06 transverse righting moment
TRANSCRIPT
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
N a v a
l A r c
h i t e c t u r e
& O c e a n
E n g
i n e e r i n g
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
- Ship Stability -
Part.1-II Righting Force and Moment
2009 Fall
Prof. Kyu-Yeul Lee
Department of Naval Architecture and Ocean Engineering,Seoul National University
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@
2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Righting Moment
Overview of Ship Stability
Force & Moment on a Floating Body
Newtons 2nd
LawEuler Equation
Stability Criteria
Damage Stability- MARPOL regulation
Pressure Integration Technique
Calculation Method to find GZwith respect to IMO regulation
sinGZ GM = , GM KB BM KG= +
sin L LGZ GM = , L LGM KB BM KG= +
- Overview of Ship Stability
BF GZ Transverse RightingMoment :
B LF GZ Longitudinal RightingMoment :
GZ Calculation
( )G BGZ y y= +( ) L G BGZ x x= +
Z
K
z
O
CL
y
z M
restoring
e
G
FG
B B1
N
FB
1 B y
G y
F B: Buoyancy force : Angle of Heel, : Angle of Trim( xG ,yG ,zG) : Center of mass in waterplane fixed frame( x B ,y B ,z B) : Center of buoyancy in waterplane fixed frame
y' G , y' B in body fixed frame
Rotational Transformation! yG , y B in waterplane fixed frame
Fundamental of Ship Stability
Properties which is related to hullform 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 Stability
Ch.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 Moment
Ch.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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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
N a v a
l A r c
h i t e c t u r e
& O c e a n
E n g
i n e e r i n g
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
Ch.6 Transverse Righting Moment- Sec.1 Calculation of Center of Buoyancy -
2009
Prof. Kyu-Yeul Lee
Department of Naval Architecture and Ocean Engineering,Seoul National University
- Ship Stability -
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2009 Fall, Ship Stability
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Sec.1 Calculation of Center of BuoyancySec.2 Calculation of BM, GZ in Wall Sided Ship
Sec.3 Inclining TestSec.4 Transverse Stability of ship (Unstable condition)Sec.5 Transverse Righting Moment due to Movement of CargoSec.6 Calculation of Heeling Angle due to Shift of Center of Mass
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Sec.1 Calculation of Center of Buoyancy- Rotational Transformation of Point and Frame- Example) Calculation of Center of Buoyancy of Ship with
Constant SectionMethod Direct calculating center of buoyancy in waterplane fixed frameMethod Transformation center of buoyancy from body fixed frame to
waterplane fixed frame
-
Calculation of Center of Buoyancy of Ship with Various Station
Shape 6/124
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SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
(Review)Transverse Righting Moment
Z
K
z
O,O'
y
z
)(+
CL
y
z M
righting
e
G
FG
B B1
N
FB
1 B y
G y
- Overview of Ship Stability
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
Righting Moment : Moment toreturn the ship to the upright floatingposition (Righting moment, Momentof statical stability))
BGZ F = i Transverse Righting moment
1( )righting G B B y y F = + i
Righting arm
Righting Arm ( GZ )
1G BGZ y y= +From direct calculation
We should know yG , y B1 in waterplane fixed frame
From geometrical figure withassumption that M does not changewithin small angle of heel (about 10 )
sinGZ GM = GM is related to below equation bygeometrical figure
GM KB BM KG= +
How to calculate inwaterplane fixed frame ?
1 1, B B y z
x,x'
G: Center of mass B: Center of buoyancy B1: Changed center of buoyancyK : Keel F G : Weight of ship F B : Buoyant force acting on ship
Z : The intersection of the line of buoyant force through B1 with the transverse linethrough G
M : The intersection of the line of buoyant force through B1 with the centerline of the ship7
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B
K
G
O,O'
y
z
Base Line
F B
F G
z
y
e
External moment ( e) is applied on theship in clockwise. A ship is heeled aboutorigin O through an angle of
- Transverse Righting Moment
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
Method 1. calculate center of the buoyancy(B 1) with respect to waterplane fixed frame
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
x,x'
y
z( )+
j
k
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Method 1. calculate center of the buoyancy(B 1) with respect to waterplane fixed frame
dydzdA = = dA A= ydA M y A, = zdA M z A,
A, Mz, My (with respect to waterplane fixed frame)
Integral value(area and 1 st moment of area) haveto be calculated for every position when position of ship is changed.
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
O,O'
y
z
F B
F G
e
B
G
B11 B
y
1 B z
Center of buoyancy is changed from B to B1.
External moment ( e) is applied on theship in clockwise. A ship is heeled aboutorigin O through an angle of
How to calculate center of buoyancy B1 withrespect to waterplane fixed frame?
Method 1. Calculate Center of buoyancy B1 withrespect to waterplane fixed frame directly
- Transverse Righting Moment
O'x'y'z' : Body fixed frame
Oxyz : Waterplane fixed frame
x,x'
y
z( )+
j
k
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Integral value(area and 1 st moment of area) haveto be calculated for every position when position of ship is changed.
y
z
B
G
B1
e
1 B y
1 B z
position change
y
z
Wetted surfaceat present
z
y
Area and moment of areahave to be calculated again
Wetted surfacewith changed position
=
A
M
A
M z y z A y A B B
,, ,),(11
Center of buoyancy with respect towaterplane fixed frame
- Transverse Righting Moment
O'x'y'z' : Body fixed frame
Oxyz : Waterplane fixed frame
Method 1. calculate center of the buoyancy(B 1) with respect to waterplane fixed frame
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
dydzdA = = dA A= ydA M y A, = zdA M z A,
A, Mz, My (with respect to waterplane fixed frame)
How to calculate center of buoyancy B1 withrespect to waterplane fixed frame?
Method 1. Calculate Center of buoyancy B1 withrespect to waterplane fixed frame directly
Center of buoyancy is changed from B to B1.
External moment ( e) is applied on theship in clockwise. A ship is heeled aboutorigin O through an angle of
O,O'
x,x'
y
z( )+
j
k
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y
z
Wetted surfaceat present
O
y
z
B
K
G
z
y B1
e
How to calculate center of buoyancy B1 withrespect to waterplane fixed frame?
Method 2. Calculate center of buoyancy B1with respect to body fixed frame, thentransform B1 to waterplane fixed frame
with respect to body fixed frame, ,, , A y A z A M M
' 'dA dy dz= = dA A, ' ' A y M y dA= , ' ' A z M z dA=
Integral value could be used as it is exceptintersection region with waterplane area whenposition of ship is changed.
( ) ( )+
1 B y 1 B z
- Transverse Righting Moment
O'x'y'z' : Body fixed frame
Oxyz : Waterplane fixed frame
Method 2. calculate center of the buoyancy(B 1) with respect to body fixed framethen transform frame from body fixed frame to waterplane fixed frame
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
O,O' x,x'
y
z( )+
j
k
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
y
z
B
K
G
z
y
e
y
zWetted surfaceat present
( ) ( )+ position change y
Only changed area andmoment of area have to calculated.
