Seakeeping: Complex numbers
Compute:
(3  4i )  (2  2i ) 
(3  4i )  (2  2i ) 
(3  4i )  (2  2i ) 
3  4i

2  2i
Euler's formula brings trigonometry and exponential function together:
eix  cos x  i sin x
where i   1 .
 
Show for one example that Re Aˆ eix  A sin( x   ) , where  is the complex amplitude with real part Re()
and imaginary part Im() and absolute value A = Â and the phase shift is tan    Re( Aˆ ) / Im( Aˆ ) . Take as
an example  = 3+4i.
Solution
We use elementary mathematical operations and get:
(3  4i )  (2  2i )  5  2i
(3  4i )  (2  2i )  1  6i
(3  4i )  (2  2i )  6  8i  6i  8  14  2i
3  4i 3  4i 2  2i  2  14i



 0.25  1.75i
2  2i 2  2i 2  2i
8
Compute the phase shift:
  arctan
Re( Aˆ )
Re( 3  4i )
3
 arctan
 arctan  36.87°
Im( 3  4i )
4
Im( Aˆ )
The (real valued) amplitude is:
A  Aˆ  Re( Aˆ )2  Im( Aˆ )  32  42  5
 
Re Aˆ eix  Re(3  4i )(cos x  i sin x)  3 cos x  4 sin x
and
A sin( x   )  5 sin( x  36.87)  5(sin x  cos 36.87  sin 36.87  cos x)  4 sin x  3 cos x
The two expressions are equal.
Seakeeping: Celerity of wave
A surf board travels with Froude number Fn = 0.5 on a deep-water wave of 4 m length. How long must the
surf board be to ‘ride’ on the wave, i.e. to have the same speed as the wave?
Solution
The speed of the surf board is: V  Fn  gL
The celerity of the wave is: c 
g
2
Both velocities must be the same for surf riding:
V  c  Fn  gL 
g
2
 L

2  F
2
n

4
 2.5 m
2  0.52
Seakeeping: Velocity and pressure in elementary wave
Consider a regular deep-water wave of wave length  =100 m and wave height 5 m.
a) What is the orbital velocity in 10 m depth?
b) What is the amplitude of the pressure fluctuation (absolute and in percent of the hydrostatic
pressure)? The pressure fluctuation is (linearised): p f   t
Solution
a) The amplitude of the wave is half the wave height: h = 2.5 m.
2
 0.0628 m1
 100
The frequency is:   g  k  9.81  0.0628  0.785 s1
The wave number is: k 
2

The orbital velocity is the maximum of the x-velocity (for deep waves). The x-velocity is:

v x   x  Re    h  e  kz e i (t  kx )

Thus the amplitude is:
v x    h  e kz  0.785  2.5  e 0.062810  1.05 m/s
b) The pressure fluctuation (=pressure without hydrostatic part) is given by:

p f  t  g  Re h  e kz e i (t kx )

The amplitude of the pressure fluctuation is then:
p f  gh  e  kz  1025  9.81  2.5  e 0.062810  13.4 kPa
The hydrostatic pressure is
p stat   gz  1025  9.81  10  100.6 kPa
Thus the pressure fluctuation is 13% of the hydrostatic pressure.
Seakeeping: Velocity and acceleration in shallow-water wave
The potential of a regular wave on shallow water is given by:
  Re  ich sinh( kH ) cosh( k ( z  H ))ei (t  kx ) 
wave length
 = 100 m
wave amplitude h = 3 m
water depth
H = 30 m
Determine velocity and acceleration field at a depth of z = 20 m below the water surface!
The following parameters are given:
Solution
The velocity is derived by differentiation of the shallow-water potential:
  h

 h
vx  x  Re 
cosh( k ( z  H ))ei (t  kx )  
cosh( k ( z  H )) cos(t  kx)
 sinh( kH )
 sinh( kH )
  ih

h
vz  z  Re 
sinh( k ( z  H ))ei (t  kx )  
sinh( k ( z  H )) sin( t  kx)
 sinh( kH )
 sinh( kH )
This derivation used the relation c = /k. The individual values are:
2 2
k

 0.06283 m1
 100
  gk tanh( kH)  9.81 0.06283  tanh( 0.06283  30)  0.767 s1
This yields:
vx  0.861  cos(0.767t  0.06283x)
vz  0.480  sin( 0.767t  0.06283x)
The accelerations are obtained by differentiating the velocities with respect to time:
ax  xt  0.661  sin( 0.767t  0.06283x)
az  zt  0.368  cos(0.767t  0.06283x)
The velocities have a phase shift of 90° and different amplitudes. The particles will thus trace an ellipse.
Seakeeping: Wave breaking
Wave breaking occurs theoretically when particle velocity in x-direction is larger than the celerity c of the
wave. In practice, wave breaking occurs for wave steepness h/ exceeding 1/14. What is the theoretical and
practical limit for h concerning wave breaking in a deep-water wave of  = 100 m?
Solution
Practically the limit is h = /14 = 100/14 = 7.143 m.
2
 0.0628 m1
 100
The maximum particle velocity in x-direction is: vx, max    h  ekh
The wave number is: k 
The celerity is: c 
2


