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Stability and Metacentric Height

The conditions for the stability of floating bodies and ships with solid loads. Includis an introduction to Metacentric heights and centre of buoyancy

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Contents

Introduction
Centre Of Buoyancy And Stability
The Stability Of Fully Submerged Bodies
The Stability Of Partially Submerged Bodies
The Determination Of Metacentric Height
Page Comments

Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\inline \tau$ ) = $\inline I\omega\Omega$ (In the limit).

$\inline I$ = Moment of Inertia.
$\inline \omega$ = Angular velocity
$\inline \Omega$ = Angular velocity of precession.

Introduction

In 1628 the Swedish warship Vasa was launched in Stockholm harbour.

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At some time during her construction it had been decided to increase the size and weight of the cannons on the upper gun deck. At the time of her launch she was ballasted but was not fully loaded. A full load would have increased her stability but whether it would have been enough to prevent a catastrophe, is not known.What is known is that she sailed a few yards, heeled over and sank. Sadly the ship builders of the time did not understand the requirements for a stable ship. This page, the first of three, examines these requirements and includes worked examples where the load or ballast is fixed. The two further pages will consider floating bodies with liquid loads or ballast and the period of roll.

Centre Of Buoyancy And Stability

Center of gravity refers to the mean location of the gravitational force acting on a body.

The Buoyancy Force act through the Centre of Gravity of the Displaced Fluid and is called The Centre of Buoyancy

There are three Types of Equilibrium:

Stable. The body returns to it's original position if given a small angular displacement.
Neutral. The body remains in a new position if given a small angular displacement.
Unstable. The body heals further over if given a small angular displacement.

The Stability Of Fully Submerged Bodies

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Let:

$\inline V$ = Volume of Body.
$\inline w$ = Specific weight of the fluid.
$\inline W$ = Mass of the Body.
$\inline G$ is the Centre of Gravity.
$\inline B$ is the Centre of Buoyancy and is the centre of gravity of the displaced liquid.

If:

$\inline B$ and $\inline G$ are coincident then the Body will be in Neutral equilibrium.
$\inline B$ is below $\inline G$ then the Body is in Unstable equilibrium.
$\inline B$ is above $\inline G$ then the body is in Stable equilibrium.

This last case has a righting couple of $\inline W\;B\;G\;\sin \theta$ . Where $\inline \theta$ is the angle of tilt. Note that the Position of $\inline B$ and $\inline G$ relative to the Body does not change when the body is rotated.

The Stability Of Partially Submerged Bodies

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$\inline G$ is the c. of g., $\inline B$ is the c. of b. and the line $\inline O$ $\inline O$ is the original water surface.

After tilting $\inline O'$ $\inline O'$ is the new water line and the angle of Tilt is $\inline \theta$ .

$\inline G$ remains in the same position relative to the ship but the Centre of Buoyancy moves to $\inline B_1$ .

$\inline M$ is the "META CENTRE" and is defined as the point where the vertical through the new Centre of Buoyancy meets the original vertical through the Centre of Gravity after a very small angle of rotation.

$\inline MG$ is called the METACENTRIC HEIGHT.

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Therefore for stable equilibrium for a floating Partially Submerged Body the Meta centre must be above the Centre of Gravity $\inline G$ . If the Metacentric height is zero the Body will be in Neutral equilibrium.

In ship design the choice of the Metacentric height is a compromise between stability and the amount that the ship rolls. In British Dreadnaught Battle ships, for instanace, the metacentric height was so great that they had a tendency to roll badly, even with large bilge keels.

The Righting couple $\inline =\:W\;GM\;\sin \theta$

The Determination Of Metacentric Height

Experimentally

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Let $\inline W$ be the weight of the Boat plus it's Load.

A small load $\inline w$ is moved a distance $\inline x$ and causes a tilt of angle $\inline \theta$ . The Boat is now in a new position of equilibrium with $\inline B'$ and $\inline G'$ lying along the Vertical through $\inline M$ .

