Buoyancy and stability
Archimedes' principle is one line and takes ten minutes to believe. Deciding whether the floating or submerged body it describes will stay upright, or roll over the instant something nudges it, takes the rest of this article — and turns out to hinge on a single signed number that a naval architect will check before almost anything else about a hull.
Buoyancy is a special case of hydrostatic force on a surface — the surface just happens to close up on itself and surround a body entirely. That closure is what makes the result so clean: sum the pressure force over every face of a submerged body and almost everything cancels, leaving one upward force with a simple size and a fixed point of action. Getting that force is the easy part of this article. Whether a body stays the right way up once it has it is the harder, more interesting part, and it takes up most of what follows.
Archimedes’ principle
Take a body submerged in a fluid at rest and consider the pressure acting on it, face by face. Every horizontal component cancels by symmetry — whatever pushes the body one way sideways, an identical pressure at the same depth on the opposite face pushes back. What does not cancel is vertical: the pressure on the underside of the body acts at greater depth than the pressure on its top, and greater depth means greater pressure. The imbalance is a net upward force, and because pressure depends on nothing but depth in a given fluid, that force can be found without knowing anything about the body’s shape beyond the volume it occupies:
This is Archimedes’ principle: the buoyant force on a submerged or floating body equals the weight of the fluid it displaces, full stop, whatever the body is made of. A steel hull and a block of the same shape carved from balsa feel identical buoyant forces if they displace the same volume; what differs entirely is their own weight, and it is the contest between the two that decides whether either one floats.
The point of application matters just as much as the size, and it is not automatically the body’s own centre of mass. The buoyant force acts through the centroid of the displaced volume — the centre of buoyancy, conventionally labelled \(B\) — which is a purely geometric point fixed by the shape of the submerged region, with no reference to how the body’s mass happens to be distributed inside it. A body of uniform density has \(B\) coincide with its own centre of gravity. A body with ballast, an engine room, or an air pocket in one corner does not, and the gap between the two points — not either point alone — is what the rest of this article is about.
Pressure pushes on every face of a submerged body — smaller above, bigger below.
Floating and submerged bodies
A fully submerged body displaces exactly its own volume, and nothing about that changes with depth: lower it further into the same fluid and \(V_{\text{displaced}}\) stays fixed, so \(F_B\) stays fixed too. Whether such a body rises, sinks, or hovers depends entirely on comparing that fixed \(F_B\) against its own weight \(W\). Heavier than the fluid it displaces, and it sinks, all the way to the bottom, its buoyancy never growing enough to help further. Lighter, and it accelerates upward for as long as it remains fully submerged.
Break the surface and everything changes, because a floating body is only partly submerged, and the volume it displaces is no longer fixed — it is whatever fraction of the body happens to sit below the waterline, and that fraction adjusts itself, rising or sinking the body until the buoyant force exactly matches its weight:
Rearranged, the submerged fraction of a floating body’s volume is set by nothing but a ratio of densities:
Ice, at roughly 917 kg/m³ against seawater’s 1025, floats with about 90% of its bulk underwater — the famous nine-tenths of an iceberg that never shows above the surface. A steel ship, whose material density is several times that of water, still floats, because \(\rho_{\text{body}}\) in this formula means the density of the whole hull as built — mostly air, by volume, once cargo holds and living spaces are counted — not the density of the steel it is welded from. Confusing the two is the standard wrong turn taken when this idea is first met: a solid block of steel sinks: a steel shape enclosing a great deal of empty space does not, for exactly the reason a cupped hand displaces more water than a clenched fist of the same mass.
Fully submerged, a body displaces its own entire volume, whatever the depth.
Stability of a submerged body
Floating or sinking is a question about size — does \(F_B\) exceed \(W\) or not. Stability is a different question entirely, about what happens when a body already in equilibrium is disturbed by some small tilt, and it turns out to hinge on the relative position of two points: the centre of buoyancy \(B\), fixed by shape, and the centre of gravity \(G\), fixed by how mass is distributed.
For a fully submerged body, \(B\) does not move when the body rotates slightly, because the displaced volume is the whole body and rotating a fixed shape about its own interior does not change that shape’s centroid. \(G\), rigidly attached to the body, rotates right along with it. Tilt the body a small angle and, unless \(G\) and \(B\) happened to coincide, the two are no longer stacked on the same vertical line — weight, acting down through \(G\), and buoyancy, acting up through \(B\), now form a couple.
