Hydrostatics & Stability

Displacement is the simplest idea on the paper and the one most candidates rush past. Δ = ρ∇: the ship's weight equals the weight of the water it displaces, and ∇ (the volume of displacement) is read off the curves of form or the displacement scale for the current mean draught. The real difficulty is the unit convention — a long ton of sea water occupies 35 cubic feet, a long ton of fresh water occupies 36, and mixing those two constants is one of the most common numerical slips on this part of the paper.

The curves of form present displacement, TPC, MT1, KB, KM and the form coefficients against draught, all read from the same vertical line. Treat them as a lookup table, not something to be derived from first principles under time pressure — the paper supplies the handbook, and the skill being tested is picking the right curve at the right draught and reading it correctly, not recalling how it was built.

The form coefficients describe how a hull's underwater volume compares with a rectangular block of the same overall dimensions. The block coefficient C_B relates the actual displaced volume to length, beam and draught; the midship coefficient C_M does the same for the midship section alone; the prismatic coefficient C_P (= C_B/C_M) strips the midship shape out and describes how full the ends are relative to the midship section; and the waterplane coefficient C_W describes how full the waterplane is relative to a rectangle of the same length and beam.

TPI (tons per inch immersion) tells you how many long tons of weight, added or removed, change the mean draught by one inch, on the assumption that the waterplane area is essentially constant over that small change — true enough for the size of problem this paper asks. It comes directly from the waterplane area at the current draught and the density of the water: a larger, fuller waterplane resists sinkage more, so it takes more weight to move the draught the same amount.

The constant 420 is not arbitrary — it falls out of 35 ft³ per long ton and 12 inches per foot: a 1-inch layer of sea water over the waterplane area A_W weighs A_W/12 cubic feet worth of water, and dividing by 35 ft³ per long ton gives A_W/420 long tons. Knowing where 420 comes from means the SI form is not a separate thing to memorise: TPC = A_W·ρ/100, with A_W in m² and ρ in t/m³, follows the same logic with metric constants.

MT1 (moment to change trim one inch) is the trimming equivalent: the moment, in foot-tons, needed to change trim by one inch. It depends on the ship's longitudinal metacentric height GM_L, which for a normal hull form is large enough that BM_L (and so GM_L) is dominated by the waterplane's longitudinal second moment rather than by KG — which is why MT1 barely moves across a normal range of loading conditions and is usually just read off the curves rather than recalculated from scratch.

Sinkage from an added weight is uniform: w/TPI, applied equally at every point along the length. Trim is different — it is generated only by the part of the weight's moment acting about the centre of flotation (LCF), and once the total trim is known it has to be split between the forward and after perpendiculars in proportion to each one's distance from the LCF, not divided equally. The LCF is rarely at amidships, and treating it as if it were is the most common way marks are lost on a trim problem.

The metacentre M is the point about which a ship appears to rotate for small angles of heel, and GM — the height of G below M — is the single number that governs initial stability: a healthy positive GM means a small heel produces a righting lever that grows quickly; a small or negative GM means it does not, or grows the wrong way.

KB and BM both come off the curves of form or are computed from the hull geometry; KG comes from the loading condition — lightship KG plus every item aboard, taken about the keel. GM by itself is rarely the final answer this paper wants, because it is almost always asked alongside a correction — free surface, or a shift from a suspended weight — since the point being tested is whether a candidate remembers what has to be deducted, not just the base formula.

GM only describes the ship at small angles. Beyond roughly 10°–15° the assumption that the metacentre stays fixed breaks down, and the righting arm GZ has to be read from the cross curves of stability (KN curves), pre-computed for a range of displacements and heel angles against an assumed KG. Because the assumed KG is virtually never the ship's actual KG, GZ has to be corrected every time it is used:

Plotting GZ against heel angle gives the GZ curve — its initial slope reflects GM, the angle where the curve crosses zero is the angle of vanishing stability, and the area under it up to a given angle measures the dynamical (energy) stability available to resist a heeling moment, such as wind or a hard turn, up to that angle.