The Moment due to the movement of the load is given by:

Moment due to movement of $\inline C$ of $\inline G=$

$wx = W\times GM\;\theta$

$\therefore\;\;\;\;\;\;GM = \frac{w\;x}{W\;\theta }$

Theory

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The Ship tilts from it's old waterline $\inline O$ $\inline O$ to a new waterline $\inline O'$ $\inline O'$ as it moves through an angle $\inline \theta$ . Due to the movement of the wedge of water from $\inline A'$ $\inline A$ $\inline C$ to $\inline D'$ $\inline D$ $\inline C$ , the Centre of Buoyancy moves from $\inline B$ to $\inline B'$ .

The Change in the moment of the buoyancy Force = $\inline wV$ $\inline X$ $\inline B$ $\inline B'$

$= w\;V\times BM\;\theta \;\;\;\;$

where $\inline \theta$ is small

The Volume of the Wedge $\inline A$ $\inline C$ $\inline A'$

$= \frac{1}{2}\times \frac{b}{2}\times \frac{b}{2}\theta \times L = \frac{b^2\,L}{8}\;\theta$

Therefore the Moment of the Couple due to the movement of the wedge $\inline w\times \displaystyle\frac{w b^2L\,\theta }{8}\times \displaystyle\frac{2}{3}b$

$wV\times BM\;\theta = \frac{wb^3\;L}{12}\;\theta$

$\therefore\;\;\;\;\;BM = \frac{b^3\,L}{12\,V} = \frac{I}{V}$

(2)

Where $\inline I$ is the Second Moment of Area of the Water Plane Section and $\inline V$ is the volume of water Displaced.

Thus if the positions of $\inline B$ and $\inline G$ are known or can be calculated , then the distance $\inline GM$ can be determined since:

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$\inline GM = BM + BG = BM + BO - OG$

There are in fact two Metacentric heights of a ship. One for Rolling and the other for Pitching. The former will always be less than the latter and unless otherwise stated, the Metacentric given will be for Rolling.

Example:

[imperial]

Example - Example 1

Problem

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A Pontoon measuring 20ft.by 12ft. and 4ft deep, weighs 12 tons. It carries a load of 8 tons. The Pontoon sits in sea water with a density of 64 lb/cu. ft.

Find it's metacentric height and establish the angular tilt which will result if the load is moved by one ft. sideways.

Workings

Taking Moments about the Base:

$(12\;+\;8)\times OG\;=\;12\times OG_P\;+\;8\times OG_L$

$\therefore\;\;\;\;\;OG\;=\;\frac{12\times 2\;+\;8\times 5}{(12\;+\;8)}\;=\;3.2ft.$

The Volume of water displaced $V$ = $\displaystyle\frac{(12\;+\;8)\times 2240}{64}\;=\;700 ft^3$

The Depth of immersion $h$ = $\displaystyle\frac{700}{20\times 12}\;=\;2.197\; ft.$
The height,

$OB\;=\;\frac{h}{2}\;=\;1.458\;ft.$

And:

$BM\;=\frac{I}{V}\;=\;\frac{L\;b^3}{12\;V}\;=\;\frac{20\times 12^3}{12\times 700}\;=\;4.143\;ft.$

But the Metacentric height

$=\;BM\;+\;OB\;-\;OG\;=$

$=4.143\;+\;1.458\;-\;3.2\;=\;2.401\;ft.$

The Moment due to the Movement of the Load = 8 ft. tons

The Moment due to the movement of the $C$ of $G$ = $W\timesGM\;\theta$ = 20 X 2.40

$\therefore\;\;\;\;\;\theta \;=\;\frac{8}{48}\;=\;\frac{1}{6}\;rds.\;=\;9^0\;31'$

Solution

The metacentric heigh is $\;2.401\;ft.$
The angle is $\theta=\;9^0\;31'$