Ship Stability

Ship Stability

September 2013

Myung-Il Roh

Department of Naval Architecture and Ocean Engineering Seoul National University

Planning Procedure of Naval Architecture and Ocean Engineering

Ship Stability

þ Ch. 1 Introduction to Ship Stability þ Ch. 2 Review of Fluid Mechanics þ Ch. 3 Transverse Stability

þ Ch. 4 Initial Transverse Stability þ Ch. 5 Free Surface Effect

þ Ch. 6 Inclining Test

þ Ch. 7 Longitudinal Stability

þ Ch. 8 Curves of Stability and Stability Criteria

þ Ch. 9 Numerical Integration Method in Naval Architecture þ Ch. 10 Hydrostatic Values

þ Ch. 11 Introduction to Damage Stability þ Ch. 12 Deterministic Damage Stability

þ Ch. 13 Probabilistic Damage Stability (Subdivision and Damage

Stability, SDS)

Ch. 7 Longitudinal Stability

Static Equilibrium

Longitudinal Stability

Longitudinal Stability in Case of Small Angle of Trim Longitudinal Righting Moment Arm

Derivation of Longitudinal Metacentric Radius (BM _L )

Moment to Trim One Degree and Moment to Trim One

Centimeter (MTC)

Static Equilibrium

L

1

W

1

F G

F B G

B

Static Equilibrium

Static Equilibrium

I w ^& = å t

② Euler equation

When the buoyant force( F _B ) and the gravitational force( F _G ) are on one line,

the total moment about the transverse axis for the ship to be in static equilibrium

0 = å t , ( ^Q w ^& = 0)

① Newton’s 2 ^nd law

ma = å F F G

= - + F _B

for the ship to be in static equilibrium

0 = å F , ( ^Q a = 0)

G F B

F

\ =

G: Center of mass of a ship F

_G

: Gravitational force of a ship

B: Center of buoyancy at initial position F

_B

: Buoyant force acting on a ship I: Mass moment of inertia

w : Angular velocity t : Moment

m: Mass of a ship

a: Acceleration of a ship

W₁L₁: Waterline at initial position

.

A P F P .

Longitudinal Stability

Longitudinal Stability - Stable Equilibrium

② Then, release the external moment by moving the weight to its original position.

③ Test whether it returns to its initial equilibrium position.

Z

_L

: Intersection point of the vertical line to the waterline W

₂

L

₂

through the changed position of the center of buoyancy (B

₁

) with the horizontal line parallel to the waterline W

₂

L

₂

through the center of mass of the ship(G)

Return to its initial equilibrium position

B₁: Changed position of the center of buoyancy after the ship has been trimmed

P

A. F. P

F

G

F

B

B t

e

W₁L₁: Waterline at initial position W₂L₂ : Waterline after trim

F

G

F

B

B

1

① Produce an external trim moment by moving a weight in the longitudinal direction ( t

_e

).

W

2

L

2

Z

L

G

A.P : after perpendicular, F.P : forward perpendicular

t

r

Longitudinal righting GZ

_L

Æ Longitudinal

L

1

W

1

G

1

q

Reference Frames

F

G

F

B

B t

e

F

G

F

B

B

1

W

2

L

2

Z

L

G

t

r

L

1

W

1

q

t

e

B

1

q

B G

L

1

W

1

Z

^L

L

2

W

2

F

G

Rotation of the water plane fixed frame while the body(ship) fixed frame is fixed

Rotation of the body(ship) fixed frame with

respect to the water plane fixed frame

Same in view of the Mechanics!!

F

B

, O O¢

x z

Longitudinal stability of a ship - Stable Condition (1/3)

B 1

z¢

x¢

q F G

, z¢

, x¢

t t

Trim moment

F B

G

B

① Apply an external trim moment to the ship

② Then release the external moment

③ Test whether it returns to its initial equilibrium position.

stem

stern

Longitudinal stability of a ship - Stable Condition (2/3)

B

B 1

x¢

z¢

x z

, O O¢

t t

Trim moment

F B r ^B

¹

= r G

Resultant moment about y-axis through point O ( ) : τ ^e

´F G +r _B

₁

´F _B τ e

, ,

, ,

, ,

( )

( )

( )

_{G G z G G y}

G G z G G x

G G y G G x

y F z F x F z F

x F y F

× - ×

+ - × + ×

+ × - ×

= i j k

1 1

1 1

1 1

, ,

, ,

, ,

( )

