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PE Naval Architecture practice questions

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Question 1Hydrostatics and Stability

A vessel floating in seawater of density 1.025 t/m³ has a displacement of 8000 t. What is its volume of displacement?

  1. 7620 m³
  2. 8000 m³
  3. 8200 m³
  4. 7805 m³ Correct answer.

Explanation

A floating body displaces a volume of water whose mass equals its own mass, so volume of displacement equals displacement divided by water density. Here 8000 ÷ 1.025 = 7804.9 m³, rounded to 7805 m³.

Question 2Hydrostatics and Stability

Which statement best defines the reserve buoyancy of a ship?

  1. The watertight volume of the hull below the waterline
  2. The total volume of the hull measured to the freeboard deck
  3. The watertight volume of the hull above the waterline Correct answer.
  4. The displacement divided by the density of the water

Explanation

Reserve buoyancy is the intact watertight volume above the waterline that can be immersed before the ship loses its buoyancy. It is what allows a ship to survive added weight or flooding.

Question 3Hydrostatics and Stability

A box-shaped pontoon floats on an even keel at draught d. Its centre of buoyancy lies at what height above the keel?

  1. One half of the draught Correct answer.
  2. Twice the draught
  3. A height equal to the moulded depth
  4. One third of the draught

Explanation

The centre of buoyancy is the centroid of the immersed volume. For a box-shaped hull the immersed volume is a rectangular prism of uniform section, so its centroid is at mid-draught, d/2.

Question 4Hydrostatics and Stability

The initial metacentric height of a ship is obtained from which expression?

  1. KM minus KG Correct answer.
  2. KM plus KG
  3. KG minus KM
  4. KB minus BM

Explanation

Metacentric height is the distance from the centre of gravity up to the metacentre, so GM = KM − KG, where KM = KB + BM. A positive value indicates initial stability.

Question 5Hydrostatics and Stability

A vessel has a negative initial metacentric height. How will it behave?

  1. Stable, returning to an even keel
  2. Unstable, and it will loll to a heel Correct answer.
  3. Neutral, holding any angle of heel
  4. Stable, but only at large angles of heel

Explanation

Negative GM means the centre of gravity lies above the transverse metacentre, so any small heel produces an upsetting moment. The vessel heels until a positive righting lever develops, which is the angle of loll.

Question 6Hydrostatics and Stability

The block coefficient of a ship is defined as the ratio of which two quantities?

  1. Waterplane area to the area of its enclosing rectangle
  2. Immersed volume to the volume of the enclosing block Correct answer.
  3. Midship section area to the area of its enclosing rectangle
  4. Displacement to the volume of displacement

Explanation

The block coefficient compares the immersed volume with the volume of the rectangular block formed by length, breadth and draught, so CB = ∇/(L·B·d). It is a measure of hull fullness.

Question 7Hydrostatics and Stability

At small angles of heel, the righting lever GZ is approximately given by which expression?

  1. GM multiplied by the sine of the heel angle Correct answer.
  2. GM divided by the angle of heel
  3. KB multiplied by the sine of the heel angle
  4. BM multiplied by the cosine of the heel angle

Explanation

For small angles the metacentre can be treated as fixed, so the righting lever is GZ = GM·sinθ. This is why GM alone governs initial stability.

Question 8Hydrostatics and Stability

What is the effect of a free liquid surface in a partly filled tank on a ship's stability?

  1. It reduces the effective metacentric height Correct answer.
  2. It increases the effective metacentric height
  3. It raises the centre of buoyancy
  4. It has no effect on the vessel's stability

Explanation

Liquid that can flow across a tank shifts its centre of gravity towards the low side as the ship heels, producing an upsetting moment. This is treated as a virtual rise of the ship's centre of gravity, so the effective GM is reduced.

Question 9Hydrostatics and Stability

A box-shaped vessel has a waterplane area of 1200 m² and floats in seawater of density 1.025 t/m³. What is its tonnes per centimetre immersion?

  1. 12.0 t/cm
  2. 1.23 t/cm
  3. 12.3 t/cm Correct answer.
  4. 123 t/cm

Explanation

TPC is the mass of water displaced by a 1 cm increase in draught, so TPC = A_w × ρ ÷ 100. Here 1200 × 1.025 ÷ 100 = 12.3 t/cm.

Question 10Hydrostatics and Stability

The transverse metacentric radius BM of a vessel is obtained by which relationship?

