Stability questions on this paper reward a candidate who can run a moments table without losing a decimal place, and who knows which correction — free surface, draught, or centre of flotation — the examiner is actually testing.
Every stability calculation on this paper begins with a number you have to go and read off the ship, not one handed to you in the question: the draught. A draught survey takes readings at the forward, aft and (where fitted) midships marks on both sides, corrects them for the ship's trim and any hog or sag in the hull, and reduces them to a single true mean draught. That draught is then looked up in the ship's hydrostatic tables, which give the corresponding underwater volume, ∇, at that waterline for the vessel's even-keel condition, with further corrections applied if the ship is trimmed or listed when the readings were taken.
Multiply that volume by the density of the water the ship is actually sitting in — not a nominal 1.025 unless you are genuinely in full-strength seawater — and you have the displacement, Δ. This is the figure a bunker survey or a cargo-by-difference calculation depends on: you take a displacement before loading and one after, and the difference is the weight you put on board. Get the draught reading wrong by a centimetre or use the wrong density and that error carries through every calculation that follows it, including the GM check at the end.
Displacement is not assumed — it is measured, corrected, and looked up. Treat the draught reading with the same care you'd give a sounding you were about to certify.
KG is not a single fact you look up — it is the weighted average vertical position of every weight on board, and you build it the same way every time: list each weight group with its own Kg (height above the keel), multiply weight by Kg to get a moment, sum the weights and sum the moments, then divide one total by the other. Weights that are removed or consumed — bunkers burnt, ballast pumped out — go into the same table as a negative entry, or equivalently are subtracted from the running total, since taking a weight off the ship changes KG exactly as adding one does.
KM, by contrast, is a property of the hull form and the waterline it is floating at, not of what is loaded — it comes straight out of the hydrostatic tables against the mean draught you established in the survey. Because the underwater shape changes as draught changes, KM is not constant: it typically rises then falls again across the draught range of a given hull. Subtracting your calculated KG from the KM read at the correct draught gives the solid GM, the metacentric height before any free surface correction is applied.
A GM that looks comfortable on the loading computer screen is only as good as the Kg figures that went into it, so when you check a condition by hand, check the table first — a transposed Kg or a weight entered against the wrong tank is the most common reason a hand check and the computer disagree.
A tank that is completely full or completely empty behaves, for stability purposes, like a solid weight fixed at its centre of gravity — the liquid cannot move relative to the tank. A slack tank is different: as the ship heels, the liquid surface stays roughly horizontal while the tank tilts with the hull, so the liquid's own centre of gravity shifts towards the low side. That shift adds a heeling moment of its own, on top of whatever moment caused the heel in the first place, and the net effect on the stability calculation is identical to raising the ship's actual centre of gravity — a virtual rise, not a physical one, but one that costs real GM.
The size of the correction depends on the tank's surface area and shape — specifically on its length and, far more strongly, on its breadth, because breadth enters the formula cubed. Halving a tank's breadth with a longitudinal wash bulkhead does not halve the free surface loss, it cuts it to an eighth; this is exactly why double bottom and deep tanks are commonly divided down the centreline. Crucially, the correction does not depend on how much liquid is in the tank — a tank at 10% full and one at 90% full give the same FSC, because the calculation only cares about the free surface dimensions, not the volume beneath it. Only very near empty or very near full, where the liquid surface presses against the tank's curved boundaries and the effective breadth shrinks, does the loss reduce — a refinement usually left to the loading computer rather than a hand calculation.
Free surface is deducted because a tank is slack, not because of what is in it. Topping a tank up to full, or stripping it dry, removes the correction; leaving it slack at any level in between does not.
List and trim are the same physical idea worked in two different directions — a moment displaces the ship's centre of gravity off the centreline (transversely) or off the centre of flotation (longitudinally), and the hull rotates until the righting moment catches up. For a one-off transverse shift — cargo breaking loose, ballast pumped asymmetrically, a lifeboat swung out — the heeling moment is simply the weight times the distance it moved. The ship settles at the angle where the righting moment, Δ·GM·tan θ for small angles, equals that heeling moment.
The GM in that formula has to be the fluid GM — solid GM with free surface already deducted — because the free surface effect is present whether or not the cargo shift ever happens. It is easy in an exam to calculate a tidy solid GM, plug it straight into the list formula and move on; the mark scheme is specifically checking whether you remembered the correction from the previous step. Note too that this is a static, one-off shift being asked for an equilibrium angle — it is a different question from a permanent list caused by an asymmetric loading condition (an off-centre KG from the outset), though the same tan θ relation applies to both once you have the transverse GG′ or the equivalent moment.
Trim problems trip up candidates who treat them as a longitudinal version of list and pivot everything about amidships. The ship does not trim about amidships — it trims about the centre of flotation, F, the centroid of the waterplane area at that draught, because F is the one point where adding a small weight changes the mean draught with no rotation at all. F is rarely exactly amidships; on many hull forms it sits a metre or several aft of it, and the stability booklet or hydrostatic tables will give its position for the draught in question.
A weight added anywhere on the ship first produces a bodily, parallel change of draught — sinkage or rise — independent of where the weight went, found from the tonnes-per-centimetre figure for that draught.
Only the weight's distance from F produces a trimming effect, and that distance is measured from F — not from the perpendiculars, not from amidships.
The resulting change of trim then has to be shared between the forward and aft draughts in proportion to each perpendicular's own distance from F — a ship with F well aft of amidships will show a noticeably larger draught change at the bow than at the stern for the same trimming moment.
