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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 1
Ship Stability
Ch. 5 Initial Longitudinal Stability
Spring 2018
Myung-Il Roh
Department of Naval Architecture and Ocean Engineering Seoul National University
Lecture Note of Naval Architectural Calculation
Naval Architectural Calculation, Spring 2018, Myung-Il Roh 2
Contents
þ Ch. 1 Introduction to Ship Stability þ Ch. 2 Review of Fluid Mechanics
þ Ch. 3 Transverse Stability Due to Cargo Movement þ Ch. 4 Initial Transverse Stability
þ Ch. 5 Initial Longitudinal Stability þ Ch. 6 Free Surface Effect
þ Ch. 7 Inclining Test
þ Ch. 8 Curves of Stability and Stability Criteria
þ Ch. 9 Numerical Integration Method in Naval Architecture þ Ch. 10 Hydrostatic Values and Curves
þ Ch. 11 Static Equilibrium State after Flooding Due to Damage þ Ch. 12 Deterministic Damage Stability
þ Ch. 13 Probabilistic Damage Stability
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Ch. 5 Initial Longitudinal Stability
1. Longitudinal Stability
2. Longitudinal Stability in Case of Small Angle of Trim 3. Longitudinal Righting Moment Arm
4. Derivation of Longitudinal Metacentric Radius (BM L ) 5. Moment to Trim One Degree and Moment to Trim One Centimeter (MTC)
6. Example of Longitudinal Stability
Naval Architectural Calculation, Spring 2018, Myung-Il Roh 4
1. Longitudinal Stability
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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 (FB)and the
gravitational force(FG) are on one line, the total momentabout the transverse axis through any pointbecomes 0.
It does not matter where the axis of rotation is selected if the two forces are on one line.
for the ship to be in static equilibrium
0 = å t , ( Q w & = 0)
① Newton’s 2ndlaw
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 FG: Gravitational force of a ship B: Center of buoyancy at initial position
FB: 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 .
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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.
ZL: Intersection point of the vertical line to the waterline W2L2through the changed position of the center of buoyancy (B1) with the horizontal line parallel to the waterline W2L2through the center of mass of the ship(G) Return to its initial equilibrium position
Stable
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 (te ).
W 2
L 2
Z L
G
A.P : after perpendicular, F.P : forward perpendicular
t r
Longitudinal righting moment(τr) GZL Æ Longitudinal
righting arm
L 1
W 1 G 1 q
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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
t r
q
B G
L 1
W 1 2 Z L L 2
W
F B
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
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In this figure, the angle of inclination, , is negative, which means that the ship is trimmed by the stern.
q 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
equilibrium position.
stem
stern
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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
B z
F F F
é ù é ù ê ú ê ú
= ê ú = ê ú ê ú ê ú ë û D
ë û
F
( ) 1
( - x G × - W - x B )
= j × D
( ) 1
( x G D x B )
= j - × - - × D
( x G x B 1 ) D
= j × - W = D
If
F G r G q
G
In this figure, the angle of inclination, , is negative, which means that the ship is trimmed by the stern.
q
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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 GZL, where ZL is the intersection point of the line of buoyant force(D) through the new position of the center of buoyancy(B1) 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
Stable!!
= 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¢
t r
Restoring moment
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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 while the water plane fixed frame is fixed
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 수없습
x¢
z¢
x z
, O O¢
F B
F G
q
G
Emerged volume Submerged
volume The ship is rotated.
The water plane is rotated.
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2. Longitudinal Stability
in Case of Small Angle of Trim
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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
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Longitudinal Metacenter (ML) (1/3)
Longitudinal Metacenter (ML) The intersection pointof 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(B1) 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 2 1
W
L 2
1
M L
B B 1
G
※ Metacenter “M” is valid for small angle of inclination.
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q
P
A. F. P
F B
W 2
L 2
2
M L
3 G W
L 3
MLremains at the same positionfor small angle of trim, up to about 3~5 degrees.
F B
As the ship is inclined with a small trim angle, Bmoves on the arc of circle whose center is at ML.
The BML is longitudinal metacentric radius.
B 2 B 1 B
The GML is longitudinal metacentric height.
1
M L
Longitudinal Metacenter (M
L) (2/3)
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q
P
A. F. P
3 G W
L 3
F B
W 4
L 4
ML does not remain in the same positionfor large trim angles over 5 degrees.
F B
B 3
B 2 B 1 B
Thus, the longitudinal metacenter, ML, is only valid fora small trim angle.