  1. Transverse waterplane inertia divided by the volume of displacement Correct answer.
  2. Volume of displacement divided by waterplane inertia
  3. Waterplane area divided by the volume of displacement
  4. KB plus the transverse waterplane inertia

Explanation

The metacentric radius is the distance from the centre of buoyancy to the metacentre, given by BM = I/V. A wide waterplane gives a large second moment and therefore a large BM.

Question 11Hydrostatics and Stability

The purpose of an inclining experiment on a new vessel is to determine which quantities?

  1. The waterplane area and the TPC
  2. The lightship KG and displacement Correct answer.
  3. The height of the centre of buoyancy
  4. The transverse metacentric radius BM

Explanation

Known weights are shifted transversely and the resulting small heel is measured, allowing KG to be calculated from tanθ = w·d/(Δ·GM). The test also establishes the lightship displacement and hence the lightship KG used for all later stability calculations.

Question 12Hydrostatics and Stability

A vessel with negative initial GM is heeled by a small external moment and then released. What happens?

  1. It heels slightly and then returns upright
  2. It remains upright in neutral equilibrium
  3. It capsizes immediately at the slightest disturbance
  4. It lolls to the angle where GZ becomes zero Correct answer.

Explanation

With negative GM the upright is unstable, so the vessel heels until the righting lever has grown back to zero. It then rests at the angle of loll in unstable equilibrium, able to fall to either side.

Question 13Hydrostatics and Stability

The free surface correction applied to a vessel's metacentric height is obtained by which expression?

  1. Free surface moment divided by the displacement Correct answer.
  2. Free surface moment divided by waterplane inertia
  3. Displacement divided by the free surface moment
  4. Free surface moment multiplied by the displacement

Explanation

The free surface moment is the product of liquid density and the second moment of the tank's free surface. Dividing by the ship's displacement converts this moment into an equivalent rise of the centre of gravity, which is subtracted from GM.

Question 14Hydrostatics and Stability

A weight w is shifted a distance d along the length of a vessel. The change of trim is given by which expression?

  1. w × d × MCT1cm
  2. w × d ÷ MCT1cm Correct answer.
  3. w ÷ (d × MCT1cm)
  4. MCT1cm ÷ (w × d)

Explanation

The trimming moment is w × d, and dividing it by the moment to change trim one centimetre converts that moment into a change of trim in centimetres. The units must be consistent, with w in tonnes, d in metres and MCT1cm in tonne-metres per centimetre.

Question 15Hydrostatics and Stability

A weight is added forward of the centre of flotation of a vessel floating freely. What is the result?

  1. It sinks bodily without any change of trim
  2. It trims by the bow and increases the mean draught Correct answer.
  3. It trims by the bow and reduces the mean draught
  4. It trims by the stern only

Explanation

Any added weight causes a parallel sinkage equal to w/TPC, increasing the mean draught. Because the weight acts forward of the centre of flotation, it also creates a trimming moment that brings the bow down.

Question 16Hydrostatics and Stability

A vessel moves from seawater of density 1.025 t/m³ into fresh water of density 1.000 t/m³ with no change in displacement. What happens to its draught?

  1. It decreases, because fresh water is less dense
  2. It stays unchanged, because the displacement is constant
  3. It increases, because a larger volume must be displaced Correct answer.
  4. It rises briefly and then returns to its original value

Explanation

The ship must always displace its own mass of water, so in the less dense fresh water a greater volume is needed. The extra volume is gained by sinking deeper, so the draught increases.

Question 17Hydrostatics and Stability

A vessel of 10 000 t displacement has a TPC of 20 t/cm in seawater. What is its fresh water allowance, given by Δ/(4 × TPC)?

  1. 250 mm
  2. 125 mm Correct answer.
  3. 63 mm
  4. 500 mm

Explanation

The fresh water allowance is the increase in draught when moving from seawater to fresh water, computed as displacement divided by four times TPC in millimetres. Here 10 000 ÷ (4 × 20) = 125 mm.

Question 18Hydrostatics and Stability

For a rectangular waterplane of length L and breadth B, what is the second moment of area about a transverse axis through its centroid?

  1. B L³ / 12 Correct answer.
  2. L B³ / 12
  3. B L³ / 3
  4. L B³ / 3

Explanation

About a transverse axis the breadth B is the dimension perpendicular to the axis and the length L is measured along it, giving I = B·L³/12. This large longitudinal second moment is why a ship is much stiffer in trim than in heel.