Wetted surfacewith changed position
z ( )+( )
- Transverse Ri htin Moment
O'x'y'z' : Body fixed frame
Oxyz : Waterplane fixed frame
Method 2. calculate center of the buoyancy(B 1) with respect to body fixed framethen transform frame from body fixed frame to waterplane fixed frame
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
How to calculate center of buoyancy B1 withrespect to waterplane fixed frame?
Method 2. Calculate center of buoyancy B1with respect to body fixed frame, thentransform B1 to waterplane fixed frame
with respect to body fixed frame, ,, , A y A z A M M
' 'dA dy dz= = dA A, ' ' A y M y dA= , ' ' A z M z dA=
Integral value could be used as it is exceptintersection region with waterplane area whenposition of ship is changed.
O,O' x,x'
y
z( )+
j
k
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SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
y
z
Wetted surfaceat present
( ) ( )+ y
Only changed area andmoment of area have to calculated.
Wetted surfacewith changed position
z
y
z
B
K
G
z
y
e
position change( )+
( )position change
y
zWetted surfaceat present z
y
Wetted surfacewith changed position
Is area invariant with respect to referenceframe?
Wetted surface area with respect to waterplane fixed frame
Area is invariant with respect toreference frame.
- Transverse Ri htin Moment
O'x'y'z' : Body fixed frame
Oxyz : Waterplane fixed frame
Method 2. calculate center of the buoyancy(B 1) with respect to body fixed framethen transform frame from body fixed frame to waterplane fixed frame
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
How to calculate center of buoyancy B1 withrespect to waterplane fixed frame?
Method 2. Calculate center of buoyancy B1with respect to body fixed frame, thentransform B1 to waterplane fixed frame
Integral value could be used as it is exceptintersection region with waterplane area whenposition of ship is changed.
with respect to body fixed frame, ,, , A y A z A M M
' 'dA dy dz= = dA A, ' ' A y M y dA= , ' ' A z M z dA=
O,O' x,x'
y
z( )+
j
k
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
y
z
B
K
G
z
y B1
e
y
z
Wetted surfaceat present
( ) ( )+ position change y
Only changed area andmoment of area have to calculated.
Wetted surfacewith changed position
1 B y 1 B z
1 1
, ' , '( ' , ' ) , A y A z B B M M
y z A A
=
Center of buoyancy in body fixed frame
- Transverse Ri htin Moment
O'x'y'z' : Body fixed frame
Oxyz : Waterplane fixed frame
Method 2. calculate center of the buoyancy(B 1) with respect to body fixed framethen transform frame from body fixed frame to waterplane fixed frame
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
How to calculate center of buoyancy B1 withrespect to waterplane fixed frame?
Method 2. Calculate center of buoyancy B1with respect to body fixed frame, thentransform B1 to waterplane fixed frame
Integral value could be used as it is exceptintersection region with waterplane area whenposition of ship is changed.
with respect to body fixed frame, ,, , A y A z A M M
' 'dA dy dz= = dA A, ' ' A y M y dA= , ' ' A z M z dA=
( )+( )
1 1
1 1
'cos sin
sin cos ' B B
B B
y y
z z
=
Center of buoyancy in waterplane fixedframe : Rotational Transformation
O,O' x,x'
y
z( )+
j
k
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y
z
B
G
B1
1 B y
1 B z
O
y
z
B
K
G
z
y
B1
=
A M
A M z y z A y A B B
,, ,),(11
Center of buoyancy with respect towaterplane fixed frame
dydzdA = = dA A= ydA M y A, = zdA M z A,
with respect to waterplane fixedframe
y z M M A ,,
1 1
, ' , '( ' , ' ) , A y A z B B M M
y z A A
=
1 1
1 1
'cos sinsin cos '
B B
B B
y y z z
=
Center of buoyancy with respect to waterplanefixed frame
Rotational transformation
Method 1. Calculate Center of buoyancy B1 withrespect to waterplane fixed frame directly
Method 2. Calculate center of buoyancy B1 withrespect to body fixed frame, then transform B1 towaterplane fixed frame
Same
with respect to body fixed frame, ' , ', , A y A z A M M
' 'dA dy dz = , ' ' A y M y dA= , ' ' A z M z dA=
Convenient
1 B y1 B
z
- Transverse Ri htin Moment
1 B y
1 B z
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
Comparison between Method 1 and Method 2
Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?
O,O'
x,x'
y
z( )+
j
k
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Rotational Transformation of Point and Frame(A) Rotation of the point (B) Rotation of the frame
Given : Coordinate of P with respect to oyz frameFind : Coordinate of Q which is rotated coordinate of Pabout origin O in the yz plane through an angle of .
P
Pr
y
z
o
QQr
=P
P
Q
Q
z
y
z
y
cossinsincos
Given: Coordinate of P with respect to oyz frameFind: Coordinate of P with respect to oyz which is rotatedframe about origin O ' from o ' y' z ' through an angle of .
' yP y
P y
P z
P z
cos sinsin cos
P P
P P
y y
z z
=
Rotational transformation of pointRotational transformation of frame
P
Pr
y
z
,o o
' z
- Transverse Ri htin Moment
cos sinsin cos
P P
P P
y y
z z
=
...(1) ...(2 1)
...(2 2)
Rotational transformation of frame16
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Rotational Transformation of Point and Frame
Given : Coordinate of P with respect to oyz frameFind : Coordinate of Q which is rotated coordinate of Pabout origin O in the yz plane through an angle of .
P
Pr
y
z
o
QQr
=P
P
Q
Q
z y
z y
cossinsincos
P y
P z
Rotational transformation of point through an angle of equals to
rotational transformation of frame through an angle of - .
Rotational transformation of point
y
z
,o o
- Transverse Ri htin Moment
Given: Coordinate of P with respect to oyz frameFind: Coordinate of P with respect to oyz which is rotatedframe about origin O ' from o ' y' z ' through an angle of - .
(A) Rotation of the point (B) Rotation of the frame
cos sinsin cos
P P
P P
y y z z
= ...(2 1)
If we substitude - into
cos sinsin cos
P P
P P
y y
z z
=
...(2 1)
Pr
y
z
P y
P z P
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(Proof) Rotational Transformation of Point
sincoscossin)sin( +=+Summation formula of trigonometric function
sinsincoscos)cos( =+
Coordinate of point P, Q is expressed byan angle
)sin()cos(
+=+=
QQ
QQ
z y
rr
sincos
PP
PP
z y
rr
==
( ) ( )
sincos
sinsincoscos
sinsincoscos
)cos(
PP
PP
QQ
QQ
z y
y
==
=
+=
rr
rr
r
Let coordinate of Q be expressed by difference formulaof trigonometric function.
( ) ( )
sincos
sincoscossin
sincoscossin
)sin(
PP
PP
QQ
QQ
y z
z
+=+=
+=
+=
rr
rr
r
)(, QP rr =
)(, QP rr =
=
P
P
Q
Q
z
y
z
y
cossinsincos
In the matrix form
P
Pr
y
z
o
QQr
- Transverse Righting Moment
Given : Coordinate of P with respect to oyz frameFind : Coordinate of Q which is rotated coordinate of P about origin O in the yz plane through anangle of .
