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

HNC NAUTICAL SCIENCE

Group Award Code: G8F5 15

Unit Code: F0LD 34 Ship Stability

Outcome 3 – Statical Stability

3.5 Free Surface Effect

& Correction of and

Aims

To give the student an understanding of:



the creation of Free Surface Effect (FSE);

Aims

To give the student an understanding of:



how FSE can be reduced/eliminated;

Objectives

The Student will be able to:



Describe with the aid of sketches the

effects of FSE in part filled compartments

containing fluids;



calculate the reduction in GM caused by

FSE, as an adjustment to KG/GM (Single

Weights), or by the inclusion of FSM‟s into

the Moment about the Keel table (Multiple

Weights);

Objectives

The Student will be able to:



Calculate the new FSE if a compartment is

subdivided.



Describe the correct procedures required

to carry out the correction of and angle of

loll without putting the vessel in further

Free Surface Effect

Showing the

vessel at rest

with a part

filled undivided

double bottom

tank. The GM

shown is the

GM

, all of

which are on

the centreline

of the vessel.

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

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

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)

Calculating FSE



The stability information required by law to be

supplied to a vessel must include information on

the effect of free surface of liquid in the tanks

and also how to correct the GM for this effect.



Information is usually supplied for each tank in

the form of "Free Surface Moments".



FSE = Free surface moment

or

FSM

Calculating FSE



If there are several Free Surface Moments

involved, then they should all be added,

then divided by the displacement.



FSE = Σ Free surface moments

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.

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 GM_SOLID 0.31 m

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

Subdivisions



FSM and therefore FSE can be reduced to the

fitting of equally spaced longitudinal divisions in

the tank.



To Calculate the subdivided value the FSM or

FSE is divided by the new number of

compartments (n) squared



So:



FSM

= FSM

or

FSE

= FSE

Example 1

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

(i) A single longitudinal bulkhead,

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

Example 2

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

(i) A single longitudinal bulkhead,

(ii) A further two longitudinal bulkheads.

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

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

Subdivisions

FSE

= FSE

Can be combined with

n

FSE = FSM

or

ƩFSM

Δ

Δ

To give

FSE

= FSM

or

ƩFSM

(Δ x n

)

(Δ x n

)

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.

Example 3

(i) The tank is undivided

FSE = FSM = 2880 = 0.36m Δ 8000

Example 3

(i) The tank is undivided

FSE = FSM = 2880 = 0.36m Δ 8000

(ii) There is a centreline division

FSE_SUB = FSM = 2880 = 0.09 tm (Δ x n2_{) (8000 x 4}2₎

2 1

Example 3

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

FSE_SUB = FSM = 2880 = 0.023 tm (Δ x n2_{) (8000 x 4}2₎ 4 3 2 1

FSE and the Angle of Loll



FSE causes a virtual rise in G



If the vessel is tender she will have a small

GM



If the FSE is greater than the GM

then the

vessel will have a negative GM

and will be

in unstable equilibrium.



An unstable vessel could capsize, but more

FSE and the Angle of Loll



The best way to avoid this is to keep the number

of slack tanks to a minimum during the voyage.

Wherever possible tanks should be either empty

or pressed up.



Whilst the vessel is on passage she will use

FW, DO & FO, so some slack tanks cannot be

avoided.



To avoid an angle of loll due to FSE the vessel‟s

GM

must be large enough to withstand any

anticipated rise in G during the voyage.

List vs Angle of Loll

Angle of List

1.

+ve GM

2.

Stable Equilibrium.

G off the Centreline.

Corrected by moving

G back to the

Centreline – by

moving/loading

weights towards the

“high side”.

Angle of Loll

-ve GM

Unstable Equilibrium.

G on the Centreline.

Corrected by

lowering G below M

Correcting an angle of Loll



Lowering G below M to make the vessel stable

will correct an angle of Loll.



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.



Filling an empty tank will introduce FSE causing

a further virtual rise of G, so this must be done

with caution and adopting the following

Correcting an angle of Loll

1.

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.

Correcting an angle of Loll

4. Start with the smallest tank on the LOW side first.

(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.

Correcting an angle of Loll

5.

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

FSM and Moments about the

Keel



Since Moments about the Keel and Free Surface

Moments are both Vertical Moments, they can

be combined into the same table to calculate

KG.



The KG calculated automatically will be the

KG



The FSM can just be added to the Loaded

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 GM_f 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

Example 1

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

Loaded Discharged Loaded Discharged

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

Example 1

KG_f = Ʃ Moments about the Keel = 237 942.82

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