[2009] [09] [10] Innovative ship design Naval Architecture & Ocean … · 2018. 1. 30. · Department of Naval Architecture and Ocean Engineering, Seoul National University of College

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[2009] [09] [10]

Innovative ship design

- Ship Motion & Wave Load -

April, 2009

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering,Seoul National University of College of Engineering

서울대학교 조선해양공학과 학부4학년 “창의적 선박설계” 강의 교재

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Chap.1 Loads acting on a ship

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Ship Structural Design Ship Structural Design

what is designer’s major interest?

Safety : Won’t ‘IT’ fail under the load?

L

x

y( )f x

( )f x

x

y

reactF

( )V x( )M x

( )y x

Differential equations of the defection curve4

4

( ) ( )d y xEI f xdx

= −

what is our interest?

: ( ): ( )

: ( )

Shear Force V xBending Moment M xDeflection y x

: ( )Load f xcause

( ) ( )dV x f xdx

= − ( ), ( )dM x V xdx

=

2

2

( ), ( )d y xEI M xdx

=

‘relations’ of load, S.F., B.M., and deflection

Safety : Won’t it fail under the load?

Geometry :How much it would be bent under the load?

, acty i

M MwhereI y Z

σ = =act lσ σ≤

Stress should meet :a shipa stiffenera plate

global

local

σact : Actual Stressσt : Allowable Stress

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a shipa stiffenera plate

L

x

y( )f x

global

local

a ship

Actual stress on midship section should be less than allowable stress

lact σσ ≤.

Allowable stress by Rule (for example):2

1, 175 [ / ]l f N mmσ =

., S Wact

mid

M Mσ

Z+

= Hydrostatics

L

x

y( )f x

( )f x

x

y

reactF

( )V x( )M x

( )y x


4


= −


: ( ): ( )

: ( )


: ( )Load f xcause

( ) ( )dV x f xdx

= − ( ), ( )dM x V xdx

=

2

2


=




, acty i

M MwhereI y Z

σ = =

Stress should meet :


act lσ σ≤

MS = Still water bending momentMW = Vertical wave bending moment Hydrodynamics

z

x

what kinds of load cause hull girder moment?

f

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global

local

a ship

L

x

y( )f x

( )f x

x

y

reactF

( )V x( )M x

( )y x


4


= −


: ( ): ( )

: ( )


: ( )Load f xcause

( ) ( )dV x f xdx

= − ( ), ( )dM x V xdx

=

2

2


=




, acty i

M MwhereI y Z

σ = =



act lσ σ≤

( )Sf x

., S Wact

mid

M Mσ

Z+

=lact σσ ≤.

: load in still water

0( ) ( )

x

S SV x f x dx= ∫( )SV x

( )SM x

0( ) ( )

x

S SM x V x dx= ∫

Hydrostatics Hydrodynamics

.F K

diffractionadded mass

mass inertiadamping

: still water shear force

: still water bending moment

, MS = Still water bending momentMW = Vertical wave bending moment

what kinds of load f cause hull girder moment?

weight

buoyancy

fS(x) : load in still water= weight + buoyancy

( )Wf x : load in wave

0( ) ( )

x

W WV x f x dx= ∫( )WV x

( )WM x

0( ) ( )

x

W WM x V x dx= ∫: wave shear force

: vertical wave bending moment

fW(x) : load in wave= added mass + diffraction

+ damping + Froude-Krylov + mass inertia

( ) ( ) ( )S Wf x f x f x= +( ) ( ) ( )S WV x V x V x= +( ) ( ) ( )S WM x M x M x= +

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1. Loads acting on a ship (1)- Loads in still water

z

xx

z( )w x w(x):weight

x

z( )b x+

b(x):bouyancy

In still water( ) ( ) ( )Sf x b x w x= −

what kinds of load cause ?sMSf

( ) ( ) ( )S Wf x f x f x= +

x

z( )Sf x

f S(x)= b(x) – w(x) : Load

=b(x) : buoyancy distribution in longitudinal directionw(x) = LWT(x) + DWT(x)

- w(x) : weight distribution in longitudinal direction- LWT(x) : lightweight distribution- DWT(x) : deadweight distribution

(MS : Still water bending Moment in midship)

f(x) : distribution load in longitudinal directionfS(x) : distribution load in longitudinal direction in still waterfW(x) : distribution load in longitudinal direction in wave

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1. Loads acting on a ship (2) - Example of 3,700TEU Container carrier

Load Curve, fS(x)Still water

Shear Force, VS(x)

Still water

Bending Moment, MS(x)Weight, w(x)Buoyancy, b(x)

- Frame space : 800mm

Example of 3,700 TEU Container Ship in Homogeneous 10t Scantling Condition

- Principal dimensions & drawings

- Loading Condition (Sailing state) in homogeneous 10t scantling condition

0( ) ( )

x

S SV x f x dx= ∫ 0( ) ( )

x

S SM x V x dx= ∫

- principal dimension - profile & plan drawing - midship section

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LIGHTWEIGHT DISTRIBUTION DIAGRAM

FR. NO

0

25

50

74

99

125

150

175

200

226

251

276

301

326

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

200.0

220.0

240.0

TONNES

Crane

Bow ThrusterEmergency

Pump

Engine

(Example of 3,700TEU Container carrier)

1. Loads acting on a ship (2) - Lightweight

AP FP

E/R

A.P F.P


Shear Force, VS(x)

Still water


0( ) ( )

x

S SV x f x dx= ∫ 0( ) ( )

x

S SM x V x dx= ∫

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- Loading Plan in homogenous 10t scantling condition

1. Loads acting on a ship (3) - Deadweight

(Example of 3,700TEU Container carrier)Deadweight distribution in longitudinal directionin homogenous 10t scantling condition

- Deadweight distribution curve in homogenous 10t scantling condition

A.P F.PFR.Space : 800 mm


Shear Force, VS(x)

Still water


0( ) ( )

x

S SV x f x dx= ∫ 0( ) ( )

x

S SM x V x dx= ∫

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(2) 면적 길이 방향으로 적분하여 부피계산

(1) 수선면 아래의 선박단면적계산

x

A

'y

'z

Buoyancy Curve in Homogeneous 10ton Scantling Condition

1. Loads acting on a ship (4) - Buoyancy curve


100 ton

FR.NoA.P F.PFR.Space : 800 mm

Buoyancy 계산방법


Shear Force, VS(x)

Still water


0( ) ( )

x

S SV x f x dx= ∫ 0( ) ( )

x

S SM x V x dx= ∫

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LIGHTWEIGHT DISTRIBUTION DIAGRAM

FR. NO

0

25

50

74

99

125

150

175

200

226

251

276

301

326

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

200.0

220.0

240.0

TONNES

=Lightweight + Deadweight

Load Curve

Still Water shear Force

Load Curve

=Weight+ Buoyancy

+

+

=

Bouyancy Curve

Weight curve

( )Sf x

1. Loads acting on a ship (5) - Load/Shear/Moment curve


Lightweight Distribution Curve Deadweight Distribution Curve

FR.No

A.P F.P A.P F.PA.P F.PFR.Space : 800 mm

100 ton

FR.NoA.P F.P

FR.Space : 800 mmA.P

F.P

= Weight w(x) + Buoyancy b(x)

Load Curve, fS(x)Still Water

Shear Force, VS(x)

Still Water


0( ) ( )

x

S SV x f x dx= ∫ 0( ) ( )

x

S SM x V x dx= ∫

in homogenous 10t scantling condition

in homogenous 10t scantling conditi onin homogenous 10t scantling condition

in homogenous 10t scantling condition

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Shear force

Permissible shear force

Still Water

Shear Curve

Bending Moment

Permissible Bending Moment

Still Water

Bending

Moment Curve

Load Curve

Actual still water shear force is lower than permissible Shear.

→ O.K

Actual still water bending moment is lower than permissible bending.

→ O.K

( ) ( ) ( )Sf x b x w x= −

0( ) ( )

x

S SV x f x dx= ∫

1. Loads acting on a ship (6) - Load/Shear/Moment curve


0( ) ( )

x

S SM x V x dx= ∫


Shear Force, VS(x)

Still water


0( ) ( )

x

S SV x f x dx= ∫ 0( ) ( )

x

S SM x V x dx= ∫

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33 3 33 3( ) ( ) ( )Rf x a x b xξ ξ= − −

How to know ? 3 3,ξ ξ

z

x

z

x

In still water

In wave

( ) ( ) ( )Sf x b x w x= −

( ) ( )( ) S Wf x f x f x= +

3 .( ) ( ) ( ) (( ) ( )) D F K Rm x f x f x f xb x w x ξ− + + += −

x

z( )Wf x

?

2. Loads in wave

• for example, consider heave motion

• from 6DOF motion of ship

[ ]T654321 ,,,,, ξξξξξξ=x

3ξ

x

z( )Sf x

f S(x)= b(x) – w(x) : Load

+

fD(x) : Diffraction force in a unit lengthfR(x) : Radiation force in a unit lengthfF.K(x) : Froude-Krylov force in a unit length

Where,

Roll ξ4

Pitch ξ5

Hea

ve ξ

3

Yaw ξ6

O

x

y

z

ref.> 6 DOF motion of ship

additional loads in wave

Loads in wave

In order to know loads in wave,we have to know 3 3,ξ ξ

f(x) : Distribution load in longitudinal directionfS(x) : Distribution load in longitudinal direction in still waterfW(x) : Distribution load in longitudinal direction in wave

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3. 6DOF Equation of motion of ship

R

D

I

ΦΦΦ : Incident wave velocity potential

: Diffraction wave velocity potential

: Radiation wave velocity potential

FF.K : Froude-krylov forceFD : Diffraction forceFR : Radiation force

matrixcoeff.restoring66:matrixcoeff.damping66:matrixmassadded66:

×××

CBM A

선박에 작용 하는 유체력

.Static F K D R+ + +F F F F+GravityF=xM

RestoringF Wave exciting F R = − −F Ax Bx

, ,( )Gravity Static Wave exciting External dynamic External static= + + − − + +Mx F F F Ax Bx F F

( ) , ,Wave exciting External dynamic External static+ + + = + +M A x Bx Cx F F F

Linearization , ( ( ) )Restoring Gravity StaticF = + ≈ −F F Cx

addedmass

DampingCoefficient

Surface forceBody force

, ,External dynamic External static+ +F F

Wave exciting force를 제외한외력 (ex. 제어력 등)

( )B B B

I D RFluid S S S

P dSρgz dS dSt t t

ρ ∂Φ ∂Φ ∂Φ= = − − + +

∂ ∂ ∂∫∫ ∫∫ ∫∫F n n

선박의 6자유도 운동방정식

Newton’s 2nd Law==∑FxM

Gravity Fluid= +F F

FBody + FSurface

External+F

선박의 Surface force로 작용

How to know ? 3 3,ξ ξ

FluidPρgzt

ρ ∂Φ= − −

∂

∂Φ∂

+∂Φ∂

+∂Φ∂

−−=ttt

ρgz RDIρ

By solving equation of motion, we could know the velocities, accelerations!!

선박에 작용하는 압력

Static= F .F K D R+ + +F F F

LinearizedBernoulli Eq.By solving equation of motion,

we could know the velocities, accelerations.

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Chap 2. 6 DOF Equations of Ship Motion

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Coordinate system[ ]T61 ,, ξξ =x(변위 : )

O-xyz : Global coordinate system

① right-handed coordinate system (O-x,y,z)

origin in the plane of the undisturbed free-surface

② if body moves with a mean forward speed,

coordinate moves with the same speed

③ body have the x-z plane as a plane of symmetry

O’-x’ y’ z’ : Body-fixed coordinate system

G : Position of center of gravity

y

z

G

O

cy′

cz′

xc,yc,zc: distance from O’-x’y’z’ to center of gravity

U Roll ξ4

Pitch ξ5H

eave

ξ3

Yaw ξ6

O G

x

y

z z′

y′2ξ

3ξ

4ξ

O′

x

z

Ox′

z′

GO′

cx′

cz′

1ξ

3ξ

5ξ

1) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1977, pp 285~290

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Uncoupled Heave motion equation

Heave 운동 방정식 유도

B

W

restoreFdampingF

z

Z

X

평형상태에서더 들어간 부피 RDKFstaticgravity FFFFF ++++= .

Fluidgravity

SurfaceBody

FFFFFM

+=

+==∑3ξ

exciting,3F

333333 ξξ BA −−30 ξρρ ⋅− wpgAgV

Mg−

3333333,30 )( ξξξρρ BAFgAgVMg excitingwp −−+−+−=

3,3333333)( excitingwp FgABAM =+++∴ ξρξξ

Surge 운동 방정식 유도(Heave에서 복원력 성분만 제외)

1,111111)( excitingFBAM =++∴ ξξ

Sway 운동 방정식 유도(Heave에서 복원력 성분만 제외)

2,222222 )( excitingFBAM =++∴ ξξ

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Uncoupled Heave motion equation 풀이

ex) For Heave motion

( ) 3,333333333 excitingFCBAM =+++ ξξξ

tiAtiAtiAtiA efeCeiBeAM ωωωω ηξξωξω 3033333332

33 )()())(( =++−+

tiAet ωξξ 33 )( =tiAeit ωξωξ 33 )( =

tiAet ωξωξ 32

3 )( −=

tiAtiAexciting efeFF ωω η 3033, ==

( : Wave Amplitude, Real), (ξ3A : Amplitude of heave motion, Complex)0η

( : 1m 파고에 대한 Wave exciting force Amplitude, Complex)Af3

Assumption : 시간이 충분히 흘러 steady 상태에서, 선박이 외력의 주파수 ω 와 같은 운동을 함

(Harmonic Motion) -> 초기 Transient Motion은 고려하지 않음.

{ } tiAtiA efeCBiAM ωω ηξωω 3033333332 )( =+++−

D=

{ } AA fCBiAM 3033333332 )( ηξωω =+++− 1

303−= Df AA ηξ 1

30

3 −= DAA

fηξ

RAO(Response Amplitude Operator): 1m wave Amplitude 를 가지는주파수 ω인 wave에 대한heave운동 변위의 진폭

( Complex)1) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1977, pp 307~310

동일한 항으로 정리가능

( : Wave exciting frequency)ω

,( A33 : heave motion에 의한 heave 방향 added mass),( B33 : heave motion에 의한 heave 방향 damping coefficient)

,( C33 : heave motion에 의한 heave 방향 복원력 coefficient)

,( M :Mass of ship)

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Uncoupled roll motion equation

Roll 운동 방정식 유도

Fluidgravity

SurfaceBodyxx

MMMMMI

+=

+==∑4ξ

RDKFstaticgravity MMMMM ++++= .

exciting,4M

444444 ξξ BA −−kr ∆×1B

kr WG ×

4444444, ξξ BAMGZ exciting −−+⋅∆=

1, BG yyGZ +−=

O

y

z

τ

)(+

B

G

z′

y′

KCL

y

z

F∆

B1

1Br

g1

gW

MT

eτ

GrZ

W

GrGγ

O

restoringτ

4ξ

4ξ

F∆

1Br

1Bγ

γπ −O

γγπ sin)sin( =−

+ πγ

π2−γπ −

xsin

x

MT : B1을 지나는 부력 작용선과 선체 중심선과의 교점coordinateGlobaloyz

coordinatefixedBodyzoy:

:''

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Uncoupled Roll motion equation

Roll 운동 방정식 유도

Fluidgravity

SurfaceBodyxx

MMMMMI

+=

+==∑4ξ


exciting,4M

444444 ξξ BA −−kr ∆×1B

kr WG ×

4444444, ξξ BAMGZ exciting −−+⋅∆=

4444444,4sin ξξξ BAMGM exciting −−+⋅∆−=

4444444,4 ξξξ BAMGM exciting −−+⋅∆−≈

44 44 4 44 4 4 ,4( ) T excitingI A B GM Mξ ξ ξ∴ + + + ∆ ⋅ =

Pitch 운동 방정식 유도(Roll 운동 방정식과 동일)

Yaw 운동 방정식 유도(RollHeave에서 복원력 성분만 제외)

4sin

GZ

GM ξ= −

44sin ξξ ≈

55 55 5 55 5 5 ,5( ) L excitingI A B GM Mξ ξ ξ∴ + + + ∆ ⋅ = 66 66 6 66 6 ,6( ) excitingI A B Mξ ξ∴ + + =

O

B

G

z′

y′

KCL

y

z

F∆

B1

g1

gW

MT

Z

W

F∆

1Br

1Bγ

γπ −O

γγπ sin)sin( =−

+ πγ

π2−γπ −

xsin

x

MT : B1을 지나는 부력 작용선과 선체 중심선과의 교점coordinateGlobaloyz

coordinatefixedBodyzoy:

:''

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Uncoupled Pitch motion equation

For Pitch motion

( )55 55 5 55 5 55 5 ,5excitingA I B C Fξ ξ ξ+ + + =

255 55 5 55 5 55 5 0 5( )( ) ( ) ( )A i t A i t A i t A i tA I e B i e C e f eω ω ω ωω ξ ωξ ξ η+ − + ⋅ + ⋅ =

5 5( ) A i tt e ωξ ξ=

5 5( ) A i tt i e ωξ ωξ=

25 5( ) A i tt e ωξ ω ξ= −

,5 5 0 5A i t A i t

excitingF F e f eω ωη= =( : Wave Amplitude, Real), (ξ5

A : Amplitude of pitch motion, Complex)0η( : 1m 파고에 대한 Wave exciting force Amplitude, Complex)5

Af

Assumption : 시간이 충분히 흘러 steady 상태에서, 선박이 외력의 주파수 ω 와 같은 운동을 함


{ }255 55 55 55 5 0 5( ) A i t A i tA I i B C e f eω ωω ω ξ η− + + + =

D=

{ }255 55 55 55 5 0 5( ) A AA I i B C fω ω ξ η− + + + = 1

5 0 5A Af Dξ η −= 15

50

AAf

ξη

−= D

RAO(Response Amplitude Operator): 1m wave Amplitude 를 가지는주파수 ω인 wave에 대한pitch운동 변위의 진폭

( Complex)


동일한 항으로 정리가능

( : Wave exciting frequency)ω

,( A55 : pitch motion에 의한 pitch 방향 added mass),( B55 : pitch motion에 의한 pitch 방향 damping coefficient)

,( C55 : pitch motion에 의한 pitch 방향 복원력 coefficient)

,( I55 : Mass moment of inertia of ship with respect to y axis)

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Motion of any point on the body- Acceleration of any point

Position vector of any point about the Origin(O)

T R′ ′= + + ×xξxξx

Translatory displacements of body fixed coordinate in the x-,y-, and z-directions with respect to the origin(O)

Angular displacements of rotational motion in the x-,y-, z-axis with respect to the origin(O)

4 5 6R ξ ξ ξ= + +ξi j k

,(Surge, Sway, Heave)

,(Roll, Pitch, Yaw)

4 5 6, R

x y zξ ξ ξ

′× = ′ ′ ′

i j kξx

x-,y-,z- acceleration are coupled with other motions

G′

G5ξ x′

z′

'OO

z

x′

Heave motion caused by pitch motionSimilarly, roll motion cause heave motion

( ) ( ) ( )1 5 6 2 4 6 3 4 5z y z x y xξ ξ ξ ξ ξ ξ ξ ξ ξ′ ′ ′ ′ ′ ′= + − + − + + + −x i j k

ex.) Acceleration of heave motion :

x′ G

G′

5ξO

5 1cx zξ ≈ k

s1z

If ξ5 is small,

1s z≈ k

ik

( )3 4 5y xξ ξ ξ′ ′+ − k

( )3 4 5y xξ ξ ξ′ ′+ − k

① : heave acceleration ② heave acceleration by roll③ : heave acceleration by pitch

① + ② + ③

Find : Acceleration of any pointx

Motion of equation :

1 2 3T ξ ξ ξ= + +ξi j k

( : position vector with respect to O’-x’y’z’ coordinate)

( : position vector with respect to O-xyz coordinate)

x y z′ ′ ′ ′= + +x i j kx y z= + +x i j k

restoring exciting radiationM = + +x F F F

T R R′ ′ ′= + + × + ×xξxξxξx

Velocity vector

Acceleration vector

( )2T R R R R

T R R R

′ ′ ′ ′ ′= + + × + × + × + ×

′ ′ ′ ′= + + × + × + ×

xξxξxξxξxξx

ξxξxξxξx

선박을 강체로 가정하면, 선박 위의 한 점은 시간에 따라

변하지 않는다

T R ′= + ×xξξx

( ) ( ) ( )1 5 6 2 4 6 3 4 5z y z x y xξ ξ ξ ξ ξ ξ ξ ξ ξ′ ′ ′ ′ ′ ′= + − + − + + + −i j k


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Motion of any point on the body- (참조) Acceleration of any point Find : Acceleration of any pointx




T R′ ′= + + ×xξxξx

( Translatory displacements : ξT=ξ1i+ ξ2j+ ξ3k , (Surge, Sway, Heave) )

( Angular displacements : ξR=ξ4i+ ξ5j+ ξ6k , (Roll, Pitch, Yaw) )

4 5 6, R

x y zξ ξ ξ

′× = ′ ′ ′

i j kξx

OT O′ ′= +xξR x


6 61

6 62

cos sinsin cos

x xy y

ξ ξξξ ξξ

′− = + ′

6 61

6 62

cos sinsin cos

x yxx yy

ξ ξξξ ξξ

′ − = + ′ ′+

If we consider 2-dimensional motion

If we consider 2-dimensional motion

16

2

0 00

x xy y x y

ξξ

ξ ′

= + + ′ ′ ′

i j k

61

62

yx xxy y

ξξξξ

′′ − = + + ′+′ 61

62

x yxy xy

ξξξξ

′ ′− = + ′ ′+

Linearize (cosθ→1, sinθ→θ)

x′

O′

y′

P′

Inertial frame(O-frame)

y

O x

[ ], Tx y′ ′ ′=x

Oi

OjOk

O′iO′j

O′k

6ξ[ ], Tx y=x

[ ]1 2, TT ξ ξ=ξ (병진운동)

(회전운동)

x′O′

y′P′


y

O x

[ ], Tx y′ ′ ′=x

Oi

OjOk

O′iO′j

O′k

6ξ[ ], Tx y=x

[ ]1 2, TT ξ ξ=ξ (병진운동)

(회전운동)

′×ξxP

If summing the body motions ξj are small and neglecting,

Linearize

Rξ ′x

R ′×ξx

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Motion of any point on the body- (참조) Acceleration of any point Find : Acceleration of any pointx




T R′ ′= + + ×xξxξx

T R R′ ′ ′= + + × + ×xξxξxξx

Velocity vector

Acceleration vectorT R R R R′ ′ ′ ′ ′= + + × + × + × + ×xξxξxξxξxξx

OT O′ ′= +xξR x

O OT O O

O OT O O

ddt

ω

′ ′

′ ′

′ ′= + +

′ ′= + × +

x ξR x R x

ξR x R x

( ) ( )2

2O O O O O

T O O O O Oddt

ω ω ω′ ′ ′ ′ ′′ ′ ′ ′ ′= + × + × + × + +x ξR x R x R x R x R x


Velocity vector

Acceleration vector

①E-frame 에 대한점 P의 가속도

( Translatory displacements : ξT=ξ1i+ ξ2j+ ξ3k , (Surge, Sway, Heave) )

( Angular displacements : ξR=ξ4i+ ξ5j+ ξ6k , (Roll, Pitch, Yaw) )

