2018-08-07

Lecture Note of Naval Architectural Calculation

Ship Stability

Ch. 7 Inclining Test

Spring 2018

Myung-Il Roh

Department of Naval Architecture and Ocean Engineering Seoul National University

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

Contents

þ Ch. 1 Introduction to Ship Stability þ Ch. 2 Review of Fluid Mechanics þ Ch. 3 Transverse Stability Due to Cargo Movement þ Ch. 4 Initial Transverse Stability þ Ch. 5 Initial Longitudinal Stability þ Ch. 6 Free Surface Effect þ Ch. 7 Inclining Test þ Ch. 8 Curves of Stability and Stability Criteria þ Ch. 9 Numerical Integration Method in Naval Architecture þ Ch. 10 Hydrostatic Values and Curves þ Ch. 11 Static Equilibrium State after Flooding Due to Damage þ Ch. 12 Deterministic Damage Stability þ Ch. 13 Probabilistic Damage Stability

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How can you get the value of the KG?

K: Keel G: Center of gravity

Ch. 7 Inclining Test

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The Problem of Finding an Accurate Vertical Center of Gravity (KG)

The problem of finding an accurate KG for a ship is a serious one for the ship’s designer.

FG

G ü Any difference in the weight of structural parts, equipment, or welds in different ship will produce a different KG.

K

There is an accurate method of finding KG for any particular ship and that is the inclining test.

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Required Values to Find the KG (2/3)

Heeling moment produced by total weight

Righting moment produced by buoyant force

Static equilibrium of moment t =F × GZ Inclining test formula r B

6 Naval Architectural Calculation, Spring 2018, Myung-Il Roh Required Values to Find the KG (1/3) GZ» GM ×sinf (at small angle f ) GM= KB +BM - KGKG The purpose of the inclining test is to determine the position of the center of mass of the ship in an accurately known condition. z z¢ t e

G: Center of total mass (gravity) B: Center of buoyancy at initial position * Angle of inclination should be within the range of inclinations G1: New position of center of total mass (gravity) after the weight has been moved where the metacenter (M) is not changed (about 7°~10°). B1: New position of center of buoyancy after the ship has been inclined Z: The intersection of vertical line through the center of buoyancy with the transversally parallel line to a waterline through center of total mass (gravity) M: The intersection of a vertical line through the center of buoyancy at previous position(B)

with a vertical line through the present position of the center of buoyancy(B1) after the τe : Heeling moment ship has been inclined transversally through a small angle Required values to find the KG τr : Righting moment M f d • Draft FG

Z • Total weight (FG) G 1 y • Hydrostatic values (KB, BM) • Weight (w) y¢ • Distance (d) B 1 • Angle of inclination (f )*

FB

t r

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t r=F B × GZ GZ» GM ×sinf (at small angle f ) GM= KB +BM - KGKG

z ¢ z w× d t e GG1 = FG M f d

th=F G × GG1 cos f Shift of center of total mass (gravity) FG Z G 1 y tr=F B × GZ » F B × GM sin f

y¢ FGB× GG1 cosf = F × GM sin f B1 (QStatic equilibrium: FGB= F )

FB

t r

3 Inclining experiment formula 2018-08-07

The angle of inclination can be measured when we perform the inclining test.

Derivation of Inclining Test Formula (1/5)

t r=F B × GZ GZ» GM ×sinf (at small angle f ) 8 Naval Architectural Calculation, Spring 2018, Myung-Il Roh GM= KB +BM - KGKG

z w× d KG z¢ GM = t e FG × tanf

M f d KG= KB + BM - GM FG Z w× d G Required Values to Find the KG (3/3) 1 y =KB + BM - FG × tanf

y¢

B1

FB

t r

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t r=F B × GZ GZ» GM ×sinf (at small angle f ) GM= KB +BM - KGKG

d y ¢ d z ¢ z B B B z n B n z¢ 1 M ¢ ¢ KBsinf dyB cos f dzB sin f f

KG, N0 NZ,

Suppose that the center of gravity is located at K. Then the y y KN represents the righting arm.

f d y ¢ B B B y¢ B1 FG F d z ¢ G F B B FB FB w KG, N NZ, KG, 0

GZ = KN

=KNNN0 + 0 ¢ ¢ = KB ×sinf +d yBB ×cosf + d z × sin f

These terms have positive effect to the restoring moment arm.

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t r=F B × GZ GZ» GM ×sinf (at small angle f ) 10 Naval Architectural Calculation, Spring 2018, Myung-Il Roh GM= KB +BM - KGKG ¢ KG= d zG z z z¢ z¢ M M

f f

FG

G DerivationZ of Inclining Test Formula (2/5) y y

d z ¢ G f d y ¢ d y ¢ B B B B y¢ y¢ B1 B1 FG ¢ FG ¢ d zB d zB FB FB FGB ' FB KG, NZ, N N0 K N ¢ d0zG sin f GZ = KN GZ =KNKG - ' =KNNN + ¢ ¢ 0 0 = KB ×sinf +d yBB ×cosf + d z × sin f ¢ ¢ = KB ×sinf +d yBB ×cosf + d z × sin f

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This term has negative effect to the restoring moment arm.

t r=F B × GZ GZ» GM ×sinf (at small angle f ) GM= KB +BM - KGKG ¢ ¢ KG= d zG G d yG z zn GG= d y ¢ z¢ Z z¢ 1 G M M G1 ¢ Z1 dyG cos f f f d FG G ' G1 ' N FG FG G Z G Z ¢G Z1 y d yG 1 y

d z ¢ G f f d y ¢ d yB B B B y¢ y¢ B1 B1 FG d z ¢ d zB B ¢ G ' dyG cos f FGB ' FB FGB ' 1 FB Derivation of Inclining Test Formula (3/5) K N¢ N K N¢ N dzG0sin f dzG 0sin f

GZ =KNKG - ' GZ1 1 =KNKGGG -''' - 1 ¢ ¢ ¢ ¢ = KB ×sinf +d yBB ×cosf + d z × sin f = KB ×sinf +d yBB ×cosf + d z × sin f ¢ ¢ -dzG ×sin f -dzG ×sin f

This term has negative effect to the restoring moment arm.

