Naval Architecture and Ocean Engineering I: Hydrostatics, Stability, Resistance and Propulsion

Section 4 of the Naval Architecture and Marine Engineering (NM) paper is the longest in the syllabus and is really two subjects: the ship as a floating, moving body in calm water — its geometry, hydrostatics, stability, resistance and propulsion — and the ship and the offshore structure in a seaway — manoeuvring, waves and motions, structural strength, and ocean engineering. It is therefore written as two chapters, along that line. This first chapter takes the calm-water half: ship geometry and main particulars, Archimedes and the laws of flotation, form coefficients and hydrostatic calculations with Simpson’s rules, TPC and MCT1cm; statical stability at small angles, the inclining experiment, the shift of G by adding, removing, moving and suspending masses, and the free-surface effect; stability at large angles, the curve of statical stability, dynamical stability, the angle of loll, damage stability, floodable length, grounding and docking; the components of resistance, the form factor, roughness, model testing and the ITTC-1957 extrapolation, tank-wall effects, series data and advanced vehicles; and the screw propeller — its geometry, theories, open-water coefficients, hull–propeller interaction, cavitation, series, types, materials and manufacture. Sea water is taken at 1025 kg/m³ (1.025 t/m³) throughout.

1. Ship geometry, flotation and hydrostatic calculations

A ship is described by its main particulars: the length between perpendiculars L_BP (from the forward perpendicular at the stem to the after perpendicular, usually at the rudder stock), the length on the waterline, the moulded breadth B, the moulded depth D and the draught T; freeboard is D − T. The hull is drawn as a lines plan — body plan (transverse sections), half-breadth plan (waterlines) and profile (buttocks) — and tabulated as a table of offsets, the half-breadths at stations and waterlines. A hull that is streamlined, with a fine entrance, a gradual run aft and no abrupt changes of section, keeps the flow attached to the stern, lowers the viscous pressure resistance and delivers an even wake to the propeller.

Archimedes’ principle and the laws of flotation: a floating ship displaces a mass of water equal to its own mass, Δ = ρ∇, and it floats in equilibrium only when the buoyancy through the centre of buoyancy B and the weight through the centre of gravity G act in the same vertical line. The fullness of the underwater body is expressed by the form coefficients: the block coefficient C_B = ∇/(LBT), the midship-section coefficient C_M = A_M/(BT), the prismatic coefficient C_P = ∇/(A_M L) = C_B/C_M and the waterplane coefficient C_WP = A_W/(LB). A ship 100 m × 16 m × 6 m with C_B = 0.70 displaces ∇ = 0.70 × 100 × 16 × 6 = 6720 m³, that is Δ = 1.025 × 6720 = 6888 t in sea water. A tanker has C_B above 0.8, a fast container ship about 0.6 and a warship nearer 0.5.

Hydrostatic calculations integrate the offsets numerically, most often by Simpson’s first rule, ∫y dx ≈ (h/3)(y₀ + 4y₁ + 2y₂ + 4y₃ + … + y_n) for an even number of intervals. A waterplane with half-breadths 0, 4, 6, 4, 0 m at stations 10 m apart has a half-area of (10/3)(0 + 16 + 12 + 16 + 0) = 146.7 m² and an area A_W = 293.3 m²; the same rule on sectional areas gives the volume, and on moments gives the centroids. The results are drawn as hydrostatic curves against draught: displacement, KB, KM, LCB, LCF, TPC and MCT1cm. The tonnes per centimetre immersion TPC = ρA_W/100 (ρ in t/m³) is the mass that sinks the ship 1 cm in parallel; the moment to change trim by one centimetre MCT1cm = Δ·GM_L/(100L), in t·m per cm, turns a trimming moment into trim.

