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Seakeeping: Complex numbers
Compute:
=−++ )22()43( ii
=−−+)22()43( ii
=−⋅+ )22()43( ii
=−+
i
i
22
43
Euler's formula brings trigonometry and exponential function together:
xi xeix sincos +=where −=i !"how for one example that )sin(#$e ε += x Ae A ix % where A# is the complex amplitude with real part$e() and imaginary part &m() and absolute alue A A# and the phase shift is )#&m()#$e(tan A A−=ε !
*a+e as an example A# 3,4i.
Solution
-e use elementary mathematical operations and get:iii 2.)22()43( +=−++ iii /)22()43( +=−−+
iiiii 240/0/)22()43( +=+−+=−⋅+
ii
i
i
i
i
i
i1.!2.!
0
42
22
22
22
43
22
43 +−=+−=++⋅
−+=
−+
Compute the phase shift:
==++
== 43
arctan)43&m(
)43$e(arctan)#&m(
)#$e(arctan i
i
A
Aε 3/!01
*he (real alued) amplitude is:
.43)#&m()#$e(# 222 =+=+== A A A A
[ ] x x xi xie A ix sin4cos3)sin)(cos43($e#$e +=+−=and
x x x x x x A cos3sin4)cos01!3/sin01!3/cos(sin.)01!3/sin(.)sin( +=⋅°+°⋅=°+=+ ε
*he two expressions are eual!
8/18/2019 Seakeeping Exercises
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Seakeeping: Celerity of wave
5 surf board traels with 6roude number F n !. on a deep7water wae of 4 m length! 8ow long must the
surf board be to 9ride on the wae% i!e! to hae the same speed as the wae;
Solution
*he speed of the surf board is: gL F V n ⋅=
*he celerity of the wae is:π
λ
2
g c =
8/18/2019 Seakeeping Exercises
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Seakeeping: Velocity and pressure in elementary wave
Consider a regular deep7water wae of wae length λ m and wae height . m!
a) -hat is the orbital elocity in m depth;
b) -hat is the amplitude of the pressure fluctuation (absolute and in percent of the hydrostatic
pressure); *he pressure fluctuation is (linearised): t f p ρφ −=
Solution
a) *he amplitude of the wae is half the wae height: h 2!. m!
*he wae number is: /20!
22 === π λ
π k m−
*he freuency is: 10.!/20!0!= =⋅=⋅= k g ω s−
*he orbital elocity is the maximum of the x7elocity (for deep waes)! *he x7elocity is:
( ))($e kxt ikz x x eehv −−⋅⋅−== ω ω φ
*hus the amplitude is:.!.!210.! /20! =⋅⋅=⋅⋅= ⋅−− eehv kz x ω ms
b) *he pressure fluctuation (pressure without hydrostatic part) is gien by:
( ))($e kxt ikz t f eeh g p −−⋅⋅−=−= ω ρ ρφ
*he amplitude of the pressure fluctuation is then:
4!3.!20!=2. /20! =⋅⋅⋅=⋅= ⋅−− ee gh p kz f ρ +>a*he hydrostatic pressure is
/!0!=2. =⋅⋅=−= gz p stat ρ +>a*hus the pressure fluctuation is 3? of the hydrostatic pressure!
8/18/2019 Seakeeping Exercises
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Seakeeping: Velocity and acceleration in shallow-water wave
*he potential of a regular wae on shallow water is gien by:
( ))())(cosh()sinh($e kxt ie H z k kH ich −−−= ω φ *he following parameters are gien: wae length λ m
wae amplitude h 3 mwater depth H 3 m
@etermine elocity and acceleration field at a depth of A 2 m below the water surfaceB
Solution
*he elocity is deried by differentiation of the shallow7water potential:
)cos())(cosh()sinh(
))(cosh()sinh(
$e )( kxt H z k kH
he H z k
kH
hv kxt i x x −−
−=
−
−== − ω
ω ω φ ω
)sin())(sinh()sinh(
))(sinh()sinh(
$e )( kxt H z k kH
he H z k
kH
hiv kxt i
z z −−=
−
−
==
− ω ω ω
φ ω
*his deriation used the relation c = ω /k ! *he indiidual alues are:
/203!
22 === π λ
π k m−
1/1!)3/203!tanh(/203!0!=)tanh( =⋅⋅⋅== kH gk ω s−
*his yields:
)/203!1/1!cos(0/! xt v x −⋅−=)/203!1/1!sin(40! xt v
z −⋅−=*he accelerations are obtained by differentiating the elocities with respect to time:
)/203!1/1!sin(//! xt a xt x −⋅== φ
)/203!1/1!cos(3/0! xt a zt z −⋅−== φ *he elocities hae a phase shift of = and different amplitudes! *he particles will thus trace an ellipse!
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Seakeeping: Wave breaking
-ae brea+ing occurs theoretically when particle elocity in x7direction is larger than the celerity c of the
wae! &n practice% wae brea+ing occurs for wae steepness h/ λ exceeding 4! -hat is the theoretical and
practical limit for h concerning wae brea+ing in a deep7water wae of λ m;
Solution
>ractically the limit is h λ 4 4 1!43 m!
*he wae number is: /20!
