### Transport Publication TP 14609 E

## 3.1. Density, specific gravity, mass and weight

The notions of density and specific gravity (or relative density) are very useful for finding the mass of a known volume of solid or liquid. They are also useful for finding the volume occupied by a given mass of liquid or solid.

### 3.1.1. Density

**Density** is the mass of a substance (expressed in
kg
) per unit of volume. The standardized unit of volume is the cubic metre (
m^{3}
). The density unit is therefore
kg
/
m^{3}
.

Density is a property that is unique to each type of matter. Liquids, solids and gases all have their own density.

A few examples of densities:

- Pure water has a density of 1,000
kg
/
m
^{3}. This means that one cubic metre of pure water has a mass of 1,000 kg . - Saltwater has a mean density considered to be 1,025
kg
/
m
^{3}. - Steel has a density (depending on its composition) of about 7,430
kg
/
m
^{3}. If one metric tonne represents 1,000 kg , then 1 cubic metre of steel has a mass of 7.43 metric tonnes. - The density of wood turpentine can be 650
kg
/
m
^{3}.

Bodies with a density lower than that of water will float, whereas the others will sink.

Density is expressed by the Greek letter ρ (rho).

### 3.1.2. Specific gravity

Specific gravity is an expression that simplifies the use of density. Specific gravity is defined as the ratio of the density of a substance to the density of pure water (1,000
kg
/
m^{3}
).

The specific gravity of a substance is always the ρ of the substance divided by 1,000. Note that no units follow the transformation of density into specific gravity. The value of specific gravity is shown without units.

A few examples of the specific gravities of liquids:

- Pure water = 1
- Lubricating oil = 0.9 (can vary)
- Fuel = 0.72 to 0.75
- Average fuel = 0.86 to 0.92
- Heavy fuel = 0.92 to 0.95
- Fuel oil = 0.95 to 0.99

A few examples of the specific gravities of solids:

- Coal = 1.3
- Gypsum (ground) = 1.5
- Cement = 1.5
- Wheat = 0.77
- Copper = 8.93
- Lead = 11.34
- Mercury = 13.59
- Steel = 7.5 (variable depending on its composition)

### 3.1.3. Mass and weight

**Mass** and **weight**: The term "mass" (with its unit, the kilogram) is used regularly right from the start of this text. A distinction must be made between the mass and the weight of a body.

The **mass** of a body or a substance is a value that represents the quantity of matter of the body or substance. The mass will not change if the body is placed on the moon, the top of a mountain or at sea level.

The force exerted by this mass when it is subject to the Earth's gravitational attraction is the **weight** of the body or substance. Since gravitational attraction is variable (it is lower at higher altitudes than at sea level, the Moon's gravitational attraction is weaker than the Earth's,
etc.
), the weight of a body or substance will vary according to its location, but its mass will always remain the same.

In the International System of Units ( SI ), the force exerted on a mass that is subject to the Earth's gravitational attraction is expressed in a Newtons. The weight of a body or substance should thus be expressed in Newtons, which is rarely the case in practice. However, in theory, this distinction is very important.

The value of the weight of a body or substance can be found using the formula

**
F
=
m
×
a
**

The force (in Newtons) will be produced by a mass (in kilograms) that is subject to the acceleration created by the Earth's gravitational attraction (in
m/s^{2}
), which is an average value of 9.81
m/s^{2}
at sea level.

Newton =
kg
× (
m/s^{2}
)

Mass and weight

F
=
m
×
a

= 10 × 9.81

= 98.1
N

### 3.1.4. Solved problems

#### 3.1.4.a. Example 1

Find the mass and weight of a block of wood with a density of 720
kg
/
m^{3}
and with the following dimensions: length of 25
cm
, width of 10
cm
, and height of 5
cm
.

- First find the volume of the block of wood in
m
^{3}:- Volume of a rectangular block = length × width × height
- Volume of the block = 0.25
m
× 0.10
m
× 0.05
m
= 0.00125
m
^{3}

- If wood has a density of 720
kg
per
m
^{3}and we have 0.00125 m^{3}, then the mass- = 720
kg
/
m
^{3}× 0.00125 m^{3}= 0.9 kg

- = 720
kg
/
m
- The mass of the block of wood is 0.9 kg
- The weight of the block is:
F
= 0.9
kg
× 9.81
m/s
^{2}= 8.83 Newtons

#### 3.1.4.b. Example 2

A block of cast iron with a mass of 30
kg
has the following dimensions:

20
cm
× 20
cm
× 10
cm
. What is the density of the cast iron?

