Question

(a) What factors influence the frictional resistance of a ship and what formula is used to calculate this resistance. (6)

(b) A ship of 12000 tonne displacement has a rudder 15m² in area, whose centre is 5m below the waterline. The metacentric height of the ship is 0.3m and the centre of buoyancy is 3.3m below the waterline. When travelling at 20 knots the rudder is turned through 30°. Find the initial angle of heel if the force Fn perpendicular to the plane of the rudder is given by: Fn = 577 Av² sinα N

Allow 20% for the race effect. (10)

Answer 1

Part (a)

Factors influencing frictional resistance, and the formula used.

Frictional (skin) resistance is caused by the shearing of water over the hull surface. The important factors are:

Speed (V): within normal service ranges frictional resistance rises approximately as V^1.8 to V^1.85, so speed strongly increases resistance.
Wetted surface area (S): the larger the wetted hull area the higher the resistance; it depends on length, breadth, draught, displacement and fullness of the ends.
Surface roughness and fouling: marine growth (barnacles, slime), corrosion pitting, weld beads, protruding plating and deteriorated antifouling all raise the friction coefficient sharply. Regular hull cleaning and good antifouling coatings reduce this.
Density of the water: resistance is proportional to density, so it is higher in sea water than in fresh water.
Viscosity and temperature: cooler, more viscous water gives a slightly higher frictional coefficient.
Hull length: the frictional coefficient varies with the length and Reynolds number.

The formula used is the frictional resistance line. In Admiralty-style form:

Rf = f and S and V^n

or the simplified relation R = 0.45 S V^1.83 (newtons)

Where f is the frictional coefficient, S the wetted surface area (m2) and V the ship speed. Modern practice uses the ITTC 1957 friction line, Rf = 0.5 Cf rho S V^2, where Cf = 0.075/(log10 Re - 2)^2 and Re is the Reynolds number; this permits accurate model-ship scaling.

Part (b)

Rudder heeling angle.

Data: displacement Delta=12000 t; rudder area A=15 m2; rudder centre 5 m below the waterline; GM=0.3 m; centre of buoyancy 3.3 m below the waterline; speed 20 knots; helm angle alpha=30 deg; apply 20% race allowance.

Speed converted: v=20 x 0.5144 = 10.29 m/s, so v2=105.84.

Normal rudder force Fn=577 A v2 sin(alpha) = 577 x 15 x 105.84 x 0.5 = 458,035 N.

Race effect (propeller race increases water velocity over the rudder) 20%: F=1.2 x 458,035 = 549,642 N.

The heeling couple is the force times its lever about the centre of lateral resistance, taken as the vertical distance of the rudder centre below the waterline, lever=5 m:

Heeling moment = 549,642 x 5 = 2,748,210 Nm.

Balancing against the ship's righting moment (displacement as a weight, W=12000 tonne-force):

tan(theta) = (F x lever)/(W x GM) = (549,642 x 5)/(12000 x 9810 x 0.3) = 2,748,210/35,316,000 = 0.07782.

Hence theta = atan(0.07782) = 4.45 deg.

Answer: the initial angle of heel is about 4.5 deg towards the rudder (outboard) side.

✓ Verified Model Answer (Text Solution)

Factors influencing frictional resistance, and the formula used.

Rudder heeling angle.