22 === π λ
π k m−
*he maximum particle elocity in x7direction is:kh
x ehv ⋅⋅=ω max%
*he celerity is:k
c ω =
*he condition c = v x,max yields:kh
ekh
=
*he aboe expression is soled iteratiely for kh, using khekh −= ! *he iteration starts with the practicallimit: k 0h 1!43 ⋅ !/20 !4400% then yields:
2 3 4 . / 1 0 =
!4400 !/304 !.20 !.0=1 !..4. !.144 !./3 !./=. !./.0 !./1=
2 3 4 . / 1 0
!.//1 !./14 !./1 !./12 !./12 !./1 !./1 !!!
2/!=/20!
./1!./1! ==→= hkh m
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Seakeeping: Wave with two buoys
*he position of two buoys is gien by their ( x,y) coordinates in mD as s+etched below! *he buoys are excited
by a regular wae of λ /2!0 m% amplitude h m% and angle µ −3 to the x7axis!
a) -hat is the maximum ertical relatie motion between the two buoys if they follow exactly the
waes;
b) -hat is the largest wae length λ to achiee maximum ertical relatie motion of twice the wae
amplitude;
Solution
a) *he wae number of this wae is: !0!/2
22===
π
λ
π k m−
*ransform coordinates in local ξ 7system! >oints on ξ const! hae same ζ 7alues for regular waes!
*he two buoys hae then coordinates:
ξ 1
ξ x2 ⋅ cos 3 − y2 ⋅ sin 3 3 ⋅ cos 3 − ⋅ sin 3 2 m
*he relatie difference between the wae eleations is:
( ( (( ))()(2 22 $e$e$e ξ ω ξ ω ξ ω ζ ζ ik t ik t ik t i eeheheh −−− −⋅=⋅−⋅=−
*he amplitude ( maximum) of this relatie motion is gien by:
( ) ( ) ( ) 222
2
))sin()cos( 2 ξ ξ ξ ω k k heeh ik t i +−⋅=−⋅ −
( ) ( ) 22 )2!sin()2!cos( ⋅+⋅−⋅= !13 m
b) *he maximum difference between wae eleations is twice the wae amplitude if the buoys are
spaced by an odd multiple of half the wae length! *he longest wae is obtained for a spacing of
half the wae length:
22=
λ m → λ 42 m
5ll other wae lengths fulfilling the criterion are een shorter!
8/18/2019 Seakeeping Exercises
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Seakeeping: Encounter freuency
5 ship traels with 20!20 +n in deep sea in regular sea waes! *he ship traels east% the waes come from
southwest! *he wae length is estimated to be between m and 2 m! *he encounter period ! e is
measured at 3!42 s!
a) -hat is the length of the seaway;
b) *wo days before% a storm started in an area . +m southwest of the ship's position!
Can the waes hae their origin in this storm area;
Solution
*he speed is V 20!20 ⋅ !.44 ms 4!.. ms!
*he encounter period yields the encounter freuency: 2!42!3
22===
π π ω
e
e!
s−
*he alue of τ is: 2=//!0!=
..!42! =⋅== g
V eω τ
*here are three possible freuencies which could excite this encounter freuency:
( ) ( ) 23!4.cos2=//!44.cos..!42
0!=cos4
cos2 =°⋅⋅++°⋅⋅
=++= µ τ µ
ω V
g 8A
( ) ( ) //04!4.cos2=//!44.cos..!42
0!=cos4
cos22 =°⋅⋅−+°⋅⋅
=−+= µ τ µ
ω V
g 8A
( ) ( ) 20.4!4.cos2=//!44.cos..!42
0!=cos4
cos23 =°⋅⋅−−°⋅⋅
=−−= µ τ µ
ω V
g 8A
*he corresponding wae lengths are:
4=23!
0!=2222
=⋅
== π
ω
π λ
g m
30//04!
0!=2222
2
2 =⋅
== π
ω
π λ
g m
1./20.4!
0!=2222
3
3 =⋅
== π
ω
π λ
g m
nly λ 30 m fits the obsered bandwidth of wae lengths!
-ae groups trael with group elocity! 6or deep ocean water% this is
34!12
300!=
2
22
=
⋅==
π π
λ g c g" ms
5t this speed% waes can trael within two days:
s 40 h ⋅ / minh ⋅ / smin ⋅ 1!34 ms 2/0 +m F . +m
*he waes can thus not originate from the storm area!
8/18/2019 Seakeeping Exercises
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Seakeeping: !ndamped free heave oscillation
*he differential euation of a free undamped heae oscillation is:
( ) 33 =++ cz z mm m is the displacement of the ship% m## the added mass for heae motion! c is the restoring force coefficient!
*he (circular) natural freuency is subseuently:
33mm
c z +=ω
Consider the ship with the following particulars:
L 2. m% $ 1 m% ! 1 m% % $ !1.% % &p !0% m## !=⋅m!
a) @etermine the (circular) natural freuency and natural period of heaeB
b) &nitial conditions are: 5t t we hae z and z !. ms! -hat is the maximal load a winch of 4 +g mass exerts on its foundation;
Solution
a) *he natural freuency follows from:
)()( 333333 mm! %
% g
mm! $ L%
$ L% g
mm
c
$
&p
$
&p
z +
⋅=
+⋅⋅⋅⋅⋅
⋅⋅⋅⋅=
+=
ρ
ρ ω
0=!)=!(11.!
0!0!==
+⋅⋅⋅
= 8A
0!10=!
22===
π
ω
π
z
z ! s
b) *he winch exerts its maximal load when graity g and amplitude of heae oscillation are
superposed! -e write the heae motion as:
t i z e ' z ω #$e=2
# i' ' ' += is determined by the initial conditions:
( ) ( ) #$e#$e ===== ' ' e ' z z it ω
( ) ./!0=!