k
The condition c = vx,max yields:
1
 e kh
kh
The above expression is solved iteratively for kh, using kh  e  kh . The iteration starts with the practical
limit: k0h = 7.143  0.0628 = 0.4488, then yields:
1
0.4488
11
0.5667
2
0.6384
12
0.5674
3
0.5281
13
0.5670
kh  0.5671  h 
4
0.5897
14
0.5672
5
0.5545
15
0.5672
0.5671
 9.026 m
0.0628
6
0.5744
16
0.5671
7
0.5631
17
0.5671
8
0.5695
18
...
9
0.5658
10
0.5679
Seakeeping: Wave with two buoys
The position of two buoys is given by their (x,y) coordinates in [m] as sketched below. The buoys are
excited by a regular wave of  = 62.8 m, amplitude h = 1 m, and angle  = 30° to the x-axis.
a) What is the maximum vertical relative motion between the two buoys if they follow exactly the
waves?
b) What is the largest wave length  to achieve maximum vertical relative motion of twice the wave
amplitude?
Solution
a) The wave number of this wave is: k 
2


2
 0.1 m1
62.8
Transform coordinates in local ξ-system. Points on ξ = const. have same ζ-values for regular waves.
The two buoys have then coordinates:
ξ1 = 0
ξ2 = x2  cos 30°  y2  sin 30° = 30  cos 30°  10  sin 30° = 21 m
The relative difference between the wave elevations is:
 1   2  Re h  ei (t  k )   Re h  ei (t  k
1
2)
  Reh  e 1  e
it
 ik 2 )

The amplitude (= maximum) of this relative motion is given by:


h  e it 1  e ik 2 )  h 
1  cos( k 2 ) 2  sin( k 2 ) 2
 1  1  cos(0.1  21)  sin( 0.1  21)
2
2
= 1.73 m
b) The maximum difference between wave elevations is twice the wave amplitude if the buoys are
spaced by an odd multiple of half the wave length. The longest wave is obtained for a spacing of
half the wave length:

2
 21 m

 = 42 m
All other wave lengths fulfilling the criterion are even shorter.
Seakeeping: Encounter frequency
A ship travels with 28.28 kn in deep sea in regular sea waves. The ship travels east, the waves come from
southwest. The wave length is estimated to be between 100 m and 200 m. The encounter period Te is
measured at 31.42 s.
a) What is the length of the seaway?
b) Two days before, a storm started in an area 1500 km southwest of the ship's position.
Can the waves have their origin in this storm area?
Solution
The speed is V = 28.28  0.5144 m/s = 14.55 m/s.
The encounter period yields the encounter frequency:

The value of  is:
eV
g

e 
2
2

 0.2 s1
Te 31.42
0.2  14.55
 0.2966
9.81
There are three possible frequencies which could excite this encounter frequency:












g
9.81
1  1  4 cos  
1  1  4  0.2966  cos 45  1.123 Hz
2V cos 
2  14.55  cos 45
g
9.81
2 
1  1  4 cos  
1  1  4  0.2966  cos 45  0.6684 Hz
2V cos 
2  14.55  cos 45
g
9.81
3 
1  1  4 cos  
1  1  4  0.2966  cos 45  0.2854 Hz
2V cos 
2  14.55  cos 45
1 
The corresponding wave lengths are:
2  9.81
 49 m

1.1232
2g 2  9.81
2  2 
 138 m
2
0.66842
2g 2  9.81
3  2 
 756 m
3
0.28542
1 
2g
2
1

Only 2 = 138 m fits the observed bandwidth of wave lengths.
Wave groups travel with group velocity. For deep ocean water, this is
cgr 
1 g 1 9.81  138

 7.34 m/s
2 2 2
2
At this speed, waves can travel within two days:
s = 48 h  60 min/h  60 s/min  7.34 m/s = 1268 km < 1500 km
The waves can thus not originate from the storm area.
Seakeeping: Undamped free heave oscillation
The differential equation of a free undamped heave oscillation is:
m  m33 z  cz  0
m is the displacement of the ship, m33 the added mass for heave motion. c is the restoring force coefficient.
The (circular) natural frequency is subsequently:
z 
c
m  m33
Consider the ship with the following particulars:
L = 125 m, B = 17 m, T = 7 m, CB = 0.75, Cwp = 0.8, m33 = 0.9m.
a) Determine the (circular) natural frequency and natural period of heave!
b) Initial conditions are: At t = 0 we have z = 0 and z =0.5 m/s. What is the maximal load a winch of
4000 kg mass exerts on its foundation?
Solution
a) The natural frequency follows from:
z 
c

m  m33
  g  Cwp  L  B
  CB  L  B  T  (1  m33 / m)

g  Cwp
CBT (1  m33 / m)
9.81  0.8
 0.89 Hz
0.75  7  (1  0.9)
2
2
Tz 

 7.08 s
 z 0.89

b) The winch exerts its maximal load when gravity g and amplitude of heave oscillation are
superposed. We write the heave motion as:

z  Re Kˆ ei z t

Kˆ  K1  iK 2 is determined by the initial conditions:
z  Re Kˆ ei z 0  Re Kˆ  K  0
t 0
zt  0
   