Question 19Hydrostatics and Stability

A weight of 50 t is shifted 10 m transversely on a vessel of 5000 t displacement with a metacentric height of 1.0 m. What is the approximate resulting angle of heel?

  1. 5.7° Correct answer.
  2. 2.9°
  3. 11.3°
  4. 0.6°

Explanation

The heel from a transverse shift is given by tanθ = w·d/(Δ·GM). Here tanθ = 500/(5000 × 1.0) = 0.1, so the heel is about 5.7°.

Question 20Hydrostatics and Stability

For a wall-sided vessel, which expression gives the righting lever at a large angle of heel θ?

  1. GM tanθ + ½ BM sin²θ
  2. sinθ (GM + ½ BM tan²θ) Correct answer.
  3. GM sinθ + BM sin²θ
  4. GM cosθ + ½ BM tanθ

Explanation

The wall-sided formula recognises that the metacentre rises as the vessel heels, adding a term proportional to tan²θ. It reduces to GM·sinθ at small angles, as it must.

Question 21Hydrostatics and Stability

A weight resting on deck is lifted clear of the deck by a crane on the same vessel. What is the immediate effect on stability?

  1. The centre of buoyancy rises to the crane head
  2. GM is unchanged, because the weight stays on board
  3. GM increases, because the weight is raised
  4. The centre of gravity rises to the crane head Correct answer.

Explanation

Once the weight hangs freely, its whole mass acts at the point of suspension, so the vessel's centre of gravity moves vertically towards the crane head. The resulting rise in KG reduces GM for as long as the weight is suspended.

Question 22Hydrostatics and Stability

A vessel's GZ curve starts at zero on an even keel, becomes negative, and crosses the axis at 12° before rising to a maximum at 40°. What does this tell you?

  1. The vessel's range of stability is 12°
  2. The vessel will capsize beyond 40°
  3. The maximum righting lever occurs at 12°
  4. The vessel is initially unstable and will loll to about 12° Correct answer.

Explanation

A negative slope at the origin means negative GM, so the upright is unstable and the ship will settle at the first zero crossing, the angle of loll. Beyond that angle the levers are positive, so the vessel is stable up to the eventual vanishing angle.

Question 23Hydrostatics and Stability

A rectangular bulkhead is 6 m wide and extends 8 m from the tank top to its upper edge, and is flooded with fresh water over that full depth. Taking ρ = 1000 kg/m³ and g = 9.81 m/s², what is the total hydrostatic thrust on it?

  1. 1.88 MN Correct answer.
  2. 1.41 MN
  3. 3.77 MN
  4. 941 kN

Explanation

The thrust equals pressure at the centroid multiplied by the area, P = ρ·g·A·ȳ. The area is 6 × 8 = 48 m² and the centroid is 4 m below the surface, so P = 1000 × 9.81 × 48 × 4 ≈ 1.88 MN.

Question 24Hydrostatics and Stability

For a vessel heeled to a large angle, which statement about the transverse metacentre is correct?

  1. It is not a fixed point; GZ comes from the true hull form Correct answer.
  2. It moves along the locus of the centre of buoyancy
  3. It stays fixed relative to the keel at every angle
  4. It coincides with the centre of gravity at the loll angle

Explanation

The fixed metacentre is an artifice valid only for small angles, where successive buoyancy normals intersect at a nearly constant point. At large angles the intersection migrates, so righting levers must be found from the actual immersed shape, for example by integration of the hull form.

Question 25Hydrostatics and Stability

Using Simpson's first rule with ordinates 0 m, 5 m, 6 m, 5 m and 0 m spaced 2 m apart, what is the enclosed waterplane area?

  1. 26.0 m²
  2. 34.7 m² Correct answer.
  3. 17.3 m²
  4. 52.0 m²

Explanation

Simpson's first rule weights end ordinates by one, alternate ordinates by four and the remainder by two, then multiplies by the common spacing divided by three. Here (2/3)[0 + 4(5) + 2(6) + 4(5) + 0] = (2/3)(52) = 34.7 m².

Question 26Hydrostatics and Stability

How is the reserve buoyancy of a ship best described?