.!.!.!#$e 22
−=−=−=→=−=== z
z
i
z t ' ' e ' i z z
ω ω ω ω m
*hen the amplitude of the heae acceleration is gien by:
4.!./!0=! 222 =⋅== ' z z ω ms2
*he maximal load of the winch is then:
4)4.!0!=(4)( =+⋅=+ z g m&inch +G
8/18/2019 Seakeeping Exercises
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Seakeeping: "ower reuirements for a wave maker
5 wae ma+er is to be designed for a towing tan+ of width $ 4 m and depth H 2!. m! *he wae ma+er
shall be designed for a wae of . m length and !2 m amplitude!
a) -hat is the power reuirement for the motor for the wae ma+er if we assume 3? total efficiency
between motor and wae; (6or the considered wae length% the depth can be regarded as 9deep!
*he power reuirement of the wae is (energymeter in direction of wae propagationD) H group
elocity% as the energy in a wae is transported with group elocity!)
b) 5fter switching the wae ma+er off% for a long time there is still a wae motion with period 4 s
obsered in the tan+! 8ow long is the tan+ if the motion is due to the lowest natural freuency of
the tan+;
Solution
a) *he group elocity is:
3=1!2
.0!=
2
22
2
=
⋅===
π π
λ g cc g" ms
*he aerage energy per area is:
2!=/2!0!=2
2
22 =⋅⋅⋅== gh ( ρ Gm
*hus the power of the wae is:
=/3=1!42!=/ =⋅⋅=⋅⋅= g" c $ ( ) -*he power of the motor then needs to be:
/.!33! == ) ) m +-
b) *he wae length is now long compared to the depth! *hus we hae to use finite water depth
expressions:
g kH k
2
)tanh( ω =
-ith .1!2
==!
π ω 8A% this yields: 2..!
0!=
.1!)tanh(
2
==kH k m−
*his euation has to be soled iteratiely! 5s the conergence is slow% it is useful to start with a good
estimate! 6or small x% tanh( x) ≈ x! *his would yield 312!.!22..! ==k !
-e sole the problem using Gewton's iteration:
)('
)(
k f
k f k k −=
-ith 2..!).!2tanh( −⋅= k k f and ).!2(cosh).!2tanh(' 2 k k k f += % we obtain within
1 iterations:
=031.!
2231.! ===→= π π λ
k k m!
*he lowest natural freuency is at λ 2 L% i!e! the tan+ has a length of == m!
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Seakeeping: Wave maker for reuired wave length and amplitude
5 wae ma+er consists of a cylinder which moes up and down at a wall! *he wae ma+er has a cross
section of half a Iewis section with % m !0!
*he draft of the section is ! m% the half7width is $2 !/ m! @etermine the amplitude and freuency of
motion to create waes of . m length and !2 m amplitude!
Gote: *he ratio between wae amplitude and body motion amplitude is z
A !
Solution
*he freuency corresponding to λ . m is:
.!3.
0!=22=
⋅==
π
λ
π ω
g 8A
&mportant parameters for Iewis section cures:
1.4!
0!=
/!.!3
2
22
=⋅
=⋅
g
$ω
/!
/!
2===
!
$ H
*he cures for Iewis sections gie z
A !.2! *hus:
30!.2!
2!3 ===
z A
h* m
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Seakeeping: Wave maker with #ewis section moves floating body
5 wae ma+er consists of a cylinder which moes up and down at a wall! *he wae ma+er has a cross
section of half a Iewis section with % m !0! *he draft of the section is ! m% its half7width $2 !/ m!
*he wae ma+er performs heae motions with !/ m amplitude and ω 3!. 8A! *he basin is m wide!
*he mass of the wae ma+er itself is negligibly small! Jse linear theory in your solution approach!
a) -hat is the amplitude of the force needed to moe the wae ma+er;
b) -hat is the aerage power supply if an efficiency of 1? is assumed;
c) &n the middle of the basin is a cylinder of % m !0% $ !2 m% ! m% L . m% arranged with axis
parallel to the wae ma+er axis! *he cylinder has all six degrees of freedom suppressed! -hat is the
exciting force on the cylinder;
d) &f the cylinder in the middle of the basin is free to heae% what is its heae amplitude;
5ssume deep water!
Solution
a) *he basic euation for heae motion is:
( ) f *cnimm −=⋅+++− 33333332 ω ω
(*he force is negatie as now the body excites the water and not the water the body!) 5ll uantities
are in this euation per length% i!e! for the final computation they need to be multiplied by the tan+
width! *he indiidual uantities are:
1.4!0!=
/!.!3
2
22
=⋅=⋅ g
$ω
/!
/!
2===
!
$ H
*he cures for Iewis sections gie % z !/% .2!= z A ! *hus:
33=0
2!/!
0
22
33 =⋅⋅⋅=⋅⋅⋅= π π ρ $
% m z +gm
0!/.!3
0!=.2!