Hydrostatic quantities and their working formulas
QuantityFormulaWorked value
DisplacementΔ = ρ C_B L B T1.025 × 0.7 × 100 × 16 × 6 = 6888 t
Tonnes per cm immersionTPC = ρ A_W/1001.025 × 1400/100 = 14.35 t/cm
Moment to change trim 1 cmMCT1cm = Δ GM_L/(100 L)6888 × 120/(100 × 100) = 82.66 t·m/cm
Change of trimtrim = w d/MCT1cm50 t moved 40 m: 2000/82.66 = 24.2 cm
Transverse metacentric radiusBM = I_T/∇; box: B²/(12T)12²/(12 × 4) = 3 m
🧠 Heel and trim use the same machinery
A weight moved transversely heels the ship through tan θ = w d/(Δ GM); moved longitudinally, it trims the ship by w d/MCT1cm, the trim being shared between the ends about the centre of flotation (the centroid of the waterplane). Because GM_L is of the order of the ship’s length and GM_T of a metre or so, a ship trims by centimetres where it would heel by degrees.

2. Statical stability at small angles, the inclining experiment and the shift of G

When a ship heels through a small angle φ, the centre of buoyancy moves towards the immersed side and the new buoyancy line meets the centre line at the transverse metacentre M. The righting lever is GZ = GM sin φ, and the equilibrium is stable if M is above G (GM > 0), neutral if they coincide and unstable if M is below G. The metacentric height is built up from the keel: GM = KB + BM − KG, with BM = I_T/∇, I_T being the second moment of the waterplane about the centre line. For a box barge of breadth B and draught T, KB = T/2 and BM = B²/(12T); with B = 12 m, T = 4 m and KG = 4.5 m, GM = 2 + 3 − 4.5 = 0.5 m. A large GM gives a stiff ship with a short, uncomfortable roll; a small GM a tender ship with a long, easy roll and little reserve.

Shift of G. Adding a mass w at height Kg moves G to KG₁ = (Δ·KG + w·Kg)/(Δ + w), towards the added mass; removing it moves G away. A mass shifted a distance d moves G parallel to the shift by GG₁ = w d/Δ; shifted transversely, it heels the ship to tan φ = w d/(Δ·GM). A suspended mass acts as if it were at its point of suspension, because as soon as the ship heels it swings out to hang below that point: lifting a cargo off the deck with the ship’s own derrick raises its effective centre instantly to the derrick head, which is why heavy lifts are a stability case in their own right.

The free-surface effect: a slack tank whose liquid surface is free to move shifts its liquid to the low side as the ship heels, which reduces the righting lever as though G had risen by the free-surface correction FSC = ρ_l i/(ρ_sw ∇) = ρ_l i/Δ, where i = l b³/12 is the second moment of the tank’s free surface about its own centre line and ρ_l the density of the tank liquid. The correction depends on the tank’s breadth cubed and not on how much liquid it holds, and a longitudinal division into two halves reduces it to a quarter. An oil tank 10 m long and 8 m wide (i = 426.7 m⁴, ρ_l = 0.9 t/m³) in a ship of 6888 t costs 0.9 × 426.7/6888 = 0.056 m of GM; the fluid GM is GM − FSC.

The inclining experiment finds the KG of a completed ship, which no calculation can give reliably. With the ship upright, free of free surfaces, moorings slack and all weights accounted for, known masses w are moved across the deck a distance d and the heel is read from long pendulums (deflection a over length l, tan φ = a/l) or a U-tube; then GM = w d/(Δ tan φ), and KG = KM − GM with KM from the hydrostatics. Moving 10 t through 8 m in a 5000 t ship, with a 0.12 m deflection on a 6 m pendulum (tan φ = 0.02), gives GM = 80/(5000 × 0.02) = 0.8 m. The lightship KG found this way is the starting point of every loading condition.

⚠️ Free surface depends on breadth, not on the amount of liquid
A tank a quarter full and a tank three-quarters full have the same free-surface correction (while the surface stays clear of the top and bottom); only a pressed-full or empty tank has none. Doubling the breadth multiplies the correction by eight. The correction uses the density of the liquid in the tank and the displacement of the ship — dividing by the ship’s volume without the ratio of densities is the usual slip.

