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SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr

Seoul NationalUniv.

2009 Fall, Ship Stability

Nav

al A

rch

itec

ture

& O

cean

En

gin

eeri

ng


Seoul NationalUniv.

- Ship Stability -

Part.1-IV Pressure Integration Technique

2009

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering,Seoul National University


Righting Moment

Overview of “Ship Stability”

Force & Moment on a Floating BodyNewton’s 2nd Law Euler Equation

Stability Criteria

Damage Stability- MARPOL regulation

Pressure Integration Technique

Calculation Method to find GZ with respect to IMO regulation

sinGZ GM φ= ,GM KB BM KG= + −

sinL LGZ GM θ= , L LGM KB BM KG= + −

- Overview of Ship Stability

BF GZ×Transverse Righting Moment :

B LF GZ×Longitudinal Righting Moment :

<Method ②>

GZ Calculation

( )G BGZ y y= − +

( )L G BGZ x x= − +

<Method ①>

Z≡

K

z′

O

CL

y

z M

φ

restoringτ

eτ

G

FG

B B1

≡ NFB

≡

1By

Gy

φ

φ

FB: Buoyancy forceφ : Angle of Heel, θ : Angle of Trim(xG,yG,zG) : Center of gravity in waterplane fixed frame(xB,yB,zB) : Center of buoyancy in waterplane fixed fra

y'G , y'B in body fixed frame

Rotational Transformation!yG , yB in waterplane fixed frame

Fundamental of Ship Stability

• Properties which is related to hull form of the ship

Hydrostatic Values

Intact Stability- IMO Requirement (GZ)- Grain Stability- Floodable Length

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- Contents -Part.1-I Fundamentals of ship stability

Ch.1 Overview of Ship StabilityCh.2 Physics for Ship StabilityCh.3 Hydrostatic Pressure, Force and Moment on a floating bodyCh.4 Concept of Righting MomentCh.5 Hydrostatic Values

Part.1-II Righting MomentCh.6 Transverse Righting MomentCh.7 Longitudinal Righting MomentCh.8 Heeling Moment caused by Fluid in Tanks

Part.1-III Stability CriteriaCh.9 Intact StabilityCh.10 Damage Stability

Part.1-IV Pressure Integration TechniqueCh.11 Calculation of Static Equilibrium PositionCh.12 Governing Equation of Force and Moment with Immersion, Heel and TrimCh.13 Partial derivatives of force and moments with immersion, heel, and trim

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- Ship Stability -

Ch.11 Calculation of Static Equilibrium Position

2009

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering,Seoul National University


Seoul NationalUniv.


Sec.1 Example : Immersion in Static Equilibrium

after Cargo Being Loaded

Sec.2 Example : Immersion and Heel in Static Equilibrium



after Cargo Hold Being Flooded

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CLx

y′

, ′x

yPlan View

300L m=

50B m=O

w

Example : Immersion in Static Equilibrium after Cargo Being Loaded (1)

Given : loading 300,000 [kN] cargo on deck above the intersection point centerline and LCFMass of ship : 1,350,000 [kN]

ξ3 : Immersion

Find : ship’s attitude in static equilibrium

Example ) Find a ship’s attitude in Static Equilibrium after Cargo Being Loaded.( Origin of waterplane fixed frame is located in the intersection point

between centerline and LCF )

CL

,

( ) ( )

( ) ( )

( ) ( )

Gravity G

Buoyancy B

Ext Static Ext

F F

F F

F F

→

→

→

r r

r r

r r

w

Assumption : considering force with immersion only( due to cargo being loaded on the intersection point

between centerline and LCF. No heel, no trim)

3, ( [0,0, ,0,0,0] )Tξ=r

LCF

x′ y′∇

, ′z z

O

Section View

,yx

*3 ?ξ =

ξ3* : immersion in

static equilibriumO-xyz frame: Waterplane fixed frameO'-x'y'z' frame: Body fixed frame

Position vector of the origin of the body fixed frame with respect to thewaterplane fixed frame

26.5626.56 10m40m

50m

( )03 0ξ =

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w


Given : loading 300,000 [kN] cargo on deck above the intersection point centerline and LCFMass of ship : 137,755 [Mg]

Section View

ξ3 : Immersion

Find : ship’s attitude in static equilibrium

Example ) Find a ship’s attitude in Static Equilibrium after Cargo Being Loaded.( Origin of waterplane fixed frame is located in the intersection point between centerline and LCF )

x′y′

∇

, ′z z

x( )03 0ξ =

,y

CL

O

<Attitude : ξ3(0) >

(0) (0) ( (00)3

)33 3( ) ( ) ( )( )z G ExtBF F F Fξ ξξ ξ += +

Initial condition with cargo being loaded

<Attitude : ξ3* >

Static equilibrium condition ( = Total forces and moments must be ‘zero’)

*3( ) 0zF ξ =

* * * *3 3 3 3, ( ) ( ) ( ) ( )z G B Extwh ere F F F Fξ ξ ξ ξ= + +

We want to find *3ξ

Assumption : considering force with immersion only 3, ( [0,0, ,0,0,0] )Tξ=r

w

CL

x′ y′,O′∇

, ′z z

x

Section View

,yO

*3 ?ξ =

Given : (0)3

(0)3( ), ExtF ξξ

,

( ) ( )

( ) ( )

( ) ( )

Gravity G

Buoyancy B

Ext Static Ext

F F

F F

F F

→

→

→

r r

r r

r r

(0)3( )ExtF wξ = −

( )03 0ξ =

!

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ξ3 : Immersion

<Example> in case of force with immersion onlyGoverning equation

Given : Find : (0) (0) (0) (0)3 3 3 3, ( ) ( ) ( ) ( )z G B Extwh ere F F F Fξ ξ ξ ξ= + +


*3ξ(0) (0)

3 3, ( )ExtFξ ξ

(0) (0)* (0) (0) (0)3 3

3 3 3 33 3

( ) ( )( ) ( ) B Extz z

F RFF F ξ ξξ ξ ξ ξξ ξ

∂ ∂= + ∆ + ∆ +

∂ ∂

*3( ) 0zF ξ =

Why Taylor series expansion ? Nonlinear equation

* (0) (0)3 3 3, ( )ξ ξ ξ+ ∆≈

* (0) (0)3 3 3, ( )ξ ξ ξ= + ∆

Linearization(0) (0)

* (0) (0) (0)3 33 3 3 3

3 3

( ) ( )( ) ( ) B Extz z

F FF F ξ ξξ ξ ξ ξξ ξ

∂ ∂+ ∆ + ∆

∂ ∂≈

Approximation

(0)* (0) (0)3

3 3 33

)( ) ( ) zz z

F RF F ξξ ξ ξξ

∂ (= + ∆ +

∂

(0) (0) (0)* (0) (0) (0) (0)3 3 3

3 3 3 3 33 3 3

( ) ( ) ( )( ) ( ) G B Extz z

F FF F RFξ ξ ξξ ξ ξ ξ ξξ ξ ξ

∂ ∂ ∂= + ∆ + ∆ + ∆ +

∂ ∂ ∂

w

CL

x′ y′

∇

, ′z z

x( )03 0ξ =

Section View

,y

*3 ?ξ =

Taylor series expansion

Assumption : 33

, ( )BWP

F g Aρ ξξ

∂= − ⋅

∂Buoyancy force is proportional to AWP

(AWP : waterplane area of ship)

Opposite direction

3

, 0ExtFξ

∂=

∂

Weight of cargo does not change with respect to immersion

Known :

* (0) (0) (0)3 3 3 3, ( ) ( ) ( )z z WPwhere F F g Aξ ξ ρ ξ ξ= − ⋅ ⋅∆Governing

equation*

3( ) 0zF ξ = We want to find *3ξ

FExt is independent of ξ3

( FG is not changed with immersion)

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w


(0) (0) (0) (0)3 3 3 3( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + + : Total force

* (0) (0) (0)3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆


Given : Find :

Governing equation

(0) (0)3 3, ( )ExtFξ ξ *

3ξ

*3( ) 0zF ξ =

1

1

1,350,000[ ] 1,350,000[ ] 300,000[ ]kN kN kN= − −

(0)3( ) 300,000 [ ]ExtF kNξ = −

300,000 [ ]kN= −

(0)3

(0)3

( )

( )BV

F g dVξ

ξ ρ= ∫∫∫

(0) (0) (0) (0)3 3 3 3( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

300,000ExtF w

kN= −

= −* (0) (0)3 3 3( )ξ ξ ξ+ ∆≈

x′y′

∇

, ′z z

x ( )03 0ξ =

Section View

,y

ξ3 : Immersion

CLx

y′

, ′x

yPlan View

LCF w

ξ3*←ξ3

(0)

10m40m

50m

(0) (0) (0)3 3 3

1 ( ( ) 4 0 )( ) ( )2 sg B T Lρ ξ ξ ξ = + ⋅ ⋅

(0)3(section area) ( )g Lρ ξ= ⋅ ⋅

( )110 50 40 10 300 1,350,000[ ]2

kN = ⋅ + ⋅ ⋅ =

(0)3( ) 1,350,000[ ]GF kNξ = −

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w


( )03( )WPA

g dx dyξ

ρ= − ⋅ ∫∫

* (0) (0) (0)3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆


Given : Find :

Governing equation

(0) (0)3 3, ( )ExtFξ ξ *

3ξ

*3( ) 0zF ξ =

2

( )03( )WPg Aρ ξ− ⋅2

3, ( 10 [kN/m ])gρ ≈

* (0) (0)3 3 3( )ξ ξ ξ+ ∆≈

x′y′

∇

, ′z z

x ( )03 0ξ =

Section View

,y

ξ3 : Immersion

( )03( )WPA ξ

CLx

y′

, ′x

yPlan View

LCF w

300L m=

( )03( ) 50sB mξ =

ξ3*←ξ3

(0)

WPArea A=

310[ / ] 50[ ] 300[ ]kN m m m= − ⋅ ⋅

150,000 [ / ]kN m= −

( ) ( )0 03 3( ) ( )sg B Lρ ξ ξ= − ⋅ ⋅


B(breadth of AWP) : dependent on ξ3L (length of AWP) : independent of ξ3 → constant

( )03( )sg B Lρ ξ= − ⋅ ⋅

( )03( ) 50B mξ =

300,000ExtF w

kN= −

= −

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w

x′y′

w


* (0) (0) (0)3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆


Given : Find :

Governing equation

(0) (0)3 3, ( )ExtFξ ξ *

3ξ

*3( ) 0zF ξ =

3

* (0) (0)3 3 3( )ξ ξ ξ+ ∆≈x′

y′∇

, ′z z

x ( )03 0ξ =

Section View

,y

ξ3 : Immersion

ξ3*←ξ3

(0)

(AWP : waterplane area of ship)(0)3( ) 300,000 [ ]zF kNξ = −1

1 2

(0)3( ) 150,000 [ / ]WPg A kN mρ ξ− ⋅ = −2

3

( )( )

* (0)0 3 3

3 03

( ) ( )( )

z z

WP

F FgA

ξ ξξρ ξ

−∆ =

−

( ) ( )0 03 3 0[m] 2[m] 2 [m]ξ ξ+ ∆ = − = −

(0)3ξ∆ : change of immersion

ξ3* ←ξ3

(0) +∆ξ3(0)

∇

, ′z z

x

Section View

,y( )0 (0)

3 3ξ ξ+ ∆

( ) ( ) ?0 0

3 3( ) 0zF ξ ξ+ ∆ =Static equilibrium? Check!


( ) ( )0 0*3 3 3

?ξ ξ ξ+ ∆=

0 ( 300,000[ ]) 2 [m]150,000[ / ]

kNkN m

− −= = −

−

300,000ExtF w

kN= −

= −

If we assume that Fz(ξ3*) =0

( )0(1) (0)3 3 3, ( ( ) ( ) 10[m] ( 2[m]) 12.0[ ])T T mξ ξ ξ= − ∆ = − − =

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( ) ( )0 0(0) (0)3 3 3 3( ) ( ) ( )G B ExtF F Fξ ξ ξ ξ= + + ∆ +

( ) ( ) ( ) ( ) ( ) ( )0 0 0 0 0 03 3 3 3 3 3( ) ( ) ( )z G BF F Fξ ξ ξ ξ ξ ξ+ ∆ = + ∆ + + ∆

6,000[ ]kN=

( ) ( )0 03 3( )BF ξ ξ+ ∆

( ) ( )0 03 3( )GF ξ ξ+ ∆

( ) ( )0 03 3( )zF ξ ξ+ ∆

( ) ( )0 03 3( )ExtF ξ ξ+ ∆ (0)

3( )ExtF ξ=

(0)3( )GF ξ=

( )13( )g Vρ ξ= ⋅

* (0) (0) (0)3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆


Given : Find :

Governing equation

(0) (0)3 3, ( )ExtFξ ξ *

3ξ

*3( ) 0zF ξ =

* (0) (0)3 3 3( )ξ ξ ξ+ ∆≈4

( ) ( )0 03 3( ) 0zF ξ ξ+ ∆ =

Static equilibrium?Check!

( ) ( )0 0*3 3 3ξ ξ ξ+ ∆=

( ) ( )0 03 3( )ExtF ξ ξ+ + ∆

( ) ( )0 03 3( ) 0zF ξ ξ+ ∆ ≠ ( ) ( )0 0*

3 3 3ξ ξ ξ+ ∆≠

( ) ( )0 0(1)3 3 3ξ ξ ξ+ ∆≡

x′y′

w

ξ3* ←ξ3

(0) +∆ξ3(0)

∇

, ′z z

x

Section View

,y( )0 (0)

3 3ξ ξ+ ∆

? ?


Re-define

* (1) (1) (1)3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆

How can we do?

※If the ship has vertical wall, the displacement volume is 1,650,000 [kN]. → No iteration ξ*=ξ3

(0) +∆ξ3(0)

26.56

※ displacement volume V( ξ3(1))

( )1 (1)3 3( ) 40 2 ( ) tan 26.56

40 2 12 tan 26.56 52[ ]

B T

m

ξ ξ= + ⋅ ⋅

= + ⋅ ⋅ =

( ) ( )1 1 (1)3 3 3

3

( ) 0.5 (40 ( )) ( ) 300

0.5 (40 52) 12 300 165,600 [ ]

V B T

m

ξ ξ ξ= ⋅ + ⋅ ⋅

= ⋅ + ⋅ ⋅ =4

300,000ExtF w

kN= −

= −

1,350,000[ ] 1,656,000[ ] 300,000[ ]kN kN kN= − + −

( )13( )sB ξ

(1)3( )T ξ

4012.0=

3 310[ / ] 165,600[ ]kN m m= × 1,656,000[ ]kN=(1)3( )

12.0T ξ=

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x′y′

w* (1) (1) (1)

3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆


Given : Find :

Governing equation

(0) (0)3 3, ( )ExtFξ ξ *

3ξ

*3( ) 0zF ξ =Section View

ξ3 : Immersion

ξ3* ←ξ3

(1)


1

∇

, ′z z

x ,y(1)3ξ

CL

(1) (1) (1) (1)3 3 3 3( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +1

2 ( )13( )WPgAρ ξ−

310[ / ] 52[ ] 300[ ]kN m m m= − ⋅ ⋅

,(in previous calculation)

CLx

y′

, ′x

yPlan View

LCF w

300L m=

( )13( ) 52sB mξ =

WPArea A=( )1

3( )WPA

g dx dyξ

ρ= − ⋅ ∫∫( )1

3( )sg B Lρ ξ= − ⋅ ⋅

2

156,000 [ / ]kN m= −

3, ( 10 [kN/m ])gρ ≈

6,000 [ ]kN=


26.56

300,000ExtF w

kN= −

= −

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=

(in previous calculation)

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CL

x′y′

w* (1) (1) (1)

3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆


Given : Find :

Governing equation

(0) (0)3 3, ( )ExtFξ ξ *

3ξ

*3( ) 0zF ξ =Section View

ξ3 : Immersion

ξ3* ←ξ3

(1)

1

∇

, ′z z

x ,y(1)3ξ

CL

2

3

( )( )

* (1)1 3 3

3 13

( ) ( )( )

0.034 [ ]

z z

WP

F FgA

m

ξ ξξρ ξ

−∆ =

−=

( ) ( ) ( )2 1 13 3 3 2[ ] 0.034[ ]

1.966 [ ]m m

mξ ξ ξ= + ∆ = − +

= −

(1)3ξ∆

(1)3( ) 6,000 [ ]zF kNξ =1

(1)3( ) 156,000 [ / ]WPgA kN mρ ξ− = −2

x′y′

w

Section Viewξ3* ←ξ3

(2)

∇

, ′z z

x ,y(2)3ξ

Static equilibrium? Check!


3

( )2*3 3ξ ξ=

? ( )23( ) 0zF ξ =

?


(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


0 6,000[ ]156,000[ / ]

kNkN m

−=

−

If we assume that Fz(ξ3*) =0

( )2(2) (0)3 3 3, ( ( ) ( ) 10[ ] ( 1.966[ ]) 11.966[ ])T T m m mξ ξ ξ= − = − − =

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* (1) (1) (1)3 3 3 3( ) ( ) ( )z z WPF F gAξ ξ ρ ξ ξ= − ⋅∆


Given : Find :

Governing equation

(0) (0)3 3, ( )ExtFξ ξ *

3ξ

*3( ) 0zF ξ =

4

( )23( ) 0zF ξ =


( )2*3 3ξ ξ=

( )23( )zF ξ ε< , 1,000ε ≡

( )2*3 3ξ ξ=

End of iteration

(2)3( )BF ξ

(2)3( )GF ξ

(2)3( )ExtF ξ

(2)3( )zF ξ

(1)3( )GF ξ=

(1)3( )ExtF ξ=

(2) (2) (2) (2)3 3 3 3( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

(1) (2) (1)3 3 3( ) ( ) ( )G B ExtF F Fξ ξ ξ= + +

640 [ ]kN=attitude in static equilibrium

, ( 0,1, 2,3...)k =

CL

x′y′

w

Section Viewξ3

* ←ξ3(2)

∇

, ′z z

x ,y(2)3ξ

? ?


※If the ship has vertical wall, The displacement volume is 1,650,000 [kN]. → No iteration ξ*=ξ3

(2)

ε : Tolerance

300,000ExtF w

kN= −

= −

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=

※ displacement volume V( ξ3(2))

4

(2) (2)3 3( ) 40 2 ( ) tan 26.56

40 2 11.966 tan 26.56 51.963[ ]B T

mξ ξ= + ⋅ ⋅

= + ⋅ ⋅ =

( )1(2) (1)3 3 3

3

( ) 0.5 (40 ( )) ( ) 3000.5 (40 51.963) 11.966 300 165,064 [ ]

V B Tm

ξ ξ ξ= ⋅ + ⋅ ⋅

= ⋅ + ⋅ ⋅ =

( )13( )sB ξ

(1)3( )T ξ

4012.0=

( )23( )g Vρ ξ= ⋅

3 310[ / ] 165,064[ ]kN m m= × 1,650,640[ ]kN=

1,350,000[ ] 1,650,640[ ] 300,000[ ]kN kN kN= − + −

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ξ3* ←ξ3

(0)300,000

ExtF wkN

= −= −

x′y′

∇

w

, ′z z

x ( )03 0ξ =

Section View

,y

CL

x′y′

w

∇x ,y

(1)3 2ξ = −

CL

, ′z zξ3* ←ξ3

(1)

x′y′

w∇

, ′z z

x ,y(2)3 1.966ξ = −

CL

ξ3* ←ξ3

(2)

(0) (0)3( [0,0, ,0,0,0])ξ=r

(1) (1)3( [0,0, ,0,0,0])ξ=r

(2) (2)3( [0,0, ,0,0,0])ξ=r

Immersion

Immersion

-300,000 -1,350,000 1,350,000 -300,000

[ ]ZF kN [ ]GF kN [ ]BF kN [ ]ExtF kN

6,000 -1,350,000 1,656,000 -300,000


640 1,350,000 1,650,640 -300,000


( )23( ) 640zF εξ = < , ( 1,000)ε =

( )2*3 3ξ ξ=

attitude in static equilibrium

End of iteration

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w g

y

z

LC

,O O′

z′

Example : Immersion and Heel in Static Equilibrium after Cargo Being Loaded (1)

Given : loading 40,000 [kN] cargo on bottom shell in right-side of LCFWeight of ship : 360,000 [kN]

Find : Immersion and trim of ship in static equilibrium

Example ) Find a ship’s attitude in Static Equilibrium after Cargo Being Loaded.( Origin of waterplane fixed frame is located in the intersection point

between centerline and LCF(Longitudinal center of floating) )

Assumption : considering force and transverse momentwith immersion and heel

( due to cargo being loaded on LCF. No trim) 3 4, ( [0,0, , ,0,0] )Tξ ξ=r

O-xyz frame: Waterplane fixed frameO'-x'y'z' frame: Body fixed frame

Position vector of the origin of the body fixed frame with respect to the waterplane fixed frame

Rotation vector of the origin of the body fixed frame with respect to thewaterplane fixed frame

y

z

LC

Section view(Front view)

OBaseline

K

G

, z′

, y′Bw g

,O O′Immersion

y′

G

w g B ξ3* : immersion in

static equilibrium

y

z

LC

,O O′

z′

y′

G

w g B

*4 ?ξ

Plan view (Top view)

x

y

LC

20m

,O O′G , x′

, y′

, B

100m ξ4* : heel in

static equilibrium

Heel

*3 ?ξ

18/120


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G : Center of gravity of ship

B : Center of buoyancy of ship

Cargo information- w (weight) : 40,000 kN

- g (center of gravity of cargo) : (0,-10,-4.5)

100L m=

x

zElevation view (port-view)

Baseline,O O′

G

APFP

w, x′

, z′

g , B

40sB m=

30D m=9T m= y

z

LC

Section View(Front view)

,O O′Baseline

K

G

w

, z′

, y′g B


Principal Dimension

- L (length) : 100 m, Bs (breadth) : 40m, T (draft) : 9m, D (depth) : 30m

20m

Plan view(Top View)

x

y

LC,O O′ G

w, x′

, y′

, B

O-xyz : Waterplane fixed frame

O'-x'y'z': Body fixed frame

g

19/120


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( FG is not changed with immersion)


<Example> in case of force with immersion and heel Governing equation of forceGiven : Find : (0) (0) (0) (0) (0) (0) (0) (0)

3 4 3 4 3 4 3 4, ( , ) ( , ) ( , ) ( , )z G B Extwh ere F F F Fξ ξ ξ ξ ξ ξ ξ ξ= + +

* * * * * * * *3 4 3 4 3 4 3 4, ( , ) ( , ) ( , ) ( , )z G B Extwh ere F F F Fξ ξ ξ ξ ξ ξ ξ ξ= + +

*3ξ(0) (0) (0) (0)

3 4 3 4, , ( , )ExtFξ ξ ξ ξ

(0) (0)(0) (0)* * (0) (0) 3 43 4

3 4 3 4(0) (0)3 3

3 3

( , )( , )( , ) ( , ) ExtB

z z RFFF F ξ ξξ ξξ ξ ξ ξ ξ ξ

ξ ξ∂∂

= + ∆ + ∆ +∂ ∂

* *3 4( , ) 0zF ξ ξ =

* (0) (0)3 3 3, ( )ξ ξ ξ+ ∆≈

* (0) (0)3 3 3, ( )ξ ξ ξ= + ∆

Linearization(0) (0)(0) (0)

* * (0) (0) 3 43 43 4 3 4

(0) (0)3 3

3 3

( , )( , )( , ) ( , ) ExtB

z zFFF F ξ ξξ ξ

ξ ξ ξ ξ ξ ξξ ξ

∂∂= + ∆ + ∆

∂ ∂

Approximation

(0) (0) (0) (0)* * (0) (0) 3 4 3 4

3 4 3 4(0) (0)3 4

3 4

( , ) ( , )( , ) ( , ) z z

z zF FF F Rξ ξ ξ ξ

ξ ξ ξ ξ ξ ξξ ξ

∂ ∂= + ∆ + ∆ +

∂ ∂(0) (0) (0) (0)(0) (0)

* * (0) (0) 3 4 3 43 43 4 3 4

(0) (0) (0)3 3 3

3 3 3

( , ) ( , )( , )( , ) ( , ) G ExtB

z zF FFF RF ξ ξ ξ ξξ ξ

ξ ξ ξ ξ ξ ξ ξξ ξ ξ

∂ ∂∂= + ∆ + ∆ + ∆ +

∂ ∂ ∂


Assumption : 3 43 4

3

( , ), ( , )BWP

F gAξ ξ ρ ξ ξξ

∂= −

∂Buoyancy force is proportional to AWP


Opposite direction

3 4

3

( , ), 0ExtF ξ ξξ

∂=

∂

Weight of cargo does not change with respect to immersion

Known :

* * (0) (0) (0)3 4 3 4 4

(0) (0)3 3( , ) ( , ), ( , )z z WPwhere F F gAξ ξ ξ ξ ξρ ξ ξ= − ⋅ ∆

Governing equation of force

* *3 4( , ) 0,zF ξ ξ =

FExt is independent of ξ3

This term is not considered in this example

Given : Find : (0) (0) (0) (0) (0) (0) (0) (0)3 4 3 4 3 4 3 4, ( , ) ( , ) ( , ) ( , )z G B Extwh ere F F F Fξ ξ ξ ξ ξ ξ ξ ξ= + + *

3ξ(0) (0) (0) (0)3 4 3 4, , ( , )ExtFξ ξ ξ ξ

20/120


Seoul NationalUniv.