3
M L M L 1 , M L 2
Longitudinal Metacenter (ML) (3/3)
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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
①A small trim angle (3~5°) About which point a box-shaped ship rotates, while the submerged volume
and the emerged volumeare to be the same?
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Longitudinal Stability for a General Ship
What will happen if the hull form of a ship is not symmetricabout the transverse (midship section) plane through point O?
The submerged volume and the emerged volume are not same!
Assumption①A 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 draftmustbe adjustedto maintain same displacement.
O
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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) 20
Naval Architectural Calculation, Spring 2018, Myung-Il Roh
= Emerged volume (vf) Submerged volume (va)
.
. ( 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
What does this equation mean?
V x¢ ( ) ( ) A x dx ¢ ¢
= ò
tan x y ¢ ¢ q 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 W1L1 at any position
F O
From now on, the adjusted waterline is indicated as W2L2.
W1L1: Waterline at initial position W2L2 : Adjusted waterline
x¢
x z¢
y¢
z
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 21 .
F A P x y dx ¢ ¢ ¢
= ò ×
That means the point Flieson 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
A P x y dx ¢ × ¢ ¢ = F x y dx ¢ × ¢ ¢
ò ò
Longitudinal moment of the after water plane area from Fabout 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 Fabout 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 pointFis zero.
<Plan view of water plane>
Longitudinal Center of Floatation (LCF)(2/3)
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Therefore, for the ship to incline under the condition that the submerged volume and emerged volume are to be the same,
the ship rotatesabout 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)
O q Z L
x¢
x z¢
y¢
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3. Longitudinal Righting Moment Arm
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B M L
Longitudinal Righting Moment Arm (GZL)
: Vertical center of buoyancy at initial position
: Longitudinal metacentric height
: Longitudinal metacentric radius
: Vertical center of mass of the ship
K
GM L =
KB ≒ 51~52% draft Vertical center of mass
of the ship How can you get the value of the BML?
q
B 1
F B
Z L
F G
: Longitudinal Metacenter
q F G
Longitudinal Righting Moment From geometrical configuration
L L sin
GZ @ GM × q
with assumption that MLremains at the same position within a small angle of trim (about 3~5°)
x¢
x
z¢
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4. Derivation of Longitudinal Metacentric Radius (BM L )
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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 BB1 equals to
1
v × gg Ñ
q B¢ d x B ¢ z B
d ¢
B
, ,
( )
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 BML 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 gg1.
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
= ¢
¢
, ,
( ) ...(1)
B v a v f
x v x x
d ¢ = × ¢ + ¢ Ñ
, ,
( ) ...(2)
B v a v f
z v z z
d ¢ = × ¢ + ¢ Ñ
: Longitudinal translation of B to B1
: Vertical translationof B to B1
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
①A small trim angle (3~5°)
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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) and (2) into and
q x B
d ¢ z B
d ¢
B M L
q
B 1 2 B¢
B
P F.
q
P A.
W
2L
2M L
q
K
L
1W
1F
t
rg¢
,
z¢ v f ,
x¢ v f ,
x¢ v a
B
21B B¢ B
O
g g 1
,
z¢ v a
F z¢
y¢ x¢
28 Naval Architectural Calculation, Spring 2018, Myung-Il Roh
( , , , , )
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) : Momentof 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 2 1
B B¢ B
,
v x¢ × v f .
( )
F P F A x ¢ x dx ¢ ¢
= ò ×
. ( ) 2
tan F P
F x y dx
q ¢ ¢ ¢
= ò ×
. tan
F P
F x y ¢ ¢ q x dx ¢ ¢
= ò × × ×
y: Breadth of waterline W1L1at 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 (BML)(3/8)
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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)
,
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 × × ×
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) : Momentof the submerged volume
about y’-z’ plane
tan q
( ) ( y ¢ x ¢ tan q )
= × ×
( ) A x¢
tan x y ¢ ¢ q
= × ×
F
L
C
B 2 1
B B¢ B
z¢
y¢
Derivation of Longitudinal Metacentric Radius (BM
L)(4/8) 30
Naval Architectural Calculation, Spring 2018, Myung-Il Roh
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
( )
2 . 2
tan 2
F P
F x y dx
q ¢ ¢ ¢
= ò ×
: Momentof 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)
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x¢
x z¢
Derivation of Longitudinal Metacentric Radius (BML)(6/8)
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) : Momentof the submerged volume about x’-y’ plane
x¢
tan q
( ) ( y ¢ x ¢ tan q )
= × ×
( ) A x¢
tan x y ¢ ¢ q
= × ×
F
L
C
1 tan
z ¢ = 2 x ¢ × q
B 1
B¢ B B 2
y¢
32 Naval Architectural Calculation, Spring 2018, Myung-Il Roh Derivation of Longitudinal Metacentric Radius (BML)(7/8)
1 1 3
tan . tan
tan q I L 2 q I L
q
æ ö
= Ñ × ç è × + × ÷ ø
1 2
1 tan
2
L
BM L = I æ ç + q ö ÷
Ñ è ø
By substituting (A), (B), (C), and (D) into the BML 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
tan 2
2
F
AP x y dx
q ¢ ¢ ¢
= ò ×
(C)
,
v z¢ × v f tan 2 ( ) 2
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 ¢ ¢
IL : 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
¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢
= Ñ × ç è × ò × + × ò × + ç 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 1 2
B B¢ B
O
x¢
x
z¢
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 33
Derivation of Longitudinal Metacentric Radius (BM L ) (8/8)
L L
BM = I Ñ
which is generally known as BML.