- Find the volume of the block of cast iron in
m
^{3}:

Volume of the block = 0.20 m × 0.20 m × 0.10 m = 0.004 m^{3} - We know that 0.004
m
^{3}of cast iron has a mass of 30 kg . What is the mass of 1 m^{3}? Using the rule of three:

(30 kg ÷ 0.004 m^{3}) × 1 m^{3}= 7,500 kg - 1
m
^{3}of cast iron has a mass of 7,500 kg , therefore the ρ of the cast iron = 7,500 kg / m^{3}

#### 3.1.4.c. Example 3

A rectangular double bottom tank 20 m long, 10 m wide and 1 m high is filled with fuel with a specific gravity of 0.95. What is the mass of the fuel in the tank?

- Find the volume of fuel in the tank:

20 m × 10 m × 1 m = 200 m^{3}of fuel in the tank - The fuel's specific gravity of 0.95 means that it has a density of 950
kg
/
m
^{3} - There are 200
m
^{3}of fuel, therefore the mass of the fuel

= 200 m^{3}× 950 ( kg / m^{3}) = 190,000 kg = 190 tonnes

One cubic metre of liquid always contains 1,000 litres, regardless of the type of liquid. We can therefore state that the volume of 200
m^{3}
in the tank corresponds to 200,000 litres.

#### 3.1.4.d. Example 4

A barrel 80 cm in diameter and 1.2 m high is three quarters full of lubricating oil with a specific gravity of 0.88. What is the mass of the oil in the barrel?

- Find the volume of the barrel:

Volume of a cylinder = base area × height ( m^{2}× m = m^{3})

= (π D^{2}÷ 4) × H = (π × 0.8^{2}÷ 4) × 1.2 = 0.603 m^{3} - Find the volume corresponding to three quarters of the barrel

0.603 m^{3}× ¾ = 0.452 m^{3} - Alternately, three quarters of the volume of the barrel also represent a height of 1.2
m
× ¾ = 0.9
m

The volume for three quarters of the barrel = (π × 0.8^{2}÷ 4) × 0.9 = 0.452 m^{3} - Find the mass of the oil corresponding to 0.452
m
^{3}:

1 m^{3}of oil has a mass of 880 kg / m^{3}, therefore 0.452 m^{3}will have a mass of 880 kg / m^{3}× 0.452 m^{3}= 397.75 kg

#### 3.1.4.e. Example

A ship loads 150 tonnes of fuel (density = 0.987) that must be stored in a rectangular tank with the following dimensions:

Length of 17.5 m , width of 7 m and height of 1.25 m . What will the fuel level be in the tank?

- Find the volume filled by 150 tonnes of fuel:

Fuel has a density of 987 kg / m^{3}. We have 150,000 kg .

Using the rule of three: (1 m^{3}÷ 987) × 150,000 = 152 m^{3}

150 tonnes of fuel therefore fill a volume of 152 m^{3} - These 152
m
^{3}fill the same volume in the tank.

152 m^{3}= length × width × height of the liquid

152 m^{3}= 17.5 m × 5 m × height of the liquid in metres

Height of the liquid in the tank = 152 ÷ (17.5 × 7) = 1.24 m

### 3.1.5. Problems to solve

#### 3.1.5.a. Problem 1

Find the mass and weight of a block of copper (
ρ
= 8,930
kg
/
m^{3}
) that is 50
cm
× 50
cm
× 25
cm
.

Answer: 558 kg and 5,475 N

#### 3.1.5.b. Problem 2

Find the density ( ρ ) of a block of alloy steel that has a mass of 50 kg . The dimensions of the block are 16 cm × 16 cm × 30 cm .

Answer: 6,510.5
kg
/
m^{3}

#### 3.1.5.c. Problem 3

Lubricating oil is stored in a tank that measures 1.5 m × 1.5 m and the level of the oil in the tank is 2.25 m . If the specific gravity of the oil is 0.91, what is the mass of the oil in the tank?

Answer: 4,607 kg or 4.6 tonnes

#### 3.1.5.d. Problem 4

110 tonnes of fuel are to be transferred into an empty double bottom tank that is 10
m
long and 8
m
wide. If the density of the transferred fuel is 985
kg
/
m^{3}
, what will the fuel level be in the tank after the transfer?

Answer: 1.39 m

## 3.2. Buoyancy and flotation

### 3.2.1. Definitions

**Displacement**: Mass of the volume of water that a ship displaces. This mass is equal to the ship's mass. Displacement is expressed in tonnes. Symbol: Δ**Displacement volume or underwater volume**: Volume of the underwater part of a ship. It is expressed in m^{3}. Symbol:**Draft**: Depth of the underwater part of a ship. There is forward draft, aft draft and mean draft. It is expressed in metres or centimetres. Symbol: d**Deadweight**: The mass that a ship can carry. This mass represents the cargo, fuel, water and everything required for proper operation of the ship. Specifically,**cargo deadweight**represents the mass of the cargo that can be loaded.**Lightship displacement**: Mass of a ship in light condition.**Loaded displacement**: Mass of a fully loaded ship ready for sea. Loaded displacement equals lightship displacement plus deadweight.**Waterplane area**: Area at the intersection of the surface of the water and the waterline of a ship. It can vary according to the ship's draft. Symbol: A_{w}.**Amidships**: Amidships is the midship section of a ship taken at its widest breadth. This is the reference for transverse stability calculations. It also allows you to visualize the transverse structural members of the hull.

