Why a ship comes back upright

The previous rung settled how deep a floating body sits and was completely silent about which way up it sits. Those are different questions, and the second one is not answered by making the first one more precise. A log floats at the same draft on its side and on its end; only one of those is what actually happens.

Everything on this page follows from the observation in that caption. When the hull tilts, the shape of the underwater part changes, so the centre of that shape moves — and it moves sideways, which is what makes a moment possible.

Two points and a line between them #

A floating body has exactly two forces on it. Its weight acts downward through its centre of gravity, $G$, which is a property of how the mass is arranged and does not move when the hull tilts. The upthrust acts upward through the centre of the displaced volume, $B$, which is a property of the shape underwater and moves whenever that shape changes.

Upright and symmetric, $B$ sits directly below $G$ and the two forces are collinear: no moment, and the body is in equilibrium. Tilt it, and the submerged shape becomes asymmetric — more of the low side is under water and less of the high side — so $B$ shifts toward the low side while $G$ stays put. Now the two lines of action are parallel and separated, which is a couple.

Which way that couple acts is the entire question. If $B$ has moved far enough to end up outboard of $G$, the couple pushes the hull back; if it has not, the couple carries it further over.

In the language the rest of this collection uses, an equilibrium at the bottom of a well returns when disturbed and one balanced on a hill does not — and nothing about the depth of the float distinguishes the two. What distinguishes them is the curvature of the landscape at the point where the body is sitting, which is a second derivative and not a position. A hull can float exactly as it was designed to and still be sitting on a hill.

The point that does not move #

There is a construction that turns this from a case-by-case question into a number. Extend the new line of action of the buoyancy upward until it crosses the hull’s original centreline. That crossing point is called the metacentre, $M$.

For small heels, $M$ is very nearly fixed — it stays in the same place whatever the angle, which is what makes it useful. And its height above $B$ has a closed form:

$$\overline{BM} = \frac{I}{V},$$

where $V$ is the displaced volume and $I$ is the second moment of area of the waterplane — the flat slice the water surface cuts through the hull — taken about the axis the hull rolls around.

That formula is the whole subject in one line, and the quantity in the numerator is worth dwelling on. It is not the hull’s mass distribution, not its cross-section, not its volume. It is the second moment of the shape at the waterline, which is the same quantity that decides how much a beam resists bending and which is a cousin of the moment of inertia that decides which body wins a rolling race. In all three cases a number that looks as though it should depend on the material turns out to depend only on how the area is arranged.

For a rectangular waterplane of beam $b$ and length $L$, $I = Lb^3/12$. The cube is the thing to notice. Widening a hull by ten per cent increases $I$ by thirty-three per cent, and widening it by a half more than triples it.

The condition, and its three terms #

Stability requires $M$ to be above $G$, and the distance between them is the metacentric height:

$$\overline{GM} = \overline{KB} + \overline{BM} - \overline{KG},$$

measured from the keel, $K$. Positive and the hull rights itself; negative and it does not.

Three terms, and they say three different things. $\overline{KB}$ is how deep the centre of the displaced volume sits, which is roughly half the draft. $\overline{BM}$ is $I/V$, decided by the waterplane. $\overline{KG}$ is where the mass is, which is the only one a designer controls by adding ballast.

The common explanation of stability — weight low down pulls the hull upright — is therefore one term out of three, and not the one with the cube in it. A hull can be made stable by widening it while adding no ballast whatever, which is what a catamaran does and why a raft is hard to capsize. The dependence is on an arrangement rather than on a strength, which is the same shape of result as an exponent that belongs to the geometry. And no quantity of ballast rescues a hull whose waterplane is too narrow, which is why a tall spar buoy needs an enormous amount of it and why a floating cylinder on its end is a difficult object.

What the waterplane is, and why it is the only shape that matters #

It is worth being explicit about which slice of the hull the stability depends on, because it is a narrower thing than most people assume and the narrowness is the useful part.