The direction of that couple is the entire answer. With \(G\) below \(B\), a small tilt swings \(G\) out to one side while \(B\) stays put directly above where it started, and the resulting couple rotates the body back toward upright — restoring, and therefore stable. With \(G\) above \(B\), the same geometry produces a couple in the opposite sense, one that increases the tilt rather than undoing it — unstable, and a submarine or a submerged buoy built this way will roll over given the smallest provocation and settle upside-down instead. This is precisely why a submarine carries its heaviest machinery low and a diving buoy carries a weighted keel: not for any reason of trim, but because it is the only way to put \(G\) below \(B\) and make the equilibrium one that survives contact with reality.
A fully submerged body, upright, with ballast keeping its centre of gravity low.
Stability of a floating body
A floating body complicates this picture in a way that, at first look, seems to threaten the whole result of the previous section: most ships float quite happily with \(G\) sitting above \(B\), which by the submerged-body argument ought to be unstable. It is not, and the reason is that a floating body’s centre of buoyancy is not fixed the way a submerged body’s is — because a floating body’s submerged shape changes as it tilts.
Tip a floating hull a small angle and one side sinks a little deeper while the other rises a little clear of the water. The submerged cross-section is no longer symmetric about the hull’s own centreline: a wedge of new volume has gone under on the low side, and an equal wedge has emerged on the high side. The centroid of that new, lopsided submerged shape — the new \(B\) — has shifted horizontally toward the side that went down, even though \(G\) has not moved relative to the hull at all.
Follow the buoyant force’s new line of action, now passing through the shifted \(B\), back up until it crosses the hull’s original vertical centreline — the line the hull sat on before it was disturbed. That crossing point is the metacentre, \(M\), and for small tilt angles its position on the centreline barely moves as the angle changes, which is what makes it useful as a fixed reference rather than a fresh calculation at every angle of heel.
The stability test is now a comparison against \(M\) rather than against \(B\) directly. If \(M\) sits above \(G\), the couple formed by weight at \(G\) and buoyancy along its new, shifted line is restoring, and the hull is stable even with \(G\) comfortably above \(B\). If \(M\) sits below \(G\), the same disturbance grows instead of decaying. A ship’s hull is shaped precisely to buy itself a metacentre well clear of its centre of gravity, and everything about how a hull’s cross-section widens at the waterline is in service of that one requirement.
Upright, G and B both sit on the hull's own centreline.
Metacentric height
Turn the geometric picture of the previous section into a single number and the whole stability question is answered by its sign. The metacentre sits a fixed distance above the centre of buoyancy, a distance called the metacentric radius:
where \(V\) is the displaced volume and \(I\) is the second moment of area of the waterplane — the shape traced out at the waterline, looking down on the hull from above — about the axis the hull tilts around. This is the same second moment of area used to locate the centre of pressure two articles ago, doing yet another job: here it measures how strongly the waterplane’s own shape resists being tilted, geometrically, independent of anything about the hull’s loading.
Subtract the distance from \(B\) up to \(G\) — call it \(BG\), positive when \(G\) sits above \(B\), as it usually does for a loaded hull — and what is left is the metacentric height:
\(GM > 0\) means \(M\) sits above \(G\): stable. \(GM < 0\) means the opposite: unstable, and the vessel needs no external disturbance at all to begin capsizing — it will do so on its own, given enough time. \(GM = 0\) is the marginal case, neutrally stable, indifferent to small tilts in a way no vessel is actually built to be. The size of a positive \(GM\), not merely its sign, also matters in practice: a small positive value rights the vessel slowly and gently, while a very large one produces a stiff, fast, uncomfortable roll, which is why naval architects target a range rather than simply the largest number achievable.
A worked case shows how sharply this depends on hull shape. A barge 20 m long and 8 m in beam, drawing 1.5 m, has a waterplane second moment of area \(I = Lb^3/12 = 20(8)^3/12 = 853.3\ \text{m}^4\) about its length-wise tilt axis, and a displaced volume \(V = 20 \times 8 \times 1.5 = 240\ \text{m}^3\), giving
With a loading that puts \(G\) 2.2 m above \(B\), the metacentric height comes out to \(GM = 3.56 - 2.2 = +1.36\) m — comfortably stable. The formula’s dependence on beam is cubic, not linear, because \(I\) for a rectangular waterplane goes as the beam cubed: halve the beam of the same barge and \(BM\) falls by a factor of eight, easily enough to turn a comfortably stable vessel into a dangerously tender one. It is the single largest reason a wide, shallow barge is a stable machine and a narrow, deep-drafted boat is a comparatively delicate one, and why widening a vessel's beam is so often the cheapest fix available for a stability problem that trimming its loading cannot solve.