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(Proof) Rotational Transformation of Frame
sincoscossin)sin( +=+Summation formula of trigonometric function
sinsincoscos)cos( =+
Coordinate of point P is expressed by an angle
cos( )sin( )
P P
P P
y z
= + = +
rr
cossin
P P
P P
y z
==
rr
( ) ( )
cos( )cos cos sin sincos cos sin sin
cos sin
P P
P P
P P
P P
y
y z
= += =
=
rr rr r
Let coordinate of P be expressed by difference formula of trigonometric function.
( ) ( )
sin( )
sin cos cos sin
sin cos cos sin
cos sin
P P
P P
P P
P P
z
z y
= += += += +
r
r r
r r
cos sinsin cos
P P
P P
y y
z z
=
In the matrix form
yP y
P y
P zP z
P
Pr
y
z
o
z
- Transverse Righting Moment
Given: Coordinate of P with respect to oyz frameFind: Coordinate of P with respect to oyz which is rotated frame about origin O ' from o ' y' z ' throughan angle of - .
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Sec.1 Calculation of Center of Buoyancy- Rotational Transformation of Point and Frame- Example) Calculation of Center of Buoyancy of Ship with
Constant SectionMethod Direct calculating center of buoyancy in waterplane fixed frameMethod Transformation center of buoyancy from body fixed frame to
waterplane fixed frame
- Calculation of Center of Buoyancy of Ship with Various Station
Shape 20/124
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Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Direct calculating center of buoyancy in waterplane fixed frame
G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y B , z B) in Waterplane fixed frame
z
y
O
30 B
K
B1
2020 z
y P
Q
R
S
Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin O through anangle of -30 . Calculate center of buoyancy with respect to waterplane fixed frame
z,z
y,y O
B
K
20
10
20
QP
R S
- Transverse Righting Moment
Section view
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G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
z
yO
30 B
K
B1
2020
z
y P
Q
R
SQuestion) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame
Sol.) P, Q, R,S with respect to waterplanefixed frame
( , )P PP x y is calculated by rotational transformation from
Q( xQ ,yQ), R( x R ,y R), S ( xS ,yS ) are calculated in the same way.
=
A
M
A
M z y z A y A B B
,, ,),(11
cos(30) sin(30) 10 3.66sin(30) cos(30) 10 13.66
Q
Q
x y = =
cos(30) sin(30) 10 13.66
sin(30) cos(30) 10 3.66 R
R
x
y
= =
cos(30) sin(30) 10 3.66
sin(30) cos(30) 10 13.66S
S
x
y
= =
cos(30) sin(30)sin(30) cos(30)
0.866 0.5 10 13.660.5 0.866 10 3.66
P P
P P
x x
y y
= = =
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Direct calculating center of buoyancy in waterplane fixed frame
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z
( , )P PP x y
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Sol.) Area
y
z
=
A
M
A
M z y z A y A B B
,, ,),(11
P
Q
R
S
2 11 1
2 1
( ) z z
z z y y y y
=
Intersection point between straight line and y
axis have to be calculated in order to know area.Equation of straight line P1S1 have to becalculated in order to know intersection point.
Equation of straight line through two points isas follows.
Substituting two points of P1, S1 into equation of straight line
13.66 ( 3.66)( 3.66) ( ( 13.66))
( 3.66) ( 13.66) z y
=
(-13.66,-3.66)
(-3.66, 13.66)
(13.66,3.66)
(3.66,-13.66)
1.732 20 z y = +
T
Intersection point T ( 11.55,0)T
Equation of straight line of Q1R 1 is obtained in the same way
1.732 20 z y =
U
Intersection point U (11.55,0)U
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Direct calculating center of buoyancy in waterplane fixed frame
z
yO
30 B
K
B1
2020
z
y P
Q
R
S
G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z
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l ) l l f f f h h
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Sol.) Area
y
z
=
A
M
A
M z y z A y A B B
,, ,),(11
P
Q
R
S
Divide area into 4 part , A1,A
2,A
3,A
4.
Then calculate area of A1.
(-13.66,-3.66)
(-3.66, 13.66)
(13.66,3.66)
(3.66,-13.66)
T (-11.55,0) U(11.55,0)A1 A2
A3A4
1
111.55 ( 13.66) 3.66
23.867
A Area = =
Area of A2, A3, A4 can be calculated in the same way.
2 55.66, A Area =3
86.6 A Area =
453.87 A Area =
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Direct calculating center of buoyancy in waterplane fixed frame
z
yO
30 B
K
B1
2020 z
y P
Q
R
S
G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Sol.) Centroid
y
z
=
A
M
A
M z y z A y A B B
,, ,),(11
P
Q
R
S
(-13.66,-3.66)
(-3.66, 13.66)
(13.66,3.66)
(3.66,-13.66)
T (-11.547,0) U(11.547,0)A1 A2
A3A4
1 _
211.54 ( 11.54 ( 13.66))
312.25
c A x = +
=
1 _
23.66 ( ) 2.44
3c A y = =
Centroid of A1 can be calculated as follows1 1 _ _ ( , )c A c A x y
Centroids of A2, A3, A4 are calculated in the same way.
2 2 _ _ ( , ) ( 3.96, 1.83)c A c A x y =
3 3 _ _ ( , ) ( 2.11, 6.99)c A c A x y =
4 4 _ _ ( , ) (6.29, 4.55)c A c A x y =
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Direct calculating center of buoyancy in waterplane fixed frame
z
yO
30 B
K
B1
2020 z
y P
Q
R
S
G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Sol.) 1st moment of area
=
A
M
A
M z y z A y A B B
,, ,),(11
1st
moments of area are calculated with areasand centroids which are calculated in previous.
(13.66,3.66)
y
z
P
Q
R
S
(-13.66,- 3.66)
(-3.66, 13.66)
(3.66,-13.66)
T (-11.547,0) U(11.547,0)A1 A2
A3A4Area y c zc Area yc Area zc
A1 3.87 -12.25 -2.44 -47.38 -9.44
A2 55.66 -3.96 -1.83 -220.24 -101.85
A3 86.60 -2.11 -6.99 -183.01 -605.62
A4 53.87 6.29 -4.55 338.78 -245.28
Sum 200.00 -111.85 -962.19
Centoid of total area is calculated as follows.