 Re i Kˆ e    K 
1
i z 0
z
2
z
 0.5  K 2  
Then the amplitude of the heave acceleration is given by:
z  z2 K2  0.892  0.56  0.45 m/s2
The maximal load of the winch is then:
mwinch ( g  z)  4000  (9.81  0.45)  41 kN
0.5
z

0.5
 0.56 m
0.89
Seakeeping: Power requirements for a wave maker
A wave maker is to be designed for a towing tank of width B = 4 m and depth H = 2.5 m. The wave maker
shall be designed for a wave of 5 m length and 0.2 m amplitude.
a) What is the power requirement for the motor for the wave maker if we assume 30% total efficiency
between motor and wave? (For the considered wave length, the depth can be regarded as ‘deep’.
The power requirement of the wave is (energy/meter [in direction of wave propagation]) * group
velocity, as the energy in a wave is transported with group velocity.)
b) After switching the wave maker off, for a long time there is still a wave motion with period 40 s
observed in the tank. How long is the tank if the motion is due to the lowest natural frequency of
the tank?
Solution
a) The group velocity is:
1
1 g 1 9.81  5
cgr  c 

 1.397 m/s
2
2 2 2
2
The average energy per area is:
E
1
1
gh 2   1000  9.81  0.22  196.2 N/m
2
2
Thus the power of the wave is:
P  E  B  cgr  196.2  4  1.397  1096 W
The power of the motor then needs to be:
Pm  P / 0.3  3.65 kW
b) The wave length is now long compared to the depth. Thus we have to use finite water depth
expressions:
k tanh( kH ) 
With  
2
g
2
 0.157 Hz, this yields:
T
k tanh( kH ) 
0.157 2
 0.002515 m1
9.81
This equation has to be solved iteratively. As the convergence is slow, it is useful to start with a
good estimate. For small x, tanh(x)  x. This would yield k  0.002515 / 2.5  0.03172 .
We solve the problem using Newton's iteration:
f (k )
f ' (k )
With f  k  tanh( 2.5k )  0.002515 and f '  tanh( 2.5k )  k / cosh 2 (2.5k ) , we obtain within 7
kk
iterations:
k  0.03175   
2
2

 198 m.
k
0.03175
The lowest natural frequency is at /2 = L, i.e. the tank has a length of 99 m.
Seakeeping: Wave maker for required wave length and amplitude
A wave maker consists of a cylinder which moves up and down at a wall. The wave maker has a cross
section of half a Lewis section with Cm = 0.8.
The draft of the section is T = 1 m, the half-width is B/2 = 0.6 m. Determine the amplitude and frequency of
motion to create waves of 5 m length and 0.2 m amplitude.
Note: The ratio between wave amplitude and body motion amplitude is Az .
Solution
The frequency corresponding to  = 5 m is:

2g


2  9.81
 3.51 Hz
5
Important parameters for Lewis section curves:
2  B
3.512  0.6
 0.754
2g
9.81
B 0.6
H

 0.6
2T
1
The curves for Lewis sections give Az = 0.52. Thus:
h
0.2
u3 

 0.38
Az 0.52

m
Seakeeping: Wave maker with Lewis section moves floating body
A wave maker consists of a cylinder which moves up and down at a wall. The wave maker has a cross
section of half a Lewis section with Cm = 0.8. The draft of the section is T = 1 m, its half-width B/2 = 0.6 m.
The wave maker performs heave motions with 0.6 m amplitude and  = 3.51 Hz. The basin is 10 m wide.
The mass of the wave maker itself is negligibly small. Use linear theory in your solution approach.
a) What is the amplitude of the force needed to move the wave maker?
b) What is the average power supply if an efficiency of 70% is assumed?
c) In the middle of the basin is a cylinder of Cm = 0.8, B = 1.2 m, T = 1 m, L = 5 m, arranged with axis
parallel to the wave maker axis. The cylinder has all six degrees of freedom suppressed. What is the
exciting force on the cylinder?
d) If the cylinder in the middle of the basin is free to heave, what is its heave amplitude?
Assume deep water.
Solution
a) The basic equation for heave motion is:
  m  m
2
33
  in33  c33  u3   f
(The force is negative as now the body excites the water and not the water the body.) All quantities
are in this equation per length, i.e. for the final computation they need to be multiplied by the tank
width. The individual quantities are:
2  B
3.512  0.6
 0.754
2g
9.81
B 0.6
H

 0.6
2T
1

The curves for Lewis sections give Cz = 0.6, Az  0.52 . Thus:
B2
1.22
 0.6  1000   
 339 kg/m
8
8
g 2
1000  9.812
n33  Az2  3  0.522 
 601.8 kg/ms

3.513
c33  gB  1000  9.81  1.2  11770 N/m2
m33  Cz     
The mass of the wave maker is negligible: m = 0
Thus:
 3.51  (0  339)  i  3.51 601.8  11770 0.6   f
2

f  (4556  1267i) N/m
This force is for waves radiated in both directions, i.e. so far we assumed symmetry. The total force
amplitude is then given by:
| F |
1
1
| f | BTank 
45562  1267 2  10  23.6 kN
2
2
b) The power needed to drive the wave maker is given by:
P
E  BTank  cgr


gh2
c 1
 BTank   kN
2
2 
The wave amplitude is:
h | u3 |  Az  0.6  0.52  0.312 m
The celerity is:
c
Thus:
g