$I$ is the second moment of the region the water surface cuts out — a flat, two-dimensional shape, at one particular height, about the fore-and-aft axis. Nothing above the waterline enters it. Nothing below it enters it either, except through $V$. Two hulls with utterly different underwater shapes and the same waterplane and the same displacement have the same $\overline{BM}$.

That is why beam is the dominant lever. In $I = Lb^3/12$ the length appears once and the beam three times, so a metre of extra beam is worth far more than a metre of extra length. It is also why hull forms that concentrate area away from the centreline are so effective: a catamaran’s waterplane is two narrow strips a long way out, and the parallel-axis contribution — area times the square of the distance to the axis — dwarfs anything a single hull of the same total area can manage. A catamaran is stable for the same reason a wide-flanged beam is stiff, and the two calculations are the same calculation.

The quantity the previous rung computed is worth drawing again for contrast. The density ratio fixes the draft and therefore the displaced volume, and it says nothing whatever about the waterplane above it. Two vessels can share that picture exactly — same displacement, same draft, same freeboard — and differ completely in whether they stay upright, because the second moment of the waterplane is a quantity the first calculation never asks for.

There is a corollary that catches designers of small craft. Because $\overline{BM} = I/V$, loading a vessel more deeply increases $V$ and reduces $\overline{BM}$ — while usually raising $\overline{KG}$ as well, if the cargo goes anywhere but the bilges. Both terms move the wrong way at once, which is why deck cargo is dangerous out of proportion to its weight and why loading rules are about placement rather than tonnage.

Reading the arm off the geometry #

The couple’s size is the weight times the horizontal separation of the two lines of action. That separation is called the righting arm, $GZ$, and for small angles

$$GZ = \overline{GM} \sin\varphi.$$

The figures on this page do not use that formula to draw. They solve for the waterline at each angle by requiring the submerged area to be unchanged, take the centroid of the resulting polygon, and measure the arm off it. That matters because the small-angle result then becomes something the picture can be checked against rather than something it was built from.

For a hull with vertical sides the exact result is

$$GZ = \sin\varphi \left( \overline{GM} + \tfrac12 \overline{BM} \tan^2\varphi \right),$$

and the drawn curve reproduces it without having been given it. The departure from the straight line is therefore a fact about the geometry rather than a numerical artefact, and it is a comfortable direction to be wrong in.

That the two routes agree is the strongest check available on the figure. The closed form is a textbook result derived by expanding the shift of the centre of buoyancy in powers of the heel; the drawing knows nothing of that derivation and finds the waterline by bisection, the centroid by the shoelace formula, and the arm by projection. Two independent routes to one curve is the arrangement this collection asks of every figure that claims a number, and it is the reason the departure from the straight line can be quoted as a physical effect rather than as a wobble in the plotting.

Why more is not better #

A large metacentric height gives a large righting moment, which sounds unambiguously good and is not. The hull’s roll is an oscillation, and its period depends on the stiffness of the restoring couple in exactly the way a pendulum’s period depends on gravity:

$$T \approx 2\pi \sqrt{\frac{k^2}{g \, \overline{GM}}},$$

with $k$ the radius of gyration. A larger $\overline{GM}$ means a shorter period — a snappier, more violent roll. A vessel with too much stability rolls hard and fast, throws its cargo about, and is exhausting and dangerous to be aboard. Warships are built stiff on purpose and are notoriously uncomfortable; passenger ships are deliberately made tender, trading righting moment for a roll slow enough to be tolerable.

A rolling hull performs an ordinary exchange: energy moves between the potential stored in the displaced water and the kinetic energy of the roll, and the stiffness of the restoring couple sets how fast the trade happens. That is why the roll period is a measurement of the metacentric height — a stiff ship rolls quickly and an uncomfortable one slowly — and why a naval architect can be handed a stopwatch reading and told the stability from it.