( )1 1
, , 111.85 962.( , ) , , 0.56, 4.81200 200
A y A z B B
M M y z
A A
= = =
, A y M , A z M
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Direct calculating center of buoyancy in waterplane fixed frame
z
yO
30 B
K
B1
2020 z
y P
Q
R
S
G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
Sol.) Centroid
Centroid of total area before heel
_ _ ( , ) (0, 5)c A c A y z =
Centroids of changed area after heel
1 1 _ _
2 1( , ) ( 10, 10 tan 30 )
3 3
( 6.67, 1.92)
c A c A y z =
=
2 2 _ _ 2 1
( , ) ( 10, 10 tan 30 )3 3
(6.67,1.92)
c A c A y z = =
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Transformation of center of buoyancy from body fixed frame to waterplane fixed frame
z
y
y
z
O 30
B
K
B1
20
20
QP
R S
A1
A2
A
z
y
y
z
O 30
B
K
B1
20
20
QP
R S
G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
Sol.) 1st moment of area
= A
M
A
M z y z A y A B B
',', ','
)','(11
1st moments of area are calculated with areas
and centroids which are calculated in previous.Area yc zc Area y Area z
A 200.00 0.00 -5.00 0.00 -1000.00A1 -28.87 -6.67 -1.92 192.45 55.56A2 28.87 6.67 1.92 192.45 55.56
Sum 200.00 384.90 -888.89
, A y M , A z M
Centroid of total area after heel with respect to body fixed frame is as follows( )
1 1
, , 384.90 888.89( , ) , , 1.92, 4.44200 200
A y A z B B
M M y z
A A
= = =
Rotationaltransformation
1 1
1 1
cos sin 1.92 0.56cos(30 ) sin(30 )sin cos 4.44 4.81sin(30 ) cos(30 )
B B
B B
y y
z z
= = =
Transform coordinate of centroid of total area with respect to body fixed frame tocentroid with respect to waterplane fixed frame by rotational transformation
Same result
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Transformation of center of buoyancy from body fixed frame to waterplane fixed frame
z
y
y
z
O 30
B
K
B1
20
20
QP
R S
A1
A2
A
z
y
y
z
O 30
B
K
B1
20
20
QP
R S
G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy
Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame
Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
1
1
0.56
4.81 B
B
y
z
=
1
1
0.56( )
4.81 B
B
y
z
=
Same
Calculation Result
Calculation Result
- Transverse Righting Moment
Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Transformation of center of buoyancy from body fixed frame to waterplane fixed frame
z
yO
30 B
K
B1
2020 z
y P
Q
R
S
z
y
y
z
O 30
B
K
B1
20
20
QP
R S
Method Direct calculating center of buoyancy inwaterplane fixed frame
Method Transformation of center of buoyancyfrom body fixed frame to waterplane fixed frame
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Sec.1 Calculation of Center of Buoyancy- Rotational Transformation of Point and Frame- Example) Calculation of Center of Buoyancy of Ship with
Constant SectionMethod Direct calculating center of buoyancy in waterplane fixed frameMethod Transformation center of buoyancy from body fixed frame to
waterplane fixed frame
- Calculation of Center of Buoyancy of Ship with Various StationShape
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Calculation of Center of Buoyancy of Ship with Various Station Shape
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Calculation of Center of Buoyancy of Ship with Various Station Shape- Introduction
1 Br 1 B z
1 B y
' y
y
' z z
G
B
K 1 B
Lc
O O
-
1' B y
1' B z1 B
z
1 B y
1 Br
' z z
G
B
K
' y
y
Lc
dA
Method Direct calculating center of buoyancy in waterplane fixed frame
Method Transformation center of buoyancy from body fixed frameto waterplane fixed frame
How to calculate center of buoyancy of ship with various sections ?
- Transverse Righting Moment
How to calculate inwaterplane fixed frame ?
1 1, B B y z
Righting Moment : Moment toreturn the ship to the upright floatingposition (Restoring moment, Momentof statical stability))
BGZ F = i Transverse Restoring moment
1( )righting G B B y y F = + i
Righting arm
Righting Arm ( GZ )
1G BGZ y y= +
From direct calculation
We should know yG , y B1 in waterplane fixed f rame
From geometrical figure withassumption that M does not changewithin small angle of heel (about 10 )
sinGZ GM = GM is related to below equation bygeometrical figure
GM KB BM KG= +
Z
K
z
O,O'
y
z
)(+
CL
y
z M
righting
e
G
FG
B B1
N
FB
1 B y
G y
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
x,x'
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VCB(Vertical Center of Buoyancy)
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
VCB(Vertical Center of Buoyancy)Step Area, 1 st Moment of Area
z
y
z'
dA
y'
dz'
dy'
c y
c z G
1
n
ii
A dA dy dz
A=
= =
=
Area , A
dA dy dz=
Differential element of area, dA
, A y
c
M y dA y dy dz
y A
= =
=
1st
moment of area with respect to z axis, M A,x
, A z
c
M z dA z dy dz
z A
= =
=
1st moment of area with respect to y axis, M A,y
( ), ,, , A y A z c c M M
y z A A
= =
G
Centroid G
(Ai : Area of i-th element)
( , )c c y z : center of A
1) James M. Gere, Mechanics of materials, THOMSON, pp.828-850, 6 th Edition
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VCB(Vertical Center of Buoyancy)
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
VCB(Vertical Center of Buoyancy)Step Sectional Area (A M), Displacement Volume
Displacement volume
( ) ' A x dx =
== '' dzdydA ASectional Area
L
formhullr under wate
B
T
A
x y
z
( ) ''''''
dxdzdy
dzdydxdV
===
( ) A x
After calculation of each station area,displacement volume can be calculated byintegral of section area over the length of ship
Each stationarea Integral
in longitudinaldirection
Volume
A
Area of this Curve Volume( )
A.P x L
A
A1
A2
A3 An..
Oxyz : Waterplane fixed frame
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VCB(Vertical Center of Buoyancy)
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
VCB(Vertical Center of Buoyancy)Step Vertical Moment of Volume, Vertical Center of Buoyancy(VCB)
Vertical Center of Buoyancy
, , ( ) z A z M M x dx =
, z M VCB =
Vertical Moment of Volume
( )
, z M z dV
z dx dy dz
z dy dz dx
=
=
=
, ( ) A z M x M A,z : Vertical moment of area about y axis
After calculation of each vertical moment of stationarea about the y axis( M A,z), vertical moment of
displaced volume can be calculated by integral of vertical moment of section area over the length of ship
L
formhull
r under wate
B
T
x y
z A
vertical momentof station areaabout the y axis
Integralin longitudinaldirection
Verticalmoment of volume
, ' A z M , ' z M
Area of this Curve 1st Moment of Volume( )
, z M
A.P x L M A1, z
M A2, z
M A3, z
M An, z .
