Ship Stability - Seoul National University · 2018. 9. 13. · Seoul National University Lecture...

2018-08-07

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1Naval Architectural Calculation, Spring 2018, Myung-Il Roh

Ship Stability

Ch. 9 Numerical Integration Method in Naval Architecture

Spring 2018

Myung-Il Roh

Department of Naval Architecture and Ocean EngineeringSeoul National University

Lecture Note of Naval Architectural Calculation


Contents

þ Ch. 1 Introduction to Ship Stabilityþ Ch. 2 Review of Fluid Mechanicsþ Ch. 3 Transverse Stability Due to Cargo Movementþ Ch. 4 Initial Transverse Stabilityþ Ch. 5 Initial Longitudinal Stabilityþ Ch. 6 Free Surface Effectþ Ch. 7 Inclining Testþ Ch. 8 Curves of Stability and Stability Criteriaþ Ch. 9 Numerical Integration Method in Naval Architectureþ Ch. 10 Hydrostatic Values and Curvesþ Ch. 11 Static Equilibrium State after Flooding Due to Damageþ Ch. 12 Deterministic Damage Stabilityþ Ch. 13 Probabilistic Damage Stability

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Ch. 9 Numerical Integration Method in Naval Architecture

1. Simpson’s Rule2. Gaussian Quadrature3. Green’s Theorem4. Calculation of Hydrostatic Values by Using Simpson’s Rule5. Calculation of Hydrostatic Values by Using Gaussian Quadrature and Green’s Theorem


1. Simpson’s Rule

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Simpson’s 1st and 2nd Rules

s

y0 y1 y2

s

y0 y1 y2 y3

Simpson’s 1st Rule Simpson’s 2nd Ruley y

x x

Simpson’s 1st and 2nd Rules


5·8·-1, 3·10·-1, and 7·36·-3 Rules

5·8·-1 Rule 3·10·-1 Rule

s

y0 y1 y2

)1103(241

2102 yyysM y -+=

y

x

)3367(120

1210

3 yyysI y -+=

7·36·-3 Rule

5·8·-1, 3·10·-1, and 7·36·-3 Rules

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Derivation of Simpson’s 1st Rule (1/4)

Simpson’s 1st Rule:Approximate the function y by a parabola (quadratic polynomial curve) whose

equation has the form

The parabola is represented by three points defining this curve.The three points (y0, y1, y2) are obtained by dividing the given interval into equal

subintervals “s”.

s

y0

y1 y2y

x

2210 xaxaay ++=

Parabola :

00:0 ayx ==

The relation between the coefficients a0, a1, a2 (“Find”) and y0, y1, and y2 are

22101: sasaaysx ++==

22102 42:2 sasaaysx ++==



00 ay =2

2101 sasaay ++=2

2102 42 sasaay ++=

①

2210 xaxaay ++=

②

③

0102

21 =-++ yysasa

042 202

21 =-++ yysasa

4 x ② - ③:

0432 2101 =+-+ yyysa

)43(21

2101 yyys

a -+-=\

③ - 2 x ②:

022 2102

2 =-+- yyysa

)2(21

21022 yyys

a +-=\

s

y0

y1 y2y

x

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2210 xaxaay ++=

,00 ya = ),43(21

2101 yyys

a -+-= )2(21

21022 yyys

a +-=

Integrate the area A from 0 to 2s. (Definite Integral)

221022100 )2(

21)43(

21 xyyy

sxyyy

syy +-+-+-+=

dxxyyys

xyyys

y

ydxA

s

s

ò

ò+-+-+-+=

=

2

0

221022100

2

0

)2(21)43(

21

s

y0

y1 y2y

x



dxxyyys

xyyys

yAs

ò +-+-+-+=2

0

221022100 )2(

21)43(

21

s

xyyys

xyyys

xy2

0

32102

22100 )2(

61)43(

41

+-+-+-+=

32102

22100 )2)(2(

61)2)(43(

41)2( syyy

ssyyy

ssy +-+-+-+=

syyysyyysy )2(34)43(2 2102100 +-+-+-+=

)141(3 210 yyysA ++=\

s

y0

y1 y2y

x

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Derivation of Simpson’s 2nd Rule (1/4)

Simpson’s 2nd rule :Approximate the function by a cubic polynomial curve whose equation has the

form

The cubic polynomial curve is represented by four points defining this curve.The four points (y0, y1, y2, y3) are obtained by dividing the given interval into equal


33

2210 xaxaxaay +++=

Cubic polynomial curve:

00:0 ayx ==

The relation between the coefficients a0, a1, a2, a3 (“Find”) and y0, y1, y2, and y3 are

33

22101: sasasaaysx +++==

322102 842:2 ssasaaysx +++==

322103 2793:3 ssasaaysx +++==

s

y0

y1 y2 y3y

x



33

2210 xaxaxaay +++=

The unknown coefficients, a0, a1, a2, and a3 lead to

s

y0

y1 y2 y3y

x

,00 ay = 2 31 0 1 2 3 ,y a a s a s a s= + + +

,842 322102 ssasaay +++= 32

2103 2793 ssasaay +++=

00 ya =

)291811(61

32101 yyyys

a +-+-=

)452(21

321022 yyyys

a -+-=

)33(61

321033 yyyys

a +-+-=

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33

2210 xaxaxaay +++=

,00 ya = ),291811(61

32101 yyyys

a +-+-=

),452(21

321022 yyyys

a -+-= )33(61

321033 yyyys

a +-+-=

Integrate the area A from 0 to 3s.

òò +++==ss

dxxaxaxaaydxA3

0

33

2210

3

0)(

s

xaxaxaxa3

0

4332210 432

+++=

43

32

210 4

813

27293 sasasasa +++=

s

y0

y1 y2 y3y

x


)33(83

3210 yyyysA +++=\


33

2210 xaxaxaay +++=

,00 ya = ),291811(61

32101 yyyys

a +-+-=

),452(21

321022 yyyys

a -+-= )33(61

321033 yyyys

a +-+-=

43

32

210 4

813

27293 sasasasaA +++=

By substituting a0, a1, a2 and a3 into the equation, the Area “A” leads to

432103

332102

232100

)33(61

481)452(

21

327

)291811(61

293

syyyys

syyyys

syyyys

syA

+-+-×+-+-×+

+-+-×+=

s

y0

y1 y2 y3y

x

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Derivation of 5·8·-1 Rule (1/4)

5·8·-1 Rule:Approximate the function y by a parabola whose equation has the form

The parabola is represented by three points defining this curve.The three points (y0, y1, y2) are obtained by dividing the given interval into equal


2210 xaxaay ++=

Parabola :

00:0 ayx ==

The relation between the coefficients a0, a1, a2 (“Find”) and y0, y1, and y2 are

22101: sasaaysx ++==

22102 42:2 sasaaysx ++==

s

y0

y1 y2y

x



s

y0

y1 y2y

x

00 ay =2

2101 sasaay ++=2

2102 42 sasaay ++=

①

2210 xaxaay ++=

②

③

0102

21 =-++ yysasa

042 202

21 =-++ yysasa

4 x ② - ③:

0432 2101 =+-+ yyysa

)43(21

2101 yyys

a -+-=\

③ - 2 x ②:

022 2102

2 =-+- yyysa

)2(21

21022 yyys

a +-=\

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s

y0

y1 y2y

x

2210 xaxaay ++=

,00 ya = ),43(21

2101 yyys

a -+-= )2(21

21022 yyys

a +-=

Integrate the area A from 0 to s.