2

2 2O O O OT O O O O

ddt

ω ω ω ω′ ′ ′ ′′ ′ ′ ′= + × + × × + × +x ξR x R x R x R x

( )2T R R R′ ′ ′ ′= + + × + × + ×xξxξxξxξx

x′

O′

y′

P′


y

O x

[ ], Tx y′ ′ ′=x

Oi

OjOk

O′iO′j

O′k

6ξ[ ], Tx y=x

[ ]1 2, TT ξ ξ=ξ (병진운동)

(회전운동)

x′O′

y′P′


y

O x

[ ], Tx y′ ′ ′=x

Oi

OjOk

O′iO′j

O′k

6ξ[ ], Tx y=x

[ ]1 2, TT ξ ξ=ξ (병진운동)

(회전운동)

′×ξxP

Linearize

② E-frame 에 대한A-frame의 원점A의 가속도

⑥ A-frame 를 고정시켜 놓았을 때,A-frame 에 대한 점 P의 가속도

③A-frame이 각 가속도를가지고 회전하고 있을 때, 점 P의 접선 방향의 가속도

④A-frame이 회전하고 있을 때, 점 P의 회전 중심방향의 가속도 (구심력)

⑤ Coriolis Acceleration (Coriolis Effect)회전 좌표계에서 기술된 움직이는 점을고정 좌표계에서 바라봤을 때,발생하는 효과

O O O O OT O O O O Oω ω ω ω ω′ ′ ′ ′ ′′ ′ ′ ′ ′= + × + × × + × + × +ξR x R x R x R x R x ⑥ A-frame 를 고정시켜 놓았을 때,

A-frame 에 대한 점 P의 가속도

⑤ Coriolis Acceleration (Coriolis Effect)회전 좌표계에서 기술된 움직이는 점을고정 좌표계에서 바라봤을 때,발생하는 효과

③A-frame이 각 가속도를가지고 회전하고 있을 때, 점 P의 접선 방향의 가속도

② E-frame 에 대한A-frame의 원점A의 가속도

①E-frame에대한 점 P의가속도

(선형화한 가정으로 인해 구심가속도성분이 나타나지 않음)

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( )2 64 4 66 6 = yaw cM x I Iτ ξ ξ ξ′ + + k

( )1 2

1 3 55 5

= +

= pitch pitch pitch

c cM z M x I

τ τ τ

ξ ξ ξ′ ′− + j

( )2 44 4 46 6 = roll cM z I Iτ ξ ξ ξ′− + + i

2 1 3 = pitch c cz M x Mτ ξ ξ× + ×k i i k

Moment in pitch motion

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for Windows”, Delft University of Technology, 2003, pp38~422) 구종도 역, 선체와 해양구조물의 운동학, 연경문화사, pp68~71

Moment in pitch motion (with respect to y-axis)

Find : Moment

restoring exciting radiationI M M Mω = + +

Iω

Motion of equation for roll, pitch, yaw

x′

z′

O′

G

cx′

cz′

3Mξ

1Mξ5yyI ξ

1

55 5

= pitch I

I

τ

ξ=

ω

j

Moment by mass moment of inertia

( )1 3= c cM z M xξ ξ− j moment arm force

Moment by inertia force

, ( I : mass moment of inertia ) , ( : Angular acceleration ) ω

Moment in pitch motion

Moment in roll motion(with respect to x-axis, yc=0 according to Lateral Symmetric)

Moment in yaw motion(with respect to z-axis , yc=0 according to Lateral Symmetric)

[ ]T654321 ,,,,, ξξξξξξ=x

heaveswaysurge

:::

3

2

1

ξξξ

yawpitchroll

:::

5

4

3

ξξξ

x'c,y’c,z’c: distance from O’-x’y’z’ to center of gravity

O

z

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Mass moment of inertia

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for Windows”, Delft University of Technology, 2003, pp38~422) 구종도 역, 선체와 해양구조물의 운동학, 연경문화사, pp68~71

mass moment of Inertia2 2

44 ( )V

I y z dVρ= +∫∫∫2 2

55, ( )V

I x z dVρ= +∫∫∫2 2

55, ( )V

I x y dVρ= +∫∫∫

46V

I xzdVρ= ∫∫∫ 45, ...V

I xydVρ= ∫∫∫

관성 모멘트는 계산하기 위해 많은 정보가 필요하여, 계산이 복잡하다.따라서 자료가 주어지거나 추정식을 사용하여 구한다. (m=ρ▽)

2 2 244 44 55 55 66 66, , ,I k I k I kρ ρ ρ= ∇ = ∇ = ∇

- 추정식①1)

44

55

66

0.30 to 0.400.22 to 0.280.22 to 0.28

k B Bk L Lk L L

≈ ≈ ≈

- 추정식②1) (Proposal of Bureau Veritas)

2

4420.289 1.0 KGk B

B

≈ +

- 추정식③2) K44

- 객선 : 0.38 ~ 0.43- 화물선 : 0.32 ~ 0.35(만재)

0.375 ~ 0.4(Ballast)- 석탄운반선 : 0.31~0.33(만재)

0.35~0.39(Ballast)- 전함 : 0.34 ~ 0.38- 순양함 : 0.39 ~ 0.42- 어선 : 0.38 ~ 0.44

Find :Mass moment of inertia I

관성모멘트를 Radius of gyration ( 관동반경 k44, k55, k66 )을 이용하여 표현하면 다음과 같다.

관성모멘트 추정식

restoring exciting radiationMx F F F= + + Equation of motion :


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Restoring Force & Moment1)

Restoring force & Moment in “Hydrostatic by pressure integration technique”

이 때, 미소 경사시에 수선면적이 x-z평면에 대하여 대칭(Lateral symmetric)이라 하면,수선면의 x축에 대한 1차 모멘트(TWP) 와 products of inertia(IP)는 ‘0’ 이므로, TWP=0, IP=0 이다.

(k)를 생략하여 나타내고, 초기자세를 정적 평형상태 기준으로 표현하면 (ΔFk= Fk-0 =Fk , Δξk=ξk-0=ξk)

선박의 운동(자세 변화)시 복원력이 생기는 자세 : Heave(ξ3), Roll(ξ4), Pitch(ξ5)

( )3( )4( )5

k

k

k

ξξξ

∆ = ⋅ ∆

∆

( )3( )4

( ) ( ) ( ) ( )5 5

( )( )

( ) ( )

kWP

kP

k k k kL B G

gLgI

gI gV z mg z

ρ ξρ ξ

ρ ξ ρ ξ− − + ⋅

( )3( )4

( )5

( )( )

( )

kWP

kWP

kWP

gAgT

gL

ρ ξρ ξρ ξ

−−

( )3

( ) ( ) ( ) ( )4 4

( )5

( )( ) ( )

( )

kWP

k k k kT B G

kP

gTgI gV z mg z

gI

ρ ξρ ξ ρ ξ

ρ ξ

−− − + ⋅

3

4

5

FFF

∆ ∆ ∆

3

4

5

ξξξ

= ⋅

0WP

L B G

gL

gI gV z mg z

ρ

ρ ρ− − + ⋅0

WP

WP

gA

gL

ρ

ρ

− 0

0T B GgI gV z mg zρ ρ− − + ⋅

3

4

5

FFF

x

y

zLCF LC

Δξj 대신 ξj 가 쓰인 이유?

: 선박의 자세변화가 임의의 자세로부터 변한 것이 아닌,

정적 평형상태로 부터 변한 것으로 고려했기 때문

ex) 3 3 0ξ ξ∆ = −

3ξ=

(k) : k번째 자세

1) 이규열, 2009년 창의적선박설계 강의자료, Hydrostatics Calculation by Pressure Integration Technique, 2009년 5월

AWP : 수선 면적

TWP : x축에 대한, 수선면의 폭 뱡향 1차 모멘트

LWP :y축에 대한, 수선면의 종 뱡향 1차 모멘트

IT : x축에 대한, 수선면의 횡 뱡향 2차 모멘트

IL : y축에 대한, 수선면의 종 뱡향 2차 모멘트

IP : x축 및 y축에 대한 수선면의 모멘트

xG ,yG,, zG : 무게 중심

xB ,yB, zB : 부력 중심

Find : Restoring Force & moment

restoring exciting radiationMx F F F= + +


Equation of motion :

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Restoring Force & Moment1)

Restoring force & Moment in “Hydrostatic by pressure integration technique”

,Restoringk j k j jF C ξ= −

33 WPC gAρ=

35 53 WPC C gLρ= = −

44 T B GC gI gV z mg zρ ρ= + − ⋅

55 L B GC gI gV z mg zρ ρ= + − ⋅

Restoring Force & Moment

( Ckj: j방향 운동(자세변화)에 의한 k방향의 복원력계수)

( Fkj : j방향 운동(자세변화)에 의한 k방향의 복원력)

( C53 : heave motion에 의한 pitch 방향 복원력 coefficient)

( C33 : heave motion에 의한 heave 방향 복원력 coefficient)

( C35 : pitch motion에 의한 heave 방향 복원력 coefficient)

( C44 : roll motion에 의한 roll 방향 복원력 coefficient)

( C55 : pitch motion에 의한 pitch 방향 복원력 coefficient)

AWP : 수선 면적

TWP : x축에 대한, 수선면의 폭 뱡향 1차 모멘트

LWP : y축에 대한, 수선면의 종 뱡향 1차 모멘트

IT : x축에 대한, 수선면의 횡 뱡향 2차 모멘트

IL : y축에 대한, 수선면의 종 뱡향 2차 모멘트

IP : x축 및 y축에 대한 수선면의 모멘트

3

4

5

ξξξ

= ⋅

0WP

L B G

gL

gI gV z mg z

ρ

ρ ρ− − + ⋅0

WP

WP

gA

gL

ρ

ρ

− 0

0T B GgI gV z mg zρ ρ− − + ⋅

3

4

5

FFF

xG , yG,, zG : 무게 중심

xB , yB, zB : 부력 중심

V : displaced volume of water

GMT : transverse metacentric height

GML : longitudinal metacentric height

Fk j 형태로 나타내면

Find : Restoring Force & moment


1) 이규열, 2009년 창의적선박설계 강의자료, Hydrostatics Calculation by Pressure Integration Technique, 2009년 5월


Equation of motion :

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3Mξ

3 Body Surface

gravity Fluid

M F F FF F

ξ = = +

= +∑

( )3 4 5c cM y xξ ξ ξ+ −

Coupled Heave-pitch motion equation 유도

Uncoupled Heave motion Equation

Find :Coupled Heave motion eq.

3 restoring exciting radiationM F F Fξ = + + Coupled heave Motion equation

RDKFstaticgravity FFFFF ++++= .

① Acceleration of ship

( ) ( )( )

1 5 6 2 4 6

3 4 5

c c

c

z y z x

y x

ξ ξ ξ ξ ξ ξ

ξ ξ ξ

= + − + − + +

+ −

x i j

k

연성된 가속도를 고려하면exciting,3F


Mg−

①

② ② Restoring Force & Moment

C33, C35 are related with heave motion

33 WPC gAρ= 35, WPC gLρ= −

gravity staticF F+

Mg− 30 ξρρ ⋅− wpgAgV

운동방정식을 위해 선박의 무게 중심의 가속도를 고려

( , , ) ( , , )c c cx y z x y z→

연성된 복원력 성분

Coupled Heave motion Equation 유도

③


3Mξ

② Restoring Force & Moment

gravity staticF F+

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③ Radiation Force

6

,1

A i tR k j k j

kF F e ωξ

=

=∑6

1( )j k j j k j

kA Bξ ξ

=

= − −∑

6

3 31( )j j j j

jA Bξ ξ

=

− −∑ Heave 운동 경우

( )( )

1 31 2 32 3 33 4 34 5 35 6 36

1 31 2 32 3 33 4 34 5 35 6 36

A A A A A A

B B B B B B

ξ ξ ξ ξ ξ ξ

ξ ξ ξ ξ ξ ξ

= − − − − − −

+ − − − − − −

선박의 형상이 Lateral Symmetric이라면, resultant horizontal force(y방향) = 0 이다.

따라서 A33, B32=0 이다. 또한 압력분포도Lateral symmetric이므로 A34, B34,A36, B36=0



( ) ( )1 31 3 33 5 35 1 31 3 33 5 35A A A B B Bξ ξ ξ ξ ξ ξ= − − − − − −

From the definition of Radiation Force

( Fjk : j방향 운동으로 인해 나타나는 k방향 힘 )

( Akj : j방향 운동으로 인해 나타나는 k방향 added mass )

( Bkj : j방향 운동으로 인해 나타나는 k방향 damping coefficient )

Fluidgravity

SurfaceBody

FFFFFM

+=

+==∑3ξ

Uncoupled Heave motion Equation


exciting,3F


Mg−

①

②

Coupled Heave motion Equation 유도

③

③ Radiation Force

RF

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6

3 31( )j j j j

jA Bξ ξ

=

− −∑ Heave 운동



( ) ( )1 31 3 33 5 35 1 31 3 33 5 35A A A B B Bξ ξ ξ ξ ξ ξ= − − − + − − −

Uncoupled Heave motion Equation Coupled Heave motion Equation 유도

Fluidgravity

SurfaceBody

FFFFFM

+=

+==∑3ξ


exciting,3F


Mg−

①

② ③

③ Radiation Force

3Mξ ( )3 4 5c cM y xξ ξ ξ+ −


② Restoring Force & Moment

gravity staticF F+


6

,1

A i tR k j k j

kF F e ωξ

=

=∑

( ) ( ) ( ) ( )3 4 5 1 31 3 33 5 35 1 31 3 33 5 35 33 3 35 5 3i t

cM y x A A A B B B C C F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ+ − = − − − + − − − + − − +

Coupled Heave motion of Equation

( ) ( ) ( ) ( )3 4 5 1 31 3 33 5 35 1 31 3 33 5 35 33 3 35 5 3i t

cM y x A A A B B B C C F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ+ − + + + + + + + + =

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Coupled Heave-pitch motion equation 유도Find :Coupled Pitch motion eq.


Coupled Pitch Motion equation

Uncoupled pitch motion Equation Coupled pitch motion Equation 유도

55 5Iτ ξ=

( )1 3 55 5c cM z M x Iτ ξ ξ ξ= − + j

① Acceleration of ship55 5 Body Surface

gravity Fluid

I M M MM M

ξ = = +

= +∑


exciting,5 ,M 55 5 55 5A Bξ ξ− − 5

,B ×∆r k,G W×r k

,4 44 4 44 4L excitingGZ M A Bξ ξ= ∆ ⋅ + − −

① Moment in pitch motion

55 5I ξ

② Restoring Moment


gravity staticM M+

gravity staticF F+

Mg− 30 ξρρ ⋅− wpgAgV


연성된 복원력 성분

C53, C55 are related with heave motion

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Coupled Heave-pitch motion equation 유도Find :Coupled Pitch motion eq.


Coupled Pitch Motion equation

55 5 Body Surface

gravity Fluid

I M M MM M

ξ = = +

= +∑


exciting,5 ,M 55 5 55 5A Bξ ξ− − 5


,5 55 5 55 5L excitingGZ M A Bξ ξ= ∆ ⋅ + − −

③ Radiation Moment

6

,1

A i tR k j k j

kF F e ωξ

=

=∑6

1( )j k j j k j

kA Bξ ξ

=

= − −∑


6

5 51( )j j j j

jA Bξ ξ

=

− −∑ Pitch 운동 경우

( ) ( )1 51 3 53 5 55 1 51 3 53 5 55A A A B B Bξ ξ ξ ξ ξ ξ= − − − + − − −

선박의 형상이 Lateral Symmetric이라면, resultant horizontal force(y방향) = 0 이다.

따라서 A52, B52=0 이다. 또한 압력분포도Lateral symmetric이므로 A54, B54,A56, B56=0

( )( )

1 51 2 52 3 53 4 54 5 55 6 56

1 51 2 52 3 53 4 54 5 55 6 56

A A A A A A

B B B B B B

ξ ξ ξ ξ ξ ξ

ξ ξ ξ ξ ξ ξ

= − − − − − −

− − − − − −


,R R kM F=

( Fjk : j방향 운동으로 인해 나타나는 k방향 힘 )

( Akj : j방향 운동으로 인해 나타나는 k방향 added mass )

( Bkj : j방향 운동으로 인해 나타나는 k방향 damping coefficient )

Uncoupled pitch motion Equation Coupled pitch motion Equation 유도

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Coupled Pitch motion of Equation

Coupled Heave-pitch motion equation 유도Find :Coupled Pitch motion of eq.


Coupled Pitch Motion of equation

( )1 3 55 5c cM z M x Iξ ξ ξ− + j

① Moment in pitch motion55 5 Body Surface

gravity Fluid

I M M MM M

ξ = = +

= +∑


exciting,5 ,M 55 5 55 5A Bξ ξ− − 5


,5 55 5 55 5L excitingGZ M A Bξ ξ= ∆ ⋅ + − − ③ Radiation Moment

6

,1

A i tR k j k j

kF F e ωξ

=

=∑6

1( )j k j j k j

kA Bξ ξ

=

= − −∑


6

5 51( )j j j j

jA Bξ ξ

=

− −∑ Pitch 운동 경우

( ) ( )1 51 3 53 5 55 1 51 3 53 5 55A A A B B Bξ ξ ξ ξ ξ ξ= − − − + − − −

( ) ( ) ( ) ( )1 3 55 5 5 1 51 3 53 5 55 1 51 3 53 5 55 53 3 55 5i t

cM z x I F e A A A B B B C Cωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ− + = − + + − + + − +

( ) ( ) ( ) ( )1 3 55 5 1 51 3 53 5 55 1 51 3 53 5 55 33 3 35 5 5i t

cM z x I A A A B B B C C F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ− + + + + + + + + + =


55 5I ξ

gravity staticM M+

55 53,WP WPC gA C gLρ ρ= = −

Uncoupled pitch motion of Equation Coupled pitch motion of Equation 유도

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Coupled Roll-Sway-Yaw motion equationFind :Coupled roll motion of eq.


Coupled heave Motion of equation

Coupled Roll motion of Equation ( ) ( ) ( ) ( )2 44 4 46 6 2 42 4 44 6 46 2 42 4 44 6 46 44 4 4

i tcM z I I A A A B B B C F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ− + + + + + + + + + =

Fluidgravity

SurfaceBodyxx

MMMMMI

+=

+==∑4ξ


exciting,4M

44 4 44 4RF A Bξ ξ= − − kr ∆×1B

kr WG ×

Uncoupled Roll motion of Equation Coupled Roll motion of Equation 유도

( )2 44 4 46 6cM z I Iξ ξ ξ− + + j

① Moment in pitch motion


6

,1

A i tR k j k j

kF F e ωξ

=

=∑6

1( )j k j j k j

kA Bξ ξ

=

− −∑

From the definition of Radiation ForceRoll

( ) ( )2 42 4 44 6 46 2 42 4 44 6 46A A A B B Bξ ξ ξ ξ ξ ξ= − − − + − − −

② Restoring Moment44 4I ξ

gravity staticM M+

44T B GgI gV z mg z Cρ ρ+ − ⋅ =

Coupled Sway motion of Equation ( ) ( ) ( )2 4 5 2 22 4 24 6 26 2 22 4 24 6 26 2

i tc cM z x A A A B B B F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ− + + + + + + + =

Coupled Yaw motion of Equation ( ) ( ) ( )2 46 4 66 6 2 62 4 64 6 66 2 62 4 64 6 66 6

i tcM x I I A A A B B B F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ+ + + + + + + + =

1) Kreyszig, Advanced engineerng mathematics, Wiley, Systems of Ordinary differential equations,pp.125~165

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(ex) 좌우 대칭 선박

y

x

x

z z

y

6DOF Equation of ship Motion

( ) excitingFCxxBxAM =+++

Assumption

1. Small amplitude water wave (파장에 비해 파고가 작음)

2. small amplitude motion (선박의 운동이 작음)

3. Slender body (선박의 길이에 비해 폭이 작음, Strip theory에서 자세히 설명) 물체의 전후 동요(Surge)는 독립적으로 취급 (Coupling 고려 안함)

4. Lateral symmetry (symmetric about xz-plane) 물체 운동이 종운동(Longitudinal motion) 과 횡운동(Transverse motion)으로 나뉨

sway,roll,yawsurge,heave,pitch서로 영향을 주지 않음

6DOF Equations of Ship Motion : 6 coupled equation

surge , pitch

sway , roll , yaw

( )Matrix66:,,, ×CBAM[ ]T61 ,, ξξ =x(변위 : )

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for Windows”, Delft University of Technology, 2003, pp38~422) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 307~3113) Journee, J.M.J. , Massie, W.W. , Offshore Hydrodynamics, Delft University of Technology, 2001,pp8-1~4

(단, 계수 A,B,C 및 외력 Fexciting은주어져 있다고 가정)Given Find

xxx ,,

Heave운동에 의해 영향을 받는 운동은?