3. Stability at large angles, loll, dynamical stability and damage

Beyond about 10° the metacentre no longer stays fixed and GZ must be computed from the actual inclined waterplanes (by cross curves, the KN curves, with GZ = KN − KG sin φ). For a wall-sided ship, whose sides are vertical over the range of immersion, the exact result is GZ = sin φ (GM + ½BM tan²φ); with GM = 1.0 m, BM = 4 m and φ = 20°, GZ = 0.342 × (1 + 2 × 0.1325) = 0.433 m. The curve of statical stability (GZ against φ) is read for its initial slope (equal to GM per radian), its maximum GZ and the angle at which it occurs, the angle of vanishing stability where GZ returns to zero, and the range of stability. The area under it up to an angle is the dynamical stability, the work Δ∫GZ dφ needed to heel the ship to that angle, and it is what resists a sudden gust or a wave impact.

A ship with a small negative GM is not necessarily capsized: as it heels, the wall-sided term raises GZ until, at the angle of loll tan φ = √(2|GM|/BM), the lever is zero again and the ship lies there, flopping from one side to the other. With GM = −0.1 m and BM = 5 m, tan φ = √(0.04) = 0.2 and φ = 11.3°. Loll is distinguished from a list caused by an off-centre weight (which has G off the centre line and a positive GM) by the response: the cure for loll is to lower G — by filling low tanks one at a time, the smaller and lower first, never by moving weight across, which would throw a lolling ship over to the other side.

Damage stability asks whether the ship survives flooding. The deterministic approach floods a prescribed number of adjacent compartments and requires the damaged waterline to stay below the margin line (76 mm below the bulkhead deck) with a minimum residual GM and GZ; flooding is computed by lost buoyancy (the damaged volume is removed and the ship sinks until the intact part supports it; Δ and KG unchanged) or by added weight (the flood water is a mass added; both give the same final waterline). A compartment’s permeability μ is the fraction of its volume that water can fill — about 0.95 for accommodation, 0.85 for machinery spaces and lower for cargo. The floodable length curve gives, at each point along the ship, the greatest length centred there that can flood without submerging the margin line; multiplied by the factor of subdivision it gives the permissible length of a compartment. The probabilistic approach sums, over every damage case, the probability of that damage times the probability of surviving it, into an attained subdivision index A, which must be at least the required index R.

Grounding and docking. When a ship touches the blocks in dry dock or takes the ground, an upward force P acts at the keel. It can be treated as a mass P removed at the keel, which raises G by GG₁ = P·KG/(Δ − P), or as a fall of the metacentre by MM₁ = P·KM/Δ; the second, simpler form gives the virtual loss of GM. With P = 100 t, KM = 8 m and Δ = 5000 t the loss is 0.16 m. P grows as the water level falls, so the critical instant is just before the ship takes the blocks along its whole length; a ship with too small a GM, or trimmed heavily by the stern so that P is large before she settles, can fall over on the blocks. The same reasoning applies to a ship aground on a falling tide.

4. Ship resistance: components, model testing and the ITTC-1957 extrapolation

The calm-water resistance of a ship is divided into frictional resistance, the tangential shear on the wetted surface, which dominates slow full ships; viscous pressure (form) resistance, from the thickened boundary layer and any separation at the stern; wave-making resistance, the energy carried away by the ship’s wave system, which dominates at high Froude number and oscillates with humps and hollows as the bow and stern waves interfere; plus air resistance of the above-water body and appendage resistance of rudders, bilge keels, shaft brackets and stabilisers. In a seaway the ship also suffers added resistance in waves, and hull roughness and fouling add to the friction as the ship ages. The form factor k expresses viscous resistance as (1 + k) times the flat-plate friction, and is found from low-speed model tests where wave resistance is negligible (the Prohaska method).

A model cannot match both the Froude number Fn = V/√(gL) and the Reynolds number Re = VL/ν, so Froude’s method runs the model at the corresponding speed V_m = V_s/√λ (equal Fn) and splits the total resistance coefficient C_T = R_T/(½ρSV²) into a frictional part that depends on Re and a residuary part C_R that depends on Fn and is the same for model and ship. The friction line now used is the ITTC-1957 model–ship correlation line, C_F = 0.075/(log₁₀Re − 2)². The steps: C_R = C_Tm − C_Fm at the model’s Re; C_Ts = C_R + C_Fs at the ship’s Re (plus a correlation allowance, where one is used); R_Ts = ½ρ_s S_s V_s² C_Ts, with S_s = λ²S_m; and the effective power P_E = R_T·V, the power needed to tow the bare hull. The 1978 ITTC method refines this with the form factor, scaling (1 + k)C_F rather than C_F alone.