221022100 )2(

21)43(

21 xyyy

sxyyy

syy +-+-+-+=

0

20 0 1 2 0 1 220

1 1( 3 4 ) ( 2 )2 2

s

s

A ydx

y y y y x y y y x dxs s

=

= + - + - + - +

ò

ò



s

y0

y1 y2y

x

20 0 1 2 0 1 220

1 1( 3 4 ) ( 2 )2 2

sA y y y y x y y y x dx

s s= + - + - + - +ò

2 30 0 1 2 0 1 22

0

1 1( 3 4 ) ( 2 )4 6

s

y x y y y x y y y xs s

= + - + - + - +

2 30 0 1 2 0 1 22

1 1( ) ( 3 4 )( ) ( 2 )( )4 6

y s y y y s y y y ss s

= + - + - + - +

0 0 1 2 0 1 21 1( 3 4 ) ( 2 )4 6

y s y y y s y y y s= + - + - + - +

0 1 2(5 8 1 )12sA y y y\ = + -

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Derivation of 3·10·-1 and 7·36·-3 Rules

2 30 1 20 0 0

20 1 2

1 (3 10 )24

s s s

y LM M xdA xydx a x a x a x dx

s y y y

= = = = + +

= + -

ò ò ò

3·10·-1 Rule: The first moment of area about y axis

7·36·-3 Rule: The second moment of area about y axis

0

s

y LM M xdA= = ò2

0

s

y LI I x dA= = òs

y0

y1 y2y

x

2 2 3 4 50 1 20 0 0

30 1 2

1 (7 36 3 )120

s s s

y LI I x dA x ydx a x a x a x dx

s y y y

= = = = + +

= + -

ò ò ò

,00 ya = ),43(21

2101 yyys

a -+-= )2(21

21022 yyys

a +-=

Å

,00 ya = ),43(21

2101 yyys

a -+-= )2(21

21022 yyys

a +-=

Å


2. Gaussian Quadrature

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t1 = -0.7745966692

n Node

t2 = 0

t3 = 0.7745966692

3

jt

Coefficients

jA

A1 = 0.5555555556

A2 = 0.8888888889

A3 = 0.5555555556

In the case of cubic Gaussian quadrature,

1

1 1 2 2 3 31( ) ( ) ( ) ( )f t dt A f t A f t A f t

-» × + × + ×ò

Calculation of Area by Using Gaussian Quadrature

1- 10

)(tf

t

1( )f t3( )f t

1t 2t 3t

2( )f t

Find: Integration of f(t) at a given interval [-1, 1]

Given: Function f(t)

1

1( )f t dt

-òGaussian quadrature:

t1 = -0.8611363115

n Node

t2 = -0.3399810435

t3 = 0.33998104354(Quartic)

jt

Coefficients

jA

A1 = 0.3478548451

A2 = 0.6521451548

A3 = 0.6521451548

t4 = 0.8611363115A4 = 0.3478548451

t1 = -0.9061798459

t2 = -0.5384693101

t3 = 0.05(Quintic)

A1 = 0.2369268850

A2 = 0.4786286704

A3 = 0.6521451548

t4 = 0.5384693101A4 = 0.4786286704

t5 = 0.9061798459A5 = 0.2369268850

1

11

( ) ( )n

j jj

f t dt A f t-

=

» ×åò

1

1 1 2 2 3 3 4 41( ) ( ) ( ) ( ) ( )f t dt A f t A f t A f t A f t

-» × + × + × + ×ò

In the case of quartic Gaussian quadrature,

In the case of quintic Gaussian quadrature,

1

1 1 2 2 3 3 4 4 5 51( ) ( ) ( ) ( ) ( ) ( )f t dt A f t A f t A f t A f t A f t

-» × + × + × + × + ×ò


3. Green’s Theorem*

* Erwin Kreyszig, Advanced Engineering Mathematics, 9th Edition, pp.439-445, 2006

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Calculation of Area by Using Green’s Theorem

( )òòò +=÷÷ø

öççè

æ¶¶

-¶¶

CR

NdyMdxdxdyy

MxN

x

y

RC

M, N: The functions of x and y. And M, N, dM/dy, and dN/dx are continuous on R.

Surface Integral Line Integral

( )ò -=\C

ydxxdyA2

Adxdydxdyyy

xx

dxdyy

MxN

RRR

22)()( ==÷÷ø

öççè

æ-

¶¶

-¶¶

=÷÷ø

öççè

æ¶¶

-¶¶

òòòòòò

L.H.S = (A: Area)

( ) ( ) ( )òòò -=+-=+CCC

ydxxdyxdyydxNdyMdx

R.H.S =

xNyM =-= ,

( )ò -=C

ydxxdyA21

üCalculation of area

( )ò òò== dxdydAA

If

Green's theorem gives the relationship between a line integral around a simple closed curve Cand a double integral over the plane region D bounded by C.

ø The region should be the left-hand of the curve.


Calculation of First Moment of Areaby Using Green’s Theorem (1/2)

2

,22A y

C

xM dy xydxæ ö

\ = -ç ÷è øòÑ

2

,( ) ( ) 2 22 A y

R R R

N M xdxdy xy dxdy xdxdy Mx y x y

æ öæ ö¶ ¶ ¶ ¶- = - - = =ç ÷ç ÷¶ ¶ ¶ ¶è ø è ø

òò òò òò

L.H.S =

( ) òòò ÷÷ø

öççè

æ-=÷÷

ø

öççè

æ+-=+

CCC

xydxdyxdyxxydxNdyMdx22

22

R.H.S =

2

,2xM xy N= - =

ò ÷÷ø

öççè

æ-=

CyA xydxdyxM

221 2

,

üFirst moment of area about the y-axis in x direction

( )ò òò== xdxdyxdAM yA,

If

( )òòò +=÷÷ø

öççè

æ¶¶

-¶¶

CR

NdyMdxdxdyy

MxN




x

y

RC


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Calculation of First Moment of Areaby Using Green’s Theorem (2/2)

2

,22A x

C

yM xydy dxæ ö

\ = -ç ÷è øòÑ

2

,( ) ( ) 2 22 A x

R R R

N M ydxdy xy dxdy ydxdy Mx y x y

æ öæ ö¶ ¶ ¶ ¶- = - - = =ç ÷ç ÷¶ ¶ ¶ ¶è ø è ø

òò òò òò

L.H.S =

( ) òòò ÷÷ø

öççè

æ-=÷÷

ø

öççè

æ+-=+

CCC

dxyxydyxydydxyNdyMdx22

22

R.H.S =

xyNyM =-= ,2

2

ò ÷÷ø

öççè

æ-=

CxA dxyxydyM

221 2

,

üFirst moment of area about the x-axis in y direction

( )ò òò== ydxdyydAM xA,

If

( )òòò +=÷÷ø

öççè

æ¶¶

-¶¶

CR

NdyMdxdxdyy

MxN




x

y

RC



32

,12 3A y

C

xI dy x ydxæ ö

= -ç ÷è øòÑ

Calculation of Second Moment of Areaby Using Green’s Theorem (1/2)

