**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

_{SOLID}

### , all of

### which are on

### the centreline

### of the vessel.

**K**

**B**

**G**

**W**

**L**

**M**

**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 b_{1} in
the vessel, moving the
overall buoyancy of the
vessel along a parallel
line from B to B_{1}.

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 g_{1 }

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

line to a new position
of G_{1}.

This reduces the

righting lever to G_{1}Z_{1}.
**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 G_{1}
to the centreline of the

vessel, the GZ can be redrawn between the

centreline and the BM line.
This gives G_{2}Z_{2} which is
equal to G_{1}Z_{1}.

The distance along the centreline measured

between G and G_{2} is the
“virtual loss of GM”.

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

**M**
**G2** **Z2**
**G _{1}**

**G**

_{Z}**Z**

_{1}**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

_{SUB }

### = FSM

### or

### FSE

_{SUB }

### = 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_{ 2}2
(ii) FSM_{SUB }**= FSM = 3586 = 224.13 tm**
n2_{ 4}2
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_{ 2}2

(ii) FSE_{SUB }**= FSE = 0.26 = 0.016 tm **
n2_{ 4}2
1 2
4
3
2
1

**Subdivisions **

### FSE

_{SUB }

### = FSE

### Can be combined with

### n

2### FSE = FSM

### or

### ƩFSM

### Δ

### Δ

### To give

### FSE

_{SUB }

### = FSM

### or

### ƩFSM

### (Δ x n

2### )

### (Δ x n

2### )

**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

_{SOLID. }

### If the FSE is greater than the GM

_{SOLID}

### then the

### vessel will have a negative GM

_{FLUID}

### 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

_{SOLID }

### 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.

3.### G off the Centreline.

4.### Corrected by moving

### G back to the

### Centreline – by

### moving/loading

### weights towards the

### “high side”.

### Angle of Loll

1.### -ve GM

2.### Unstable Equilibrium.

3.### G on the Centreline.

4.### 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.

### 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.

**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.

### now fill the opposite tank on the high side.

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

_{FLUID}

### 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 **