Manoeuvring: Ferry in turning – rudder 5°
Determine the turning circle radius R for the ferry in Tables 6.1 and 6.2 at 3 kn speed with rudder at  = -5°.
Nonlinear coefficients can be neglected. R follows from the relation V = rR, where r   is the yaw rate.
Solution
The non-dimensional motion equations using only linear coefficients for steady turning are:
 Y ' 
D' u '    
 N  '
 Yr 'm' 
  Yv '

 Nv '  N r 'm' xG '
with D'  
We compute some necessary quantities: L    Lm  16  8.725  139.6 m
 Yr 'm'  (1901  6765)  10 6  4864  10 6
 N r ' m' xG '  (2579  116)  10 6  2463  10 6

  5 
 0.0873
180
12095 4864 v'
 3587 
v'  0.136
    
Thus: 
  0.0873     


 3919 2463 r '
 1621
r '  0.274 
The radius R of the turning circle is given by:
V  r  R  r '
V
L 139.6
R  R  
 510 m
L
r ' 0.274
Note that the turning radius does not depend on speed within this simplified model.
Manoeuvring: Ferry in turning – rudder 20°
Determine the turning circle radius R for the ferry in Tables 6.1 and 6.2 at 20 kn speed with rudder at  = 20°.
Nonlinear coefficients can be neglected. R follows from the relation V = rR, where r   is the yaw rate.
Solution
The non-dimensional motion equations using only linear coefficients for steady turning are:
 Y ' 
D' u '    
 N  '
 Yr 'm' 
  Yv '

 Nv '  N r 'm' xG '
with D'  
We compute some necessary quantities: L    Lm  16  8.725  139.6 m
 Yr 'm'  (1901  6765)  10 6  4864  10 6
 N r ' m' xG '  (2579  116)  10 6  2463  10 6

  20 
 0.349
180
12095 4864 v'
 3587 
v'  0.544 
    
Thus: 
  0.349     


 3919 2463 r '
 1621
r '  1.095
The turning rate is negative due to the different sign convention for rudder angle and yaw angle.
The radius R of the turning circle is given by:
R
L 139.6

 127.4 m
r ' 1.095
Note that the turning radius does not depend on speed within this simplified model.
Manoeuvring: Series 60 in turning circle
A side thruster is installed in the Series-60 ship in Tables 6.1 and 6.2. The thruster is installed 4 m aft of F.P.
and yields 300 kN side thrust at zero speed. The zero-speed efficiency of the thruster is  = 0.8. Which
turning rate and radius of turning circle are achieved for the ship with side thruster alone (rudder in central
position) for 2 kn speed? Assume linear correlation between hull forces and motions. R follows from the
relation V = rR, where r   is the yaw rate.
Forward speed changes the thrust of a bow thruster T from the value at zero speed T0:
Solution
The zero-speed thrust is corrected for forward speed reading from the diagram for V = 2 kn:
T
 0.8  T  0.8  T0  0.8  3.0  105  3.0  105 N
T0
The motion equations using only linear coefficients and steady turning with zero rudder angle are:
  T' 
D' u '  

T 'xt 
 Yr 'm' 
  Yv '

 Nv '  N r 'm' xG '
with D'  
We compute some necessary quantities: L    Lm  26  7.034  182.9 m
V = 2 kn = 2  0.5144 = 1.029 m/s
The ‘unit’ force (used to make forces non-dimensional) is:

1025
 1.0292  182.92  1.815  107 N
2
2
L
182.9
x 87.45
xt   4 m =
 4 m = 87.45 m  xt '  t 
 0.478
2
2
L 182.9
T
2.4  105
T ' xt '  13223  106  0.478  6321  106
T '

 13223  10 6 and
F0 1.815  107
F0 
 V 2  L2 
 Yr 'm'  (4330  11432)  106  7102  106
 N r ' m' xG '  (2900  57)  106  2957  106
16630 7102 v' 13223
v'  2.30
 6570 2957  r '   6321   r '   7.25 

   

  

