The topic that rewards structure over memory. Know the design spiral, the weight groups, and how a hull is actually built, and most questions become straightforward.
Ship design cannot be done in one pass, because every decision invalidates an earlier estimate. Choose dimensions, and the weight changes. Estimate weight, and the required displacement changes, which changes the dimensions. The process is therefore iterative, and it is drawn as a spiral converging inward.
Two habits of thought come out of it. First, the first pass is deliberately crude — empirical formulas and coefficients from similar ships, not calculation from first principles. Second, the constraints usually bind before the optimisation does: canal beam limits, port draught limits, berth length, cargo density, and the required deadweight all fence the design in before anyone starts refining.
The classic starting point is a basis ship — an existing vessel of similar type and size, whose known dimensions, weights and powering are scaled to the new requirement. Almost every empirical coefficient in this chapter exists to make that scaling possible.
Displacement divides in two, and the two halves belong to different people.
Steel weight scales with the enclosed volume of the structure, which is why the cubic number L·B·D is the usual base — with depth, not draught, because the structure runs to the deck regardless of where the waterline sits. Watson's method refines this with an equipment numeral and a block coefficient correction.
Outfit weight scales with area rather than volume: accommodation, hatch covers, deck machinery, piping, insulation and paint all grow with L×B.
Machinery weight scales with installed power, at a fraction under one — a bigger engine is heavier but not proportionately so.
A quick check on any weight estimate is the deadweight coefficient, DWT/Δ:
Steel goes with volume, outfit with area, machinery with power. If a weight estimate lands outside the deadweight coefficient band for the ship type, one of those three is wrong.
Three quantities that sound similar and measure entirely different things.
Freeboard is the distance from the waterline to the freeboard deck — reserve buoyancy, expressed as a minimum. It is not chosen by the designer; it is computed from the Load Line rules by ship type and length, then corrected for depth, superstructure extent, sheer and block coefficient. The load line marks on the side of the ship are the result.
Tonnage is volumetric and dimensionless. Gross tonnage measures total enclosed volume; net tonnage measures the volume useful for carrying cargo.
Tonnage matters because so much regulation is pegged to it: port and canal dues, manning scales, and the thresholds at which conventions apply to a ship at all.
Capacity is the actual volume available for cargo — grain capacity measured to the shell, bale capacity measured inside the frames and beams, which is smaller. Which one applies depends on whether the cargo flows into the corners.
Modern shipbuilding is assembly, not construction. Steel is cut and formed in the shop, welded into panels, panels into blocks, blocks into grand blocks, and only then erected on the berth.
The reason is economic and worth stating in an answer: work done in the shop costs a fraction of the same work done on the berth, where access is poor, weather intervenes and the building dock is the yard's bottleneck. So blocks are outfitted before erection — pipework, cable trays, ladders, even machinery installed while the block is still upside down and accessible.
Two disciplines make it work:
Too little heat risks a hard, crack-prone heat affected zone; too much causes grain growth, loss of toughness and distortion. That is why a welding procedure specifies a range, not a maximum.
Ship steel is specified by grade: A, B, D and E in normal strength, and AH, DH, EH in higher strength. The letters denote notch toughness — the temperature at which the steel still absorbs energy rather than fracturing in a brittle manner. Higher-toughness grades go where the stresses concentrate: hatch corners, sheer strake, bilge strake.
Higher tensile steel allows thinner scantlings for the same strength, saving weight. Two costs follow, and both are examinable: the structure becomes more flexible, and because fatigue crack growth is nearly independent of yield strength, the higher working stresses make fatigue detailing more critical, not less.
Non-destructive testing splits cleanly by what it can find:
The pairing is a favourite one-mark question: dye penetrant will never find a subsurface inclusion, and radiography is poor at detecting a tight planar crack lying parallel to the beam.
Design decisions are justified in money, and the vocabulary is small enough to learn in five minutes.
Simple payback is the figure that gets quoted, and it ignores the cost of capital, the remaining life of the ship, maintenance of whatever was installed, and off-hire during installation. Quote it, then state the assumptions it rests on — a payback with no stated fuel price and no stated number of sea days is not an answer.
A weight estimate, a tonnage calculation, and the commercial justification an owner actually asks for.
A bulk carrier has L = 150 m, B = 22 m, moulded depth D = 12 m, design draught T = 8.5 m and C_B = 0.75. Steel weight may be taken as 0.055 t per m³ of cubic number, outfit as 0.40 t per m² of deck area (L×B), and machinery weight is 600 t. Find the displacement, the lightweight and the deadweight, and check the result against typical practice.
L = 150 m, B = 22 m, D = 12 m, T = 8.5 m C_B = 0.75, ρ = 1.025 t/m³ Steel 0.055 t/m³ of L·B·D; outfit 0.40 t/m² of L·B Machinery 600 t
Find the displacement, the lightweight and the deadweight, and check the result against typical practice
Displaced volume from the block coefficient.
Then displacement.
Steel weight from the cubic number.
Note that it uses depth, not draught — the structure exists all the way to the deck whether or not that part is immersed.
