# Ship Stability, Statical Stability, Free Surface Effect ,Correction of and Angle of Loll.

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### HNC NAUTICAL SCIENCE

Group Award Code: G8F5 15

## & Correction of and

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SOLID

K B G W L M

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### Free Surface Effect

When the vessel in stable equilibrium is inclined by an external force, buoyancy is lost on the raised side and an equal amount

created on the submerged side.

This creates a shift of buoyancy from b to b1 in the vessel, moving the overall buoyancy of the vessel along a parallel line from B to B1.

This creates a righting lever of GZ. K B G W L M W1 L1 b b1 B1 Z Δ Δ

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### Free Surface Effect

As the ballast moves to the low side this causes a shift of weight of g to g1

This causes a shift of the overall centre of gravity of the vessel G along a parallel

line to a new position of G1.

This reduces the

righting lever to G1Z1. K B G W L M W1 L1 B1 Z g g1 G1 Z1

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### Free Surface Effect

If a perpendicular line is drawn upwards through G1 to the centreline of the

vessel, the GZ can be redrawn between the

centreline and the BM line. This gives G2Z2 which is equal to G1Z1.

The distance along the centreline measured

between G and G2 is the “virtual loss of GM”.

This is also known as the Free Surface Effect (FSE).

M G2 Z2 G1 G Z Z1 θ° Vir tua l Lo ss o f G M ( FS E)

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### Key Points

 FSE does not depend upon the weight of liquid in the

tank, providing the area of the free surface remains unchanged.

 FSE does not depend upon the position of the tank

within the ship.

 FSE is zero if a tank is full or empty

 Every slack tank contributes it‟s own FSE to the total

FSE for the ship therefore to reduce FSE keep the number of slack tanks to a minimum.

 If it is decided to improve stability by filling a DB tank

then FSE will worsen the situation before the increased bottom weight is sufficient to bring G down. If at an angle of loll then fill the smallest tank, on the lowest side first.

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### Execise 1

A vessel has a KM of 5.13m, KG = 4.82m and the FSE = 0.11m. Calculate the effective (fluid) GM.

KM 5.13 m

KG - 4.82 m GMSOLID 0.31 m

FSE - 0.11 m (FSE is always negative) GMFLUID 0.20 m GMFLUID is the effective GM

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SUB

SUB

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### Example 1

A tank has a FSM of 3586 tm. Calculate the FSM if the tank is fitted with:

(ii) A further two longitudinal bulkheads. (i) FSMSUB = FSM = 3586 = 396.50 tm n2 22 (ii) FSMSUB = FSM = 3586 = 224.13 tm n2 42 1 2 4 3 2 1

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### Example 2

A tank has a FSE of 0.26 m. Calculate the FSE if the tank is fitted with:

(ii) A further two longitudinal bulkheads.

(i) FSESUB = FSE = 0.26 = 0.065m n2 22

(ii) FSESUB = FSE = 0.26 = 0.016 tm n2 42 1 2 4 3 2 1

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SUB

2

SUB

2

2

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### Example 3

A vessel displacing 8000 tonne has a DB tank half full, it has a free surface moment 2880 tm. Calculate the free surface effect if:-

i) the tank is undivided

ii) there is a centreline division

iii) there is a centreline division and two equally spaced longitudinal bulkheads.

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### Example 3

(i) The tank is undivided

FSE = FSM = 2880 = 0.36m Δ 8000

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### Example 3

(i) The tank is undivided

FSE = FSM = 2880 = 0.36m Δ 8000

(ii) There is a centreline division

FSESUB = FSM = 2880 = 0.09 tm (Δ x n2) (8000 x 42)

2 1

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### Example 3

(iii) There is a centreline division and two equally spaced longitudinal wash bulkheads

FSESUB = FSM = 2880 = 0.023 tm (Δ x n2) (8000 x 42) 4 3 2 1

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SOLID.

SOLID

FLUID

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SOLID

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### This can be achieved by:

 Moving cargo to a lower position;

 Jettisoning top-weight (in an emergency);

 Reducing FSE by pressing up/emptying tanks;  Filling low ballast spaces such as DB tanks.

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### Correcting an angle of Loll

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top up tanks that are already slack.

2. calculate the FSE which will arise before pumping into empty tanks. This will ensure that the rise of G during the operation is acceptable.

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### Correcting an angle of Loll

(If a tank on the high side is filled first, the ship will start to right herself but will then tend to roll over suddenly in an uncontrolled fashion as she passes through the

upright. She will then „whip‟ through to a larger angle of loll on the other side. She may even capsize if the

momentum gathered is sufficient.)

When the low side is filled first, the angle of list will increase initially, but in a slow and controlled fashion. After some time, the weight of the ballast water added will be sufficient to lower the ship‟s COG (despite the extra FSE), to cause the angle of list to decrease. By this method the inclining motions of the v/l take place in a gradual and controlled manner.

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### Correcting an angle of Loll

5.

now fill the opposite tank on the high side.

6. fill tanks alternately, low side first, until the v/l returns to positive GM.

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FLUID

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### Example 1

A vessel of  = 17,922 tonnes is initially upright, KG = 12.66m, KM = 14.24m. The FSM‟s of the various tanks add up to 1225tm. Calculate the GMf after the following cargo operations if KM is constant.

Weight (t) Kg (m) Discharge: 624 14.88 1,296 8.71 Load: 3,042 6.69 312 13.27 397 14.88

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### Example 1

Weight (t) KG (m) Moment about the Keel (tm)

17 922 12.66 226 892.52 624 14.88 9 285.12 1 296 8.71 11 288.16 3 042 6.69 20 350.98 312 13.27 4 140.24 397 14.88 5 907.36 FSM 1225.00 21 673 - 1 920 1 920 258 516.10 - 20 573.28 20 573.28 Ʃ19 753 Ʃ237 942.82

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### Example 1

KGf = Ʃ Moments about the Keel = 237 942.82

Ʃ Weights 19 753.00 KGf = 12.045m KM 14 240 m KGf - 12 045 m GMf 2.195 m The Final GMf is 2.20m

References

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