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6DOF Equation of ship MotionFind : 6 DOF equation of ship motion


Coupled heave Motion of equation

Roll-Sway-Yaw Motion of Equation

( ) ( ) ( ) ( )2 44 4 46 6 2 42 4 44 6 46 2 42 4 44 6 46 44 4 4i t

cM z I I A A A B B B C F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ− + + + + + + + + + =

( ) ( ) ( )2 4 5 2 22 4 24 6 26 2 22 4 24 6 26 2i t

c cM z x A A A B B B F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ− + + + + + + + =

( ) ( ) ( )2 46 4 66 6 2 62 4 64 6 66 2 62 4 64 6 66 6i t

cM x I I A A A B B B F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ+ + + + + + + + =

( ) ( ) ( ) ( )3 5 1 31 3 33 5 35 1 31 3 33 5 35 33 3 35 5 3i tM x A A A B B B C C F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ− + + + + + + + + =

( ) ( ) ( ) ( )1 3 55 5 1 53 3 53 5 55 1 53 3 53 5 55 33 3 35 5 5i t

cM z x I A A A B B B C C F e ωξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ− + + + + + + + + + =

Heave-Pitch Motion of Equation

Surge motion of Equation

( ) ( ) ( )1 5 1 11 3 13 5 15 1 11 3 13 5 15 1i t

cM z A A A B B B F e ωξ ξ ξ ξ ξ ξ ξ ξ+ + + + + + + =

11 13 15 11

22 24 26 22

31 33 35 33 353

42 44 46 444

51 53 55 53 555

62 64 66 6

0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0

0 0 0 0 0 0 00 0 0 0 0 0 0 0

0 0 0 0 0 0 00 0 0 0 0 0 0 0 0

B B BB B B

B B B C CB B B C

B B B C CB B B

ξξξξ

ξξξξ

+ +

1

2

3 3

4 4

5 5

6 6

FFFFFF

ξξξξ

=

11 13 151

22 24 262

31 33 353

44 46 42 44 464

55 51 53 555

64 66 62 66

0 0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0

0 0 0 0 0 00 0 0 0 0

C

C C

C

C

C C

C

M Mz A A AM Mz Mx A A A

M Mx A A AMz I I A A A

Mz Mx I A A AMx I I A A

ξξξξξξ

− −

+ − − −

−

1

2

3

4

5

4 66 60 A

ξξξξξξ

( ) excitingFCxxBxAM =+++ : 6DOF Equation of ship motion

Matrix from

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6DOF Equations of Ship Motion: 3 kinds of coupled motions (surge / heave-pitch / sway-roll-yaw)

=

666462

555351

464442

353331

262422

151311

000000

000000

000000

AAAAAA

AAAAAA

AAAAAA

A

=

666462

555351

464442

353331

262422

151311

000000

000000

000000

BBBBBB

BBBBBB

BBBBBB

B

=

0000000000000000000000000000000

5553

44

3533

CCC

CCC

44 46

55

64 66

0 0 0 00 0 00 0 0 00 0 0

0 0 00 0 0

C

C C

C

C

C C

C

M MzM Mz Mx

M MxMz I I

Mz Mx IMx I I

− −

= − − −

−

M


=

6

5

4

3

2

1

ξξξξξξ

x

=

6

5

4

3

2

1

FFFFFF

excitingF

운동 방정식 :

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[ ]T654321 ,,,,, ξξξξξξ=x

Sway

Heave

Roll

Surge Pitch

Yaw

44 46

55

64 66

0 0 0 00 0 00 0 0 00 0 0

0 0 00 0 0

C

C C

C

C

C C

C

M MzM Mz Mx

M MxMz I I

Mz Mx IMx I I

− −

= − − −

−

M


=

666462

555351

464442

353331

262422

151311

000000

000000

000000

AAAAAA

AAAAAA

AAAAAA

A

=

666462

555351

464442

353331

262422

151311

000000

000000

000000

BBBBBB

BBBBBB

BBBBBB

B

=

0000000000000000000000000000000

5553

44

3533

CCC

CCC

heave-pitchmotion of equation :

=

6

5

4

3

2

1

FFFFFF

excitingF

(가정3. Surge운동은 독립적)

운동 방정식 :

=

+

+

++−+−+

5

3

5

3

5553

3533

5

3

5553

3533

5

3

5553

3533

FF

CCCC

BBBB

IAAMxAMxAM

yyC

C

ξξ

ξξ

ξξ

6DOF Equation of ship Motion : Heave-Pitch Equation of Motion

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=

666462

555351

464442

353331

262422

151311

000000

000000

000000

AAAAAA

AAAAAA

AAAAAA

A

=

0000000000000000000000000000000

5553

44

3533

CCC

CCC


sway-roll-yaw :

=

+

+

++−++−++−++−+

6

4

2

6

4

2

44

6

4

2

666462

464442

262422

6

4

2

666462

464442

262422

00000000

FFF

CBBBBBBBBB

AIAIAmxAIAIAmzAmxAmzAm

zzzxc

xzxxc

cc

ξξξ

ξξξ

ξξξ

=

6

5

4

3

2

1

FFFFFF

excitingF

44 46

55

64 66

0 0 0 00 0 00 0 0 00 0 0

0 0 00 0 0

C

C C

C

C

C C

C

M MzM Mz Mx

M MxMz I I

Mz Mx IMx I I

− −

= − − −

−

M

=

666462

555351

464442

353331

262422

151311

000000

000000

000000

BBBBBB

BBBBBB

BBBBBB

B

[ ]T654321 ,,,,, ξξξξξξ=x

좌우동요변위

상하동요변위

횡동요변위

전후동요변위

종동요변위

선수동요변위

6DOF Equation of ship Motion : Sway-Roll-Yaw Motion of Equation

운동 방정식 :

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(풀이) Coupled Heave-pitch motion of equation in frequency domain

Given : Coupled heave-pitch motion of equation 의 계수

,333 35 33 35 33 35 33 3

,553 55 55 53 55 53 55 55 5

extC

extC

FM A Mx A B B C CFMx A A I B B C C

ξξ ξξξ ξ

+ − + + + = − + +

Find : heave, pitch 운동 변위, 속도, 가속도

tiA

tiA

etet

ω

ω

ξξ

ξξ

55

33

)(

)(

=

=

Motion amplitude(complex)

파고(real) Wave exciting force amplitude(complex)

ω : Wave frequency

−==

−==tiAtiA

tiAtiA

eteit

eteitωω

ωω

ξωξωξξ

ξωξωξξ

52

555

32

333

)(,)(

)(,)(

① 운동 방정식에 변위,속도,가속도 대입

=

+

+

−−

++−+−+

tiA

tiA

tiA

tiA

tiA

tiA

tiA

tiA

yyC

C

eFeF

ee

CCCC

eiei

BBBB

ee

IAAMxAMxAM

ω

ω

ω

ω

ω

ω

ω

ω

ηη

ξξ

ωξωξ

ξωξω

50

30

3

3

5553

3533

5

3

5553

3533

52

32

5553

3533

AA53 ,ξξ 만 구하면 됨

= tiA

tiA

eFeFω

ω

ηη

50

30

(변위)(속도)(가속도)1) Bhattacharyya,R., Dynamics of Marine Vehicles, John Wiley & Sons, 1978, pp183-2072) Kreyszig, Advanced engineerng mathematics, Wiley, Systems of Ordinary differential equations,pp.125~165

Assumption : 시간이 충분히 흘러, 선박이 외력의 주기와 같은 운동을 함


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① 운동 방정식에변위,속도,가속도 대입

233 35 33 35 33 353 3 3 0 3

253 55 55 53 55 53 555 5 3 0 5

A i t A i t A i t A i tC

A i t A i t A i t A i tC

M A Mx A B B C Ce i e e F eMx A A I B B C Ce i e e F e

ω ω ω ω

ω ω ω ω

ω ξ ωξ ξ ηω ξ ωξ ξ η

+ − + − + + = − + + −

(continue)

33 35 33 35 33 352 3 3 3 0 3

53 55 55 53 55 53 555 5 3 0 5

A A A AC

A A A AC

M A Mx A B B C C Fi

Mx A A I B B C C Fξ ξ ξ η

ω ωξ ξ ξ η

+ − + − + + = − + +

② 양변을 로 나눔tie ω

33 35 33 35 33 352 3 0 3

53 55 55 53 55 53 55 3 0 5

A AC

A AC

M A Mx A B B C C Fi

Mx A A I B B C C Fξ η

ω ωξ η

+ − + − + + = − + +

2 233 33 33 35 35 35 3 0 3

2 253 53 53 55 55 55 55 3 0 5

( ) ( )( ) ( )

A AC

A AC

M A i B C Mx A i B C FMx A i B C A I i B C F

ω ω ω ω ξ ηω ω ω ω ξ η

− + + + − − + + += − − + + + − + + +

③ 로 묶어서 정리

A

A

5

3

ξξ

=

A

A

A

A

FF

SRQP

50

30

3

3

ηη

ξξ 2

33 33 332

35 35 352

53 53 532

55 55 55 55

( )

( )

( )

( )

C

C

P M A i B C

Q Mx A i B C

R Mx A i B C

S I A i B C

ω ω

ω ω

ω ω

ω ω

= − + + +

= − − + + + = − − + + + = − + + +

1) Bhattacharyya,R., Dynamics of Marine Vehicles, John Wiley & Sons, 1978, pp183-2072) Kreyszig, Advanced engineerng mathematics, Wiley, Systems of Ordinary differential equations,pp.125~165

(풀이) Coupled Heave-pitch motion equation in frequency domain

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(continue)

=

A

A

A

A

FF

SRQP

50

30

3

3

ηη

ξξ

③ 로 묶어서 정리

A

A

5

3

ξξ

④ 역행렬을 곱하여변위 를 구함

AA53 ,ξξ

−

−−

=

=

−

A

A

A

A

A

A

FF

PRQS

QRPSFF

SRQP

50

30

50

301

5

3 1ηη

ηη

ξξ

( )( )

+−−

−=

PFRFQFSF

QRPS AA

AA

530

5301ηη

−−−−

=

QRPSRFPF

QRPSQFSF

AA

AA

350

530

η

η

QRPSRFPF

QRPSQFSF

AAA

AAA

−−

=

−−

=∴

35

0

5

53

0

3

ηξ

ηξ

⑤ 1m 파고에 대한Heave-Pitch 연성운동 변위 (RAO*)

※ RAO(Response Amplitude Operator) : 1m v에 대한 선박의 운동 응답

233 33 33

235 35 35

253 53 53

255 55 55 55

( )

( )

( )

( )

C

C

P M A i B C

Q Mx A i B C

R Mx A i B C

S I A i B C

ω ω

ω ω

ω ω

ω ω

= − + + +

= − − + + + = − − + + + = − + + +

1) Bhattacharyya,R., Dynamics of Marine Vehicles, John Wiley & Sons, 1978, pp183-2072) Kreyszig, Advanced engineerng mathematics, Wiley, Systems of Ordinary differential equations,pp.125~165

(풀이) Coupled Heave-pitch motion equation in frequency domain

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6DOF Equation of ship Motion : Solving 6DOF Equation of motion in Frequency domain



General Case

,

6

5

4

3

2

1

6

5

4

3

2

1

tiAti

A

A

A

A

A

A

ee ωω

ξξξξξξ

ξξξξξξ

xx =

=

= ,tiAei ωωxx = ,2 tiAe ωω xx −= tiAti

A

A

A

A

A

A

exciting ee

ffffff

ωω ηη fF 0

6

5

4

3

2

1

0 =

=

( )( ) ( ) ( ) tiAtiAtiAtiA eeeie ωωωω ηωω fxCxBxAM 02 =++−+

D=

RAO(Response Amplitude Operator): 1m wave Amplitude 를 가지는주파수 ω인 wave에 대한선박의 6자유도 운동 변위( ){ } tiAtiA eei ωω ηωω fxCBAM 0

2 =+++−

( ){ } AAi fxCBAM 02 ηωω =+++− AA fDx 1

0−=η A

A

fDx 1

0

−=η


Complex amplitude

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6DOF Equation of ship Motion : 6DOF RAO(Response Amplitude Operator)

1) 그림 출처 : Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1977, pp 47

<Roll>

<Pitch>

Example of RAO

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Nav

al A

rchi

tect

ure

& O

cean

Eng

inee

ring


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Chap.3 Surface Force & Moment

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Force acting on a ship in fluid

: 하나의 유체 입자가선박 표면에 가하는 힘

02 =Φ∇0

21 2 =+Φ∇++

∂Φ∂ ρgzPt

ρρ

tρgzP

∂Φ∂

−−= ρ

Bernoulli Equation

Linearization

RDI Φ+Φ+Φ=Φ

Laplace Equation

∂Φ∂

+∂Φ∂

+∂Φ∂

−−=ttt

ρgz RDIρ

staticP=

dSPd nF =

dS

dS

Fd

: 미소 면적

n : 미소 면적의 Normal 벡터

RDKFstatic FFFF +++= .

RDKF PPP +++ .유체입자가 선체표면에작용하는 압력

선박의 침수 표면 전체에 대하여 적분 (표면력)(유체입자가 선박에 작용하는 힘과 모멘트)

R

D

I




dynamicP

( : wetted surface)BS

유체입자 하나에 작용하는 Body force 와 Surface force로부터 구한 압력을 선박의 침수 표면 전체에 대해 적분하여 선박에 작용하는 유체력을 계산함

yz

Linear combination of basis solutions Basis solutions

FluidP

FluidP

∫∫=BSFluid dSnF

FF.K: Froude- krylov forceFD: Diffraction forceFR: Radiation force

How to know fluid pressure on Ship’s body?

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Force & moment acting on the surface- 계산방법

∫∫=BS

dSPnF

Fluid forceacting on the surface

성분별로나눠쓰면,

( )∫∫ −=BS

dSznnyPM 231

( )∫∫ −=BS

dSxnnzPM 312

( )∫∫ −=BS

dSynxnPM 123

성분별로나눠쓰면,

= ∫∫

BS

dSPnF 44,

= ∫∫

BS

dSPnF 55,

= ∫∫

BS

dSPnF 66,

∫∫=BS

jj dSPnF

( )∫∫ ×=BS

dSP nrM

Fluid momentacting on the surface

[ ][ ]

=

=T

T

zyx

nnn

,,

,, 321

r

n

−+−+−==× )()()( 123123

321

ynxnxnnzznnynnnzyx kjikji

nr

∫∫=BS

dSPnF 11

∫∫=BS

dSPnF 22

∫∫=BS

dSPnF 33

)6,,1( =j

[ ]( )Tnnn 321 ,,=n

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Nav

al A

rchi

tect

ure

& O

cean

Eng

inee

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- Fruode-Krylov Force & Diffraction Force- Radiation Force

Froude-Krylov Force & Diffraction Force

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Froude-Krylov Force & Diffraction Force (1)

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for Windows”, Delft University of Technology, 2003, pp36~382) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 300~307

+

입사파에 의한 힘

Froude-Kriloff Force Diffraction Force

산란파에 의한 힘

tiII ezyxtzyx ωφ ),,(),,,( =Φ

Incident Wave & Diffraction Wave Velocity Potential

tiDD ezyxtzyx ωφ ),,(),,,( =Φ

0( , , ) ikx kzI

gx y z e eφ ηω

− = −

)( BSonnn

ID

∂∂

−=∂∂ φφ

Body B.C. :

Froude Krylov Force & Diffraction Force

,),,( tiI

IFK eizyxρ

tρP ωωφ−=

∂Φ∂

−= tiD

DD eizyxρ

tρP ωωφ ),,(−=

∂Φ∂

−=

( )∫∫ +=+BS DFKDFK dSPP nFF

Considerkth component

(k=3이면, Heave Force)( )∫∫ +=+

BS kDFKkDkFK dSnPPFF ,,( )∫∫ +−=

BS kti

DI dSnie ωφφρ ω

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(Continue)

( )∫∫ +−=+BS k

tiDIkDkFK dSnieFF ωφφρ ω

,,

( )∫∫ ∂∂

+−=BS

ktiDI dS

neρ φφφ ω

kk ni

nωφ

=∂∂

∫∫

∂∂

+∂∂

−=BS

kD

kI

ti dSnn

ρe φφφφω

∫∫

∂∂

+∂∂

−=BS

Dk

kI

ti dSnn

ρe φφφφω

Green’s 2nd Theorem 사용

∂∂

=∂∂

∫∫∫∫BB S

DkS

kD dA

ndA

nφφφφ

Diffraction Wave velocity potential Body B.C.

)( BSonnn

ID

∂∂

−=∂∂ φφ

∫∫

∂∂

−∂∂

−=BS

Ik

kI

ti dSnn

ρe φφφφω

Haskind relations:

( 1, ,6)k

kφ

=

: Radiation potential

Radiation wave velocity potential Body B.C.

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(Continue)

∫∫

∂∂

−∂∂

−=+BS

Ik

kI

tikDkFK dS

nnρeFF φφφφω

,,

kzyxikI eeg )sincos(

0µµη

ωφ −−−=

kth radiation wave velocity potential : kφ

Incident wave velocity potential :

Already found

대입

∫ ∫

∂∂

−∂∂

−=L C

Ik

kI

ti dxdlnn

ρex

φφφφω

∫ +−=L kk

ti dxhfρe )(ω

∂∂

−=

∂∂

=

∫

∫

x

x

CI

kk

Ck

Ik

dln

h

dln

f

φφ

φφ : 단면에 작용하는 Froude-Krylov force

: 단면에 작용하는 Diffraction force

( 1, ,6)k

kφ

=


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- Froude-Krylov Force & Diffraction Force- Radiation Force

Radiation Force

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Radiation Force (FR) (1)정수중 선박의 강제

운동에 의해 발생한 힘

Radiation Force

∫∫=BS RR dSP nF

Radiation Wave Velocity Potential

tiRR ezyxtzyx ωφ ),,(),,,( =Φ

Radiation Force

tρP R

R ∂Φ∂

−= ti

jj

Aj ezyxρi ωφξω∑

=

−=6

1),,(

∫∫ ∑

−=

=BS kti

jj

Aj dSnieρ ωφξ ω

6

1

∫∫ ∑∫∫

−==

=BB S kj

tij

AjS kRkR dSnezyxρidSnPF

6

1, ),,( ωφξω

Considerkth component

(k=3이면, Heave Force)

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theroy program“ Seaway for Windows”, Delft University of Technology, 2003, pp30~332) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 295~300

∑=

=6

1),,(),,(

jj

AjR zyxzyx φξφ - B.C.과 Laplace Eq. 으로부터 구한 것

- 변위 는 주어진 값Ajξ ( )tiA

jj et ωξξ =)(

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( j방향 운동으로 인해 나타나는 k방향 힘 )

Radiation Force (FR) (2)

(Continue)

∫∫ ∑

−=

=BS kti

jj

AjkR dSnieρF ωφξ ω

6

1,

∫∫ ∑ ∂∂

−=

=BSkti

jj

Aj dS

neρ φφξ ω

6

1

대입kk ni

nωφ

=∂∂

∂∂

++∂∂

+∂∂

−= ∫∫∫∫∫∫BBB S

ktiA

SktiA

SktiA dS

nedS

nedS

neρ φφξφφξφφξ ωωω

662211

∑ ∫∫=

∂∂

−=6

1jS

ktij

Aj

B

dSn

eρ φφξ ω

ti

jS

kj

Aj edS

nρ

B

ωφφξ∑ ∫∫=

∂∂

−=6

1

6

1

A i tj k j

jF e ωξ

=

=∑B

kk j jS

F dSnφρ φ ∂

= −∂∫∫ 로 치환

:k jF

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theroy program“ Seaway for Windows”, Delft University of Technology, 2003, pp30~332) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 295~300

( 1, ,6)k

kφ

=


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(Continue)6

,1

A i tR k j k j

jF F e ωξ

=

=∑

B

kk j jS

F dSnφρ φ ∂

= −∂∫∫ ∫ ∫

∂∂

−=L

ck

j dxdlnx0

φφρ0

L

k jf dx= ∫

x

kk j jc

f dlnφρ φ ∂

= −∂∫

xC

(단면에서 j방향 운동으로 인해 나타나는 k방향 힘 )

(단면에서 구한 힘 또는 모멘트를 길이 방향으로 적분하여선박 전체에 작용하는 힘 또는 모멘트를 계산 = “Strip Theory”)

(If Slender body)

2k j k ja i bω ω= −

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theroy program“ Seaway for Windows”, Delft University of Technology, 2003, pp30~33


(단면의Added mass)

(단면의Damping Coefficient)

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Radiation Force (FR) (4)1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for

Windows”, Delft University of Technology, 2003, pp30~332) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 295~300

0

L

k j k jF f dx= ∫

jkjk bia ωω −= 2

(Continue)

2k j k jA i Bω ω= −

(Added mass) (Damping Coefficient)

6

,1

A i tR k j k j

jF F e ωξ

=

=∑

2

0 0

L L

k j k ja dx i b dxω ω= −∫ ∫

62

1( )A i t

j k j k jj

e A i Bωξ ω ω=

= −∑6

2

1( )A i t A i t

j k j j k jj

e A i e Bω ωξ ω ξ ω=

= −∑6

1( )j k j j k j

jA Bξ ξ

=

= − −∑

(가속도에 비례) (속도에 비례)

tiAjj

tiAjj

tiAjj

e

ei

e

ω

ω

ω

ωξξ

ωξξ

ξξ

2−=

=

=

y

3ξ

변위 :

속도 :

가속도 :

z

2ξ4ξ

jξ− jξ−

x

kk j jc

f dlnφρ φ ∂

= −∂∫

xC

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=×

6661

2221

161211

66

AA

AAAAA

A

=×

6661

2221

161211

66

BB

BBBBB

B

- Added mass matrix - Damping coefficientmatrix

- Radiation wave velocity potential

654321 ,,,,, φφφφφφ jkjkck

jjk biadln

fx

ωωφφρ −=∂∂

−= ∫ 2대입

( 선박의 j방향 운동변위가 1일 때 Velocity Potential):jφ


1) Journee, J.M.J. , Massie, W.W. , Offshore Hydrodynamics, Delft University of Technology, 2001,8-5~10

단면의 정보로 부터 선박의 added mass와 Damping Coefficient 구하기 위해서는

각 단면의 를 구한 뒤, 길이 방향으로 적분한다. (Strip Theory)jkjk ba , )6,,1,( =kj

How to find added mass and damping coefficient ???

∫=L

jkjk dxaA0 ∫=

L

jkjk dxbB0

- Added mass component - Damping coefficient component6개 Velocitypotential을모두 구해야Matrix를구할 수 있음

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x

y

z

1x y

3ξ

z

2ξ4ξ

xC

Strip Theory

: 각 2차원 단면의 유체력 계수 (Added mass, Damping Coefficient) 및 Wave exciting force를구한 후, 이를 길이 방향으로 적분하여 전체의 유체력을 구하는 근사적 방법

Assumption

(1) Resulting motion will be small

(2) The hull is slender

(4) The frequency of encounter should not be too low or too high

(3) Forward speed of the ship should be relatively low

(5) The hull sections are wall-sided at the waterline


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tiAet ωξξ 22 )( =

y

3ξy축 병진 운동 :

z

2ξ4ξ

xC tiAet ωξξ 33 )( =tiAet ωξξ 44 )( =

z축 병진 운동 :

x축 회전 운동 :

다음 중 2-D 단면에서 구할 수 있는 것은?

1φ 2φ 3φ 4φ 5φ 6φ

( 선박의 j방향 운동변위가 1일 때 Velocity Potential):jφ

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x

z

1x

(-) Moment

3φ

31φx−x

y

1x

(+) Moment

2φ

21φx



은 어떻게 구할 수 있을까? ( 선박의 j방향 운동변위가 1일 때 Velocity Potential):jφ651 ,, φφφ

x

y

z

1x

tiAet ωξξ 33 )( =z축 병진 운동 : 315 φφ x−=,3φ

216 φφ x=tiAet ωξξ 22 )( =y축 병진 운동 : ,2φ

※ 은 일반적인 2-D strip theory로 구할 수 없다.따라서, 경험식 또는 길이 방향 단면을 사용하여 계산함

1φ

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x

y

z

1x


y

3ξy축 병진 운동 :

z

2ξ4ξ

xC tiAet ωξξ 33 )( =tiAet ωξξ 44 )( =

z축 병진 운동 :

x축 회전 운동 :

Conclusion

2-D 단면의 세 velocity potential

을 구하고,

의 관계식을 사용하여

다른 velocity potential을 구한다.

432 ,, φφφ ,35 φφ x−=

26 φφ x=

즉, 2-D 단면의 만구하면 된다.

432 ,, φφφ

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Radiation Force (FR) (10)1) Journee, J.M.J. , Massie, W.W. , Offshore Hydrodynamics, Delft University of Technology, 2001,8-5~10

( )3333223

323

35

555)()( biaxdl

nxdl

nxxdl

nf

xxx cccωωφφρφφρφφρ −=

∂∂

−=∂−∂

−−=∂∂

−= ∫∫∫

,222222

222 biadln

fxc


−= ∫ 333323

333 biadln

fxc


−= ∫

,444424

444 biadln

fxc


−= ∫

( )2222222

222

26

666)()( biaxdl

nxdl

nxxdl

nf

xxx cccωωφφρφφρφφρ −=

∂∂

−=∂

∂−=

∂∂

−= ∫∫∫

242424

224 biadln

fxc


−= ∫

( )333323

33

35

335 )()( biaxdln

xdlnxdl

nf

xxx cccωωφφρφφρφφρ −−=

∂∂

−−=∂−∂

−=∂∂

−= ∫∫∫

( )26 2 226 2 2 2 22 22

( )x x xc c c

xf dl dl x dl x a i bn n nφ φ φρ φ ρ φ ρ φ ω ω∂ ∂ ∂

= − = − = − = − −∂ ∂ ∂∫ ∫ ∫

Given : 432 ,, φφφ

만 알고 있으면,

나머지 added mass 및 damping coefficient를 구할 수 있다.