3
22
3
22
33 =⋅
⋅=⋅=ω
ρ g An z +gms
112!0!=33 =⋅⋅== g$c ρ Gm2
*he mass of the wae ma+er is negligible: m
*hus:
)2/14../(/!110!/.!3)33=(.!3 2 i f f i +=→−=⋅+⋅⋅++⋅− Gm*his force is for waes radiated in both directions% i!e! so far we assumed symmetry! *he total force
amplitude is then gien by:
/!232/14../2
KK
2
KK 22 =⋅+=⋅= !ank $ f F +G
b) *he power needed to drie the wae ma+er is gien by:
η
ρ
η
22
2
⋅⋅⋅=⋅⋅
= c $ ghc $ (
) !ank g" !ank
+G
*he wae amplitude is:
32!.2!/!KK 3 =⋅=⋅= z A*h m*he celerity is:
1=!2
.!3
0!====
ω
g c ms
*hus:
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.!=1!
2
1=!2
2
32!0!= 2=⋅⋅⋅⋅⋅= ) +-
Gow we chec+ the result% using a different approach! *he power needed is due to the damping force
times the elocity! *he mass force and the restoring force are phase shifted by = to the elocity
and thus do not contribute to the power! 5gain we ta+e into account that we hae only half a cross
section:
.!=1!
)/!.!3(
2
2/1
2
)#(
2
)&m(
2
=⋅⋅⋅
⋅=⋅⋅⋅
⋅=
η ω z $
f ) !ank +-
c) -e read from the diagrams for Iewis sections:
(2!=032!2!0!=2.!2.!2.!
#$e% =⋅⋅⋅⋅=⋅=→= g$h f
g$h
f " e
e ρ ρ
Gm
(2!=02.!
#&m% =→= ie
e f g$h
f
ρ Gm
*his then gies the absolute alue for the three7dimensional force:
.!/.2!=02!=0 222%2% =⋅+=⋅+= L f f F ie" ee +G
d) *he basic euation for heae motion is:
( ) e f *cnimm =⋅+++− 33333332 ω ω
*he mass (per length) follows from 5rchimedes:
=/!2!0! =⋅⋅⋅=⋅⋅⋅= ! $% m m ρ +gm5s before% we hae then:
( )[ ] )2!=02!=0(110!/.!333==/.!3 32 i*i +=⋅+⋅⋅++−[ ] )2/!01!()2!=02!=0(3!220!4233 33 i*i*i −−=→+=⋅+− m
214!)2/!01!22
3 =+=* m
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Seakeeping: Cylinder with #ewis cross section in regular waves
5 cylinder of Iewis cross section ( L m% $ m% ! !4 m% % m !0) floats parallel to the wae crests
of regular waes λ . m% h !2. m) which excite heae motions! 5ssume that the form of the free surface
is not changed by the cylinder! -hat is the amplitude of relatie motion between cylinder and free surface
predicted by linear theory;
Solution
6undamental euation for heae motion:
( ) e f *cnimm =⋅+++− 33333332 ω ω
*he indiidual uantities needed are:
.!3.
0!=22=
⋅==
π
λ
π ω
g 8A
/20!0!=2
.!3
2
22
=⋅
⋅=
⋅ g
$ω
2.!4!2
2=
⋅==
!
$ H
*hen the cures for Iewis sections yield: % z !/2% /0!= z A % 3=!H =e" f % 30!
H =ei f ! *his in turngies:
2430
/2!
0
22
33 =⋅⋅⋅=⋅⋅⋅= π π ρ $
% m z +gm
324!!0! =⋅⋅⋅=⋅⋅⋅= ! $% m m ρ +gm
2=.!3
0!=/0!
3
22
3
22
33 =⋅
⋅=⋅=ω
ρ g An z +gms
=0!0!=33 =⋅⋅== g$c ρ Gm2
( ) iihi f f g$ f eie" e !=32.!=./2.!)30!3=!(!0!=HH +=⋅+⋅⋅⋅=+= ρ Gm
-e assume ρ +gm3 here for conenience! 5s ρ appears in all terms% the final result is not influenced
by the choice of ρ !
*his yields the heae motion amplitude:
( )[ ] i*i !=32.!=./=02=.!332243.!3 32 +=⋅+⋅⋅++−
ii
i* 3//!201!
3/22014
!=32.!=./3 −=+
+= m
*he amplitude of relatie motion is determined from:
( ) 223 3//!31!3//!31!2.!3//!201! +=−=−−=−= iih*" !.2 m
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Seakeeping: #ewis section in forced heave
5n infinite cylinder of width $ 2 m% draft ! m and with Iewis cross section of coefficient % m !0
floats in euilibrium in fresh water! *hen a harmonic force per length with period ! e 3!4 s and amplitude
f e Gm is applied! -hat motion results after a long time (when the initial start7up has decayed);
-hat wae length and what wae amplitude are generated% assuming a 9deep basin;
Solution
*he motion has only one degree of freedom% namely heae! *he basic euation is then:
( ) e
f *cnimm =⋅+++− 33333332 ω ω
*# is the complex amplitude of heae motion! *he indiidual uantities needed are:
24!3
22 === π π ω e!
8A
40!
0!=2
22
2
22
=⋅
⋅=
⋅
g
$ω
2
2
2=
⋅==
!
$ H
*hen the cures for Iewis sections yield: % z !1% .!= z A ! *his in turn gies:
0
21!
0
22
33 =⋅⋅⋅=⋅⋅⋅= π π ρ $
% m z +gm
/!!20! =⋅⋅⋅=⋅⋅⋅= ! $% m m ρ +gm
32=2
0!=.!