32

,23A y

C

xI dy x ydxæ ö

\ = -ç ÷è øòÑ

32 2

,( ) ( ) 2 23 A y

R R R

N M xdxdy x y dxdy x dxdy Ix y x y

æ öæ ö¶ ¶ ¶ ¶- = - - = =ç ÷ç ÷¶ ¶ ¶ ¶è ø è ø

òò òò òò

L.H.S =

( ) òòò ÷÷ø

öççè

æ-=÷÷

ø

öççè

æ+-=+

CCC

ydxxdyxdyxydxxNdyMdx 233

2

33

R.H.S =

3,

32 xNyxM =-=

üSecond moment of area about the y-axis in x direction

( )2 2,A yI x dA x dxdy= =ò òò

If

( )òòò +=÷÷ø

öççè

æ¶¶

-¶¶

CR

NdyMdxdxdyy

MxN




x

y

RC


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32

,12 3A x

C

yI xy dy dxæ ö

= -ç ÷è øòÑ

Calculation of Second Moment of Areaby Using Green’s Theorem (2/2)

32

,23A x

C

yI xy dy dxæ ö

\ = -ç ÷è øòÑ

32 2

,( ) ( ) 2 23 A x

R R R

N M ydxdy xy dxdy y dxdy Ix y x y

æ öæ ö¶ ¶ ¶ ¶- = - - = =ç ÷ç ÷¶ ¶ ¶ ¶è ø è ø

òò òò òò

L.H.S =

( ) òòò ÷÷ø

öççè

æ-=÷÷

ø

öççè

æ+-=+

CCC

dxydyxydyxydxyNdyMdx33

322

3

R.H.S =

23

,3

xyNyM =-=

üSecond moment of area about the x-axis in y direction

( )2 2,A xI y dA y dxdy= =ò òò

If

( )òòò +=÷÷ø

öççè

æ¶¶

-¶¶

CR

NdyMdxdxdyy

MxN




x

y

RC



[Example] Calculation of Area, First Moment of Area, and Centroid with Respect to the Inertial Frame (1/10)

y

z'z

'y

4p

- 11-

1-

1

1B A

' '::

oy z Body fixed coordinateoyz Water plane fixed coordinate

2

22221

=××=A

1 1

2( , ) (0, )3B By z = - 1B

1:2

üArea A

ò -=C

zdyydz21

Green’s theorem

y

z

02,2)(,)( ££---== tttztty

①

②

③

òò -÷øö

çèæ -=-

0

221

21 dt

dtdyz

dtdzyzdyydz

①

( )ò- ×----=0

21)2()1(

21 dttt

12221

2212

21 0

2

0

2

==

==--ò tdt

ò òò== dydzdAA

Using the chain rule, convert the line integral for y and z into the integral for only one parameter “t”.

Cf: From the geometry of the triangle, the area and the centroid can be obtained as follows;

Segment ①:

( )ò -=C

ydxxdyA21

“Water plane fixed coordinate”

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Segment ②:

òò ÷øö

çèæ -=-

2

021

21 dt

dtdyz

dtdzyzdyydz

②

( )ò ×--×=2

01)2(1

21 dttt

2

0

2

02

212

21 tdt == ò

20,2)(,)( ££-== tttztty

121

=-ò① zdyydz

Segment ①:

ò -=C

zdyydzA21


y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③


üArea A

1 1

2( , ) (0, )3B By z = -


òò-

÷øö

çèæ -=-

2

221

21 dt

dtdyz

dtdzyzdyydz

③

( ) 010021 2

2=×-×= ò

-dtt

121

=-ò② zdyydz

Segment ③:

22,0,)( ££-== tztty

① ② ③

ò -=C

zdyydzA21


y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③


Segment ②:

Segment ①:

121

=-ò① zdyydz

üArea A

1 1

2( , ) (0, )3B By z = -

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16


Green’s theorem

ò -=C

yzdydzy22

1 2

òò - ÷÷ø

öççè

æ-=-

0

2

22

221

221 dt

dtdyyz

dtdzyyzdydzy

①

ò- ÷÷ø

öççè

æ×----=

0

2

2

1)2()1(22

1 dtttt

ò- ÷÷ø

öççè

æ+=

0

2

2

222

1 dttt0

2

23

22

621

-

úû

ùêë

é+= tt

ò òò== ydydzydAM zA,


y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③


üFirst moment of area about the z-axisin y direction MA,z

02,2)(,)( ££---== tttztty

Segment ①:

1 1

2( , ) (0, )3B By z = -

ò ÷÷ø

öççè

æ-=

CyA xydxdyxM

221 2

,


32

221 2

-=-ò① yzdydzy

òò ÷÷ø

öççè

æ-=-

2

0

22

221

221 dt

dtdyyz

dtdzyyzdydzy

②

ò ÷÷ø

öççè

æ×--×=

2

0

2

1)2(122

1 dtttt

ò ÷÷ø

öççè

æ+-=

2

0

2

222

1 dttt

2

0

23

22

621

úû

ùêë

é+-= tt

ò -=C

zA yzdydzyM22

1 2

,


y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③



Segment ①:

Segment ②:

20,2)(,)( ££-== tttztty

1 1

2( , ) (0, )3B By z = -

2018-08-07

17


32

221 2

-=-ò① yzdydzy

32

221 2

=-ò② yzdydzy

òò-

÷÷ø

öççè

æ-=-

2

2

22

221

221 dt

dtdyyz

dtdzyyzdydzy

③

010022

1 2

2

2

=÷÷ø

öççè

æ××-×= ò

-dttt

①

② ③

ò -=C

zA yzdydzyM22

1 2

,[Example] Calculation of Area, First Moment of Area, and Centroid with Respect to the Inertial Frame (6/10)

y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③



Segment ③:

Segment ②:

Segment ①:

22,0,)( ££-== tztty

1 1

2( , ) (0, )3B By z = -


ò -=C

dyzyzdz22

1 2

òò - ÷÷ø

öççè

æ-=-

0

2

22

221

221 dt

dtdyz

dtdzyzdyzyzdz

①

ò- ÷÷ø

öççè

æ×

------=

0

2

2

12

)2()1)(2(21 dtttt

ò- ÷÷ø

öççè

æ ++-+=

0

2

22

22222

21 dttttt

ò- ÷÷ø

öççè

æ-=

0

2

2

122

1 dtt0

2

3

621

-úû

ùêë

é-= tt

ò -=C

yA dyzyzdzM22

1 2

,


y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③


üFirst moment of area about the y-axisin z direction MA,y

Green’s theorem

02,2)(,)( ££---== tttztty

Segment ①:

1 1

2( , ) (0, )3B By z = -

ò ÷÷ø

öççè

æ-=

CxA dxyxydyM

221 2

,

2018-08-07

18


ò ÷÷ø

öççè

æ×

--×-=

2

0

2

12

)2(1)2(21 dtttt

ò ÷÷ø

öççè

æ +---=

2

0

22

22222

21 dttttt

ò ÷÷ø

öççè

æ-=

2

0

2

122

1 dtt2

0

3

621

úû

ùêë

é-= tt

òò ÷÷ø

öççè

æ-=-

2

0

22

221

221 dt

dtdyz

dtdzyzdyzyzdz

②

ò -=C

yA dyzyzdzM22

1 2

,[Example] Calculation of Area, First Moment of Area, and Centroid with Respect to the Inertial Frame (8/10)

y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③



Segment ①:

Segment ②:

20,2)(,)( ££-== tttztty32

221 2

-=-ò① dyzyzdz

1 1

2( , ) (0, )3B By z = -


012010

21 2

2

2

=÷÷ø

öççè

æ×-××= ò

-dtt

òò-

÷÷ø

öççè

æ-=-

2

2

22

221

221 dt

dtdyz

dtdzyzdyzyzdz

③

① ② ③

ò -=C

yA dyzyzdzM22

1 2

,


y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③



Segment ③:

Segment ②:

Segment ①:

22,0,)( ££-== tztty

32

221 2

-=-ò① dyzyzdz

32

221 2

-=-ò② dyzyzdz

1 1

2( , ) (0, )3B By z = -

2018-08-07

19


022

1 2

, =-= òC

zA yzdydzyM

322

221 2

, -=-= òC

yA dyzyzdzM

1 1

,,( , ) , A yA zB B

MMy z

A Aæ ö

= ç ÷è ø

÷÷ø

öççè

æ÷÷ø

öççè

æ-×=

322

21,

20

221

=-= òC

zdyydzA[Example] Calculation of Area, First Moment of Area, and Centroid with Respect to the Inertial Frame (10/10)

y

z'z

'y

4p

- 11-

1-

1

1B A

' '::


2

22221

=××=A1B

1:2

y

z

①

②

③


üArea A



üCentroid

1 1

2( , ) (0, )3B By z = -


Using the chain rule, convert the line integral for y’ and z’ into the integral for only one parameter “t”.