),(),,(),,(),,( 4444333324242222 babababa

: 2-D 단면에서 j 방향 운동에의해 나타나는 k 방향 힘

jkf

y

3ξ

z

2ξ4ξ

xC※

∂∂

−= ∫xc

kjjk dl

nf φφρ

∫∫ ∂∂

=∂∂

xx ccdl

ndl

n2

44

2φφρφφρ

이므로, 42244224 , bbaa ==

※

( )424222

42

46

446)( biaxdl

nxdl

nxdl

nf

xxx cccωωφφρφφρφφρ −−=

∂∂

−=∂

∂−=

∂∂

−= ∫∫∫

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y

z1xx =

0

Added Mass (Heave 운동의 경우) Review : strip theory & 선박 운동 방정식 유도 과정)

)( 133 xa단면의 상하 동요로 인한z방향 added mass :

331135 )( axxa −=단면의 상하 동요로 인한y축 방향 회전 added mass :

535333 ξξ aa −− 5331333 ξξ axa +−=단면에 작용하는added mass force : ( )51333 ξξ xa −−=( )3335 xaa −=

Added Mass & Damping Coefficients

Damping Coefficient(Heave 운동의 경우)

x

y

z

1x

y

z1xx =

0

33 1( )b x단면의 상하 동요로 인한z방향 Damping Coefficeitns :

35 1 1 33( )b x x b= −단면의 상하 동요로 인한y축 방향 회전 damping Coefficient :

33 3 35 5b bξ ξ− − 33 3 1 33 5b x bξ ξ= − + 단면에 작용하는

damping coefficientforce : ( )33 3 1 5b xξ ξ= − − ( )35 33b xb= −

(a33,b33) 를 알고 있으면, (a35,b35), (a53,b53), (a55,b55)

added mass 및 damping coefficient를 구할 수 있다.

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Added Mass & Damping Coefficient- Heave & Pitch motion

:,,:

:

Ajk

AjkA

j

bax

hω

선박의 전진 속도

유체의 밀도

Wave amplitude

Sectional Froude Krylov force (jth mode)

Sectional Diffraction force (jth mode)

Encounter wave frequency

Values at the aftermost section

A

LbUdxaA 3323333 ω

−= ∫A

LUadxbB 333333 += ∫

AAAL

aUbxUBUdxxaA 332

2

33203323335 ωωω

−+−−= ∫

AAAL

bUaUxUAdxxbB 332

2

330333335 ω

−−+−= ∫

AALbxUBUdxxaA 332

03323353 ωω++−= ∫

AALaUxUAdxxbB 33

0333353 −−−= ∫

AA

AAL

axUbxUAUdxaxA 332

2

3320332

2

332

55 ωωω+−+= ∫

AA

AAL

bxUaUxBUdxbxB 332

2

3320

332

2

332

55 ωω+++= ∫

( ) A

Lh

iUdxhfF 3333 ω

ραρα ++= ∫

( ) AALhx

iUdxh

iUhfxF 33335 ω

ραω

ραρα −

++−= ∫

::::

jf

U

αρ

== ∫∫ L jkjkL jkjk dxbBdxaA 00 ,

(a33,b33)를 알고 있으면, (A33,B33),(A35,B35), (A53,B53), (A55,B55)

added mass 및 damping coefficient를 구할 수 있다.

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:,,:

:

*4B

bax

h

Ajk

AjkA

j

ω선박의 전진 속도

유체의 밀도

Wave amplitude





::::

jf

U

αρ

Roll Damping

A

LbUdxaA 2222222 ω

−= ∫A

LUadxbB 222222 += ∫

A

LbUdxaAA 242244224 ω

−== ∫A

LUadxbBB 24244224 +== ∫

AAAL

aUbxUBUdxxaA 222

2

22202222226 ωωω

+−+= ∫AA

ALbUaUxUAdxxbB 222

2

220222226 ω

++−= ∫A

LbUdxaA 4424444 ω

−= ∫*44444444 BUadxbB A

L++= ∫

AAAL

aUbxUBUdxxaA 242

2

24202422446 ωωω

+−+= ∫AA

ALbUaUxUAdxxbB 242

2

240242446 ω

++−= ∫

AA

AAL

axUbxUAUdxaxA 222

2

222

20222

2

222

66 ωωω+−+= ∫

AA

AAL

bxUaUxBUdxbxB 222

2

2220

222

2

222

66 ωω+++= ∫

AALbxUBUdxxaA 222

02222262 ωω−−= ∫

AALaUxUAdxxbB 22

0222262 ++= ∫

AALbxUBUdxxaA 242

02422464 ωω−−= ∫

AALaUxUAdxxbB 24

0242464 ++= ∫

== ∫∫ L jkjkL jkjk dxbBdxaA 00 ,Added Mass & Damping Coefficient

- Sway & Roll & Yaw motion(a22,b22), (a24,b24) , (a44,b44)를 알고 있으면added mass 및 damping coefficient를 구할 수 있다.

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Sway & Roll & Yaw Force

( ) A

Lh

iUdxhfF 2222 ω

ραρα ++= ∫

( ) AALhx

iUdxh

iUhfxF 22226 ω

ραω

ραρα +

++= ∫

( ) A

Lh

iUdxhfF 4444 ω

ραρα ++= ∫

:

:,,:

:

*4B

bax

h

Ajk

AjkA

j

ω선박의 전진 속도

유체의 밀도

Wave amplitude





::::

jf

U

αρ

Roll Damping

== ∫∫ L jkjkL jkjk dxbBdxaA 00 ,

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∫=L

jkjk dxfF0

jkjk bia ωω −= 2

ti

jjk

AjkR eFF ωξ∑

=

=6

1, ∑

=

−−=6

1)(

jjkjjkj BA ξξ

∫ ∂∂

−=xc

kjjk dl

nf φφρ

Froude-Krylov Force & Diffraction Force

Radiation Force

(We don’t need to calculate ) ),,( zyxDφ

nk

D ∂∂φφ

nD

k ∂∂φφ

nI

k ∂∂

−φφ

Green’s2nd theorem Body B.C.

∫ ∫

∂∂

−∂∂

−=+L C

Ik

kI

tikDkFK dxdl

nnρeFF

x

φφφφω,,

+: Radiation wavevelocity potentialRΦ

∑=

==Φ6

1),,(),,,(

j

tij

tiRR eezyxtzyx ωω φφ

+

Fixed

: Incident wavevelocity potentialIΦ

: Diffraction wavevelocity potentialDΦ

tiII ezyxtzyx ωφ ),,(),,,( =Φ


jkjk BiA ωω −= 2

Radiation Force (FR) (10)-Summary

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Chap.4 Velocity Potential

- Incident Wave Velocity Potential- Problem in Infinitely Long Horizontal Circular Cylinder- Radiation Wave Velocity Potential- Diffraction Wave Velocity Potential

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Force acting on a ship in fluid

: 하나의 유체 입자가선박 표면에 가하는 힘

02 =Φ∇0

21 2 =+Φ∇++

∂Φ∂ ρgzPt

ρρ

tρgzP

∂Φ∂

−−= ρ

Bernoulli Equation

Linearization

RDI Φ+Φ+Φ=Φ

Laplace Equation

∂Φ∂

+∂Φ∂

+∂Φ∂

−−=ttt

ρgz RDIρ

staticP=

dSPd nF =

dS

dS

Fd

: 미소 면적

n : 미소 면적의 Normal 벡터


RDKF PPP +++ .유체입자가 선체표면에작용하는 압력

선박의 침수 표면 전체에 대하여 적분 (표면력)(유체입자가 선박에 작용하는 힘과 모멘트)

R

D

I




dynamicP

( : wetted surface)BS

유체입자 하나에 작용하는 Body force 와 Surface force로부터 구한 압력을 선박의 침수 표면 전체에 대해 적분하여 선박에 작용하는 유체력을 계산함

yz

Linear combination of basis solutions Basis solutions

FluidP

FluidP

∫∫=BSFluid dSnF

FF.K: Froude- krylov forceFD: Diffraction forceFR: Radiation force

How to know fluid pressure on Ship’s body?

How to know velocity potential?

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선형화1)된 wave로 분해

+

+

입사파가 선박에 의해 교란되지 않는다고 가정함 입사파에 의한 velocity potential

선박의 존재로 인하여 교란된 파(wave). 물체 고정 산란파에 의한 velocity potential

정수 중에서 선박의 강제 진동으로 인해 발생하는 파(wave) 기진력에 의한 파의 velocity potential

I D RΦ = Φ +Φ +Φ Total Velocity Potential

Fixed

Incident wave velocity potential ( )IΦ

Diffraction wave velocity potential ( )DΦ

Radiation wave velocity potential ( )RΦ

1) Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, Ch 12.1 (pp 535~538)

Superposition TheoremLaplace equation은 선형방정식이므로, 각의 해를더한 것 (superposition)도해가 된다1).

파랑 중 선박 주위 유체의 운동: 유체장의 운동으로 인해 유체 입자의 속도,가속도,압력이변하게 되고, 선박 표면의 유체 입자가 선박에 가하는 압력도변하게 된다.

교란정지상태

yz

yz

y

z

y

z

y

z

R

D

I

ΦΦΦ : Incident Wave V.P.

: Diffraction Wave V.P.

: Radiation Wave V.P.

Velocity Potential- Decomposition of Velocity Potential

TPρgzt

ρ∂Φ

= − −∂

∫∫BS

dSPn=FluidF


How to know velocity potential?

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+

+


선박의 존재로 인하여 교란된 파(wave). 물체 고정 산란파에 의한 velocity potential



교란정지상태

Fixed

Incident wave velocity potential ( )IΦ





yz

yz

y

z

y

z

y

z

R

D

I




Velocity Potential : Decomposition of Velocity Potential

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73/2031) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 285~290

Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, Ch 12.1 (pp 535~538)

Decomposition of Velocity potential

),,,(),,,(),,,(),,,( tzyxtzyxtzyxtzyx RDIT Φ+Φ+Φ=Φ

{ } tiRDI ezyxzyxzyx ωφφφ ),,(),,(),,( ++=

Time Independent Term (Complex)

시간이 많이 지나 Steady 상태에서Harmonic Motion(Transient motion 고려안함)

{ }),,,(Re),,,( tzyxtzyx TΦ=Φ

{ } taI ωcosRe =Φ

ex) ),,( zyxIφ is not a complex (real)If ),,( zyxIφ is a complexIf

axI =)(φLet

( )( )( ) ( )tatbitbta

titibaωωωω

ωωsincossincos

sincos++−=

++=

(or Take an Imaginary term)

tiII ex ωφ )(=Φ

)sin(cos tita ωω +=

tiata ωω sincos +=

(Euler 공식)

{ } )cos(sincosRe εωωω −=−=Φ tctbtaI

ibaxI +=)(φLet

tiII ex ωφ )(=Φ (Euler 공식)

Phase가 나타남

R

D

I




Phase가 안나타남

Wave의 Phase차이를 고려해 주어야 한다.

Velocity Potential : Superposition of Velocity potential

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※ ∑=

=+++++=6

1665544332211),,(

jj

Aj

AAAAAAR zyx φξφξφξφξφξφξφξφ

Decomposition of Velocity potential

),,,(),,,(),,,(),,,( tzyxtzyxtzyxtzyx RDIT Φ+Φ+Φ=Φ

{ }( , , ) ( , , ) ( , , ) i t i tI D R Tx y z x y z x y z e eω ωφ φ φ φ= + + =

Time Independent Term (Complex)

시간이 많이 지나 Steady 상태에서Harmonic Motion(Transient motion 고려안함)

1) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 285~290Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, Ch 12.1 (pp 535~538)

6

1( , , ) ( , , ) ( , , )A

I D j jj

x y z x y z x y zφ φ ξ φ=

= + +∑

선박의 j방향 운동변위가 1일 때 Velocity Potential:jφ:A

jξ 선박의 j방향 운동변위의 크기 ( j = 4,5,6에서는 rotational angle in Radian)

선박의 운동변위(Given)tiA

jj et ωξξ =)(

크기(Amplitude)

파진폭(waveamplitude)

파고(waveheight)

R

D

I




Velocity Potential : Superposition of Velocity potential

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Nav

al A

rchi

tect

ure

& O

cean

Eng

inee

ring


Seoul NationalUniv.


- Incident Wave Velocity Potential- Problem of Infinitely Long Horizontal Circular Cylinder- Radiation Wave Velocity Potential- Diffraction Wave Velocity Potential

Incident Wave Velocity Potential

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+

+



교란정지상태




yz

yz

y

z

R

D

I



: Radiation Wave V.P.Incident Wave Velocity Potential1)

선박의 존재로 인하여 교란된 파(wave). 물체 고정 산란파에 의한 velocity potentialFixed


y

z


Incident wave velocity potential ( )IΦy

z

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Wave Equation

① Governing Equation :

02 =Φ∇

Lateral B.C.

Lateral B.C.

Bottom B.C.u

w

Dynamic Free Surface B.C. (경계면 사이에서 압력의 변화가 없음즉, 두 매질의 경계면에서 압력은 동일)

Kinematic Free Surface B.C. (No flow across the interface 경계면에서 속도가 동일함)

② Boundary condition(B.C.) :

zx

h

1) Dean, R.G. , Water wave mechanics for engineers and scientists, Prentice-Hall,Inc, 1984, pp.41-77Incident Wave Velocity Potential1)

: Superposition of Velocity potential

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① Kinematic Free Surface B.C.(KFSBC) : 경계면 사이에 유동(flow)이 없기 위해서는 경계면에서 입자의 속도가 동일해야 한다(z축 방향 속도가 같음)

zx

Lateral B.C.Lateral B.C.

Bottom B.C.u

w

Dynamic Free Surface B.C. Kinematic Free Surface B.C.

Boundary condition(B.C.)

( )txz ,η=

η* : z방향 변위

h

유체입자의 z 방향 속도 :

z

wz η=

∂Φ=∂

자유표면의 z 방향 속도 :

( , )

z

d x t dx udt t x dt t x η

η η η η η

=

∂ ∂ ∂ ∂= + = +∂ ∂ ∂ ∂

dwdtη

∴ = uz t x

η η∂Φ ∂ ∂= +

∂ ∂ ∂at z=η

자유표면(free surface)의 x축 방향속도는유체 입자의 x방향 속도성분 u와 같다.


: Boundary Condition

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1t t=

zx


Bottom B.C.u

w



( )txz ,η=


h

자유표면의 z 방향 속도 :

( , )d x t dxdt t x dt

η η η∂ ∂= +∂ ∂

① ②

①②의 의미는?

양변에 dt를 곱하면,

( , )d x t dt dxt xη ηη ∂ ∂

= +∂ ∂

① 위치가 고정일 때, 시간 변화에 따른 η의 변화량

① ②

② 시간이 고정일 때, 위치 변화에 따른 η의 변화량

z

x기울기 :

xη∂∂

dx

dxxη∂∂dt

tη∂∂

2t t= 2t t=



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zx

uw

( )txz ,η=η* : z방향 변위

h

ddtη

w

속도가다르면 ?

경계면에서 유체입자가서로 이동함즉, 경계가 없음

속도가같으면 ?

경계면을 유지함

즉, Kinematic F.S.B.C.은두 매질간에 경계를 만드는 조건

Kinematic Free Surface Boundary condition의 의미(F.S.B.C.)



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Incident Wave Velocity Potential1)

: Boundary condition

zx


Bottom B.C.u

w



② Bottom B.C. (BBC)

: 바닥면에서 유체가 스며들거나 바닥으로 침투하지 않는다면(Impermeable) 다음 조건이 성립

( )txz ,η=


h

만약, 바닥이 수심 z=-h에서평평하다고 가정하면(Horizontal bottom)

0=∂Φ∂

−= hzz

(바닥면의 속도) = (바닥면의 유체의 속도)

바닥면은 고정되어 있으므로,(바닥면의 속도) = 0

→ (좌변) :

hzn −=∂Φ∂

=•= nV(바닥면 유체의 속도)

→ (우변) :

0=∂Φ∂

∴−= hzn

1) Dean, R.G. , Water wave mechanics for engineers and scientists, Prentice-Hall,Inc, 1984, pp.41-77

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zx


Bottom B.C.u

w



③ Dynamic Free Surface B.C. (DFSBC) : 경계면에서 유체의 압력은 대기압과 같아야 함

( )txz ,η=


h

한편, 경계면에서 표면의 압력은 대기압과 같으므로,

21 (on z )2

Surface atmP Pgt

η ηρ ρ

∂Φ+ + ∇Φ + = =

∂

Wave가 생성되었을 때, Bernoulli Equation에 의해 표면에서 유체의 압력은,

Wave가 생성되기 전 상태를 고려하면, 이므로,)0z(on,0 ===Φ∇= atmPPV

Bernoulli Equation

21 ( )2

P g z F tt ρ

∂Φ+ + ∇Φ + =

∂

(at z )η=

021 2 =+Φ∇+

∂Φ∂ ηgt

/ ( )atmP F tρ =

( ))z(on η== atmSurface PP

212

atm atmP Pgt

ηρ ρ

∂Φ+ + ∇Φ + =

∂


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zx

uw

( )txz ,η=η* : z방향 변위

h

airP

fluidP

airP

fluidP

압력의 차이만큼수면의 높낮이(z)가 바뀜즉, 경계면이 ‘자유’롭게 변형됨

즉, Dynamic F.S.B.C.은‘자유’ 표면을 만드는 조건

(압력이 동일하도록 자유 표면의 모양이 변함)

Dynamic Free Surface Boundary condition의 의미(F.S.B.C.)

Bernoulli Equation21 ( )

2P g z F t

t ρ∂Φ

+ + ∇Φ + =∂

수면 위 압력은대기압으로항상 동일함2)

Bernoulli equation에 의해유체의 유동이 생기면속도에 따라 압력이 달라짐

( )air fluidP P≠

fluidP

airP 수면 위 압력은대기압

유동이 없는상태

( )air fluidP P=

유동발생

2) 실제로는 수면의 높이차이에 의해 대기압도 변하지만, 그 차이가 작다고 보고, 수면에서 대기압은 모두 동일한 것으로 가정함

( )air fluidP P=

0z =


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zx


Bottom B.C.u

w



④ Lateral B.C. (DFSBC)

( )txz ,η=


h

),,(),,(),,(),,(

tzLxtzxTtzxtzx

+Φ=Φ+Φ=Φ

파의 주기(wave period)를 T, 파장(wave length)을 L 이라고 하면, 다음이 성립한다.

: 주기가 일정하다는 조건에 의해 Periodic lateral B.C를 적용한다.

L


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zx


Bottom B.C.u

w



( )txz ,η=


h

L

④ Lateral B.C.

),,(),,(),,(),,(

tzLxtzxTtzxtzx

+Φ=Φ+Φ=Φ

① Kinematic Free Surface B.C.(KFSBC)

0uz t x

η η∂Φ ∂ ∂− − =

∂ ∂ ∂

③ Dynamic Free Surface B.C. (DFSBC)

021 2 =+Φ∇+

∂Φ∂ ηgt


0=∂Φ∂

−= hzz

<Summary of the 2-D periodic water wave boundary condition>

)z(on η= )z(on η=



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uz t x

η η∂Φ ∂ ∂= +

∂ ∂ ∂)z(on η=

0 0

. .z z z

u u u H O Tt x t x z t xη

η η η η η ηη= = =

∂ ∂ ∂ ∂ ∂ ∂ ∂ + = + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂

우변을 Tayler series로 전개하면,

여기서 파장에 비해 파고가 작다고 가정했으므로, 1, 1, 1u wη << << <<

1,10

00

0<<

∂Φ∂

=<<∂Φ∂

==

==

=z

zz

z zw

xu

작은 항이 두 개 이상 곱해진 경우를 무시하면,

z tη∂Φ ∂

=∂ ∂

=> Linearized Kinematic Free Surface B.C.(KFSBC)

(High Order Term)

0ztη

=

∂ = ∂

(at z 0)=

2) kundu,P.K., Fluid Mechanics, Academic Press, 2008, pp.219-2231) Dean, R.G. , Water wave mechanics for engineers and scientists, Prentice-Hall,Inc, 1984, pp.41-77

(For infinitesimally small waves, eta is small, and therefore it is assumed that velocities and pressures are small) 1)

uw

1<<1<<η 1<<z

x

dη

dx xη∂∂

기울기 : 1<<


: Linearization(선형화) of boundary condition

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021 2 =+Φ∇+

∂Φ∂ ηgt )z(on η=

0..21

21

21

0

2

0

22 =+

+Φ∇+∂Φ∂

∂∂

+

+Φ∇+∂Φ∂

=

+Φ∇+∂Φ∂

===

TOHgtz

gt

gt zzz

ηηηηη

Tayler series로 전개하면,

여기서 파장에 비해 파고가 작다고 가정했으므로, 1, 1, 1u wη << << <<

1,10

00

0<<

∂Φ∂

=<<∂Φ∂

==

==

=z

zz

z zw

xu

작은 항이 두 개 이상 곱해진 경우를 무시하면,

00

=

+∂Φ∂

=z

gt

η

=> Linearized Dynamic Free Surface B.C.(DFSBC) tg ∂Φ∂

−=1η )0z(on =

(High Order Term)

2) kundu,P.K., Fluid Mechanics, Academic Press, 2008, pp.219-2231) Dean, R.G. , Water wave mechanics for engineers and scientists, Prentice-Hall,Inc, 1984, pp.41-77

(For infinitesimally small waves, eta is small, and therefore it is assumed that velocities and pressures are small)1)

uw

1<<1<<η 1<<z

x

dη

dx xη∂∂

기울기 : 1<<


: Linearization(선형화) of boundary condition

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<Summary of the 2-D periodic water wave boundary condition>

zx


Bottom B.C.u

w



④ Lateral B.C.

( )txz ,η=


h

),,(),,(),,(),,(

tzLxtzxTtzxtzx

+Φ=Φ+Φ=Φ

L

① Kinematic Free Surface B.C.(KFSBC) ③ Dynamic Free Surface B.C. (DFSBC)


0=∂Φ∂

−= hzz

)0z(on = )0z(on =tg ∂Φ∂

−=1η

tz ∂∂

=∂Φ∂ η




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zx


Bottom B.C.u

w



( )txz ,η=


h

L



)0z(on =

)0z(on =tg ∂Φ∂

−=1η

tz ∂∂

=∂Φ∂ η

t로 미분

2

21tgt ∂Φ∂

−=∂∂η

2

21tgz ∂Φ∂

−=∂Φ∂

02

2

=∂Φ∂

+∂Φ∂

zg

t( )0=Φ+Φ ztt g

)0z(on =

=> Linearized Free Surface B.C.1) Dean, R.G. , Water wave mechanics for engineers and scientists, Prentice-Hall,Inc, 1984, pp.41-77



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Boundary condition(B.C.) η* : z방향 변위

zx


uw


( )txz ,η=

h

L

0=∂Φ∂z

)0(on =z )0(on =ztg ∂Φ∂

−=1η

tz ∂∂

=∂Φ∂ η

)(on hz −=

Bottom B.C.

),,(),,(),,(),,(

tzLxtzxTtzxtzx

+Φ=Φ+Φ=Φ

0=Φ+Φ ztt gLinearized Free Surface B.C.

)0(on =z




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zx

Lateral B.C.

Lateral B.C.

uw

( )txz ,η=

h

L

0=∂Φ∂z )(on hz −=

Bottom B.C.

),,(),,(),,(),,(

tzLxtzxTtzxtzx

+Φ=Φ+Φ=Φ

0=Φ+Φ ztt g

Linearized Free Surface B.C.