, A z M
Oxyz : Waterplane fixed frame
- Transverse Righting Moment
z
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Area of Triangle by Vector
0 0( , ) x y
1 1( , ) x y
2 2( , ) x y
x
y
1 0r r
2 0r r
o
2r
1r
0r
1 0 2 01( ) ( ) ( )2
Area = r r r r r
1 0 2 0 2 0 1 01 ( )( ) ( )( )2
x x y y x x y y=
1 0 1 0
2 0 2 0
0
0
x x y y
x x y y
=
i j k
1)Cross Product
( )sin =a b a b n: Area of parellelogram sinh = ba
b
b
a
: Area of triangle with side a and b
A = a b
12
A = a b
Given : Position vector r 0, r 1, r 2, of vertex of triangle Find : Area of triangle
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Area of Triangle
1) Zill,C. Advanced Engineering Mathematics, p319, Jones and Bartlett, 3 rd
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
Area of Triangleby Rotational Transformation
)'')(''()'')(''(21
01020201 y y x x y y x x =
1 0 2 0 2 0 1 0
1( )( ) ( )( )2 x x y y x x y y =
1 0 2 01( ) ( ) ( )2
Area = r r r r r
1 0 2 01( ') ( ' ' ) ( ' ' )2
Area = r r r r r
[ ] [ ]0 1 2 0 1 2cos sin' ' ' sin cos = r r r r r r
=210
210
210
210
cossinsincos
''''''
y y y x x x
y y y x x x
' = r R r
1 1 0 0 2 2 0 0
2 2 0 0 1 1 0 0
1 (( cos sin ) ( cos sin ))(( sin cos ) ( sin cos ))2
(( cos sin ) ( cos sin ))(( sin cos ) ( sin cos ))
x y x y x y x y
x y x y x y x y
= + +
+ +
)cos)(sin))((sin)(cos)((
)cos)(sin))((sin)(cos)((21
01010202
02020101
y y x x y y x x
y y x x y y x x
+
+=
2 21 0 2 0 1 0 2 0 2 0 1 0 1 0 2 0
2 21 0 2 0 2 0 1 0 1 0 2 0 1 0 2 0
1(( )( ) cos sin ( )( ) cos ( )( ) sin ( )( ) sin cos )
2
( )( ) cos sin ( )( ) cos ( )( ) sin ( )( ) sin cos )
x x x x x x y y x x y y y y y y
x x x x x x y y x x y y y y y y
= +
+ +
{ }{ }2 2 1 0 2 0 1 0 2 01
cos sin ( )( ) ( )( )2 x x y y y y y y = +
Is a area changed after rotation aboutorigin O? Given : Area before rotation Find : Area after rotation
0r
1r
2r
x0'r
1'r
2'r
y
- Transverse Righting Moment
,(R: )
o
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Area of Triangleby Rotational Transformation
)'')(''()'')(''(21
01020201 y y x x y y x x =
1 0 2 0 2 0 1 0
1( )( ) ( )( )2 x x y y x x y y =
( )( ) ( )
Area
Area Area
= =
Rr
r r( R : Rotationaltransformation matrix)
1 0 2 01( ) ( ) ( )2
Area = r r r r r
1 0 2 01( ') ( ' ' ) ( ' ' )2
Area = r r r r r
{ }{ }2 2 1 0 2 0
1 0 2 0 1 0 2 0
1 0 2 0
1cos sin ( )( ) ( )(
1( )( )
)
(
2
)( )2
x x y y y y y
x x y y y y y y
y = +
=
Area is invariant with respect to frame.
Given : Find :
0r
1r
2r
x0'r
1'r
2'r
y
- Transverse Righting Moment
( )( ) a Area Are = rr
o
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1st Moment of Area1) Zill,C. Advanced Engineering Mathematics, p319, Jones and Bartlett, 3 rd
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
1 Moment of Areaby Rotational Transformation
1 0 2 01( ) ( ) ( )
2
Area = r r r r r
( ) C A= M r r
0 0( , ) x y
1 1( , ) x y
2 2( , ) x y
x
y
C C r ( , ) x y
1 2 31
( )3C
= + +r r r r
A
B C
O
G
Position vector of centroid oftriangle ABC.
a
b
c
g
Let cent of side BC as D
( )12OD b c= +
Now, G is the point of internal division with ratio of 2 to 1
2 1 22 1 3
OD OA OD OAOG
+ += =+
( )1 123 2g b c a = + + =
( )13 a b c+ +
2
1
1 2 3
3 x x x x xdA xdxdy A+ += = = M
1 2 3
3 y y y y
ydA ydxdy A+ += = = M
Given : Position vector of vertex of triangle Find : 1 st moment of area of triangle
Position vector of centroid of triangle r 1r 2r 2
Area of triangle r 1r 2r 2
1st moment of area of triangle r 1r 2r 2 with respect to x andy axis
(*) ( ) , 10- ,40th ,2005, ,pp.15- Transverse Righting Moment
o
2r
0r1r
D39
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1st Moment of Area
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
1 2 31 ( ' ' ' )3C = + +r r r r
1 2 31 ( )3
= + +R r R r R r
1 2 31 ( )3
= + +R r r r
C A=R r( ) Moment = R r
( ) C Moment A=r r1 2 3
1 ( )3C
= + +r r r r
( ') 'C Moment A=r r
1 0 2 01( ) ( ) ( )2
Area = r r r r r
1st moment of area of triangle before rotation
( ') ( ) , ( ) Area Area A A= =r r1st moment of area of triangle after rotation
Given : Position vector of triangle, angle of rotation Find : 1st moment of area of triangle after rotationabout origin O
C = R rC C =r R r
( )
C
Moment
A=r
r[ ] [ ]0 1 2 0 1 2
cos sin' ' '
sin cos
= r r r r r r
=210
210
210
210
cossinsincos
''''''
y y y x x x
y y y x x x
' = r R r
0r
1r
2r
x 0'r
1'r
2'r
y
C
'C C r
- Transverse Righting Moment
1 Moment of Areaby Rotational Transformation
C r
o
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
C C =r R r
C A= R r( ) Moment = R r
( ) ( )
C
Moment Moment
A
==
R r R r
R r( R : Rotationaltransformation matrix)
0r
1r
2r
x 0'r
1'r
2'r
y
C
'C C r
( ) C Moment A=r r1 2 3
1 ( )3C
= + +r r r r
( ') 'C oment A=M r r
1 0 2 01( ) ( ) ( )2
Area = r r r r r
1st moment of area of triangle before rotation
( ') ( ) , ( ) Area Area A A= =r r1st moment of area of triangle after rotation
1st moment of area is invariant withrespect to frame.
[ ] [ ]0 1 2 0 1 2cos sin
' ' 'sin cos
= r r r r r r
=210
210
210
210
cossinsincos
''''''
y y y x x x
y y y x x x
' = r R r
Given : Position vector of triangle, angle of rotation Find : 1st moment of area of triangle after rotationabout origin O
- Transverse Righting Moment
1 Moment of Areaby Rotational Transformation
C r
o
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Sec.1 Calculation of Center of Buoyancy- Rotational Transformation of Point and Frame- Example) Calculation of Center of Buoyancy of Ship with
Constant SectionMethod Direct calculating center of buoyancy in waterplane fixed frameMethod Transformation center of buoyancy from body fixed frame to
waterplane fixed frame
- Calculation of Center of Buoyancy of Ship with Various StationShape
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Example) Calculation of Center of Buoyancy of Ship with Various Station
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
a p e) Ca cu at o o Ce te o uoya cy o S p w t Va ous Stat oShape
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixed
frame.