( )

( )

B B z B B y

B B z B B x

B B y B B x

y F z F

x F z F

x F y F

+ × - ×

+ - × + ×

+ × - ×

i j k

, , ,

G G G

G x G y z

G

G

G

x y z

F F F

´ =

i j k

r F

, ,

, ,

, ,

( )

( )

( )

G G z G G y

G G z G G x

G G y G G x

y F z F

x F z F

x F y F

× - ×

+ -

= × + ×

+ × - ×

i j k

, ,

( - x

_G

× F

_{G z}

+ z

_G

× F

_{G x}

)

= j

1

,

1

,

( x _B F _{B z} z _B F _{B x} )

+ - j × + ×

, , ,

0 0

G x

G G y

G z

F F

F W

é ù é ù

ê ú ê ú

= ê ú = ê ú ê ú ê ë - ú û

ë û

F

, ,

0 0

B x

B B y

F F

é ù é ù ê ú ê ú

= ê ú = ê ú F

( )

₁

( - x

_G

× - W - x

_B

)

= j × D

( )

₁

( x _G D x _B )

= j - × - - × D W = D

If

F G r _G

q

G

Longitudinal stability of a ship - Stable Condition (3/3)

B

B 1

x z

, O O¢

F B r ^B

¹

F G

G

r G

x¢

z¢

q The moment arm induced by the

buoyant force and gravitational force is expressed by GZ

_L

, where Z

_L

is the intersection point of the line of buoyant force(D) through the new position of the center of buoyancy(B

₁

) with a transversely parallel line to a waterline through the center of the ship’s mass(G).

r GZ L

t = D ×

• Longitudinal Righting Moment

= r G

Resultant moment about y-axis through point O ( ) : τ ^e

´F G +r _B

₁

´F _B τ e

GZ L

= × D × j

( x _G x _B

1

) D

= j × -

Z L

x G

B

1

x

, z¢

, x¢

Position and Orientation of a Ship

with Respect to the Water Plane Fixed and Body(Ship) Fixed Frame

Rotation of the water plane fixed frame while the body(ship) fixed frame is fixed Rotation of the body(ship) fixed frame with

respect to the water plane fixed frame

Same in view of

the Mechanics!!

B B 1

x¢

z¢

x z

, O O¢

F

B

F

G

q

G

B

관계 ID가 rId28인 이미지 부분을 파일에서 찾을 수 없습니다.

x¢

z¢ z

, O O¢

F

B

F

G

q

G

Emerged volume

Submerged volume

The ship is rotated.

The water plane is rotated.

Longitudinal Stability

in Case of Small Angle of Trim

Assumptions for Small Angle of Trim

Assumptions

① Small angle of inclination (3°~5° for trim)

② The submerged volume and the emerged volume are to be the same.

q

P

A. F. P

F

G

F

B

Emerged volume Submerged volume

t

e

W₁L₁: Waterline at initial position W₂L₂ : Waterline after trim

F

G

F

B

L

1

W

1

W

2

L

2 L

G

Z

B B

1

t

r

Longitudinal Metacenter (M _L ) (1/3)

Longitudinal Metacenter (M _L )

The intersection point of

a vertical line through the center of buoyancy at a previous position (B)

with a vertical line through the center of buoyancy at the present position (B ₁ ) after the ship has been trimmed.

q

P

A. F. P

F

B

W₁L₁: Waterline at previous position W₂L₂ : Waterline after trim

F

B

L

1

W

²1

W

L

2

1

M L

B B

1

G

※ Metacenter “M” is valid for small angle of inclination.

q

P

A. F. P

F

B

W

2

L

2

2

M L

3

G W

L

3

M _L remains at the same position for small angle of trim, up to about 2~5 degrees.

F

B

As the ship is inclined with a small trim angle, B moves on the arc of circle whose center is at M

_L

.

The BM

_L

is longitudinal metacentric radius.

B

2

B B

₁

The GM

_L

is longitudinal metacentric height.

L 1

M

Longitudinal Metacenter (M _L ) (2/3)

q

P

A. F. P

3

G W

L

3

F

B

W

4

L

4

M _L does not remain in the same position for large trim angles over 5 degrees.

F

B

B

3

B

2

B B

₁

Thus, the longitudinal metacenter, M _L , is only valid for a small trim angle.