9.81
 2.79 m/s
3.51
P
1000  9.81  0.3122
2.79 1
 10 

 9.5 kW
2
2 0.7
Now we check the result, using a different approach. The power needed is due to the damping force
times the velocity. The mass force and the restoring force are phase shifted by 90° to the velocity
and thus do not contribute to the power. Again we take into account that we have only half a cross
section:

1  Im( f )
1 1  1267
1

P  
 BTank   (  zˆ)   
 10   (3.51  0.6) 
 9.5 kW
2 2
 2 2
0.7


c) We read from the diagrams for Lewis sections:
 
Re fˆe
 0.25 
gBh
Im fˆe
 0.25 
gBh
 
f e, r  0.25  gBh  0.25  1000  9.81  1.2  0.312  918.2 N/m
f e,i  918.2 N/m
This then gives the absolute value for the three-dimensional force:
Fe 
f e2,r  f e2,i  L  918.22  918.22  5  6.5 kN
d) The basic equation for heave motion is:
  m  m
2
33
  in33  c33  u3 
fe
The mass (per length) follows from Archimedes:
m    Cm  B  T  1000  0.8  1.2  1.0  960 kg/m
As before, we have then:
 3.51 960  339  i  3.51  601.8  11770 u
2
 4233.8  2112.3i u3  (918.2  918.2i)
u3  0.0872  0.2602 )  0.274 m
3
 (918.2  918.2i )
 u3  (0.087  0.260i) m
Seakeeping: Cylinder with Lewis cross section in regular waves
A cylinder of Lewis cross section (L = 10 m, B = 1 m, T = 0.4 m, Cm = 0.8) floats parallel to the wave crests
of regular waves  = 5 m, h = 0.25 m) which excite heave motions. Assume that the form of the free surface
is not changed by the cylinder. What is the amplitude of relative motion between cylinder and free surface
predicted by linear theory?
Solution
Fundamental equation for heave motion:
  m  m
2
33
  in33  c33  u3 
fe
The individual quantities needed are:
2  9.81
 3.51 Hz

5
 2  B 3.512  1

 0.628
2g
2  9.81
B
1
H

 1.25
2T 2  0.4

2g

Then the curves for Lewis sections yield: Cz = 0.62, Az  0.68 , f er*  0.39 , f ei*  0.38 . This in turn gives:
B2
12
 0.62  1000     243 kg/m
8
8
m    Cm  B  T  1000  0.8  1.0  0.4  320 kg/m
m33  Cz     
n33  Az2 
2
g 2
2 1000  9.81

0
.
68

 1029 kg/ms
3
3.513
c33  gB  1000  9.81  1.0  9810 N/m2


f e  gB f er*  f ei*i h  1000  9.81  1.0  (0.39  0.38i )  0.25  956.5  932.0i N/m
We assume  = 1000 kg/m3 here for convenience. As  appears in all terms, the final result is not influenced
by the choice of .
This yields the heave motion amplitude:
 3.51 243  320  i  3.51 1029  9810 u
2
3
u3 
 956.5  932.0i
956.5  932.0i
 0.287  0.0366i m
2874  3612i
The amplitude of relative motion is determined from:
r  u3  h  0.287  0.0366i   0.25  0.037  0.0366i  0.0372  0.03662 = 0.052 m
Seakeeping: Lewis section in forced heave
An infinite cylinder of width B = 2 m, draft T = 1 m and with Lewis cross section of coefficient Cm = 0.8
floats in equilibrium in fresh water. Then a harmonic force per length with period Te = 3.14 s and amplitude
fe=1000 N/m is applied. What motion results after a long time (when the initial start-up has decayed)?
What wave length and what wave amplitude are generated, assuming a ‘deep’ basin?
Solution
The motion has only one degree of freedom, namely heave. The basic equation is then:
  m  m
2
33
  in33  c33  u3 
fe
u3 is the complex amplitude of heave motion. The individual quantities needed are:
2
2

 2 Hz
Te 3.14

2  B
22  2

 0.408
2g
2  9.81
B
2
H

1
2T 2  1
Then the curves for Lewis sections yield: Cz = 0.7, Az  0.51 . This in turn gives:
B2
22
m33  Cz     
 0.7  1000     1100 kg/m
8
8
m    Cm  B  T  1000  0.8  2.0  1.0  1600 kg/m
n33  Az2 
2
g 2
2 1000  9.81

0
.
51

 3129 kg/ms
3
23
c33  gB  1000  9.81  2.0  19620 N/m2
This yields the heave motion amplitude:
 2 1600  1100  i  2  3129  19620 u
2
3
u3 
1000
 0.0754  0.0535i m
8820  6258i
The amplitude is:
u3  0.07542  0.05352  0.0925 m
The wave length follows from:
2
g