  1. The mass of cargo that may still be loaded before the load line is reached
  2. The enclosed watertight volume above the waterline, usually expressed as a percentage of the total volume Correct answer.
  3. The volume of the hull occupied by machinery, stores and permanent ballast
  4. The volume of the hull below the waterline, usually expressed as a percentage of the total volume

Explanation

Reserve buoyancy is the watertight volume of the hull above the waterline, and it is what allows the ship to take on additional weight before being overwhelmed. It is normally quoted as a percentage of the total enclosed volume, so a full hull form with little sheer has less of it.

Question 27Hydrostatics and Stability

Which statement correctly distinguishes heel from list?

  1. Heel is produced by a shift of weight within the ship; list is produced by an external moment such as wind.
  2. Heel occurs only at large angles of inclination; list occurs only at small angles of inclination.
  3. Heel is measured in metres as a linear displacement; list is measured in degrees of inclination.
  4. Heel is a temporary inclination caused by an external moment; list is a permanent inclination caused by an internal weight shift. Correct answer.

Explanation

Heel is the inclination produced by an external heeling moment such as wind or turning, and the ship returns upright when the moment is removed. List is a steady inclination caused by an internal transverse shift or eccentric addition of weight, which persists with no external moment acting.

Question 28Hydrostatics and Stability

The midship section coefficient Cm of a ship is defined as which ratio?

  1. Waterplane area divided by the product of length and breadth
  2. Displaced volume divided by the product of length, breadth and draught
  3. Immersed midship area divided by the product of maximum breadth and draught Correct answer.
  4. Immersed midship area divided by the product of waterline breadth and length

Explanation

The midship section coefficient compares the actual immersed midship section area with the rectangle formed by the maximum breadth and the draught, so Cm = Am/(B x d). It describes how full the midship section is and always lies between zero and one.

Question 29Hydrostatics and Stability

A vessel has a transverse radius of gyration of 6 m and a natural rolling period of 10 s. Taking g = 9.81 m/s^2, what is its metacentric height?

  1. 1.15 m
  2. 1.75 m
  3. 1.45 m Correct answer.
  4. 2.05 m

Explanation

The rolling period is T = 2 x pi x k / sqrt(g x GM), so GM = 4 x pi^2 x k^2 / (g x T^2). Substituting k = 6 m and T = 10 s gives GM = (4 x 9.87 x 36)/(9.81 x 100), which is about 1.45 m.

Question 30Hydrostatics and Stability

A vessel of length between perpendiculars 100 m has a trim of 0.40 m by the stern, with its centre of flotation 2 m aft of midships. What is the resulting change in draught at the after perpendicular compared with the even-keel condition?

  1. An increase of about 0.40 m
  2. An increase of about 0.21 m Correct answer.
  3. A decrease of about 0.21 m
  4. An increase of about 0.19 m

Explanation

The draught change at a perpendicular equals the trim multiplied by the distance from the centre of flotation to that perpendicular, divided by the length between perpendiculars. The after perpendicular is 50 + 2 = 52 m from the centre of flotation, so the draught aft increases by 0.40 x 52/100 = 0.21 m.

Question 31Hydrostatics and Stability

A weight of 200 t is lowered 6 m within a vessel of 8000 t displacement whose initial metacentric height is 0.60 m. What is the new metacentric height?

  1. 0.75 m Correct answer.
  2. 1.05 m
  3. 0.90 m
  4. 0.45 m

Explanation

Moving a weight vertically shifts the centre of gravity by w x d / displacement, in this case 200 x 6 / 8000 = 0.15 m. Lowering the weight moves G downwards, which increases GM, so the new value is 0.60 + 0.15 = 0.75 m.

Question 32Hydrostatics and Stability

On a curve of statical stability, what does the area under the curve up to a given angle of heel represent?

  1. The righting moment acting at that particular angle of heel
  2. The dynamical stability, that is the work needed to heel the vessel to that angle Correct answer.
  3. The free surface correction applicable at that angle of heel
  4. The reserve of dynamical stability expressed in tonne-metres

Explanation

The ordinate of the GZ curve is the righting lever, so the area beneath it represents the integral of righting lever with angle, which is the dynamical stability or work required to incline the ship. It is commonly expressed in metre-radians or, when multiplied by displacement, in tonne-metre-radians.

Question 33Hydrostatics and Stability

A wall-sided vessel has a negative metacentric height of 0.15 m and a transverse metacentric radius BM of 4.0 m. At what angle of loll will it float in equilibrium?