L 3

M ^M

^L1

^{, M}

^L2

Longitudinal Metacenter (M _L ) (3/3)

② The submerged volume and the emerged volume are to be the same.

Longitudinal Stability for a Box-Shaped Ship

① Apply an external trim moment( t

_e

) which results in the ship to incline with a trim angle q .

② For the submerged volume and the emerged volume are to be the same, the ship rotates about the transverse axis through the point O.

: Midship

q

P

A. F. P

F

G

K F

B

Emerged volume Submerged volume

t

e

t

r

W₁L₁: Waterline at initial position

W₂L₂: Waterline after a small angle of trim

F

G

F

B

L

1

W

1

W

2

L

2

O

B B

1

G

Assumption

① A small trim angle

(3°~5° )

About which point a box-shaped ship rotates, while the submerged volume

and the emerged volume are to be the same?

Longitudinal Stability for a General Ship

What will happen if the hull form of a ship is not symmetric about the transverse(midship section) plane through point O?

The submerged volume and the emerged volume are not same!

Assumption

① A small angle of inclination

(3°~5° for trim)

② The submerged volume and the emerged volume are to be the same.

The hull form of a general ship is not symmetric about the transverse(midship section) plane.

q

P

A. F. P

F

G

K F

B

B

Emerged volume Submerged volume

F

G

F

B

B

1

L

1

W

1

W

2

L

2

G

So, the draft must be adjusted to maintain same displacement.

O

The intersection of the initial waterline(W

₁

L

₁

) with the adjusted waterline(W

₃

L

₃

) is a point F, on which the submerged volume and the emerged volume are supposed to be the same.

So, the draft must be adjusted to maintain same displacement.

q

P

A. K F. P

B

F G

F B

B 1

L 1

W 1 ^G

W 3

L 3

Submerged volume

Emerged volume

W 2

L 2

F O

Expanded view

What we want to find out is the point F. How can we find the point F ?

W₁L₁: Waterline at initial position

W₂L₂ : Waterline after a small angle of trim W₃L₃ : Adjusted waterline

x¢

W 1

x

q

Rotation Point (F)

= Emerged volume(v

_f

) Submerged volume(v

_a

)

.

.

( tan ) ( tan )

F F P

A P

y ¢ × x ¢ × q dx ¢ =

F

y ¢ × x ¢ × q dx ¢

ò ò

. .

F F P

A P

x y dx ¢ × ¢ ¢ =

F

x y dx ¢ × ¢ ¢

ò ò

( ) V x¢

( ) A x dx ¢ ¢

= ò

L

1

P F.

W

1

W

2

L

2

q

P A.

q

K

F

  ^

 ^



^

tan  q

( ) ( ^{y ¢ x ¢ tan q} )

= × ×

( ) A x¢

tan x y ^{¢ ¢} q

= × ×

L

C

x¢

G

B

1

B

Longitudinal Center of Floatation (LCF) (1/3)



^

: Longitudinal coordinate from the temporary origin F



^

: Breadth of waterline W

₁

L

₁

at any position 

^

F _O

From now on, the adjusted waterline is indicated as W

₂

L

₂

.

W₁L₁: Waterline at initial position W₂L₂ : Adjusted waterline

x¢

x z¢

y¢

z

. F

A P

x y dx ¢ ¢ ¢

= ò ×

That means the point F lies on the transverse axis through the centroid of the water plane, called longitudinal center of floatation(LCF).

a f

v = v

^.

.

( tan ) ( tan )

F F P

A P

y ^¢ × x ^¢ × q dx ^¢ =

F

y ^¢ × x ^¢ × q dx ^¢

ò ò

.

F F P

x y dx ¢ × ¢ ¢ = x y dx ¢ × ¢ ¢

ò ò

Longitudinal moment of the after water plane area from F about the transverse axis(y’) through the point F

dx¢ y¢

dA x¢

F x¢

y¢

. F

A P

x dA ¢

ò

Longitudinal moment of the forward water plane area from F about the transverse axis(y’) through the point F

. F P

F

x dA ¢

ò = ò

F^{F P.}

x y dx ¢ × ¢

Since these moments are equal and opposite, the longitudinal moment of the entire water plane area about the transverse axis through point F is zero.

<Plan view of waterplane>

Longitudinal Center of Floatation (LCF) (2/3)

Therefore, for the ship to incline under the condition that the submerged volume and emerged volume are to be the same,

the ship rotates about the transverse axis through the longitudinal center of floatation (LCF).