3
22
3
22
33 =⋅
⋅=⋅=ω
ρ g An z +gms
=/2!20!=33 =⋅⋅== g$c ρ Gm2
*his yields the heae motion amplitude:
( )[ ] =/232=2/2 32 =⋅+⋅⋅++− *i
ii
* .3.!1.4!/2.0002
3 −=+= m
*he amplitude is:
=2.!.3.!1.4! 223 =+=* m
*he wae length follows from:
4!.22
2
2
==→=ω
π λ
λ
π ω g
g m
*he wae amplitude follows from z A :
41!=2.!.!3 =⋅=⋅= * Ah z m
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Seakeeping: "ower generator
5 cylinder for power generation is moored in regular waes such that the cylinder axis is parallel to the wae
crests! *he cylinder dimensions are L . m% $ m% ! . m% the cross section is a Iewis section with
% m !0! *he waes hae λ . m and h m! *he mooring adds a restoring force coefficient c 2!.⋅/
Gm and a damping (from the generator) of + !.⋅
/
+gs! *he mooring remains at all times under tensionas the buoyancy of the cylinder is much larger than its weight due to its mass of / +g! *he sea water has a
density of 2. +gm3!
-hat is the amplitude of heae motions for the cylinder;
-hat would be the amplitude without generator;
Solution
*he basic euation of motion is:
( ) e Lf * Lcc Ln+ i Lmm =⋅+++++− 3333333
2 )()(ω ω
*he indiidual uantities needed are:
!.
0!=22 =⋅== π λ π ω g 8A
/20!0!=2
!
2
22
=⋅
⋅=⋅ g
$ω
.2
2=
⋅==
!
$ H
*hen the cures for Iewis sections yield: % z !/% /.!= z A % 3=!H =e" f % 3/!
H =ei f ! *his in turngies:
/22
33 221.!0
2./!.
0⋅=⋅⋅⋅⋅=⋅⋅⋅⋅=⋅ π π ρ
$% Lm L z +g
/3
2
23
2
233 .23.!
!0!=2./.!. ⋅=⋅⋅⋅=⋅⋅=⋅ ω
ρ g A Ln L z +gs
/
33 21.!.0!=2.. ⋅=⋅⋅⋅=⋅=⋅ g$ Lc L ρ Gm
( ) /HH )0!=/!()30!3=!(0!=2.. ⋅+=⋅+⋅⋅⋅⋅=+⋅=⋅ iihi f f g$ L f L eie" e ρ G*hen the basic euation yields:
( )[ ] i*i 0!=/!)21.!..!2().23.!.!(!221.!! 32 +=⋅+++⋅++−
ii
i* /!4.3!
3./!3103!4
0!=/!3 −=+
+= m
*he amplitude is:
4/!/!4.3! 223 =+=* m*he basic euation modifies for the case of no generator (i!e! + ):
( ) e Lf * Lcc Lni Lmm =⋅++++− 33333332 )()(ω ω
( )[ ] i*i 0!=/!)21.!..!2(.23.!!221.!! 32 +=⋅++⋅++−
ii
i* 20!403!
/=!103!4
0!=/!3 −=+
+= m
.3!20!403! 223 =+=* m
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Seakeeping: Catamaran in waves
5 raft consists of two cylinders with m diameter and L m length! *he two cylinders hae a
distance of 2e 3 m from centre to centre! *he raft has no speed and is located in regular waes coming
directly from abeam! *he waes hae λ 3π m and wae amplitude h !. m! 5ssume the two cylinders to
be hydrodynamically independent% i!e! waes created by one cylinder are not reflected at the other cylinder!
*he raft has a draft of !. m% centre of graity in the centre of the connecting plate% radius of moment of
inertia for rolling is k x m! *he density of the water is +gm3! *he % m π4 is close enough to !0 to
use the Iewis section cures for this % m! Geglect sway motion! -hat is the maximum roll angle;
8int: *he centre of graity of a semi7circle is 4" (3π) from the flat baseline% where " is the radius!
Solution
*he mass of the raft is: 10.420
2
0
22
=⋅⋅⋅
⋅=⋅⋅ π π
ρ L ,
+g
*he mass moment of inertia is: 10.410.422 =⋅=⋅=Θ xk m +gm2
-ae number: ///1!3
22===
π
π
λ
π k m−
-ae freuency: ./!2
3
0!=22 =⋅==π
π
λ
π ω
g s−
*he metacentric height is:
( ) ,
L ,
,Le L ,! '- $. '$-. −
⋅++⋅
−=−+=
4
22
3
4
2
23
π π
( )
23!.4
2.!2.!
3
4
2
23
=−⋅⋅
⋅⋅⋅+⋅+⋅
−=
π π m
>arameters for Iewis cures: .!2
!
2=
⋅=
!
$and 333!
0!=2
!./!2
2
22
=⋅
⋅= g
$ω
*hen the cures for Iewis sections yield: % z !0% 4!= z A % /!H =e" f % 2.!