<Example> in case of transverse moment with immersion and heel Governing equation of Transverse moment

Given : Find : (0) (0) (0) (0) (0) (0) (0) (0)3 4 3 4 3 4 3 4,( , ) ( , ) ( , ) ( , ), T TG TB T Extwhere M M M Mξ ξ ξ ξ ξ ξ ξ ξ= + +

* * * * * * * *3 4 3 4 3 4 , 3 4, ( , ) ( , ) ( , ) ( , )T TG TB T Extwhere M M M Mξ ξ ξ ξ ξ ξ ξ ξ= + +

*4ξ(0) (0) (0) (0)

3 4 , 3 4, , ( , )T ExtMξ ξ ξ ξ

* *3 4( , ) 0TM ξ ξ =

* (0) (0)4 4 4, ( )ξ ξ ξ+ ∆≈

* (0) (0)4 4 4, ( )ξ ξ ξ= + ∆

Linearization(0) (0)(0) (0) (0) (0)

3 4* * (0) (0) 3 4 3 43 4 3 4

,(0) (0) (0)4 4 4

4 4 4

( , )( , ) ( , )( , ) ( , ) T ExtTG TB

T T

MM MM M

ξ ξξ ξ ξ ξξ ξ ξ ξ ξ ξ ξ

ξ ξ ξ∂∂ ∂

= + ∆ + ∆ + ∆∂ ∂ ∂

Approximation

(0) (0) (0) (0)* * (0) (0) 3 4 3 4

3 4 3 4(0) (0)3 4

3 4

( , ) ( , )( , ) ( , ) T T

T TM M

M M Rξ ξ ξ ξξ ξ ξ ξ ξ ξ

ξ ξ∂ ∂

= + ∆ + ∆ +∂ ∂

(0) (0)(0) (0) (0) (0)3 4* * (0) (0) 3 4 3 4

3 4 3 4,(0) (0) (0)

4 4 44 4 4

( , )( , ) ( , )( , ) ( , ) T ExtTG TB

T T

MM MM M R

ξ ξξ ξ ξ ξξ ξ ξ ξ ξ ξ ξ

ξ ξ ξ∂∂ ∂

= + ∆ + ∆ + ∆ +∂ ∂ ∂



This term is not considered in this example.

3 43 4 3 4 3 4

4

( , )( , ) ( , ) ( , ),TB

B B TM F z gIξ ξ

ξ ξ ξ ξ ξ ξρξ

∂= − ⋅ −

∂

3 43 4 3 4

4

( , )( , ) ( , ),TG

G GM F zξ ξ

ξ ξ ξ ξξ

∂= − ⋅

∂

3 43 4 3 4

,

4

( , )( , ) ( , )T Ext

Ext Ext

MF z

ξ ξξ ξ ξ ξ

ξ∂

= − ⋅∂

zG: Z coordinate of the center of mass of the ship IT: 2nd moment of waterplane area with respect to x axis of the waterplane fixed framezB: Z coordinate of the center of buoyancy

zExt: Z coordinate of the point acting the external force

* * * * * * * *3 4 3 4 3 4 3 4,( , ) ( , ) ( , ) ( , ), T TG TB T Extwhere M M M Mξ ξ ξ ξ ξ ξ ξ ξ= + +Governing equation

of transverse moment* *

3 4( , ) 0TM ξ ξ =

Given : (0) (0) (0) (0)3 4 , 3 4, , ( , )T ExtMξ ξ ξ ξ Find : *

4ξ(1) (0) (0)

4 4 4, ( )ξ ξ ξ+ ∆=

(0) (0) (0) (0) (0) (0) (0) (0)3 4 3 4 3 4 3 4* * (0) (0)

3 4 3 4 (0) (0) (0) (0) (0) (0)3 4 3 4 3 4

(0)4

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

( , ) ( , ) ( , )( , ) G G B B

T TT Ext Ext

F z F zM M

gI F zξ ξ ξ ξ ξ ξ ξ ξ

ξ ξξ ξ ξ ξ ξ ξ

ξ ξ ξρ

− ⋅ − ⋅= + ∆ − − ⋅

21/120



Seoul NationalUniv.

(ref.) Transverse moment of gravity with heel

Transverse moment of gravity

3 4 4( , )TGM ξ ξ ξ+ ∆

( )3 4 3 4 4 3 4 4( , ) ( , ) cos ( , ) sinG G GF y zξ ξ ξ ξ ξ ξ ξ ξ= ⋅ ⋅ ∆ − ⋅ ∆

3 4 3 4 3 4 4( , ) ( , ) ( , )TG G GM F zξ ξ ξ ξ ξ ξ ξ= − ⋅ ⋅ ∆3 4( , )TGM ξ ξ

If is small4ξ∆

Transformation

3 4 4 3 4 4( , ) ( , )G GF yξ ξ ξ ξ ξ ξ= + ∆ ⋅ + ∆

3 4 4 3 4

3 4 4 3 4 4 3 4 4

3 4 4 3 4 4 3 4 4

( , ) ( , )

( , ) ( , )cos ( , )sin

( , ) ( , )cos ( , )sin

G G

G G G

G G G

x xy y zz z y

ξ ξ ξ ξ ξ

ξ ξ ξ ξ ξ ξ ξ ξ ξ

ξ ξ ξ ξ ξ ξ ξ ξ ξ

+ ∆ =

+ ∆ = ∆ − ∆ + ∆ = ∆ + ∆

3 4 4 3 4

3 4 4 4 4 3 4

3 4 4 4 4 3 4

( , ) 1 0 0 ( , )( , ) 0 cos sin ( , )( , ) 0 sin cos ( , )

G G

G G

G G

x xy yz z

ξ ξ ξ ξ ξξ ξ ξ ξ ξ ξ ξξ ξ ξ ξ ξ ξ ξ

+ ∆ + ∆ = ∆ − ∆ + ∆ ∆ ∆

3 4 3 4 3 4 3 4 4( , ) ( , ) ( , ) ( , )G G G GF y F zξ ξ ξ ξ ξ ξ ξ ξ ξ= ⋅ − ⋅ ⋅ ∆( )3 4 3 4 3 4 4( , ) ( , ) ( , )G G GF y zξ ξ ξ ξ ξ ξ ξ≈ ⋅ − ⋅ ∆

3 4 3 4 4( , ) ( , )G GF yξ ξ ξ ξ ξ= ⋅ + ∆

4

TGMξ

∂∂

=

y

z

LC

z′

y′

4ξ∆

,O O′

Section view

GF

3 4 4( , )G ξ ξ ξ+ ∆3 4( , )G ξ ξ

22/120


Seoul NationalUniv.



Governing equation* *

3 4( , ) 0TM ξ ξ = * * * * * * * *3 4 3 4 3 4 3 4,( , ) ( , ) ( , ) ( , ), T TG TB T Extwhere M M M Mξ ξ ξ ξ ξ ξ ξ ξ= + +

(0) (0) (0) (0) (0) (0) (0) (0)3 4 3 4 3 4 3 4* * (0) (0)

3 4 3 4 (0) (0) (0) (0) (0) (0)3 4 3 4 3 4

(0)4

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

( , ) ( , ) ( , )( , ) G G B B

T TT Ext Ext

F z F zM M



ξ ξ ξρ

− ⋅ − ⋅= + ∆ − − ⋅

zG: Z coordinate of the center of mass of the ship

IT: 2nd moment of waterplane area with respect to x axis of the waterplane fixed frame

zB: Z coordinate of the center of buoyancyzExt: Z coordinate of the point acting the external force

≡

K

O

CL

y

z,

φ

G Z

FG

B B1

≡ N

FB

≡

restoringτ

M

O'

Comparison with classical hydrostatics

restoring BF GZτ = ⋅

sinBF GM φ≈ ⋅ ⋅

[ ] sinBF KB BM KG φ= ⋅ + − ⋅

KB OB OK= −

For example

5[ ]Bz OK m= − =(0,0,6), (0,0, 5), 10[ ]G B T m− =

( )KG OG OK= −

O'

CL

z, z′

K

B

G

BaseLine

O y, y′

KBKG

( )Gz OK= −

( ) ( ) sinB B GF z OK BM z OK φ = ⋅ − + − − ⋅

[ ] sinB B GF z BM z φ= ⋅ + − ⋅continue…

assume that is smallφ

( ) ( ) sinBF OB OK BM OG OK φ = ⋅ − + − − ⋅

5[ ] ( 1 0 )[ ] 5[ ]m m m= − − − =

6[ ] ( 10[ ]) 16[ ]m m m= − − =

OK zB OB=

zGOG =

23/120


Seoul NationalUniv.


Governing equationof transverse moment

* *3 4( , ) 0TM ξ ξ = * * * * * * * *

3 4 3 4 3 4 3 4,( , ) ( , ) ( , ) ( , ), T TG TB T Extwhere M M M Mξ ξ ξ ξ ξ ξ ξ ξ= + +(0) (0) (0) (0) (0) (0) (0) (0)3 4 3 4 3 4 3 4* * (0) (0)

3 4 3 4 (0) (0) (0) (0) (0) (0)3 4 3 4 3 4

(0)4

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

( , ) ( , ) ( , )( , ) G G B B

T TT Ext Ext

F z F zM M



ξ ξ ξρ

− ⋅ − ⋅= + ∆ − − ⋅


zG: Z coordinate of the center of mass of the ship

IT: 2nd moment of waterplane area with respect to x axis of the waterplane fixed frameV: Displacement volume


≡

K

O'

CL

y

z,

φ

G Z

FG

B B1

≡ N

FB

≡

restoringτ

M

O



[ ] sinB B GF z BM z φ= ⋅ + − ⋅

sinTB B G

IF z zV

φ = ⋅ + − ⋅

TIBMV

=

sinTB B B G

IF z g V F zV

ρ φ = ⋅ + ⋅ − ⋅ ⋅ in static equilibrium

0G BF F+ =

[ ] sinB B T G GF z g I F zρ φ= ⋅ + ⋅ + ⋅ ⋅

Linearizationsinφ φ≈[ ]G G B B TF z F z g Iρ φ≈ ⋅ + ⋅ + ⋅ ⋅

In similar way, if FExt are applied, then FExt⋅ zExt would be included

[ ]G G B B T Ext ExtF z F z g I F zρ φ= ⋅ + ⋅ + ⋅ + ⋅ ⋅Match! except sign?

BF gVρ=

B GF F∴ = −

[ ] sinG G B B TF z F z g Iρ φ= ⋅ + ⋅ + ⋅ ⋅

24/120


Seoul NationalUniv.



* *3 4( , ) 0TM ξ ξ = * * * * * * * *


3 4 3 4 (0) (0) (0) (0) (0) (0)3 4 3 4 3 4

(0)4

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

( , ) ( , ) ( , )( , ) G G B B

T TT Ext Ext

F z F zM M



ξ ξ ξρ

− ⋅ − ⋅= + ∆ − − ⋅


Comparison with classical hydrostatics ( (+), (-) sign )

Defined that clockwise direction is positive.

In classical hydrostatics In pressure integration technique

Defined that counter-clockwise direction is positive.

≡

K

O'

CL

y

z,

-∆ξ4

G Z

B0 B1

≡ N

M

O


[ ] 4( )G G B B T Ext ExtF z F z g I F zρ ξ= ⋅ + ⋅ + ⋅ + ⋅ −∆⋅

[ ] ( )4G G B B T Ext ExtF z F z g I F zρ ξ= − ⋅ − ⋅ − ⋅ ⋅ ∆− ⋅Match!

if we consider opposite definition of heel angle ( )4φ ξ= −∆

+ +

≡

K

O'

CL

y

z,

+φG Z

B0 B1

≡ N

M

O

restoringτ restoringτ

[ ]G G B B T Ext ExtF z F z g I F zρ φ= ⋅ + ⋅ + ⋅ + ⋅ ⋅

25/120


Seoul NationalUniv.



m: Mass of the ship, zG: Z coordinate of the center of mass of the ship

IT: 2nd moment of Waterplane area with respect to x axis of the waterplane fixed frameV: Displacement volume



[ ] ( )4G G B B T Ext ExtF z F z g I F zρ ξ= − ⋅ − ⋅ − ⋅ ⋅ ∆− ⋅

But, external forces are not always negative or positive z direction.

ex) wind, current, mooring…

so, sign of FExt is determined, after a kind of FExt is determined.

Buoyancy force FB is always positive in z direction, so we could let BF gVρ=

and gravity force of ship FG is always negative in z direction, so GF mg= −

[ ] ( )4G G B B T Ext ExtF z F z g I F zρ ξ= − ⋅ − ⋅ − ⋅ ⋅ ∆− ⋅

Governing equation


* *3 4( , ) 0TM ξ ξ = * * * * * * * *


3 4 3 4 (0) (0) (0) (0) (0) (0)3 4 3 4 3 4

(0)4

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

( , ) ( , ) ( , )( , ) G G B B

T TT Ext Ext

F z F zM M



ξ ξ ξρ

− ⋅ − ⋅= + ∆ − − ⋅

26/120


Seoul NationalUniv.



<Notation>* * (0) (0) (0)

3 4 3 4 4(0) (0)3 3( , ) ( , ), ( , )z z WPwhere F F gAξ ξ ξ ξ ξρ ξ ξ= − ⋅ ∆


* *3 4( , ) 0,zF ξ ξ =

Given : Find : (0) (0) (0) (0) (0) (0) (0) (0)3 4 3 4 3 4 3 4, ( , ) ( , ) ( , ) ( , )z G B Extwh ere F F F Fξ ξ ξ ξ ξ ξ ξ ξ= + + *

3ξ(0) (0) (0) (0)3 4 3 4, , ( , )ExtFξ ξ ξ ξ

(0) (0)3 4( , )zF ξ ξ (0,0)

3,4( )zF ξ

<Definition of notation>

(2) (1) (1)3 4 5( , , )zF ξ ξ ξ (2,1,1)

3,4,5( )zF ξ (2,1,1)3,4,5( )zF ξ

Immersion2nd Step

Attitude

Heel1st Step Trim

1st Step

* (0,0) (0,0) (0)3,4 3,4 3,4 3( ) ( ) ( ), z z WPwhere F F gAξ ξ ξρ ξ= − ⋅∆


*3,4( ) 0,zF ξ =

Given : Find : (0,0) (0,0) (0,0) (0,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( ), , z G B Extwh ere F F F Fξ ξ ξ ξ= + + *

3ξ(0) (0) (0,0)3 4 3,4( ), , ExtF ξξ ξ


* * * *3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( ), T TG TB T Extwhere M M M Mξ ξ ξ ξ= + +Governing equation

of transverse moment*3,4( ) 0TM ξ =

Given : (0) (0) (0,0)3 4 , 3,4( ), , T ExtM ξξ ξ Find : *

4ξ

(0,0) (0,0) (0,0)3,4 3,4 3,4* (0,0) (0)

3,4 3,4 4(0,0) (0,0) (0,0)3,4 3,4 3,4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B B

T TT Ext Ext

mg z F zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξξ

ρ

⋅ − ⋅= + ∆ − − ⋅

By similar way, governing equation of transverse moment is below

27/120


Seoul NationalUniv.


ξ3 : immersionξ4 : heel



* * * *3,4 3,4 3,4 3,4( ) ( ) ( ) ( ), z G B Extwh ere F F F Fξ ξ ξ ξ= + +

Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

1

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

0 step, Immersion

(0,0) (0,0) (0,0) (0,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

: Total force

1

(0,0) 43,4( ) 4.0 10 [ ]ExtF w kNξ = − = − ×

y

z

LC40 m

30 m

9m=

Section view

,O O′Baseline

G

w

, z′

, y′g B

Plan view

x

y

LC

20m

,O O′G

w

, x′

, y′

, B

44.0 10w kN= ×

(0,0)3,4

(0,0)3,4

( )

( )BV

F g dVξ

ξ ρ= ⋅ ∫∫∫

(0,0) 53,4( ) 3.6 10 [ ]GF kNξ = − ×

5 5 43.6 10 [ ] 3.6 10 [ ] 4.0 10 [ ]kN kN kN= − × + × − ×44.0 10 [ ]kN= − ×

100m

(0,0)3,4( )ExtF wξ = −

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

Immersion(0)3ξ∆ 0 step

(0,0) (0,0) (0,0)3,4 3,4 3,4( ) ( ) ( )sg L B Tξ ξ ξρ= ⋅ ⋅ ⋅

10[ ] 100[ ] 40[ ] 9[ ]m m m m= ⋅ ⋅ ⋅53.6 10 [ ]kN= ×

(0,0)3,4( )T ξ

g(0,0) (0,0) (0,0) (0,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

28/120


Seoul NationalUniv.




Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

(0,0)3,4( )WPg A ξρ−


2

0 step, Immersion

2y

z

LC40[ ]m

9m

Section view

,O O′Baseline

G

w

, z′

, y′g B

Plan view

x

y

LCj i

20m,O O′ G

w, x′

, y′

, B

(0,0)3,4( )WPA

g dx dyξ

ρ= − ⋅ ∫∫

(0,0) (0,0)3,4 3,4( ) ( )sg B Lξ ξρ= − ⋅ ⋅

WPArea A=

WPArea A=

310[ / ] 40[ ] 100[ ]kN m m m= − ⋅ ⋅

44.0 10 [ / ]kN m= − ×

(0,0)3,4( ) 100L mξ =

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r


(0,0)3,4( ) 20sB mξ =


29/120


Seoul NationalUniv.




Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

y

z

LC

Section view

,O O′

, z′

y′Baselinew 1g

K


3

0 step, Immersion

y

z

LC

Section view

,O O′Baseline

G

w

, z′

, y′g B

(0,0) 43,4( ) 4.0 10 [ ]zF kNξ = − ×1

(0,0) 43,4( ) 4.0 10 [ / ]WPgA kN mξρ− = − ×2

* (0,0)3,4 3,4(0)

3 (0,0)3,4

( ) ( )

( )z z

WP

F FgA

ξ ξ

ξξ

ρ−

∆ =−

(0)3ξ∆3 : change of immersion

1 2

( ) ( ) ( )1 0 03 3 3 0 1 1 [ ]mξ ξ ξ= + ∆ = − = −



9m=

( )1*3 3ξ ξ=

? (1,0)3,4( ) 0zF ξ =

?10m

If we assume that Fz(ξ3,4*) =0

4

4

0[ ] ( 4.0 10 )[ ] 1[ ]4.0 10 [ ]

kN kN mkN

− − ×= = −

− ×

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r


1G

1B

(0,0)3,4( )T ξ


30/120


Seoul NationalUniv.




Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

Baseline

G

w 1g


0 step, Immersion

y

z

LC

Section view

,O O′

, z′

y′B

(1,0)3,4( ) 0zF ξ =


( )1*3 3ξ ξ=

(0)3ξ∆

(1,0)3,4( )BF ξ

(1,0)3,4( )GF ξ

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= ++

(1,0)3,4( )ExtF ξ (0,0) 4

3,4( ) 4.0 10 [ ]ExtF kNξ= = − ×

(0,0) 53,4( ) 3.6 10 [ ]GF kNξ= = − ×

(1,0)3,4( )g V ξρ= ⋅

3 4 310[ / ] 4.0 10 [ ]kN m m= × ×

(1, 0)3, 4

(1,0)3,4

( )

( )V

V dVξ

ξ = ∫∫∫

54.0 10 [ ]kN= ×

(1,0)3,4( )zF ξ (1,0) (1,0) (1,0)

3,4 3,4 3,4( ) ( ) ( )G B ExtF F Fξ ξ ξ= ++5 5 43.6 10 [ ] 4.0 10 [ ] 4.0 10 [ ]kN kN kN= − × + × − ×

0[ ]kN=

4(1,0)3,4( ) 0zF ξ = ( )1*

3 3ξ ξ=

If ξ4(1) is also ξ4

*, then the iteration ends.

