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The major points discuss in these assignment notes are: Definitive Maneuvers, Zigzag Maneuver, Schematic Diagram, Maneuverability, Maneuvering Equations, Sway and Yaw, Rudder Working, Steady Turning Radius, Independent of Forward Speed, Rudder Derivatives
Typology: Exercises
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Topic : Definitive Maneuvers
Question 1
Draw a schematic diagram of a 20/10 zigzag maneuver and identify all the terms.
Answer:
The figure is shown below and the terms like (time of) 1
st ,
nd ,
rd executes, (time to)
reach, period, 1
st and 2
nd overshoot etc. are explained on the diagram.
Question 2
(i) The sway and yaw maneuvering equations with rudder working are as follows:
v v r r
v v r Z r
Y v m Y v Y mU r Y r Y
N v N v N r I N r N
δ
δ
& &
& &
From this, show that the steady turning radius for a given ship at a given rudder
angle is independent of forward speed but its yaw rate is in proportion to forward
speed.
(ii) For a ship of L =110m, B =18m, T =4.1m, CB =0.68, the hydrodynamic and
rudder derivatives are as follows:
' 3 ' 3 ' 3 ' 3
' 3 ' 3
v r v r
δ δ
− − − −
− −
Find its turning radius, drift angle and yaw rate for 16 knots at 35 deg. rudder.
Answer:
(i)
During steady turning phase, the time derivatives are all zero, and therefore the
sway and yaw equations reduces to (obtained by making v &^ = r &= 0 ) :
v (^ r )
v r
Y v Y mU r Y
N v N r N
δ
δ
In non-dimensional form:
' ' ' ' ' '
' ' ' ' '
v r
v r
Y v Y m r Y
N v N r N
δ
δ
From this we get by solving for
' v and
' r :
' ' ' ' ' ' ' ' ' ' '
r r
v r v r
N Y m Y N v Y N N Y m
δ δ δ
' ' ' ' ' ' ' ' ' ' ( )
v v
v r v r
r Y N N Y m
δ δ δ
We have (from definition of non-dimensionalization),
' r = rL / V. Also, V = rR.
Thus,
' r = rL / V = rL / rR = L / R
Question 3
(i)
Find an approximate expression for heel angle during steady turn.
(ii)
A vessel turns in a radius of 300 m. at a speed of 16 knots under the action of rudder
force of 2 MN. If the draft of the ship if 8m, KG is 9m and GM is 3m, find the
approximate agle of heel during the steady turn
Answer:
(i)
During steady turn, let Fh and Fr are the hull and rudder forces respectively.
Referring to the figure below, by equating forces, we have
2
h r
Rg
− = where ∆ is
mass, R is turning radius, and V is speed during turn.
If Fh and Fr acts at the points H and E respectively as shown in the diagram, the
heeling moment is given by
h r r h
h r r
h r r
( K represent point on keel)
For most ships, both E and H are approximately same and at the half draft level.
With this approximation, heeling moment becomes ( )( ) h r
If the heel angle is α , we have
2
sin ( (^) h r )( ) ( )
Rg
2 V GE
Rg GM
(ii)
For this problem, V = 16 knots, and R = 450 m.
Assuming that the rudder force acts at half draft level, we have
GE = KG − T / 2 = 9 − 8 / 2 = 5 m
Thus,
2 (16 0.5144) 5 0.0384 2.2 deg. (300)(9.8) 3