There is a sharper version of the same trap. If the roll period matches the period at which waves arrive, the response builds in exactly the way a driven oscillator’s does at resonance, and a modest sea produces enormous rolls. Choosing $\overline{GM}$ is therefore partly a matter of putting the natural period somewhere the weather is unlikely to sit.

Measuring it on a vessel that already exists #

The three terms can be computed from drawings, and two of them can be computed well. $\overline{KB}$ and $\overline{BM}$ come from the hull form, which is known exactly. $\overline{KG}$ is the difficult one, because it is a sum over every item aboard — structure, machinery, fuel, stores, paint — and an error of a few centimetres in it is an error of a few centimetres in $\overline{GM}$, which for many vessels is a large fraction of the whole.

So it is measured instead, by an experiment that is one line of arithmetic. Move a known weight $w$ a known distance $d$ across the deck, and read the angle $\varphi$ the vessel settles at. The heeling moment $wd$ is balanced by the righting moment $W\overline{GM}\sin\varphi$, so

$$\overline{GM} = \frac{w d}{W \tan\varphi}$$

for small angles. Everything on the right is measured: the weight, the distance, the displacement from the draft marks, and the angle from a long pendulum hung inside the hull. Every vessel of any size has this done before it enters service, and it is repeated after major modification. It is the same manoeuvre as weighing a body in and out of water: arrange the measurement so that the quantity wanted is the only thing left standing.

The experiment is fussy in a way worth noting, because the fussiness is all about the assumptions on this page. It has to be done in still water, with the vessel free of moorings, with all tanks either full or empty to kill the free surface effect, and with the crew standing still — every one of those a term the derivation quietly set to zero. An inclining experiment is therefore a good example of what it takes to measure a quantity whose definition contains four idealisations.

Where the model stops #

The angles are small. The metacentre is only fixed near upright. At large heel it wanders, the righting arm curve peaks and then falls, and eventually crosses zero at the angle of vanishing stability — beyond which the couple works the other way and the vessel does not come back. A full stability assessment is that whole curve rather than one number at the origin.

The hull is wall-sided. The exact formula above assumes the sides are vertical over the range of heel considered. Real hulls flare or tumble home, the deck edge eventually immerses and the bilge emerges, and both events change the waterplane discontinuously. The deck edge going under is usually the point at which the righting arm curve turns over.

Nothing is free to move. A liquid in a partly-filled tank flows to the low side as the hull heels, moving mass in exactly the wrong direction. The effect is equivalent to raising $G$, it is called the free surface effect, and its size depends on the tank’s waterplane by the same $I$ that governs the hull’s — so a wide shallow tank is far worse than a narrow deep one holding the same liquid. Vessels are subdivided lengthwise for this reason and not only for flooding.

The water is flat and still. A wave whose crest is amidships changes the waterplane and the buoyancy distribution together, and can reduce stability substantially for as long as it is there. Parametric rolling, where a ship’s stability varies periodically as waves pass and drives a growing roll at half the encounter frequency, is a real and modern failure mode — a parametric instability rather than the ordinary driven resonance described above, and it is much harder to design against.

The waterplane is a single connected shape. Once water is over the deck, or the vessel is partly submerged, the effective waterplane can split or vanish. A fully submerged body has no waterplane at all, so $\overline{BM}$ is zero and the condition collapses to requiring $B$ above $G$ outright — a far harder constraint, and the reason a submarine’s internal arrangement is fixed by stability in a way a ship’s is not.

The body is rigid and intact. Flooding changes both $\overline{KG}$ and the waterplane at once, and usually for the worse.

The log, finished #

The page opened with a log floating at the same draft on its side and on its end, and it is worth closing that, because the arithmetic is short and the three terms behave very differently in the two cases.