A tank that is completely full or completely empty behaves, for stability purposes, as a fixed weight at a fixed point — its own centre of gravity does not move as the ship heels. A tank that is slack, part-full with a free surface, behaves differently: as the ship heels, the liquid surface stays horizontal while the tank tilts with the hull, so the liquid's centre of gravity shifts toward the low side. That shift acts exactly like a virtual rise of the ship's centre of gravity, reducing the effective GM without any weight actually moving.

The fact worth remembering under exam pressure is that FSC does not depend on how much liquid is in the tank — only on the free surface's breadth cubed and its length. A tank that is 10% full and one that is 90% full produce essentially the same correction, provided both are slack. This is why the danger is greatest whenever a tank sits anywhere in the middle of its range, and why operators track how many tanks are slack at once rather than how full any single one of them is.

Because the correction scales with b³, subdividing a wide tank with a centreline bulkhead is disproportionately effective — halving the breadth of each free surface cuts that tank's contribution to FSC to one-eighth, not one-half. This is the practical reason cargo and fuel tanks on many ships are split down the centreline rather than left as one wide space.

Damage stability asks a different question than intact stability: given that sea water has entered one or more compartments, does the ship still float with an adequate residual margin, and in what condition? The paper treats this in outline rather than asking for a full probabilistic subdivision calculation — the concepts that come up are lost buoyancy, permeability, and what the resulting stability picture looks like, not the detailed index calculations behind a full subdivision assessment.

The lost-buoyancy method treats the flooded compartment as open to the sea and no longer contributing buoyancy; the ship sinks and trims until the remaining intact volume again equals the displacement, with the waterplane area reduced over the flooded length. This changes TPI and MT1 for the damaged condition, because the intact waterplane area is smaller — worth remembering if a problem asks for sinkage or trim after damage rather than before it.

Permeability μ is the fraction of a compartment's volume that can actually be occupied by water — a cargo hold full of machinery or stores floods less completely than an empty tank, and a ballast tank that is already full of water effectively floods by zero. Where a problem gives a permeability figure for a space, it is telling you to scale the flooded volume by μ before working out the added weight or lost buoyancy, rather than treating the whole compartment as open sea.

The general shape of an adequate damaged condition is: positive freeboard at the final waterline, a positive GZ range beyond the point of equilibrium heel, and no progressive down-flooding through openings that are not weathertight. Exactly where those thresholds sit is set out in the applicable subdivision and damage stability rules for the vessel type in question — worth having the concept clear, without trying to memorise clause numbers that the reference material will supply on the day.

The inclining experiment exists because KG cannot be measured directly — it can only be inferred from how the ship responds to a known heeling moment. A known weight is moved a known distance across the deck, the resulting angle of heel is measured, usually with a plumb line and batten or a U-tube, and GM is calculated from the ratio of the applied moment to the displacement and the tangent of the heel angle.

Every precaution taken before and during the experiment exists to protect that one small measured angle from anything that is not the inclining weight. Mooring lines are slacked so the ship is genuinely free to heel; loose weights, cranes and derricks are secured in a fixed position; tanks are pressed full or empty, never left slack, to remove any free surface; personnel not needed for the test stay off the deck and out of the way during a reading, because a person's own shift in position is itself an uncontrolled heeling moment; and the weather, wind and current are watched for the same reason, because they add their own moment on top of the one being measured.

The GM produced directly by the calculation is a solid GM — it reflects whatever was aboard, in whatever tank condition, at the moment of the test. If any tank was slack during the experiment, its free surface correction has to be added back before quoting the ship's true KG, and every subsequent loading condition then has its own free surface deducted again for whatever happens to be slack in that condition. The inclining experiment measures a moment in time; every stability calculation that follows still has to apply FSC for itself.

The three problems below chain the ideas above together the way the exam does — a sinkage-and-trim problem, an inclining experiment with a free surface correction, and a large-angle GZ problem set against a wind heeling arm — each needing more than one formula applied in the right order, not a single substitution.