Outfit weight scales with deck area.
Accommodation, hatch covers, piping, deck machinery, paint.
Lightweight is the sum of the three groups: what the yard hands over.
| Group | Basis | Weight (t) |
|---|---|---|
| Steel | 0.055 × 39 600 m³ | 2178 |
| Outfit | 0.40 × 3300 m² | 1320 |
| Machinery | given | 600 |
| LWT | 4098 |
Empty.
Deadweight is everything the owner can then put aboard.
Cargo, fuel, fresh water, stores, crew, ballast.
Now check it.
The deadweight coefficient is the fraction of displacement available for earning revenue, and for a bulk carrier it should land between about 0.80 and 0.86.
A sense of proportion is worth carrying: a bulk carrier is around 0.83.
A tanker 0.86, a container ship nearer 0.70 (light cargo, heavy machinery), a passenger ship as low as 0.25. If your answer falls outside the range for the ship type, one of the weight groups is wrong.
AnswerΔ = 21 563 t, LWT = 4098 t, DWT = 17 465 t, deadweight coefficient 0.81
The trap: using draught instead of depth in the cubic number. It gives a steel weight around 30 % light here, and the deadweight comes out correspondingly — and impossibly — high.
A ship has a total enclosed volume of 25 000 m³. Find her gross tonnage. Her displacement is 21 500 t and her deadweight 17 500 t. Explain how the three figures differ and what each is used for.
V = 25 000 m³ (total enclosed volume) Δ = 21 500 t, DWT = 17 500 t GT = K₁·V, K₁ = 0.2 + 0.02·log₁₀V
Find her gross tonnage
Explain how the three figures differ and what each is used for
Find the coefficient K₁.
It grows slowly with volume, so a larger ship is measured slightly less generously per cubic metre.
Apply the formula.
Notice that the answer carries no unit.
Gross tonnage is a dimensionless index derived from volume, not a mass. Writing “7199 tonnes” is wrong, and examiners look for exactly that.
Now the three quantities side by side.
And what each is for.
Displacement is the naval architect's quantity — it drives stability, strength and resistance. Deadweight is the commercial one — it is what a charterer buys. Gross tonnage is the regulatory one — it decides port dues, canal fees, manning scales, and which conventions apply to the ship at all.
A worked consequence: a car carrier has an enormous enclosed volume and light cargo.
So its GT is high relative to its deadweight, and it pays port dues out of proportion to the cargo it lifts. A tanker is the opposite. This is why comparing two ships by tonnage alone tells you very little.
AnswerGT = 7199 (dimensionless)
The trap: treating gross tonnage as a mass, or confusing it with deadweight. They measure different things in different units — one is volume-derived and dimensionless, the other is mass in tonnes.
A fuel-saving device costs US$ 450 000 installed. The ship burns 28 t/day of fuel at sea, sails 280 days a year, and the device is expected to save 4 %. Fuel costs US$ 600/t. Find the annual saving and the simple payback period, and state what would change the decision.
Capital cost = US$ 450 000 Consumption 28 t/day, 280 sea days/year Saving 4 %, fuel price US$ 600/t
Find the annual saving and the simple payback period, and state what would change the decision
Daily saving in tonnes.
Convert to money per day, then per year.
Simple payback.
State the assumptions.
| Assumption | Base case | Changed to | Payback |
|---|---|---|---|
| Fuel price | $600/t | $400/t | 3.6 years |
| Sea days | 280/year | 200/year | 3.3 years |
| Saving achieved | 4 % | 2 % | 4.8 years |
Because that is where the marks are in a management-level question. Three of them move the answer materially.
And the ones simple payback ignores entirely: maintenance cost of the device.
The remaining life of the ship, the cost of capital, off-hire during installation, and any effect on the ship's CII rating — which may be worth more than the fuel if it moves the ship out of a poor band.
The decision rule an owner actually applies.
A payback comfortably shorter than the remaining life of the ship, with a margin for the saving being optimistic. Two and a half years on a ship with ten years to run is a straightforward yes; the same payback on a ship due for sale in eighteen months is a no.
AnswerAnnual saving US$ 188 160; simple payback 2.4 years
The trap: quoting the payback and stopping. The number is only as good as the assumed saving, and vendor-quoted savings are measured in ideal conditions. State the assumptions and the sensitivity, or the answer is half an answer.
Δ = LWT + DWTLWT = steel + outfit + machinery∇ = C_B·L·B·TFirst dimensional check of any design passCN = L·B·DCubic number — uses depth, not draughtW_steel ≈ k·CNWatson refines with equipment numeral and C_BW_outfit ≈ k·L·BOutfit scales with areaW_mach ≈ k·(MCR)^0.7Machinery scales with power, at less than 1:1C_DWT = DWT/ΔTanker 0.86, bulker 0.83, container 0.70, passenger 0.30GT = K₁·V, K₁ = 0.2 + 0.02·log₁₀VDimensionless — volume, not massHeat input = η·V·I/vJ/mm; a procedure specifies a rangePayback = capital/annual savingState the assumptions or it is half an answer