1 2
1 tan
2
L
BM L I
æ q ö
= ç + ÷
Ñ è ø
If θ is small, tan2θ» θ2=0 The BMLdoes not considerthe change of center of buoyancyin vertical direction.
In order to distinguish between them, the two will be indicated as follows:
(Consideringthe change of center of buoyancy in vertical direction) (Without consideringthe 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 1 2
B B¢ B
O z¢
x¢
x
34 Naval Architectural Calculation, Spring 2018, Myung-Il Roh
5. Moment to Trim One Degree and Moment to Trim One Centimeter
(MTC)
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 35
Moment to Heel One Degree
B 1
M t e
F B
Definition It is the moment of external forcesrequired to produce one degree heel.
Assumption Small inclination where the metacenter is not changed (about 7~10°)
f
When the ship is heeled and in static equilibrium, Moment to heel = Righting moment
Moment to heel one degree = F GM B × × sin1 ° t r
F GZ B
= ×
(at small angle ) f Z
f = 1 °
Substituting G: Center of mass
B: Center of buoyancy at initial position FG: Gravitational force of a ship FB: Buoyant force acting on a ship B1: New position of center of buoyancy after the ship has been inclined Z : The intersection of vertical line through the center of buoyancy with the transversally parallel line to a waterline through center of mass M: The intersection point of a vertical line through the center of
buoyancy at initial position(B) with a vertical line through the new position of the center of buoyancy(B1) after the ship has been inclined transversally through a small angle
F G
36 Naval Architectural Calculation, Spring 2018, Myung-Il Roh
LCF: Longitudinal center of floatation ZL: 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 ML: 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 3~5°)
When the ship is trimmed and in static equilibrium, Moment to trim = Righting moment
Moment to trim one degree = F GM B × L × sin1 °
B L
F GZ
= ×
(at small angle ) q
1 q = ° τ r
q
Z L
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 37
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.
MTC:Moment to trim 1 cm
sin tan
BP
t q » q = L
Trim (t): da – df, LBP: Length between perpendiculars [m]
da: Draft after df: Draft forward
P P F.
A. L BP
L 2
( If q is small ) 1 °
LBP
A.P F.P
t da
df q
B.L
F
L BP
q
Moment to Trim One Centimeter (MTC) (1/2)
1 t = cm
38 Naval Architectural Calculation, Spring 2018, Myung-Il Roh 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 GMLthat BMLcan be substituted for GML.
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 ×
G B τ e
F B 1
F B
M L
IL: 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
2018-08-07
Naval Architectural Calculation, Spring 2018, Myung-Il Roh 39
6. Example of Longitudinal Stability
Naval Architectural Calculation, Spring 2018, Myung-Il Roh 40
W 1
L 1
[Example] Calculation of Draft Change Due to Fuel Consumption (1/4)
During a voyage, a cargo ship uses up 320ton of consumable stores (H.F.O: Heavy Fuel Oil), located 88m forward of the midships.
Before the voyage, the forward draft marks at forward perpendicular recorded 5.46m, and the after marks at the after perpendicular, recorded 5.85m.
At the mean draft between forward and after perpendicular, the hydrostatic data show the ship to have LCF after of midship = 3m, Breadth = 10.47m, moment of inertia of the water plane area about transverse axis through point F = 6,469,478m4, 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=1.0ton/m3).
P P F.
A.
F
BP 195
L = m
3 m
88 m
Consumable Stores
5.46m
5.85m
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 41
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) 42Naval Architectural Calculation, Spring 2018, Myung-Il Roh
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 F.
P A.