H =ei f
*he basic euation of roll motion is: 44444444
2
)( e f *cnim =+++Θ− ω ω *he hydrodynamic added mass m and damping n (moments) are deried from the added mass for heae(forces) times leer e:
( ) e*m*m ⋅⋅⋅= 333444 2 *he factor 2 is due to the two cylinders! *he term in parentheses is the ertical force! -ith 43 *e* ⋅= % weget for the whole 37d raft:
431.!0!0
!2
02 2
22
2
44 =⋅⋅
⋅
⋅⋅⋅=⋅⋅
⋅⋅= π π ρ e L% $m z +gm2
43.!./!2
0!=4!22
2
3
222
3
22
44 =⋅⋅
⋅⋅⋅=⋅⋅
⋅= e L
g An z
ω
ρ ρ +gm2s
*he restoring force constant for roll motion is:4323!.0!=10.444 =⋅⋅=⋅= -. mg c +Gm
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*he exciting moment has to consider the phase shift in the wae between the two hulls! *he force on the left
floater is then:
( ) ikeeie" 0 e L g$hif f f −⋅⋅⋅+= ρ HH#
*he force on the right floater is:
( ) ikeeie" " e L g$hif f f ⋅⋅⋅+= ρ HH#
*he phase shift is thus in the last term! Combine the two forces with leer e to get:
( ) ( )ikeikeeie" e ee L g$hif f e f −⋅⋅⋅+⋅= − ρ HH4
( ) ( )).!///1!sin(2.!!0!=2.!/!.! ⋅−⋅⋅⋅⋅⋅⋅+⋅= ii (3=./ − 142=3 i) +gm2s2
*hen our basic euation becomes:
)142=33=./(4343./!2)43110.4(./!2 42 i*i −=⋅+⋅⋅++−
[ ] i*i*i 200!2!)2=!14=/!3(1!.=!2.0 44 −=→−=⋅+*he (real) roll amplitude is then
°==+= .!/200!200!2! 224*
*he assumption of linearity is thus no longer Lustifiable!
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Seakeeping: Strip method roll
5 circular cylinder (length L 2 m% m% radius of moment of inertia k xx !4 m% mass m 0 +g)
floats in water of density ρ = +gm3! *he centre of graity is ! m under the centre of the circle! *he
cylinder is excited by regular waes with wae crests parallel to the cylinder axis! *he waes hae amplitude
h !. and length λ 3!4 m! *he following hydrodynamic properties are +nown:
a) hydrodynamic mass for sway: !4 m
b) hydrodynamic damping for sway gien by ratio of amplitudes 3!= y A
c) real part of horiAontal exciting force: −!0 orbital elocity of wae times damping coefficient as
gien in b)
d) imaginary part of horiAontal exciting force: −!/ wae steepness times cylinder weight
*he coordinate system should hae its origin in the centre of the circle! *hen the resultant force of the
hydrodynamic forces passes through the origin!
@etermine the complex amplitude of roll motionB
Solution
-e use a coordinate system with z pointing down% y pointing right in direction of wae propagation! x is then
the cylinder axis!
*he displacement of the fully submerged cylinder is:
/=24
4
22
=⋅⋅⋅
=⋅⋅ π ρ
π L
, +g
5s the cylinder mass is 0 +g% the draft of the cylinder is ! 2 !. m! *he freuency is:
43!44!3
0!=22
=⋅
== π
λ
π
ω
g
8A
*he orbital elocity inoles differentiation with respect to time (factor ω )% the steepness with respect to y
(factor k )! *he hydrodynamic uantities 9hidden in a) to d) are then:
3204!4!22 =⋅=⋅= mm +g
24243!4
0!==3!
3
22
3
22
22 =⋅⋅⋅=⋅⋅= L g An y
ω
ρ +gs
0424.!43!40!0!$e 22 −=⋅⋅⋅−=⋅⋅⋅−= nh F e ω G430.!43!4/!/!/!&m 22 −=⋅⋅⋅−=⋅⋅⋅−=⋅⋅⋅⋅−= mhm g hk F e ω G
-e can use the formulae for strip method and simplify them! 5s the waes come Lust from abeam% no
9symmetric motions (heae% surge% pitch) are excited (with linear theory)! 5lso% as the cylinder has always
the same cross section% no yaw motion is excited! *he fundamental euation of motions is:
[ ] (h*1 2 i A . =⋅+++− ω ω )(2"way motion does not excite roll motion% neither does roll motion excite sway for a circular cylinder with
origin as chosen! *here is no exciting roll moment either as in potential theory only normal pressures exist
and for the circular cylinder all normal ectors pass through the origin% i!e! excite no moment! "imilarly%
there is no radiation moment% i!e! no damping and added mass for roll! "way has no restoring term! *he
metacentre of a circular cylinder is the centre of the circle! *hus - ! m! *hus we can write for the two
degrees of freedom:
−
−=
−
−⋅=
Θ−
−=
200
002
xx g
g
z g
g
k z
z m
mz
mz m .
⋅=
=
32
22m
m A
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⋅=
=
24
22m
n 2
⋅=
=
1040
m
mg-. 1
*hus:
+−
=
+
+
−
−
430
1040
=31
2.2.1
.12=0
4
2 i
*
*i
+−
=
−
+−
430
1212.1
.1=312=0
4
2 i
*
*i
*he solution of this system of euations is:
+−+
=
i
i
*
*
/3!2!
/=!2!
4
2
*hus the roll amplitude is 224 /3!2! +=* !/3 3!/!
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Seakeeping: "ontoon with crane in waves
Consider a pontoon with a heay7lift derric+ as s+etched! *he pontoon has L m% $ 2 m% m%
m p 1 +g! *he load at the derric+ has mass m
/ +g! *he height of the derric+ oer dec+ is m! *his is
where the load can be considered to be concentrated in one point! *he longitudinal position of the derric+ is
2 m before amidships! 5 force 6 / G acts on the forward corner! -e assume homogeneous mass
distribution in the pontoon!
a) Consider the pontoon 9in air (without hydrodynamic masses) and determine the acceleration ector
* B b) Consider the pontoon statically in water and determine *B
Solution
5ll numbers are gien in standard units! -e use the coordinate system as in the boo+ with z pointing down!
rigin is at ' !
a) *he general /7component force ector is:
⋅−
−−
=
⋅−
⋅−−
=
.