  1. About 14.5 degrees
  2. About 11.0 degrees Correct answer.
  3. About 6.5 degrees
  4. About 8.2 degrees

Explanation

For a wall-sided vessel the angle of loll satisfies tan^2(theta) = -GM / BM when GM is negative. Here tan^2(theta) = 0.15/4.0 = 0.0375, so tan(theta) = 0.194 and theta is approximately 11 degrees.

Question 34Hydrostatics and Stability

A box-shaped vessel 60 m long and 10 m wide floats in fresh water at a draught of 3.0 m with its centre of gravity 4.0 m above the keel. What is its metacentric height?

  1. 0.28 m Correct answer.
  2. 0.78 m
  3. 2.78 m
  4. 1.28 m

Explanation

For a box-shaped vessel KB is half the draught, 1.5 m, and BM is B^2/(12 x d) = 100/36 = 2.78 m, giving KM = 4.28 m. Subtracting KG of 4.0 m gives GM = 0.28 m.

Question 35Resistance, Propulsion, and Powering

The Froude number used to characterise ship resistance is defined as:

  1. V/√(gL) Correct answer.
  2. √(gL)/V
  3. V/(gL)
  4. V²/(gL)

Explanation

The Froude number expresses the ratio of inertial to gravitational forces and is written as V/√(gL), where V is speed and L is a reference length such as length on the waterline. It is the similarity parameter that governs wave-making resistance and is held equal between model and ship.

Question 36Resistance, Propulsion, and Powering

The ITTC 1957 model-ship correlation line gives the skin friction coefficient as a function of Reynolds number in the form:

  1. 1.328/√Re
  2. 0.455/(log₁₀Re)^2.58
  3. 0.075/(log₁₀Re − 2)² Correct answer.
  4. 0.074/Re^0.2

Explanation

The ITTC 1957 line is Cf = 0.075/(log₁₀Re − 2)², adopted so that model and ship friction are treated by one internationally agreed formula. It is applied to the equivalent flat-plate wetted area of the hull.

Question 37Resistance, Propulsion, and Powering

At very low Froude numbers, below about 0.15, the dominant component of a ship's total calm-water resistance is:

  1. Frictional resistance Correct answer.
  2. Appendage resistance
  3. Air resistance
  4. Wave-making resistance

Explanation

At low speed the ship generates almost no wave system, so wave-making resistance becomes negligible. The resistance is then dominated by tangential shear stresses over the wetted surface, i.e. frictional resistance.

Question 38Resistance, Propulsion, and Powering

The real slip ratio of a propeller is given by:

  1. (P·n)/V_A
  2. (P·n − V_A)/(P·n) Correct answer.
  3. V_A/(P·n)
  4. (P·n + V_A)/(P·n)

Explanation

Real slip compares the theoretical advance per revolution, P·n, with the actual advance velocity V_A, so slip = (P·n − V_A)/(P·n). It expresses the fraction of pitch speed lost in giving the water a rearward velocity.

Question 39Resistance, Propulsion, and Powering

In terms of thrust deduction fraction t and wake fraction w, the hull efficiency is:

  1. (1 − w)/(1 − t)
  2. (1 − t)/(1 − w) Correct answer.
  3. (1 − t)(1 − w)
  4. 1 − t − w

Explanation

Hull efficiency is the ratio of the power to tow the hull to the power delivered into the wake, which reduces to (1 − t)/(1 − w). The thrust deduction raises the required thrust while the wake lowers the velocity the propeller sees.

Question 40Resistance, Propulsion, and Powering

The quasi-propulsive coefficient of a ship is obtained as:

  1. ηH × ηR
  2. ηO × ηR
  3. ηO × ηH
  4. ηO × ηH × ηR Correct answer.

Explanation

The quasi-propulsive coefficient is the product of open-water efficiency, hull efficiency and relative rotative efficiency. It converts effective power into delivered power at the propeller.

Question 41Resistance, Propulsion, and Powering

The thrust deduction fraction t is defined as:

  1. T/(T − R)
  2. R/T
  3. (T − R)/T Correct answer.
  4. (T + R)/T

Explanation

The propeller must supply more thrust than the resistance of the towed hull because its suction increases the hull resistance. The thrust deduction fraction is therefore (T − R)/T, where R is the hull resistance and T the required thrust.