A worked Froude extrapolation (λ = 25)
StepModel (4 m, 2 m/s, ν = 1.0 × 10⁻⁶ m²/s)Ship (100 m, 10 m/s, ν = 1.2 × 10⁻⁶ m²/s)
Reynolds number8.0 × 10⁶, log = 6.9038.33 × 10⁸, log = 8.921
ITTC-1957 C_F0.075/4.903² = 0.0031200.075/6.921² = 0.001566
Total coefficientmeasured C_Tm = 0.00520C_Ts = 0.00208 + 0.001566 = 0.003646
Resistance and powerC_R = 0.00520 − 0.00312 = 0.00208R_T = ½ × 1025 × 2500 × 100 × 0.003646 = 467 kN; P_E = 4.67 MW

Tank-wall (blockage) effects: in a towing tank of finite breadth and depth the model’s flow is constrained, the water past it speeds up and the measured resistance is too high unless the model is small against the tank section or a blockage correction is applied; shallow water raises resistance for the same reason. Where no model is tested, resistance is estimated from methodical series — families of systematically varied hull forms such as the Taylor standard series and Series 60, presented as C_R against Fn for each C_P and B/T — or from regression methods fitted to many test results, such as Holtrop–Mennen. Advanced marine vehicles escape the displacement-hull wave barrier by lifting the hull out: planing craft ride on dynamic pressure at high Fn, hydrofoil craft lift the hull clear on foils, air-cushion vehicles and surface-effect ships on a cushion of air, and SWATH and catamaran forms trade wave resistance for more wetted surface.

⚠️ The ITTC line takes log to base 10, and the ship needs its own Reynolds number
Using ln Re in 0.075/(log Re − 2)² gives a friction coefficient several times too small. And the frictional coefficient is the one thing that is NOT carried from model to ship: C_R is carried unchanged, C_F is recomputed at the ship’s Reynolds number, which is roughly a hundred times larger, so C_Fs is about half C_Fm. Carrying C_Tm unchanged to the ship overestimates the ship’s resistance badly.

5. The screw propeller: geometry, theory, coefficients, interaction, cavitation and types

Geometry. A screw propeller is defined by its diameter D, number of blades Z, pitch P (the advance per revolution of the blade’s helical face, often quoted as P/D at 0.7R), the blade area ratio (expanded blade area over disc area), and the rake, skew, section shape and boss diameter. Theories: the momentum (actuator-disc) theory treats the propeller as a disc that accelerates the flow and gives the ideal efficiency η_i = 2/(1 + √(1 + C_T)), C_T = T/(½ρA V_A²) — so a lightly loaded, large-diameter propeller is the more efficient; blade-element theory integrates the lift and drag of each blade section; and lifting-line and lifting-surface theories model the blades as bound vortices shedding helical trailing vortices, which is how modern propellers are designed.

Open-water characteristics. Tested alone in uniform flow (the open-water test), a propeller of diameter D turning at n rev/s and advancing at V_A is described by the advance coefficient J = V_A/(nD), the thrust coefficient K_T = T/(ρn²D⁴), the torque coefficient K_Q = Q/(ρn²D⁵) and the open-water efficiency η₀ = TV_A/(2πnQ) = (J/2π)(K_T/K_Q). A propeller of 5 m diameter at 2 rev/s advancing at 6 m/s has J = 0.6; with K_T = 0.2 and K_Q = 0.03, η₀ = 0.6 × 0.2/(2π × 0.03) = 0.64, and the thrust is 0.2 × 1025 × 2² × 5⁴ = 512.5 kN. K_T and K_Q fall as J rises, and η₀ peaks just before K_T reaches zero.