V
1.029
 0.041 rad/s
Thus the turning rate is: r  r '  7.25 
L
182.9
L 182.9
 25 m
The radius R of the turning circle is given by: R  
r ' 7.25
Thus:
The ship is almost turning on the spot. The transverse velocity is that at the origin of the coordinate system,
i.e. amidships. The ship is turning around a point ahead of amidships, thus amidships is moving in the
opposite direction of the waterjet thrust.
Note: The Series-60 ship is in addition only very weakly yaw stable. Thus linear prediction fails already at
moderate yaw rates and transverse velocities. This inherent weakness increases as the forward speed
decreases. Thus in this case a nonlinear simulation would have been highly advisable.
Manoeuvring: Ferry in turning circle with thruster
Determine the turning circle radius R for the ferry in Tables 6.1 and 6.2 at 3 kn speed with rudder in central
position, using only its side thruster. The side thruster is located 5% L aft of F.P. It has 2 m diameter and
3000 kW power. The efficiency of the bow thruster is  = 0.8. Nonlinear coefficients can be neglected. R
follows from the relation V = rR, where r   is the yaw rate.
Forward speed changes the thrust of a bow thruster T from the value at zero speed T0:
Solution
The area of the bow thruster is
A 
D2
22
 
 3.14 m2
4
4
The thrust of the transverse bow thruster at zero speed follows from the relations:
P
1 
  A  v3
 2
 v3
P   2 3 3000  10 3  0.8  2

 11.42 m/s
A
1025  3.14
T0    A  v 2  1025  3.14  11.42 2  4.20  10 5 N
This is corrected for forward speed reading from the diagram for V = 3 kn:
T
 0.5  T  0.5  T0  0.5  4.20  10 5  2.10  10 5 N
T0
The motion equations using only linear coefficients and steady turning with zero rudder angle are:
  T' 
D' u '  

T 'xt 
 Yr 'm' 
  Yv '

 Nv '  N r 'm' xG '
with D'  
We compute some necessary quantities: L    Lm  16  8.725  139.6 m
V = 3 kn = 3  0.5144 = 1.543 m/s
The ‘unit’ force (used to make forces non-dimensional) is:
F0 

 V 2  L2 
1025
 1.543 2  139.6 2  2.379  10 7 N
2
2
x
xt '  t  0.45
L
T
2.10  10 5
T'

 8827  10 6
F0 2.379  10 7
and
T ' xt '  8827  10 6  0.45  3972  10 6
 Yr 'm'  (1901  6765)  10 6  4864  10 6
 N r ' m' xG '  (2579  116)  10 6  2463  10 6
Thus:
12095 4864 v' 8827
v' 0.226
 3919 2463  r '  3972  r '  1.254 

   

  

The radius R of the turning circle is given by:
R
L 139.6

 111 m
r ' 1.254
Manoeuvring: Containership in turning circle with rudder 5°
Determine the turning circle radius R for the container ship in Tables 6.1 and 6.2 at 15 kn speed with rudder
at  = -5°.
Nonlinear coefficients can be neglected. R follows from the relation V = rR, where r   is the yaw rate.
Solution
The non-dimensional motion equations using only linear coefficients for steady turning are:
 Y ' 
D' u '    
 N  '
 Yr 'm' 
  Yv '

 Nv '  N r 'm' xG '
with D'  
We compute some necessary quantities: L    Lm  34  8.029  273 m
 Yr 'm'  (2840  6399)  106  3559  106
 N r ' m' xG '  (1960  127)  106  1833  106

  5 
 0.0873
180
Thus:
8470 3559 v'
 1660 
v'  0.250
3800 1833   r '   753  0.0873  r '   0.553 

  


  

The radius R of the turning circle is given by:
R
L
273

 493 m
r ' 0.553
Manoeuvring: Containership in turning circle with thruster
The container ship in Tables 6.1 and 6.2 sails in a turning circle with rudder in centre position using just its
bow thruster at maximum power. The bow thruster is located 4 m aft of the forward perpendicular. The
power of the thruster is 4000 kW. The pipe diameter is 2.5 m. The ship speed is 5 kn. The efficiency of the
bow thruster is  = 0.8. Compute the radius of the turning circle R assuming linear correlation between hull
forces and motions.
R follows from the relation V = rR, where r   is the yaw rate.
Forward speed changes the thrust of a bow thruster T from the value at zero speed T0:
Solution
The area of the bow thruster is:
A 
D2
2.52
 