F Longitudinal Center of Floatation (LCF)

L

1

P F.

W

1

W

2

L

2

P A.

M L

q

K

F ^G

B

1

B

Longitudinal Center of Floatation (LCF) (3/3)

q

L

O Z

x¢

x z¢

y¢

Longitudinal Righting Moment Arm

B

M

L

Longitudinal Righting Moment Arm (GZ _L )

: Vertical center of buoyancy at initial position



_

: Longitudinal metacentric height



_

: Longitudinal metacentric radius

GM L = KB + BM _L - KG

KB ≒ 51~52% draft Vertical center of mass of the ship

How can you get the value of the BM

_L

?

q

B

1

F B

Z

L

F G

q F ^G

L B

GZ

= × F

Longitudinal Righting Moment

From geometrical configuration

L L sin

GZ @ GM × q

with assumption that M

_L

remains at the same position within a small angle of trim (about 2°~5°)

x¢

x

z¢

Derivation of Longitudinal Metacentric

Radius (BM _L )

Derivation of Longitudinal

Metacentric Radius (BM _L ) (1/8)

Because the triangle gg'g

₁

is similar with BB'B

₁

1

1

BB BB gg gg

= ¢

¢

1 1

BB = v × gg Ñ

The distance BB

₁

equals to v

₁

gg Ñ ×

q

, ,

( )

B

v a v f

x

x x

d ¢

= ¢ + ¢ (

_{, ,}

)

B

v a v f

z

z z

d ¢

= ¢ + ¢

x'

_v,f

, z'

_v,f

: longitudinal and vertical distance of the emerged volume from F x'

_v,a

, z'

_v,a

: longitudinal and vertical distance

of the submerged volume from F

B₁: Changed position of the center of buoyancy

B: Center of buoyancy at initial position Ñ: Displacement volume

v: Submerged / Emerged volume

Let’s derive longitudinal metacentric radius BM

_L

when a ship is trimmed.

, g: Center of the emerged volume , g₁: Center of the submerged volume

The center of buoyancy at initial position(B) moves parallel to gg

₁

.

M L

q

B 1

P F.

q

P A.

W

2

L

2

M L

B

1

q

K

g

L

1

W

1

F

t

r

g¢

,

z¢ v f ,

z¢ v a

,

x¢ v f ,

x¢ v a

1 1

1 1

BB B B gg g g

= ¢

¢

B¢ B

1

O

g x¢

x z¢

y¢

F z¢

y¢ x¢

② The submerged volume and the emerged volume are to be the same.

Assumption

① A small trim angle

(3°~5° )

Derivation of Longitudinal

Metacentric Radius (BM _L ) (2/8)

tan

2

L

B

B M × q = B

2 L

tan BM BB

= q

2 2

BB = BB ¢ + B B ¢

(

2

)

1

tan BB B B

q ^{¢ ¢}

= +

( )

1 tan

tan d x

^B

d z

^B

q

q ^{¢ ¢}

= +

, , , ,

1 ( ) ( ) tan

tan

^{v a v f v a v f}

v v

x x z z q

q

æ ö

¢ ¢ ¢ ¢

= ç × + + × + ÷

Ñ Ñ

è ø

(

, , , ,

)

1 ( ) tan

tan

^{v a v f v a v f}

L

v x v x v

BM z v z q

q ^{¢ ¢ ¢ ¢}

= × + × + × + ×

Ñ ×

Find!

, ,

( ) ...(1)

B v a v f

x v x x

d ^¢ = × ^¢ + ^¢ Ñ

, ,

( ) ...(2)

B v a v f

z v z z

d ¢ = × ¢ + ¢ Ñ

Substituting (1), (2) into 

_^

and 

_^

q

x B

d ^¢ z B

d ^¢

B M L

q

B 1 2 B¢

B

P F.

q

P A.

W2

L2

M

L

q

K

L1

W1

F

tr

g¢

,

z¢

v f ,

x¢

v f ,

x¢

v a

B¹2

B B¢ B

O

g g1

,

z¢

v a

F z¢

y¢ x¢

(

^{, , , ,}

)

1 ( ) tan

tan

^{v f v a v f v a}

L

v x v x v

BM z v z q

q ^{¢ ¢ ¢ ¢}

= × + × + × + ×

Ñ × (A)

(A) : Moment of the emerged volume

about y’-z’ plane

P F.

q

P A.