4

Check ξ4(1)

10m=

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r


4 3100[ ] 40[ ] 10[ ] 4.0 10 [ ]m m m m= ⋅ ⋅ = ×44.0 10w kN= ×

(1,0)3,4( )ExtF wξ = −

?

?

(1,0)3,4( )T ξ

(1,0) (1,0) (1,0)3,4 3,4 3,4( ) ( ) ( )sL B Tξ ξ ξ= ⋅ ⋅

In this example


31/120


Seoul NationalUniv.


* * * *3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( ), T TG TB T Extwhere M M M Mξ ξ ξ ξ= + +


*3,4( ) 0TM ξ =


0 step

(0) (0) (0)3 4( [0,0, , ,0,0])ξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel

LC

O′

(0,0)3,4( )G ξ

w

,z z′

,y y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ

O -40,000 -360,000 360,000 -40,000

-400,000 0 0 -400,000


[ ]TM kN m⋅ [ ]TGM kN m⋅ [ ]TBM kN m⋅ , [ ]T ExtM kN m⋅


Governing equationof force

*3,4( ) 0zF ξ =

0 -360,000 400,000 -40,000

-400,000 0 0 -400,000



(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=rNext step

End of Immersion 0 step.

Start Heel 0 step.

(1,0)3,4( )G ξ

w(1,0)3,4( )g ξ

y

,z z′

LC

O

y′(1,0)3,4( )B ξ

(0)3ξ∆

O′

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r



32/120


Seoul NationalUniv.



y

z

LC

,O O′

(0,0)3,4( )G ξ

w

, z′

, y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ9[ ]m=

(0,0)3,4( )T ξ

y

,z z′

LC

O (0)3ξ∆

y′w(1,0)3,4( )g ξ

O′10m=

(1,0)3,4( )T ξ

(0,0)3,4(0,0)3,4(0,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

Transformation G, gto waterplane fixed frame

(0,0)3,4( ) (0,0,6)G ξ =

(0,0)3,4( ) (0, 10, 4.5)g ξ = − −

(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5.5)g ξ = − −

0 step, Heel(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

=

(0)3

00

ξ

+ ∆

(0,0)3,4(0,0)3,4(0,0) (0)3,4 3

( )( )( )

00

G

G

G

xyz

ξ

ξ

ξ ξ

= + ∆

(1,0)3,4(1,0)3,4(1,0)3,4

( )( )( )

G

G

G

xyz

ξ

ξ

ξ

(1) (0) (0)3 3 3ξ ξ ξ= + ∆

0 00 06 1

= + −

(0,0)3,4(0,0)3,4(0,0) (0)3,4 3

( )( )( )

00

g

g

g

xyz

ξ

ξ

ξ ξ

= + ∆

(1,0)3,4(1,0)3,4(1,0)3,4

( )( )( )

g

g

g

xyz

ξ

ξ

ξ

0 010 04.5 1

= − + − −

005

=

0105.5

= − −

(0,0)3,4( ) (0,0, 4.5)B ξ = − (1,0)

3,4( ) (0,0, 5)B ξ = −

(1,0)3,4( )B ξ

(1,0)3,4( )G ξ

For center of gravity of ship (1,0)3,4( )G ξ

For center of weight of cargo (1,0)3,4( )g ξ

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

Heel(0)4ξ∆ 0 step


to calculate (1,0)3,4( )TM ξCalculate (1,0) (1,0) (0,0)

3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ0 step, Heel

33/120


Seoul NationalUniv.



y

z

LC

,O O′

(0,0)3,4( )G ξ

w

, z′

, y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ9[ ]m=

(0,0)3,4( )T ξ

y

,z z′

LC

O (0)3ξ∆

y′w(1,0)3,4( )g ξ

O′10m=

(1,0)3,4( )T ξ

(0,0)3,4( ) (0,0,6)G ξ =

(0,0)3,4( ) (0, 10, 4.5)g ξ = − −

(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5.5)g ξ = − −

(1,0)3,4( )

00

/ 2T ξ

= −

(1,0)3,4(1,0)3,4(1,0)3,4

( )( )( )

B

B

B

xyz

ξ

ξ

ξ

00

10 / 2

= −

005

= −

(0,0)3,4( ) (0,0, 4.5)B ξ = − (1,0)

3,4( ) (0,0, 5)B ξ = −

(1,0)3,4( )B ξ

(1,0)3,4( )G ξ

For center of Buoyancy is calculated by(1,0)3,4( )B ξ

in this example, is located in center of draft.(1,0)3,4( )B ξ

(1,0) (1,0) (1,0), 3,4 , 3,4 , 3,4

(1,0) (1,0) (1,0)3,4 3,4 3,4

( ) ( ) ( )

( ) ( ) ( )( , , )V x V y V zM M M

V V Vξ ξ ξ

ξ ξ ξ

,V xM :1st moment of volume V about x axis in y direction in waterplane fixed frame

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r



0 step, Heel

(0,0)3,4(0,0)3,4(0,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


0 step, Heel(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

=

(0)3

00

ξ

+ ∆ (1) (0) (0)

3 3 3ξ ξ ξ= + ∆


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ

34/120


Seoul NationalUniv.



Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )T B

T TG Ext Ext

gI g V zM M

mg z F z

ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

− − ⋅ ⋅= + ∆ + ⋅ − ⋅

Governing equation of transverse moment

*3,4( ) 0TM ξ =

1

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( )T TG TB T ExtM M M Mξ ξ ξ ξ= + +1

(1,0) (1,0) (1,0)3,4 3,4 3,4

5

( ) ( ) ( )

0[ ] ( 3.6 10 )[ ] 0 [ ]TG G GM y F

m kN kN

ξ ξ ξ=

= × − × =(1,0) (1,0) (1,0)3,4 3,4 3,4

5

( ) ( ) ( )

0[ ] (4.0 10 )[ ] 0[ ]TB B BM y F

m kN kN

ξ ξ ξ=

= × × =(1,0) (1,0) (1,0)

, 3,4 3,4 3,4

(1,0) (1,0)3,4 3,4

4 5

( ) ( ) ( )

( ) ( )

[ ] [ ] [ ]10 ( 4.0 10 ) 4.0 10

T Ext Ext Ext

g Ext

m kN kN m

M y F

y F

ξ ξ ξ

ξ ξ

⋅

=

=

= − × − × = ×

(1,0)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×(1,0) 43,4( ) 4.0 10 [ ]ExtF kNξ = − ×

(1,0) 53,4( ) 3.6 10 [ ]G NF kξ = − ×

5

5

0 0 4.0 10 [ ]4.0 10 [ ]

kN mkN m

= + + × ⋅

= × ⋅

0 step, Heel

Baseline

G

w g

y

,z z′

LC

Section view

O

y′B

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5.5)g ξ = − −

(1,0)3,4( ) (0,0, 5)B ξ = −

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( )T TG TB T ExtM M M Mξ ξ ξ ξ= + +

(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


35/120


Seoul NationalUniv.


Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =


0 step, Heel2

(1, 0)3, 4

(1,0) (1,0) 33,4 3,4(1,0) 2

3,4( )

( ) ( )( )

12WP

sT

A

L BgI g y dx dy g

ξ

ξ ξξρ ρ ρ= =∫∫

(1, 0)3, 4

(1,0) (1,0) (1,0)3,4 3,4 3,4

( )

( ) ( ) ( )B BV

gV z g dV zξ

ξ ξ ξρ ρ⋅ = ⋅ ⋅∫∫∫

(1,0) 53,4( ) 3.6 10 [ ] 5[ ]Gmg z kN mξ⋅ = × ×

m: Mass of the shipzG: Z coordinate of the center of mass of the ship


V: Displacement volume


2

33 6100[ ] 40[ ]10[ / ] 5.333 10 [ ]

12[ ]m mkN m kN m

m×

= × = × ⋅

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )s Bg L B T zξ ξ ξ ξρ= ⋅ ⋅ ⋅ ⋅

310[ / ] 100[ ] 40[ ] 10[ ] ( 5[ ])kN m m m m m= × × × × −62.0 10 [ ]kN m= − × ⋅

61.8 10 [ ]kN m= × ⋅

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4

5( ) 4.0 10 [kN]BF ξ = ×(1,0) 43,4( ) 4.0 10 [kN]ExtF ξ = − ×

(1,0) 53,4( ) 3.6 10 [kN]GF ξ = − ×

(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5.5)g ξ = − −

(1,0)3,4( ) (0,0, 5)B ξ = −

,y y′

LC

Plan view

(1,0)3,4( )

40SB

m

ξ

=

,O O′ G

w, B

(1,0)3,4( ) 100L mξ =

,x x′

g

Baseline

G

w g

y

,z z′

LC

Section view

O

y′B(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


36/120


Seoul NationalUniv.



0 step, Heel

m: Mass of the shipzG: Z coordinate of the center of mass of the ship




2(1,0)3,4( ), (0, 10, 5.5)g ξ = − −

(1,0) (1,0)3,4 3,4( ) ( )Ext ExtF zξ ξ⋅

(1,0) (1,0) (1,0)3,4 3,4 3,4

(1,0) (1,0) (1,0)3,4 3,4 3,4

( ) ( ) ( )

( ) ( ) ( )G B

T Ext Ext

mg z g V z

gI F z

ξ ξ ξ

ξ ξ ξ

ρ

ρ

⋅ − ⋅ ⋅

− − ⋅

6 6

6 5

6

1.8 10 [ ] 2.0 10 [ ]5.333 10 [ ] 2.2 10 [ ]1.7533 10 [ ]

kN m kN mkN m kN mkN m

= × ⋅ + × ⋅

− × ⋅ − × ⋅

= − × ⋅

(1,0) 43,4( ) 4.0 10 [ ]ExtF kNξ = − ×

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0) (1,0)3,4 3,4( ) ( )Ext gF zξ ξ= ⋅

( )44.0 10 [ ] 5.5 [ ]kN m= − × × −52.2 10 [ ]kN m= × ⋅

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

2

Baseline

G

w g

y

,z z′

LC

Section view

O

y′B(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′

,y y′

LC

Plan view

(1,0)3,4( )

40SB

m

ξ

=

,O O′ G

w, B

(1,0)3,4( ) 100L mξ =

,x x′

g


37/120


Seoul NationalUniv.


Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =


0 step, Heel2 3

1

1(1,0) 53,4( ) 4.0 10 [ ]TM kN mξ = × ⋅

2 (1,0) (1,0) (1,0)3,4 3,4 3,4

(1,0) (1,0) (1,0)3,4 3,4 3,4

6

( ) ( ) ( )

( ) ( ) ( )

1.7533 10 [ ]

G B

T Ext Ext

mg z g V z

gI F z

kN m

ξ ξ ξ

ξ ξ ξ

ρ

ρ

⋅ − ⋅ ⋅

− − ⋅

= − × ⋅* (1, 0)

3,4 3, 4

(1,0) (1,0) (1,0)3,4 3,4 3,4

(1,0) (1,0) (1,0)3,4 3,4 3,4

(0)4

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

T T

G B

T Ext Ext

M M

mg z g V z

gI F z

ξ ξ

ξ ξ ξ

ξ ξ ξ

ξρ

ρ

−=

∆ ⋅ − ⋅ ⋅ − − ⋅

3



( )1*4 4ξ ξ=

? (1,1)3,4( ) 0TM ξ =

?

5

6

0[ ] 4.0 10 [ ] 0.23rad 13.071.7533 10 [ ]

kN m kN mkN m

⋅ − × ⋅= = =

− × ⋅

If we assume that MT(ξ3,4*) =0

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


y

z

LC

G y′

g B

(0)4ξ∆

w

O

z′

O′

(1,1)3,4( )G G ξ=(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1) (0) (0)4 4 4 0 13.07 13.07ξ ξ ξ= + ∆ = + ° = °

Baseline

G

w g

y

,z z′

LC

Section view

O

y′B(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


38/120


Seoul NationalUniv.



y

,z z′

LC

OO′

y′w g10m

immersion(0,0)3,4(0,0)3,4(0,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

(1,0)3,4( ) (0,0,5)G ξ =(1,0)3,4( ) (0, 10, 5.5)g ξ = − −

0 step, Immersion(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

=

(0)3

00

ξ

+ ∆

(1,0)3,4

(0) (0) (1,0)4 4 3,4(0) (0) (1,0)4 4 3,4

( )( )( )

1 0 00 cos sin0 sin cos

G

G

G

xyz

ξ

ξ

ξ

ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(1,1)3,4(1,1)3,4(1,1)3,4

( )( )( )

G

G

G

xyz

ξ

ξ

ξ

(1) (0) (0)3 3 3ξ ξ ξ= + ∆

01.13

4.87

= −

(1,1)3,4(1,1)3,4(1,1)3,4

( )( )( )

g

g

g

xyz

ξ

ξ

ξ

(0) (0)4 4(0) (0)4 4

1 0 0 00 cos sin 100 sin cos 5.5

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆ −

08.5

7.62

= − −

B

G

(1,0)3,4( ) (0,0, 5)B ξ = −

0 step, Heel

(0) (0)4 4(0) (0)4 4


ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(1,1)3,4(1,1)3,4(1,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


(1) (0) (0)4 4 4ξ ξ ξ= + ∆

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


heel

(0) (0)4 4(0) (0)4 4

1 0 0 00 cos sin 00 sin cos 5

ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(1,1)3,4( ) (0, 1.13,4.87)G ξ = −

(1,1)3,4( ) (0, 8.5, 7.62)g ξ = − −

(1,0)3,4

(0) (0) (1,0)4 4 3,4(0) (0) (1,0)4 4 3,4

( )( )( )


g

g

g

xyz

ξ

ξ

ξ

ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(0)4, ( 13.07 )ξ∆ =

(0)4, ( 13.07 )ξ∆ =

Next

y

z

G

z′

y′

gB

(0)4ξ∆

w

O

(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

(1,1)3,4( )G G ξ=(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=


O′

For center of gravity of ship, (1,1)3,4( )G ξ

For center of weight of cargo, (1,1)3,4( )g ξ



(1,1) (1,1) (1,1), 3,4 , 3,4 , 3,4

(1,1) (1,1) (1,1)3,4 3,4 3,4

( ) ( ) ( )

( ) ( ) ( )( , , )V x V y V zM M M

V V Vξ ξ ξ

ξ ξ ξ,V xM

:1st moment of volume about x axisin y direction in waterplane fixedframe

in this example, is simply calculated as next page(1,1)3,4( )B ξ


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ

39/120


Seoul NationalUniv.


Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

y

z,z z′ ′′

y′

(0)4 13.07ξ∆ =

(0)4ξ∆

A2

A3

A1

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ

(1,1) (1,1)3,4 3,4( ) ( )( , )c cc y zξ ξCentroid of section area

=

Area of A1 in O''-x''y''z'' frame

1A sArea B T′′ ′′= ⋅ 240[ ] 10[ ] 400[ ]m m m= ⋅ =

10[ ]T m′′ =Draft in O''-x''y''z'' frame

(0)3ξ∆

T ′′

sB′′

2 3( )A AArea Area=

{ }0.5 (40 / 2[ ]) (40 / 2)[ ] tan(13.07)m m= ⋅ ⋅ ⋅246.44[ ]m=

Area of A3 (=A2)in O''-x''y''z'' frame

A3

/ 2sB′′

(0)4( / 2) tan( )sB ξ′′ ⋅ ∆

(0)4ξ∆

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B ξ

{ }(0)40.5 ( / 2) ( / 2) tan( )s sB B ξ′′ ′′= ⋅ ⋅ ∆

y

z

G

z′

y′

gB

(0)4ξ∆

w

OO′

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=

Section view

Calculate centroid of section area in O''-x''y''z'' frame,then transformation centroid of section area to waterplane fixed frame.


Define new frame O''-x''y''z''Section view

y′′

<Transverse moment of >(1,1)3,4( )ξ

,O O′′

O′

(1,1)3,4of ξ

40/120


Seoul NationalUniv.



0 step, Heel

3 3( , )A Ay z′′ ′′ (1)

42 1, tan( )

2 3 2 3s sB B ξ′′ ′′ = ⋅ − ∆ ⋅

( )(40 / 2) (2 / 3), (40 / 2) tan(13.07) (1/ 3)= ⋅ − ⋅ ⋅

Centroid of A3 in O''-x''y''z'' frame,


1 1( , )A Ay z′′ ′′

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


( )13.33, 1.55= −Centroid of A2 in O''-x''y''z'' frame,

2 2( , )A Ay z′′ ′′in similar way ( )13.33,1.55= −


( )1*4 4ξ ξ=

?

A3

/ 2sB′′

(0)4

( / 2)

tan( )sB

ξ

′′

⋅ ∆

(0)4ξ∆

3Ay′′ 3Az′′

10(0, ) (0, ) (0, 5)2 2

T ′′= − = − = −

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

4

y

z

G

z′

y′

gB

(0)4ξ∆

w

OO′

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=

Section view

(1,1)3,4( ) 0TM ξ =

?



(1,1)3,4of ξ

y

z,z z′ ′′

y′

(0)4 13.07ξ∆ =

(0)4ξ∆

A2

A3

A1

(0)3ξ∆

T ′′

B′′


y′′,O O′′

O′

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(1,1)3,4( )B ξ


41/120


Seoul NationalUniv.



0 step, Heel

A1 400.00 0.00 -5.00 0.00 -2000.00

A2 46.44 -13.33 1.55 -619.14 71.88

A3 -46.44 13.33 -1.55 -619.14 71.88

Sum 400.00 -1,238.29 -1,856.25

3 3

, ,1 1

3 3

1 1

( , )

,Ai Ai

c c

A y A zi i

Ai Aii i

y z

M M

Area Area

′′ ′′= =

= =

′′ ′′ =

∑ ∑

∑ ∑

, AiA yM ′′ :1st moment of area Ai about x axis in y direction in O''-x''y''z'' frame

, AiA zM ′′ :1st moment of area Ai about y axis in z direction in O''-x''y''z'' frame

( )( , ) 0, 3.1, 4.64c cy z′′ ′′∴ = − −

iAy′′

, ( 1, 2,3)i =

,( )

AiA yM

′′

iAi AArea y′′×,

( )AiA z

M′′

iAi AArea z′′×

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


1,238.29 1,856.25,400 400

− − =


( )1*4 4ξ ξ=

?


AiArea

Centroid of section area in O''-x''y''z'' frame,

( )3.1, 4.64= − −

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

4

y

z

G

z′

y′

gB

(0)4ξ∆

w

OO′

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=

Section view

(1,1)3,4( ) 0TM ξ =

?



(1,1)3,4of ξ

y

z,z z′ ′′

y′

(0)4 13.07ξ∆ =

(0)4ξ∆

A2

A3

A1

(0)3ξ∆

T ′′

B′′


y′′,O O′′

O′

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(1,1)3,4( )B ξ


iAz′′

42/120


Seoul NationalUniv.



0 step, Heel

( )( , ) 3.1, 4.64c cy z′′ ′′∴ = − −

(1,1)3,4(1,1)3,4(1,1)3,4

( )( )( )

c

c

c

xyz

ξ

ξ

ξ

=

(1,1)3,4( ) (0, 1.97, 5.22)B ξ∴ = − −

(0) (0)4 4(0) (0)

4 4

1 0 0 00 cos sin 3.10 sin cos 4.64

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆ −

(0)4, ( 13.07 )ξ∆ =

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )1*4 4ξ ξ=

?

Calculate centroid of section area in O''-x''y''z'' frame,

then transformation centroid of section area to waterplane fixed frame.

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )c c cx y zξ ξ ξ

01.975.22

= − −

(0) (0)4 4(0) (0)

4 4


c

c

c

xyz

ξ ξξ ξ

′′ ′′= ∆ − ∆

′′ ∆ ∆

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

4

y

z

G

z′

y′

gB

(0)4ξ∆

w

OO′

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=

Section view

(1,1)3,4( ) 0TM ξ =

?

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , ) (0, 1.97, 5.22)c c cx y zξ ξ ξ∴ = − −


(1,1)3,4of ξ

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(1,1)3,4( )B ξ


y

z,z z′ ′′

y′

(0)4 13.07ξ∆ =

(0)4ξ∆

A2

A3

A1

(0)3ξ∆

T ′′

B′′


y′′,O O′′

O′

43/120


Seoul NationalUniv.



(1,1)3,4( )BF ξ

(1,1)3,4( )GF ξ

(1,1)3,4( )ExtF ξ (0,0)

3,44( ) 4.0 10 [ ]ExtF kNξ= = − ×

(0,0)3,4

5( ) 3.6 10 [ ]GF kNξ= = − ×

3 4 310[ / ] 4.0 10 [ ]kN m m= × ×

(1,1)3,4( )gV ξρ=

54.0 10 [ ]kN= ×

(1,1)3,4

(1,1)3,4

( )

( )V

V dVξ

ξ = ∫∫∫

0 step, Heel

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )1*4 4ξ ξ=

?

(1,1)3,4( )section area L ξ= ×

1

2[ ]400A mArea =

2

2[ ]46.44A mArea =

3

2[ ]46.44A mArea = −2

_ [ ]400A Total mArea =

2 4 3400[ ] 100[ ] 4.0 10 [ ]m m m= × = ×

(1,1)3,4_ ( )Area LA Total ξ= ×

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

4

y

z

G

z′

y′

gB

(0)4ξ∆

w

OO′

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=

Section view

(1,1)3,4( ) 0TM ξ =

?

Calculate force (1,1) (1,1) (1,1)3,4 3,4 3,4( ), ( ), ( )G B ExtF F Fξ ξ ξ



44/120


Seoul NationalUniv.



0 step, Heel

stImmersion 1 step

<Transverse moment of (ξ3(1),ξ4

(1)) >

(1,1)3,4

(1,1) (1,1)3,4 3,4( ) () )(TB B BM y Fξ ξξ= ⋅

(1,1)3,4

(1,1) (1,1)3,4 3,4( ) () )(TG G GM y Fξ ξξ= ⋅

(1,1) (1,1) (1,1)3,4 3,4 3,4, ( ) ( ) ( )T Ext Ext ExtM y Fξ ξ ξ= ⋅

(1,1)3,4( )TM ξ

4

(1,1)3,4( ) 0TM ξ ≠ ( )1*

4 4ξ ξ≠

(1,1)3,4( ) (0, 1.13,4.87)G ξ = −(1,1)3,4( ) (0, 1.97, 5.22)B ξ = − −(1,1)3,4( ) (0, 8.5, 7.62)g ξ = − −

(1,1)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×

(1,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×

(1,1)3,4

4( ) 4.0 10 [ ]ExtF kNξ = − ×

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,1)3,4

(1,1)3,4

(1,1)3,4 ,( () )( ) T ExBG tT TM M Mξ ξξ= ++


( )1*4 4ξ ξ=

?