[Example] Calculation of Area, First Moment of Area, and Centroid with Respect to the Body Fixed Frame (1/10)

ò -=C

dyzdzy ''''21

2:1

34

32

32

34)

31,

31()','(

11-=BB zy

( )ò- ×--×=1

11)1(0

21 dtt

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::


üArea A

Green’s theorem

Segment ①:

ò òò== ''dzdydAA

òò -÷øö

çèæ -=-

1

1''

21''''

21 dt

dtdyz

dtdzydyzdzy

①

11,1)(',)(' ££--== ttztty

( )ò -=C

ydxxdyA21

“Body fixed coordinate”

2018-08-07

20



( )ò- ×-×=1

1011

21 dtt

121 1

1==

-t

2:1

34

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::


üArea A

Segment ②:

Segment ①:

ò -=C

dyzdzyA ''''21

1''''21

=-ò① dyzdzy

11,)(',1)(' ££-== tttzty

òò -÷øö

çèæ -=-

1

1''

21''''

21 dt

dtdyz

dtdzydyzdzy

②


2011''''21

=++=-=\ òC

dyzdzyA

① ② ③


2:1

34

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::


üArea A

ò -=C

dyzdzyA ''''21

Segment ③:

Segment ②:

Segment ①:

1''''21

=-ò① dyzdzy

1''''21

=-ò② dyzdzy

11,)(',)(' ££-== tttztty

òò-

÷øö

çèæ -=-

1

1''''''

21 dt

dtdyz

dtdzydyzdzy

③

( ) 011111

1=×-×= ò

-dt

2018-08-07

21



2:134

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::


Green’s theorem

üFirst moment of area about the z’-axisin y’ direction MA,z’

Segment ①:

''''' ', ò òò== dzdyydAyM zA

ò -=C

dyzydzy ''''2'

21 2

11,1)(',)(' ££--== ttztty

òò - ÷÷ø

öççè

æ-=-

1

1

22 ''''2'

21''''

2'

21 dt

dtdyzy

dtdzydyzydzy

①

ò- ÷÷ø

öççè

æ×--×=

1

1

2

1)1(022

1 dttt

041

21 1

1

21

1===

--ò ttdt

ò ÷÷ø

öççè

æ-=

CyA xydxdyxM

221 2

,



2:1

34

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::



Segment ①:

Segment ②:

ò -=C

zA dyzydzyM ''''2'

21'

2

',

0''''2'

21 2

=-ò① dyzydzy

11,)(',1)(' ££-== tttzty

òò - ÷÷ø

öççè

æ-=-

1

1

22 ''''2'

21''''

2'

21 dt

dtdyzy

dtdzydyzydzy

②

ò- ÷÷ø

öççè

æ××-×=

1

1

2

01121

21 dtt

21

41

21

21 1

1

1

1===

--ò tdt

2018-08-07

22


① ② ③


2:134

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::



ò -=C

zA dyzydzyM ''''2'

21'

2

',

Segment ③:

Segment ②:

Segment ①:

òò - ÷÷ø

öççè

æ-=-

1

1

22 ''''2'

21''''

2'

21 dt

dtdyzy

dtdzydyzydzy

②

ò- ÷÷ø

öççè

æ××-×=

1

1

2

1122

1 dtttt61

12221

1

1

31

1

2

-=-=÷÷ø

öççè

æ-=

--ò

tdtt

2

, '1 '' ' ' ' '2 2

1 1 202 6 3

A zC

yM dz y z dy\ = -

= + - =

òÑ

0''''2'

21 2

=-ò① dyzydzy

21''''

2'

21 2

=-ò② dyzydzy

11,)(',)(' ££-== tttztty



2:1

34

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::


üFirst moment of area about the y’-axisin z’ direction MA,y’

Green’s theorem

Segment ①:

''''' ', ò òò== dzdyzdAzM yA

ò -=C

dyzdzzy '2''''

21 2

11,1)(',)(' ££--== ttztty

òò - ÷÷ø

öççè

æ-=-

1

1

22 '2''''

21'

2''''

21 dt

dtdyz

dtdzzydyzdzzy

①

ò- ÷÷ø

öççè

æ×

--×-=

1

1

2

12

)1(0)1(21 dtt

21

41

21

21 1

1

1

1-=-=÷

øö

çèæ-=

--ò tdt

2018-08-07

23



2:134

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::



Segment ①:

Segment ②:

ò -=C

yA dyzdzzyM '2''''

21'

2

',

21'

2''''

21 2

-=-ò① dyzdzzy

11,)(',1)(' ££-== tttzty

òò - ÷÷ø

öççè

æ-=-

1

1

22 '2''''

21'

2''''

21 dt

dtdyz

dtdzzydyzdzzy

②

ò- ÷÷ø

öççè

æ×-××=

1

1

2

02

1121 dttt

041

21 1

1

21

1===

--ò ttdt


① ② ③

2

,1 '' ' ' ' '2 2

1 1 202 6 3

A yC

zM y z dz dy\ = -

= - + = -

òÑ


2:1

34

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::



ò -=C

yA dyzdzzyM '2''''

21'

2

',

Segment ③:

Segment ②:

Segment ①:

òò - ÷÷ø

öççè

æ-=-

1

1

22 '2''''

21'

2''''

21 dt

dtdyz

dtdzzydyzdzzy

③

ò- ÷÷ø

öççè

æ×-××=

1

1

2

12

121 dtttt

61

12221

1

1

31

1

2

===-

-òtdtt

21'

2''''

21 2

-=-ò① dyzdzzy

0'2''''

21 2

=-ò② dyzdzzy

11,)(',)(' ££-== tttztty

2018-08-07

24



2:134

32

32

34)

31,

31()','(

11-=BB zy

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

2:1


22221

=××=A

' '::


üArea A



üCentroid

2''''21

=-= òC

dyzdzyA

32''''

2'

21'

2

', =-= òC

zA dyzydzyM

32'

2''''

21'

2

', -=-= òC

yA dyzdzzyM

÷÷ø

öççè

æ=

AM

AM

zy yAzABB

',', ',

')','(

11

÷÷ø

öççè

æ÷øö

çèæ-××=

32

21,

32

21

÷øö

çèæ -=

31,

31


32' ', =zAM

[Example] Calculation of Area, First Moment of Area, and Centroid- Transform the Position Vectors with Respect to the Inertial Frame

2=A

32' ', -=yAM

úúú

û

ù

êêê

ë

é

-úúúú

û

ù

êêêê

ë

é

÷øö

çèæ-÷

øö

çèæ-

÷øö

çèæ--÷

øö

çèæ-

=úû

ùêë

é=

31

31

4cos

4sin

4sin

4cos

1

1

1 pp

pp

B

BB z

yr

ü Transform the center of buoyancy in oy’z’ frame intooyz frame by rotating the point about the negative x’-axis with an angle of . Then the result is the same asthe calculation result of centroid in the inertial frame.