)0(on =z

{ }tiezx ωφ ),(Re=Φ),( zxφ( : Complex amplitude of the velocity potential)

에서 시간에 대한 주기 함수로),,( tzxΦ=Φ

가정하고, 시간 항을 분리하면,

위 Velocity potential을 지배 방정식과경계 조건에 대입하면,

① 지배 방정식

( ) ( ) 0),(),( 222 =∇=∇=Φ∇ zxeezx titi φφ ωω

=

∂∂

+∂∂

=∇ 0,0 2

2

2

22

zxφφφ

② Linearized Free Surface B.C.

( ) 02 =+−=Φ+Φ tiz

tiztt egeg ωω φφω

tie ω 로 나누면,

02 =+−∴ zgφφω

③ Bottom B.C.

0=∂∂

=∂Φ∂

ze

zti φω 0=

∂∂

∴zφ )(on hz −=

)0(on =z

④ Lateral B.C.

),,(),,( tzLxtzx +Φ=Φ

( , , ) ( , , )i t i tx z t e x L z t eω ωφ φ= +

),(),( zLxzx +=φφ1) Dean, R.G. , Water wave mechanics for engineers and scientists, Prentice-Hall,Inc, 1984, pp.41-77



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zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

0=∂Φ∂z )(on hz −=

Bottom B.C.

),,(),,(),,(),,(

tzLxtzxTtzxtzx

+Φ=Φ+Φ=Φ

0=Φ+Φ ztt g


)0(on =z


zx


uw

( )txz ,η=

h

L

)(on hz −=

Bottom B.C.


)0(on =z02 =+− zgφφω

0=∂∂

zφ ),(),( zLxzx +=φφ




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1) Dean, R.G. , Water wave mechanics for engineers and scientists, Prentice-Hall,Inc, 1984, pp.41-772) Erwin Kreyszig, Advanced Engineering Mathematics 9th ,Wiley,2005,Ch12.5(p552~562)

공학 수학 Chapter 12.51)

(2-D Heat equation ofSteady state )

)(),( xfbxu =

0)0,( =xu

0),( =yau0),0( =yu

a

b

Rx

y

02

2

2

22 =

∂∂

+∂∂

=∇yu

xuu

② Boundary condition :


3학년 해양파 역학(Wave Equation)


02 =∇ φ

동일한 방정식을 푸는데 각각 다른 경계 조건이 적용됨

② Boundary condition(B.C.) :


zx

uw

( )txz ,η=

h

L



)(on hz −=

Bottom B.C.

0=∂∂


=

∂∂

+∂∂ 02

2

2

2

yxφφ

( Dirichlet B.C.1) )

( Robin B.C.1) )

계산결과로 이동


: Analytic solution

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zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


02 =∇ φ

Laplace Equation

: Governing Equation

① Velocity potential 는 의 함수 이므로,변수 분리법(separation of variables)에 의해

φ zx,

)()( zGxF ⋅=φ 로 둘 수 있다.

② Laplace Equation에 대입하면,

02

2

2

2

2

2

2

22 =+=

∂∂

+∂∂

=∂∂

+∂∂

=∇ zzxx FGGFzGFG

xF

zxφφφ

0=+∴ zzxx FGGF 0=+FF

GG xxzz p

FF

GG xxzz =−=

FG 로 나눔

(∵ x와 z만의 함수가 같은 것은 상수뿐)

00

=−=+

pGGpFF

zz

xx


: Analytic solution

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③ 의 부호에 따른 방정식의 해를 계산해 보면,p

02 =− FFxx ν

00

=−=+

pGGpFF

zz

xx

(i) 일 때,0<p 2ν−=p

Lateral B.C.에 의해 x에 대한 주기함수여야 하는데 exponential 함수는 주기함수가 아님.

따라서 해가 될 수 없음.

xx BeAeF νν −+=

(ii) 일 때,0=p

0=xxF BAxF +=

이라 하면,


z

xLateral B.C.

Lateral B.C.

uw

( )txz ,η=

h

L

)(on hz −=

Bottom B.C.



0=∂∂


BLxALxFBAxxF ++=+=+= )()()(

BxF =∴ )(

wave는 x에 따라서 주기적으로 ‘변’하는 데, 이를 만족하지 않음.

따라서 해가 될 수 없음

( )txz ,η=Incident Wave Velocity Potential1)

: Analytic solution

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: Analytic solution

00

=−=+

pGGpFF

zz

xx

(iii) 일 때,0>p 2kp = 이라 하면,

02 =+ FkFxx ( ) ikx ikxF x Ae Be−= +


z

xLateral B.C.

Lateral B.C.

uw

( )txz ,η=

h

L

)(on hz −=

Bottom B.C.



0=∂∂


( )txz ,η=

(Euler 공식에 의해 kxikxeikx sincos += 이므로,x에 대한 주기 함수의 성질을 가짐 -> Lateral B.C. 만족)

한편, 0=− pGGzz 에서

02 =− GkGzzkzkz DeCezG −+=)(

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z

xLateral B.C.

Lateral B.C.

uw

( )txz ,η=

h

L

)(on hz −=

Bottom B.C.



0=∂∂


( )txz ,η=

kxBkxAxF sincos)( ′+′=

ikxikx BeAeF −+=

기저 변환

)sin()cos()( kLkxBkLkxALxF +′++′=+=

2kL nπ∴ =

2n

nk kLπ

∴ = =

( ) n nik x ik xn n nF x A e B e−∴ = +

( 1, 2, )n =

( ) n nk z k zn n nG z C e D e−= +


: Analytic solution

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④ Bottom B.C. 적용

( )n nk z k zn nn n n n n n

GF F C k e D k ez zφ −∂ ∂

= = −∂ ∂

( ) 0n nk h k hnn n n n n

z h

F C k e D k ezφ −

=−

∂= − =

∂

0n nk h k hn n n nC k e D k e− − = 2 nk h

n nC D e=


zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


( )2 2( ) n n n n n n n nk z k z k h k z k z k h k z k zn n n n n nG z C e D e D e e D e D e e e− − −= + = + = +

( , ) ( ) ( )n n

n n

n n nik x ik x

n n nk z k z

n n n

x z F x G z

F A e B e

G C e D e

φ−

−

= ⋅

= + = + nF 나눔

( )( , ) 2 cosh ( )n n nik x ik x k hn n n n nx z A e B e D e k z hφ −∴ = + +

( ) ( )( ) ( )n n n n n n n nk h k z k h k z k h k h k z h k z hn nD e e e D e e e+ − − + − += + = +

( ) ( )

22

n nn

k z h k z hk h

ne eD e

+ − + +=

2 cosh ( )nk h

n nD e k z h= +


: Analytic solution

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1t t=


zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


Progressive wave인 경우 (방향 +x)

i tn n e ωφΦ = ⋅

( )( ) ( ) ( )n ni k x t i k x tn n nA e B e G zω ω+ − −= + ⋅

2t t=

파의 진행방향

sin( )

sin

n

nn

k x t

tk xk

ω

ω

+

= +

sin( )

sin

n

nn

k x t

tk xk

ω

ω

−

= −

x

z

x

0nA∴ =

After Δt( )ni k x tAe ω+ ( )ni k x tBe ω−

( , ) 2 cosh ( )n nik x k hn n n nx z B D e e k z hφ −∴ = ⋅ ⋅ +

( )( , ) 2 cosh ( )n n nik x ik x k hn n n nx z A e B e De k z hφ −= + +

( , ) ( ) ( )

2 cosh ( )

n n

n

n n nik x ik x

n n nk h

n n n

x z F x G z

F A e B e

G D e k z h

φ−

= ⋅

= + = +


: Analytic solution

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zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


⑤ Linearized Free Surface B.C. 적용

2 0nn g

zφω φ ∂

− + =∂

2 0nn n n n

dGF G gF

dzω− + =

2 0nn

dGG g

dzω− + =

( ) 2 cosh ( )

2 sinh ( )

khn n

khnn

G z D e k z hdG kD e k z hdz

= + = +

2 cosh ( ) sinh ( ) 0cosh cosh

n nn n

n n

k z h k z hgkk h k h

ω + +− + =

2 sinh tanhcosh

nn n n n

n

k hgk gk k hk h

ω = =

2 tanhn n ngk k hω∴ =

=> dispersion relation

( ) ( )n n nF x G zφ = 대입

nF 로 양변을 나눠줌

, nn

dGG

dz대입

0=z 대입

( , ) ( ) ( )

2 cosh ( )

n

n

n n nik x

n nk h

n n n

x z F x G z

F B e

G D e k z h

φ−

= ⋅

= = +


: Analytic solution

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2 tanhn n n ngk k h gkω = ≈

Deep sea 일 때, ( h→∞)

h→∞

( )22

limsinhlimkhkhkh

hh

eeekh =−

=−

∞→∞→

( )22

limcoshlimkhkhkh

hh

eeekh =+

=−

∞→∞→

( ) 12/2/

coshsinhlimtanhlim ===

∞→∞→ kh

kh

hh ee

khkhkh

2n T

πω =

2n

nkLπ =

Lg

Tππ 22 2

=

Lg

T=2

2π

L :파장

T :주기

=> 파장(길이)과 주기(시간)와의 관계식

즉, 장파일수록 주기가 길고, 단파일수록 주기가 짧아짐을 알 수 있다.


: Reference, Dispersion Relation의 의미

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zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


Dynamic Free Surface B.C. 적용

Dynamic Free Surface B.C. :

Dynamic Free Surface B.C.

)0(on =ztg ∂Φ∂

−=1η

2 cosh ( )n n nik x k h i tnn n n

i B D e e k z h eg

ωω −= − ⋅ ⋅ + ⋅

0z = 일 때 이므로,

( )on 0z =

2 coshn n nik x k h i tnn n n n

i B D e e k h eg

ωωη −= − ⋅ ⋅ ⋅

( )2 coshn n nk h i k x tnn n ni B D e k h e

gωω − −

= − ⋅ ⋅ ⋅

( )ˆ n ni k x tni e ωη − −=

Amplitude

Wave amplitude

ˆ 2 coshnk hnn n n nB D e k h

gωη = − ⋅ ⋅

1ˆ2cosh

nk hn n n

n n

gB D ek h

ηω

= − ⋅ ⋅

( , ) 2 cosh ( )n nik x k hn n n nx z B D e e k z hφ −= ⋅ ⋅ +

cosh ( )ˆcosh

nik x nn n

n n

k z hg ek h

φ ηω

− +∴ = − ⋅ ⋅ ⋅

cosh ( )ˆcosh

n nik x i tnn n

n n

k z hg e ek h

ωηω

− +Φ = − ⋅ ⋅ ⋅

( )( , )1 1 n

n

i tn i tn n

n n

x z e ie

g t g t g

ωω

φ ωη φ

∂∂Φ= − = − = −

∂ ∂


: Wave Amplitude


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<Summary of the wave equation>

,cosh ( )ˆ( , )

coshnik x n

I n nn

k z hgx z ek h

φ ηω

− += − ⋅ ⋅ ⋅


zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


, ,( , , ) ni tI n I nx z t e ωφΦ =

Irregular wave의 경우,

,1

( , , ) ( , , )I I nn

x z t x z t∞

=

Φ = Φ∑

Regular wave의 경우,

( , , ) i tI Ix z t e ωφΦ =

cosh ( )ˆ( , )cosh

ikxI

g k z hx z ekh

φ ηω

− += − ⋅ ⋅ ⋅

(regular wave를 합성하여 irregular wave 만들 수 있음.)


: Summary


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<Summary of the wave equation>


zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


If Deep water,

cosh ( )cosh

kzk z h ekh+

=

)( ∞→h

( ) ( ) ( )

cosh ( )limcosh

lim lim

h

k z h k z h k z hkz

kh kh khh h

k z hkh

e e e ee e e

→∞

+ − + +

−→∞ →∞

+

− = = = −

ˆ ikx kzg e eηω

−= −),( zxIφ

Regular wave의 경우,

( , , ) i tI Ix z t e ωφΦ =

cosh ( )ˆ( , )cosh

ikxI

g k z hx z ekh

φ ηω

− += − ⋅ ⋅ ⋅


: Summary


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zx

Lateral B.C.

Lateral B.C.

uw

( )tx,η

h

L

)(on hz −=

Bottom B.C.



0=∂∂


Step1.해상에서 파의 파고 및 주파수 계측

ωη ,0⇒

Step3. Velocity Potential식에 대입

kzikxI eegzx −−−= 0),( η

ωφ

Step2.Dispersion Relation으로 부터 Wave number 계산 dispersion relation : khgk tanh2 =ω

(해양파를 Regular wave로 가정)


: Summary, ϕI 계산과정

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Nav

al A

rchi

tect

ure

& O

cean

Eng

inee

ring


Seoul NationalUniv.



Problem of Infinitely Long Horizontal Circular Cylinder

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Nav

al A

rchi

tect

ure

& O

cean

Eng

inee

ring


Seoul NationalUniv.




< Analytic Solution >

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원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (analytic solution)- Problem description

Governing equation

Boundary Condition :

- Motion of the ship : 3 3( ) A i tt e ωξ ξ=

23 3 3 0rr rr r θθφ φ φ+ + =

1) Faltinsen, O.M. , Sea loads on ships and offshore structures, Cambridge Univ. Press, 1998, Ch3 (pp 50~55)

- 원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 문제를 고찰하여 보자.(2차원 문제)

θ r

R3ξ

y

z

r∂∂φ

w θ

0=φ 0=φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

0=φ

ωθξφ A

r 3cos−=∂∂

0θ =

*2πθ θ = −

① Free surface condition

( )3 0, rφ = → ∞

=−=

2,

2πθπθ3 0,φ =

④ Body boundary condition : ( )3 cos ,i r Rrφ θ ω∂

= − =∂

Wave가 없음

,(Laplace Equation in Polar Coordinate)

, (물체로 부터 무한히 먼 곳에서 파가 발생하지 않음)

, ( , ) ( co s , sin ),y z r rθ θ= ,2 2

r R π πθ = − ≤ ≤

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θr

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

3 0,φ = 3 0,φ =

=

2πθ

−=

2πθ

( )Rr =

( ), r →∞

03 =φ

−=∂∂ ωθφ i

rcos3

*2πθ θ = −

0θ =

경계 조건

1) Faltinsen, O.M. , Sea loads on ships and offshore structures, Cambridge Univ. Press, 1998, Ch3 (pp 50~55)

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (analytic solution)- Boundary Condition

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nVn

=∂Φ∂ 3

θ r

RtiAe ωξξ 33 =

y

z

n∂Φ∂

w θ

LHS = RHS

[ ][ ]32 ,

cos,sinnn=−= θθn

n2n

3n

33 3 3A i t A i te i e n

nω ωφξ ξ ω∂=

∂3

3 cosi n inφ

ω θ ω∂

∴ = = −∂

3 3 3A i te ωξ φΦ = 3 3

3A i te

n nωφ

ξ∂Φ ∂

=∂ ∂

LHS :

RHS : normal방향 속도 성분

3A i ti e ωξ ω=

면에 Normal한 속도 성분

( ) ( ) ( )( )2 3

3 3

0, , , 0,n

A i t

V w n n w

i e nωξ ω

= = ⋅ =

=

w n w

3wtξ∂

=∂

θ

( )( )

1 2 3

1 2 3

1 2 3

, , , ( , , )

, , ( , , )n

u v w n n n

V u v w n n nu n v n wn

= =

= = ⋅

= + +

V n

V n

( , ) ( co s , sin ),y z r rθ θ=

≤≤−=

22, πθπRr

0θ =

*2πθ θ = −

물체 표면 경계 조건

- 물체 표면의 normal 속도는 그 점에서 유체입자의 표면에 normal한 속도와 동일함(Body Boundary Condition)

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (analytic solution)- Boundary Condition

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23 3

2 2

1 1 0rr r r r

φ φθ

∂ ∂∂ + = ∂ ∂ ∂

01123

2

223

23 =

∂∂

+∂∂

+∂∂

θφφφ

rrrr

023

23

23

22 =

∂∂

+∂∂

+∂∂

θφφφ

rrr

03332 =++ θθφφφ rrr rr

)()(),(3 θθφ GrFr = 라 하면,

02 =++ θθFGGrFGFr rrr

23 3 3 0rr rr r θθφ φ φ+ + =

Governing Equation

1,2) Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, 1) Ch. 2.5, p69-73, 2) Ch.2.4, p61~64

22

kG

GF

rFFr rrr =−=+ θθ

,022 =−+ FkrFFr rrr 02 =+ GkGθθ

(Euler-Cauchy Equation1)), (Undamped System2))

Step1. Separation of Variable (assumption)

( , ) ( ) ( )r F r Gφ θ θ=Step2. Separate into two O.D.E’s ( ) ( ) 0rrF r kF r− = 2, ( ) ( ) 0G k Gθθ θ θ− =

Step2.1. Solve (Satisfying Boundary Conditions)( )G θ

Step2.2. Solve (with same Eigenvalues)( )F r

EigenvaluesEigenfunctions

Step3. Find Coefficients (Applying Initial Conditions)

Step2.3. Assemble ( , ) ( ) ( )r F r Gφ θ θ=

θ r

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

03 =φ 03 =φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

03 =φ

−=∂∂ ωθφ i

rcos30θ =

*2πθ θ = −

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (analytic solution)- Finding analytic solution of Velocity Potential

,(Laplace Equation in Polar Coordinate)

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1,2) Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005,

1) Ch. 2.5, p69-73, 2) Ch.2.4, p61~64

02 =+ GkGθθ

three cases according to k

1) k2 = 0

0Gθθ =

( ) 1 2G c cθ θ= +

( ) ( )3 ( , )r F r Qφ θ θ= ⋅

( ) ( )3 ( , / 2 ) / 2 0r F r Qφ π π= ⋅ =

( ) ( )3 ( , / 2 ) / 2 0r F r Qφ π π− = ⋅ − =

Boundary Condition

1 2 02 2

G c cπ π = + =

1 2 02 2

G c cπ π − = − + =

1 2 0c c= =

( ) 0G θ∴ =

<Trivial Solution>

2) k2 = -p2 < 0

2 0G p Gθθ − =

( ) 3 4px pxG c e c eθ −= +

Boundary Condition

2 23 4 0

2p p

G c e c eπ ππ − = + =

2 23 4 0

2p p

G c e c eπ ππ − − = + =

if c3+c4=0, then c3=-c4

2 23 ( ) 0

p pc e e

π π−

− =

2 2if ( ) 0p p

e eπ π

−− =

2 2 0p p

e eπ π

−− =

1pe π =0,p = 0Gθθ =

3 0,c =4 0c =

( ) 0G θ =

( ) 2 23 4 ( ) 0

p pc c e e

π π−

+ − =2 2( 0)

p pe e

π π−

− ≠

<Trivial Solution><Trivial Solution>

(Undamped System2))








θ r

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

03 =φ 03 =φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

03 =φ

−=∂∂ ωθφ i

rcos30θ =

*2πθ θ = −


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02 =+ GkGθθ

three cases according to k

3) k2 = p2 > 0

( ) ( )3 ( , )r F r Qφ θ θ= ⋅

( ) ( )3 ( , / 2 ) / 2 0r F r Qφ π π= ⋅ =

( ) ( )3 ( , / 2 ) / 2 0r F r Qφ π π− = ⋅ − =

( ) cos sinG C kx D kxθ = +

( ) cos sin2 2 2

G C k D kπ π π= +

Boundary Condition

( ) cos sin2 2 2

G C k D kπ π π− = −

( )0 and cos 0, k 1,3,5...2n

nk

D Cπ

= = =

( )0 and sin 0, k 2,4,6...2n

nk

C Dπ

= = =

Case ①

Case ②

(Undamped System2))

,022 =−+ FkrFFr rrr

(Euler-Cauchy Equation1))

( ) mF r r= 이 해가 되려면

( )2 0am b a m c+ − + =2 2 0m k− =

m k= ±

( ) k kF r Ar Br−= +








θ r

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

03 =φ 03 =φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

03 =φ

−=∂∂ ωθφ i

rcos30θ =

*2πθ θ = −


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,022 =−+ FkrFFr rrr 02 =+ GkGθθ

(Euler-Cauchy Equation1))

( ) n nk kn n nF r A r B r−= + , ( ) cos sinn n nG C k D kθ θ θ= +

(Undamped System2))

( ) ( )3 ( , ) cos sinn nk kn nr Ar Br C k D kφ θ θ θ−= + ⋅ +

θ r

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

03 =φ 03 =φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

03 =φ

−=∂∂ ωθφ i

rcos30θ =

*2πθ θ = −


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r→∞ 일때 이 되려면, A=0이어야 한다.0φ =

( ) ( )3 cos sinnkn nBr C k D kφ θ θ−∴ = ⋅ +


( ) n nk kn n nF r A r B r−= +

( ) cos sinn n n n nG C k D kθ θ θ= +

물체 표면의 경계 조건을 대입하면,

3 cosirφ ω θ∂

= −∂

( ), r R=

( ) ( )13 cos sin cosnkn n n

r R

Bk R C k D k irφ

θ θ ω θ− −

=

∂= − ⋅ + = −

∂








θ r

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

03 =φ 03 =φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

03 =φ

−=∂∂ ωθφ i

rcos30θ =

*2πθ θ = −


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0 and C = sin 0,2 nD kπ =

Case 1:

2,4nk =

( ) sinn n nG D kθ θ=

1sin ,nk

n nn

d r kφ θ∞

−

=

= ∑

1

1( ) sin cosnk

n n nnr R

d k R k irφ θ ω θ

∞− −

==

∂= − = −

∂ ∑만족 시킬 수 없음!!




Let us try to find such a combination which does satisfy the initial/boundary condition. In order to allow all possible n’s we write an infinite series.









( ), 2, 4n n n nd D B k= ⋅ =

θ r

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

03 =φ 03 =φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

03 =φ

−=∂∂ ωθφ i

rcos30θ =

*2πθ θ = −


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Case 2:

1,3nk =

( )( ) cosn n nG C kθ θ=

1cos ,nk

n nn

c r kφ θ∞

−

=

= ∑

1

1( ) cos cosnk

n n nnr R

c k R k irφ θ ω θ

∞− −

==

∂= − = −

∂ ∑

0 andD = cos 0,2 nC kπ =

Let us try to find such a combination which does satisfy the initial/boundary condition. In order to allow all possible n’s we write an infinite series.