 4.91 m2
4
4
The thrust of the transverse bow thruster at zero speed follows from the relations:
P
1 
P   2 3 4000  103  0.8  2
  A  v3  v  3

 10.83 m/s
 2
A
1025  4.91
T0    A  v 2  1025  4.91  10.832  5.90  105 N
This is corrected for forward speed reading from the diagram for V = 5 kn:
T
 0.3  T  0.3  T0  0.3  5.90  105  1.77  105 N
T0
The non-dimensional linear motion equations for steady turning with zero rudder angle are:
  T' 
D' u '  

T 'xt 
 Yr 'm' 
  Yv '

 Nv '  N r 'm' xG '
with D'  
We compute some necessary quantities: L    Lm  34  8.029  273 m
V = 5 kn = 5  0.5144 = 2.572 m/s
The 'unit' force (used to make forces non-dimensional) is:

1025
 2.5722  2732  2.527  108 N
2
2
L
273
x 132.5
xt   4 m =
 4 m = 132.5 m  xt '  t 
 0.485
2
2
L
273
T
1.77  105
T ' xt '  700  106  0.485  340  106
T '

 700  10 6
and
F0 2.527  107
F0 
 V 2  L2 
 Yr 'm'  (2840  6399)  106  3559  106
 N r ' m' xG '  (1960  127)  106  1833  106
8470 3559 v' 700
v' 0.0365
3800 1833   r '  340  r '   0.110 

    
  

L
273
 2481 m
The radius R of the turning circle is given by: R  
r ' 0.110
Thus:
Manoeuvring: Nomoto equation
a) A ship lays rudder according to sine function over time alternating between port and starboard with
amplitude 10° and 2 minutes period. The ship performs course changes of 20°. The maximum
course deviation to port occurs 45 seconds after the maximum rudder angle to port has been reached.
Determine from these data the parameters of the Nomoto equation:
T     K
The ship is yaw stable for K > 0. Is the ship yaw stable?
b) The ship speed is reduced to 50% of the value in a). The rudder action is the same as in a). How
large is the amplitude of course changes and what is the delay between maximum rudder angle and
maximum course deviation? Hint: The non-dimensional constant T' depends only on hull
characteristics, non-dimensional K' only on rudder characteristics.
Solution
a) The rudder angle changes as:
  Re 10eit 
with  


The course angle changes as1:   Re  20ei (t  45s )
Inserting these expressions in the Nomoto equation yields:





2 2

 0.0524 s1
T 120

T  Re 20 2ei (t  45s )  Re 20iei (t  45s )   K Re 10eit
20T 2  20i  10 K  e i45

With ei 45  ei 0.052445  0.707  0.707i , we get (first imaginary part, then real part):
 20  10 K  0.707  K  2 / 0.707  0.148 s1
20T 2  10K  0.707  T  K  0.3535 /  2  19.1 s
K is positive, thus the ship is yaw stable.
b) The non-dimensional values K' = KL/U and T'=TU/L remain unchanged as they just depend on the
characteristics of hull and rudder. Thus K is now half as large (K = 0.074 s1) and T is twice as large
(T = 38.2 s).
Let us write the yaw angle as function of time with yet unknown amplitude and time shift:
  Re   0ei (t  t s ) s
The Nomoto equation then yields:







T  Re  0 2ei (t  t s )  Re   0iei (t  t s )   K Re 10eit

it s
 0 (T  i )  10 Ke

Sorting in real and imaginary parts yields two equations:
2
cos(ts )  
 0T 2
10 K
;
sin( t s )  
 0
10 K
Using the relation cos2+sin2=1, we get:
0 
10K
T 2 4   2

10  0.074
38.2 2  0.0524 4  0.0524 2
 6.31
For ts, we must make sure that we find the correct value in the right quadrant, using both sin and
cos information:
cos(t s )  
6.31 38.2  0.05242
 0.894 ;
10  0.074
sin( t s ) 
6.31  0.0524
 0.447
10  0.074
The solution is thus in the second quadrant:
ts  arccos( 0.894)  2.678  ts 
1
2.678