20
CL
20
10
O,O ' y,y' z,z'
x,x'
10
10
20 20
10
20
20
20
20
20
2010
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O y z
A1
Example) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
y dz dy dx
dz dy dx
=
y dz dy dx =
dz dy dx =
O y
x
A3
A2
- Transverse Righting Moment
Sectional area of each section(Body fixed frame)
( ) A x dA dy dz = =
1st moment of area of each section(Body fixed frame)
, A y M y dy dz
=
, A z M z dy dz =
Displacement volume(Body fixed frame)
dV =
1st moment of displacement volume(Body fixed frame)
, y M
, z M
.
.( )
F P
A P A x dx =
Center of buoyancy(Body fixed frame )
,c y
.
.( ( ))
F P
c A P y A x dx =
, ,( , )c c y z : Center of displaced volume
.
..
.
( )
( )
F P
c A PF P
A P
A x y dx
A x dx
=
, y M =
,c z , z M =
z dz dy dx
dz dy dx
=
.
..
.
( )
( )
F P
c A PF P
A P
A x z dx
A x dx
=
, ,
, ,
cos sinsin cos
c c
c c
y y
z z
=
Center of Buoyancy(Waterplane fixed frame )
z dz dy dx = .
.( ( ))
F P
c A P z A x dx =
( )c y A x =
( )c z A x =
( ) A x : Sectional area at x( , )c c y z : Centroid of section at x'
p ) y y pwith Various Station Shape
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Example) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
O,O ' y,y' z,z'
x,x'
)
y'
y
z' z
< Section A 1 >
< Section A 2 >
< Section A 3 >
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
2
CL
y
z
20
10
O y z
x
10
10
20 20
10
y'
y
z' z
y'
y
z' z
- Transverse Righting Moment
p ) y y pwith Various Station Shape
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Example) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
) In the same way of previous example, calculate center of buoyancy with respect to bodyfixed frame at first, then calculate center of buoyancy with respect to waterplane fixedframe by rotational transformation.
Coordinate of P 1, Q1 is known as (-5,0), (5,0) by geometric shape.Coordinate of P 1, P2, Q 1, Q 2 of section A3
< Section A 3 >
O y z
x
A3
A2
A1
z
y
y
z
O30
B
K
B1
20
20 Q2
Q1
P1P2
QP
Calculate equations of straight line PK, KQ in order to know P 2,Q2
The equation of straight line PK 2 10 z y =
The equation of straight line KQ 2 10 z y =
The equation of line of waterplane with respect to body fixed frame is as follows,
because waterplane is inclined through an angle of 30 .tan 30 0.5774 z y y = =
Intersection point P 2,Q2 betwwen waterplane and straight line PK, KQ can be calculated as followsP2(-3.88, -2.24), Q 2(7.03, 4.06)
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
Area below waterplane can be calculated also by Gaussian Quadrature.
p ) y y pwith Various Station Shape
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
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Example) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
) -A3 : Sectional area of A 3O y z
x
A3
A2
A1
z
y
y
z
O30
B
K
B1
20
20 Q2
Q1
P1P2
QP
A3_2 A3_1
A3_0
3_ 0 0.5 10 10 50 A = =
3_1 1 2
1( ) ( )
2
0 5 0
0 3.88 2.245.60
A =
=
=
P O P O
i j k
3_ 2 1 2
1( ) ( )
2
0 5 0
0 7.03 4.0610.15
A =
=
=
Q O Q O
i j k
Total sectional area before heel
Changed area after heel
Total sectional area after heel
3 50 5.60 10.15 54.55 A = + =
If a ship is not a wall sided ship,area below waterplane is different.
P2(-3.88, -2.24), Q 2(7.03, 4.06)P1(-5,0), Q 1(5,0)
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
( )
( )i
A x dA
dy dz
A x
= =
=
( )k A x: Sectional area at x
_ ( )k i A x : Partial area of section at x'
p ) y y pwith Various Station Shape
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 3 >
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Example) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
) Centroid of A3O y z
x
A3
A2
A1
z
y
y
z
O30
B
K
B1
20
20 Q2
Q1
P1P2
QP
A3_2 A3_1
A3_0
P2(-3.88, -2.24), Q 2(7.03, 4.06)P1(-5,0), Q 1(5,0)
3_ 0
1Cetroid of (0, ( 10))3
10(0, )3
A =
=
3_10 5 3.88 0 0 2.24Cetroid of ( , )
3 3
( 2.96, 0.75)
A + =
=
3_ 20 7.03 5 0 0 4.06Cetroid of ( , )
3 3(4.01, 1.35)
A+ + + +=
=
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
p ) y y pwith Various Station Shape
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 3 >
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Example) Calculation of Center
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2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
)
Calculate 1 st moment of area in order to know centroid of section A 3 with respect to body fixed frame
Area Area*y'c_3_i Area*z'c_3_i A3_0 50.00 0.00 -3.33 0.00 -166.67 A3_1 5.60 -2.96 -0.75 -16.57 -4.18 A3_2 10.15 4.01 1.35 40.69 13.73
- + 54.55 57.26 -148.76
- A3 : 1st moment of area of section A 3
Centroid of section A 3 with respect to body fixed frame is calculated as follows
( )
3 3
3 3
, , _ 3 _ 3( , ) ,
57.26 148.76, 1.05, 2.73
54.55 54.55
A y A zc c
A A
M M y z
Area Area =
= =
3 , A y M 3, A z M
3, A y
M : 1
st moment of area of section A 3 about z ' axis 3
, A z M : 1st moment of area of
section A 3 about y ' axis
O y z
x
A3
A2
A1
y
yO30
B
K
B1
20 Q2
Q1
P1P2
QP
A3_2 A3_1
A3_0
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
z 20
- Transverse Righting Moment
,k A y M y dy dz
=
_ _ _ ( )c k i k i y A x = _ ( )c k k y A x =
of Buoyancy of Ship with Various Station Shape
_3 _ c i y _ 3_ c i z
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 3 >
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Example) Calculation of Center
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2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
)
Calculate 1st
moment of area in order to know centroid of section A 2 with respect to body fixed frame
Area Area*y'c_2_i Area*z' c_2_i A3_0 200.00 0.00 -5.00 0.00 -1000.00 A3_1 -28.87 -6.67 -1.92 192.45 55.56 A3_2 28.87 6.67 1.92 192.45 55.56Sum 200.00 384.90 -888.89
-A2, - A2 : A2 1
Centroid of section A 2 with respect to body fixed frame is calculated as follows
( )3 32 2
, , _ 2 _ 2
384.90 888.89( , ) , , 1.92, 4.44
200 200 A y A z
c c A A
M M y z
Area Area = = =
y'
z'
y
z
A2_2 A2_1
A2_0
2 , A y M
: 1st moment of area of section A 2 about z ' axis
2 , A z M
: 1st moment of area of section A 2 about y ' axis
O y z
x
A3
A2
A1
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
of Buoyancy of Ship with Various Station Shape
2 , A y M
2, A z M
_ 2 _ c i y _ 2_ c i z
,k A y M y dy dz
=
_ _ _ ( )c k i k i y A x = _ ( )c k k y A x =
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 2 >
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zExample) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
)
Coordinate of R 1, S1 is known as (-7.5,0), (7.5,0) by geometric shape.