( ), 1,3n n n nc C B k= ⋅ =


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1cos ,nk

n nn

c r kφ θ∞

−

=

=∑1

1( ) cos cosnk

n n nnr R

c k R k irφ θ ω θ

∞− −

==

∂= − = −

∂ ∑1 21 1 1

1 1 1 2 2 21

( ) cos ( ) cos ( ) cos

cos

nk k kn n n

nc k R k c k R k c k R k

i

θ θ θ

ω θ

∞− − − − − −

=

− = − + − +

= −

∑

21 cos cosc R iθ ω θ−− = − 2

1c i Rω=

( )1 1k =

2

( , ) cosRr ir

ωφ θ θ∴ =

Case 2: ( ) ( )3 ( , ) cos sinn nk kn nr Ar Br C k D kφ θ θ θ−= + ⋅ +



1( , ) cosnk

n nn

r c r kφ θ θ∞

−

=

= ∑ , (k=1일때만 만족함)








θ r

RtiAe ωξξ 33 =

y

z

r∂∂ 3φ

w θ

03 =φ 03 =φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

03 =φ

−=∂∂ ωθφ i

rcos30θ =

*2πθ θ = −


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V1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for Windows”, Delft University of Technology, 2003, pp30~332) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 295~3003) Faltinsen, O.M. , Sea loads on ships and offshore structures, Cambridge Univ. Press, 1998, Ch3 (pp 50~55)


ex) 원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동에의한 Velocity potential이 주어져 있다고 했을 때, Heave 방향Added mass 및 Damping Coefficient를 구하시오

θ r

R3ξ

y

z

r∂∂φ

w θ

0=φ 0=φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

0=φ

ωθξφ A

r 3cos−=∂∂

tiAti eir

Rertr ωω θωξθφθ ⋅==Φ cos),(),,(2

333

- 선박의 운동 변위 :

- Radiation Wave Velocity Potential :

sol) )(,3333 Rrondl

nf

xc=

∂∂

−= ∫φφρ

θωθφ cos),(2

3 ir

Rr =

θωφφ cos2

233 i

rR

rn−=

∂∂

=∂∂

−×

=

∂∂ θωθωφφ coscos 2

223

3 irRi

rR

n

θω 2223

4

cosirR

−= θω 223

4

cosrR

=

∫ ⋅−=xc

dlRf θωρ 2233 cos

θω 22 cosR= )( Rron =

tρgzP T

∂Φ∂

−−= ρ

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (analytic solution)- Finding analytic solution of Added Mass

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1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for Windows”, Delft University of Technology, 2003, pp30~332) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 295~3003) Faltinsen, O.M. , Sea loads on ships and offshore structures, Cambridge Univ. Press, 1998, Ch3 (pp 50~55)

tiAet ωξξ 33 )( =θ r

R3ξ

y

z

r∂∂φ

w θ

0=φ 0=φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

0=φ

ωθξφ A

r 3cos−=∂∂

tiAti eir


333



sol) ∫ ∂∂

−=xc

dln

f 3333φφρ

∫−=xc

dlR θωρ 22 cos

∫− ⋅−=2/

2/

2233 cos

π

πθθωρ RdRf

θRddl =

∫−−=2/

2/

222 cosπ

πθθωρ dR

∫−+

−=2/

2/

22

22cos1π

πθθωρ dR

2/

2/

22

42sin

2

π

π

θθωρ−

+−= R

222 πωρR−=

−= ρπω

2

22 R

33a (반원 단면의 질량과 동일함)



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tiAti eir


333



θ r

R3ξ

y

z

r∂∂φ

w θ

0=φ 0=φ

=

2πθ

−=

2πθ

( )Rr =

( )∞→r

0=φ

ωθξφ A

r 3cos−=∂∂

−= ρπω

2

22

33Rf

−== ρπωξξ ωω

2

22

333333RefeF tiAtiA

1) Journee, J.M.J. , Adegeest, L.J.M. ,Theoretical Manual of Strip Theory program“ Seaway for Windows”, Delft University of Technology, 2003, pp30~332) Newman, J.N. , Marine Hydrodynamics, The MIT Press, Cambridge, 1997, pp 295~3003) Faltinsen, O.M. , Sea loads on ships and offshore structures, Cambridge Univ. Press, 1998, Ch3 (pp 50~55)

sol)

tiA

tiA

tiA

et

eit

et

ω

ω

ω

ωξξ

ωξξ

ξξ

233

33

33

)(

)(

)(

−=

=

=

변위 :

속도 :

가속도 :

3ξ=

333aξ=

33a=



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Nav

al A

rchi

tect

ure

& O

cean

Eng

inee

ring


Seoul NationalUniv.




< Approximate Solution >By Singularity distribution method

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123/2031) Young, F.D. , Munson, B.R., Okiishi, T.H, Fluid Dynamics, Wiley, 2004, Ch 6.5.2

Source & Sink1)

y-z 평면에 normal하고 원점을 지나는 선에서반경방향으로 흐르는 유동을 고려해 보자.

constantφ =constantψ =

: streamlineψ

y

z

rv

θrm을 그 선(단위 길이 당)에서 흘러나오는 체적유량이라 하면,

질량보존법칙으로부터

( )2 rr v mπ =

2rmv

rπ=또는,

이 유동은 반경 방향만의 유동(vθ=0)이므로, 이에 대한 속도 포텐셜은

1, 02m

r r rφ φ

π θ∂ ∂

= =∂ ∂

을 적분하여 얻을 수 있다.

ln2m rφπ

=if m>0, 유동은 반경방향 → Source

if m<0, 유동은 원점방향 → Sink

m : Strength of Source (or Sink)r : Source로부터의 거리

- 원점에서 떨어진 지점에서의 몇몇 실제 유동은 Source & Sink를 이용하여 근사화 가능- 실제 유동에 존재하지 않는 가상적인 Velocity potential , 수학적인 특이점(singularity)- 이 가상적인 유동에 대한 Velocity potential(Source or Sink)는 실제의 유동을근사적으로 표현하는 다른 기본 velocity potential들의 중첩으로 구해짐

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (approximate solution)- Source & Sink

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Source:net outward flow

Sink:net inward flow

0))((div >PF 0))((div <PF

PP

0=⋅∇ F

0≠⋅∇ F

:incompressible flow

:compressible flow

Generate a body shape by using

Source and Sink


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PPGenerate a body-like

shape by using Source and Sink

Uniform Flow

Source

Half Body

- Uniform Flow + Source

Stagnation Point

Dividing Streamline


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Uniform Flow

Source

Rankine Ovoid

- Uniform Flow + Source + Sink

Stagnation Point

Sink

Dividing Streamline


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Singularity Distribution Method2) (2-D)

특이점 (source, doublet, vortex)을 물체 경계면

을 생성하도록 분포시키고, 그 특이점의 강도

(Strength)를 결정한다. 그리고 특이점의 강도를 통

하여 전체 유장의 velocity potential을 구하는 방법

Velocity Potential Pressure Surface Force to Hull

C21 2 =+Φ∇++

∂Φ∂ zgPt

ρρρ

Laplace Equation을 만족함 02 =Φ∇

∫∫BS

dSPn=FluidF


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정수중 선박의 강제운동에 의해 발생한 파

Radiation Wave

Given 02 =∇ jφ2 0j

j gzφ

ω φ∂

− + =∂

)6,,1( =j

±∞→∝ yase ikyj ,φ j

j nin

ωφ

=∂

∂ )( BSon- Governing Equation :

- Boundary Condition :

Find : jφ )6,,1( =j

- Motion of the ship : tiAjj e ωξξ = )6,,1( =j

How to solve ?

Lewis Conformal Mapping1) (2-D)

3311:

ζζaaw ++=Mapping

Function


LineSource

각 Line마다 Source 분포.경계 조건을 만족하도록

Source Strength 구함.

y

z

0j

z hzφ

=−

∂=

∂

1) Journee, J.M.J. , Massie, W.W. , Offshore Hydrodynamics, Delft Univ. of Technology, 2001, Ch 7-22~232) Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, Ch 17,Ch183) 이승건, 선박운동 조종론, 부산대학교 출판부, 2004, pp93~101, pp91~93

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (approximate solution)- 2 methods for finding solution

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Lewis Conformal Mapping1) (2-D)

1) Journee, J.M.J. , Massie, W.W. , Offshore Hydrodynamics, Delft Univ. of Technology, 2001, Ch 7-22~232) Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, Ch 17,Ch183) 이승건, 선박운동 조종론, 부산대학교 출판부, 2004, pp93~101, pp91~93

3311:

ζζaaw ++=Mapping

Function

planew− planez −

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (approximate solution)- Lewis Conformal Mapping for finding solution

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1) Journee, J.M.J. , Massie, W.W. , Offshore Hydrodynamics, Delft Univ. of Technology, 2001, Ch 7-30~362) Faltinsen, O.M. , Sea loads on ships and offshore structures, Cambridge Univ. Press, 1998, Ch3 (pp 104~108)3) 이승건, 선박운동 조종론, 부산대학교 출판부, 2004, pp93~101


특이점 (source, doublet, vortex)을 물체 경계면

을 생성하도록 분포시키고, 그 특이점의

강도(Strength)를 결정한다. 그리고 특이점의

강도를 통하여 전체 유장의 velocity potential을

구하는 방법

Laplace Equation을 만족함011

2

2

2 =∂∂

+

∂∂

∂∂

θφφ

rrr

rr

,1rr

=∂∂φ

rln=φ

,1=∂∂

rr φ ,0=

∂∂

∂∂

rr

rφ

02

2

=∂∂θφ

※ Laplace equation on polar coordinate

Let 2-D source

0= 0=

Given : N개의 특이점분포

(source, doublet, vortex)

Find : 특이점의 강도(Strength)

\∑=

=N

mmmq

1φφ

GivenFind

N개의 미지수가 존재함

N개의 경계조건으로 부터 방정식 구함

nVnφ∂=

∂

( ): distance from sourcer

(특이점 하나로는 식을 만족 못하기 때문에, basis function의 Linear combination으로 해를 가정한다.)

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (approximate solution)- Singularity Distribution Method for finding solution

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R y

z

ex) 반원이 로 운동 중일 때,Velocity potential을 구하시오.

tiet ωξ ⋅=1)(3

tiet ωξ ⋅=1)(3

1S

2S

3S4S

5S

6S

7S

가 구해짐3φ

※물체의 경계면을 따라 source를 분포 시킨다.

( ):, mm zy 1+mmSS 의 중점

1개의 Source 만으로는 다양한 물체의형상을 표현하기 어려움. =>여러 개의 소스

를 분포 시킨다.

Governing Equation인 Laplace Equation은 Linear Equation이므로, 각 해를 더한것(Superposition)도 Laplace Equation의해가 된다.

※ 2차원 source에 의한 Velocity potential :

rq ln , r( : Distance from source)


개념을 이해하기 위해 다음의 간단한 예제를살펴 보자.

지금까지는 해석적인 방법을 이용하여 velocity potential을 구하였다. 다음의 형상에 대하여 Singularity distribution method를 이용하여 Velocity potential 을 계산해보자.

Singularity distribution method

∑=

=N

mmmq

1φφ

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (approximate solution)- Singularity Distribution Method for finding solution

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R y

z

ex) 다음과 같은 단면이 로 운동 중일때, Velocity potential을 구하시오.

tiet ωξ ⋅=1)(3


tiet ωξ ⋅=1)(31S

2S

3S

가 구해짐3φ

Step 1. 선체에 Source 3개(S1,S2,S3)를 분포

② 임의의 점에서의 Velocity potential은 3개의source에 의한 Velocity potential의 합과 같다.

1 1( , )η ζ

2 2( , )η ζ

3 3( , )η ζ

( , )y z2 2

3 1 1 1

2 22 2 2

2 23 3 3

ln ( ) ( )

ln ( ) ( )

ln ( ) ( )

q y z

q y z

q y z

φ η ζ

η ζ

η ζ

= − + −

+ − + −

+ − + −

1 1lnq r ( ) ( )2 21 1 1lnq y zη ζ= − + −

① 으로 부터 r1만큼 떨어진위치 (y,z)에서속도 포텐셜( )1 1,η ζ

3개의 미지수 3개의 식 필요함



원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 (approximate solution)- Example 1. Finding solution by Singularity Distribution Method

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R y

z


tiet ωξ ⋅=1)(3


2S

3S

가 구해짐3φ

1 1( , )η ζ

2 2( , )η ζ

3 3( , )η ζ

3 3( , )y z

2 2( , )y z

1 1( , )y z

2 23 1 1 1 1 1 1 1

2 22 1 2 1 2

2 23 1 3 1 3

( , ) ln ( ) ( )

ln ( ) ( )

ln ( ) ( )

y z q y z

q y z

q y z

φ η ζ

η ζ

η ζ

= − + −

+ − + −

+ − + −

2 23 2 2 1 2 1 2 1

2 22 2 2 2 2

2 23 2 3 2 3

( , ) ln ( ) ( )

ln ( ) ( )

ln ( ) ( )

y z q y z

q y z

q y z

φ η ζ

η ζ

η ζ

= − + −

+ − + −

+ − + −

Step 2. 물체 경계 조건(Body boundary condition)

물체 표면의 점의 속도와 유체의 속도가 동일해야 함

3개의 점에 대해서 조건을 적용하면,3개의 방정식이 구해짐

2 23 3 3 1 3 1 3 1

2 22 3 2 3 2

2 23 3 3 3 3

( , ) ln ( ) ( )

ln ( ) ( )

ln ( ) ( )

y z q y z

q y z

q y z

φ η ζ

η ζ

η ζ

= − + −

+ − + −

+ − + −

① 먼저 물체 표면의 3개 점에 대해 Velocity potential을 구해 보자.





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R y

z


tiet ωξ ⋅=1)(3


2S

3S

가 구해짐3φ

1 1( , )η ζ

2 2( , )η ζ

3 3( , )η ζ

3 3( , )y z

2 2( , )y z

1 1( , )y z ( )( )( )

2 21 1 1 1 1

2 22 1 2 1 2

2 23 1 3 1 3

ln ( ) ( )

ln ( ) ( )

ln ( ) ( )

q y zn

q y zn

q y zn

η ζ

η ζ

η ζ

∂= − + −

∂∂

+ − + −∂∂

+ − + −∂




식3개 미지수 3개 풀 수 있음!

1 1

3

( , )y znφ∂∂

1 1

3

( , )y znφ∂∂

33i n

nφ ω∂

=∂

② Body boundary condition :

1 13 ( , )at y zi nω= ⋅ ⋅

( )2 21 2 1 2 1ln ( ) ( )q y z

nη ζ∂

= − + − +∂

2 2

3

( , )y znφ∂∂

2 23 ( , )at y zi nω= ⋅ ⋅

( )2 21 3 1 3 1ln ( ) ( )q y z

nη ζ∂

= − + − +∂

3 3

3

( , )y znφ∂∂

3 33 ( , )at y zi nω= ⋅ ⋅

given :

Find :

, ; , ( 1, 2,3)m m m my z mη ζ =

1 2 3, ,q q q

적용

③




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1 11 2 12 3 13 1q I q I q I b+ + =

( ) ( ) ( ) ( )1 1

1 1 1 1

2 2 2 21 1 1 1 1 3 1 3 1 3 3 ( , )

( , ) ( , )ln ln at y z

y z y zq y z q y z i n

n nη ζ η ζ ω∂ ∂ − + − + + − + − = − ⋅ ⋅ ∂ ∂

11I 13I

1 21 2 22 3 23 2q I q I q I b+ + =

1 31 2 32 3 33 3q I q I q I b+ + =

Step 3. 방정식을 Matrix 형태로 나타내면,

=Aq b11 12 13

21 22 23

31 32 33

,I I II I II I I

=

A1

2

3

,qqq

=

q1

2

3

bbb

=

b

bAq 1−=

R y

z


2S

3S

1 1( , )η ζ

2 2( , )η ζ

3 3( , )η ζ

3 3( , )y z

2 2( , )y z

1 1( , )y z



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R y

z


tiet ωξ ⋅=1)(3


2S

3S

가 구해짐3φ

1 1( , )η ζ

2 2( , )η ζ

3 3( , )η ζ

3 3( , )y z

2 2( , )y z

1 1( , )y z

1 1

3

( , )y znφ∂∂




3개의 점에 대해서만 조건을 만족물체 표면의 다른 점들은 알 수 없음

더 많은 점에서 경계조건을 적용하면 더 정확한해를 얻을 수 있다.

식3개 미지수 3개인 문제로 풀 수 있음





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R y

z


tiet ωξ ⋅=1)(3

tiet ωξ ⋅=1)(3

1S

2S

3S4S

5S

6S

7S

가 구해짐3φ

Step 1. 경계면을 6개의 Line으로 근사화 한다.

Step 2. 등분된 점과 점 사이를 선으로 연결( ):, mm zy 1+mmSS 의 중점

1개의 Source 만으로는 다양한 물체의형상을 표현하기 어려움.

Governing Equation인 Laplace Equation은 Linear Equation이므로, 각 해를 더한것(Superposition)도 Laplace Equation의해가 된다.

따라서, 더 많은 Line으로 근사화 하면, 더 정밀한 계산결과를 얻을 수 있음.(현재는 6개 line으로 근사화)





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② 미소구간에서의

세기:

( )11, zy

1S

2S

R y

z


tiet ωξ ⋅=1)(3


2S

3S4S

5S

6S

7S

가 구해짐3φ


Step 3. 각 Line segment에 물체 경계면을생성하도록 Line source를 분포시킨다. 한 Line에 분포된 source는 같은 강도(strength)를 가진다고 가정한다

1 1lnq s r∆

( ),y z

( ) ( )2 21 1 1lnq s y zη ζ= ∆ − + −

( )1 1,η ζ

1q① 단위길이당 세기:

1r1q s∆

③ 으로 부터 r1만큼 떨어진위치 (y,z)에서속도 포텐셜

④ Line 을 n등분하였을 경우

s∆1q s∆

( )1 1,η ζ

( ) ( )

( ) ( )

( ) ( )

2 21 1 1 1

2 21 2 2

2 21

ln

ln

ln n n

q s y z

q s y z

q s y z

φ η ζ

η ζ

η ζ

∆ = ∆ − + −

+ ∆ − + −

+ + ∆ − + −

( ) ( )2 21 ln

n

n nq s y zη ζ= ∆ − + −∑

( ) ( )1 2

2 21 1 ln ( ) ( )

S Sq y s z s dsφ η ζ∆ = − + −∫

⑤ Line 을 무한히 등분한 것으로 가정하고,Integral 형태로 나타냄

임의의위치





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Rtiet ωξ ⋅=1)(3 y

z

1S

2S

3S4S

5S

6S

7S

가 구해짐3φ


tiet ωξ ⋅=1)(3


Step 4. 나머지 5개의 line segment에도Line source를 분포시킨다.

( ) ( )1 2

2 21 1 ln ( ) ( )

S Sq y s z s dsφ η ζ∆ = − + −∫1 2 :S S

( ) ( )2 3

2 22 2 ln ( ) ( )

S Sq y s z s dsφ η ζ∆ = − + −∫2 3 :S S

( ) ( )

6 7

2 26 6 ln ( ) ( )

S Sq y s z s dsφ η ζ∆ = − + −∫6 7 :S S

( ) ( )1

6

31

62 2

1

( , )

ln ( ) ( )m m

mm

m S Sm

y z

q y s z s ds

φ φ

η ζ+

=

=

= ∆

= − + −

∑

∑ ∫

q1~q6까지 6개의 미지수

6개의 미지수 6개의 식 필요함

Velocity potential은 6개 Source에 의한포텐셜의 합과 같다.





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Line segment들의 중점에 대해, Body boundary condition을 적용

6개의 line segment가 존재하므로, 6개의 방정식이 구해짐

,(Body boundary condition)

점 에 대해서,velocity potential을 구해 보면 다음과 같다.

하나의 line source에 의한 velocity potential

( ) ( )

( ) ( )

( ) ( )

1 2

2 3

6 7

2 23 1

2 22

2 26

( , ) ln ( ) ( )

ln ( ) ( )

ln ( ) ( )

m m m mS S

m mS S

m mS S

y z q y s z s ds

q y s z s ds

q y s z s ds

φ η ζ

η ζ

η ζ

= − + −

+ − + −

+

+ − + −

∫

∫

∫


z

( )11, zy

1S

2S

3S4S

5S

6S

7S

( )22 , zy

( )33, zy( )44 , zy

( )55 , zy

( )66 , zy

가 구해짐3φ


tiet ωξ ⋅=1)(3

1φ∆

n∂∆∂ )( 1φ

33 ni

nωφ

=∂∂

( , )m my z



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Line segment들의 중점에 대해, Body boundary condition을 적용

6개의 line segment가 존재하므로, 6개의 방정식이 구해짐

33 ni

nωφ

=∂∂

),(cos

mm zyi θω−=3niω

( ) ( )

( ) ( )

( ) ( )),(

226

),(

222

),(

221

),(

3

76

32

21

)()(ln

)()(ln

)()(ln

mm

mm

mmmm

zySS

zySS

zySSzy

dsszsyn

q

dsszsyn

q

dsszsyn

qn

−+−

∂∂

+

+

−+−

∂∂

+

−+−

∂∂

=∂∂

∫

∫

∫

ζη

ζη

ζηφ

LHS:

RHS :


z

( )11, zy

1S

2S

3S4S

5S

6S

7S

( )22 , zy

( )33, zy( )44 , zy

( )55 , zy

( )66 , zy

가 구해짐3φ


tiet ωξ ⋅=1)(3

n∂∆∂ )( 1φ

,(Body boundary condition)



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6개 Line segment에 대해 Body boundary condition을 적용해 보면다음과 같다.

( ) ( ) ( ) ( ) ),(),(

226

),(

221 11

1176

1121

cos)()(ln)()(ln zyzySSzySS

idsszsyn

qdsszsyn

q θωζηζη −=

−+−

∂∂

++

−+−

∂∂

∫∫

( ) ( ) ( ) ( ) ),(),(

226

),(

221 22

2276

2221


idsszsyn

qdsszsyn

q θωζηζη −=

−+−

∂∂

++

−+−

∂∂

∫∫

( ) ( ) ( ) ( ) ),(),(

226

),(

221 66

6676

6621


idsszsyn

qdsszsyn

q θωζηζη −=

−+−

∂∂

++

−+−

∂∂

∫∫

61 ,, qq 미지수 : 6개

방정식 : 6개

Now we can find the solution !!!


z

( )11, zy

1S

2S

3S4S

5S

6S

7S

( )22 , zy

( )33, zy( )44 , zy

( )55 , zy

( )66 , zy



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1166111 bIqIq =++


z

( )11, zy

1S

2S

3S4S

5S

6S

7S

( )22 , zy

( )33, zy( )44 , zy

( )55 , zy

( )66 , zy

( ) ( ) ( ) ( ) ),(),(

226

),(

221 11

1176

1121


idsszsyn

qdsszsyn

q θωζηζη −=

−+−

∂∂

++

−+−

∂∂

∫∫

11I 16I

2266211 bIqIq =++

6666611 bIqIq =++

Step 6. 방정식을 Matrix 형태로 나타내면,

bAq = ,

6661

1611

=

II

II

A ,

6

1

=

q

qq

=

6

1

b

bb

bAq 1−=2

( , ) cosRr ir

ωφ θ φ θ=



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z

( )11, zy

1S

2S

3S4S

5S

6S

7S

( )22 , zy

( )33, zy( )44 , zy

( )55 , zy

( )66 , zy

( )32 ,,),( nnzf

yff

nzyf

•

∂∂

∂∂

=•∇=∂

∂ n

( ) ( )),(

22

1

)()(lnjj

kk zySSjk dsszsyn

I

−+−

∂∂

= ∫+

ζη

( ) ( )

( ) ( )

−+−−

=∂∂

−+−−

=∂∂

22

22

)()()(

)()()(

szsysz

zf

szsysy

yf

ζηζ

ζηη

( ) ( ) ( ) ( ) dsszsy

szn

szsysy

nkk SS

jj

j

jj

j∫+

−+−

−+

−+−

−=

1223222 )()(

)()()(

)(ζη

ζζη

η

( ) ( )22 )()(ln),( szsyzyf ζη −+−=

(참고) 의 계산jkI

라 하면,



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Nav

al A

rchi

tect

ure

& O

cean

Eng

inee

ring


Seoul NationalUniv.