2.678
 51.1 s
0.0524
Rudder angle usually is taken as positive to port, course change negative to port.
Manoeuvring: Nomoto equation with rudder change
A ship lays rudder according to sine function over time alternating between port and starboard with
amplitude 10° and 2 minutes period. The rudder height is 6 m. The ship performs course changes of 20°.
The maximum course deviation to port occurs 45 s after the maximum rudder angle to port has been reached.
The Nomoto equation then yields K = 0.148 s1 and T = 19.1 s.
Now the rudder height is kept constant at 6 m, but the rudder area is increased from 18 m2 to 24 m2. The
speed is kept as before.
Remember that K is the 'rudder effectiveness', T a hull characteristic.
a) Determine the rudder lift coefficient before and after the modification!
b) Determine the new K and T after the modification!
c) Determine the course amplitude  and the delay between maximum rudder angle and maximum
course deviation ts after the modification!
Solution
Originally the rudder aspect ratio was  
6
6
 2 . Now  
 1.5 .
18 / 6
24 / 6
We estimate the lift coefficients, assuming CQ  1 (as usual):
  (  0.7)
 sin   sin 2   cos 
2
(  1.7)
2  (2  0.7)
C L , 2  2
 sin 10  sin 2 10  cos10  0.46
For  = 2,  = 10°:
(2  1.7) 2
1.5  (1.5  0.7)
 sin 10  sin 2 10  cos 10  0.38
For  = 1.5,  = 10°: C L ,1.5  2
2
(1.5  1.7)
C L  2
The rudder force depends on the lift coefficient times the rudder area (and the speed, which is not changed
here). Thus the new K (rudder force per rudder angle) is given by:
K  1.5  K   2 
C L ,1.5 Anew
0.38 24

 0.149 

 0.163 s1
C L , 2 Aold
0.46 18
T remains at 19.1s. Then we can use the same formulae as in the previous exercise to derive:

10 K
4
2

10  0.163
 22
T  
19.1  0.0524 4  0.0524 2
T 2
24.7  19.1  0.0524 2
cos(t s )  

 0.708
10 K
10  0.183
 24.7  0.0524
sin( t s ) 

 0.707
10 K
10  0.183
2.357 2.357
t s  arccos( 0.708)  2.357  t s 

 45.0 s

0.0524
2
2
Manoeuvring: System identification using Norrbin equation
A ship follows the 'Norrbin' equation: T     3   K
The ship performs a pull-out manoeuvre and the following curve for  is recorded:
a) Determine the constants T and !
b) Sketch a corresponding curve of  for the case that at t = 0 we have  = 0.002 rad/s instead of
0.02 rad/s.
Solution
a) Consider long term behaviour of ship. Then the rudder angle is  = 0 in a pull-out manoeuvre. The
yaw rate  does no longer change, i.e.  =0. Then the Norrbin equation yields:
T     3   K     3  0  1   2  0
   2  0.0072  20408 s2
Consider now the initial behaviour. At t = 0, we have  = 0. A tangential to the curve (measured
from curve) determines   0.0077 rad/s2 at t = 0. This determines T using the Norrbin equation
for t = 0:
T     3   K  T  (0.0077)  0.02  20408  0.023  0  T  18.6 s
b)  remains the same and thus the long-term behaviour remains the same. T remains the same. This
determines  at t = 0:
T     3   K
  18.6   0.002  20408  0.0023  0    0.0001 rad/s2
This is an almost horizontal tangential at t = 0. The ship turns much slower ( much smaller). So
the curve may look like:
Manoeuvring: Waterjet versus rudder
A motor yacht of 10 t displacement is equipped with a 1 m2 profile rudder with  = 1.2. The yacht is a twinscrew ship with central rudder. The rudder lies outside the propeller slip stream. The yacht has a speed of
13.33 m/s. For the central position of the rudder we can assume a velocity of 0.75 ship speed due to the
wake. The ‘glide ratio’ (ratio of propeller thrust to ship weight) is  = 0.15.
The yacht shall be converted to waterjet propulsion. For this purpose propeller and rudder shall be
dismantled and waterjets be installed. Waterjets are used to manoeuvre the ship by turning the jets by a
maximum of 35°, just as previously the maximum rudder angle was 35°. The speed may be assumed to be
unaffected by the conversion.
Will the yacht react faster or slower after its conversion? Why?
Solution
The rudder lies in an area of V = 10 m/s, i.e. ship speed. At a rudder angle of 35° a profile rudder with  =
1.2 has typically CL  1.2. The rudder force is then:
1
1
Lrud  CL  q  AR  CL     (V  0.75) 2  AR  1.2   1000  (13.33  0.75) 2  1  60000 N
2
2
The thrust of the waterjets is at straight-ahead course: T    m  g  0.15  10000  9.81  14715 N
L jet  T  sin 35  14715  sin 35  8440 N
The transverse force at 35° angle of the jet is:
Thus the manoeuvring force is reduced by a factor of 7! The yacht will react much slower after the
conversion to waterjets.
Also, yaw stability will be reduced as the rudder increases yaw stability like all fins in the aftbody.
Manoeuvring: Rudder lift force of a semi-balanced rudder
Determine the rudder lift at 10° rudder angle for the semi-balanced rudder behind a propeller and hull. The
ship speed is 30 kn, the propeller thrust 4000 kN, the wake fraction w = 0.25. Correct the rudder lift for the
propeller loading according to Söding2 as given in the book. Take dimensions from the sketch. Following
Goodrich and Molland3 we assume that at 10° rudder angle the depicted semi-balanced rudder may have
79.2% of the lift of a ‘normal’ rudder.
Solution
b 2 62