W

2

L

2

M L

q

K

g

g

1

L

1

W

1

t

r

g¢

,

z¢ v f ,

z¢ v a

,

x¢ v f ,

x¢ v a

x¢

F

(B) (C) (D)

B

¹2

B B¢ B

,

v x¢ × v f

.

( )

F P

F

A x ¢ x dx ¢ ¢

= ò ×

. F P .

tan

F P

F

x y ^{¢ ¢} q x dx ^{¢ ¢}

= ò × × ×

y: Breadth of waterline W

₁

L

₁

at any distance x from point F[m]

x¢

x z¢

y¢

F

  ^

 ^



^

tan 

q ⁼ ( ) ( ^{y ¢ × x ¢ × tan q} )

( ) A x¢

tan x y ^{¢ ¢} q

= × ×

L

C

Derivation of Longitudinal

Metacentric Radius (BM _L ) (3/8)

x¢

x

(

^{, , , ,}

)

1 ( ) tan

tan

^{v f v a v f v a}

L

v x v x v

BM z v z q

q ^{¢ ¢ ¢ ¢}

= × + × + × + ×

Ñ × (A)

F

A(x) y

x tan q

q ⁼ ( ^{2 y} ) ( ^{× x × tan q} )

( ) A x

2 x y tan q

= × ×

,

v x¢ × v a

.

( )

F

A P

A x ¢ x dx ¢ ¢

= ò ×

( )

²

tan

^F.

A P

x y dx

q ¢ ¢ ¢

= ò ×

( )

.

tan

F

A P

x y ¢ ¢ q x ¢ dx ¢

= ò × × ×

x

L

C

P F.

q

P A.

W

2

L

2

M L

q

K

g

g

1

L

1

W

1

t

r

g¢

,

z¢ v f ,

z¢ v a

,

x¢ v f ,

x¢ v a

x¢

F

(B) (C) (D)

(B) : Moment of the submerged volume

about y’-z’ plane

  ^

 ^



^

tan  q

( ) ( ^{y ¢ x ¢ tan q} )

= × ×

( ) A x¢

tan x y ^{¢ ¢} q

= × ×

 ^

L F C

B

¹2

B B¢ B

z¢

y¢

Derivation of Longitudinal

Metacentric Radius (BM _L ) (4/8)

x¢

x z¢

F

  ^

 ^



^

tan 

q ⁼ ( ) ( ^{y ¢ × x ¢ × tan q} )

( ) A x¢

tan x y ^{¢ ¢} q

= × ×

,

v z¢ × v f

(C)

.

( )

F P

F

A x ¢ z dx ¢

= ò ×

( )

.

tan tan

2

F P F

x y ¢ ¢ q ^æ x ^¢ q ^ö dx ¢

= × × ç ÷

è ø

ò

: Moment of the emerged volume about x’-y’ plane

1 tan

z ^¢ = 2 x ^¢ × q

L

C

P F.

q

P A.

W

2

L

2

M L

q

K

g

g

1

L

1

W

1

t

r

g¢

,

z¢ v f ,

z¢ v a

,

x¢ v f ,

x¢ v a

F

(

^{, , , ,}

)

1 ( ) tan

tan

^{v f v a v f v a}

L

v x v x v

BM z v z q

q ^{¢ ¢ ¢ ¢}

= × + × + × + ×

Ñ × (A) (B) (C) (D)

2

x¢

B

¹

B

B¢ B

y¢

Derivation of Longitudinal

Metacentric Radius (BM _L ) (5/8)

x¢

x z¢

F

A(x) 2y

x tan q

q ⁼ ( ^{2 y} ) ( ^{× x × tan q} )

( ) A x

2 x y tan q

= × ×

x

Derivation of Longitudinal

Metacentric Radius (BM _L ) (6/8)

L

C

P F.

q

P A.

W

2

L

2

M L

q

K

g

g

1

L

1

W

1

t

r

g¢

,

z¢ v f ,

z¢ v a

,

x¢ v f ,

x¢ v a

F

(

^{, , , ,}

)

1 ( ) tan

tan

^{v f v a v f v a}

L

v x v x v

BM z v z q

q ^{¢ ¢ ¢ ¢}

= × + × + × + ×

Ñ × (A) (B) (C) (D)

,

v z¢ × v a

.