F
L 3
W 3 88 m
H.F.O Tank [Example] Calculation of Draft Change
Due to Fuel Consumption (3/4)
2018-08-07
Naval Architectural Calculation, Spring 2018, Myung-Il Roh 43
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
§ Changed Draft at A.P : draft – parallel rise + draft change due to trim
5.85[ ] 0.1567[ ] 0.4113[ ] m m m
= - +
F
195 m 3 m
P P F.
A.
6.1046[ ] m
= 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 = la : ?)
la lb
(195 : 0.8488 = lb : ?)
44 Naval Architectural Calculation, Spring 2018, Myung-Il Roh
Reference Slides
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 45
[Appendix] Derivation of Longitudinal Metacentric Radius (BM L ) by Using the
Origin of the Body Fixed Frame
Naval Architectural Calculation, Spring 2018, Myung-Il Roh 46
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 inclination passes through the point “F” (longitudinal center of floatation).
K Z L
B 1
x¢
z¢
F B
W1 L1
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
①A small trim angle (3~5°)
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 47
Derivation of BM L (2/12)
B 1 : 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 g 1 : 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
, where
v
is the each volume of the submerged and emerged volume.
Ñ
is total volume of the ship.
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
①A small trim angle (3~5°)
48 Naval Architectural Calculation, Spring 2018, Myung-Il Roh
Derivation of BML(3/12)
K
B 1
x¢
z¢
G
O O¢
x
q F
M L
z¢ v
x¢ v
x B
d ¢ B
d z ¢ 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
, ,
( ) /
B v f f v a a
x x v x v
d ¢ = ¢ × + ¢ × Ñ
, ,
( ) /
B v f f v a a
z z v z v
d ¢ = ¢ × + ¢ × Ñ
Submerged volume(vf) Emerged
volume(va)
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 Ñ
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
①A 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
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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 49
( , , , , )
1 ( ) tan
tan v a a v f f v a a v f f
L x v x v
BM z v z v q
q ¢ ¢ ¢ ¢
= × + × + × + ×
Ñ ×
Derivation of BML(4/12)
, , , ,
1 1 1
( ) ( ) tan
tan x v a v a x v f v f z v a v a z v f v f q q
æ ¢ ¢ ¢ ¢ ö
= ç è Ñ × × + × + Ñ × × + × ÷ ø
Find!
( )
1 tan
tan B
L x z B
BM d d q
q ¢ ¢
= +
Substituting (1), (2) into d x'B,d z'B
, ,
( ) / ...(1)
B v a a v f f
x x v x v
d ¢ = ¢ × + ¢ × Ñ
, ,
( ) / ...(2)
B v a a v f f
z z v z v
d ¢ = ¢ × + ¢ × Ñ
: Moment about the yt axis through the point F due to the force in tz direction : Moment about the yt axis through the point F due to the force in tx direction
K
B 1
x¢
z¢
G
O O¢
x
q F
M L
x B
d ¢
z B
d ¢ B 2
,
x¢ v a x¢ v f ,
Submerged volume(vf) Emerged
volume(va)
g 1
g
,
z¢ v a
,
z¢ v f
B BM L
B cos x
d ¢ q
B sin d ¢ z q
B 3
sin 3
B M L q = B B
3 L sin BM BB
= q
( )
1 cos sin
sin d x B q d z B q
q ¢ ¢
= +
( )
1 tan
tan d x B d z B q
q ¢ ¢
= +
cos sin
sin q x B z B cos q
d d
q q
æ ö
¢ ¢
= ç + ÷
è ø
50 Naval Architectural Calculation, Spring 2018, Myung-Il Roh Derivation of BML(5/12)
(A): Moment about transverse axis through point Oof the submerged volume
, v f f
x ¢ × v
fore port F star x dv ¢ f
= ò ò ×
( )
tan q F fore star port x ¢ x ¢ x F ¢ dy dx ¢ ¢
= × ò ò × - × ×
( ) tan
fore port
F star x ¢ x ¢ x F ¢ q dy dx ¢ ¢
= ò ò × - × × ×
(A)(B)(C)(D)
( , , , , )
1 ( ) tan
tan v f f v a a v f f v a a
L x v x v
BM z v z v q
q ¢ ¢ ¢ ¢
= × + × + × + ×
Ñ ×
K
x¢
z¢
G
O O¢
x
q F
M L
,
x¢ v a x¢ v f ,
Submerged volume(vf) Emerged
volume(va)
g 1
g
,
z¢ v a
,
z¢ v f
B BM L
x¢ F x¢
x ¢ - x F ¢
F
L
C
q
x'-x'
Fdy'
dx'
dv f
(x'-x'F)tanθ