1
The rudder has an aspect ratio of  
AR 36
We assume CQ = 1. Then the lift coefficient for the rudder (79.2% of ‘normal’ rudder) is:
   (  0.7)

CL  2
 sin   CQ  sin 2   cos    0.792
2
(  1.7)


 1  (1  0.7)

 2
 sin 10  1  sin 2 10  cos10  0.792  0.225
2
(1  1.7)


The ship speed is V0 = 30  0.5144 = 15.432 m/s. The inflow velocity is VA  (1  0.25)  V0  0.75  15.432 =
11.574 m/s. The propeller diameter is taken from the sketch as DP = 4 m, i.e. AP  0.25DP2
 0.25  4 2  12.566 m2.
T
4000000
 1
 4.64
2
  VA AP 2  1025  11.5742  12.566
1
1
The rudder lift is: L  CL  q  AR  CL     VA2  AR  0.225   1025  11.5742  36  556 kN
2
2
CTh 
1
2
This lift does not take into account the propeller loading. Following Söding, we add a correction:

1
L  T  1 

1  CTh


1

 sin   4000  1 
 sin 10  987 kN

1  4.64 


Thus the total transverse rudder force is 556+987 = 1543 kN.
2
3
Söding, H. (1998). Limits of potential theory in rudder flow predictions. Ship Technology Research 45, pp. 141–155.
Goodrich, G.J., Molland, A.F. (1979). Wind tunnel investigation of semi-balanced ship skeg rudders. Trans RINA 121
Manoeuvring: Autopilot influence on resistance
A twin-screw ship has a central rudder outside the propeller slipstream. The ship has a wake fraction
w = 0.25. The rudder has an effective side ratio  = 2.5 and an area AR = 40 m2. The ship sails with 20 knots
speed. The rudder fluctuates around the mean rudder angle 0° due to seaway and wind. The variance (square
root of average of square of fluctuations) of the rudder angle is 10°. By improving the autopilot, the variance
is reduced to 5°. The lift and drag coefficients are given by (CQ  1):
   (  0.7)

C L  2
 sin   sin 2   cos  
2
(  1.7)


2
C
C D  L  (sin  ) 3  C D ,0

How much is the resulting reduction in effective power (= variance in power) if the speed is constant at 20
knots? Assume  = 1019 kg/m3.
Solution
The lift coefficient is:
 2.5  (2.5  0.7)

C L,10  2
 sin 10  sin 2 10  cos10  0.5246
2
(2.5  1.7)


 2.5  (2.5  0.7)

C L,5  2
 sin 5  sin 2 5  cos 5  0.2559
2
(2.5  1.7)


