A ship of 12000 tonne displacement has a rudder 15 m2 in area, whose centre is 5 m below the waterline. The meta centric height of the ship is 0.3 m and the centre of buoyancy is 3.3 m 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: Fa=577 A v2 sin𝜶 N, Allow 20% for the race effect.
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Given
Δ = 12000 t
A = 15 m2
GM = 0.3 m
COB = 3.3 m below water line
V = 20 knots
α = 300
To find
1. Initial angle of heel , θ
Solution
Consider that rudder is turned by an angle α and let
Fn = Rudder force normal to the plane of rudder
Ft = Transverse rudder force
Consider Triangle ABD
Sin α = Fn / F
Fn = F sin α --------------------------(1)
Consider Triangle ABC
Cos α = Ft / Fn
Ft = Fn cos α -------------------------(2)
Substituting Fn value from (1)
Ft = F cos α sin α
But in the question above it is given that Rudder force Fn = 577 A x V2 x Sin α x N
So substituting in (2) ---------------->
Ft = 577 A x v2 x Sin α x cos α N
race is the additional flow past the rudder that is cause by the propeller turning at a faster speed than the vessel is moving through the water
Consider that race effect is 20%. This will affect ship speed.
So ship speed = V x 1852/3600
Considering race effect
Ship speed = 1.2 x V x 1852/3600
= 1.2 x 20 x 1852/3600
v = 12.35 knots
Ft = 577 A x v2 x Sin α x cos α N
= 577 x 15 x 12.352 x sin 30 x cos 30 N
= 571.61 kN
NL = Distance from COB to centre of the rudder
= 5 - 3.3
= 1.7 m
Heeling moment = Ft x NL x cos θ ----------------------------(3)
= 571.61 x 1.7 x cos θ
= 971.7 x cos θ kN m
Righting moment = Δ x g x GM x sin θ -----------------------(4)
When heel is steady
Heeling moment = Righting moment
Ft x NL x cos θ = Δ x g x GM x sin θ
971.7 x cos θ = 12000 x 9.81 x 0.3 x sin θ
971.7 x cos θ = 35316 x sin θ
sin θ / cos θ = 0.0275
tan θ = 0.0275
Angle of heel θ = 10 571