5( 1.13) ( 3.6 10 )= − ⋅ − ×54.0709 10 [ ]kN m= × ⋅

5( 1.97) (4.0 10 )− ⋅ ×=57.8638 10 [ ]kN m= − × ⋅

(1,1) (1,1)3,4 3,4( ) ( )g Exty Fξ ξ= ⋅

4( 8.5) ( 4.0 10 )= − ⋅ − ×53.3988 10 [ ]kN m= × ⋅

[ ]kN m⋅5 5 54.0709 10 [ ] 7.8638 10 [ ] 3.3988 10kN m kN m= × ⋅ − × ⋅ + ×43.9415 10 [ ]kN m= − × ⋅

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0)3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

4

y

z

G

z′

y′

gB

(0)4ξ∆

w

OO′

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=

Section view

(1,1)3,4( ) 0TM ξ =

?


45/120


Seoul NationalUniv.



0 step

(0) (0) (0)3 4( [0,0, , ,0,0])ξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel

-40,000 -360,000 360,000 -40,000

-400,000 0 0 -400,000



0 -360,000 400,000 -40,000

-400,000 0 0 -400,000



(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=r 0 -360,000 400,000 -40,000

-39,415 407,094 -786,388 339,880



Start Immersion 1st step.End of Heel 0 step.

Immersion

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


y

z

G

z′

y′

g B

( )04ξ∆

w

OO′

1st step



*3,4( ) 0TM ξ =



*3,4( ) 0zF ξ =

LC

O′

(0,0)3,4( )G ξ

w

,z z′

,y y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ

O

(1,0)3,4( )G ξ

w(1,0)3,4( )g ξ

y

,z z′

LC

O

y′(1,0)3,4( )B ξ

(0)3ξ∆

O′


46/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

w

OO′

Section view* (1,1) (1,1) (1)3,4 3,4 3,4 3( ) ( ) ( ), z z WPwhere F F gAξ ξ ξρ ξ= − ⋅∆


Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


1st step, Immersion

(1,1)3,4( )BF ξ

(1,1)3,4( )GF ξ

(1,1) (1,1) (1,1) (1,1)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

(1,1)3,4( )ExtF ξ 44.0 10 [ ]kN= − ×

53.6 10 [ ]kN= − ×

3 4 310[ / ] 4.0 10 [ ]kN m m= × ×

(1,1)3,4( )gV ξρ=

54.0 10 [ ]kN= ×

(1,1) (1,1) (1,1) (1,1)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

5 5 43.6 10 [ ] 4.0 10 [ ] 4.0 10 [ ]kN kN kN= − × + × − ×0=

Total force is not changed during heel.

1

1

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


47/120


Seoul NationalUniv.




Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=

1


* (1,1)3,4 3,4

(1,1)3,4

(1)3

( ) ( )

( )z z

WP

F Fg A

ξ ξ

ξξ

ρ−

∆ =− ⋅

(1,1)3,4( ) 0zF ξ =1

3

3

(1)3 0ξ∆ =

4

(2 ,1)3,4( ) 0zF ξ = ( )2*

3 3ξ ξ=


*, then iteration end.

Check ξ4(2)

4

(2) (1) (1) (1)3 3 3 3ξ ξ ξ ξ= + ∆ =

(2 ,1) (1,1)3,4 3,4( ) ( ) 0z zF Fξ ξ= =


1st step, Immersion

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(1,1)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×(1,1)3,4

4( ) 4.0 10 [ ]ExtF kNξ = − ×

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 4.0 10 [ ]ExtF kNξ = − ×

(1,1)3,4( )

0 (0) 0WPg A ξρ

−= =

− ⋅

*3ξ

y

z

G

z′

y′

gB

w

OO′

Section view

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


48/120


Seoul NationalUniv.


0 -360,000 400,000 -40,000

-39,415 407,094 -786,388 339,880


1st step

2nd step

(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0])ξ ξ=r

(2) (2) (2)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel



0 -360,000 400,000 -40,000

-39,415 407,094 -786,388 339,880



End of Immersion 1st step.

Start Heel 1st step.

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r




*3,4( ) 0TM ξ =



*3,4( ) 0zF ξ =

y

z

(1,1)3,4( )G ξ

z′

y′

(1,1)3,4( )g ξ

(1,1)3,4( )B ξ

w

OO′

y

z

(2,1)3,4( )G ξ

z′

y′

(2,1)3,4( )g ξ

(2,1)3,4( )B ξ

w

OO′


49/120


Seoul NationalUniv.



0 step, Heel

(1) (0) (0)4 4 4ξ ξ ξ= + ∆

1st step, Immersion

(0) (0)4 4(0) (0)4 4


ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(1,1)3,4(1,1)3,4(1,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


(2) (1) (1)3 3 3ξ ξ ξ= + ∆

heel

Immersion

(1)3

00ξ

+ ∆

(2,1)3,4(2,1)3,4(2,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

=

(1,1)3,4(1,1)3,4(1,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

For center of gravity of ship, G (1)3, ( 0)ξ∆ =

(1,1)3,4(1,1)3,4(1,1) (1)3,4 3

( )

( )

( )

00

G

G

G

xyz

ξ

ξ

ξ ξ

= + ∆

(2,1)3,4(2,1)3,4(2,1)3,4

( )

( )

( )

G

G

G

xyz

ξ

ξ

ξ

0 01.13 0

4.87 0

= − +

01.13

4.87

= −

(1,1)3,4( ) (0, 1.13,4.87)G ξ = −

(1,1)3,4( ) (0, 1.97, 5.22)B ξ = − −

(1,1)3,4( ) (0, 8.5, 7.62)g ξ = − −

(2,1)3,4( ) (0, 1.13,4.87)G ξ = −

(2,1)3,4( ) (0, 1.97, 5.22)B ξ = − −

(2,1)3,4( ) (0, 8.5, 7.62)g ξ = − −

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


For center of gravity of weight, g (1)3, ( 0)ξ∆ =

(1,1)3,4(1,1)3,4(1,1) (1)3,4 3

( )

( )

( )

00

g

g

g

xyz

ξ

ξ

ξ ξ

= + ∆

(2,1)3,4(2,1)3,4(2,1)3,4

( )

( )

( )

g

g

g

xyz

ξ

ξ

ξ

0 08.5 0

7.62 0

= − + −

08.5

7.62

= − −

(1,1) (2,1)3,4 3,4( ) ( )(0, 1.97, 5.22)B Bξ ξ= − − =


y

z

G

z′

y′

gB

w

O

(1,1)3,4( )G G ξ=(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

O′

y

z

G

z′

y′

gB

w

O

(2,1)3,4( )G G ξ=(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

O′



(1,1) (1,1) (1,1), 3,4 , 3,4 , 3,4

(1,1) (1,1) (1,1)3,4 3,4 3,4

( ) ( ) ( )

( ) ( ) ( )( , , )V x V y V zM M M

V V Vξ ξ ξ

ξ ξ ξ,V xM

:1st moment of volume about x axisin y direction in waterplanefixed frame

in this step, is not changed(1,1)3,4( )B ξ


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ1st step, Heel

50/120


Seoul NationalUniv.


Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =


1st step, Heel

-Total transverse moment

(2,1)3,

(2,1) (2,1)3,4 3

(2,1)3 ,4,44 ,( ) (( )( ) )TT T ExBTG tM MMMξ ξξ ξ+= +1

,(in previous calculation)

2

(2,1) (2,1)3,4 3,4( ) ( )BgV zξ ξρ ⋅

x

y

LC

Plan view

(2,1)3,4( )sB ξ

O

G

w

(2,1)3,4( ) 100L mξ =

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


12

(2,1)3,4( ), (0, 1.13,4.87)G ξ = −(2,1)3,4( ), (0, 1.97, 5.22)B ξ = − −(2,1)3,4( ), (0, 8.5, 7.62)g ξ = − −

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 4.0 10 [ ]ExtF kNξ = − ×

(2,1)3,4( )TM ξ (1,1)

3,4( )TM ξ=

43.9415 10 [ ]kN m= − × ⋅

54.0 10 [ ] 5.22[ ]kN m= − × ×

62.0882 10 [ ]kN m= − × ⋅

y

z

G

z′

y′

gB

(0)4ξ∆

w

OO′

Section view

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


(2,1)3,4( )Gmg z ξ⋅

53.6 10 [ ] 4.8704[ ]kN m= × ×

61.7533 10 [ ]kN m= × ⋅

51/120


Seoul NationalUniv.



(2,1) (2,1)3,4 3,4( ) ( )Ext ExtF zξ ξ⋅

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1) (2,1) (2,1)3,4 3,4 3,4

( ) ( ) ( )( ) ( ) ( )

G B

T Ext Ext

mg z g V zgI F z

ξ ξ ξξ ξ ξ

ρρ

⋅ − ⋅ ⋅− − ⋅

6 6

6 5

6

1.7533 10 [ ] 2.0882 10 [ ]5.7704 10 [ ] 3.0477 10 [ ]

2.2335 10 [ ]

kN m kN mkN m kN m

kN m

= × ⋅ + × ⋅− × ⋅ − × ⋅

= − × ⋅

2

1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


( )44.0 10 [ ] 7.62 [ ]kN m= − × × −

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

12

x

y

LC

Plan view

(2,1)3,4( )sB ξ

O

G

w

(2,1)3,4( ) 100L mξ =

y

z

G

z′

y′

gB

w

OO′

Section view

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 4.0 10 [ ]ExtF kNξ = − ×


(2,1)3,4( ), (0, 1.13,4.87)G ξ = −(2,1)3,4( ), (0, 1.97, 5.22)B ξ = − −(2,1)3,4( ), (0, 8.5, 7.62)g ξ = − −

(2,1) (2,1)3,4 3,4(2,1)

3,4

3( ) ( )( )

12s

T

L BgI g

ξ ξξρ ρ=

( )( )(1,0) (1,0)3,4 3,4

304( ) ( ) / cos

12sL B

gξ ξ ξ

ρ∆

=

( )33 100[ ] 40[ ] / cos13.07

10[ / ]12

m mkN m

× °= ×

65.7704 10 [ ]kN m= × ⋅

53.0477 10 [ ]kN m= × ⋅

52/120


Seoul NationalUniv.


Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =



[ ]* (2,1)3,4 3,4(1)

4 (2,1) (2,1) (2,1)3,4 3,4 3,4

(2,1) (2,1) (2,1)3,4 3,4 3,4

( ) ( )( ) ( ) ( )

( ) ( ) ( )

T T

G B

T Ext Ext

M Mmg z gV z

gI F z

ξ ξξ

ξ ρ ξ ξ

ρ ξ ξ ξ

−∆ =

⋅ − ⋅ − − ⋅

(2) (1) (1)4 4 4ξ ξ ξ= + ∆

3



( )2*4 4ξ ξ=

? (2,2)3,4( ) 0TM ξ =

?

1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


12 3

4

6

0[ ] ( 3.9415 10 [ ])2.2335 10 [ ]

kN m kN mkN m

⋅ − − × ⋅=

− × ⋅0.0176 rad 1.01= − = −

13.07 1.01= −

12.06=

x

y

LC

Plan view

(2,1)3,4( )sB ξ

O

G

w

(2,1)3,4( ) 100L mξ =

y

z

G

z′

y′

gB

w

OO′

Section view

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 4.0 10 [ ]ExtF kNξ = − ×


(2,1)3,4( ), (0, 1.13,4.87)G ξ = −(2,1)3,4( ), (0, 1.97, 5.22)B ξ = − −(2,1)3,4( ), (0, 8.5, 7.62)g ξ = − −

53/120


Seoul NationalUniv.



(2,2)3,4( ) (0, 1.04,4.89)G ξ = −

(2,2)3,4( )B ξ =(2,2)3,4( ) (0, 8.63, 7.74)g ξ = − −

z

G

z′

g

y

w

O

BO′

y′

1 step, Immersion

1 step, Heel

(1) (1)4 4(1) (1)4 4


ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(2,2)3,4(2,2)3,4(2,2)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

(2,1)3,4(2,1)3,4(2,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


(2) (1) (1)4 4 4ξ ξ ξ= + ∆

heel

(2,1)3,4

(1) (1) (2,1)4 4 3,4(1) (1) (2,1)4 4 3,4

1 0 0 ( )0 cos sin ( )0 sin cos ( )

G

G

G

xyz

ξξ ξ ξξ ξ ξ

= ∆ − ∆ ∆ ∆

(2,2)3,4(2,2)3,4(2,2)3,4

( )( )( )

G

G

G

xyz

ξξξ

01.04

4.89

= −

For center of gravity of ship G

For center of weight of cargo g(2,2)3,4(2,2)3,4(2,2)3,4

( )( )( )

g

g

g

xyz

ξξξ

(1) (1)4 4(1) (1)4 4

1 0 0 00 cos sin 8.50 sin cos 7.62

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆ −

08.637.74

= − −

(1) (1)4 4(1) (1)4 4

1 0 0 00 cos sin 1.130 sin cos 4.87

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆

(2,1)3,4

(1) (1) (2,1)4 4 3,4(1) (1) (2,1)4 4 3,4

1 0 0 ( )0 cos sin ( )0 sin cos ( )

g

g

g

xyz

ξξ ξ ξξ ξ ξ

= ∆ − ∆ ∆ ∆

(1)4, ( 1.01 )ξ∆ = −

(1)4, ( 1.01 )ξ∆ = −

Next

y

z

G

z′

y′

gB

(1)4ξ∆

w

O

(2,1)3,4( )G G ξ=(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

O′

(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=


(2,2) (2,2) (2,2), 3,4 , 3,4 , 3,4

(2,2) (2,2) (2,2)3,4 3,4 3,4

( ) ( ) ( )

( ) ( ) ( )( , , )V x V y V zM M M

V V Vξ ξ ξ

ξ ξ ξ,V xM

:1st moment of volume about x axisin y direction in waterplanefixed frame


(2,1)3,4( ) (0, 1.13,4.87)G ξ = −(2,1)3,4( ) (0, 1.97, 5.22)B ξ = − −(2,1)3,4( ) (0, 8.5, 7.62)g ξ = − −


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



54/120


Seoul NationalUniv.


Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =


4

1st step, Heel

A1 400.00 0.00 -5.00 0.00 -2000.00

A2 42.73 -13.33 1.42 -569.75 60.86

A3 -42.73 13.33 -1.42 -569.75 60.86

Sum 400.00 -1,139.49 -1,878.27

3 3

, ,1 1

3 3

1 1

( , )

,Ai Ai

c c

A y A zi i

Ai Aii i

y z

M M

Area Area

′′ ′′= =

= =

′′ ′′ =

∑ ∑

∑ ∑

(2)4 12.06ξ =

, AiA yM ′′ :1st moment of area Ai about x axis in y direction in O''-x''y''z'' frame,

, AiA zM ′′ :1st moment of area Ai about x axis in z direction in O''-x''y''z'' frame,

( )( , ) 2.85, 4.7c cy z′′ ′′∴ = − −

iAy′′iAz′′

,( )AiA yM

i iA AArea y′′×,( )

AiA zMi iA AArea z′′×

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?

iAArea

Centroid of section area inO''-x''y''z'' frame,

( )0, 2.85, 4.7= − −

, ( 1, 2,3)i =

1,139.49 1,878.27,400 400

− − =

z

G

z′

g

y

w

O

BO′

y′( )14ξ∆

(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

(2,2)3,4( )G G ξ=

<Transverse moment of >(2,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,2)3,4( )B ξ



(2,2)3,4( )ξ

y

z,z z′ ′′

y′(2)

4ξA2

A3

A1

(0)3ξ∆

T ′′

B′′


y′′,O O′′

O′

(2,2)3,4of ξ

55/120


Seoul NationalUniv.


z

G

z′

g

y

w

O

BO′

y′( )14ξ∆


1st step, Heel

(2,2)3,4(2,2)3,4(2,2)3,4

( )( )( )

c

c

c

xyz

ξξξ

=

(2,2)3,4( ) (0, 1.8, 5.19)B ξ∴ = − −

(2) (2)4 4(2) (2)

4 4

1 0 0 00 cos sin 2.850 sin cos 4.7

ξ ξξ ξ

= − − −

(2)4 1, ( )2.06ξ =

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?

01.825.19

= − −

(2) (2)4 4(2) (2)

4 4


c

c

c

xyz

ξ ξξ ξ

′′ ′′= −

′′

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

(2,2)3,4( )G G ξ=

( )( , ) 2.85, 4.7c cy z′′ ′′∴ = − −Calculate centroid of section area in body fixed frame,


(2 ,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )c c cx y zξ ξ ξ

<Transverse moment of >(2,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,2)3,4( )B ξ

(2 ,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , ) (0, 1.8, 5.19)c c cx y zξ ξ ξ = − −


(2,2)3,4( )ξ

(2)4 12.06ξ =

y

z,z z′ ′′

y′(2)

4ξA2

A3

A1

(0)3ξ∆

T ′′

B′′


y′′,O O′′

O′

(2,2)3,4of ξ

56/120


Seoul NationalUniv.



1st step, Heel

(2,2)3,4( )BF ξ

(2,2)3,4( )GF ξ

(2,2)3,4( )zF ξ

(2,2)3,4( )ExtF ξ (0,0)

3,44( ) 4.0 10 [ ]ExtF kNξ= = − ×

(0,0)3,4

5( ) 3.6 10 [ ]GF kNξ= = − ×

3 4 310[ / ] 4.0 10 [ ]kN m m= × ×

(2,2)3,4( )gV ξρ=

54.0 10 [ ]kN= ×

<Transverse moment of >

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?

( 2 ,2)3,4

(2,2)3,4

( )

( )V

V dVξ

ξ = ∫∫∫(2,2)3,4( )section area L ξ= ×

1

2[ ]400A mArea =

2

2[ ]43.13A mArea =

3

2[ ]43.13A mArea = −2

_ [ ]400A Total mArea =

2 2 4 2400[m ] 100[m ] 4.0 10 [ ]m= × = ×

(2,2)3,4_ ( )Area LA Total ξ= ×

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

z

G

z′

g

y

w

O

BO′

y′( )14ξ∆

(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

(2,2)3,4( )G G ξ=


(2,2)3,4( ) (0, 1.8, 5.19)B ξ = − −


(2,2)3,4( )ξ

(2,2)3,4( ) (0, 1.04,4.89)G ξ = −

(2,2)3,4( ) (0, 8.63, 7.74)g ξ = − −

57/120


Seoul NationalUniv.


z

G

z′

g

y

w

O

BO′

y′( )14ξ∆


1st step, Heel

<Transverse moment of (ξ3(2) ,ξ4

(2))>

(2,2)3,4

(2,2) (2,2)3,4 3,4( ) () )(TB B BM y Fξ ξξ ⋅=

(2,2)3,4

(2,2) (2,2)3,4 3,4( ) () )(TG G GM y Fξ ξξ ⋅=

(2,2) (2,2) (2,2), 3,4 3,4 3,4( ) ( ) ( )T Ext g ExtM y Fξ ξ ξ= ⋅

(2,2)3,

(2,2) (2,2)3,4 , 3,4

(2,2)3 44 ,( )) (( )( )TT T ExBTG tM MMMξ ξξ ξ+= +

[ ]kN m⋅

(2) (2)3 4( ),TM ξ ξ ε< , 1,000ε ≡

( )2*4 4ξ ξ=

End of iteration

ε : Tolerance


( )2*4 4ξ ξ= ( )2*

3 3,ξ ξ=

(2,2)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×

(2,2)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×

(2,2)3,4

4( ) 4.0 10 [ ]ExtF kNξ = − ×

(2,2)3,4( ) (0, 1.04,4.89)G ξ = −(2,2)3,4( ) (0, 1.8, 5.19)B ξ = − −(2,2)3,4( ) (0, 8.63, 7.74)g ξ = − −

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?

4

5( 1.04)[ ] ( 3.6 10 )[ ]m kN= − ⋅ − ×53.7609 10 [ ]kN m= × ⋅

5( 1.8)[ ] (4.0 10 )[ ]m kN= − ⋅ ×57.2189 10 [ ]kN m= − × ⋅

4( 8.63)[ ] ( 4.0 10 )[ ]m kN= − ⋅ − ×53.4521 10 [ ]kN m= × ⋅

5 5 53.7609 10 [ ] 7.2189 10 [ ] 3.4521 10kN m kN m= × ⋅ − × ⋅ + ×

603 [ ]kN m= − ⋅

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1)3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( )G B

T TT Ext Ext

mg z g V zM M

gI F z

ξ ξ ξξ ξ

ξ ξ ξ

ρξ

ρ

⋅ − ⋅ ⋅= + ∆ − − ⋅


*3,4( ) 0TM ξ =

4

(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

(2,2)3,4( )G G ξ=


58/120


Seoul NationalUniv.


-40,000 -360,000 360,000 -40,000

-400,000 0 0 -400,000


(0) (0) (0)3 4( [0,0, , ,0,0])ξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0])ξ ξ=r

(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel



0 -360,000 400,000 -40,000

-400,000 0 0 -400,000



0 -360,000 400,000 -40,000

-39,415 407,094 -786,388 339,880


[ ]TM kN m⋅ [ ]TGM kN m⋅ [ ]TBM kN m⋅ , [ ]T ExtM kN m⋅Immersion

0 step

1st step



*3,4( ) 0TM ξ =



*3,4( ) 0zF ξ =

y

z

G

z′

y′

g B

( )04ξ∆

w

OO′

LC

O′

(0,0)3,4( )G ξ

w

,z z′

,y y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ

O

(1,0)3,4( )G ξ

w(1,0)3,4( )g ξ

y

,z z′

LC

O

y′(1,0)3,4( )B ξ

(0)3ξ∆

O′

59/120


Seoul NationalUniv.