.
1
1
/
m F
m F
F F
*he mass matrix needs the following expressions:1/1 ! ⋅=+=+= 0 p mmm
1/1
%% 1)2().( ⋅−=−⋅+−⋅=⋅+⋅=⋅ 0 g 0 p g p g z m z m z m1/1
%% 22 ⋅=⋅+⋅=⋅+⋅=⋅ 0 g 0 p g p g z m xm xm
*he moments of inertia are computed for pontoon and load:
+=
+=
+⋅
⋅=+=Θ ∫ 4
2
3
4343)(
22
122
3322
%
$ ,
m $ , , $
Lm+ y z
p p
p xx
ρ
!//1 ⋅ =
=4/2
%% 4!2 ⋅=⋅=⋅=Θ 0 g 0 0 xx z m
*hus=
%% /1! ⋅=Θ+Θ=Θ 0 xx p xx xx ! Correspondingly we get:% ==Θ ∫ m+ xz p xz
=/%%% 4!2)2( ⋅−=⋅−⋅=⋅⋅=Θ 0 g 0 g 0 0 xz z xm
*hus=
%% 4! ⋅−=Θ+Θ=Θ 0 xz p xz xz !
+=
+=
+⋅
⋅=+=Θ ∫ 4
3
4343)(
22
122
3322
%
L ,
m L , , L
$m+ z x
p p
p yy
ρ
0!//1 ⋅ =
( ) =22/2%2
%% 0!)22( ⋅=+⋅=+⋅=Θ 0 g 0 g 0 0 yy z xm
*hus=
%% 4/1!= ⋅=Θ+Θ=Θ 0 yy p yy yy
( ) ( ) ( )221
223322
% 22
22)( +=+=+
⋅=+=Θ ∫ L $
m $L L$
,m+ y x
p p
p zz
ρ
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0!//1 ⋅ =
( =22/2%2
%% 4!)2( ⋅=+⋅=+⋅=Θ 0 g 0 g 0 0 yy y xm
*hus=
%% /1!= ⋅=Θ+Θ=Θ 0 zz p zz zz
*his yields the mass matrix:
−−
−
−
⋅=
=/142
=4/121
4/11
221
1
/ .
Gow we can use the fundamental euation * F ⋅= % where F and * are generalised /7component
ectors! "ince we sole manually% it is adisable to decouple the /×/ system of euations into 3×3
systems for symmetric and anti7symmetric degrees of freedom:
.
3
.3
.3
.
!!
3!
.
=4/1212
1
−=−=
−=
→−−
==
=
−−−
−
**
*
*****
**
/
4
2
/42
/42
/42
=/142
4/11
21
=
=
=
→−
=
=
=
*
*
*
***
***
***
b)
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Seakeeping: "robability of exceeding operational limit
5 ship sails in a natural seaway of approximately ! e / s encounter period between ship and waes! *he
bridge of the ship is located forward near the bow! @uring one hour% / times a downward acceleration
occurred exceeding graity acceleration g ! 6or downward accelerations exceeding !. g % seere inLuries for
the crew hae to be expected! -hat is the probability for this happening if the ship continues sailing with the
same speed in the same seaway for 2 hours;
Solution
*he number of amplitudes encountered per hour (3/ s) is:
//
3/3/===
e! n
*he actual number of amplitudes for acceleration exceeding g in one hour was /! *hus the probability that
one amplitude exceeding g occurs is 4 ( g ) // !! n the other hand% we can write this probability
formally:2
)2()( =!!)( 2
g me g 4 m g =→== −
*he probability for an amplitude exceeding !. g is correspondingly:.)=!2(.!)2().!( 2=!3).!(
2
2 −⋅−⋅− ⋅=== ee g 4 m g
*he number of amplitudes encountered in 2 h is !12/22 =⋅=⋅ n *hus the probability for oneamplitude exceeding !. g in 2 h is:
.2=!312 −⋅⋅=4 !24 24 ?
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Seakeeping: $oll motion of cargo ship in waves
5 new cargo ship type features relatiely large beam and extreme sectional flare at the ship ends to allow
many container on dec+! *he ship has the following main data: L4L =4m% $ / m% 0!4 m% ! . m%
radius of inertia (including added mass) k5 !4 $ = /!4 m! *he serice speed is V !0 +n!
&nestigate whether (respectiely when) there is a danger of capsiAing due to resonance in wae induced
rolling for possible metacentric heights (!3 m F - F ! m)! Consider the following scenarios:
a) waes from abeam (λ F . m)
b) following waes (λ F . m)
&f there is a danger of capsiAing% specify the conditions which are most critical! 5ssume deep water!
Solution
a) waes from abeam
$esonance appears for ω e ω n! 6or elementary waes on deep water this yields:
-. -. -.
k
k
-. g g 4!2.14!/2'2
'
2 22
2 =
⋅=
⋅=→
⋅= π π λ
λ
π
*hus for !3 m F MN F ! m only wae lengths 2.1 m F λ F 0.0 m lead to resonance! 6or the
gien range below . m% there is no danger of capsiAing!
b) following waes
$esonance appears for2
⋅= ne
n
ω
ω with n % 2% 3% !!!