úúú

û

ù

êêê

ë

é

-úúúú

û

ù

êêêê

ë

é

-=

31

31

22

22

22

22

)31,

31()','(

11-=BB zy

úú

û

ù

êê

ë

é

-=

32

0

)32,0(),(

11-=\ BB zy

÷øö

çèæ -

31,

31

y

z'z

'y

4p

- 11-

1-

1

1B A

÷÷ø

öççè

æ-

32,0

Body fixed frame Inertial frame

4p

'y

yz

4p

11-

1-

1

A

①

②

③

1B

'z

üCalculation of centroid (Center of buoyancy B1)

in the body fixed frame and inertial frame

' '::


2018-08-07

25


4. Calculation of Hydrostatic Valuesby Using Simpson’s Rule


What is a “Hull form”?

þ Hull formn Outer shape of the hull that is streamlined in order to satisfy requirements of a

ship owner such as a deadweight, ship speed, and so onn Like a skin of human

þ Hull form designn Design task that designs the hull form

Hull form of the VLCC(Very Large Crude oil Carrier)

Wireframe model Surface model

2018-08-07

26


Lines of a 320,000ton VLCC

Body Plan

Water Plan Sheer Plan


Station

þ Stations are ship hull cross sections at a spacing of LBP/20.þ The station 0 is located at the aft perpendicular and the station 20

is at the forward perpendicular. And the station 10 therefore represents the midship section.

AP FP

Sheer Plan (Elevation View)

0

Station No.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Midship

• Station spacing = LBP / 20• X position of the Station “A” = Station No. of “A” ´ Station spacing

2018-08-07

27


Section Line and Body Plan

St. 19.75

St. 1xy

z

y

z

Section line or Station line

þ Section line is a curve located on a cross section.

þ In general, because the section lines are located at each station, they are called “station lines”.

þ Section lines make up the lines plan (Body plan).


Buttock Line and Sheer Plan (Buttock Plan)

þ Buttock line is a curve located on a profile (lateral) section (x-z plane).

þ Buttock lines make up the sheer plan or buttock plan of lines.

section line (station)

Example of water line of a 320K VLCC

AP FP

DLWL (Design Load Water Line)Æ Design Draft

Sheer Plan (Elevation View)

2018-08-07

28


Water Line and Water Plan (Half-Breadth Plan)

þ Water line is a curve located on a water plane (vertical) section (x-y plane).

þ Water lines make up the water plan or half-breadth plan of lines.

section line (station)

Water Plan (Plan View)

Example of water line of a 320K VLCC

DLWL (Design Load Water Line)Æ Design Draft

AP FP


Example of Offsets Table of a 6,300TEU Container Ship

Waterline

Stations

Half-Breadth

* Unit: mm

2018-08-07

29


Relationship Between Lines and Offsets Table (1/2)

Generation of offsets tablefrom the lines

Lines

Offsets table


Relationship Between Lines and Offsets Table (2/2)

Waterline at 18m

Half-breadth for each stationat 18m waterline

Waterline at 18m

Half-breadth for St. 19

7036

Half-breadth for St. 18

13033

2018-08-07

30


Calculation of Sectional Area

St.15St.19St.19.75

Z

Y

z = 5 m

Æ

Æ

z

y

HB0

HB1

Half-Breadth(HB)

HB2

HB3HB4 HB5

0 1 2 3 4 5

Simpson’s 1st Rule (S1)

Simpson’s 2nd Rule (S2)

w

z

y

HB0

HB1

Half-Breadth(HB)

HB2

HB3HB4 HB5

0 1 2 3 4 5

Simpson’s 1st Rule (S1) with half spacing



HB0.5

w


Calculation of the First Moment of Sectional Area

1 0 1 2 0 1 21 1( 4 ) ( 4 )3 3

Area dA s y y y w HB HB HB= = + + = + +ò

2 0 1 2 3

2 3 4 5

3 ( 3 3 )8

3 ( 3 3 )8

Area dA s y y y y

w HB HB HB HB

= = + + +

= + + +

ò

Simpson’s 1st Rule

Simpson’s 2nd Rule

Calculation of Sectional Area

z

y

HB0

HB1

Half-Breadth(HB)

HB2

HB3HB4 HB5

0 1 2 3 4 5



w

Calculation of the First Moment of Sectional Area (about y axis)

( ) ( ),1 0 1 2 0 1 2 0 1 21 1 1( 4 ) 1 (0 ) 4 ( ) 1 (2 ) 1 (0 ) 4 ( ) 1 (2 )3 3 3yM zdA s Y Y Y s y s y s y w HB w HB w HB= = + + = × × + × × + × × = × × + × × + × ×ò

( )

( )

,2 0 1 2 3 0 1 2 3

2 3 4 5

3 3( 3 3 ) 1 (0 ) 3 ( ) 3 (2 ) 1 (3 )8 8

3 1 (2 ) 3 (3 ) 3 (4 ) 1 (5 )8

yM zdA s y y y y s y s y s y s y

w w HB w HB w HB w HB

= = + + + = × × + × × + × × + × ×

= × × + × × + × × + × ×

ò



1 2Area Area Area\ = +

,1 ,2y y yM M M\ = +

Distance of each ordinate from y axis

Distance of each ordinate from y axis

2018-08-07

31


Calculation of the First Moment of Sectional Area (about z axis)

Calculation of the First Moment of Sectional Area

z

y

HB0

HB1

Half-Breadth(HB)

HB2

HB3HB4 HB5

0 1 2 3 4 5



w

( )

( )

,1 0 1 2

0 0 1 1 2 2

0 0 1 1 2 2

1 ( 4 )3

1 1 (( / 2) ) 4 (( / 2) ) 1 (( / 2) )31 1 (( / 2) ) 4 (( / 2) ) 1 (( / 2) )3

zM zdA s Y Y Y

s y y y y y y

w HB HB HB HB HB HB

= = + +

= × × + × × + × ×

= × × + × × + × ×

ò

( )

( )

,2 0 1 2 3

0 0 1 1 2 2 3 3

2 2 3 3 4 4 5 5

3 ( 3 3 )8

3 1 (( / 2) ) 3 (( / 2) ) 3 (( / 2) ) 1 (( / 2) )83 1 (( / 2) ) 3 (( / 2) ) 3 (( / 2) ) 1 (( / 2) )8

zM zdA s y y y y

s y y y y y y y y

w HB HB HB HB HB HB HB HB

= = + + +

= × × + × × + × × + × ×

= × × + × × + × × + × ×

òSimpson’s 1st Rule


,1 ,2z z zM M M\ = +

Distance of each ordinate from z axis

Distance of each ordinate from z axis


Calculation of the Centroid of Sectional Area

Calculation of the Centroid

z

y

HB0

HB1

Half-Breadth(HB)

HB2

HB3HB4 HB5

0 1 2 3 4 5



w

0 1 2 2 3 4 51 3( 4 ) ( 3 3 )3 8

Area w HB HB HB w HB HB HB HB\ = + + + + + +

( ),1 0 1 21 1 (0 ) 4 ( ) 1 (2 )3yM w HB w HB w HB= × × + × × + × ×

( ),2 2 3 4 53 1 (2 ) 3 (3 ) 3 (4 ) 1 (5 )8yM w w HB w HB w HB w HB= × × + × × + × × + × ×