Coordinate of R 1, R 2, S1, S2 of section A 1O y z
x
A3
A2
A1
Calculate equations of straight line RR 3, SS 3 in order to know R 2,S2
The equation of straight line RR 3 4 30 z y =
The equation of straight line SS 3 4 30 z y =
The equation of line of waterplane with respect to body fixed frame is asfollows, because waterplane is inclined through an angle of 30 .tan 30 0.5774 z y y = =
Intersection point R 3,S3 between waterplane and straight line RR 3, SS 3 can be calculated as followsP2(-6.55,-3.78), Q 2(8.77,5.06)
z
y
y
z
O30
B
K
B1
20
20
SR
S2
S1
R 1
R 2
R 3 S3
10
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
with Various Station Shape
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 1 >
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zExample) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
) - A1 : Sectional area of section A 1
A1_2 A1_1
A1_0
1_ 0 0.5 (15 10) 10 125 A = + =
1_1 1 2
1( ) ( )
2
0 7.5 0
0 6.55 3.7814.19
A =
=
=
P O P O
i j k
1_ 2 1 2
1( ) ( )
2
0 7.5 0
0 8.77 5.0618.98
A =
=
=
Q O Q O
i j k
1 125 14.19 18.98 129.79 A = + =If a ship is not a wall sided ship,area below waterplane is different.
P2(-6.55,-3.78), Q 2(8.77,5.06)P1(-7.5,0), Q 1(7.5,0)
O y z
x
A3
A2
A1
z
y
y
z
O30
B
K
B1
20
20
SR
S2
S1
R 1
R 2
R 3 S3
10
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
_
( )
( )
k
k ii
A x dAdy dz
A x
= ==
with Various Station Shape
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 1 >
Total sectional area before heel
Changed area after heel
Total sectional area after heel
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OzExample) Calculation of Center of Buoyancy of Ship
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
) Centroid of section A 1
3_ 0 (15 10) ( 5) (0.5 2.5 10) 10 (2 / 3)Cetroid of (0, )125(0, 4.67)
A + =
=
3_10 7.5 6.55 0 0 3.78Cetroid of ( , )
3 3( 4.68, 1.26)
A + =
=
3_ 20 7.5 8.77 0 0 5.06Cetroid of ( , )
3 3(5.42, 1.69)
A+ + + +=
=
P2(-6.55,-3.78), Q 2(8.77,5.06)P1(-7.5,0), Q 1(7.5,0)
A1_2 A1_1
A1_0
O y z
x
A3
A2
A1
z
y
y
z
O30
B
K
B1
20
20
SR
S2
S1
R 1
R 2
R 3 S3
10
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
with Various Station Shape
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 1 >
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Oz
Example) Calculation of Center
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
Calculate 1 st moment of area in order to know centroid of section A 1 with respect to body fixed frame
Area Area*y'c_1_iArea*z'c_1_i A1_0 125.00 0.00 -4.67 0.00 -583.34 A1_1 14.19 -4.68 -1.26 -66.48 -17.90 A1_2 18.98 5.42 1.69 102.90 32.02
- + 129.79 169.38 -533.42
- A1 : 1st moment of area of section A 1
Centroid of section A 1 with respect to body fixed frame is calculated as follows
1 , A y M
: 1st moment of area of section A 1 about z ' axis 1 , A z
M
: 1st moment of area of section A 1 about y ' axis
)
A1_2 A1_1
A1_0
O y z
x
A3
A2
A1
z
y
y
z
O30
B
K
B1
20
20
SR
S2
S1
R 1
R 2
R 3 S3
10
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
, A y M y dy dz
=
_ ( )c i i y A x = ( )c y A x =
of Buoyancy of Ship with Various Station Shape
1 , A y M 1, A z M
_1_ c i y _1_ c i z
( )
1 1
1 1
, , _1 _1( , ) ,
169.38 533.42, 1.31, 4.11
129.79 129.79
A y A zc c
A A
M M y z
Area Area =
= =
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
< Section A 1 >
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Example) Calculation of Center of Buoyancy of Ship
O yz
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
with Various Station Shape
)
Area of this curve Volume( )
Displacement volume can be calculated by
integral of sectional area in longitudinal direction50
0( ) 7,304 A x dx = =
Displacement Volume
x' 0
Area
A1 A2 A3
54.55
200
127.79
25 50
' Adx =
O y z
x
A3
A2
A1
20
CL
y
z
20
10
O y z
x
10
10
20 20
10
- Transverse Righting Moment
dV dx dy dz = = dz dy dx =
.
.( )
F P
A P A x dx =
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
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Example) Calculation of Center of Buoyancy of Ship
O y z
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
with Various Station Shape
)
Area of this curve Transverse moment
of volume
x'
169.38
25 50
384.9
57.26
0After calculation of each transverse moment of sectional area about the z ' axis( M A,y' ), transverse moment of displaced volume can be calculated byintegral of transverse moment of section area over the length of ship
, A y M
-1 : Transverse moment of displaced volume
50
, ,012,455 y A y M M dx = =, A yM : Transverse moment of area about z ' axis
, A z M
: Vertical moment of area about y ' axis
1 , A y M
2 , A y M
3 , A y M
, ' , ' ' y A y M M dx=
O y z
x
A3
A2
A1
20
CL
y
z
20
10
O y
x
10
10
20 20
10
- Transverse Righting Moment
y dy dz dx = , y M .
.( ( ))
F P
c A P y A x dx =
,c y=
(Body fixed frame)
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
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Example) Calculation of Center of Buoyancy of Ship
O y z
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
with Various Station Shape
)-2 : Vertical moment of displaced volume
, A z M
Area of this curve Vertical moment
of volume
After calculation of each vertical moment of sectional area about the yaxis( M A,z), vertical moment of displaced volume can be calculated byintegral of vertical moment of section area over the length of ship
50
, ,030,794 z A z M M dx = =
x'
-148.76
-888.89
-533.42
25 50
1 , A z M
2 , A z M
3 , A z M
, ' , ' ' z A z M M dx=
O y z
x
A3
A2
A1
20
CL
y
z
20
10
O y
x
10
10
20 20
10
- Transverse Righting Moment
, z M z dy dz dx = .
.( ( ))
F P
c A P z A x dx =
,c z=
(Body fixed frame)
0
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
, A yM
: Transverse moment of area about z ' axis
, A z M
: Vertical moment of area about y ' axis
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Example) Calculation of Center of Buoyancy of Shiph h O y
z
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
with Various Station Shape
)Center of buoyancy (Body fixed frame)
Center of buoyancy can be calculated if we divide transverse, vertical moment of displaced volume by displaced volume.