6or following waes (angle of encounter µ ) we hae: V g
e ⋅−=2
ω ω ω
-ith λ π ω 2 g = we get:λ
π
λ
π ω
V g e
22−=
*he first resonance lies at ω n O ω e ! *he natural freuency of roll is'k
-. g n
⋅=ω
*hus we get for the first resonance:λ
π
λ
π V g
k
-. g −=
⋅ 22
'
"oling for - yields:2
222
3
232 ''2
2
'
λ
π
λ
π
λ
π
g
V k V k
g
k -. +−=
*a+e all uantities in standard units (m and s)% to get with V !0 +n /!1 ms:
232
222
3
2323!.0!/2.34!/4
0!=
1!/4!/1!/4!/
0!=
2
2
4!/
λ λ λ λ
π
λ
π
λ
π +−=
⋅⋅⋅
+⋅
−⋅
=-.
*his is ealuated for arious wae lengths in the gien interal:
λ mD 3 4 . / 0 2 .
MN mD !3 != !3 !. !1 !1 !1 !/
*he case MN P !3 m is neer reached! *hus there is no danger of capsiAing for n !
*he second resonance lies at ω n ω e! @ue to the uadratic relation% we hae thus four times thealues for - :
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λ mD 3 4 . / 0 2 .
MN mD ! !3. !. !/ !/1 !/0 !// !/2
*hus for most wae lengths% there is a danger of resonance for smaller - alues!
*he third resonance lies at ω n !. ω e!
λ mD 3 4 . / 0 2 .
MN mD ! !3. !. !/ !/1 !/0 !// !/2
*he third resonance is also reached% but only for wae lengths much shorter than the ship length!
*hus the expected fluctuation in righting leers will be smaller as for second resonance!
Correspondingly the probability of capsiAing will be lower!
8igher resonances will shift resonance to een shorter waes which will be uncritical!
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Seakeeping: $oll motion of submarine in waves
Consider a surfaced submarine with speed V / +n! Geglect the sail and model the submarine hull as a
circular cross section! Geglect damping!
a) -hat is the expected roll amplitude in elementary waes from abeam;
b) Ealuate the danger of parametric roll excitation in elementary following waesB
"hip data: L 1! m% 0! m (diameter)% ! 1! m% '- 3!/ m% k5 3! m (radius of the
transerse mass moment of inertia incl! added mass)
-ae data (deep water): λ 1 m (wae length)Q h !1. m (wae amplitude)
Solution:
a) waes from abeam
-ae circular freuency: =303.!1
0!=22=
⋅==
π
λ
π ω
g e s
−
6or a circular cylinder% the metacentre is always in the centre of the circle:
4!/!32
!0
2=−=−=−= '- , '- '. -. m
*he natural freuency for roll is: //3!!3
4!0!=
' 22 =
⋅=
⋅=
k
-. g nω s
−
6or 42!//3!
=303.! ==n
e
ω
ω the roll excitation lies in the super7critical region! *he roll amplitude
for undamped roll is thus (with k 2πλ ):
°==⋅⋅−
=⋅⋅−
== 03!0.4/!1.!1
2
42!
)(
K#KKK
224
π
ω ω ϕ hk *
ne
b) following waes
Encounter freuency ω e is then:
( )
//3!1
.44!/2
1
0!=2222=
⋅⋅−
⋅=−=⋅−=
π π
λ
π
λ
π ω ω ω
V g V
g e s
−
5lthough the roll natural freuency ω n is close to encounter freuency ω e% there is no parametric
excitation% because for a circular cylinder there is no fluctuation of righting leers! *he righting leer
is always h - ⋅ sin ϕ % with - !4 m const!
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Seakeeping: Cruise ship in waves from abeam
5 small cruise ship is designed! 5s the destination island has no deep7water port% the passengers will
disembar+ ia small boats! *o allow normal disembar+ation% the metacentric height should be always
sufficiently high to hae maximum roll amplitude ϕ / in regular waes from abeam up to wae steepness
δ max 4!. up to wae length of λ max / m! 5ssume a linear restoring moment! 6or the small roll anglesconsidered% damping can be neglected!
*he radius of inertia including added mass is k5 4 m!
-hat is the resulting reuirement for the metacentric height;
Solution
*he reuirement ϕ ≤ / for δ max 4!. means for the amplification function: 33!.!4
/
max
==≤δ
ϕ "ati6V
*he euation for V yields then limiting ratios of freuency:
1.!)(
2.!)(33!
)(
2
2
2 ≥≤
→≤−
=ne
ne
ne
"ati6V ω ω
ω ω
ω ω
*hus the freuency interal to be aoided is 1.!)(2.!2 ≤≤ ne ω ω ! -ith the relations for a deep water
regular wae:
λ
π ω
g e
2=
and for the natural freuency of roll
2'k
-. g n
⋅=ω
we get for the interal to be aoided:
1.!'2
2.!2
≤⋅⋅
≤-.
k
λ
π
6or arious wae lengths% we can then determine the interal to be aoided:
λ interal to aoid
2 m 2!01 m ≤ - ≤ 2! m
4 m !44 m ≤ - ≤ !. m
/ m !=/ m ≤ - ≤ /!1 m
ne can uic+ly see that short wae lengths pose no problem! *he limiting alue is - ≤ !=/ m! *his
reuirement will hardly be possible for a passenger ship due to damaged stability reuirements!