0 -360,000 400,000 -40,000

-39,415 407,094 -786,388 339,880


(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0])ξ ξ=r

(2) (2) (2)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel



0 -360,000 400,000 -40,000

-39,415 407,094 -786,388 339,880



0 -360,000 400,000 -40,000

-603 376,091 -721,898 345,205



End of Heel 1st step. ξ3*=ξ3

(2), ξ4*=ξ4

(2) → END

1st step

2nd step



*3,4( ) 0TM ξ =



*3,4( ) 0zF ξ =

y

z

(1,1)3,4( )G ξ

z′

y′

(1,1)3,4( )g ξ

(1,1)3,4( )B ξ

w

OO′

y

z

(2,1)3,4( )G ξ

z′

y′

(2,1)3,4( )g ξ

(2,1)3,4( )B ξ

w

OO′

z

(2,2)3,4( )G ξ

z′

(2,2)3,4( )g ξ

y

w

O(2,2)3,4( )B ξ

y′( )14ξ∆

60/120


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61/120



Seoul NationalUniv.

v g

y

z

LC

,O O′

z′

Example : Immersion and Heel in Static Equilibrium after Cargo Hold Being Flooded (1)

Given : Cargo hold dimensionWeight of ship : 360,000 [kN]

Find : Immersion and trim of ship in static equilibrium

Example) Find a ship’s attitude in Static Equilibrium after cargo hold in right side of midship being flooded. (Origin of waterplane fixed frame is located in the intersection point between centerline and LCF(Longitudinal center of floating))

Assumption : considering force and transverse momentwith immersion and heel

( due to cargo being loaded on LCF. No trim) 3 4, ( [0,0, , ,0,0] )Tξ ξ=r

O-xyz frame: Waterplane fixed frameO'-x'y'z' frame: Body fixed frame

Position vector of the origin of the body fixed frame with respect to the waterplane fixed frame

Rotation vector of the origin of the body fixed frame with respect to thewaterplane fixed frame

y

z

LC

Section view(Front view)

OBaseline

K

G

, z′

, y′Bv g

,O O′Immersion

y′

G

v g B ξ3* : immersion in

static equilibrium

y

z

LC

,O O′

z′

y′

G

w g B

*4 ?ξ

ξ4* : heel in

static equilibrium

Heel

*3 ?ξ

Plan view (Top view)

x

y

LC

20m

,O O′G , x′

, y′

, B

100m

v : flooded volume

62/120


Seoul NationalUniv.


G : Center of gravity of ship

B : Center of buoyancy of ship

Cargo hold dimension- l (length) : 20 [m]

- b (breadth) : 20 [m]

- h (height) : 30 [m]- v (flooded volume)- g (center of gravity of cargo) : (0,-10,-4.5)

40sB m=

30D h m= =

9T m= y

z

LC

Section View(Front view)

OBaseline

K

G

v

, z′

, y′g B


Principal Dimension

- L (length) : 100 m, Bs (breadth) : 40m, T (draft) : 9m, D (depth) : 30m

20b m=

Elevation view (port-view)

x

z

Baseline

,O O′G

F.PA.P

v, x′

, z′

B

100L m=

g

Plan view(Top View)

x

y

LC O G

v

, x′

, y′

, B

O-xyz : Waterplane fixed frame

O'-x'y'z': Body fixed frame

g

20l m=63

/120


Seoul NationalUniv.


In case of force with immersion and heel




Given : Find :

*3,4( ) 0,zF ξ = * * * *

3,4 3,4 3,4 3,4( ) ( ) ( ) ( ), z G B Extwh ere F F F Fξ ξ ξ ξ= + +

*3ξ(0) (0) (0,0)

3 4 3,4( ), , ExtF ξξ ξ

(0,0) (0,0)3,4 3,4* (0,0) ( )

3,4 3,4 33 3

( ) ( )( ) ( ) B Ext k

z z

F FF F

ξ ξξ ξ ξ

ξ ξ+

∂ ∂= + ∆ ∂ ∂

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

, ( 0,1, 2,3...)k =

assumption :

3,43,4

3

( )( )B

WP

Fg A

ξξρ

ξ∂

= −∂

① Buoyance force is proportional to AWP


② External force is proportional to aWP

(aWP : waterplane area of flooding volume) 3,43,4

3

( )( )Ext

WP

Fg a

ξξρ

ξ∂

=∂

*3,4( ) 0,zF ξ = * * * *

3,4 3,4 3,4 3,4( ) ( ) ( ) ( ), z G B Extwh ere F F F Fξ ξ ξ ξ= + +

( )* (0,0) (0,0) (0,0) (0)3,4 3,4 3,4 3,4 3( ) ( ) ( ) ( )z z WP WPF F g A g aξ ξ ξ ξρ ρ ξ+= − + ∆

Given : (0) (0) (0,0)3 4 3,4( ), , ExtF ξξ ξ Find : *

3ξ (1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

, ( 0,1, 2,3...)k =

(Refer to detail derivation in Ch.16.2)

Opposite direction

Same direction

64/120


Seoul NationalUniv.


In case of force with immersion and heel




Given : Find :

*3,4( ) 0,TM ξ = * * * *

3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( ), z TG TB T Extwhere F M M Mξ ξ ξ ξ= + +

*4ξ(0) (0) (0,0)

3 4 3,4( ), , ExtF ξξ ξ

(0,0) (0,0) (0,0)3,4 3,4 , 3,4* (0,0) ( )

3,4 3,4 44 4 4

( ) ( ) ( )( ) ( ) TG TB T Ext k

T T

M M MM M

ξ ξ ξξ ξ ξ

ξ ξ ξ+

∂ ∂ ∂= + + ∆ ∂ ∂ ∂

(1) (0) (0)4 4 4( )ξ ξ ξ+ ∆=

, ( 0,1, 2,3...)k =

assumption :

*3,4( ) 0,TM ξ = * * * *

3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( ), z TG TB T Extwhere F M M Mξ ξ ξ ξ= + +(0,0) (0,0) (0,0)3,4 3,4 3,4(0,0)

3,4 (0,0) (0,0) (0,0) (0,0)3,4 3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( )G B

T TT Ext Ext T

mg z g V zM M

g I F z g i

ξ ξ ξξ ξ

ξ ξ ξ ξ

ρξ

ρ ρ

⋅ − ⋅ ⋅= + ∆ − ⋅ − ⋅ + ⋅

Given : (0) (0) (0,0)3 4 3,4( ), , ExtF ξξ ξ Find : *

3ξ (1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

, ( 0,1, 2,3...)k =

3,4 3,43,4

3,4 3,4 3,4 3,44

( )( ) ( ) ( ) ( ) ( ) ( )TB

B B T B T

MF z g I g V z g I

ξξ ξ ξ ξ ξ ξρ ρ ρ

ξ=

∂= − ⋅ − ⋅ − ⋅ ⋅ − ⋅

∂

3,43,4 3,4 3,4

4

( )( ) ( ) ( ),TG

G G G

MF z mg z

ξξ ξ ξ

ξ=

∂= − ⋅ ⋅

∂

, 3,43,4 3, 44 ,

43

( )( ) ( ) ( )T Ext

Ext TExt

MF gz i

ξξ ξξ ρ

ξ+

∂= − ⋅ ⋅

∂

65/120


Seoul NationalUniv.




( )* (0,0) (0,0) (0,0) (0)3,4 3,4 3,4 3,4 3( ) ( ) ( ) ( )z z WP WPF F g A g aξ ξ ξ ξρ ρ ξ= + − + ∆


Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

0 step, Immersion

(0,0) (0,0) (0,0) (0,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + + : Total force1

(0,0)3,4

(0,0) (0,0) (0,0) (0,0)3,4 3,4 3,4 3,4

( )

( ) ( ) ( ) ( )Extv

F g dv g l b Tξ

ξ ξ ξ ξρ ρ ⋅ ⋅= − = − ⋅∫∫∫

y

z

LC40 m

30 m

9m=

Section view

,O O′Baseline

G

v

, z′

, y′g B

Plan view

x

y

LC(0,0)3,4( )

20b

mξ

=

,O O′G

v

, x′

, y′

, B

(0,0)3,4

(0,0)3,4

( )

( )BV

F g dVξ

ξ ρ= ⋅ ∫∫∫

(0,0) 53,4( ) 3.6 10 [ ]GF kNξ = − ×

5 5 43.6 10 [ ] 3.6 10 [ ] 3.6 10 [ ]kN kN kN= − × + × − ×43.6 10 [ ]kN= − ×

(0,0)3,4( ) 100L mξ =

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r


(0,0) (0,0) (0,0)3,4 3,4 3,4( ) ( ) ( )sg L B Tξ ξ ξρ= ⋅ ⋅ ⋅

10[ ] 100[ ] 40[ ] 9[ ]m m m m= ⋅ ⋅ ⋅53.6 10 [ ]kN= ×

1

(1,0)3,4( )T ξ

g

(0,0)3,4( )

20l

mξ

=

(0,0)3,4( )

40sB

mξ

=

(0,0)3,4( )G G ξ=(0,0)3,4( )B B ξ=(0,0)3,4( )g g ξ=

310 [kN/m ] 20[ ] 20[ ] 9[ ]m m m= − × × ×43.6 10 [kN]= − ×

(0,0) (0,0) (0,0) (0,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

66/120


Seoul NationalUniv.


(0,0) (0,0)3,4 3,4( ) ( )WP WPg A g aξ ξρ ρ+− ⋅ ⋅





Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

2

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

0 step, Immersion

y

z

LC40 m

30 m

9m=

Section view

,O O′Baseline

G

v

, z′

, y′g B

Plan view

x

y

LC(0,0)3,4( )

20b

mξ

=

,O O′G

v

, x′

, y′

, B

(0,0)3,4( ) 100L mξ =

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r


(0,0)3,4( )T ξ

g

(0,0)3,4( )

40sB

mξ

=(0,0)3,4( )

20l

mξ

=

2

(0,0)3,4( )WPg A ξρ− ⋅

(0,0)3,4( )WPA

g dx dyξ

ρ= − ⋅ ∫∫

(0,0) (0,0)3,4 3,4( ) ( )sg B Lξ ξρ= − ⋅ ⋅

310[ / ] 40[ ] 100[ ]kN m m m= − ⋅ ⋅

44.0 10 [ / ]kN m= − ×

(0,0)3,4( )WPg a ξρ ⋅

(0,0)3,4( )WPa

g dx dyξ

ρ= ⋅ ∫∫

(0,0) (0,0)3,4 3,4( ) ( )g b lξ ξρ= ⋅ ⋅

310[ / ] 20[ ] 20[ ]kN m m m= ⋅ ⋅

34.0 10 [ / ]kN m= ×

(0,0)3,4( )G G ξ=(0,0)3,4( )B B ξ=(0,0)3,4( )g g ξ=

(0,0) (0,0)3,4 3,4( ) ( )WP WPg A g aξ ξρ ρ+− ⋅ ⋅

4 34.0 10 [ / ] 4.0 10 [ / ]kN m kN m= − × + ×43.6 10 [ / ]kN m= − ×

67/120


Seoul NationalUniv.


GF





Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

3

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

0 step, Immersion

y

z

LC40 m

30 m

9m=

Section view

,O O′Baseline

G

v

, z′

, y′g B

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r


(0,0)3,4( )T ξ

1 2

(0,0) 43,4( ) 3.6 10 [kN]zF ξ = − ×1

2

(0)3ξ∆ : change of immersion3

* (0,0)3,4 3,4(0)

3 (0,0) (0,0)3,4 3,4

( ) ( )

( ) ( )z z

WP WP

F Fg A g a

ξ ξ

ξ ξξ

ρ ρ−

∆ =− ⋅ + ⋅

4

4

0[ ] ( 3.6 10 )[ ] 13.6 10 [ / ]

kN kN mkN m

− − ×= = −

− ×

If we assume that Fz(ξ3*,ξ4

*) =0

( ) ( ) ( )1 0 03 3 3 0 1 1 [ ]mξ ξ ξ= + ∆ = − = −We want to find *

3ξ

Static equilibrium? Check!( )1*

3 3ξ ξ=? (1,0)

3,4( ) 0zF ξ =?

y

z

LC

Section view

,O O′

, z′

y′Baselinew g(0,0)

3,4( )T ξ10 m=

G

B

(0,0) (0,0)3,4 3,4( ) ( )WP WPg A g aξ ξρ ρ+− ⋅ ⋅ 43.6 10 [ / ]kN m= − ×

(0,0)3,4( )G G ξ=(0,0)3,4( )B B ξ=(0,0)3,4( )g g ξ=

(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

68/120


Seoul NationalUniv.






Given : Find :


(0) (0) (0,0)3 4 3,4, , ( )ExtFξ ξ ξ *

3ξ

*3,4( ) 0,zF ξ =

4

(1) (0) (0)3 3 3( )ξ ξ ξ+ ∆=

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4( )BF ξ

(1,0)3,4( )GF ξ

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= ++

(1,0)3,4( )ExtF ξ (1,0)

3,4( )g v ξρ= − ⋅

(0,0) 53,4( ) 3.6 10 [ ]GF kNξ= = − ×

(1,0)3,4( )g V ξρ= ⋅

3 4 3 510[ / ] 4.0 10 [ ] 4.0 10 [ ]kN m m kN= × × = ×

(1, 0)3, 4

(1,0)3,4

( )

( )V

V dVξ

ξ = ∫∫∫

(1,0)3,4( )zF ξ (1,0) (1,0) (1,0)


0[ ]kN=

100[ ] 40[ ] 10[ ]m m m= ⋅ ⋅

(1,0) (1,0) (1,0)3,4 3,4 3,4( ) ( ) ( )sL B Tξ ξ ξ= ⋅ ⋅

In this example

3 4 3 410[ / ] 4.0 10 [ ] 4.0 10 [ ]kN m m kN= − × × = − ×

0 step, Immersion

(1,0)3,4( ) 0zF ξ =


( )1*3 3ξ ξ=

4(1,0)3,4( ) 0zF ξ = ( )1*

3 3ξ ξ=



Check ξ4(1)

?

?

y

z

LC

Section view

,O O′

, z′

y′(1,0)3,4( )T ξ

10m

G

B

(0)3ξ∆

4 34.0 10 [ ]m= ×

(1, 0)3, 4

(1,0)3,4

( )

( )v

v dvξ

ξ = ∫∫∫

20[ ] 20[ ] 10[ ]m m m= ⋅ ⋅

(1,0) (1,0) (1,0)3,4 3,4 3,4( ) ( ) ( )l b Tξ ξ ξ= ⋅ ⋅

In this example

3 34.0 10 [ ]m= ×

Baselinev g

69/120


Seoul NationalUniv.




*3,4( ) 0TM ξ =


0 step

(0) (0) (0)3 4( [0,0, , ,0,0])ξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel

LC

O′

(0,0)3,4( )G ξ

v

,z z′

,y y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ

O -40,000 -360,000 360,000 -36,000

-360,000 0 0 -360,000





*3,4( ) 0zF ξ =

0 -360,000 400,000 -40,000



(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=rNext step

End of Immersion 0 step.

Start Heel 0 step.

(1,0)3,4( )G ξ

v(1,0)3,4( )g ξ

y

,z z′

LC

O

y′(1,0)3,4( )B ξ

(0)3ξ∆

O′

(0) (0) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r



70/120


Seoul NationalUniv.



y

z

LC

,O O′

(0,0)3,4( )G ξ

v

, z′

, y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ9[ ]m=

(0,0)3,4( )T ξ

y

,z z′

LC

O (0)3ξ∆

y′v(1,0)3,4( )g ξ

O′10m=

(1,0)3,4( )T ξ

(0,0)3,4(0,0)3,4(0,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


(0,0)3,4( ) (0,0,6)G ξ =(0,0)3,4( ) (0, 10, 4.5)g ξ = − −

(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

=

(0)3

00

ξ

+ ∆

(0,0)3,4(0,0)3,4(0,0) (0)3,4 3

( )( )( )

00

G

G

G

xyz

ξ

ξ

ξ ξ

= + ∆

(1,0)3,4(1,0)3,4(1,0)3,4

( )( )( )

G

G

G

xyz

ξ

ξ

ξ

(1) (0) (0)3 3 3ξ ξ ξ= + ∆

0 00 06 1

= + −

005

=

(0,0)3,4( ) (0,0, 4.5)B ξ = −

(1,0)3,4( )B ξ

(1,0)3,4( )G ξ

immersion

For center of gravity of ship (1,0)3,4( )G ξ

next

0 step, Heel to calculate (1,0)3,4( )TM ξCalculate (1,0) (1,0) (1,0)

3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r



71/120


Seoul NationalUniv.



y

z

LC

,O O′

(0,0)3,4( )G ξ

v

, z′

, y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ9[ ]m=

(0,0)3,4( )T ξ

y

,z z′

LC

O (0)3ξ∆

y′v(1,0)3,4( )g ξ

O′10m=

(1,0)3,4( )T ξ

Transformation G, g to waterplane fixed frame

(0,0)3,4( ) (0,0,6)G ξ =

(0,0)3,4( ) (0, 10, 4.5)g ξ = − −

(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5)g ξ = − −

0 step, Heel

(1,0)3,4(1,0)3,4

( )( )

0/ 2/ 2

bT

ξ

ξ

= − −

(1,0)3,4(1,0)3,4(1,0)3,4

( )( )( )

g

g

g

xyz

ξ

ξ

ξ

020 / 210 / 2

= − −

0105

= − −

(1,0)3,4( )

00

/ 2T ξ

= −

(1,0)3,4(1,0)3,4(1,0)3,4

( )( )( )

B

B

B

xyz

ξ

ξ

ξ

00

10 / 2

= −

005

= −

(0,0)3,4( ) (0,0, 4.5)B ξ = − (1,0)

3,4( ) (0,0, 5)B ξ = −

(1,0)3,4( )B ξ

(1,0)3,4( )G ξ


in this example, is located in center of draft.(1,0)3,4( )B ξ

in this example, is located in center of draft.(1,0)3,4( )g ξ

(1,0) (1,0) (1,0), 3,4 , 3,4 , 3,4

(1,0) (1,0) (1,0)3,4 3,4 3,4

( ) ( ) ( )

( ) ( ) ( )( , , )V x V y V zM M M

V V Vξ ξ ξ

ξ ξ ξ

,V xM :1st moment of displacement volume V about x axis in y direction in waterplane fixed frame

For center of weght of cargo is calculated by(1,0)3,4( )g ξ

(1,0) (1,0) (1,0), 3,4 , 3,4 , 3,4

(1,0) (1,0) (1,0)3,4 3,4 3,4

( ) ( ) ( )

( ) ( ) ( )( , , )v x v y v zM M M

v v vξ ξ ξ

ξ ξ ξ

,v xM :1st moment of flooded cargo volume v about x axis in y direction in waterplane fixed frame


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ0 step, Heel

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r



72/120


Seoul NationalUniv.



Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

1

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( )T TG TB T ExtM M M Mξ ξ ξ ξ= + +1

(1,0) (1,0) (1,0)3,4 3,4 3,4

5

( ) ( ) ( )

0[ ] ( 3.6 10 )[ ] 0 [ ]TG G GM y F

m kN kN

ξ ξ ξ=

= × − × =(1,0) (1,0) (1,0)3,4 3,4 3,4

5

( ) ( ) ( )

0[ ] (4.0 10 )[ ] 0[ ]TB B BM y F

m kN kN

ξ ξ ξ=

= × × =(1,0) (1,0) (1,0)

, 3,4 3,4 3,4

(1,0) (1,0)3,4 3,4

4 5

( ) ( ) ( )

( ) ( )

[ ] [ ] [ ]10 ( 4.0 10 ) 4.0 10

T Ext Ext Ext

g Ext

m kN kN m

M y F

y F

ξ ξ ξ

ξ ξ

⋅

=

=

= − × − × = ×

(1,0)3,4

5( ) 4.0 10 [kN]BF ξ = ×(1,0) 43,4( ) 4.0 10 [ ]ExtF kNξ = − ×

(1,0) 53,4( ) 3.6 10 [kN]GF ξ = − ×

5

5

0 0 4.0 10 [ ]4.0 10 [ ]

kN mkN m

= + + × ⋅

= × ⋅

0 step, Heel

G

v g

y

,z z′

LC

Section view

O

y′B

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5)g ξ = − −

(1,0)3,4( ) (0,0, 5)B ξ = −

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 , 3,4( ) ( ) ( ) ( )T TG TB T ExtM M M Mξ ξ ξ ξ= + +

(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


73/120


Seoul NationalUniv.



Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

2

2

0 step, Heel

G

v g(1,0)3,4( )

10T

mξ

=

y

,z z′

LC

Section view

O

y′B

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


Plan view

x

y

LC(1,0)3,4( )

20b

mξ

=

,O O′G

v

, x′

, y′

, B

(1,0)3,4( ) 100L mξ =

g(1,0)3,4( )

40sB

mξ

=

(1,0)3,4( )

20l

mξ

=m: Mass of the shipzG: Z coordinate of the center of mass of the ship

IT : 2nd moment of waterplane area of ship with respect to x axis of the waterplane fixed frame



iT : 2nd moment of waterplane area of flooded volume with respect to x axis of the waterplane fixed frame

(1,0)3,4

5( ) 4.0 10 [kN]BF ξ = ×(1,0)3,4

4( ) 4.0 10 [kN]ExtF ξ = − ×

(1,0)3,4

5( ) 3.6 10 [kN]GF ξ = − ×

(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5)g ξ = − −

(1,0)3,4( ) (0,0, 5)B ξ = −

(1, 0)3, 4

(1,0) (1,0) 33,4 3,4(1,0) 2

3,4( )

( ) ( )( )

12WP

sT

A

L BgI g y dx dy g

ξ

ξ ξξρ ρ ρ= =∫∫

(1, 0)3, 4

(1,0) (1,0) (1,0)3,4 3,4 3,4

( )

( ) ( ) ( )B BV

gV z g dV zξ

ξ ξ ξρ ρ⋅ = ⋅ ⋅∫∫∫(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )s Bg L B T zξ ξ ξ ξρ= ⋅ ⋅ ⋅ ⋅

310[ / ] 100[ ] 40[ ] 10[ ] ( 5[ ])kN m m m m m= × × × × −62.0 10 [ ]kN m= − × ⋅

(1,0) 53,4( ) 3.6 10 [ ] 5[ ]Gmg z kN mξ⋅ = × ×

61.8 10 [ ]kN m= × ⋅

33 6100[ ] 40[ ]10[ / ] 5.333 10 [ ]

12[ ]m mkN m kN m

m×

= × = × ⋅

74/120


Seoul NationalUniv.



Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

2

2

0 step, Heel

G

v g(1,0)3,4( )

10T

mξ

=

y

,z z′

LC

Section view

O

y′B

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


Plan view

x

y

LC(1,0)3,4( )

20b

mξ

=

,O O′G

v

, x′

, y′

, B

(1,0)3,4( ) 100L mξ =

g(1,0)3,4( )

40sB

mξ

=

(1,0)3,4( )

20l

mξ






(1,0) (1,0)3,4 3,4( ) ( )Ext ExtF zξ ξ⋅

(1,0) (1,0)3,4 3,4( ) ( )Ext gF zξ ξ= ⋅

( )44.0 10 [ ] 5.0 [ ]kN m= − × × −

52.0 10 [ ]kN m= × ⋅(1,0) (1,0) 33,4 3,4(1,0)

3,4

( ) ( )( )

12T

l bgi g

ξ ξξρ ρ

⋅=

,(parallel axis theorem)3

3 20[ ] 20[ ]10[ / ]12

m mkN m ×= ×

2(1,0) (1,0) (1,0)3,4 3,4 3,4( ) ( ) ( )

12

g b l bξ ξ ξρ ⋅ + ⋅ × ⋅

55.333 10 [ ]kN m= × ⋅

23 110[ / ] 20[ ] 20[ ] 20[ ]

2kN m m m m + × × × ×

(1,0)3,4

5( ) 4.0 10 [kN]BF ξ = ×(1,0)3,4

4( ) 4.0 10 [kN]ExtF ξ = − ×

(1,0)3,4

5( ) 3.6 10 [kN]GF ξ = − ×

(1,0)3,4( ) (0,0,5)G ξ =(1,0)3,4( ) (0, 10, 5)g ξ = − −(1,0)3,4( ) (0,0, 5)B ξ = −

75/120


Seoul NationalUniv.



Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

2

2

0 step, Heel

G

v g(1,0)3,4( )

10T

mξ

=

y

,z z′

LC

Section view

O

y′B

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


Plan view

x

y

LC(1,0)3,4( )

20b

mξ

=

,O O′G

v

, x′

, y′

, B

(1,0)3,4( ) 100L mξ =

g(1,0)3,4( )

40sB

mξ

=

(1,0)3,4( )

20l

mξ






(1,0)3,4

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4

(1,0) (1,0)3,4 3,4

( ) ( ) ( ) ( )

( ) ( ) ( )

G B T

Ext Ext T

mg z gV z g I

F z g i

ξ ξ ξ ξ

ξ ξ ξ

ρ ρ

ρ

⋅ − ⋅ − ⋅

− ⋅ + ⋅

6 6 6

5 5

1.8 10 [ ] 2.0 10 [ ] 5.333 10 [ ]2.0 10 [ ] 5.333 10 [ ]

kN m kN m kN mkN m kN m

= × ⋅ + × ⋅ − × ⋅

− × ⋅ + × ⋅

61.2 10 [ ]kN m= − × ⋅

(1,0)3,4

5( ) 4.0 10 [kN]BF ξ = ×(1,0)3,4

4( ) 4.0 10 [kN]ExtF ξ = − ×

(1,0)3,4

5( ) 3.6 10 [kN]GF ξ = − ×

(1,0)3,4( ) (0,0,5)G ξ =(1,0)3,4( ) (0, 10, 5)g ξ = − −(1,0)3,4( ) (0,0, 5)B ξ = −

76/120


Seoul NationalUniv.



Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

2

2

0 step, Heel

G

v g(1,0)3,4( )

10T

mξ

=

y

,z z′

LC

Section view

O

y′B

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

O′


Plan view (1,0)3,4

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4

(1,0) (1,0)3,4 3,4

( ) ( ) ( ) ( )

( ) ( ) ( )

G B T

Ext Ext T

mg z gV z g I

F z g i

ξ ξ ξ ξ

ξ ξ ξ

ρ ρ

ρ

⋅ − ⋅ − ⋅

− ⋅ + ⋅

1 3

1 (1,0) 53,4( ) 4.0 10 [ ]TM kN mξ = × ⋅

* (1, 0)

3,4 3, 4

(1,0)3,4

(0)4 (1,0) (1,0) (1,0) (1,0)

3,4 3,4 3,4 3,4

(1,0) (1,0)3,4 3,4

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

T T

G B T

Ext Ext T

M M

mg z gV z g I

F z g i

ξ ξ

ξ ξ ξ ξ

ξ ξ ξ

ξρ ρ

ρ

−=

∆ ⋅ − ⋅ − ⋅ − ⋅ + ⋅

3

5

6

0[ ] 4.0 10 [ ] 0.33rad 19.11.2 10 [ ]

kN m kN mkN m

⋅ − × ⋅= = =

− × ⋅


(1) (0) (0)4 4 4 0 19.1 19.1ξ ξ ξ= + ∆ = + ° = °

61.2 10 [ ]kN m= − × ⋅



( )1*4 4ξ ξ=

? (1,1)3,4( ) 0TM ξ =

?

y

z

LC

G y′

g B

(0)4ξ∆

v

O

z′

O′

(1,1)3,4( )G G ξ=(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

77/120


Seoul NationalUniv.



y

,z z′

LC

OO′

y′v g10m

immersion(0,0)3,4(0,0)3,4(0,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

(1,0)3,4( ) (0,0,5)G ξ =

(1,0)3,4( ) (0, 10, 5.5)g ξ = − −

0 step, Immersion(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

=

(0)3

00

ξ

+ ∆

(1,0)3,4

(0) (0) (1,0)4 4 3,4(0) (0) (1,0)4 4 3,4

( )( )( )


G

G

G

xyz

ξ

ξ

ξ

ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(1,1)3,4(1,1)3,4(1,1)3,4

( )( )( )

G

G

G

xyz

ξ

ξ

ξ

(1) (0) (0)3 3 3ξ ξ ξ= + ∆

01.64

4.72

= −

B

G

(1,0)3,4( ) (0,0, 5)B ξ = −

0 step, Heel

(0) (0)4 4(0) (0)4 4


ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(1,1)3,4(1,1)3,4(1,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

(1,0)3,4(1,0)3,4(1,0)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


(1) (0) (0)4 4 4ξ ξ ξ= + ∆

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


heel

(0) (0)4 4(0) (0)4 4

1 0 0 00 cos sin 00 sin cos 5

ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(1,1)3,4( ) (0, 1.64,4.72)G ξ = −

(0)4, ( 19.1 )ξ∆ =

Next

y

z

G

z′

y′

gB

(0)4ξ∆

v

O

(1,0)3,4( )G G ξ=(1,0)3,4( )B B ξ=(1,0)3,4( )g g ξ=

(1,1)3,4( )G G ξ=(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=


O′

For center of gravity of ship, (1,1)3,4( )G ξ


(1,1) (1,1) (1,1), 3,4 , 3,4 , 3,4(1,1)

3,4 (1,1) (1,1) (1,1)3,4 3,4 3,4

( ) ( ) ( )( )

( ) ( ) ( )( , , )V x V y V zBM M M

V V Vξ ξ ξ

ξξ ξ ξ

=

,V xM :1st moment of volume about x axis in y direction in waterplane fixed frame


For center of Buoyancy , and center of weight of cargo hold is calculated by

(1,1)3,4( )B ξ

(1,1)3,4( )g ξ

(1,1) (1,1) (1,1), 3,4 , 3,4 , 3,4(1,1)

3,4 (1,1) (1,1) (1,1)3,4 3,4 3,4

( ) ( ) ( )( )

( ) ( ) ( )( , , )v x v y v zg

v v v

M M Mξ ξ ξξ

ξ ξ ξ=

,v xM :1st moment of flooded volume about x axis in y direction in waterplane fixed frame


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ

78/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

(0)4ξ∆

v

OO′

Section view

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

y

z,z z′ ′′

y′

(0)4 19.1ξ∆ =

(0)4ξ∆

A2

A3

A1

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ

(1,1) (1,1)3,4 3,4( ) ( )( , )c cc y zξ ξCentroid of section area of displacement volume

=


1A sArea B T′′ ′′= ⋅ 240[ ] 10[ ] 400[ ]m m m= ⋅ =

10[ ]T m′′ =Draft in O''-x''y''z'' frame

(0)3ξ∆

T ′′

sB′′

2 3( )A AArea Area=

{ }0.5 (40 / 2[ ]) (40 / 2)[ ] tan(19.1)m m= ⋅ ⋅ ⋅269.25[ ]m=


A3

/ 2sB′′

(0)4( / 2) tan( )sB ξ′′ ⋅ ∆

(0)4ξ∆

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B ξ

{ }(0)40.5 ( / 2) ( / 2) tan( )s sB B ξ′′ ′′= ⋅ ⋅ ∆

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=




y′′


,O O′′

O′

(1,1)3,4of ξ

Center of buoyancy

79/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

(0)4ξ∆

v

OO′

Section view

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

y

z,z z′ ′′

y′

(0)4 19.1ξ∆ =

(0)4ξ∆

A2

A3

A1

(0)3ξ∆

T ′′

sB′′

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=




y′′,O O′′

O′

(1,1)3,4of ξ

3 3( , )A Ay z′′ ′′ (1)

42 1, tan( )

2 3 2 3s sB B ξ′′ ′′ = ⋅ − ∆ ⋅

( )(40 / 2) (2 / 3), (40 / 2) tan(19.1) (1/ 3)= ⋅ − ⋅ ⋅



1 1( , )A Ay z′′ ′′

( )13.33, 2.31= −Centroid of A2 in O''-x''y''z'' frame,

2 2( , )A Ay z′′ ′′in similar way ( )13.33,2.31= −

A3

/ 2sB′′

(0)4

( / 2)

tan( )sB

ξ

′′

⋅ ∆

(0)4ξ∆

3Ay′′ 3Az′′

10(0, ) (0, ) (0, 5)2 2

T ′′= − = − = −

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(1,1)3,4( )B ξ


Center of buoyancy

80/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

(0)4ξ∆

v

OO′

Section view

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

y

z,z z′ ′′

y′

(0)4 19.1ξ∆ =

(0)4ξ∆

A2

A3

A1

(0)3ξ∆

T ′′

sB′′

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=




y′′,O O′′

O′

A1 400.00 0.00 -5.00 0.00 -2000.00

A2 69.25 -13.33 2.31 -923.34 159.86

A3 -69.25 13.33 -2.31 -923.34 159.86

Sum 400.00 -1,846.69 -1,680.29

3 3

, ,1 1

3 3

1 1

( , )

,Ai Ai

c c

A y A zi i

Ai Aii i

y z

M M

Area Area

′′ ′′= =

= =

′′ ′′ =

∑ ∑

∑ ∑



( )( , , ) 0, 4.62, 4.2c c cx y z′′ ′′ ′′∴ = − −

iAy′′

, ( 1, 2,3)i =

,( )

AiA yM

′′


( )AiA z

M′′

iAi AArea z′′×

1,846.69 1,680.29,400 400

− − =

AiArea


( )4.62, 4.2= − −

(1,1)3,4of ξ

iAz′′

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(1,1)3,4( )B ξ


Center of buoyancy

81/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

(0)4ξ∆

v

OO′

Section view

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


( )( , ) 3.09, 4.64c cy z′′ ′′∴ = − −

(1,1)3,4(1,1)3,4(1,1)3,4

( )( )( )

c

c

c

xyz

ξ

ξ

ξ

=

(1,1)3,4( ) (0, 2.99, 5.48)B ξ∴ = − −

(0) (0)4 4(0) (0)

4 4

1 0 0 00 cos sin 4.620 sin cos 4.2

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆ −

(0)4, ( 19.1 )ξ∆ =



(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )c c cx y zξ ξ ξ

02.995.48

= − −

(0) (0)4 4(0) (0)

4 4


c

c

c

xyz

ξ ξξ ξ

′′ ′′= ∆ − ∆

′′ ∆ ∆

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , ) (0, 2.99, 5.48)c c cx y zξ ξ ξ∴ = − −

(1,1)3,4of ξ

y

z,z z′ ′′

y′

(0)4 19.1ξ∆ =

(0)4ξ∆

A2

A3

A1

(0)3ξ∆

T ′′

sB′′


y′′,O O′′

O′

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(1,1)3,4( )B ξ


Center of buoyancy

82/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

(0)4ξ∆

v

OO′

Section view

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


( ), ,( , ) 10.86, 3.12v c v cy z′′ ′′∴ = − −

(1,1)3,4(1,1)3,4(1,1)3,4

,

,

,

( )( )( )

v c

v c

v c

xyz

ξ

ξ

ξ

=

(1,1)3,4( ) (0, 9.24, 6.5)g ξ∴ = − −

(0) (0)4 4(0) (0)

4 4

1 0 0 00 cos sin 10.860 sin cos 3.12

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆ −

(0)4, ( 15.28 )ξ∆ =

Calculate centroid of section area of flooded volume in O''-x''y''z'' frame,


(1,1) (1,1) (1,1)3,4 3,4 3,4, , ,( ) ( ) ( )( , , )v c v c v cx y zξ ξ ξ

09.246.5

= − −

,(0) (0)

4 4 ,(0) (0)

4 4 ,


v c

v c

v c

xyz

ξ ξξ ξ

′′ ′′= ∆ − ∆ ′′ ∆ ∆

(1,1) (1,1) (1,1)3,4 3,4 3,4, , ,( ) ( ) ( )( , , ) (0, 9.24, 6.5)v c v c v cx y zξ ξ ξ∴ = − −

(1,1)3,4of ξ

y

z,z z′ ′′

y′

(0)4 19.1ξ∆ =

(0)4ξ∆

A2

A4

(0)3ξ∆

T ′′

sB′′


y′′,O O′′

O′

(1,1) (1,1) (1,1)3,4 3,4 3,4( ) ( ) ( )( , , )g g gg x y zξ ξ ξ

(1,1) (1,1)3,4 3,4, ,( ) ( )( , )v v c v cc y zξ ξCentroid of section area of flooded cargo volume

=(1,1)3,4( )g ξ


Center of flooded volume

83/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

(0)4ξ∆

v

OO′

Section view

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=



(1,1)3,4( )BF ξ

(1,1)3,4( )GF ξ

(1,1)3,4( )ExtF ξ

45.39 10 [ ]kN= − ×

(0,0)3,4

5( ) 3.6 10 [ ]GF kNξ= = − ×3 4 310[ / ] 4.0 10 [ ]kN m m= × ×(1,1)

3,4( )g V ξρ= ⋅54.0 10 [ ]kN= ×

(1,1)3,4

(1,1)3,4

( )

( )V

V dVξ

ξ = ∫∫∫ (1,1)3,4( )section area L ξ= × 1

2[ ]400A mArea =

2

2[ ]69.25A mArea =

3

2[ ]69.25A mArea = −2

_ [ ]400A Total mArea =2 4 3400[ ] 100[ ] 4.0 10 [ ]m m m= × = ×

(1,1)3,4_ ( )Area LA Total ξ= ×


(1,1)3,4

(1,1)3,4

( )

( )v

v dvξ

ξ = ∫∫∫ (1,1)3,4( )section area l ξ= ×

(1,1)3,4_ ( )Area a Total l ξ= ×

2 3 3269.25[ ] 20[ ] 5.39 10 [ ]m m m= × = ×

3 3 310[ / ] 5.39 10 [ ]kN m m= − × ×(1,1)3,4( )g v ξρ= − ⋅

1

2[ ]200a mArea =

2

2[ ]69.25a mArea =

2_ [ ]269.25a Total mArea =

84/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

(0)4ξ∆

v

OO′

Section view

Given : (1) (0) (1,0)

3 4 , 3,4( ), , T ExtM ξξ ξ Find : *4ξ (1) (0) (0)

4 4 4( )ξ ξ ξ+ ∆=

(1,0) (1,0) (1,0) (1,0)3,4 3,4 3,4 3,4(1,0)

3,4 (1,0) (1,0) (1,0)3,4 3,4 3,4

* (0)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z g IM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ − ⋅= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


0 step, Heel


( )1*4 4ξ ξ=

?

(1,1)3,4( ) 0TM ξ =

?

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


4

4

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


<Transverse moment of (ξ3,4(1,1)) >

(1,1)3,4

(1,1) (1,1)3,4 3,4( ) () )(TB B BM y Fξ ξξ= ⋅

(1,1)3,4

(1,1) (1,1)3,4 3,4( ) () )(TG G GM y Fξ ξξ= ⋅


(1,1)3,4( )TM ξ

(1,1)3,4( ) (0, 1.64,4.72)G ξ = −(1,1)3,4( ) (0, 2.99, 5.48)B ξ = − −(1,1)3,4( ) (0, 9.24, 6.5)g ξ = − −

(1,1)3,4

5( ) 4.0 10 [ ]B kNF ξ = ×

(1,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×

(1,1)3,4

4( ) 5.39 10 [ ]ExtF kNξ = − ×

(1,1)3,4

(1,1)3,4

(1,1)3,4 ,( () )( ) T ExBG tT TM M Mξ ξξ= ++

5( 1.64) ( 3.6 10 )= − ⋅ − ×55.8895 10 [ ]kN m= × ⋅

5( 2.99) (4.0 10 )− ⋅ ×=61.1953 10 [ ]kN m= − × ⋅

(1,1) (1,1)3,4 3,4( ) ( )g Exty Fξ ξ= ⋅

4( 9.24) ( 5.39 10 )= − ⋅ − ×54.9751 10 [ ]kN m= × ⋅

[ ]kN m⋅5 6 55.8895 10 [ ] 1.1953 10 [ ] 4.9751 10kN m kN m= × ⋅ − × ⋅ + ×51.088 10 [ ]kN m= − × ⋅

stImmersion 1 step

4

(1,1)3,4( ) 0TM ξ ≠ ( )1*

4 4ξ ξ≠

85/120


Seoul NationalUniv.



0 step

(0) (0) (0)3 4( [0,0, , ,0,0])ξ ξ=r

(0) (1) (0)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel

-36,000 -360,000 360,000 -36,000

-400,000 0 0 -400,000



0 -360,000 400,000 -40,000

-400,000 0 0 -400,000



(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=r

-13,850 -360,000 400,000 -53,850

-108,798 588,950 -1,195,257 497,508



Start Immersion 1st step.End of Heel 0 step.

Immersion

(0) (1) (0)3 4( [0,0, , ,0,0] )Tξ ξ=r

(0) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


1st step



*3,4( ) 0TM ξ =



*3,4( ) 0zF ξ =

LC

O′

(0,0)3,4( )G ξ

v

,z z′

,y y′(0,0)3,4( )g ξ

(0,0)3,4( )B ξ

O

(1,0)3,4( )G ξ

v(1,0)3,4( )g ξ

y

,z z′

LC

O

y′(1,0)3,4( )B ξ

(0)3ξ∆

O′


y

z

G

z′

y′

g B

( )04ξ∆

v

OO′

86/120


Seoul NationalUniv.


y

z

G

z′

y′

gB

v

OO′

Section view



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


1st step, Immersion

(1,1)3,4( )BF ξ

(1,1)3,4( )GF ξ

(1,1) (1,1) (1,1) (1,1)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

(1,1)3,4( )ExtF ξ 45.39 10 [ ]kN= − ×

53.6 10 [ ]kN= − ×

54.0 10 [ ]kN= ×

(1,1) (1,1) (1,1) (1,1)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= + +

5 5 43.6 10 [ ] 4.0 10 [ ] 5.39 10 [ ]kN kN kN= − × + × − ×

41.39 10 [ ]kN= − ×

Total force is changed during heel !

1

1

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


: Total force

(0)4ξ∆

87/120


Seoul NationalUniv.


(1,1) (1,1)3,4 3,4( ) ( )WP WPg A g aξ ξρ ρ+−

(1,1) (1,1)3,4 3,4( ) ( )WP WPg A g aξ ξρ ρ+−y

z

G

z′

y′

g B

v

OO′

Section view



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


1st step, Immersion

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


1 2

Plan view

x

y

LC(1,0)3,4( )

20b

mξ

=

,O O′G

v

, x′

, y′

, B

(1,0)3,4( ) 100L mξ =

g

(1,0)3,4( )

40sB

mξ

=(1,0)3,4( )

20l

mξ

=

2

(1,1)3,4( )WPgA ξρ−

(1,1)3,4( )WPA

g dx dyξ

ρ= − ⋅ ∫∫(1,1) (1,1)3,4 3,4( ) ( )sg B Lξ ξρ= − ⋅ ⋅

3 40[ ]10[ / ] 100[ ]cos19.1

mkN m m= − ⋅ ⋅°

44.23 10 [ / ]kN m= − ×

(1,1)3,4( )WPg a ξρ

(1,1)3,4( )WPa

g dx dyξ

ρ= ⋅ ∫∫(1,1) (1,1)3,4 3,4( ) ( )g b lξ ξρ= ⋅ ⋅

3 20[ ]10[ / ] 20[ ]cos19.1

mkN m m= ⋅ ⋅°

34.23 10 [ / ]kN m= ×

(0)4ξ∆

(1,0)3,4 (1,0)

3,4(0)4

( )( )

cossB

g Lξ

ξρξ

= − ⋅ ⋅∆

(1,0)3,4 (1,0)

3,4(0)4

( )( )

cosb

g lξ

ξρξ

= ⋅ ⋅∆

43.81 10 [ / ]kN m= − ×

4 34.23 10 [ / ] 4.23 10 [ / ]kN m kN m= − × + ×

88/120


Seoul NationalUniv.



z′

y′

O′

(1,1) (1,1)3,4 3,4( ) ( )WP WPg A g aξ ξρ ρ−

y

z

G

z′

y′

g B

v

OO′

Section view


Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


1st step, Immersion

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ

(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(1,1)3,4( )G G ξ=


1 2

2

(0)4ξ∆

1(1,1)3,4

4( ) 1.39 10 [ ]zF kNξ = − ×

43.81 10 [ / ]kN m= − ×3 : change of immersion(1)

3ξ∆

* (1,1)3,4 3,4(1)

3 (1,1) (1,1)3,4 3,4

( ) ( )

( ) ( )z z

WP WP

F Fg A g a

ξ ξ

ξ ξξ

ρ ρ−

∆ =− +


4

4

0 (1.39 10 [ ]) 0.363.81 10 [ / ]

kNkN m

− ×= = −

− ×( ) ( ) ( )2 1 1

3 3 3 1 0.36 1.36 [ ]mξ ξ ξ= + ∆ = − − = −



( )2*3 3ξ ξ=

? (2,1)3,4( ) 0zF ξ =

?

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

3

89/120


Seoul NationalUniv.


z′

y′

O′



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ


4

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

1st step, Immersion

(2,1)3,4( ) 0zF ξ = Static equilibrium?

