth of the wedge’s added-mass derivative — and contains no property of the water, the deadrise or the speed.

A slam is an inelastic collision #

Sharing momentum is not the same as sharing energy, and the difference says where part of a slam goes. Before contact the hull carries $\tfrac12 m V_0^2$. After recruiting an added mass $m_a$ the hull and the water moving with it carry $\tfrac12(m + m_a)V^2 = \tfrac12 m V_0^2 \cdot m/(m + m_a)$, which is less: a fraction $m_a/(m + m_a)$ of the hull’s energy is missing. At chine wetting that is $1/(1 + \mu)$ — half the hull’s energy for a hull as heavy as the water under its width, three-quarters for the lightest here.

The missing energy is not lost to viscosity, which the theory does not have. It leaves in the jets. Wagner’s outer solution has the water climbing the bottom and, at the edge of the wetted region, a thin sheet thrown up and out along the hull at twice the speed the contact line moves — the spray that every photograph of a slamming hull shows. The kinetic energy that sheet carries away is the inelastic collision’s deficit, and it is why a slam is a collision rather than a bounce: the force of getting going is the added mass’s reaction to being accelerated, and in a slam the acceleration is permanent and partly wasted.

A dinghy, a rigid inflatable and a lifeboat #

The mass ratio makes the regime of a real hull easy to place, and most small craft are far below one. Take as stated examples a planing dinghy of 30 kilograms per metre of length on a half-beam of 0.6 metres; a rigid inflatable of 150 kilograms per metre on a metre; and the section of a lifeboat of 1.5 tonnes per metre on 1.5 metres. Their mass ratios are 0.053, 0.095 and 0.42, and if their bottoms were 10° wedges their peak slams would be 6.0, 8.0 and 17 per cent of what the constant-speed theory puts at their chines.

That is a large factor to leave out of a design load, and it cuts both ways. A designer who uses the constant-speed load for a light boat over-designs its bottom several times over; one who uses the free-hull load must be sure the boat cannot meet the water with something heavier behind it — a crew, an engine, a second hull — that changes the effective mass, or at a deadrise so shallow that the peak moves to contact.

Why gravity and buoyancy could be left out #

Over the 10.7 milliseconds the wedge takes to reach its peak, gravity adds about ten centimetres a second to the entry speed of five metres a second, two per cent. Buoyancy, the weight of the water displaced, is at the peak a few hundred newtons per metre against a slamming force of hundreds of kilonewtons. Both matter after the slam, when the hull settles, and neither in it. The pressure that does the work is the unsteady one, $\rho\,\partial\phi/\partial t$ — the pressure that depends on the past — which for a few milliseconds is three orders of magnitude larger than the hydrostatic pressure at the same depth.

The shape decides whether mass matters #

Generalise the bottom to $f = k|x|^n$ — a wedge is $n = 1$, a parabola $n = 2$ — and Wagner’s width gives $m_a \propto h^{2/n}$, so the constant-speed force goes as $h^{2/n - 1}$. The free hull’s peak then comes where the recruited water is $(2 - n)/(4 + n)$ of the hull’s mass: a fifth for a wedge, a ninth for $n = 1.5$, and zero for a parabola. A parabola’s constant-speed force does not grow at all — Wagner’s added mass for it grows in proportion to the penetration, so its derivative is constant — and the largest force is at the first instant of contact, before any water has been recruited and before the hull has slowed.

Every round bottom is a parabola near its keel, so every round bottom is in this class. The stepped entry for four powers between 1 and 1.75 lands on $(2-n)/(4+n)$ to $2\times10^{-4}$.

Two hulls of the same mass #

Put the two shapes side by side at the same mass, $\mu = 1$, and the time histories show the difference. The round bottom hits 314 kilonewtons per metre at contact — $2\pi\rho R V_0^2$, Wagner’s circle — and the force falls from there as the hull slows. The wedge builds over 10.7 milliseconds to 181 kilonewtons per metre, 25.9 per cent of what a constant-speed calculation would put at its chine. The round bottom’s peak is the same 314 at any mass, and the wedge’s scales with the square root of it. A lighter round hull is not a softer one.

That is the opposite of the ranking a constant-speed comparison gives. At constant speed the 10° wedge’s chine force, 700 kilonewtons per metre, is more than twice the round bottom’s; freed to slow, the wedge’s peak is little more than half the round bottom’s, and its pressure is spread over a growing width rather than concentrated at the instant of contact.