Question 42Resistance, Propulsion, and Powering

The sea margin applied to trial-trip power when specifying the service power of a newbuilding is typically:

  1. 0–2%
  2. 15–20% Correct answer.
  3. 40–50%
  4. 5–8%

Explanation

A sea margin is added to the clean-hull trial power to cover fouling, weather and added resistance in service. It commonly lies in the range of about 15 to 20%.

Question 43Resistance, Propulsion, and Powering

According to the ITTC 1957 line, frictional resistance varies approximately as Vⁿ, where n is about:

  1. 3.0
  2. 1.0
  3. 1.8 Correct answer.
  4. 2.0

Explanation

The ITTC 1957 line gives a friction coefficient proportional to (log Re)^-2, which for practical purposes behaves like Re^-0.2. Since Re scales with V, frictional resistance, which is proportional to V²Cf, varies approximately as V^1.8.

Question 44Resistance, Propulsion, and Powering

A 1:25 scale model is tested by Froude scaling. The full-scale wave-making resistance is obtained by multiplying the model value by:

  1. 5
  2. 15625 Correct answer.
  3. 625
  4. 25

Explanation

Wave-making resistance scales with the cube of the linear scale ratio because it is a force and Froude similarity preserves force ratios equal to λ³. For λ = 25 the factor is 25³ = 15 625.

Question 45Resistance, Propulsion, and Powering

For fixed thrust deduction fraction, an increase in the wake fraction w causes the hull efficiency to:

  1. Decrease
  2. Become zero
  3. Increase Correct answer.
  4. Remain unchanged

Explanation

Hull efficiency is (1 − t)/(1 − w), so a larger wake fraction reduces the denominator and raises the efficiency. Physically, a stronger wake means the propeller works in slower water and produces the same thrust for less power.

Question 46Resistance, Propulsion, and Powering

A fixed-pitch propeller follows the propeller law, with power proportional to the cube of shaft speed. At 90% of the design revolutions the required power is about:

  1. 90%
  2. 95%
  3. 73% Correct answer.
  4. 81%

Explanation

Power varies as n³ under the propeller law, so 0.9³ = 0.729, i.e. about 73% of the design power. Speed through the water falls in proportion to n, while thrust falls with n².

Question 47Resistance, Propulsion, and Powering

A propeller of diameter 2.5 m turns at 2.0 rev/s with an advance velocity of 5.0 m/s. The advance coefficient J is:

  1. 0.5
  2. 2.5
  3. 1.0 Correct answer.
  4. 5.0

Explanation

The advance coefficient is J = V_A/(nD), the advance per revolution divided by diameter. Here 5.0/(2.0 × 2.5) = 1.0.

Question 48Resistance, Propulsion, and Powering

At a point on a propeller blade the local absolute pressure is 100 kPa, the vapour pressure is 3 kPa, the water density is 1000 kg/m³ and the local velocity is 5 m/s. The cavitation number is closest to:

  1. 0.13
  2. 3.9
  3. 7.8 Correct answer.
  4. 8.0

Explanation

The cavitation number is (p₀ − p_v)/(½ρV²). With p₀ − p_v = 97 000 Pa and ½ρV² = 12 500 Pa, the value is 7.8. Low values indicate a greater risk of cavitation.

Question 49Resistance, Propulsion, and Powering

A bulbous bow is fitted primarily in order to:

  1. Increase the buoyancy available forward
  2. Improve course keeping in a following sea
  3. Reduce wave-making resistance near the design speed Correct answer.
  4. Reduce frictional resistance over the hull

Explanation

The bulb generates its own wave system, which is tuned to be out of phase with the bow wave so that the two partly cancel. This lowers wave-making resistance over a range of Froude numbers around the design point.

Question 50Resistance, Propulsion, and Powering

In the Hughes–Prohaska method the form factor (1 + k) is found by:

  1. Assuming (1 + k) equals the prismatic coefficient
  2. Reading it directly from the ITTC 1957 line
  3. Extrapolating resistance data plotted as CT/CF against Fn to zero Froude number Correct answer.
  4. Measuring the model's wave pattern at the design speed

Explanation

Hughes–Prohaska plots the measured total coefficient divided by the flat-plate friction coefficient against a power of the Froude number. Extrapolating the low-speed trend back to Fn = 0 removes the wave component and leaves (1 + k).