Hull–propeller interaction. Behind the hull the propeller works in the ship’s wake, so the water reaches it at V_A = V(1 − w), w being the Taylor wake fraction; and its suction on the stern raises the hull’s resistance, so it must deliver a thrust T = R_T/(1 − t), t being the thrust-deduction fraction. The hull efficiency η_H = (1 − t)/(1 − w) is often above 1 — with t = 0.16 and w = 0.20 it is 0.84/0.80 = 1.05 — because the propeller recovers energy from the wake. The relative rotative efficiency η_R corrects the open-water torque for the non-uniform flow behind the hull. The propulsive efficiency (quasi-propulsive coefficient) is η_D = P_E/P_D = η_H η₀ η_R, and the shaft and brake powers follow by dividing by the shaft transmission and gearing efficiencies. The self-propulsion test tows a model with its own propeller running at the ship self-propulsion point to measure w, t and η_R; a resistance of 300 kN at 15 knots (7.716 m/s) needs P_E = 2315 kW and, with η_D = 0.70, a delivered power of 3307 kW.

Propeller cavitation occurs where blade pressure falls to the vapour pressure: sheet cavitation on the suction back near the leading edge, bubble and cloud cavitation further aft, tip-vortex and hub-vortex cavitation in the trailing vortices, and face cavitation on the pressure side at negative angle. Its effects are erosion of the blades, noise, vibration and hull pressure pulses, and, when extensive, a breakdown of thrust. It is avoided by giving enough blade area (the Burrill chart relates the thrust loading of the projected area to the local cavitation number), by skew, by unloading the tip, and by tests in a cavitation tunnel at the ship’s σ. Propellers are selected from series such as the Wageningen B-series, whose K_T–K_Q charts and B_p–δ diagrams give the optimum diameter and pitch for a given power and speed.

Types of propulsor and how each works
PropulsorWorking principle and use
Fixed-pitch propellerOne-piece casting; thrust varied by rpm; reversed by reversing the engine. Most merchant ships.
Controllable-pitch propellerBlades turned in the hub by a hydraulic mechanism; constant-speed engine, reversal by pitch. Ferries, tugs, ships with shaft generators.
Ducted (Kort nozzle) propellerA foil-section duct adds thrust at high loading. Tugs and trawlers at bollard pull.
Contra-rotating propellersTwo coaxial propellers turning oppositely; the aft one recovers the rotational energy of the forward one’s slipstream.
Cycloidal (Voith–Schneider) propellerVertical blades on a rotating disc, their pitch varied cyclically; thrust in any direction. Tugs, ferries.
WaterjetAn internal pump ejects a jet astern; steered and reversed by nozzle and bucket. Fast craft.
Podded and azimuthing thrustersA propeller on a steerable pod or leg, usually driven by an electric motor; propulsion and steering in one.
Unconventional propellersSurface-piercing and supercavitating propellers for very fast craft; tip-loaded and highly skewed designs for low vibration.

Materials, strength and manufacture. Most large propellers are cast in nickel–aluminium bronze or manganese bronze, which resist corrosion and cavitation erosion and cast well; stainless steel serves ice-class and some small propellers, and composites are used for quietness. The blade is checked as a cantilever from the root under the thrust and torque loads, with centrifugal and fatigue loads in the wake’s fluctuating inflow, and classification rules set a minimum root thickness. It is cast in a mould, machined or ground to the design pitch and section, finished to a specified surface class, and statically balanced before fitting.

Key takeaways

  • Δ = ρC_B LBT; TPC = ρA_W/100; MCT1cm = ΔGM_L/(100L); Simpson’s first rule (h/3)(1, 4, 2, 4, …, 1) integrates the offsets.
  • GM = KB + BM − KG with BM = I/∇; inclining: GM = wd/(Δ tan φ); free surface raises G virtually by ρ_l i/Δ with i = lb³/12.
  • Wall-sided GZ = sin φ(GM + ½BM tan²φ); loll at tan φ = √(2|GM|/BM); dynamical stability is the area under the GZ curve; docking loss MM₁ = P·KM/Δ.
  • Froude: run at V_m = V_s/√λ, carry C_R = C_Tm − C_Fm, recompute C_F = 0.075/(log₁₀Re − 2)² for the ship; P_E = R_T V.
  • J = V_A/(nD), K_T = T/(ρn²D⁴), K_Q = Q/(ρn²D⁵), η₀ = JK_T/(2πK_Q); η_H = (1 − t)/(1 − w); η_D = η_H η₀ η_R = P_E/P_D.