A drop test measures the free hull #

The distinction matters for reading experiments. Slamming loads are measured two ways: by dropping a wedge or a section into a tank from a height, or by driving it into the water at a controlled speed with a rig stiff and heavy enough to hold the speed. The second measures the constant-speed force; the first measures the free hull’s, and its peak depends on the dropped mass as computed here. Comparing a drop test with a constant-speed theory, or scaling a drop test of one mass to a hull of another, mixes the two. The mass ratio $\mu$ is the number that says how far apart they are: for a drop-test wedge of 300 kilograms per metre and a half-beam of 0.3 metres, $\mu$ is about two, and the measured peak should be 37 per cent of the constant-speed chine force rather than near it.

The impulse is the one quantity the two kinds of test agree on in a simple way. A hull that comes to the end of its slam with the water under its whole width moving with it has given up $mV_0/(1 + \mu)$ of momentum per metre whatever its shape — the shape decides how that impulse is spread in time, sharply for a round bottom, gradually for a wedge, but not how large it is. Everything about the start, except one vector made the same point for a body started impulsively in an unbounded fluid: the impulse is fixed by the end state, and the history is where the shape enters.

A heavy hull and a stepped wedge, held to their closed forms #

The entry is stepped in penetration with the closed-form Wagner width for a power-law bottom, $c = ((\pi/2)h/(kI_n))^{1/n}$ with $I_n$ the integral of $\sin^n\theta$ over a quarter-turn, which agrees with the earlier essay’s quadrature to $6\times10^{-9}$ for a wedge and $2\times10^{-5}$ for a circle at small penetration. The rate of change of the added mass’s momentum, differenced along the stepped entry, equals the force to $4\times10^{-7}$ at every step: the momentum really is shared. The peak force and depth match their closed forms, and the infinitely heavy hull returns the constant-speed entry.

The convention: per metre of hull, from contact #

Forces and masses are per metre of hull length — the two-dimensional entry of a section. Penetration is measured from first contact of the keel; the chine is where the bottom ends and Wagner’s width stops growing. The mass ratio is against the added mass of the full width, $(\pi/2)\rho b^2$. Speeds are vertical. Gravity, buoyancy and air are left out, which is right for the milliseconds of a slam and wrong for what follows it.

What the shared momentum leaves out #

The elastic hull. A real bottom panel deflects under the slam and its own mass and stiffness respond on the same millisecond scale, so the structure feels a dynamic amplification of the force that depends on its natural period against the slam’s duration — which the free-hull result changes, by lengthening the wedge’s rise time to the peak. The air. A flat or nearly flat bottom traps a cushion of air whose compression softens and delays the impact; the round bottom’s contact-instant peak is exactly where that cushion acts. Steep deadrise. Wagner’s theory is a small-deadrise approximation and over-predicts for wedges beyond about twenty degrees, as the earlier essay noted. Three dimensions and forward speed are absent: a ship’s section enters a wave while moving along it, and the water it recruits is not only beneath it. And the sea’s surface is taken flat and still. A real hull meets a wave whose own orbital velocity adds to or subtracts from the entry speed, and whose slope changes the effective deadrise — a wing skimming the same surface barely touches it because the air’s pressure is small beside the water’s inertia, but a hull meets that inertia head on.

The fifth, and who had it #

Von Kármán’s 1929 paper on seaplane floats, which the borrowed mass the boundary decides discusses, already conserved momentum between the float and the water it accelerated, and used it to estimate the maximum load on a float landing with a given weight; Wagner’s 1932 width replaced the geometric one in that balance. The ratio of hull mass to the added mass of the wetted region has been the organising parameter of seaplane landing loads since, and the peak at a fixed share of the hull’s mass is in that literature for wedges. The general share $(2-n)/(4+n)$, and the observation that it vanishes for every round bottom, follow from the same balance in a line.

Still open: a bottom that bends #

The hull here is rigid, so the only time scale besides the slam’s is none. A bottom panel is a plate with a natural period of a few milliseconds, the same order as a wedge’s rise time to its peak. The next calculation gives the bottom an elastic panel between stiff frames, couples its deflection to the Wagner entry through the pressure that depends on the past, and asks how the panel’s peak stress compares with the rigid-hull force, whether a light hull’s slower rise time lets its panels respond quasi-statically where a heavy hull’s quick one makes them ring, and so whether the free hull’s softer slam is also a gentler one for its structure.