2

 
2g
2
 15.4 m
The wave amplitude follows from Az :
h  Az  u3  0.51  0.0925  0.047 m
 1000
Seakeeping: Power generator
A cylinder for power generation is moored in regular waves such that the cylinder axis is parallel to the
wave crests. The cylinder dimensions are L = 50 m, B = 10 m, T = 5 m, the cross section is a Lewis section
with Cm = 0.8. The waves have  = 50 m and h = 1 m. The mooring adds a restoring force coefficient c =
2.5106 N/m and a damping (from the generator) of d = 1.5106 kg/s. The mooring remains at all times under
tension as the buoyancy of the cylinder is much larger than its weight due to its mass of 106 kg. The sea
water has a density of 1025 kg/m3.
What is the amplitude of heave motions for the cylinder?
What would be the amplitude without generator?
Solution
The basic equation of motion is:
  m  m
2
33

L   i (d  n33L)  (c  c33L)  u3  Lf e
The individual quantities needed are:
2g
2  9.81
 1.11 Hz

50
 2  B 1.112  10

 0.628
2g
2  9.81
B
10
H

1
2T 2  5


Then the curves for Lewis sections yield: Cz = 0.61, Az  0.65 , f er*  0.39 , f ei*  0.36 . This in turn gives:
B2
102
 50  0.61  1025   
 1.2275  106 kg
8
8
g 2
1025  9.812
L  n33  L  Az2  3  50  0.652 
 1.5235  106 kg/s
3

1.11
L  c33  L  gB  50  1025  9.81  10  5.0275  106 N/m
L  m33  L  Cz     


L  f e  L  gB f er*  f ei*i h  50  1025  9.81  10  (0.39  0.38i )  1  (1.961  1.81i )  106 N
Then the basic equation yields:
 1.11 1  1.2275  i1.11  (1.5  1.5235)  (2.5  5.0275) u
2
3
u3 
 1.961  1.81i
1.961  1.81i
 0.453  0.061i m
4.783  3.356i
The amplitude is:
u3  0.4532  0.0612  0.46 m
The basic equation modifies for the case of no generator (i.e. d = 0):
  m  m L  i (n L)  (c  c L) u  Lf
 1.11 1  1.2275  i1.11 1.5235  (2.5  5.0275) u
2
33
33
33
3
e
2
3
u3 
1.961  1.81i
 0.483  0.208i m
4.783  1.691i
u3  0.4832  0.2082  0.53 m
 1.961  1.81i
Seakeeping: Catamaran in waves
A raft consists of two cylinders with D = 1 m diameter and L = 10 m length. The two cylinders have a
distance of 2e = 3 m from centre to centre. The raft has no speed and is located in regular waves coming
directly from abeam. The waves have  = 3 m and wave amplitude h = 0.5 m. Assume the two cylinders to
be hydrodynamically independent, i.e. waves created by one cylinder are not reflected at the other cylinder.
The raft has a draft of 0.5 m, centre of gravity in the centre of the connecting plate, radius of moment of
inertia for rolling is kx = 1 m. The density of the water is 1000 kg/m3. The Cm = /4 is close enough to 0.8 to
use the Lewis section curves for this Cm. Neglect sway motion. What is the maximum roll angle?
Hint: The centre of gravity of a semi-circle is 4r/(3) from the flat baseline, where r is the radius.
Solution
The mass of the raft is: 
D 2
  12
 10  2  7854 kg
8
The mass moment of inertia is:   m  k x2  7854  12  7854 kgm2
2 2
k

 0.6667 m1
Wave number:
 3
2g
2  9.81
Wave frequency:


 2.56 s1

3
8
 L  2  1000 
The metacentric height is:


4 
D3 L / 12  DLe 2  2

GM  KB  BM  KG  1 
D
 T 
D 2 L / 4
 3 
4 
13  10 / 12  1  10  1.52  2

 1 
 1  5.23 m
  0.5 
  12  10 / 4
 3 
B
1 .0
 2 B 2.562  1.0


1
Parameters for Lewis curves:
and

 0.333
2T 2  0.5
2g
2  9.81
Then the curves for Lewis sections yield: Cz = 0.8, Az  0.4 , f er*  0.6 , f ei*  0.25




  2 (  m44 )  in44  c44 u4  f e 4
The basic equation of roll motion is:
The hydrodynamic added mass m44 and damping n44 (moments) are derived from the added mass for heave
(forces) times lever e:
m44u4  2  m33  u3   e
The factor 2 is due to the two cylinders. The term in parentheses is the vertical force. With u3  e  u4 , we
get for the whole 3-d raft:
 B 2



  1.02
2



m44  2   
 Cz   L  e  2  1000 
 0.8   10  1.52  14137 kgm2
8
 8



2
2


g 
1000  9.81 
  10  1.52  41300 kgm2/s
n44  2   Az2  3   L  e2  2   0.42 
3

2
.
56




The restoring force constant for roll motion is:
c44  mg  GM  7854  9.81 5.23  403 kNm
The exciting moment has to consider the phase shift in the wave between the two hulls. The force on the left
floater is then:


fˆl  fer*  if ei*  gBh  L  eike
The force on the right floater is:


fˆr  fer*  if ei*  gBh  L  eike
The phase shift is thus in the last term. Combine the two forces with lever e to get:



f e 4  e  f er*  if ei*  gBh  L  e ike  eike

 1.5  0.6  0.25i  1000  9.811.0  0.5 10   2i sin( 0.6667 1.5)
= (30956  74293 i) kgm2/s2
Then our basic equation becomes:
 2.56 (7854  14137)  i  2.56  41300  40300 u
 (30956  74293i)
258.9  105.7i u4  (30.96  74.29i)  u4  0.002  0.288i
2
The (real) roll amplitude is then
u4  0.0022  0.2882  0.288  16.5
The assumption of linearity is thus no longer justifiable.
4
Seakeeping: Strip method roll
A circular cylinder (length L = 20 m, D = 1 m, radius of moment of inertia kxx = 0.4 m, mass m = 8000 kg)
floats in water of density  = 1019 kg/m3. The centre of gravity is 0.1 m under the centre of the circle. The
cylinder is excited by regular waves with wave crests parallel to the cylinder axis. The waves have
amplitude h = 0.15 and length  = 3.14 m. The following hydrodynamic properties are known:
a) hydrodynamic mass for sway: 0.4 m
b) hydrodynamic damping for sway given by ratio of amplitudes Ay  0.3
c) real part of horizontal exciting force: 0.8 orbital velocity of wave times damping coefficient as
given in b)
d) imaginary part of horizontal exciting force: 0.6 wave steepness times cylinder weight
The coordinate system should have its origin in the centre of the circle. Then the resultant force of the
hydrodynamic forces passes through the origin.
Determine the complex amplitude of roll motion!
Solution
We use a coordinate system with z pointing down, y pointing right in direction of wave propagation. x is
then the cylinder axis.
The displacement of the fully submerged cylinder is:
D 2
4
L 
  12
4
 20  1019  16000 kg
As the cylinder mass is 8000 kg, the draft of the cylinder is T = D/2 = 0.5 m. The frequency is:
2g



2  9.81
 4.43 Hz
3.14
The orbital velocity involves differentiation with respect to time (factor ), the steepness with respect to y
(factor k). The hydrodynamic quantities ‘hidden’ in a) to d) are then:
m22  0.4  m  0.4  8000  3200 kg
2
g 2
2 1019  9.81

L

0
.
3

 20  2040 kg/s
3
4.433
Re Fe  0.8    h  n22  0.8  4.43  0.15  2040  1084 N
Im Fe  0.6  k  h  g  m  0.6   2  h  m  0.6  4.432  0.15  8000  14130 N
n22  Ay2 
We can use the formulae for strip method and simplify them. As the waves come just from abeam, no
‘symmetric’ motions (heave, surge, pitch) are excited (with linear theory). Also, as the cylinder has always
the same cross section, no yaw motion is excited. The fundamental equation of motions is:
  (M  A)  iN  S  u  Eh
2
Sway motion does not excite roll motion, neither does roll motion excite sway for a circular cylinder with
origin as chosen. There is no exciting roll moment either as in potential theory only normal pressures exist
and for the circular cylinder all normal vectors pass through the origin, i.e. excite no moment. Similarly,
there is no radiation moment, i.e. no damping and added mass for roll. Sway has no restoring term. The
metacentre of a circular cylinder is the centre of the circle. Thus GM = 0.1 m. Thus we can write for the two
degrees of freedom:
 mzg 
 m
 1
M 
 m

z 
 mzg
 z g
0
m
3200 0
A   22
 m

0
 0 0
 0
 z g   8000  800

k xx2   800 1280 
0
n
2040 0
N   22
 m

0
 0 0
 0
0 
0 
0
0
S
 m


0 mgGM 
0 7848
Thus:
  219800 15700  9037i 0 0
0  u2   1080  14130i 

   





 15700
 25120  0
0 0 7848 u4  
0


 219800  9037i 15700  u2   1080  14130i 
 


15700
 17272 u4  
0


The solution of this system of equations is:
u2   0.002  0.069i 
 

u4   0.002  0.063i 
Thus the roll amplitude is u4  0.0022  0.0632 = 0.063 = 3.6°.
Seakeeping: Pontoon with crane in waves
Consider a pontoon with a heavy-lift derrick as sketched. The pontoon has L = 100 m, B = 20 m, D = 10 m,
mp = 107 kg. The load at the derrick has mass ml = 106 kg. The height of the derrick over deck is 10 m. This
is where the load can be considered to be concentrated in one point. The longitudinal position of the derrick
is 20 m before amidships. A force F = 106 N acts on the forward corner. We assume homogeneous mass
distribution in the pontoon.
a) Consider the pontoon ‘in air’ (without hydrodynamic masses) and determine the acceleration vector
u !
b) Consider the pontoon statically in water and determine u!
Solution
All numbers are given in standard units. We use the coordinate system as in the book with z pointing down.
Origin is at K.
0

  0 

  0 
0

 

   F    106 
a) The general 6-component force vector is: F  

7 

F

10
m

   10 
 F  50m  5  107 

 


  0 
0
The mass matrix needs the following expressions:
m  mp  ml  107  106  1.1  107
m  zg  mp  zg , p  ml  zg ,l  107  (5)  106  (20)  7 107
m  xg  mp  xg , p  ml  zg ,l  107  0  106  20  2  107
The moments of inertia are computed for pontoon and load:
 xx , p   ( z 2  y 2 ) d m 
p  L 
3
B3  m 
B 2  107  2 202 
 B  D3  D   p  D 2 