Practice questions (23)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A ship has L = 100 m, B = 16 m, draught T = 6 m and block coefficient 0.70, and floats in sea water of density 1025 kg/m³. Its displacement (in tonnes), to the nearest integer, is ____.

    Numerical answer — type the value.

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    Answer: 6888

    ∇ = C_B LBT = 0.70 × 100 × 16 × 6 = 6720 m³; Δ = ρ∇ = 1.025 × 6720 = 6888 t. Stopping at the volume gives 6720; taking the box volume without C_B gives 9840 t.
  2. A ship floating in sea water of density 1025 kg/m³ has a waterplane area of 1400 m². Its tonnes per centimetre immersion (TPC), to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 14.35

    TPC = ρA_W/100 = 1.025 × 1400/100 = 14.35 t/cm: a 1 cm layer of the waterplane holds 14 m³ of sea water. Using fresh water gives 14.00.
  3. A ship of 100 m length and 6888 t displacement in sea water of density 1025 kg/m³ has a longitudinal metacentric height of 120 m. Its moment to change trim by one centimetre (in t·m/cm), to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 82.66

    MCT1cm = Δ·GM_L/(100L) = 6888 × 120/(100 × 100) = 82.656 ≈ 82.66 t·m/cm. Moving 50 t through 40 m would then trim the ship 2000/82.66 = 24.2 cm. Omitting the 100 gives a moment per metre of trim.
  4. Half-breadths of a waterplane at five equally spaced stations 10 m apart are 0, 4, 6, 4 and 0 m. By Simpson’s first rule, the waterplane area (in m²), to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 293.3

    Simpson multipliers 1, 4, 2, 4, 1: sum = 0 + 16 + 12 + 16 + 0 = 44; half-area = (10/3) × 44 = 146.67 m²; the full waterplane is twice that, 293.3 m². Forgetting to double gives 146.7.
  5. A box-shaped barge 60 m long and 12 m wide floats at a draught of 4 m in sea water of density 1025 kg/m³, with KG = 4.5 m. Its transverse metacentric height (in m), to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 0.5

    KB = T/2 = 2 m; BM = I/∇ = (LB³/12)/(LBT) = B²/(12T) = 144/48 = 3 m; GM = KB + BM − KG = 2 + 3 − 4.5 = 0.5 m. The density does not enter: BM is a ratio of volumes. Forgetting KB gives −1.5 m.
  6. A ship of 6888 t displacement in sea water of density 1025 kg/m³ has a rectangular slack tank 10 m long and 8 m wide containing oil of density 900 kg/m³. The free-surface correction to GM (in m), to three decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 0.056

    i = lb³/12 = 10 × 512/12 = 426.7 m⁴; FSC = ρ_l i/Δ = 0.9 × 426.7/6888 = 0.0557 ≈ 0.056 m. Using lb³/12 with the length cubed (8 × 1000/12) gives 0.087; leaving out the oil density gives 0.062.
  7. In an inclining experiment on a ship of 5000 t displacement in sea water of density 1025 kg/m³, a mass of 10 t is moved 8 m across the deck and a pendulum 6 m long deflects 0.12 m. The metacentric height (in m), to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 0.8

    tan φ = 0.12/6 = 0.02; GM = wd/(Δ tan φ) = 10 × 8/(5000 × 0.02) = 80/100 = 0.8 m. Dividing by the deflection instead of tan φ gives 0.13 m.
  8. A ship of 5000 t displacement has KG = 6.0 m. A mass of 200 t is loaded with its centre of gravity 10 m above the keel. The new KG (in m), to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 6.15

    KG₁ = (5000 × 6 + 200 × 10)/(5000 + 200) = 32 000/5200 = 6.154 ≈ 6.15 m: G rises towards the added mass. Dividing by 5000 instead of 5200 gives 6.40.
  9. A heavy lift is raised off the quay by the ship’s own derrick. As soon as it leaves the quay, the ship’s centre of gravity behaves as though the lift were located at

    1. the head of the derrick, its point of suspension
    2. its own centre of gravity
    3. the ship’s centre of buoyancy
    4. the ship’s keel
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    Answer: A — the head of the derrick, its point of suspension