The approved stability booklet is not reference material to flick through once and forget — it is the document that turns all of the above into a pass or fail decision. It carries the hydrostatic tables and KN (or cross) curves used to read KM and righting levers at any draught, the tank capacity and free-surface data for every tank on board, a set of standard or worked loading conditions, and the limiting KG or GM curves the vessel must stay within across its draught range. A loading computer applies these automatically and will normally flag a condition that fails, but the OOW still needs to be able to open the booklet, find the right curve, and sanity-check the computer's answer by hand — an instrument fault or a mis-keyed tank sounding will not flag itself.
The booklet is the authority; a hand calculation is how you prove to yourself — and to an examiner — that you understand what the booklet's curves and the loading computer are actually doing.
The three examples below run the full chain end to end: a moments table into a fluid GM check, that same GM used to find an angle of list, and a trim calculation worked properly about the centre of flotation rather than amidships.
A vessel loads and sails with the weight groups below. From the ship's hydrostatic tables, KM at the resulting mean draught is 7.62 m. One double-bottom tank is slack. Find the fluid GM and confirm whether it meets the 0.15 m minimum.
Lightship: 3500 t, Kg 5.80 m Cargo (holds): 2500 t, Kg 9.20 m Heavy fuel oil: 1200 t, Kg 3.50 m Fresh water: 500 t, Kg 11.00 m Stores: 300 t, Kg 8.00 m KM at this draught (from tables) = 7.62 m Slack tank free surface: l = 15 m, b = 8 m, ρ (fresh water) = 1.000 t/m³ Required minimum GM(fluid) = 0.15 m
Find the fluid GM and confirm whether it meets the 0.15 m minimum
Build the moments table.
| Item | w (t) | Kg (m) | w×Kg (t·m) |
|---|---|---|---|
| Lightship | 3500 | 5.80 | 20300 |
| Cargo | 2500 | 9.20 | 23000 |
| HFO | 1200 | 3.50 | 4200 |
| Fresh water | 500 | 11.00 | 5500 |
| Stores | 300 | 8.00 | 2400 |
| Total | 8000 | 55400 |
Every weight carries its own Kg; sum the weights for the displacement and sum the moments for KG in one pass.
KG follows directly from the table totals.
The slack tank has a free surface regardless of how full it is.
Only its surface dimensions matter.
Take KM off the table at the actual mean draught.
Subtract KG for the solid GM, then deduct the free surface correction for the true fluid GM.
AnswerGM(fluid) = 0.615 m, which clears the 0.15 m minimum by 0.465 m.
The trap: lifting KM from the tables at a nominal or design draught instead of the draught this loaded condition actually produces — KM moves with draught, and using the wrong figure throws off every GM that follows.
The same class of vessel is under way with the particulars below. In heavy weather a parcel of deck cargo breaks its lashings and slides 10 m transversely to starboard. Find the resulting angle of list.
Displacement Δ = 10,000 t KM at this draught (from tables) = 7.80 m KG (solid, from the loading computer) = 7.30 m One slack tank: l = 12 m, b = 10 m, ρ (fresh water) = 1.000 t/m³ Cargo parcel shifted: w = 45 t, distance moved d = 10 m
Find the resulting angle of list
Before touching the list, get the free surface correction for the slack tank.
It is already reducing GM before the cargo ever moves.
GM(solid) comes from the tables and the loading computer.
Deduct the free surface for the GM the ship is actually sailing with.
The shifted parcel creates a heeling moment about the centreline.
The ship settles where that moment balances the righting moment Δ·GM·tan θ.
Convert the tangent to an angle for the reportable figure.
AnswerThe vessel settles at a list of about 6.4° to starboard.
The trap: running the list formula on the solid GM instead of the fluid GM — skipping the free surface step here understates the list by a couple of degrees, which matters when you are deciding whether to correct it before continuing.
A vessel is floating at F 6.20 m / A 6.60 m when 300 t of cargo is loaded into a forward hold, 45 m forward of amidships. Find the new forward and aft draughts.
LBP = 120 m TPC = 15.0 t/cm MCT1cm = 150 t·m/cm Centre of flotation (F) = 3 m aft of amidships Initial draughts: forward 6.20 m, aft 6.60 m Weight loaded: w = 300 t, 45 m forward of amidships
Find the new forward and aft draughts
A weight added anywhere still causes a bodily parallel sinkage first.
That part of the answer does not care where the weight sits.
Trim is worked about F.
Not amidships, so the lever for the trimming moment is the weight's distance from F — here that means adding the 3 m offset of F to the weight's distance from amidships.
Convert the moment to a change of trim.
Then split that change between the two perpendiculars in proportion to each one's distance from F.
Apply the bodily sinkage and the trim change together.
Forward draughts increase further, aft draughts come back, because the ship is trimming by the head.
AnswerNew draughts: forward ≈ 6.90 m, aft ≈ 6.34 m — the vessel swings from 0.40 m by the stern to about 0.56 m by the head.
The trap: splitting the 96 cm change of trim evenly about amidships instead of about F — with F 3 m off amidships that error alone moves the calculated draughts by several centimetres, which is exactly the margin a canal or berth restriction is checking.
Δ = ρ∇Displacement from the draught survey and the hydrostatic tablesKG = Σ(w·Kg)/ΣwMoments about the keel — add loaded weight, subtract discharged or consumed weightGM = KM − KG − FSCKM must be read from the tables at the actual mean draughtFSC = i·ρₗ/Δ, i = l·b³/12Independent of the quantity in the tank; grows with breadth cubedtan θ(list) = (w·d)/(Δ·GM)Angle of list from a one-off transverse shift, using the fluid GMTPC = A_w·ρ/100Tonnes per centimetre immersion, from the waterplane area at that draughtSinkage (cm) = w/TPCBodily, parallel change of draught for a weight added anywhereChange of trim (cm) = w·d/MCT1cmd is measured from the centre of flotation, not amidshipsGM₀ ≥ 0.15 mMinimum from the approved stability booklet, plus its other intact-stability criteria