10 

4  3 
4  3 
4 

= 0.667  109
xx,l  ml  zg2,l  106  204  0.4  109
Thus
xx  xx, p  xx,l  1.067  109 . Correspondingly we get:
 xz , p   xz d m  0
xz ,l  ml  xg ,l  zg ,l  106  (20)  20  0.4  109
Thus
xz  xz , p  xz ,l  0.4  109 .
 yy , p   ( x 2  z 2 ) d m 
= 8.667  10

9

p  B 
3
L3  m 
L2  107  2 1002 
 L  D3  D   p  D 2   
10 

4 3 
4  3 
4 

 yy ,l  ml  xg2,l  zg2,l  106  (202  202 )  0.8  109
 yy   yy , p   yy ,l  9.467  109
Thus
 zz, p   ( x 2  y 2 ) d m 
= 8.667  109

p  D
12
LB
3

 BL3 
B
12
mp
2

 L2 

107
202  1002
12


 yy ,l  ml  xg2,l  yg2,l  106  (202  02 )  0.4  109
zz  zz, p  zz,l  9.067  109
Thus
0
0
0
 70
0 
 11
 0
11
0
70
0
20 

 0
0
11
0
 20
0 
This yields the mass matrix: M  106  

70
0 1067
0
400 
 0
 70 0  20
0
9467
0 


20
0
400
0
9067
 0
 , where F and u are generalised 6-component
Now we can use the fundamental equation F  M  u
vectors. Since we solve manually, it is advisable to decouple the 66 system of equations into 33
systems for symmetric and anti-symmetric degrees of freedom:
11u1
 70u1
11u3
 20u3
11u2
70u4
70u2 1067u4
20u2 400u4
 70u5  0
u1
 20u5   1  u3
9467u5   50
u5
  0.0367 m / s 2
  0.1014 m / s 2
  0.00577 s  2
20u6  0
u2
400u6   10  u4
9067u6  0
u6
b) Basic equation of motion is:
 0.103 m / s 2
  0.0163 s  2
 0.000492 s  2
  (M  A)  i N  S  u  F
2
e
e
e
Now we consider the quasi-static case, i.e. e = 0. F is then the ‘exciting’ force. Then: S  u  F
The hydrostatic data are: Aw  L  B  100  20  2000 m2, xw = 0, T 
GM = KB + BM  KG =
m
1.1  107
 3
 5.5 m,
Aw 10  2000
T
B2

 z g  2.75 + 3.06  6.36 = 2.45 m,
2 12T
T
L2

 z g  2.75  151.51  6.36  147.9 m
2 12T
0
0
0
0 0
0 0
0
0
0

0 0 19.62
0
0
S  106  
This gives:
0
264.4
0
0 0
0 0
0
0
15960

0
0
0
0 0
GM L 
0
0
0

0
0

0
It is impossible to make any statement on u1, u2, and u6. The remaining degrees of freedom form a
simple ‘system’ of equations:
19.62u3  1  u3  0.05 m
264.4u4  10  u4  0.038
15960u5  50  u5  0.0031
Seakeeping: Probability of exceeding operational limit
A ship sails in a natural seaway of approximately Te = 6 s encounter period between ship and waves. The
bridge of the ship is located forward near the bow. During one 1 hour, 6 times a downward acceleration
occurred exceeding gravity acceleration g. For downward accelerations exceeding 1.5g, severe injuries for
the crew have to be expected. What is the probability for this happening if the ship continues sailing with the
same speed in the same seaway for 12 hours?
Solution
The number of amplitudes encountered per hour (=3600 s) is:
n
3600 3600

 600
Te
6
The actual number of amplitudes for acceleration exceeding g in one hour was 6. Thus the probability that
one amplitude exceeding g occurs is W(1g) = 600/6 = 0.01. On the other hand, we can write this probability
formally:
W (1g )  e(1g )
2
/(2 m0 )
 0.01  m0  0.109 g 2
The probability for an amplitude exceeding 1.5g is correspondingly:
W (1.5g )  e(1.5 g )
/(2 m0 )
 3.29  105
The number of amplitudes encountered in 12 h is 12  n  12  600  7200. Thus the probability for one
2
 e1.5
2
/(20.109)
amplitude exceeding 1.5g in 12 h is:
W  7200  3.29  105 = 0.24 = 24 %
Seakeeping: Roll motion of cargo ship in waves
A new cargo ship type features relatively large beam and extreme sectional flare at the ship ends to allow
many container on deck. The ship has the following main data: LWL = 94m, B = 16 m, D = 8.40 m, T = 5 m,
radius of inertia (including added mass) k' = 0.4 B = 6.4 m. The service speed is V = 11.8 kn.
Investigate whether (respectively when) there is a danger of capsizing due to resonance in wave induced
rolling for possible metacentric heights (0.30 m < GM < 1.00 m). Consider the following scenarios:
a) waves from abeam ( < 150 m)
b) following waves ( < 150 m)
If there is a danger of capsizing, specify the conditions which are most critical. Assume deep water.
Solution
a) waves from abeam
Resonance appears for e = n. For elementary waves on deep water this yields:
2g
2  k ' 2 2  6.4 2 257.4