The drag coefficient is:
0.5246 2
 sin 3 10  C D ,0  0.040  C D ,0
  2.5
0.2559 2
C D ,5 
 sin 3 5  C D ,0  0.009  C D ,0
  2.5
Thus the resistance coefficient difference is CD  CD ,10  CD ,5  0.040  0.009  0.031
C D ,10 
The inflow velocity to the rudder is: VA  (1  w)  V  (1  0.25)  20  0.5144  7.716 m/s
This yields a difference in resistance:
1
1
R  CD    VA2  AR  0.031   1019  7.7162  40  37.6 kN
2
2
The difference in effective power relates to ship speed and is thus:
PE  R  V  37.6  (20  0.5144)  387 kW
Manoeuvring: Stopping of a tanker
A tanker of 250000 t displacement sails at 15 knots at a delivered power PD = 15000 kW. The overall
efficiency is

R U
 0 .7
PD
The tanker shall perform a soft stop manoeuvre. The engine is reducing linearly the torque Q within one
minute to zero. The next three minutes are used to brake the shaft and to prepare the engine for reverse
operation. The following two minutes the engine is started and accelerated again linearly to full reverse
torque.
How long will be the stopping distance for the tanker if it sails exactly straight ahead? Quantities not given
are to be estimated!
Hints:
1. During the short periods of acceleration and deceleration of the engine, the ship speed is virtually
constant.
2. The linear deceleration and acceleration shall be approximated by step functions covering the same
‘area’ of Q  t, where t is the time:
3. The added mass in longitudinal motion may be approximated by
m"

m 
L
3
1

/   14
L is the length,  the displacement of the ship.
4. In reverse propeller operation, the ratio of thrust to resistance shall be |T/R| = 0.945.
Solution
The initial speed is U0 = 15  0.5144 = 7.716 m/s.
The stopping constant k is: k 
R0 PD  15000  0.7


 22.860 kNs2/m2
2
3
3
U0
U0
7.716
The mass of the ship needs to be increased by the hydrodynamic mass for longitudinal motion. This is rather
small:

m  m"  m  1 
 



  250000  1 
 
L3 /   14 


1


  257500 t
3
316 / 250000  14 

1

We consider three time intervals successively:
1. time interval ( t = 60 s):
Half torque yields approximately half thrust. The ship speed will remain approximately constant
during the short period. Initially thrust T (minus thrust deduction) and resistance R are in equilibrium,
i.e. T0 = R0.
Thus:
uT  U 0 
T
1 T0
1
1
 U0 
 U0 
 7.716 
 5.456 m/s
R0
2 R0
2
2
The original equation:
t 2  t1 
m
kuT

u
u
 arctan 1  arctan 2
uT
uT




transforms to:


u
ku
7.716 22.86  5.456


u 2  uT  tan  arctan 1  T (t 2  t1 )   5.456  tan  arctan

 60 
uT
m
5.456
257500




u2  7.259 m/s
The distance travelled is:
s1 
m  u12  uT2
ln 
2k  u 22  uT2
 257500  7.716 2  5.456 2 
 
  449 m
ln 
2
2 
 2  22.86  7.259  5.456 
2. time interval ( t = 240 s):
The thrust is now zero and thus uT = 0. The original equation yields:

m  u  k  u 2  uT2


du
k
  dt
2
m
u
Integration yields:
1
1
k
1
1

  (t 2  t1 )  u 2 

 6.287 m/s
1 k
1
22.86
u1 u 2
m
 t 2  t1 

 240
u1 m
7.259 257500
m  u12  uT2  257500  7.259 2 

  1619 m
s 2 
ln 
ln 
2k  u 22  uT2  2  22.86  6.287 2 
3. time interval (300 s until full stop):
uT  U 0 
T
 7.716  0.945  7.5 m/s
R
The interval ends when the ship is fully stopped, i.e. u2 = 0. As we do not have to determine the
stopping time, we can simply compute directly:
s3 
m  u12  uT2  257500  6.287 2  7.259 2 

  2997 m
ln 
ln 
2k  u 22  uT2  2  22.86  0 2  6.287 2 
The total stopping distance is: s  s1  s 2  s3  449  1619  2997  5065 m