Check!

( )2*3 3ξ ξ=

?

?

,z z′ ′′

y′O′ y

z

g B

v

OG

Define new frame O''-x''y''z''

(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′z′′

Section view

4

new frame O''-x''y''z''

GOO′′

y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(1)4ξ

(1)3ξ∆

0.36[ ]m−Oz′ Oy′

4 (2,13,4

?)( ) 0zF ξ = (2,1) (2,1) (2,1) (2,1)

3,4 3,4 3,4 3 4

?

,( ) ( ) ( ) ( ) 0z G B ExtF F F Fξ ξ ξ ξ= + + =

( Because O'' is located in Centerline.O'' : intersection point between y axis of waterplane fixed frame, ans z' axis of body fixed frame)

If we know O'' Get (2,1)3,4( )V ξ

easier

(2,1)3,4( )BF ξGet

easier

A : Calculate O in body fixed frame

B : Calculate O'' in body fixed frame

C(2,1)3,4( )V ξ: Calculate

by equation of line OO′′

D(2,1)3,4( )BF ξ: Calculate

( is calculated in similar way)(2,1)3,4( )ExtF ξ

90/120


Seoul NationalUniv.


z′

y′

O′



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ


4

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

1st step, Immersion


Check!

( )2*3 3ξ ξ=

?

?

,z z′ ′′

y′O′ y

z

g B

v

OG


(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′z′′

Section view

4


(1) (1)3 4( ) sinOy ξ ξ′ = − −∆ ⋅

( 0.36) sin19.1= − − ⋅

0.12[ ]m=

(0) (1) (1)3 3 4( ) ( ) sinOz ξ ξ ξ′ = − ∆ − ∆ ⋅

( 1) (0.36) sin19.1= − − − ⋅

1.34[ ]m=

( , , )O O Ox y z′ ′ ′ (0,0.12,1.34)=

GO

y

z,z z′ ′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(1)4ξ

(1)3ξ∆

0.36[ ]m−Oz′ Oy′

A

4 (2,13,4

?)( ) 0zF ξ = (2,1) (2,1) (2,1) (2,1)

3,4 3,4 3,4 3 4

?

,( ) ( ) ( ) ( ) 0z G B ExtF F F Fξ ξ ξ ξ= + + =

: O B : O'' C (2,1)3,4( ): V ξ D (2,1)

3,4( ): BF ξ

Calculate Origin of waterplane fixed frame in body fixed frame ( , , )O O Ox y z′ ′ ′

91/120


Seoul NationalUniv.


z′

y′

O′



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ


4

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

1st step, Immersion


Check!

( )2*3 3ξ ξ=

?

?

,z z′ ′′

y′O′ y

z

g B

v

OG


(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′ z′′

Section view

4


1.38

9

10.38T ′′ =

So, Origin of O''-x''y''z'' frame in body fixed frame

Develop equation of line OO′′by point O, and line slop tan(-ξ4

(1)).

(1)4tan( ) ( )O Oz z y yξ′ ′ ′ ′− = − ⋅ −

( , , )O O Ox y z′ ′ ′ (0,0.12,1.34)=

1.34 tan( 19.1) ( 0.12)z y′ ′− = − ⋅ −

GOO′′

y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(1)4ξ

(1)4ξ

(1)3ξ∆

0.36[ ]m= −0.35 1.38z y′ ′∴ = − ⋅ +

( , , )O O Ox y z′′ ′′ ′′′ ′ ′ (0,0,1.38)=

Oz′ Oy′

A

4 (2,13,4

?)( ) 0zF ξ = (2,1) (2,1) (2,1) (2,1)

3,4 3,4 3,4 3 4

?

,( ) ( ) ( ) ( ) 0z G B ExtF F F Fξ ξ ξ ξ= + + =

: O B : O'' C (2,1)3,4( ): V ξ D (2,1)

3,4( ): BF ξ

92/120


Seoul NationalUniv.


z′

y′

O′



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ


4

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

1st step, Immersion


Check!

( )2*3 3ξ ξ=

?

?

,z z′ ′′

y′O′ y

z

g B

v

OG


(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′ z′′

Section view

4

A2

A3

A1


1A sArea B T′′ ′′= ⋅ 240[ ] 10.38[ ] 415.39[ ]m m m= ⋅ =

10.38[ ]T m′′ =Draft in O''-x''y''z'' frame

2 3( )A AArea Area=

{ }0.5 (40 / 2[ ]) (40 / 2)[ ] tan(19.1)m m= ⋅ ⋅ ⋅269.25[ ]m=


A3

/ 2sB′′

(0)4( / 2) tan( )sB ξ′′ ⋅ ∆

(0)4ξ∆

{ }(0)40.5 ( / 2) ( / 2) tan( )s sB B ξ′′ ′′= ⋅ ⋅ ∆

10.38T ′′ =

A

4 (2,13,4

?)( ) 0zF ξ = (2,1) (2,1) (2,1) (2,1)

3,4 3,4 3,4 3 4

?

,( ) ( ) ( ) ( ) 0z G B ExtF F F Fξ ξ ξ ξ= + + =

: O B : O'' C (2,1)3,4( ): V ξ D (2,1)

3,4( ): BF ξ

93/120


Seoul NationalUniv.


z′

y′

O′



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ


4

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

1st step, Immersion


Check!

( )2*3 3ξ ξ=

?

?

,z z′ ′′

y′O′ y

z

g B

v

OG


(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′z′′

Section view

4

A1 415.39

A2 69.25

A3 -69.25

Sum 415.39

iAArea

Calculate section area of in O''-x''y''z'' frame(2,1)3,4( )V ξ

A2

A3

A4

2_ [ ]415.39A Total mArea =

A

4 (2,13,4

?)( ) 0zF ξ = (2,1) (2,1) (2,1) (2,1)

3,4 3,4 3,4 3 4

?

,( ) ( ) ( ) ( ) 0z G B ExtF F F Fξ ξ ξ ξ= + + =

: O B : O'' C (2,1)3,4( ): V ξ D (2,1)

3,4( ): BF ξ

( 2,1)3, 4

(2,1)3,4

( )

( )V

V dVξ

ξ = ∫∫∫

2[ ]415.39 100[ ]m m= ⋅

(2,1) (2,1)3,4 3,4( ( )) ( )section area V Lξ ξ= ⋅

In this example

4 34.15 10 [ ]m= ×

(2,1)3,4( )BF ξ

(2,1)3,4( )g V ξρ= ⋅3 4 310[ / ] 4.15 10 [ ]kN m m= × ×54.15 10 [ ]kN= ×

C

D

94/120


Seoul NationalUniv.


z′

y′

O′



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ


4

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

1st step, Immersion


Check!

( )2*3 3ξ ξ=

?

?

,z z′ ′′

y′O′ y

z

g B

v

OG


(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′z′′

Section view

4

A2

A3

A4

Calculate section area of flooded volume in O''-x''y''z'' frame of(2,1)3,4ξ

1_ 4 2 22

AA flooded A A A

AreaArea Area Area Area= + = +

22[ ]

[ ]415.39 69.25

2m

m= +

2[ ]276.95 m=

4 (2,13,4

?)( ) 0zF ξ = (2,1) (2,1) (2,1) (2,1)

3,4 3,4 3,4 3 4

?

,( ) ( ) ( ) ( ) 0z G B ExtF F F Fξ ξ ξ ξ= + + =

A : O B : O'' C (2,1)3,4( ): v ξ D (2,1)

3,4( ): ExtF ξ

(in previous calculation)

( 2,1)3, 4

(2,1)3,4

( )

( )v

v dvξ

ξ = ∫∫∫

2[ ]276.95 20[ ]m m= ⋅

(2,1) (2,1)3,4 3,4( ( )) ( )section area v lξ ξ= ⋅

In this example

3 35.54 10 [ ]m= ×

(2,1)3,4( )ExtF ξ

(2,1)3,4( )g v ξρ= − ⋅3 4 310[ / ] 5.54 10 [ ]kN m m= − × ×45.54 10 [ ]kN= − ×

C

D

95/120


Seoul NationalUniv.


z′

y′

O′



Given : Find :


(1) (1) (1,1)3 4 3,4, , ( )ExtFξ ξ ξ

*3,4( ) 0,zF ξ =

(2) (1) (1)3 3 3( )ξ ξ ξ+ ∆=


(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r


*3ξ


4

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

1st step, Immersion


Check!

( )2*3 3ξ ξ=

?

?

4

(2,1)3,4( )BF ξ

(2,1)3,4( )GF ξ

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4( ) ( ) ( ) ( )z G B ExtF F F Fξ ξ ξ ξ= ++

(2,1)3,4( )ExtF ξ (2,1)

3,4( )g v ξρ= − ⋅

(0,0) 53,4( ) 3.6 10 [ ]GF kNξ= = − ×

(2,1)3,4( )g V ξρ= ⋅54.15 10 [ ]kN= ×

(2,1)3,4( )zF ξ (2,1) (2,1) (2,1)


0[ ]kN=

45.54 10 [ ]kN= − ×4(2,1)3,4( ) 0zF ξ = ( )2*

3 3ξ ξ=



Check ξ4(1)

96/120


Seoul NationalUniv.


-13,850 -360,000 400,000 -53,850

-108,798 588,950 -1,195,257 497,508


1st step

2nd step

(1) (1) (1)3 4( [0,0, , ,0,0])ξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0])ξ ξ=r

(2) (2) (2)3 4( [0,0, , ,0,0])ξ ξ=r

Immersion

Heel



0 -360,000 415,389 -55,389



End of Immersion 1st step.

Start Heel 1st step.

(1) (1) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r




*3,4( ) 0TM ξ =



*3,4( ) 0zF ξ =


y

z

G

z′

y′

g B

v

OO′

Section view

(0)4ξ∆

z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆

97/120


Seoul NationalUniv.


y

z

G

z′

y′

g B

v

OO′

Section view

(0)4ξ∆

z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆


(2,1)3,4( ) (0, 1.64,4.36)G ξ = −

(2,1)3,4( )B ξ =(2,1)3,4( )g ξ =

0 step, Heel

1st step, Immersion

(1,1)3,4(1,1)3,4(1,1) (1)3,4 3

( ) 0( ) 0( )

G

G

G

xyz

ξξξ ξ

= + ∆

(2,1)3,4(2,1)3,4(2,1)3,4

( )( )( )

G

G

G

xyz

ξξξ

01.64

4.36

= −


0 01.64 0

4.72 0.36

= − + −

(1)4, ( 0.90 )ξ∆ =

Next

(1,1)3,4( )G G ξ=(1,1)3,4( )B B ξ=(1,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1) (2,1) (2,1), 3,4 , 3,4 , 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

( ) ( ) ( )( )

( ) ( ) ( )( , , )V x V y V zBM M M

V V Vξ ξ ξ

ξξ ξ ξ

=



(2,1)3,4( )B ξ

(2,1)3,4( )g ξ

(2,1) (2,1) (2,1), 3,4 , 3,4 , 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

( ) ( ) ( )( )

( ) ( ) ( )( , , )v x v y v zg

v v v

M M Mξ ξ ξξ

ξ ξ ξ=


(1,1)3,4(1,1)3,4(1,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


(2,1)3,4(2,1)3,4(2,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

=

(1)3

00ξ

+ ∆ (2) (1) (1)

3 3 3ξ ξ ξ= + ∆

immersion

in this example, is simply calculated as next page(2,1) (2,1)3,4 3,4( ), ( )B gξ ξ

(1,1)3,4( ) (0, 1.64,4.72)G ξ = −(1,1)3,4( ) (0, 2.99, 5.48)B ξ = − −(1,1)3,4( ) (0, 9.24, 6.5)g ξ = − −


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



98/120


Seoul NationalUniv.



,z z′ ′′

y′O′ y

z

g B

v

OG

(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′ z′′

Section view

Frame O''-x''y''z''

1.38

9

1st step, Heel


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

G

OO′′

y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆ (1)

4ξ

(1)4ξ

(1)3ξ∆

To get (2,1) (2,1)3,4 3,4( ), ( )B gξ ξ

(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,1)3,4( )B ξCenter of buoyancy

Calculate centroid of section area of displacement volume in waterplane fixed frame first

a : Calculate O in body fixed frame

b : Calculate O'' in body fixed frame

c

by equation of line OO′′

d

: Calculatein O''-x''y''z'' frame

(2,1) (2,1)3,4 3,4( ) ( )( , )c cy zc ξ ξ′′′′ ′′

c

b

a

G

OO′′

y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆ (1)

4ξ

(1)4ξ

(1)3ξ∆

y′′′z′′′

d

e

: Translateto O'''-x'''y'''z''' frame

(2,1) (2,1)3,4 3,4( ) ( )( , )c cc y zξ ξ′′ ′′ ′′

(2,1) (2,1)3,4 3,4( ) ( ), ( , )c cy zc ξ ξ′′′ ′′′ ′′′

G

OO′′

y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆ (1)

4ξ

(1)4ξ

(1)3ξ∆

e

(2,1) (2,1)3,4 3,4( ) ( )( , )c cc y zξ ξ

: Rotateto waterplane fixed frame

(2,1) (2,1)3,4 3,4( ) ( )( , )c cc y zξ ξ′′′ ′′′ ′′′

(2,1) (2,1)3,4 3,4( ) ( ), ( , )c cy zc ξ ξ

(2,1)3,4( )B ξ=

a : O

b : O''

c : c''

d : c'''e : c (2,1)

3,4( )B ξ=

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



99/120


Seoul NationalUniv.



,z z′ ′′

y′O′ y

z

g B

v

OG

(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′ z′′

Section view G

OO′′

y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(1)4ξ

Frame O''-x''y''z''

1.38

9

(1)3ξ∆

0.36[ ]m−

1st step, Heel

(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,1)3,4( )B ξ


Center of buoyancy

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

1


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

Oy′Oz′

( , , )O O OO x y z′ ′ ′ (0,0.12,1.34)=

a : O b : O'' c : c'' d : c''' e : c (2,1)3,4( )B ξ=

a

Same in 1st Step, Immersion

Origin of O''-x''y''z'' frame in body fixed frame

( , , )O O Ox y z′′ ′′ ′′′ ′ ′ (0,0,1.38)=

Origin of O-xyz frame in body fixed frame

b


(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



100/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆ Given :

(2) (1) (2,1)3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *

4ξ (2) (1) (1)4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


1

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=



A1 415.39 0.00 -5.19 0.00 -2,156.85

A2 69.25 -13.33 2.31 -923.34 159.86

A3 -69.25 13.33 -2.31 -923.34 159.86

Sum 415.39 -1,846.69 -1,837.14

3 3

, ,1 1

3 3

1 1

( , )

,Ai Ai

c c

A y A zi i

Ai Aii i

c y z

M M

Area Area

′′ ′′= =

= =

′′ ′′ ′′ =

∑ ∑

∑ ∑



( )( , , ) 0, 4.45, 4.42c c cc x y z′′ ′′ ′′ ′′∴ = − −

iAy′′

, ( 1, 2,3)i =

,( )

AiA yM

′′


( )AiA z

M′′

iAi AArea z′′×

1,846.69 1,837.14,415.39 415.39

− − =

AiArea


( )4.45, 4.42= − −

iAz′′

(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,1)3,4( )B ξ


Center of buoyancy

,z z′ ′′

y′O′ y

z

gB

OG

new frame O''-x''y''z''G

OO′′y

z,z z′ ′′

y′′

y′O′z′′

Section view

A2

A3

A1(0)

4 19.1ξ∆ =

(0)4ξ∆O′′ (1)

3ξ∆

a : O b : O'' c : c'' d : c''' e : c (2,1)3,4( )B ξ=

101/120


Seoul NationalUniv.



,z z′ ′′

y′O′ y

z

g B

v

OG

(1)3ξ∆

G

OO′′y

z,z z′ ′′

y′′

y′O′ z′′

Section view G

OO′′

y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(1)4ξ

,O′′′

New frame O'''-x'''y'''z'''

1.38

9O Oy y′′′′′ ′=

0.12=

(1)3ξ∆

0.36[ ]m−

1.38O Oz z′′′′′ ′= − 1.34 1.38= − 0.04= −

O O′′ ′′′ 2 2( ) ( )O Ox y′′′ ′′′′′ ′′= + 2 2(0.12) ( 0.04)= + − 0.13=

1st step, Heel

(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,1)3,4( )B ξ


Center of buoyancy

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

1


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=

,O O′′′O′′

Oy ′′′′′

(1)4ξ

Oz ′′′′′

(1)3ξ∆0.36[ ]m= −

0.38Oy′

Oz′

( , , )O O OO x y z′ ′ ′ (0,0.12,1.34)=

a : O b : O'' c : c'' d : c''' e : c (2,1)3,4( )B ξ=

a


b

y′′′z′′′

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



102/120


Seoul NationalUniv.


G

OO′′ y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(1)4ξ

,O′′′

z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆ Given :

(2) (1) (2,1)3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *

4ξ (2) (1) (1)4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


1

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,1)3,4( )B ξ


Center of buoyancy

,z z′ ′′

y′O′ y

z

gB

OG

new frame O''-x''y''z''G

OO′′y

z,z z′ ′′

y′′

y′O′ z′′

Section view

A2

A3

A1(0)

4 19.1ξ∆ =

(0)4ξ∆O′′ (1)

3ξ∆1.38

9

0.38(1)

3ξ∆0.36[ ]m−

New frame O'''-x'''y'''z'''

Calculate centroid of section area in O''-x''y''z'' frame,then transformation centroid of section area to O'''-x'''y'''z''' frame,

( )( , , ) 0, 4.45, 4.42c c cc x y z′′ ′′ ′′ ′′ = − −


transformation centroid of section area to waterplane in O'''-x'''y'''z''' frame,

( , , )c c cc x y z′′′ ′′′ ′′′ ′′′

( )0, 4.45 0.12, 4.42 ( 0.04)= − − − − −

( , , )c c O c Ox y y z z′′′ ′′′′′ ′′ ′′ ′′ ′′= − −

( )0, 4.56, 4.38= − −

y′′′z′′′

a : O b : O'' c : c'' d : c''' e : c (2,1)3,4( )B ξ=

103/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆ Given :

(2) (1) (2,1)3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *

4ξ (2) (1) (1)4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


1

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,1)3,4( )B ξ


Center of buoyancy

,z z′ ′′

y′O′ y

z

gB

OG

Define new frame O''-x''y''z''G

OO′′y

z,z z′ ′′

y′′

y′O′z′′

Section view

A2

A3

A1(0)

4 19.1ξ∆ =

(0)4ξ∆O′′ (1)

3ξ∆

( )( , , ) 0, 4.56, 4.38c c cc x y z′′′ ′′′ ′′′ ′′′∴ = − −

(2,1)3,4(2,1)3,4(2,1)3,4

( )( )( )

c

c

c

xyz

ξ

ξ

ξ

=

(2,1)3 4( ) (0, 2.88, 5.63)B ξ∴ = − −

(0) (0)4 4(0) (0)

4 4

1 0 0 00 cos sin 4.560 sin cos 4.38

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆ −

(0)4, ( 19.1 )ξ∆ =

Calculate centroid of section area in O'''-x'''y'''z''' frame,


(2 ,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )c c cc x y zξ ξ ξ

02.885.63

= − −

(0) (0)4 4(0) (0)

4 4


c

c

c

xyz

ξ ξξ ξ

′′′ ′′′= ∆ − ∆

′′′ ∆ ∆

(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , ) (0, 2.88, 5.63)c c cc x y zξ ξ ξ∴ = − −

(1,1)3,4of ξ

a : O b : O'' c : c'' d : c''' e : c (2,1)3,4( )B ξ=

104/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆ Given :

(2) (1) (2,1)3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *

4ξ (2) (1) (1)4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


1

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , )g g gg x y zξ ξ ξ


=(2,1)3,4( )g ξ



,z z′ ′′

y′O′ y

z

gB

OG

Define new frame O''-x''y''z''G

OO′′y

z,z z′ ′′

y′′

y′O′z′′

Section view

A2

A4(0)

4 19.1ξ∆ =

(0)4ξ∆O′′ (1)

3ξ∆

(2 ,1) (2,1) (2,1)3,4 3,4 3,4, , ,( ) ( ) ( )( , , )v c v c v cx y zξ ξ ξ

09.286.68

= − −

,(0) (0)

4 4 ,(0) (0)

4 4 ,


v c

v c

v c

xyz

ξ ξξ ξ

′′′ ′′′= ∆ − ∆ ′′′ ∆ ∆

(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , ) (0, 9.28, 6.68)c c cc x y zξ ξ ξ∴ = − −

(1,1)3,4of ξ

( )_ _ _( , , ) 0, 10.95, 3.28v v c v c v cc x y z′′′ ′′′ ′′′ ′′′∴ = − −

(2,1)3,4(2,1)3,4(2,1)3,4

,

,

,

( )( )( )

v c

v c

v c

xyz

ξ

ξ

ξ

=

(2,1)3 4( ) (0, 9.28, 6.68)B ξ∴ = − −

(0) (0)4 4(0) (0)

4 4

1 0 0 00 cos sin 10.950 sin cos 3.28

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆ −

(0)4, ( 19.1 )ξ∆ =

By similar way, calculate centroid of section area of flooded volume


a : O b : O'' c : cv'' e : cv(2,1)3,4( )g ξ=d : cv'''

in O'''-x'''y'''z''' frame,

105/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆ Given :

(2) (1) (2,1)3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *

4ξ (2) (1) (1)4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel


(2,1)3,

(2,1) (2,1)3,4 3

(2,1)3 ,4,44 ,( ) (( )( ) )TT T ExBTG tM MMMξ ξξ ξ+= +1

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


1

(2,1)3,4( ), (0, 1.64,4.36)G ξ = −(2,1)3,4( ) (0, 2.88, 5.63), B ξ − −=(2,1)3,4( ) (0, 9.28, 6.68), g ξ − −=

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.15 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 5.54 10 [ ]ExtF kNξ = − ×

(2,1)3,4( )TM ξ

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=



(1)) >

(2,1)3,4

(2,1) (2,1)3,4 3,4( ) () )(TB B BM y Fξ ξξ= ⋅

(2,1)3,4

(2,1) (2,1)3,4 3,4( ) () )(TG G GM y Fξ ξξ= ⋅


5( 1.64) ( 3.6 10 )= − ⋅ − ×55.8895 10 [ ]kN m= × ⋅

5( 2.88) (5.63 10 )= − ⋅ × 61.1962 10 [ ]kN m= − × ⋅(2,1) (2,1)3,4 3,4( ) ( )g Exty Fξ ξ= ⋅