( )

F

A P

A x ¢ z dx ¢ ¢

= ò ×

( )

.

tan tan

2

F A P

x y ¢ ¢ q ^æ x ^¢ q ^ö dx ¢

= × × ç ÷

è ø

ò

( )

2 2

.

tan 2

F

A P

x y dx

q ¢ ¢ ¢

= ò ×

(D) : Moment of the submerged volume

about x’-y’ plane

x¢

  ^

 ^



^

tan  q

( ) ( ^{y ¢ x ¢ tan q} )

= × ×

( ) A x¢

tan x y ^{¢ ¢} q

= × ×

 ^

L F C

( ) ( ^{y x ¢ ¢ tan q} )

= × ×

B

1

B¢ B B

2

y¢

Derivation of Longitudinal

Metacentric Radius (BM _L ) (7/8)

1 1

3

tan . tan

tan q I

^L

2 q I

^L

q

æ ö

= ç × + × ÷

Ñ × è ø

1

I æ ö

By substituting (A), (B), (C), and (D) into the BM

_L

equation

,

v x¢ ×

v a

^tan

^F

( )

²

AP

x y dx

q ^{¢ ¢ ¢}

= ò ×

(B)

,

v x¢ ×

v f

^tan

^FP

( )

²

F

x y dx

q ¢ ¢ ¢

= ò ×

(A)

(D) v z¢ ×

v a,

( )

2 2

tan 2

F

AP

x y dx

q ¢ ¢ ¢

= ò ×

(C) v z¢ ×

_{v f,}

( )

2 2

tan 2

FP

F

x y dx

q ¢ ¢ ¢

= ò ×

( ) ( )

(

^{. 2 . 2}

)

³

(

^.

^{( )}

^{2 .}

^{( )}

²

)

1 tan

tan tan 2

F P F F P F

F

x y dx

A P

x y dx q

F

x y dx

A P

x y dx

q q

æ ö

¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢

= ç × + × + × + × ÷

Ñ × è ò ò ò ò ø

(

, , , ,

)

1 ( ) tan

tan

^{v f v a v f v a}

L

v x v x v

BM z v z q

q ^{¢ ¢ ¢ ¢}

= × + × + × + ×

Ñ × (A) (B) (C) (D)

( ) ( )

(

^{. 2 . 2}

)

,

_L ^{F P F}

F A P

I = ò x ¢ × y dx ¢ ¢ + ò x ¢ × y dx ¢ ¢

I

_L

: Moment of inertia of the entire water plane about transverse axis through its centroid F

( ) ( ) ( ) ( )

2 2

. 2 2 . 2 2

. .

1 tan tan

tan tan tan

tan 2 2

F P F F P F

F

x y dx

A P

x y dx q

F

x y dx q

A P

x y dx

q q q

q

æ æ ö ö

¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢

= ç × × + × × + ç × + × ÷ ÷

Ñ × è ò ò è ò ò ø ø

P F.

q

P A.

W

2

L

2

M

L

q

K

g

g

1

L

1

W

1

F

t

r

g¢

,

z¢

v f ,

z¢

v a

x¢

_{v a,}

x¢

^{v f,}

B

¹2

B

B¢ B

O

x¢

x

z¢

Derivation of Longitudinal

Metacentric Radius (BM _L ) (8/8)

L L

BM = I Ñ

which is generally known as BM

_L

.

1 2

1 tan

2

L

BM L I

æ q ö

= ç + ÷

Ñ è ø

If θ is small, tan

²

θ » θ

²

=0

The BM

_L

does not consider the change of center of buoyancy in vertical direction.

In order to distinguish between them, the two will be indicated as follows:

2 0

(1 1 tan ) 2

L L

BM = I + q

Ñ

L L

BM = I Ñ

(Considering the change of center of buoyancy in vertical direction)

(Without considering the change of center of buoyancy in vertical direction) P F.

q

P A.

W

2

L

2

M

L

q

K

g

g

1

L

1

W

1

F

t

r

g¢

,

z¢

v f ,

z¢

v a

,

x¢

v f ,

x¢

v a

B

¹2

B

B¢ B

O

z¢

x¢

x

Moment to Trim One Degree and Moment to Trim One Centimeter

(MTC)

LCF: Longitudinal center of floatation

Z_L: Intersection point of the vertical line to the water surface through the changed center of buoyancy with the horizontal line parallel to the water surface through the center of mass

M_L: Intersection point of the vertical line to the water surface through the center of buoyancy at previous position(B) with the vertical line to the water surface through the changed position of the center of

buoyancy(B₁) after the ship has been trimmed.