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Derivation of Inclining Test Formula (5/5)

Calculation of moment using “GZ”

12 Naval Architectural Calculation, Spring 2018, Myung-Il Roh t r=F B × GZ GZ» GM ×sinf (at small angle f ) GM= KB +BM - KGKG ¢ KG= d zG z z GG= d y ¢ z¢ z n n n b ¢ 1 G rPz=b R b r P z M M écosf- sin f ù n ê ú = Rb ()f ësinf cos f û f y¢ ü Rotational transformation matrix (b to n frame) f f y d FG FG FG z z¢ G DerivationZ of Inclining Test Formula (4/5) G Z Z1 cos(---f ) sin( f ) ¢G1 é ù n d yG y ê ú =Rb () -f yn ësin(--f ) cos( f ) û

d zG f ü Rotational transformation matrix (b to n frame) -f y d y ¢ B B y¢ y y¢ b B1 d z ¢ ¢ B dyG cos f G ' FG ' G1 ' F é¢ ù B B édyBB ù Kécosf sinN f ù d y K N N = ê ú 0 ¢ ê ú ê ú -dzG sin f ëd zB û ë-sinf cos fû ê¢ ú ëd zB û GZ1 1 =KNKGGG -''' - 1 ¢ ¢ dyBBB= d ycos f + d z sin f ¢ ¢ = KB ×sinf +d yBB ×cosf + d z × sin f ¢ ¢ ¢ ¢ d y dyGGG= d ycos f + d z sin f -dzG ×sin f -dyG ×cos f B 11 Naval Architectural Calculation, Spring 2018, Myung-Il Roh -d yG

ü In n-frame, because the forces are acting on

vertical direction, moment arm is yn-component.

w× d KG= d z ¢ GG = t r=F B × GZ G 1 F GG= d y ¢ G GZ» GM ×sinf (at small angle f ) 1 G GM= KB +BM - KGKG

z z¢ GZ =KNKGGG -''' - t e 1 1 1 ¢ ¢ =KB ×sinf + d yBB × cos f + d z × sin f M -dz ¢ ×sin f -dy ¢ ×cos f f d G G = KN -KG sinf -GG cosf F 1 G = GZ Z G f <<1 1 y = GM sinf

: Inclining test formula =GMsinf - GG1 cos f

y¢ d MGZ= D × B1 1 1

= D ×(GM sinf - GG1 cos f ) FB

GMsinf= GG1 cos f

Moment produced by total weight & buoyant force t GG w× d r \GM =1 = tanfFG × tan f

: Assume that

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Static equilibrium of moment 2018-08-07

Deflection

Mark ! Mark !

Method of Measuring the Angle of Inclination (2/2)

* Reference: http://www.mchl.fr 14 Naval Architectural Calculation, Spring 2018, Myung-Il Roh w× d Method of Measuring GM = F × tanf the Angle of Inclination (1/2) G

How can you measure the angle of inclination when you perform the inclining test? z z¢

Inclining test formula

f y

L ¢ Deflection y

L: Length of plumb line Plumb: The line which is exactly vertical or perpendicular to a level horizontal line Batten: Long and thin strip of wood Deflection \tanf = L

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Precautions to be Taken During the Inclining Test

Certain precautions must be taken to ensure that accuracy is obtained. (1) The ship must be floating upright and freely without restraint from ropes. (2) There should be no wind on the beam. (3) All loose weights should be fixed. (4) All cross-connections between tanks should be closed. (5) Tanks should be empty or pressed full. If neither of these conditions is w× d possible, the level of liquid in the tank should be such that the free surface effect is readily calculable and will remain sensibly constant GM = through the experiment. (6) The number of men on board should be kept to a minimum and they FG × tanf should be on the center line. (7) Any mobile equipment used to move the weights across the deck must return to a known positions for each set of readings.

16 Naval Architectural Calculation, Spring 2018, Myung-Il Roh KG= KB + BM - GM A ship is inclined by moving a weight of 40 tons a distance 8 m from the center line. A 12 m pendulum shows a deflection of 0.3 m. Displacement of the ship is 3,700 tons. If the KB is 5 m and BM is 14 m, what is the KG ? z z¢

Inclining test Example formula 0.3 tanf = = 0.025 M d= 8 m 12 w× d 40 × 8 GM = = = 3.46 Z FG ×tanf 3700 × 0.025 G 1 y 12m f

y¢

B1 0.3m \KG =14.54[ m ]

Solution)

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w× d Inclining Various Problems GM = test F × tanf Using the Inclining Test Formula G formula

The inclining test formula can be used in various problems as follows:

(1) To find the angle of heel f, a ship will take by moving a weight a transverse distance d.

(2) To find the weight w necessary to remove or produce a heel by moving it a transverse distance d.

(3) To find the distance d necessary to move a weight in order to remove or produce a heel.

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