    A suspended mass swings to hang vertically below its point of suspension whenever the ship heels, so its effective centre is the suspension point. The rise of G is w × (height of derrick head − original Kg)/Δ, and it occurs the moment the load is lifted.
  10. A ship has an initial metacentric height of −0.1 m and a transverse BM of 5 m, and is wall-sided over the range considered. Its angle of loll (in degrees), to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 11.3

    Setting the wall-sided GZ = sin φ(GM + ½BM tan²φ) to zero gives tan φ = √(2|GM|/BM) = √(0.2/5) = √0.04 = 0.2, so φ = 11.3°. Forgetting the 2 gives tan φ = 0.141 and 8.0°.
  11. A wall-sided ship has GM = 1.0 m and BM = 4.0 m. Its righting lever GZ at 20° of heel (in m), to three decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 0.433

    GZ = sin φ(GM + ½BM tan²φ) = 0.3420 × (1 + 0.5 × 4 × 0.1325) = 0.3420 × 1.2650 = 0.433 m. The small-angle formula GM sin φ gives only 0.342 m, understating the lever because it ignores the wall-sided gain.
  12. A ship of 5000 t displacement enters dry dock with KM = 8 m. At a certain instant the upthrust on the keel blocks is 100 t. The virtual loss of metacentric height, taken as P·KM/Δ (in m), to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 0.16

    MM₁ = P·KM/Δ = 100 × 8/5000 = 0.16 m. The loss grows as P grows with the falling water, so the critical instant is just before the ship takes the blocks along its length.
  13. Which statements about the curve of statical stability and damage stability are correct?

    1. the slope of the GZ curve at the origin equals GM per radian
    2. the area under the GZ curve measures dynamical stability
    3. in the probabilistic method the attained index A must be at least the required index R
    4. the lost-buoyancy and added-weight methods give different final waterlines
    Show answer

    Answer: A — the slope of the GZ curve at the origin equals GM per radian; B — the area under the GZ curve measures dynamical stability; C — in the probabilistic method the attained index A must be at least the required index R

    GZ ≈ GM φ near the origin, so the tangent at φ = 1 rad reaches GM; the area Δ∫GZ dφ is the work to heel the ship; A ≥ R is the probabilistic criterion. The two flooding methods give the same final waterline — they differ only in how GM is stated, because one keeps Δ and the other adds the flood water to it.
  14. A ship 100 m long runs at 10 m/s. Taking g = 9.81 m/s², its Froude number, to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 0.32

    Fn = V/√(gL) = 10/√981 = 10/31.32 = 0.319 ≈ 0.32, the speed range of a fast cargo ship, where wave-making resistance is large. Using √(L) alone without g gives 1.0.
  15. A 4 m model of a 100 m ship (scale ratio 25) is towed at 2 m/s in water of kinematic viscosity 1.0 × 10⁻⁶ m²/s. By the ITTC-1957 line C_F = 0.075/(log₁₀Re − 2)², the model’s frictional resistance coefficient, to five decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 0.00312

    Re = VL/ν = 2 × 4/10⁻⁶ = 8 × 10⁶; log₁₀Re = 6.903; C_F = 0.075/(4.903)² = 0.075/24.04 = 0.003120. Using the natural logarithm (15.9) gives 0.00039, far too small.
  16. A 4 m model of a 100 m ship (scale 25, wetted surface 4 m², towed at the corresponding speed of 2 m/s in water of ν_m = 1.0 × 10⁻⁶ m²/s) has a measured total resistance coefficient of 0.00520. The 100 m ship runs in sea water of density 1025 kg/m³ and ν_s = 1.2 × 10⁻⁶ m²/s. Using the ITTC-1957 line for both and no correlation allowance, the ship’s effective power (in MW), to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 4.7

    V_s = 2√25 = 10 m/s; C_R = 0.00520 − 0.00312 = 0.00208; Re_s = 10 × 100/1.2 × 10⁻⁶ = 8.33 × 10⁸, C_Fs = 0.075/(6.921)² = 0.001566; C_Ts = 0.003646; S_s = 25² × 4 = 2500 m²; R_T = ½ × 1025 × 2500 × 10² × 0.003646 = 467 kN; P_E = R_T V = 4.67 MW ≈ 4.7 MW. Carrying C_Tm = 0.0052 straight to the ship gives 6.7 MW.
  17. A ship’s total resistance at 15 knots is 300 kN. Taking 1 knot = 0.5144 m/s, its effective power (in kW), to the nearest integer, is ____.