Take a uniform log ten diameters long, floating half submerged. On its end, the waterplane is a circle of diameter $d$, so $I = \pi d^4/64$ against a displaced volume of $3.93\,d^3$, and $\overline{BM}$ is about a hundredth of a diameter. Meanwhile $G$ sits five diameters up and $B$ two and a half. The metacentric height is about minus two and a half diameters — not marginal, not a close-run thing, but negative by five times the log’s own width.

On its side the waterplane is the full ten-by-one rectangle. About the long axis the log is exactly neutral, and it has to be: rotating a circular cylinder about its own axis leaves the submerged shape unchanged, so $B$ cannot move, and the three terms duly cancel to zero — $\overline{KB} = 0.288d$, $\overline{BM} = 0.212d$, $\overline{KG} = 0.5d$. That cancellation is a check on the whole apparatus rather than a result. About the transverse axis, where $I = dL^3/12$ with the length cubed, $\overline{BM}$ is twenty-one diameters and the log is about as stable as anything gets.

So the log lies down, spins freely about its own axis, and refuses absolutely to stand up — three different answers about the same object at the same draft, from the same three terms, decided entirely by which slice the water happens to cut.

The same three terms, offshore #

The condition on this page is what decides the form of everything built to float in deep water, and the interesting thing is that the three terms can be traded off against each other in quite different proportions.

A spar platform is a vertical cylinder two hundred metres deep with a small waterplane, so $\overline{BM}$ contributes almost nothing — exactly the log-on-end case. It is made stable by ballasting the bottom until $G$ is below $B$ outright, which is the submerged-body condition, and it works: the long deep column also has an enormous heave period, well clear of any ocean wave.

A semi-submersible goes the other way. Most of its volume is in pontoons well below the surface, and only four or six slender columns pierce the waterplane — but they do so a long way out, and the parallel-axis term, area times the square of the distance to the axis, makes $I$ large from a very small area. It buys stability from the arrangement of the waterplane rather than from its size, which is the catamaran argument taken to its limit.

A tension-leg platform declines the question. It floats with excess buoyancy and is held down by vertical tendons to the seabed, so its righting moment comes from the tendons rather than from hydrostatics at all, and $\overline{GM}$ stops being the governing number.

All three are in service under floating wind turbines, which is a good demonstration that the choice is an engineering trade rather than a right answer: the same nacelle sits on a spar in deep Norwegian water and on a semi-submersible off Portugal.

The history, and the calculation that arrived late #

Ships were built for several thousand years before any of this was written down, and stability was a matter of experience, rules of thumb and proportions copied from vessels that had not sunk. Pierre Bouguer introduced the metacentre in 1746 in the Traité du navire, and Euler arrived at an equivalent criterion independently at about the same time.

What is striking is how long the calculation remained something that was not done. The Vasa capsized in 1628 in Stockholm harbour on her maiden voyage, in a light breeze, having sailed about 1,300 metres, because she was too narrow and too heavily armed high up — a negative metacentric height, produced by exactly the three terms above. A stability test had in fact been carried out: thirty men ran back and forth across the deck, and it was stopped after three passes because the ship was rolling alarmingly. The result was known and there was no framework in which to act on it.

That is a recurring shape in this collection: an assertion that has never rejected anything proves nothing, and a test whose result cannot be interpreted is not much better. The men running across the deck were measuring $\overline{GM}$ by an inclining experiment — the modern method, more or less exactly — eighty years before there was a quantity for it to be a measurement of.

The ladder from here #

Later rungs on this anchor: the full righting-arm curve and the angle of vanishing stability. The free surface effect and its equivalent rise of $G$. Damaged stability, where flooding is assumed and the vessel is required to survive it. The inclining experiment, which measures $\overline{GM}$ on a finished vessel by moving a known weight and reading the heel. Parametric roll, where the stability itself is the thing oscillating. And the same analysis applied to bodies that are fully submerged, where the waterplane vanishes, $\overline{BM}$ is zero, and stability requires $B$ above $G$ outright — which is why a submarine’s arrangement is a much harder constraint than a ship’s.