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+

+


교란정지상태


yz

yz

R

D

I







y

z



z


Radiation wave velocity potential ( )RΦy

z

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정수중 선박의 강제운동에 의해 발생한 파

Radiation wave

tiRR ezyxtzyx ωφ ),,(),,,( =Φ

6

1( , , )A i t

j jj

x y z e ωξ φ=

=∑ Radiation wave velocity potential

Boundary Condition1)


02 =∂

∂+−

zg j

j

φφω )0(on =z

③ Radiation Condition :‘Wave associated with the potentials must be radiating away from the body’1)

i tj e at yωφ ∝ → ±∞ , ( 1, ,6 )j =

jj ni

nω

φ=

∂

∂)( BSon

j

n

B

nVS : 침수표면

: 침수표면에 normal인 속도

: 표면의 normal vector component

④ Body boundary condition : 선박 표면의 normal velocity(Vn)와 그 점에서 유체 입

자의 normal Velocity( )가 동일함

nR V

n=

∂Φ∂

R

n∂Φ∂

단위 진폭에 대한유체 속도 성분(선체 표면에 normal성분)

단위 진폭에 대한선체 표면의 속도 성분(선체 표면에 normal성분)


y

z② Bottom B.C. (BBC)

0=∂Φ∂

−= hzz

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Radiation Velocity Potential- 원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동과 비교



02 =∂

∂+−

zg j

j

φφω )0(on =z

③ Radiation Condition : i t

j e at yωφ ∝ → ±∞ , ( 1, ,6 )j =

④ Body boundary condition

nR V

n=

∂Φ∂

jj ni

nω

φ=

∂

∂)( BSon



0=∂Φ∂

−= hzz

Radiation wave velocity potentialin Free surface1)

원형단면을 가진 무한히 긴 Cylinder의 무한 수심속 운동 문제2)

Coordinate: Polar coordinate원형 단면을 가진 경우는 Polar coordinate를 이용하

는 것이 편리하다.

( , ) ( co s , sin ),y z r rθ θ=

≤≤−=

22, πθπRr

Governing Equation : 23 0φ∇ =

03332 =++ θθφφφ rrr rr

Governing equation in polar coordinate

Boundary Condition


=−=

2,

2πθπθ3 0,φ =

( )3 0, rφ = → ∞

④ Body boundary condition

( )3 cos ,i r Rrφ

θ ω∂

= − =∂

Coordinate : Cartesian Coordinate

2) Faltinsen O M Sea loads on ships and offshore structures Cambridge Univ Press 1998 Ch3 (pp 50~55)

Motion of Ship : Motion of Ship

3 3( ) A i tt e ωξ ξ= 3 3( ) A i tt e ωξ ξ=

(자유수면에서 파가 발생하지 않음)

(물체로 부터 무한히 먼 곳에서 파가 발생하지 않음)

단순화

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따라서, 단순한 형태의 2차원 source 대신Free surface condition을 만족하는 Green function을 사용함

( )rq ln

ex) Green function introduced by Wehausen and Laitone(1960)

( ) ( )( )

*

0

( )

1( , , ) ln ln 2 cos2

sin

ik z

i z

eG z t z z PV dk tk

e t

ζ

ν ζ

ζ ζ ζ ωπ ν

ω

− −∞

− −

= − − − + ⋅ − −

∫

ηξζ iiyxz +=+= ,

)/( 2 gων =

complex notation :

Wave number :

Wehausen,J.V.,Laitone,E.V.(1960), Surface Waves,Handbuch der hysik,edited by S.Fluegge, Vol.9,Fluid Dynamics 3,Springer Verlag,Berlin,Germany,pp 446~778

y

z

Radiation Wave Velocity Potential- Introduce Green Function

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1) Frank,W., “OSCILLATION OF CYLINDERS IN OR BELOW THE FREE SURFACE OF DEEP FLUIDS”, NSRDC Report 2375, 19672) Journee, J.M.J. , Adegeest, L.J.M. , "Theoretical Manual of Strip Theory program“ Seaway for Windows”", Delft University of Technology, 2003

( ) ( )( )

* ( )

0

1( , , ) ln ln 2 cos sin2

ik zi zeG z t z z PV dk t e t

k

ζν ζζ ζ ζ ω ω

π ν

− −∞ − − = − − − + ⋅ − −

∫

Frank Close-Fit Method

: source를 대신 Boundary condition을 만족하는 아래 함수(Green function)을 사용한다.rq ln(단, 물체 표면 경계 조건은 만족하지 않음)

[ ] [ ]

( ) ( )( )

( )

0

( , ) Re Re

1 Re ln ln 2 Re2

ik zi z

G z A i B

ez z PV dk i ek

ζν ζ

ζ

ζ ζπ ν

− −∞ − −

= +

= − − − + ⋅ − −

∫

(시간항을 분리)

cos sinA t B tω ω= +

( ) ( )Re cos sinA iB t i tω ω = + × −

A B

( ) ( )Re cos sin sin cosA t B t i A t B tω ω ω ω = + + − +

( )Re i tA iB e ω− = + )/( 2 gων =

z x iyiζ ξ η

= += +source 위치 :

공간상의 위치 :

wave number :

Radiation Wave Velocity Potential- Approximate solution by Frank Close-Fit Method

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( ) ( )( )

( )

0

1( , ) Re ln ln 2 Re2

ik zi zeG z z z PV dk ie

k

ζν ζζ ζ ζ

π ν

− −∞ − −

= − − − + ⋅ − − ∫

)/( 2 gων =

z x iyiζ ξ η



wave number :

※ Green function*( , , ) ( , ) i tG z t G z e ωζ ζ −=

{ }( , , ) Re ( , ) i tx y t x y e ωφΦ =

( ) ( )2 2 2( , ) ( , ) 0i t i tx y e e x yω ωφ φ∇ Φ = ∇ = ∇ =

2 22

2 20, 0x yφ φφ

∂ ∂∴∇ = + = ∂ ∂

대신, 시간 항을 분리시킨 을 분포시킴*( , , )G z tζ ( , )G z ζ

Laplace equation에서 시간항을 분리한 변위만의 속도 포텐셜을 구하는 것임

y

4S

Rtiet ωξ ⋅=1)(3

x

( )1 1,x y

1S

2S

3S5S

6S

7S

( )2 2,x y

( )3 3,x y( )4 4,x y

( )5 5,x y

( )6 6,x y


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( ) ( )( )

0

( )

1( , ) Re ln ln 22

Re

ik z

i z

eG z z z PV dkk

i e

ζ

ν ζ

ζ ζ ζπ ν

− −∞

− −

= − − − + ⋅ −

−

∫

)/( 2 gων =

z x iyiζ ξ η



wave number :

1 21 1 ln

S Sq rdsφ∆ = ∫

※ 2차원 source :

rq ln

※ 하나의 Line source에의한 Velocity potential:

※ Green function

1 21 1 ( , )

S Sq G z dsφ ζ∆ = ∫

※ 하나의 Line source에 의한velocity potential :

※ N개의 Segment가 존재 하므로, 임의의 점에서의 Velocity potential은다음과 같다.

1

31

1

( , )

lnj j

N

jj

N

j S Sj

x y

q rds

φ φ

+

=

=

= ∆

=

∑

∑ ∫

※ N개의 Segment가 존재 하므로, 임의의 점에서의Velocity potential은 다음과같다.

1

31

1

( , )

( , )j j

N

jj

N

j S Sj

x y

q G z ds

φ φ

ζ+

=

=

= ∆

=

∑

∑ ∫N개의 미지수 N개의 미지수

y

4S

Rtiet ωξ ⋅=1)(3

x

( )1 1,x y

1S

2S

3S5S

6S

7S

( )2 2,x y

( )3 3,x y( )4 4,x y

( )5 5,x y

( )6 6,x y



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N개의 Line 중점에 대해서 body boundary condition 적용

11 1 1

31

1( , )

j j

N

j S Sjz x iy

q G z dsn nφ ζ

+== +

∂ ∂= ∂ ∂

∑ ∫

33 ni

nωφ

=∂∂

12 2 2

32

1( , )

j j

N

j S Sjz x iy

q G z dsn nφ ζ

+== +

∂ ∂= ∂ ∂

∑ ∫

1

3

1( , )

j jN N N

N

j NS Sjz x iy

q G z dsn nφ ζ

+== +

∂ ∂= ∂ ∂

∑ ∫

1 2 2 3 11 1 2 1 1( , ) ( , ) ( , )

N NNS S S S S S

q G z ds q G z ds q G z dsn n n

ζ ζ ζ+

∂ ∂ ∂= + + +

∂ ∂ ∂∫ ∫ ∫

1 11 2 12 1N Nq I q I q I= + + +

1 21 2 22 2N Nq I q I q I= + + +

1 1 2 2N N N NNq I q I q I= + + +

cos jiω θ= −

1

( , )j j

ij iS SI G z ds

nζ

+

∂=∂ ∫

j번째 segment에 분포된 source에 의한i번째 segment의 중간 지점에서의 속도

첫 번째 segment의 중간 지점에서의 속도

y

4S

Rtiet ωξ ⋅=1)(3

x

( )1 1,x y

1S

2S

3S5S

6S

7S

( )2 2,x y

( )3 3,x y( )4 4,x y

( )5 5,x y

( )6 6,x y

1, , Nq q미지수 : N개

방정식 : N개

Now we can find the solution !!!

두 번째 segment에 분포된 source에 의한첫 번째 segment의 중간 지점에서의 속도


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1 1 1

31 11 2 12 1N N

z x iy

q I q I q Inφ

= +

∂= + + +

∂

2 2 2

31 21 2 22 2N N

z x iy

q I q I q Inφ

= +

∂= + + +

∂

31 1 2 2

N N N

N N N NNz x iy

q I q I q Inφ

= +

∂= + + +

∂

1cosiω θ= −

2cosiω θ= −

cos Niω θ= −


물체 표면 경계 조건(body boundary condition) 적용

33 ni

nωφ

=∂∂ cos jiω θ= −

11 12 1 1 1

21 22 2 2 2

1 2

coscos

cos

N

N

N N NN N N

I I I q iI I I q i

I I I q i

ω θω θ

ω θ

− − = −

1

( , )j j

ij iS SI G z ds

nζ

+

∂=∂ ∫


y

4S

Rtiet ωξ ⋅=1)(3

x

( )1 1,x y

1S

2S

3S5S

6S

7S

( )2 2,x y

( )3 3,x y( )4 4,x y

( )5 5,x y

( )6 6,x y


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를 어떻게 구할 것인가?ijI

1

( , )j j

ij iS SI G z ds

nζ

+

∂=∂ ∫


( ) ( )( )

( )

0

1( , ) Re ln ln 2 Re2

ik zi zeG z z z PV dk i e

k

ζν ζζ ζ ζ

π ν

− −∞ − − = − − − + ⋅ − −

∫

※ Green function

( , )G z ζ 내부의 Principle value integral이 존재함①

ln( )z dsn

ζ∂−

∂ ∫ 의 적분②계산이 복잡하지만 구할 수 있음

{ }( ) 1( )

0

1

cos( )ln( )!

cos( ( )) sin( ( ))0 sin( )for:

2 0 !

n

i k z ny

n

n

r nrn nePV dk e x i x

k x r nix n n

ζν η

θγ

ν ξ ν ξν θ ξ θ

θ π ξ

∞

− ⋅ ⋅ −∞ =⋅ +

∞

=

⋅ ⋅+ + + ⋅ ⋅ = ⋅ ⋅ − − ⋅ ⋅ − ⋅ − − > ⋅ ⋅ ⋅ + − − < ⋅

∑∫

∑


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1

( , )j j

ij iS SI G z ds

nζ

+

∂=∂ ∫


( ) ( )( )

( )

0

1( , ) Re ln ln 2 Re2


k

ζν ζζ ζ ζ

π ν

− −∞ − − = − − − + ⋅ − −

∫

※ Green function


ln( )z dsn

ζ∂−


( )1

2 21

2 21 1 1

( ) Re ln( )

( ) ( )sin( ) ln cos ) arctan arctan

( ) ( )

ji

i

s z z

i j i j i j i ji j i j

i j i j i j i j

L n z ds

x y y yx y x x

ζ

ξ η η ηα α α α

ξ η ξ ξ

=

+

+ + +

= ⋅∇ ⋅ − ⋅ − + − − − = − ⋅ + ( + ⋅ − − + − − −

∫


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1

( , )j j

ij iS SI G z ds

nζ

+

∂=∂ ∫


( ) ( )( )

( )

0

1( , ) Re ln ln 2 Re2


k

ζν ζζ ζ ζ

π ν

− −∞ − − = − − − + ⋅ − −

∫

※ Green function


ln( )z dsn

ζ∂−


( )2

2 21

2 21 1 1

( ) Re ln( )

( ) ( )sin( ) ln cos ) arctan arctan

( ) ( )

ji

i

s z z

i j i j i j i ji j i j

i j i j i j i j

L n z ds

x y y yx y x x

ζ

ξ η η ηα α α α

ξ η ξ ξ

=

+

+ + +

= ⋅∇ ⋅ − ⋅ − + + + + = + ⋅ + ( + ⋅ − − + + − −

∫


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1

( , )j j

ij iS SI G z ds

nζ

+

∂=∂ ∫


( ) ( )( )

( )

0

1( , ) Re ln ln 2 Re2


k

ζν ζζ ζ ζ

π ν

− −∞ − − = − − − + ⋅ − −

∫

※ Green function


ln( )z dsn

ζ∂−


( )

1

1

( )

5 0

( )( )

0

( ) ( )

0

( ) Re

Re

cos( ( )) cos( (sin( )

ji

iji j

j

i j i j

i k z

i

s z z

i k zi

k y k yi j

i j

eL n ds PV dkk

d ei e d PV dkd k

e k x e kPV dk PV

k

ζ

ζζα α

ζ

η η

ν

ζζ ν

ξα α

ν

+

+

− ⋅ ⋅ −∞

=

− ⋅ ⋅ −∞⋅ +

⋅ + ⋅ +∞

= ⋅∇ ⋅ ⋅ ⋅ − = − ⋅ ⋅ ⋅ ⋅ −

⋅ ⋅ − ⋅ ⋅= + ⋅ + ⋅ −

−

∫ ∫

∫ ∫

∫

1

1

0

( ) ( )1

0 0

))

sin( ( )) sin( ( ))cos( )

i j i j

i j

k y k yi j i j

i j

xdk

k

e k x e k xPV dk PV dk

k k

η η

ξν

ξ ξα α

ν ν

+

∞ +

⋅ + ⋅ +∞ ∞ +

− ⋅ − ⋅ ⋅ − ⋅ ⋅ − − + ⋅ + ⋅ − ⋅ − −

∫

∫ ∫


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159/203



1

( , )j j

ij iS SI G z ds

nζ

+

∂=∂ ∫


( ) ( )( )

( )

0

1( , ) Re ln ln 2 Re2


k

ζν ζζ ζ ζ

π ν

− −∞ − − = − − − + ⋅ − −

∫

※ Green function


ln( )z dsn

ζ∂−


( )

{ }{ }

1

1

( )7

( ) ( )1

( ) ( )1

( ) Re

sin( ) cos( ( )) cos( ( ))

cos( ) sin( ( )) sin( ( ))

ji

i j i j

i j i j

i zi

s z z

y yi j i j i j

y yi j i j i j

L n e ds

e x e x

e x e x

ν ζ

ν η ν η

ν η ν η

α α ν ξ ν ξ

α α ν ξ ν ξ

+

+

− ⋅ ⋅ −

=

⋅ + ⋅ ++

⋅ + ⋅ ++

= ⋅∇ ⋅ ⋅

= − + ⋅ + ⋅ ⋅ − − ⋅ ⋅ −

+ + ⋅ + ⋅ ⋅ − − ⋅ ⋅ −

∫


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Diffraction Wave Velocity Potential

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+


교란정지상태



yz

yz

R

D

I







z


Radiation wave velocity potential ( )RΦy

z


+Diffraction wave velocity potential ( )DΦ

y

z

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Diffraction wave velocity potential



02 =∂∂

+−z

g DD

φφω )0(on =z

③ Radiation Condition : 파가 무한이 발산하면 진동함

i tD e as yωφ ∝ → ±∞

( ) 0=∂+∂n

DI φφ

n

B

VS : 침수표면

: 침수표면에 수직인 속도

④ Body boundary condition : 선박 표면에서 유체 입자의 속도가 Zero

)( BSon

0=nV


nnID

∂∂

−=∂∂ φφ )( BSon

Diffraction Wave

산란에 의한 파

Fixed

y

z


0D

z hzφ

=−

∂=

∂

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Given 02 =∇ Dφ

02 =∂∂

+−z

g DD

φφω

±∞→∝ yase ikyD ,φ )( BSon

- Governing Equation :

- Boundary Condition :

Find : Dφ

nnID

∂∂

−=∂∂ φφ

를 직접 구할 수도 있지만, Body Boundary Condition과

Green 2nd Theorem을 사용하여, 를 와 로 대체 가능

Dφ

Dφ Iφ Rφ

,D kφ φ are the solutions of Laplace equation.

Both satisfy 2 20 , 0D kφ φ∇ = ∇ =

∫∫∫∫ ∂∂

=∂∂

S

Dk

S

kD dA

ndA

nφφφφ

(radiation potential)

Diffraction Wave

산란에 의한 파

Fixed

y

z

Chap.3에서 계산과정 설명

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Chap 5. Load Curve, Shear Force,

Bending Moment

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Total shear force 및 Bending moment 구하기

4. Shear force curve

∫= qdxQs

5. Bending moment curve

∫= dxQM sx

Hydrostatics선박의 자세에따른 선박에 작용하는 정적인힘(모멘트) 계산

Class rule

Hydrodynamics파랑 상태 하에서 선박에 작용하는 동적인 힘(모멘트) 계산

Section Modulus 구하기

Bending Stress ≤ Allowable Bending StressShear Stress ≤ Allowable Shear Stress

종강도 부재수정

아니오

종강도 해석 종료

예

Still water shear forces, Qs,Still water bending moments, Ms,

Wave Shear force, Qw, Bending moment, Mw

3. Load curve )()()( xBxWxq +=

1. Weight curve )(xW

2. Buoyancy curve )(xB

Direct Calculation

Load Curve, Shear Force, Bending Moment- 왜 하는가? → 종강도 계산에 반영

“Vertical Wave Bending Moment에 반영됨”

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Ship Structural Design - Review of Mechanics of materials*

*Gere J.M., Mechanics of Materials, 6th edition, Thomson, 2006

What we have studied with Beam theory

L

x

y( )q x

( )q x

x

y

reactF

( )V x( )M x

( )y x

2

( )2 2

qLx qxM x = −

( )2

qLV x qx= −

qy

x

for example ,

3 2 3( ) ( 2 )24

qxy x L Lx xEI

= − − +

Differential equations of the defection curve

:young's modulus: mass moment of inertia

EI

4

4

( ) ( )d y xEI q xdx

= −


: ( ): ( )

: ( )


: ( )Load q xcause

( ) ( )dV x q xdx

= − ( ), ( )dM x V xdx

=

2

2


=




modulus)section:(, z

)y from inertia ofmoment:(, yI

iA∆

iy

σ

y

axisneutralfromdistance:iyaxisneutral:y

< section of Beam>

< stress on beam section>

, acty i

M MwhereI y Z

σ = =act allowσ σ≤


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z

x

L

x

y( )q x

global

local

a ship

stress acting on midship section should be less than allowable stress what kinds of load

cause ?midMq

lact σσ ≤.

Allowable stress by Rule:2

1, 175 [ / ]l f N mmσ =

., midact

mid

MσZ

=

*대한조선학회, 선박해양공학개론, 1993, 동명사, p138

*

x

z( )w x w(x):weight

x

z( )b x

x

z( )q x

q(x)= b(x) – w(x) : Load

+

=

b(x):bouyancy

anything else?

Hydrostatics

Hydrodynamics

L

x

y( )q x

( )q x

x

y

reactF

( )V x( )M x

( )y x


4

( ) ( )d y xEI q xdx

= −


: ( ): ( )

: ( )


: ( )Load q xcause

( ) ( )dV x q xdx

= − ( ), ( )dM x V xdx

=

2

2


=




, acty i

M MwhereI y Z

σ = =act allowσ σ≤


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(Review) 선박에 작용하는 힘

RDKFstatic FFFF +++ .+= GravityFSurfaceBody FFxM +=

( )xBxAF −−=R

동적 평형 상태

xBxAFFFFxM −−++++−= DKFstaticGravity ..0

균일 분포 하중 wz

xL길이

z

x

AR BR

수직방향 힘성분 (운동 방정식의 3행 성분)

Load Curve, Shear Force, Bending Moment- 선박에 작용하는 힘

Hydrodynamics


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선박의 6자유도 운동방정식에서 Heave-Pitch Motion유도

1) RAO(Response Amplitude Operator) : 1m 파고에 대한 선박의 운동 응답

[ ]T654321 ,,,,, ξξξξξξ=x

−−−

−−−

−−

=

zzzxCC

yyCC

xzxxCC

CC

CC

CC

IIMxMyIMxMz

IIMyMzMxMyM

MxMzMMyMzM

00000

00000

000000

M


=

666462

555351

464442

353331

262422

151311

000000

000000

000000

AAAAAA

AAAAAA

AAAAAA

A

=

666462

555351

464442

353331

262422

151311

000000

000000

000000

BBBBBB

BBBBBB

BBBBBB

B

=

0000000000000000000000000000000

5553

44

3533

CCC

CCC

heave-pitchmotion of equation :

=

6

5

4

3

2

1

FFFFFF

excitingF

( 으로 가정)0=cy

운동 방정식 :

=

+

+

++−+−+

5

3

5

3

5553

3533

5

3

5553

3533

5

3

5553

3533

FF

CCCC

BBBB

IAAMxAMxAM

xxC

C

ξξ

ξξ

ξξ

Sway

Heave

Roll

Surge Pitch

Yaw

( )[ ]tiAD

AKF eFF ωη 3,3,.03 +=F

Diffractionforce

Froude-Krylovforce


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(Review) 선박에 작용하는 힘

z

x

( ) ( ) 5353335353333,3,.03,3,530 ξξξξηξξ ω BBAAeFFFFxM tiAD

AKFGravityStaticC −−−−++++−−=

= 각 단면에 작용하는 힘을 배 길이 전체에 대해 적분한 값

ex) 보 :0

( )L

load A BF w x dx R R= + +∫ ex) 중량 : ∫−=2/

2/3, )(L

Lstatic gdxxmF

RDKFstatic FFFF +++ .+= GravityFSurfaceBody FFxM +=

( )xBxAF −−=R

동적 평형 상태

xBxAFFFFxM −−++++−= DKFstaticGravity ..0

수직방향 힘성분 (운동 방정식의 3행 성분)


분포 하중

L길이

AR BR

x

y

( )q x

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y

z1xx =

0

상하동요 운동 방정식에서 각 성분의 의미 (Review : strip theory & 선박 운동 방정식 유도 과정)

xy

z

1x

)( 133 xa단면의 상하 동요로 인한z방향 added mass :

331135 )( axxa −=단면의 상하 동요로 인한y축 방향 회전 added mass :

2x 3x

yx

z

535333 ξξ aa −− 5331333 ξξ axa +−=단면에 작용하는added mass force : ( )51333 ξξ xa −−=( )3335 xaa −=

1x 2x 3x

( ) ( ) 3,3,5353335353333,3,.0530 GravityStatictiA

DA

KFC FFBBAAeFFxM ++−−−−++−−= ξξξξηξξ ω

dxxxaA

dxxaAL

L

L

L

)(

)(2/

2/ 3335

2/

2/ 3333

∫∫

−

−

−=

=

x위치 단면의단위 길이당부가 질량

전체부가 질량

)( 51333 ξξ xa −−

)( 52333 ξξ xa −−)( 53333 ξξ xa −−

(Example of added mass force)