4( 9.28) ( 5.54 10 )= − ⋅ − ×55.1389 10 [ ]kN m= × ⋅

(2,1)3,4( )TM ξ (2,1)

3,4(2,1)3,4

(2,1)3,4 ,( () )( ) T ExBG tT TM M Mξ ξξ= ++

[ ]kN m⋅5 6 55.8895 10 [ ] 1.1962 10 [ ] 5.1389 10kN m kN m= × ⋅ − × ⋅ + ×49.3384 10 [ ]kN m= × ⋅

106/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


Plan view

x

y

LC(2,1)3,4( )b ξ

O G

v

, x′

, y′

, B

(2,1)3,4( ) 100L mξ =

g

(2,1)3,4( )sB ξ

(2,1)3,4( ) 20l mξ =

2

(2,1) (2,1)3,4 3,4(2,1)

3,4

3( ) ( )( )

12s

T

L BgI g

ξ ξξρ ρ=

( )( )(1,0) (1,0)3,4 3,4

304( ) ( ) / cos

12sL B

gξ ξ ξ

ρ∆

=

( )33 100[ ] 40[ ] / cos19.1

10[ / ]12

m mkN m

× °= ×

66.3213 10 [ ]kN m= × ⋅

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

2

(2,1) (2,1)3,4 3,4

2

( ) ( )12

Bg OO L ξ ξρ ⋅ ′′+ ⋅ × ⋅

( )

(1,0)3,4(1,0)

3,4

2

04

( )( )

12 cos

sBg OO L

ξξρ

ξ⋅ ′′+ ⋅ × ⋅ ∆

21 40[ ]0.13 100[ ]2 cos19.1

mg mρ ⋅ + ⋅ × ⋅

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.15 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 5.54 10 [ ]ExtF kNξ = − ×

(2,1)3,4( ), (0, 1.64,4.36)G ξ = −(2,1)3,4( ) (0, 2.88, 5.63), B ξ − −=(2,1)3,4( ) (0, 9.28, 6.68), g ξ − −=

O′′ (2,1)3,4( )Gmg z ξ⋅ 5

6

3.6 10 [ ] 4.36[ ]1.5701 10 [ ]

kN mkN m

= × ×

= × ⋅

, ( )OO O O′′ ′′ ′′′=107

/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

G

Section view

(1)3ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


Plan view

x

y

LC(2,1)3,4( )b ξ

,O O′G

v

, x′

, y′

, B

(2,1)3,4( ) 100L mξ =

g

(2,1)3,4( )sB ξ

(2,1)3,4( ) 20l mξ =

2 (2,1) (2,1)3,4 3,4( ) ( )BgV zξ ξρ ⋅

54.15 10 [ ] 5.63[ ]kN m= − × × −

62.3402 10 [ ]kN m= − × ⋅

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

2

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.15 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 5.54 10 [ ]ExtF kNξ = − ×

(2,1)3,4( ), (0, 1.64,4.36)G ξ = −(2,1)3,4( ) (0, 2.88, 5.63), B ξ − −=(2,1)3,4( ) (0, 9.28, 6.68), g ξ − −=

OO′′

(2,1) (2,1)3,4 3,4( ) ( )Ext ExtF zξ ξ⋅

( )45.54 10 [ ] 6.68 [ ]kN m= − × × −

53.6994 10 [ ]kN m= × ⋅

108/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

G

Section view

(1)3ξ∆O

O′′

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


2

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


Plan view

x

y

LC(2,1)3,4( )b ξ

,O O′G

v

, x′

, y′

, B

(2,1)3,4( ) 100L mξ =

g

(2,1)3,4( )sB ξ

(2,1)3,4( ) 20l mξ =

2(2,1) (2,1) 33,4 3,4(2,1)

3,4

( ) ( )( )

12T

l bgi g

ξ ξξρ ρ

⋅=

,(parallel axis theorem)

{ }33 20[ ] 20[ ] / cos19.1

10[ / ]12

m mkN m

× °= ×

2(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )

12

g OO b l bξ ξ ξρ ⋅ ′′+ ⋅ + × ⋅

56.4341 10 [ ]kN m= × ⋅

23 1 20[ ] 20[ ]10[ / ] 0.13 20[ ]

2 cos19.1 cos19.1m mkN m m + × + × × × ° °

{ }3(2,1) (1,0) (0)3,4 3,4 4( ) ( ) / cos( )

12

l bg

ξ ξ ξρ

⋅ ∆=

2(1,0) (1,0)3,4 3,4(2,1)

3,4(0) (0)4 4

( ) ( )( )

cos( cos(12 ) )

b bg OO l

ξ ξξ

ξ ξρ ⋅

∆ ∆

′′+ ⋅ + × ⋅

(2,1)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×(2,1)3,4

5( ) 4.15 10 [ ]B kNF ξ = ×(2,1)3,4

4( ) 5.54 10 [ ]ExtF kNξ = − ×

(2,1)3,4( ), (0, 1.64,4.36)G ξ = −(2,1)3,4( ) (0, 2.88, 5.63), B ξ − −=(2,1)3,4( ) (0, 9.28, 6.68), g ξ − −=

, ( )OO O O′′ ′′ ′′′=109

/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

G

Section view

(1)3ξ∆O

O′′

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


2

(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


2

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4

(2,1) (2,1) (2,1)3,4 3,4 3,4

( ) ( ) ( ) ( )

( ) ( ) ( )G B T

Ext Ext T

mg z g V z gI

F z g i

ξ ξ ξ ξ

ξ ξ ξ

ρ ρ

ρ

⋅ − ⋅ ⋅ −

− ⋅ + ⋅

6 6

6 5

5

6.3213 10 [ ] 2.3402 10 [ ]1.5701 10 [ ] 3.6994 10 [ ]6.4341 10 [ ]

kN m kN mkN m kN mkN m

= − × ⋅ + × ⋅

+ × ⋅ − × ⋅

+ × ⋅

62.1375 10 [ ]kN m= − × ⋅

110/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

G

Section view

(1)3ξ∆O

O′′

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r


(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

(2,1)3,4( )G G ξ=


2(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4

(2,1) (2,1) (2,1)3,4 3,4 3,4

( ) ( ) ( ) ( )

( ) ( ) ( )G B T

Ext Ext T

mg z g V z gI

F z g i

ξ ξ ξ ξ

ξ ξ ξ

ρ ρ

ρ

⋅ − ⋅ ⋅ −

− ⋅ + ⋅

1 (2,1) 43,4( ) 9.3384 10 [ ]TM kN mξ = × ⋅

62.1375 10 [ ]kN m= − × ⋅

21 3

* ( 2,1)

3,4 3, 4

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4

(2,1) (2,1) (2,1)3,4 3,4 3,4

(1)4

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

T T

G B T

Ext Ext T

M M

mg z g V z gI

F z g i

ξ ξ

ξ ξ ξ ξ

ξ ξ ξ

ξρ ρ

ρ

−=

∆ ⋅ − ⋅ ⋅ − − ⋅ + ⋅

3

4

6

0[ ] 9.3384 10 [ ] 0.0437 rad 2.52.1375 10 [ ]

kN m kN mkN m

⋅ − × ⋅= = − = −

− × ⋅


(2) (1) (1)4 4 4 19.1 2.5 16.6ξ ξ ξ= + ∆ = ° − ° = °



( )2*4 4ξ ξ=

? (2,2)3,4( ) 0TM ξ =

?

111/120


Seoul NationalUniv.


z′

y′

O′

y

z

gB

v

OG

Section view

(1)3ξ∆


(2,2)3,4( ) (0, 1.44,4.43)G ξ = −

(2,2)3,4( )g ξ =

(2,2)3,4( )B ξ =

1st step, Immersion

1st step, Heel

(1) (1)4 4(1) (1)4 4


ξ ξξ ξ

= ∆ − ∆ ∆ ∆

(2,2)3,4(2,2)3,4(2,2)3,4

( )

( )

( )

xyz

ξ

ξ

ξ

(2,1)3,4(2,1)3,4(2,1)3,4

( )

( )

( )

xyz

ξ

ξ

ξ


(2) (1) (1)4 4 4ξ ξ ξ= + ∆

heel

(2,1)3,4

(1) (1) (2,1)4 4 3,4(1) (1) (2,1)4 4 3,4

1 0 0 ( )0 cos sin ( )0 sin cos ( )

G

G

G

xyz

ξξ ξ ξξ ξ ξ

= ∆ − ∆ ∆ ∆

(2,2)3,4(2,2)3,4(2,2)3,4

( )( )( )

G

G

G

xyz

ξξξ

01.44

4.43

= −


(1) (1)4 4(1) (1)4 4

1 0 0 00 cos sin 1.640 sin cos 4.36

ξ ξξ ξ

= ∆ − ∆ − ∆ ∆

(1)4, ( 2.5 )ξ∆ = −

(2,1)3,4( )G G ξ=(2,1)3,4( )B B ξ=(2,1)3,4( )g g ξ=

z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆

(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

(2,2) (2,2) (2,2), 3,4 , 3,4 , 3,4(2,2)

3,4 (2,2) (2,2) (2,2)3,4 3,4 3,4

( ) ( ) ( )( )

( ) ( ) ( )( , , )V x V y V zBM M M

V V Vξ ξ ξ

ξξ ξ ξ

=



(2,2)3,4( )B ξ

(2,2)3,4( )g ξ

(2,2) (2,2) (2,2), 3,4 , 3,4 , 3,4(2,2)

3,4 (2,2) (2,2) (2,2)3,4 3,4 3,4

( ) ( ) ( )( )

( ) ( ) ( )( , , )v x v y v zg

v v v

M M Mξ ξ ξξ

ξ ξ ξ=


(0, 2.47, 5.52)= − −

(0, 9.40, 6.5)= − −(2,1)3,4( ) (0, 1.64,4.36)G ξ = −(2,1)3,4( ) (0, 2.88, 5.63)B ξ − −=(2,1)3,4( ) (0, 9.28, 6.68)g ξ − −=

Next

in this example, is simply calculated as next page(2,1) (2,1)3,4 3,4( ), ( )B gξ ξ


3,4 3,4 3,4( ) ( ), ( ),G B gξ ξ ξ

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



112/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

O(1)

4ξ (1)3ξ∆0.36[ ]m= −

(1)4ξ

(1)4ξ∆

yO′′

B

(2)4ξ

,z z′ ′′

y′O′

y

z

g B

v

OG

Section viewnew frame O''-x''y''z''

G

OO′′

y

z, zz′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(2)4ξ

New frame O''-x''y''z''

(1)3ξ∆

0.36[ ]m−

,z z′ ′′ z

y′′

( , , )O O Ox y z′ ′ ′ (0,0.12,1.34)=,(in previous calculation)

So, Origin of O''-x''y''z'' frame in body fixed frame

Develop equation of line OO′′by point O, and slop .

(2)4tan ( )O Oz z y yξ′ ′ ′ ′− = − ⋅ −

1.34 0.3 ( 0.12)z y′ ′− = − ⋅ −0.3 1.38z y′ ′∴ = − ⋅ +

( , , )O O Ox y z′′ ′′ ′′′ ′ ′ (0,0,1.38)=

(2)4tanξ

(2)4, ( 16.6 )ξ = °

Oz′ Oy′

1.38

9

a : O b : O'' c : c'' d : c''' e : c (2,2)3,4( )B ξ=

a

b

(2)4ξ

113/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

,z z′ ′′

y′O′

y

z

g B

v

OG

Section view

9

0.37


A1 414.80 0.00 -5.19 0.00 -2,150.74

A2 59.61 -13.33 1.99 -794.74 118.43

A3 -59.61 13.33 -1.99 -794.74 118.43

Sum 414.80 -1,589.48 -1,913.88

3 3

, ,1 1

3 3

1 1

( , )

,Ai Ai

c c

A y A zi i

Ai Aii i

y z

M M

Area Area

′′ ′′= =

= =

′′ ′′ =

∑ ∑

∑ ∑



( )( , , ) 0, 3.83, 4.61c c cx y z′′ ′′ ′′∴ = − −

iAy′′

, ( 1, 2,3)i =

,( )

AiA yM

′′


( )AiA z

M′′

iAi AArea z′′×

1,589.48 1,913.88,414.8 414.8

− − =

AiArea


( )3.83, 4.61= − −

iAz′′

A2 A3

A1

(2,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,2)3,4( )B ξ


Center of buoyancy

y′′

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4

a : O b : O'' c : c'' d : c''' e : c (2,2)3,4( )B ξ=

114/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

O(1)

4ξ (1)3ξ∆0.36[ ]m= −

(1)4ξ

(1)4ξ∆

yO′′

B

(1)4ξ

(1)4ξ∆

(2)4ξ =

(0) (1)4 4ξ ξ∆ + ∆

,z z′ ′′

y′O′

y

z

g B

v

OG

Section viewnew frame O''-x''y''z''

,z z′ ′′ z

O′′

O

Oy ′′′′′

O Oy y′′′′′ ′=0.12[ ]m=

Oz ′′′′′

Define new frame O'''-x'''y'''z'''

y′′′z′′′

O′′′

y′′

1.38O Oz z′′′′′ ′= −1.34 1.38= − 0.04= −

O O′′ ′′′ 2 2( ) ( )O Ox y′′′ ′′′′′ ′′= +2 2(0.12) ( 0.04)= + − 0.13=

( , , )O O Ox y z′ ′ ′(0,0.12,1.34)=

1.38

9

(2)4ξ

a : O b : O'' c : c'' d : c''' e : c (2,2)3,4( )B ξ=

(2)4ξ

G

OO′′

z, zz′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(1)4ξ19.1= °

(2)4ξ

Oz′ Oy′(1)

3ξ∆0.36[ ]m−

115/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

,z z′ ′′

y′O′

y

z

g B

v

OG

Section view

9

0.37

A2 A3

A1

Calculate centroid of section area in O''-x''y''z'' frame,then transformation centroid of section area to O'''-x'''y'''z''' frame,

( )( , , ) 0, 3.83, 4.61c c cx y z′′ ′′ ′′ = − −

Calculate centroid of section areain O''-x''y''z'' frame,

transformation centroid of section area to waterplane in O'''-x'''y'''z''' frame,

( , , )c c cx y z′′′ ′′′ ′′′

( )0, 3.83 0.12, 4.61 ( 0.04)= − − − − −

( , , )c c O c Ox y y z z′′′ ′′′′′ ′′ ′′ ′′ ′′= − −

( )0, 3.95, 4.57= − −

G

OO′′ y

z,z z′ ′′

y′′

y′O′

(0)3ξ∆1[ ]m= −

(0)4ξ∆

19.1= °

(0)4ξ∆

(1)3ξ∆

0.36[ ]m−B

y′′′z′′′

(2,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,2)3,4( )B ξ


Center of buoyancy

y′′

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4

a : O b : O'' c : c'' d : c''' e : c (2,2)3,4( )B ξ=

116/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

,z z′ ′′

y′O′

y

z

g B

v

OG

y′′

Section view

9

0.37

A2 A3

A1

( )( , , ) 0, 3.95, 4.57c c cx y z′′′ ′′′ ′′′∴ = − −

(2,2)3,4(2,2)3,4(2,2)3,4

( )( )( )

c

c

c

xyz

ξ

ξ

ξ

=

(2,2)3,4( ) (0, 2.48, 5.52)B ξ∴ = − −

(2) (2)4 4(2) (2)

4 4

1 0 0 00 cos sin 3.950 sin cos 4.57

ξ ξξ ξ

= − − −

(2) (0) (1)4 4 4, ( 16.6 )ξ ξ ξ= ∆ + ∆ =

Calculate centroid of section area in O'''-x'''y'''z''' frame,


(2 ,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )c c cx y zξ ξ ξ

02.485.52

= − −

(2) (2)4 4(2) (2)

4 4


c

c

c

xyz

ξ ξξ ξ

′′′ ′′′= −

′′′

(2,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , ) (0, 2.48, 5.52)c c cx y zξ ξ ξ∴ = − −

(2,2)3,4of ξ

(2,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )B B BB x y zξ ξ ξ


=(2,2)3,4( )B ξ


Center of buoyancy

(2)4ξ

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4

a : O b : O'' c : c'' d : c''' e : c (2,2)3,4( )B ξ=

117/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=

,z z′ ′′

y′O′

y

z

g B

v

OG

y′′

Section view

9

0.37

A2 A3

A1

(2)4ξ

(2,2) (2,2) (2,2)3,4 3,4 3,4( ) ( ) ( )( , , )g g gg x y zξ ξ ξ


=(2,2)3,4( )g ξ



( )_ _ _( , , ) 0, 10.86, 3.55v c v c v cx y z′′′ ′′′ ′′′∴ = − −

(2,2)3,4(2,2)3,4(2,2)3,4

,

,

,

( )( )( )

v c

v c

v c

xyz

ξ

ξ

ξ

=

(2) (2)4 4(2) (2)

4 4

1 0 0 00 cos sin 10.860 sin cos 3.55

ξ ξξ ξ

= − − −

(0)4, ( 16.6 )ξ∆ =

Calculate centroid of section area of flooded volume in O'''-x'''y'''z''' frame,


(2 ,2) (2,2) (2,2)3,4 3,4 3,4, , ,( ) ( ) ( )( , , )v c v c v cx y zξ ξ ξ

09.46.5

= − −

,(2) (2)

4 4 ,(2) (2)

4 4 ,


v c

v c

v c

xyz

ξ ξξ ξ

′′′ ′′′= − ′′′

(2,1) (2,1) (2,1)3,4 3,4 3,4( ) ( ) ( )( , , ) (0, 9.4, 6.5)c c cx y zξ ξ ξ∴ = − −

(2,2)3,4of ξ

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4

a : O b : O'' c : cv'' d : cv''' e : cv(2,2)3,4( )B ξ=

118/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?

(2,2)3,4( ) (0, 1.32,4.53)G ξ = −

(2,2)3,4 (0, 2.41, 5.47)( )B ξ − −=

(2,2)3,4 (0, 9.39, 6.44)( )g ξ = − −


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=


(2,2)3,4( )BF ξ

(2,2)3,4( )GF ξ

(2,2)3,4( )ExtF ξ

45.35 10 [ ]kN= − ×

(0,0)3,4

5( ) 3.6 10 [ ]GF kNξ= = − ×3 4 310[ / ] 4.11 10 [ ]kN m m= × ×(2,2)

3,4( )gV ξρ=54.15 10 [ ]kN= ×

( 2,2)3,4

(2,2)3,4

( )

( )V

V dVξ

ξ = ∫∫∫ (2,2)3,4( )section area L ξ= × 1

2[ ]414.8A mArea =

2

2[ ]59.61A mArea =

3

2[ ]59.61A mArea = −2

_ [ ]414.8A Total mArea =2 4 3414.8[ ] 100[ ] 4.15 10 [ ]m m m= × = ×

(2,2)3,4_ ( )Area LA Total ξ= ×


( 2,2)3,4

(2,2)3,4

( )

( )v

v dvξ

ξ = ∫∫∫ (2,2)3,4( )section area l ξ= ×

(2,2)3,4_ ( )Area a Total l ξ= ×

2 3 3267.01[ ] 20[ ] 5.34 10 [ ]m m m= × = ×

3 3 310[ / ] 5.35 10 [ ]kN m m= × ×(2,2)3,4( )gv ξρ=

1

2[ ]207.4a mArea =

2

2[ ]59.61a mArea =

2_ [ ]267.01a Total mArea =

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4

119/120


Seoul NationalUniv.


z′

y′

O′

y

z

g B

v

OG

Section view

(1)4ξ∆


1st step, Heel

(1) (2) (1)3 4( [0,0, , ,0,0] )Tξ ξ=r

(1) (2) (2)3 4( [0,0, , ,0,0] )Tξ ξ=r



( )2*4 4ξ ξ=

?

(2,2)3,4( ) 0TM ξ =

?


(2,2)3,4( )G G ξ=(2,2)3,4( )B B ξ=(2,2)3,4( )g g ξ=


(1)) >

(2,2)3,4

(2,2) (2,2)3,4 3,4( ) () )(TB B BM y Fξ ξξ= ⋅

(2,2)3,4

(2,2) (2,2)3,4 3,4( ) () )(TG G GM y Fξ ξξ= ⋅


(2,2)3,4( )TM ξ

(2,2)3,4

5( ) 4.15 10 [ ]B kNF ξ = ×

(2,2)3,4

5( ) 3.6 10 [ ]G NF kξ = − ×

(2,2)3,4

4( ) 5.35 10 [ ]ExtF kNξ = − ×

(2,2)3,4

(2,2)3,4

(2,2)3,4 ,( () )( ) T ExBG tT TM M Mξ ξξ= ++

5( 1.32) ( 3.6 10 )= − ⋅ − ×55.1982 10 [ ]kN m= × ⋅

5( 2.48) (4.15 10 )= − ⋅ ×61.0281 10 [ ]kN m= − × ⋅

(2,2) (2,2)3,4 3,4( ) ( )g Exty Fξ ξ= ⋅

4( 9.4) ( 5.35 10 )= − ⋅ − ×55.0181 10 [ ]kN m= × ⋅

[ ]kN m⋅5 6 55.1982 10 [ ] 1.0281 10 [ ] 5.0181 10kN m kN m= × ⋅ − × ⋅ + ×36.439 10 [ ]kN m= − × ⋅

(2,2)3,4( )TM ξ ε< , 8,000ε ≡

( )2*4 4ξ ξ=

End of iteration

ε : Tolerance


( )2*4 4ξ ξ= ( )2*

3 3,ξ ξ=

4

(2,2)3,4( ) (0, 1.32,4.53)G ξ = −

(2,2)3,4 (0, 9.4, 6.5)( )g ξ = − −

(2,2)3,4 (0, 2.48, 5.52)( )B ξ = − −

Given : (2) (1) (2,1)

3 4 , 3,4, , ( )T ExtMξ ξ ξ Find : *4ξ (2) (1) (1)

4 4 4( )ξ ξ ξ+ ∆=

(2,1) (2,1) (2,1) (2,1)3,4 3,4 3,4 3,4(2,1)

3,4 (2,1) (2,1) (2,1)3,4 3,4 3,4

* (1)3,4 4

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )G B T

T TExt Ext T

mg z g V z gIM M

F z g i

ξ ξ ξ ξξ ξ

ξ ξ ξ

ρ ρξ

ρ

⋅ − ⋅ ⋅ −= + ∆ − ⋅ + ⋅


*3,4( ) 0TM ξ =

4

120/120

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