1 0 1 21 ( 4 )3

Area w HB HB HB= + +

2 2 3 4 53 ( 3 3 )8

Area w HB HB HB HB= + + +

( )

( )

0 1 2

2 3 4 5

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

3 1 (2 ) 3 (3 ) 3 (4 ) 1 (5 )8

yM w HB w HB w HB

w w HB w HB w HB w HB

\ = × × + × × + × ×

+ × × + × × + × × + × ×

Centroid

, yzy z

MMCentroid Centroid Area Area

\ = =

( ),1 0 0 1 1 2 21 1 (( / 2) ) 4 (( / 2) ) 1 (( / 2) )3zM w HB HB HB HB HB HB= × × + × × + × ×

( ),2 2 2 3 3 4 4 5 53 1 (( / 2) ) 3 (( / 2) ) 3 (( / 2) ) 1 (( / 2) )8zM w HB HB HB HB HB HB HB HB= × × + × × + × × + × ×

( )

( )

0 0 1 1 2 2

2 2 3 3 4 4 5 5

1 1 (( / 2) ) 4 (( / 2) ) 1 (( / 2) )3

3 1 (( / 2) ) 3 (( / 2) ) 3 (( / 2) ) 1 (( / 2) )8

zM w HB HB HB HB HB HB

w HB HB HB HB HB HB HB HB

\ = × × + × × + × ×

+ × × + × × + × × + × ×

2018-08-07

32


Station No.

Calculation of Water Plane Area

Water Plan (Plan View) DLWL (Design Load Water Line)Æ Design Draft

AP FP

0

Station No.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20


Simpson’s 2nd Rule (S2) Simpson’s 1st Rule (S1)

0-0.333 -0.166

1. Generate a temporary section (e.g., -0.166)2. Perform Simpson’s 1st Rule.

Half-Breadth (HB)


Calculation of Displacement Volume

þ The displacement volume (underwater volume) at a certain draft can be calculated by integrating sectional areas in the longitudinal direction.

þ In addition, the volume can be calculated by integrating water plane areas in the vertical direction. There can be a difference between two volumes due to approximation.

x

S

AP FP

Volume integralfrom sectional areasin the longitudinal (x) direction

Simpson’s 1st Rule (S1) Simpson’s 2nd Rule (S2)

2018-08-07

33


Calculation for Wetted Surface Area

þ The wetted surface area means ship’s area which contacts with water.

þ This area can be calculated with the following approximate formula.

2 2. 6

. 41

Sta

Sta

dy dyS z dxdx dz

d æ ö æ ö= + +ç ÷ ç ÷è ø è øò

FE

G

AB

C

3=z

St. 4 St. 5St. 6

xd

xd

zd

6=z


(1)

Sta.

(4)

Sta.Ford.

(1.1)

HB6m

(1.2)

HB3m

(5)

Sta.Aft.

(4.1)

HB6m

(4.2)

HB3m

(5.1)

HB6m

(5.2)

HB3m

(2)

δy/δz

(δy/δz)2(3)

(δy/δz)2

(6)

Meanδy/δx

(7)

(δy/δx)2

(8)

Sum

(9)

(Sum)1/2

(10)

S.M

(11)

Prod.

Example of Calculation for Wetted Surface Area (1/7)

Calculate the wetted surface area of the ship from St. 1 to St. 5between 3m and 6m of waterline.

Station interval 13.94x md = =

2 2

1 y ySumx z

d dd d

æ ö æ ö= + +ç ÷ ç ÷è ø è ø

HB: Half-breadth for waterline

HBA: Half-breadth afterward

HBf: Half-breadth forward

S: Wetted surface area of the ship

FE

G

AB

C

3=z

St. 4 St. 5St. 6

xdxd

zd

6=z

z

o

x

yds

dx

St.4

St.5

St.6

6=z

3=z

F

E

G

B

A

Cdx

dz

HB, (4)-1

(1)-1

(5)-1

(4)-2

(1)-2

(5)-2

FE

G

AB

C

3=z

St. 4 St. 5St. 6

xd

xd

zd

6=z

We can find

the vertical station shape slope

and longitudinal water line slope

by using the central difference.

dydx

dydz

15.47

Wetted Surface AreaProjected tocenter plane

2018-08-07

34


(6 3) 3z md = - =

dy ydz z

dd

»

1)

1. Approximated formula for ship’s surface area:

2 2.5

.11

Sta

Sta

dy dyS z dxdx dz



2 2

1 y ySumx z

d dd d

æ ö æ ö= + +ç ÷ ç ÷è ø è ø

(1)

Sta.

(4)

Sta.Ford.

(1.1)

HB6m

(1.2)

HB3m

(5)

Sta.Aft.

(4.1)

HB6m

(4.2)

HB3m

(5.1)

HB6m

(5.2)

HB3m

(2)

δy/δz

(δy/δz)2(3)

(δy/δz)2

(6)

Meanδy/δx

(7)

(δy/δx)2

(8)

Sum

(9)

(Sum)1/2

(10)

S.M

(11)

Prod.

. . 6 . . 3W L m W L mHB HBdydz zd

= =-»

(2)

. . 6 . . 3W L m W L my HB HBd = == -

[(1.2) – (1.1)]In the table,

22. . 6 . . 3W L m W L mHB HBdy

dz zd= =-æ öæ ö »ç ÷ ç ÷

è ø è ø

(3)






15.47

FE

G

AB

C

3=z

St. 4 St. 5St. 6xd

xd

zd

6=z



(1)

Sta.

(4)

Sta.Ford.

(1.1)

HB6m

(1.2)

HB3m

(5)

Sta.Aft.

(4.1)

HB6m

(4.2)

HB3m

(5.1)

HB6m

(5.2)

HB3m

(2)

δy/δz

(δy/δz)2(3)

(δy/δz)2

(6)

Meanδy/δx

(7)

(δy/δx)2

(8)

Sum

(9)

(Sum)1/2

(10)

S.M

(11)

Prod.


2 2.5

.11

Sta

Sta

dy dyS z dxdx dz


. . 6 . . 3

12 W L m W L m

dy dy dydx dx dx= =

æ ö= +ç ÷

è ø

2)


2 2

1 y ySumx z

d dd d

æ ö æ ö= + +ç ÷ ç ÷è ø è ø

, . . 3 , . . 3

. . 3 . . 3 2A W L m F W L m

W L m W L m

HB HBdy ydx x x

dd d

= =

= =

-» =

×

[(5.2) – (4.2)]/2δx

, . . 6 , . . 6 , . . 3 , . . 312 2 2

A W L m F W L m A W L m F W L mHB HB HB HBdydx x xd d

= = = =- -æ ö» +ç ÷× ×è ø

(6)

22, . . 6 , . . 6 , . . 3 , . . 31

2 2 2A W L m F W L m A W L m F W L mHB HB HB HBdy

dx x xd d= = = =é ù- -æ öæ ö » +ê úç ÷ç ÷ × ×è ø è øë û

(7)

, . . 6 , . . 6

. . 6 . . 6 2A W L m F W L m

W L m W L m

HB HBdy ydx x x

dd d

= =

= =

-» =

×

[(5.1) – (4.1)]/2δxIn the table,

(W.L.: Waterline)






15.47

FE

G

AB

C

3=z

St. 4 St. 5St. 6

xdxd

zd

6=z


2018-08-07

35


(1)

Sta.

(4)

Sta.Ford.

(1.1)

HB6m

(1.2)

HB3m

(5)

Sta.Aft.