Moment to Trim One Degree

F

G

G B τ

e

F q

B

1

F B

F G

M L

F

B

Definition

It is the moment of external forces required to produce one degree trim.

Assumption

Small trim where the metacenter is not changed (about 2~5°)

When the ship is trimmed, the static equilibrium states:

B L sin

F GM × × q

;

Moment to trim = Righting moment

Moment to trim one degree = F GM _B × _L × sin1 ^°

B L

F GZ

= ×

(at small angle ) q

q = 1 ^°

Substituting

τ

r

q

Z

L

F G

G B τ

e

F B

1

F B

B L

sin1

F GM

^°

= × ×

Moment to trim one degree

M L

Often, we are more interested in the changes in draft produced by a trim moment than the changes in trim angle.

1

B L 100

BP

MTC F GM

= × × L

× sin tan t

q » q = L Trim(t): d

_a

– d

_f

,

P P F.

A. L

^BP

L

2

1

^°

LBP

A.P F.P

t d

_a

d

_f

q

B.L F

L

BP

q

Moment to Trim One Centimeter (MTC) (1/2)

Moment to Trim One Centimeter (MTC) (2/2)

1

B L

100 MTC F GM

= × × LBP

×

L L

GM = KB + BM - KG

As practical matter, KB and KG are usually so small compared to GM _L that BM _L can be substituted for GM _L .

1 1

100 100

B L B L

BP BP

MTC F GM F BM

L L

= × × » × ×

× ×

1 100

L

BP

MTC I r L

= × Ñ × ×

Ñ ×

100

L BP

I L

r ×

= ×

G B τ

e

F B

1

F B

M L

I

_L

: Moment of inertia of the water plane area about y’ axis B ,

F = r × Ñ _L I ^L BM = Ñ Substituting

K

L

2

F G

P A.

1

^°

L

BP

F. P

Example

W

1

L

1

Example) Calculation of Draft Change Due to Fuel Consumption (1/4)

During a voyage, a cargo ship uses up 320 tones of consumable stores (H.F.O:

Heavy Fuel Oil), located 88 m forward of the midships.

Before the voyage, the forward draft marks at forward perpendicular recorded 5.46 m, and the after marks at the after perpendicular, recorded 5.85 m.

At the mean draft between forward and after perpendicular, the hydrostatic data show the ship to have LCF after of midship = 3 m, Breadth = 10.47 m,

moment of inertia of the water plane area about transverse axis through point F

= 6,469,478 m ⁴ , Cwp = 0.8.

Calculate the draft mark the readings at the end of the voyage, assuming that there is no change in water density(r=1.0 ton/m ³ ).

P P F.

A.

F

BP 195

L = m

3 m

88 m

Consumable

Stores

5.46m

5.85m

① Calculation of parallel rise (draft change)

§ Tones per 1 cm immersion (TPC)

F

195 m 3 m

88 m

H.F.O Tank

: 1

WP 100

TPC = r × A × ^{3 2} 1

1[ / ] 1, 633.3[ ]

100[ / ]

ton m m

= × × cm m

20.4165[ ton cm / ]

=

§ Parallel rise : weight

d TPC

d = 320[ ]

20.4165[ / ] ton

ton cm

= = 15.6736[ cm ] = 0.1567[ ] m

P P F.

A.

W

1

L

1

W

2

L

2

0.1567m

2

0.8 195 10.47 1, 633.3[ ]

WP WP

A C L B m

= × ×

= × ×

=

Example) Calculation of Draft Change

Due to Fuel Consumption (2/4)

W

2

L

₂

② Calculation of trim

§ Trim moment ^: t trim _{= 320[ ton ] 88[ ] × m = 28,160[ ton m × ]}

§ Moment to trim 1 cm (MTC)

: 100

L BP

MTC I

L r ×

= ×

3

1[ / ] 4

6, 469, 478[ ] 100[ / ] 195[ ]

ton m

cm m m m

= ×

× = 331.7949[ ton m cm × / ]

§ Trim

: Trim ^trim MTC

= t 28,160[ ]

331.7949[ / ]

ton m ton m cm

= ×

× = 84.8785[ cm ] = 0.8488[ ] m

x

195 m 3 m

P P F.