( )
, ' , ', ,( , ) ,
12, 455 30,794, 1.71, 4.21
7,304 7,304
y zc c
M M y z TCB VCB
= = = = =
20
CL
y
z
20
10
O y
x
10
10
20 20
10
- Transverse Righting Moment
y dy dz dx
dx dy dz
=
,c
y , y M =
,c z , z M
=
z dy dz dx
dx dy dz
= Center of buoyancy with respect to waterplane fixed frame have
to be calculated. Rotational transformation
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
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Example) Calculation of Center of Buoyancy of Shipi h V i S i Sh O y
z
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.
with Various Station Shape
)
20
CL
y
z
20
10
O y
x
10
10
20 20
10
- Transverse Righting Moment
Center of buoyancy (Waterplane fixed frame)
,
,
cos sin 1.71sin cos 4.21
cos(30) sin(30) 1.71 0.63sin(30) cos(30) 4.21 4.50
c
c
y
z
=
= =
, ,
, ,
cos sinsin cos
c c
c c
y y
z z
=
Center of buoyancy with respect to waterplane fixed frame have to be calculated. Rotational transformation
Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame
Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.
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2009 Fall, Ship Stability
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N a v a
l A r c
h i t e c t u r e
& O c e a n
E n g
i n e e r i n g
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability
Ch.6 Transverse Righting Moment- Sec.2 Calculating BM, GZ in Wall Sided Ship -
2009
Prof. Kyu-Yeul Lee
Department of Naval Architecture and Ocean Engineering,Seoul National University
- Ship Stability -
- Transverse Righting Moment
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2009 Fall, Ship Stability
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Sec.1 Calculation of Center of BuoyancySec.2 Calculation of BM, GZ in Wall Sided ShipSec.3 Inclining TestSec.4 Transverse Stability of ship (Unstable condition)Sec.5 Transverse Righting Moment due to Movement of CargoSec.6 Calculation of Heeling Angle due to Shift of Center of Mass
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Transverse Righting Moment
Righting Moment : Moment toreturn the ship to the upright floating
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2009 Fall, Ship Stability
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Transverse Righting Moment
Z
K
z
O,O'
y
z
)(+
CL
y
z M
righting
e
G
FG
B B1
N
FB
1 B y
G y
- Overview of Ship Stability
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
return the ship to the upright floatingposition (Righting moment, Momentof statical stability))
BGZ F = i Transverse Righting moment
1( )righting G B B y y F = + i
Righting arm
Righting Arm ( GZ )
1G BGZ y y= +
From direct calculation
We should know yG , y B1 in waterplane fixed frame
From geometrical figure withassumption that M does not changewithin small angle of heel (about 10 )
sinGZ GM =
GM is related to below equation bygeometrical figure
GM KB BM KG= +
x,x'
G: Center of mass B: Center of buoyancy B1: Changed center of buoyancyK : Keel F G : Weight of ship F B : Buoyant force acting on ship
Z : The intersection of the line of buoyant force through B1 with the transverse linethrough G
M : The intersection of the line of buoyant force through B1 with the centerline of the ship
BM KB : Vertical center
of buoyancy BM : Transverse Metacenter Radius
KG : Vertical center of mass of shipis determined by the shape of ship
,KB BM KG is determined by loading condition of cargo
restoring BGZ F = sinGZ GM =
GM KB BM KG= + BM
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Calculation of BM (1)(BM T M i R di )
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2009 Fall, Ship Stability
SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability - Transverse Righting Moment M : The intersection of the line of buoyant force through B1 with the centerline of the ship B2: The intersection of the line of buoyant force through B1 with the transverse line through B
M
Z
B
G
z
y
z
K CL
B1=
== yW 2
W 1
L2
L1
g1g
B2
restoring
F B
(BM : Transverse Metacentric Radius)
yv,c
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
B1: Changed center of buoyancy
y
z( )+ j
k
Z : The intersection of the line of buoyant forcethrough B1 with the transverse line throughG
G: Center of mass B: Center of buoyancyF G : Weight of ship (=W )F B : Buoyancy (= g )
Wall sided ship: When a ship is in upright position, aship which have perpendicular side shellto waterplane is called wall sided ship .
Assumption
1. Wall sided ship Submerged volume is same with
emerged volume when the ship is
heeled. ( A ship is heel without changeof displacement volume )
2. A main deck is not flooded.
3. Center of rotation is not changed.
( M is not changed)
F G
Lets derive BM in case of simple section likea wall sided ship.
O,O' x,x'
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Calculation of BM (2)(BM T M i R di )
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2009 Fall, Ship Stability
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M
Z
B
G
z
y
z
K CL
B1=
== yW 2
W 1
L2
L1
g1g
B2
restoring
F B
(BM : Transverse Metacentric Radius)
yv,c
Relation between moving distance of centerof changed displacement volume andmoving distance of center of center of buoyancy is as follows.
y
z( )+ j
k
- Transverse Righting Moment
1 1g v
BB ggg
=
1 12v
BB Og=
The shape of displacement volume is
changed as a ship is heeled.
F G
: Displacement volumev : Changed displacement volume
BB1: Moving distance of center of buoyancy
gg 1 : Moving distance of center of changed displacement volume
1 1g BB g v gg =
1 1( 2 )gg Og=
M : The intersection of the line of buoyant force through B1 with the centerline of the ship B2: The intersection of the line of buoyant force through B1 with the transverse line through B
O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame
B1: Changed center of buoyancy
Z : The intersection of the line of buoyant forcethrough B1 with the transverse line throughG
G: Center of mass B: Center of buoyancyF G : Weight of ship (=W )F B : Buoyancy (= g )
O,O' x,x'
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dz dy dx = dV = ( ) A x dA dy dz = = 2 ,2 v c BB v y=
Calculation of BM (4)(BM T M t t i R di )
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2009 Fall, Ship Stability
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(R.H.S)
- Transverse Righting Moment
,2
v cv y
: y coordinate of center of changed displacement volume
: Changed displacement volume
: Displacement volume
,v c yv
.
.( )
F P
A P A x dx =
, ,
, ,
cos sinsin cos
v c v c
v c v c
y y
z z
=
, , ,cos sinv c v c v c y y z =
Represent center of buoyancy with respect to waterplanefixed frame as one with respect to body fixed frame., ,( , )v c v c y z
, ,co si2
s n( )v c v c y zv =
, ,
2 2cos sinv c v cv y v z =
Transverse moment of volume withrespect to body fixed frame.
Vertical moment of volume withrespect to body fixed frame.
y
yg1
L1
L2
z z
,cos
v c y
, sinv c z ,v c y
(BM : Transverse Metacentric Radius)
O,O'
x, x'
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dz dy dx = dV = ( ) A x dA dy dz = =
Calculation of BM (5)(BM T M t t i R di )
2 ,2 v c BB v y=
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2009 Fall, Ship Stability
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(R.H.S)
,2
v cv y
.
.( )
F P
A P A x dx =
, ,
2 2cos sinv c v cv y v z =
Transverse moment of volumewith respect to body fixed frame.
Vertical moment of volume withrespect to body fixed frame.