(4.1)

HB6m

(4.2)

HB3m

(5.1)

HB6m

(5.2)

HB3m

(2)

δy/δz

(δy/δz)2(3)

(δy/δz)2

(6)

Meanδy/δx

(7)

(δy/δx)2

(8)

Sum

(9)

(Sum)1/2

(10)

S.M

(11)

Prod.

2. Substituting 1) and 2) into the formula.2 2

.5

.11

Sta

Sta

y yS z dxx z

d ddd d

æ ö æ ö» + +ç ÷ ç ÷è ø è øò

2 2.5

.11

Sta

Sta

dy dyS z dxdx dz


3. By using the Simpson’s 1st and 2nd rules, calculate the ship’s surface area.


2 2.5 , . . 6 , . . 6 , . . 3 , . . 3 . . 6 . . 3

.1

112 2 2

Sta A W L m F W L m A W L m F W L m W L m W L mSta

HB HB HB HB HB HBz dxx x z

dd d d

= = = = = =æ ö- -æ ö -æ ö= + + +ç ÷ç ÷ ç ÷× × è øè øè ø

ò(9) (8)=

, . . 6 , . . 6

, . . 3 , . . 3

12 2

2

A W L m F W L m

A W L m F W L m

HB HBdydx x

HB HBx

d

d

= =

= =

-æ» ç ×è

- ö+ ÷× ø

. . 6 . . 3W L m W L mHB HBdydz zd

= =-»

1)

2)


2 2

1 y ySumx z

d dd d

æ ö æ ö= + +ç ÷ ç ÷è ø è ø

(8) 1 (7) (3)= + +






15.47

FE

G

AB

C

3=z

St. 4 St. 5St. 6xd

xd

zd

6=z



(1)

Sta.

(4)

Sta.Ford.

(1.1)

HB6m

(1.2)

HB3m

(5)

Sta.Aft.

(4.1)

HB6m

(4.2)

HB3m

(5.1)

HB6m

(5.2)

HB3m

(2)

δy/δz

(δy/δz)2(3)

(δy/δz)2

(6)

Meanδy/δx

(7)

(δy/δx)2

(8)

Sum

(9)

(Sum)1/2

(10)

S.M

(11)

Prod.


s

y0 y1 y2 y3

y

x

)33(83

3210 yyyysArea +++=


s

y0 y1 y2

y

x

)4(31

210 yyysArea ++=

Total Area:

0 1 2 2 3 4 53 8 1 1 8 1 1 8 1 11 2 4 3 38 3 3 2 3 3 2 3 3 2

Area Area x y y y y y y yd æ ö+ = × × × × + × × × + × × + + + +ç ÷è ø


( )0 1 2 3 4 53 0.444 1.778 1.444 3 3 18

x y y y y y yd= × × + + + + +

: S.M, (10)

1) Simpson’s multiplier (10)

y0 y1 y3 y4

y=(sum)1/2,(9)

Station,(1)

y2 y5

1 11/2 2 3 4 5δx

12

xd

Simpson’s 1st Rule:

( )0 1 21 11 43 2

Area x y y yd= × × + +

Area1

Simpson’s 2nd Rule:

( )2 3 4 532 3 38

Area x y y y yd= × × + + +

Area2


15.47

FE

G

AB

C

3=z

St. 4 St. 5St. 6

xdxd

zd

6=z


2018-08-07

36


(1)

Sta.

(4)

Sta.Ford.

(1.1)

HB6m

(1.2)

HB3m

(5)

Sta.Aft.

(4.1)

HB6m

(4.2)

HB3m

(5.1)

HB6m

(5.2)

HB3m

(2)

δy/δz

(δy/δz)2(3)

(δy/δz)2

(6)

Meanδy/δx

(7)

(δy/δx)2

(8)

Sum

(9)

(Sum)1/2

(10)

S.M

(11)

Prod.

2 2.5 , . . 6 , . . 6 , . . 3 , . . 3 . . 6 . . 3

.1

112 2 2

Sta A W L m F W L m A W L m F W L m W L m W L mSta

HB HB HB HB HB HBS z dxx x z

dd d d

= = = = = =æ ö- -æ ö -æ ö» + + +ç ÷ç ÷ ç ÷× × è øè øè ø

ò

233 13.94 12.84 201.36 ( )8

m= × × × =


2 2, . . 6 , . . 6 , . . 3 , . . 3 . . 6 . . 33 1. 1

8 2 2 2A W L m F W L m A W L m F W L m W L m W L mHB HB HB HB HB HBz x S M

x x zd d

d d d= = = = = =

é ùæ ö- -æ ö -æ öê ú= × × × × + + +ç ÷ç ÷ ç ÷ê ú× × è øè øè øë ûå

(9)(10)

3 Prod.8

z xd d= × × ×å

(11)


13.94 , 3x m z md d= =

2 2

1 y ySumx z

d dd d

æ ö æ ö= + +ç ÷ ç ÷è ø è ø





15.47

FE

G

AB

C

3=z

St. 4 St. 5St. 6xd

xd

zd

6=z




2201.36S m»

4. Calculate the wetted surface area of both sides of the ship

( )22 2 201.36 402.7S m= × » × =

Wetted Surface Area, Both sides

(1)

Sta.

(4)

Sta.Ford.

(1.1)

HB6m

(1.2)

HB3m

(5)

Sta.Aft.

(4.1)

HB6m

(4.2)

HB3m

(5.1)

HB6m

(5.2)

HB3m

(2)

δy/δz

(δy/δz)2(3)

(δy/δz)2

(6)

Meanδy/δx

(7)

(δy/δx)2

(8)

Sum

(9)

(Sum)1/2

(10)

S.M

(11)

Prod.






2 2

1 y ySumx z

d dd d

æ ö æ ö= + +ç ÷ ç ÷è ø è ø

15.47

FE

G

AB

C

3=z

St. 4 St. 5St. 6

xdxd

zd

6=z


2018-08-07

37


5. Calculation of Hydrostatic Valuesby Using Gaussian Quadrature and

Green’s Theorem


Description of Section Lines (1/2)

300.0 50.0 27.0 18.0 // LBP, Bmld, Dmld, T27 // Section Line Num.…1.0 11 // Station, Point Num.y0 z0 // Y coord., Z coord. y1 z1

y2 z2

…y10 z10

1.5 10…

Example of text file for describing the body plan of a ship

y

z

),( 00 zy),( 11 zy

),( 22 zy),( 33 zy

),( 44 zy

),( 55 zy),( 66 zy

),( 77 zy

),( 88 zy

),( 99 zy

),( 1010 zy

Given: Body plan of a shipFind: Text file describing the body plan of a ship

1. Make a text file for describing the body plan of a ship.

2018-08-07

38


y

z

),( 00 zy),( 11 zy

),( 22 zy),( 33 zy

),( 44 zy

),( 55 zy),( 66 zy

),( 77 zy

),( 88 zy

),( 99 zy

),( 1010 zy

)()()()()( 311

322

311

300 uNuNuNuNu DD --+×××+++= ddddr

( ) 0 1 1

( ) 30,1 1, 1

ini

j

i , , ,DN u nu j , ,K K D n

= -

=

= - = + +

d K

K

: de Boor points (control points),

: B-splines basis function of degree

: Knots, where

)()()( 11

1

11

1 uNuuuuuN

uuuuuN n

iini

nini

ini

ini

-+

+

+-

--+

-

--

+-

-=

îíì <£

= -

else if

01

)( 10 iii

uuuuN å

-

=

=1

01)(,

D

i

ni uN

Make cubic B-spline curve which passes through

the given points

Æ Refer to the Part “Curve and Surface”

(Ship Hull Form Modeling for 2nd Year Undergraduate Course)

2. Find cubic B-spline curves passing the points on the section lines.

Given: Data of the points on the section line that describes the body plan of a shipFind: Cubic B-spline curve which passes the points on the section line

Description of Section Lines (2/2)


Given: B-spline curve, the intersection points between the B-spline curves and water plane, and B-spline parameter “u” at each end point of the line segmentsFind: Sectional area and 1st moment of sectional area

Calculation of Sectional Area and 1st Moment of Sectional Area Under the Water Plane (1/4)

'y

'z

Curve #6

Curve #1Curve #4

The section is represented byCurve #0 ~ Curve #6

The sectional area and 1st moment of the sectional area under the waterline is calculatedby integration of the following line segments.

