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The metacentric height and conditions fro buoyancy of ships and ponttons with liquid loads or ballast

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Introduction

Tankers are designed to transport liquid loads but most large vessels have fuel tanks needed for their propulsion.

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The presence of a significant volume of liquid with a free surface has an effect on the metacentric height and hence the stability of the ship. This page should be read in conjunction with "Stability and Metacentric Heights"

The Metacentric Height for a Vessel with Liquid Ballast.

O O is the original Waterline.
G is the Centre of Gravity of the Ship and Ballast.
B is the Centre of Buoyancy.

The Ship is tilted through a small angle $\theta$ Clockwise giving a new waterline of O' O'

Due to the movement of the water wedge the Centre of Buoyancy of the Ship ( i.e. The C of G of the displaced liquid) moves to B'.

By the Previous Theory $BM\;=\;{\frac{I}{V}$

Where I is the second Moment of Area of the Water plane area and V is the Total displacement by the Ship and it's contents.

Due to the movement of the liquid in the Ballast tank the C of G of the Ship and Ballast moves to G'. Now the vertical through G' cuts the old vertical centre line at N. The Stability of the ship depends upon whether N is above or below M.

The Metacentric Height is now MN

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Consider the Liquid in the Ballast Tank.

Let

I' = The second moment of area of the liquid surface about it's centre line.
$=\;\frac{L'\;b'^3}{12}$
(1)
V' = the Volume of liquid in the Tank,
w' = the specific weight of the liquid.
x = The movement of the C of G in the tank due to the tilting of the vessel.

The moment of the Couple due to the movement of the Wedge = The Moment of the Couple due to the movement of the C of G.

$w'v'x = w'\times \frac{1}{2}\times \frac{b'}{2}\times \frac{b'\;\theta }{2}\times L'\times \frac{2\;b'}{3}$

(2)

$\therefore\;\;\;\;\;V'x = \frac{L'\;b'^3\;\theta }{12}$

(3)

$\therefore\;\;\;\;\;x = \frac{I'\;\theta }{V'}$

(4)

Thus the C of G of the ship and Contents moves from G to G' where G G' is given by:-

$W\times GG'\;=\;w'\;V'\times x$

(5)

$or\;\;\;\;\;w\;v\times GG'\;=\;w'\;V'\times x$

(6)

$\therefore\;\;\;\;\; GG'\;=\;\frac{w'\;V'}{w\;V}\times \frac{I}{V'}\times \theta \;=\;\frac{w'I\;\theta }{wV}$

(7)

$But\;\;\;\;\;\;GG'\;=\;GN\times \theta$

(8)

$\therefore GN\;=\;\frac{w'\;I}{w\;V}$

(9)

Therefore the Metacentric Height MN = MB + BO - OG - GN

$=\;\frac{I}{V}\;+\;BO\;-\;OG\;-\;\frac{w' I'}{wV}$

(10)

Thus the Metacentric Height has been reduced by the liquid ballast by an amount $\frac{w'I'}{wV}$

Divided Ballast Tanks.

If the Ballast Tank is divided into two by a longitudinal Partition, the the above equation becomes.

$x\;=\;\frac{w_1\;I_1}{wV}\;\;\;\;and\;\;\;\;y\;=\;\frac{w_2\;I_2}{wV}$

(11)

$wV\times GG_1\;=\;w_1V_1\times x\;+\;w_1V_2\times y$

(12)

$\therefore\;\;\;\;\;GG_1\;=\;\frac{w_1\theta }{wV}\left(I_1\;+\;I_2 \right)$

(13)

$\therefore\;\;\;\;\;GN\;=\;\frac{w_1}{wV}\left(I_1\;+\;I_2 \right)$

(14)

Comparing the two separate cases:-

1) Single Tank of width b.

$I_1\;=\;\frac{L_1\;b_1^3}{12}$

(15)

2) Same tank divided into two equal compartments.

$I_1\;+\;I_2\;=\;\frac{2\times L_1}{12}\times \left(\frac{b_1^3}{2} \right)^3\;=\;\frac{1}{4}\times \frac{L_1b_1^3}{12}$

(16)

This shows that the reduction in the Metacentric height that occurs in case two is a quarter that experienced in case one. For this reason the tanks in Oil Tankers are divided into small compartments.

The Righting Couple with a liquid Ballast.

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The Righting Couple due to the Buoyancy Force acting upwards through $B_1\;and M$ and the weight of the ship acting downwards through $G_1\;and\;N$

$=\;W\times MN\times \theta \;\;\;\;\;where\;\theta\;is\;small$

(17)

This is the same as in the simple case without a liquid balance but it must be remembered that the Metacentric Height is now MN and NOT MG

Worked Examples

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Example 1

A Pontoon 50 ft. by 20 ft. and 7 ft. deep is ballasted with 40 tons of water. The Pontoon and its load weigh 80 tons.

Find (a)The metacentric height and (b) the angle through which the pontoon will heel if 2 tons of deck cargo are moved 10 ft. from the centre to edge.

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Example 2

A rectangular pontoon 20 ft. wide ,50 ft. long and 7 ft. deep weighs 80 tons when loaded but without ballast water. A vertical diaphragm divides the pontoon longitudinally into two compartments each 10 ft. wide and 50 ft. long. 20 tons of ballast water are admitted to the bottom of each compartment, the water surface being free to move.

The centre of gravity of the pontoon, without ballast water, is 5 ft. above the bottom and on the geometrical centre of the plan .

(a) Calculate the metacentric height for rolling.

(b) If two tons of deck cargo is shifted 10 ft. laterally, find the appropriate angle of heel.

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Example 3

Define the terms "Metacentre" and "Centre of Buoyancy".

A vessel carries a tank of oil weighing $S\;lb./ft.^3$ amidships. Prove that the effect of the fluidity of the oil on the rolling stability of the vessel is equivalent to a decrease in the value of the metacentric height by an amount:-