A.

F

L

3

W

3

88 m

H.F.O Tank

Example) Calculation of Draft Change

Due to Fuel Consumption (3/4)

L

3

W

3

③ Calculation of changed draft at F.P and A.P

§ Draft change at F.P due to trim 195 / 2 3

0.8488 195

= - + ´ = - 0.4375[ ] m

§ Draft change at A.P due to trim 195 / 2 3

0.8488 195

= - ´ = 0.4113[ ] m

§ Changed Draft at F.P : draft – parallel rise - draft change due to trim 5.46[ ] 0.1567[ ] 0.4375[ ] m m m

= - - = 4.8658[ ] m

F

195 m 3 m

P P F.

A.

88 m

H.F.O Tank

F

P

A. F. P

3 m

195 m 0.8488[ ] m

Example) Calculation of Draft Change Due to Fuel Consumption (4/4)

(195 : 0.8488 = l

_a

: ?)

l

_a

l

_b

(195 : 0.8488 = l

_b

: ?)

Reference Slides

Derivation of Longitudinal Metacentric Radius (BM _L ) by Using the Origin of

the Body Fixed Frame

Derivation of BM _L (1/12)

Assumption

1. A main deck is not submerged.

Let us derive longitudinal metacentric radius “BM _L ".

2. Small angle of inclination (3°~5° for trim)

※ The ship is not symmetrical with respect to midship section. Thus to keep the same displaced volume, the axis of inclination does not stand still. In small angle of inclination, the axis of

K

Z L

B 1

x¢

z¢

F B

W ₁ L ₁

B

G

F G

O

O¢

x

q F

M L

② The submerged volume and the emerged volume are to be the same.

Assumption

① A small trim angle

(3°~5° )

Derivation of BM _L (2/12)

B₁: The center of buoyancy after inclination B: The center of buoyancy before inclination Ñ: Displacement volume

v: Submerged / Emerged volume

g: The center of the emerged volume g₁: The center of the submerged volume

K

B 1

x¢

z¢

G

O

O¢

x

q F

g 1

g

( ) / ...(1)

B v

x x v

d ¢ = ¢ × Ñ

( ) / ...(2)

B v

z z v

d = ^¢ × Ñ z¢

v

x¢

v

x

B

d ¢

z

B

d ^¢ B

2

M L

Translation of the center of buoyancy

caused by the movement of the small volume v

② The submerged volume and the emerged volume are to be the same.

Assumption

① A small trim angle

(3°~5° )

Derivation of BM _L (3/12)

K

B 1

x¢

z¢

G

O

O¢

x

q F

M L

z¢

v

x¢

v

x

B

d ¢

z

B

d ^¢ B

2

,

x¢

v a

x¢

^{v f,}

, ,

v v f v a

x ¢ = x ¢ - x ¢

, ,

v v f v a

z ¢ = z ¢ - z ¢

(3) (4)

Substituting Eq. (3), (4) into the Eq. (1), (2), respectively.

, ,

( ) /

B v f v a

x x x v

d ¢ = ¢ - ¢ × Ñ

, ,

( ) /

B v f v a

z z z v

d ^¢ = ^¢ - ^¢ × Ñ

f

,

a

v = v - = v v Submerged

volume(v

_f

)

Emerged volume(v

_a

)

g 1

g

,

z¢

v a

,

z¢

v f

( ) / ...(1)

B v

x x v

d ¢ = ¢ × Ñ

( ) / ...(2)

B v

z z v

d = ^¢ × Ñ

, where v is the each volume of the submerged Translation of the center of buoyancy

caused by the movement of the small volume v

1

1

O g

g g g

O¢ + uuur uuur = u uuur ¢

1

g

1

g O g g = O u ¢ - ¢ uuur uuur uuur

, , , ,

( x z

_g

¢ , ¢

_g

) = ( x

_{v f}

¢ , z

_{v f}

¢ ) ( - x

_{v a}

¢ , z

_{v a}

¢ )

② The submerged volume and the emerged volume are to be the same.

Assumption

① A small trim angle

(3°~5° )

B₁: The center of buoyancy after inclination B: The center of buoyancy before inclination Ñ: Displacement volume

v: Submerged / Emerged volume

g: The center of the emerged volume g₁: The center of the submerged volume