3-②Æ0-①Æ1-①ÆW-④, 1-③Æ2-①Æ6-①ÆW-②

3-②

0-①

1-①

W-④

W-⑤

1-③

2-①

6-①

3-①

4-①

5-①

6-②W-①

W-②

W-③1-②

Waterline

x

y

RC

2018-08-07

39


Given: B-spline curve, the intersection points between the B-spline curve and water plane, and B-spline parameter “u” at each end point of the line segmentsFind: Sectional area and 1st moment of section

max minmin

( 1)( )2

t u uu u+ -= +

ü Relation between the Parameter u and t

( 1)(5 0) 02

tu + -= +

1 1

1 1

1 ' ' 1'( ( )) '( ( )) ( )2 2

dz dy duy u t z u t dt f t dtdu du dt- -

æ ö- =ç ÷è øò ò

> Since the parameter ‘u’ increases monotone, the interval can be found easily.

> Using the chain rule, convert the line integral for y’ and z’ into the integral for only one parameter ‘u’.

ü In the same way, integrate the remained line segments using Green’s theorem and Gaussian quadrature.

5

0

1 ' ''( ) '( )2

dz dyy u du z u dudu du

æ ö-ç ÷è øò

5 5

0 0

1 ' ' 1'( ) '( ) ( )2 2

dz dyy u z u du g u dudu du

æ ö= - =ç ÷è øò ò

òò ¢¢=R

zdydA ( )1 ' ' ' '2 C

y dz z dy= -òÑGreen’s Theorem

<Surface integral> <Line integral >

For example, integrate the line segment 0-①For the line integral of the segment in the y‘ and z' coordinates, the interval for the integration has to be determined.

Æ To use Gaussian quadrature, convert the integration parameter ‘u’ and the interval [0, 5] into ‘t’ and [-1,1]

1t =

1t = -u=0

u=5

3-②

0-①

1-①

W-④

W-⑤

1-③

2-①

6-①

3-①

4-①

5-①

6-②

W-①

W-②

W-③1-②



5

0

1 ' ''( ) '( )2

dz dyy u du z u dudu du

-ò5

0

1 ' ''( ) '( )2

dz dyy u z u dudu du

æ ö= -ç ÷è øò

u

( )g u

50

5

0

1 ( )2

g u duò=

Using the chain rule, convert the line integral for y’ and z’ into the integral for only one parameter “u”.

1 ' ' ' '2 C

y dz z dy= -òÑ'y

'z

Convert surface integral into line integral

t

( )f t

11-

1

1

1 ' ''( ( )) '( ( ))2

dz dy duy u t z u t dtdu du dt-

æ ö-ç ÷è øò

1

1

1 ( )2

f t dt-ò=

To use Gaussian quadrature, convert the parameter and the interval into “t” and [-1,1].

1t =

u=0

u=5

3-②

0-①

1-①

W-④

W-⑤

1-③

2-①

6-①

3-①

4-①

5-①

6-②

W-①

W-②

W-③1-②

※ Procedure for calculation of the sectional area and 1st moment of sectional area under the water plane


1t = -

max minmin

( 1)( )2

t u uu u+ -= +

ü Relation between the Parameter u and t

( 1)(5 0) 02

tu + -= +

2018-08-07

40


※ Method to check whether the line segments are located under the water plane or not

( ) 0- >n X Og( ) 0- £n X Og

üCheck the location of the point by using the sign of dot product of normal vector of the water plane and position vector of the point

: The point is above the water plane.

: The point is on or below the water plane.

§ To calculate the sectional area under the water plane, it isrequired to check whether the points on the line segments arelocated under the water plane or not.

: Normal vector

2X

1X

O

n

Point:

Point:

: Origin

üPerform only line integration for the segments which are on or below the water plane.

The line segment 1-② : Æ No integration

( ) 0- >n X Og

The line segment 0-① : Æ Perform integration

( ) 0- £n X Og

The line segment 2-① : Æ Perform integration

( ) 0- £n X Og( : the middle point of the each line segment)X

In this example, the line integration is performed as follows:

n

O

: Origin

3-②

0-①

1-①

W-④

⑤

1-③

2-①

6-①

4-①

6-②W-①

W-②

W-③1-②



Calculation of Ship’s Displacement Volume, 1st Moment of Displacement Volume, LCB, TCB, and KB

x¢

A,A xM ¢

,A yM ¢

,A zM ¢

( ') 'V A x dx= Ñ = ò , ' ' , ' '( ') 'y z A y zM M x dxÑ = ò, ' ' , ' '( ') 'x z A x zM M x dxÑ = ò, ' ' , ' '( ') 'x y A x yM M x dxÑ = ò

①

② ③

④

: The integration value is 0.

DisplacementVolume

2) Generate B-spline curve which interpolates the ordinates.

3) Perform the line integration counter-clockwise using Green’s theorem and Gaussian quadrature.

1) Make the ordinate set along ship’s length by using the results for each section.

ü Calculate the displacement volume and 1st

moment of the volume by integrating the sectional areas and 1st moments of the sectional areas over ship’s length.

Ñ= Ñ '', zyM

LCB

swrD = ×Ñ

Displacement:

, ' ', x zMTCB Ñ=

Ñ, ' ', x yM

VCB Ñ=Ñ

(from waterline) dKB VCB T= +

Calculation procedure

Calculate sectional area and 1st moment of the area of each section

The results for each section

Given: Sectional areas and 1st moments of the sectional areas under the water planeFind: Displacement volume, 1st moment of displacement volume, LCB, TCB, and KB

2018-08-07

41


Given: Intersection points between the water plane and the section linesFind: Water plane area, 1st moment and 2nd moment of the water plane area

Calculation of Water Plane Area, 1st and 2nd Moment of Water Plane Area


y

o

Intersection point between the water plane area and the section lines

Water plane area

Calculate sectional area or 1st moment of area of each section

z yx

O

üGenerate the curve which interpolates the intersection points. If a section ‘x’ has no intersection point, input the point as (x, 0, 0).

üTransform the intersection points decomposed in body fixed frame into the points decomposed in water plane fixed frame (inertial frame).

üCalculate the area, 1st moment and 2nd

moment of area using Green’s theorem or Gaussian quadrature.


Given: Intersection points between the water plane and the section linesFind: Wetted surface area

Calculation of Wetted Surface Area

4) Generate B-spline curve which interpolates the ordinates.

2) Calculate the sectional area surrounded by the girth length and water plane.

3) Make the ordinate set of the sectional area.

1) Calculate the girth length of the section lines under the water plane.


x

S

Integration direction

1 1

0 0

( )t t

t ts ds t dt= =ò ò r&

5) Integrate the area along ship’s length using Green’s theorem or Gaussian quadrature.

Æ Wetted surface area is calculated.

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