GM
GM
GM
Thus for 0.30 m < GM < 1.00 m only wave lengths 257 m <  < 858 m lead to resonance. For the

g  GM
k '2
 
given range below 150 m, there is no danger of capsizing.
b) following waves
Resonance appears for
n
1
 n  with n = 1, 2, 3, ...
e
2
For following waves (angle of encounter  = 0°) we have:  e   
With   2g /  we get: e 
2g


2
g
V
2V

The first resonance lies at n = ½ e . The natural frequency of roll is  n 
Thus we get for the first resonance:
Solving for GM yields: GM 
g  GM
k'
g  GM 1 2g V


k'
2 

k ' 2
2 3 k ' 2 V  2 k ' 2 V 2


2
g
g2
3
Take all quantities in standard units (m and s), to get with V = 11.8 kn = 6.07 m/s:
GM 
  6.4 2
2 3 6.4 2  6.07  2  6.4 2  6.07 2 64.34 625.11 1518.3





2
9.81

9.81  2
2
3
3
This is evaluated for various wave lengths in the given interval:
30
40
50
60
80 100 120 150
 [m]
GM [m] 0.03 0.09 0.13 0.15 0.17 0.17 0.17 0.16
The case GM > 0.30 m is never reached. Thus there is no danger of capsizing for n = 1.
The second resonance lies at n = e. Due to the quadratic relation, we have thus four times the
values for GM:
30
40
50
60
80 100 120 150
 [m]
GM [m] 0.11 0.35 0.50 0.60 0.67 0.68 0.66 0.62
Thus for most wave lengths, there is a danger of resonance for smaller GM values.
The third resonance lies at n = 1.5 e.
30
40
50
60
80 100 120 150
 [m]
GM [m] 0.11 0.35 0.50 0.60 0.67 0.68 0.66 0.62
The third resonance is also reached, but only for wave lengths much shorter than the ship length.
Thus the expected fluctuation in righting levers will be smaller as for second resonance.
Correspondingly the probability of capsizing will be lower.
Higher resonances will shift resonance to even shorter waves which will be uncritical.
Seakeeping: Roll motion of submarine in waves
Consider a surfaced submarine with speed V = 6 kn. Neglect the sail and model the submarine hull as a
circular cross section. Neglect damping.
a) What is the expected roll amplitude in elementary waves from abeam?
b) Evaluate the danger of parametric roll excitation in elementary following waves!
Ship data: L = 70.00 m, D = 8.00 m (diameter), T = 7.00 m, KG = 3.60 m, k' = 3.00 m (radius of the
transverse mass moment of inertia incl. added mass)
Wave data (deep water):  = 70 m (wave length); h = 1.75 m (wave amplitude)
Solution:
a) waves from abeam
e 
Wave circular frequency:
2g


2  9.81
 0.93835 s1
70
For a circular cylinder, the metacentre is always in the centre of the circle:
D
8.00
 KG 
 3.60  0.40 m
2
2
g  GM
9.81  0.40
The natural frequency for roll is: n 

 0.6603 s1
2
k'
3.002
 0.93835
 1.4211 the roll excitation lies in the super-critical region. The roll amplitude
For e 
n 0.6603
for undamped roll is thus (with k = 2/):
1
1
2
|  || uˆ4 |
k h 

 1.75  0.15406  8.83
2
2
1  (e / n )
1  1.4211 70
GM  KM  KG 
b) following waves
Encounter frequency e is then:
2  9.81 2  6  0.5144

 0.6613 s1
g


70
70
Although the roll natural frequency n is close to encounter frequency e, there is no parametric
e   
2
V 
2g

2V

excitation, because for a circular cylinder there is no fluctuation of righting levers. The righting lever
is always h = GM  sin , with GM = 0.40 m = const.
Seakeeping: Cruise ship in waves from abeam
A small cruise ship is designed. As the destination island has no deep-water port, the passengers will
disembark via small boats. To allow normal disembarkation, the metacentric height should be always
sufficiently high to have maximum roll amplitude  = 6° in regular waves from abeam up to wave steepness
max = 4.5° up to wave length of max = 60 m. Assume a linear restoring moment. For the small roll angles
considered, damping can be neglected.
The radius of inertia including added mass is k' = 4 m.
What is the resulting requirement for the metacentric height?
Solution
The requirement   6° for max = 4.5° means for the amplification function: Vratio 

 max

6
 1.33
4.5
The equation for V yields then limiting ratios of frequency:
(e / n ) 2  0.25
1
Vratio 
 1.33 
(e / n ) 2  1.75
1  (e / n ) 2
Thus the frequency interval to be avoided is 0.25  (e / n ) 2  1.75 . With the relations for a deep water
regular wave:
e 
2g

and for the natural frequency of roll
n 
g  GM
k '2
we get for the interval to be avoided:
0.25 
2  k '2
 1.75
  GM
For various wave lengths, we can then determine the interval to be avoided:

20 m
40 m
60 m
interval to avoid
2.87 m  GM  20.11 m
1.44 m  GM  10.05 m
0.96 m  GM  6.70 m
One can quickly see that short wave lengths pose no problem. The limiting value is GM  0.96 m. This
requirement will hardly be possible for a passenger ship due to damaged stability requirements.