Load Curve, Shear Force, Bending Moment- 운동방정식에서 각 성분의 의미

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( ) ( ) 3,3,5353335353333,3,.0530 GravityStatictiA

DA

KFC FFBBAAeFFxM ++−−−−++−−= ξξξξηξξ ω

상하동요 운동 방정식에서 각 성분의 의미 (Review : strip theory & 선박 운동 방정식 유도 과정)

∫−=2/

2/)(

L

LdxxmM

x위치에서의단위 길이당

질량

전체 질량 dxxxaA

dxxaAL

L

L

L

)(

)(2/

2/ 3335

2/

2/ 3333

∫∫

−

−

−=

=

x위치 단면의단위 길이당부가 질량

전체부가 질량

dxxxbB

dxxbBL

L

L

L

∫∫

−

−

−=

=2/

2/ 3335

2/

2/ 3333

)(

)(

x위치 단면의단위 길이당감쇠 계수

전체감쇠 계수

( )∫− +=+2/

2/ 333,3,. )()(L

L

AD

AKF dxxhxfFF

Froude-Krylovforce

x위치 단면의단위 길이당

Froude-Krylovforce

x위치 단면의단위 길이당Diffraction

force

Diffractionforce

( )∫− −=+2/

2/3,3, )()(L

LGravityBuoyancy dxgxmxbFF

Totalbuoyancy

Totalweight

x위치 단면의단위 길이당

weight

x위치 단면의단위 길이당buoyancy

)(xb

gxm )(

( )53)( ξξ xxm −

tiDKF eff ωη )( 3,3,.0 +

)( 5333 ξξ xa −−

)( 5333 ξξ xb −−

Load Curve, Shear Force, Bending Moment- 운동방정식에서 각 성분의 의미

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하중(Load) ?

z

x

임의의 x위치(yz단면)에서 단위 길이에 작용하는 수직 방향 힘 성분

dx

( ) ( ) 5353335353333,3,.03,3,530 ξξξξηξξ ω BBAAeFFFFxM tiAD

AKFGravityStaticC −−−−++++−−=

( )gxmxb )()( −+

( )5333 )( ξξ xxb −−

( )5333 )( ξξ xxa −−

( ) tiexhxf ωη )()( 330 ++

( )53)( ξξ xxm −−=)(xq Mass inertia

Added mass force

Potential damping

Froude-Krylov + Diffraction

Hydrostatic force+ Structural weight

Heave,Pitch 변위, 가속도와 속도는 운동 방정식으로부터 계산됨

:)(xq( 단위 길이당 하중)

Load Curve, Shear Force, Bending Moment- 하중(Load) 계산

분포 하중

L길이

AR BR

wxq =)(

x

y

( )q x

dx

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( ) ( ) ( ) ( ) ( )gxmxbxbxaexhxfxxmxq ti )()()()()()( 5333533333053 −+−−−−++−−= ξξξξηξξ ω

속도, 가속도를 대입한 단위 길이당 수직 하중

선박의 heave 및 pitch 변위, 속도, 가속도 (파고 η0는 주어지는 값)

QRPSRFPF

QRPSQFSF

AAA

AAA

−−

=

−−

=

3505

5303

ηξ

ηξ

< Amplitude >

tiA

tiA

etet

ω

ω

ξξ

ξξ

55

33

)(

)(

=

=

< 변위 >

tiA

tiA

eit

eitω

ω

ωξξ

ωξξ

55

33

)(

)(

=

=

< 속도 >

tiA

tiA

et

etω

ω

ξωξ

ξωξ

52

5

32

3

)(

)(

−=

−=

< 가속도 >

대입 미분 미분

( ) ( ) ( ) ( ) ( )gxmxbexibexaexhxfexxm tiAAtiAAtitiAA )()()()()( 5333532

33330532 −+−−−+++−= ωωωω ξξωξξωηξξω

( ) ( ) ( ) ( )[ ] ( )gxmxbexibxaxhxfxxm tiAAAAAA )()()()()( 5333532

33330532 −+−−−+++−= ωξξωξξωηξξω

- Chap.4 운동방정식으로 부터 구한 Amplitude로 구함


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( ) ( ) ( ) ( )[ ] ( )gxmxbexibxaxhxfxxmxq tiAAAAAA )()()()()()( 5333532

33330532 −+−−−+++−= ωξξωξξωηξξω

단위 길이당 수직 하중

Wave에 의한 힘 및 선박의 운동과 관련(Wave load)

질량과 물에 잠긴 형상에 관련(Still water load)

gxm )(

)(xb

)(3 xf

)(3 xh( )5333 ξξ xa −−

( )53)( ξξ xxm −−( )5333 ξξ xb −−

Massinertia

Added massforce

Potentialdamping

DiffractionFroude-Krylov

Hydrostaticforce

Structuralweight

0) =tex(본래는 시간에따라 변함: )tie ω


Hydrodynamics


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33330532 −+−−−+++−= ωξξωξξωηξξω




Hydrostaticforce

Structuralweight

Massinertia

Froude-Krylov

DiffractionAdded mass forcePotential damping

..FS

..MB

z축 방향 힘의 평형 조건

0)(.. 1

0=−= ∫∑

x

Z dxxqFSF

0)().(. 1

01 == ∫x

dxxqxFSz

x

모멘트의 평형 조건(x=x1기준)

0)(.... 1

1 0=−= ∫∑ =

x

xx dxxFSMBM

∫=1

01 ).(.).(.x

dxxFSxMB

z

x

z

x(+)(-)

Massinertia

Added massforce

Potentialdamping

DiffractionFroude-Krylov

Hydrostaticforce

Structuralweight

Load Curve, Shear Force, Bending Moment- 하중(Load) 계산, 모멘트(Moment) 계산

Hydrodynamics


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)()( xqxq staticdynamic +=

∫∫∫ +== 111

......1 )()()().(.x

PA static

x

PA dynamic

x

PAdxxqdxxqdxxqxFS

∫ ∫∫ ∫∫

+

== 111

.. .... ....1 )()().(.).(.x

PA

x

PA static

x

PA

x

PA dynamic

x

PAdxdvvqdxdvvqdxxFSxMB

VWBM(Vertical Wave Bending Moment)

SWBM(Still Water Bending Moment)

적분

적분


33330532 −+−−−+++−= ωξξωξξωηξξω



(x대신 적분 변수를 v로 둠)

Load Curve, Shear Force, Bending Moment- 하중(Load) 계산, 모멘트(Moment) 계산

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Linked slide

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Green’s 2nd Theorem

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Proof) Green TheoremDivergence Theorem of Gauss1)

(Theorem 1) Divergence Theorem of Gauss(Transformation Between Triple and Surface Integrals)

Let T be a closed bounded region in space whose boundary is a piecewise smooth orientable surface S.

x

z

yR

n

n

n2S

1S

3S

γ

γ

Fig. 250. Example of a special region

),,( zyxF

∫∫∫ ∫∫ •=•∇T S

dAdV nFF

: a vector function that is continuous and has continuous first partial derivatives in T

(2)

],,,[ 321 FFF=F ]cos,cos,[cos γβα=nUsing component,

∫∫

∫∫∫∫∫++=

++=∂∂

+∂∂

+∂∂

S

ST

dxdyFdzdxFdydzF

dAFFFdxdydzzF

yF

xF

)(

)coscoscos()(

321

321321 γβα(2’)

zF

yF

xF

∂∂

+∂∂

+∂∂

=•∇ 321F1) Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, Ch10.7 (pp 458~462)

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Proof) Green Theorem

(Example 4) Let

gfgf

zgf

zg

zf

ygf

yg

yf

xgf

xg

xf

zgf

ygf

xgfgf

∇•∇+∇=

∂∂

+∂∂

∂∂

+

∂∂

+∂∂

∂∂

+

∂∂

+∂∂

∂∂

=

∂∂

∂∂

∂∂

•∇=∇•∇=•∇

2

2

2

2

2

2

2

,,)( F

gf ∇=F

ngfff∂∂

=∇•=∇•=•=• )g()g( nnFnnF

( )∫∫∫ ∫∫ ∂∂

=∇•∇+∇T S

dAngfdVgfgf 2

(1) Green’s first formula

1) Erwin Kreyszig, Advanced Engineering Mathematics, Wiley, 2005, Ch10.8 (pp 466~467)

∫∫∫ ∫∫ •=•∇T S

dAdV nFFDivergence Theorem :

LHS :

RHS :

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(Example 4) Let gf ∇=F

(2) Green’s second formula

( )∫∫∫ ∫∫

∂∂

−∂∂

=∇−∇T S

dAnfg

ngfdVfggf 22

( )∫∫∫ ∫∫ ∂∂

=∇•∇+∇T S

dAngfdVgfgf 2

Let fg∇=F

( )∫∫∫ ∫∫ ∂∂

=∇•∇+∇T S

dAnfgdVfgfg 2

①

②

① - ② :


( )∫∫∫ ∫∫ ∂∂

=∇•∇+∇T S

dAngfdVgfgf 2

(1) Green’s first formula

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(2) Green’s second formula

( )∫∫∫ ∫∫

∂∂

−∂∂

=∇−∇T S

dAnfg

ngfdVfggf 22

If gf , are the solutions of Laplace equation

Both satisfy 0,0 22 =∇=∇ gf

From Green’s 2nd formula, we can derive an equation (3)

0=

∂∂

−∂∂

∫∫S

dAnfg

ngf(3)

∫∫∫∫ ∂∂

=∂∂

SS

dAnfgdA

ngf(3’)

항을 분리하여, 두 번째 항을 우변으로 넘김


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[참고] 2변수 함수의 전미분

),( *2

*1 xxf

),( 21 xxf

),( 2*21

*1 xxxxf ∆+∆+

f∆df

)0,,( *2

*1 xx

22 dxx =∆

11 dxx =∆

1x

2x

11

dxxf

∂∂

22

dxxf

∂∂

1xf

∂∂

기울기=

f

2xf

∂∂

기울기=

22

11

dxxfdx

xfdf

∂∂

+∂∂

=

방향의 변화량1x

방향의 변화량2x

주어진 것: ),(),,( *2

*1

*2

*1 xxfxx

실제 구해야 하는 것:

fxxfxxxxf

∆+=

∆+∆+

),(

),(*2

*1

2*21

*1

근사적으로 구할 수 있는 것:

dfxxf +),( *2

*1

dff ≅∆21 , xx ∆∆ 가 아주 작다면

라 볼 수 있음

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[참고] 선박의 Heave 운동과 Spring-mass-damping system의 비교

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NonlinearityNonlinearity of the nature Nonlinear Mathematical Model

Linearization

Linear Mathematical Model Analytic Solution

Numerical Method

m =z F

Ex) Heave Motion of a Ship – step 1

Z

X

gravity= F

m : mass

gM

mg= − k

k Mass-Spring-Damper system

m ′′ =z F

①

mg

m

g

mg= k

By Newton’s 2nd law,

zk

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Nonlinearity of the nature Nonlinear Mathematical Model

Linearization


Numerical Method


Z

X

Mass-Spring-Damper system

M

gravity

mg= −

Fk

g

0

B

static staticS

P dS gVρ= =∫∫F n k

k

static+F

m : massV0 : submerged volumeSB : submerged surface area

ρ : density of sea water

0gVρ+ k

Archimedes’ Principle 0static gVρ=F k

: static equilibrium0 ( 0)z= =

m =z F

gravity= Fmg= − k

0V

0=: static equilibrium

0sm0

zm ′′ =z F

mg= k

②

0ks− k

0ks−

mgm

)0( =′′z

k

Nonlinearity

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Linearization


Numerical Method


Z

X


M

gravity

mg= −

Fk

g

B

static staticS

P dS= ∫∫F n

k

static+F

m : massV0 : submerged volumeSB : submerged surface areaAwp : waterplane area


0gVρ+ k


0 ( 0)z= =

m =z F

gravity= Fmg= − k

0V

z0s

m

0

z

kzks −− 0

mgm

③

,externalstatic

F

restoring force

0 addtional bouyancygVρ= +k F

z

,external staticF

,external static+F,external static+FwpgAρ− z

,external static+FwpgAρ= − z

,external staticF

additionalbuoyancy caused by additional displacement z

addtional bouyancy

WPgAkρ= −

= −

F

zz

, WPk gAρ=

if, z is small

,external static+Fk= − z

m ′′ =z F

0 ,external staticmg ks kz= − − +k k k F

,external statickz= − +k F0= )0( =′′z

k

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Linearization


Numerical Method


Z

X


M

gravity

mg= −

Fk

g

B

static staticS

P dS= ∫∫F n

k

static+F



0gVρ+ k


m =z F

gravity= Fmg= − k

0V z

,external staticF

,external static+F,external static+FwpgAρ− z

,external static+FwpgAρ= − z

,external staticF

, WPk gAρ=,external static+Fk= − z

z0s

m0

z

kzks −− 0

mgm

④

,externalstatic

F

m ′′ =z F

0mg ks kz= − −k k kkz= − k

,external static+F

,external static+F

0m k′′ + =z z Oscillation by the restoring force

restoring force

0 ( 0)z= =

0 addtional bouyancygVρ= +k Fadditionalbuoyancy caused by additional displacement z

addtional bouyancy

WPgAkρ= −

= −

F

zz

, WPk gAρ=

if, z is small

Linearized Restoring Force

k

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Linearization


Numerical Method


Z

X


M

gravity

mg= −

Fk

g

B

static staticS

P dS= ∫∫F n

k

static+F



0gVρ+ k


m =z F

gravity= Fmg= − k

0V z

wpgAρ− z

wpgAρ= − z

k= − z

z0s

m0

z

kzks −− 0

mgm

④

m ′′ =z F0mg ks kz= − −k k k

kz= − k

0m k′′ + =z z Oscillation by the restoring force

restoring force

Ship will oscillate forever?

0

0

WPgV gAgV k

ρ ρρ

= −= −

k zk z

Energy is dissipated by radiation wave

정수 중 선박의 강제운동에 의해 발생한 힘

Radiation Force

B

radiation radiationS

P dS= ∫∫F n

k

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Linearization


Numerical Method


Z

X


M

gravity

mg= −

Fk

g

B

static staticS

P dS= ∫∫F n

k

static+F



0gVρ+ k


m =z F

gravity= Fmg= − k

0V z

wpgAρ− z

wpgAρ= − z

k= − z

0

0

WPgV gAgV k

ρ ρρ

= −= −

k zk z


Radiation Force

B


P dS= ∫∫F n

c= − z

z0s

⑤

m

Dashpot

0

z

kzks −− 0

mgm

zc ′−

z

restoring force

z

radiation+Fc− z

radiation c= −F z

c− zc− z

opposite to velocity

c : damping coefficient

m ′′ =z Fcz′− k0ks kz− −k kmg= k

cz′− kkz− k=

k

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am− z


Linearization


Numerical Method


Z

X


M

gravity

mg= −

Fk

g

B

static staticS

P dS= ∫∫F n

k

static+F



0gVρ+ k


m =z F

gravity= Fmg= − k

0V z

wpgAρ− z

wpgAρ= − z

k= − z

0

0

WPgV gAgV k

ρ ρρ

= −= −

k zk z


Radiation Force

B


P dS= ∫∫F n

c= − zc : damping coefficient

z0s

⑤

m

Dashpot

0

z

kzks −− 0

mgm

zc ′−

z

restoring force

radiation+Fc− z

c− zc− z

opposite to velocity

opposite to acceleration

am− z

am− z

am− z

am− z

ma : added mass

z

radiation c= −F z

z


cz′− kkz− k=

k

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Linearization


Numerical Method


Z

X


M

gravity

mg= −

Fk

g

B

static staticS

P dS= ∫∫F n

k

static+F



0gVρ+ k


m =z F

gravity= Fmg= − k

0V z

wpgAρ− z

wpgAρ= − z

k= − z

0

0

WPgV gAgV k

ρ ρρ

= −= −

k zk z


radiation+Fc− z

c− zc− z

am− z

am− z

am− z

ma : added mass

+

Wave force

Froude-Kriloff Force Diffraction Force

B

wave exciting

wave excitingS

P dS= ∫∫F

n

( )excitingF=

excitingF

exciting+Fexciting+F

exciting+F

exciting+F

am− zradiation c= −F z

z z


cz′− kkz− k=cos tω+ 0F

cos tω+ 0F

k

Hydrodynamics


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Linearization


Numerical Method


Z

X


M

gravity

mg= −

Fk

g

B

static staticS

P dS= ∫∫F n

k

static+F



0gVρ+ k


m =z F

gravity= Fmg= − k

0V z

wpgAρ− z

wpgAρ= − z

k= − z

0

0

WPgV gAgV k

ρ ρρ

= −= −

k zk z


radiation+Fc− z

c− zc− z

am− z

am− z

am− z

ma : added mass

exciting+Fexciting+F

exciting+F

exciting+F

0 cosm c k tω′′ ′+ + =z z z F( )a excitingm m c k+ + + =z z z F

excitingFam− zradiation c= −F z

z z


cz′− kkz− k=cos tω+ 0F

cos tω+ 0F

k

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Laplacian in Polar Coordinates

Advanced Ship Design Automation Laboratory,Seoul National University, 20071)

Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, p.728~p.732, Johns and Bartlett, 2006

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Problems in Polar Coordinates

Laplacian in Polar Coordinates1)

Relation between polar coordinates and rectangular coordinates are given by

cos , sinx r y rθ θ= =),(),( θroryx

θx

r

y

x

y

polar coordinate(r,θ) rectangular coordinate (x,y)

2 2 2 , tan yr x yx

θ= + =

rectangular coordinate (x,y) polar coordinate(r,θ)

1) Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, p.728~p.732, Johns and Bartlett, 2006

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Laplacian in Polar Coordinates It is possible to convert the 2-D Laplacian of the

function u, into polar coordinate.

x r x xu u r uu u r ux r x x θ

θ θθ

∂ ∂ ∂ ∂ ∂= = + = +∂ ∂ ∂ ∂ ∂

2

2

xxu u r uu

x x x r x xu r u r u u

x r x r x x x x x xu r u r u r u r u

r r x r x x r x r x

θθ

θ θθ θ

θθ θ θ θ

∂ ∂ ∂ ∂ ∂ ∂ ∂ = = + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

2

2

2 2 2 2 2 2

2 2 2 2

( ) ( )rr x r x x r xx r x x x xx

ux x x

u r u r u r u r u ur x r x x r x r x x x x

u r u r u r u r u uθ θ θθ θ

θ θ θθ

θ θ θ θθ θ θ θ

θ θ θ θ

∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= + + + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + + + +

yyxx uuu +=∇2

Chain Rule

If is differentiable and and have

continuous first partial derivatives, then

Theorem 9.5

),( yxgu =),( vufz = ),( yxhv =

xv

vz

xu

uz

xz

∂∂

∂∂

+∂∂

∂∂

=∂∂

yv

vz

yu

uz

yz

∂∂

∂∂

+∂∂

∂∂

=∂∂

( , ) ( ( , ), ( , ))z f u v f g x y h x y= =

*


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x r x xu u r uθθ= +

( ) ( )xx rr x r x x r xx r x x x xxu u r u r u r u r u uθ θ θθ θθ θ θ θ= + + + + +

2 2

2 2

1 2( )2x

x xr x yx rx y∂

= + = =∂ +

2 22

1arctan ( )1 ( )

xy y y

yx x x rx

θ ∂ = = − = − ∂ +

yyxx uuu +=∇2

2 2 2 , tan yr x yx

θ= + =

Chain Rule



Theorem 9.5

),( yxgu =),( vufz = ),( yxhv =

xv

vz

xu

uz

xz

∂∂

∂∂

+∂∂

∂∂

=∂∂

yv

vz

yu

uz

yz

∂∂

∂∂

+∂∂

∂∂

=∂∂

( , ) ( ( , ), ( , ))z f u v f g x y h x y= =

*


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xxrr

= 2xyr

θ = −

yyxx uuu +=∇2

2 2 2 , tan yr x yx

θ= + =

( ) ( )2 2 2 2

3 3 32 2 2 2 2 2 2 2

1 1 1 22xx

x x x y x yr x xx r x rx y x y x y x y

∂ ∂ + − = = = + − = = ∂ ∂ + + + +

( )22 2 2 42 2

2 2( 1)xxy y xy xy

x r x x y rx yθ

∂ ∂ = − = − = − − = ∂ ∂ + +

x r x xu u r uθθ= +

Chain Rule



Theorem 9.5

),( yxgu =),( vufz = ),( yxhv =

xv

vz

xu

uz

xz

∂∂

∂∂

+∂∂

∂∂

=∂∂

yv

vz

yu

uz

yz

∂∂

∂∂

+∂∂

∂∂

=∂∂

( , ) ( ( , ), ( , ))z f u v f g x y h x y= =

*


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xxrr

= 2xyr

θ = −

yyxx uuu +=∇2

Chain Rule



Theorem 9.5

),( yxgu =),( vufz = ),( yxhv =

xv

vz

xu

uz

xz

∂∂

∂∂

+∂∂

∂∂

=∂∂

yv

vz

yu

uz

yz

∂∂

∂∂

+∂∂

∂∂

=∂∂

( , ) ( ( , ), ( , ))z f u v f g x y h x y= =

2 2 2 , tan yr x yx

θ= + =

2

3xxyrr

=4

2xx

xyr

θ =

2

2 3

2 2 4

2 2 2

2 3 3 3 4 4

( ( ))

2( ( ))( )

2

xx rr r r

r

rr r r r

x y x yu u u ur r r r

x y y xyu u ur r r r

x xy y xy y xyu u u u u ur r r r r r

θ

θ θθ θ

θ θ θθ θ

∴ = + − +

+ + − − +

= − + − + +

θθθθ urxyu

ryu

ryu

rxyu

rx

rrrr 43

2

4

2

32

2

22 +++−=

*


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function u, into polar coordinate.yyxx uuu +=∇2

ryry = 2r

xy =θ

2

3yyxrr

=4

2yy

xyr

θ = −

2 2 2

2 3 4 3 42 2yy rr r ry xy x x xyu u u u u ur r r r rθ θθ θ= + + + −

2 2 2

2 3 4 3 42 2xx rr r rx xy y y xyu u u u u ur r r r rθ θθ θ= − + + +

xxrr

= 2xyr

θ = −2

3xxyrr

= 4

2xx

xyr

θ =In the similar way for y

2 2 2 , tan yr x yx

θ= + =


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2

2 2 2 2 2 2 2 2 2

2 4 3 2 4 3

2

1 1

xx yy

rr r rr r

rr r

u u u

x y x y x y r r ru u u u u ur r r r r r

u u ur r

θθ θθ

θθ

∴∇ = +

+ + += + + = + +

= + +

θθθθ urxyu

rxu

rxu

rxyu

ryu rrrryy 43

2

4

2

32

2

22 −+++=

θθθθ urxyu

ryu

ryu

rxyu

rxu rrrrxx 43

2

4

2

32

2

22 +++−=

2 2 2 , tan yr x yx

θ= + =

: Laplacian in polar coordinate


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2 2 2 , tan yr x yx

θ= + =

22

1 1xx yy rr ru u u u u u

r r θθ∇ = + = + +

22

1 1 0rr ru u u ur r θθ∇ = + + =

Laplace equation in Polar Coordinates


[2009] [09] [10] Innovative ship design Naval Architecture & Ocean … · 2018. 1. 30. · Department of Naval Architecture and Ocean Engineering, Seoul National University of College

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