    Numerical answer — type the value.

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    Answer: 2315

    V = 15 × 0.5144 = 7.716 m/s; P_E = R_T V = 300 × 7.716 = 2315 kW. With a propulsive efficiency of 0.70 the delivered power would be 3307 kW. Multiplying by 15 without converting knots gives 4500.
  18. Which statements about ship resistance and its prediction are correct?

    1. the form factor (1 + k) is found from low-speed model tests where wave resistance is negligible
    2. blockage in a narrow towing tank makes the measured resistance too high
    3. wave-making resistance dominates for a slow, full-bodied tanker
    4. hull roughness and fouling increase frictional resistance
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    Answer: A — the form factor (1 + k) is found from low-speed model tests where wave resistance is negligible; B — blockage in a narrow towing tank makes the measured resistance too high; D — hull roughness and fouling increase frictional resistance

    At low Fn wave resistance vanishes and C_T/C_F tends to 1 + k (Prohaska); blockage accelerates the flow past the model; roughness and fouling thicken the boundary layer. A slow, full tanker at low Fn is dominated by friction and viscous pressure, not by waves.
  19. A propeller of 5 m diameter turns at 2 rev/s with a speed of advance of 6 m/s. At this advance coefficient K_T = 0.20 and K_Q = 0.030. Its open-water efficiency, to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 0.64

    J = V_A/(nD) = 6/(2 × 5) = 0.6; η₀ = (J/2π)(K_T/K_Q) = (0.6/6.283)(0.2/0.03) = 0.0955 × 6.667 = 0.637 ≈ 0.64. Omitting the 2π gives 4.0, which is impossible for an efficiency.
  20. A propeller of 5 m diameter turns at 2 rev/s at an advance coefficient where K_T = 0.20, working in sea water of density 1025 kg/m³. Its thrust (in kN), to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 512.5

    T = K_T ρ n² D⁴ = 0.20 × 1025 × 4 × 625 = 512 500 N = 512.5 kN. Using D⁵ (the torque scaling) gives 2562.5 kN; using n instead of n² gives 256.3 kN.
  21. A ship has a Taylor wake fraction of 0.20 and a thrust-deduction fraction of 0.16. Its hull efficiency, to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 1.05

    η_H = (1 − t)/(1 − w) = 0.84/0.80 = 1.05. It exceeds 1 because the propeller works in slowed wake water; inverting the ratio gives 0.95.
  22. Which statements about propellers are correct?

    1. a controllable-pitch propeller allows the ship to be reversed without reversing the engine
    2. by actuator-disc theory, a more heavily loaded propeller has a higher ideal efficiency
    3. a ducted propeller gives extra thrust at high loading such as bollard pull
    4. nickel–aluminium bronze is used because it resists corrosion and cavitation erosion
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    Answer: A — a controllable-pitch propeller allows the ship to be reversed without reversing the engine; C — a ducted propeller gives extra thrust at high loading such as bollard pull; D — nickel–aluminium bronze is used because it resists corrosion and cavitation erosion

    CPP reverses by pitch; the Kort nozzle adds duct thrust when loading is high; NAB resists corrosion and erosion. η_i = 2/(1 + √(1 + C_T)) FALLS as the loading C_T rises, which is why large, slowly turning propellers are efficient.
  23. Cavitation that appears on the suction side of a propeller blade near its leading edge, as a thin attached cavity, is called

    1. sheet cavitation
    2. face cavitation
    3. hub-vortex cavitation
    4. cloud cavitation at the trailing edge only
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    Answer: A — sheet cavitation

    Sheet cavitation forms at the suction peak near the leading edge on the back of the blade. Face cavitation is on the pressure side at negative angle of attack; hub-vortex cavitation lies in the vortex behind the boss; cloud cavitation is the unstable break-up of a sheet, not a leading-edge form.