**Lightship weight**

Real weight of an empty ship

**Deadweight** is the total mass of goods that a ship can carry at its maximum permissible draft (including fuel, fresh water, gear, provisions,
etc.
)

**Loaded displacement**

Lightship + deadweight = loaded displacement

### 3.2.2. Archimedes' principle

Any body immersed in a liquid is subject to an upward vertical force equal to the weight of the displaced mass of water.

Experiment 1

**Situation 1**

Vessels 1 and 2 are identical and empty.

The weight W is in the air.

The scale is balanced with the weights placed in the right pan.

The tank is filled to the rim with liquid.

**Situation 2**

The weight is immersed.

The liquid that spilled from the tank is in vessel 2.

The scale is unbalanced.

The weight thus appears to be buoyed up by vertical thrust.

**Situation 3**

The weight is still immersed.

Vessels 1 and 2 have been interchanged. The balance has been re-established.

Therefore, the buoyancy that caused the unbalance was equal to the weight of the volume of water displaced.

### Conclusion

If we generalize, we can say that any solid object immersed in a fluid (liquid or gas) is subject to an upward vertical buoyant force equal to the weight of the volume of fluid displaced. This is called Archimedes' principle, which he came up with around 250 BC . Legend has it that he was taking a bath at the time of his discovery and ran into the street shouting EUREKA.

Archimedes' discovery allows us to understand why a boat floats, even if it sometimes weighs an enormous amount. Let's look at what happens when we gradually immerse a vat carrying a total weight of 100 kN in a tank.

**Experiment 2**

**Situation 1**

The 100 kN vat is suspended over the water.

The dynamometer shows that F = 100 kN to hold it in this position.

**Situation 2**

The vat is partially immersed in the tank.

The weight is still 100 kN .

The dynamometer shows that F = 75 kN .

Buoyancy B from the water is 25 kN .

**Situation 3**

The vat is immersed deeper in the tank.

The weight is still 100 kN .

The dynamometer shows that F = 25 kN .

Buoyancy B from the water is 75 kN .

**Situation 4**

The vat is floating.

The weight is still 100 kN .

The dynamometer shows that F = 0 kN .

Buoyancy from the water is 100 kN .

With respect to stability, Archimedes' principle can be expressed as: the mass of a body immersed in a fluid is equal to the mass of the volume of fluid displaced.

When a ship is floating freely at rest, the mass of the ship (displacement,
Δ
) is equal to the mass of the volume of water displaced by the ship.

Δ =

×
ρ

kg
=
m^{3}
×
kg
/
m^{3}

OR

Δ
=

× specific gravity

t
=
m^{3}
×
t
/
m^{3}

#### 3.2.2.a. Solved problems

**3.2.2.a.(i) Example 1**

A block of wood measuring 1
m
× 1
m
× 1
m
is floating in fresh water (density = 1,000
kg
/
m^{3}
) with a draft of 0.5
m
. What is the mass of the block of wood?

- Find the underwater volume of the wood:

Underwater volume = 1 m × 1 m × 0.5 m = 0.5 m^{3}of immersed wood - Find the volume of water displaced:

The underwater volume of the wood is equal to the volume of water displaced = 0.5 m^{3}. - Find the mass of the volume of water displaced:

Volume of water displaced = 0.5 m^{3}; if 1 m^{3}of pure water has a mass of 1,000 kg , then a volume of 0.5 m^{3}has a mass of 500 kg .

The mass of the volume of water displaced is 500 kg . - Find the mass of the block of wood:

The mass of the floating object = mass of water displaced = 500 kg = 0.5 tonne.

Δ = × ρ

Δ = 0.5 m^{3}× 1,000 kg / m^{3}= 500 kg

**3.2.2.a.(ii) Example 2**

A rectangular barge is floating in salt water (
ρ
= 1,025
kg
/
m^{3}
) and has a draft of 2.5
m
. The length and breadth of the barge are 75
m
and 20
m
, respectively. Find the barge's displacement.

- Find the underwater volume of the barge:

Underwater volume = 75 m × 20 m × 2.5 m = 3,750 m^{3} - Find the volume of water displaced:

The volume of water displaced is equal to the underwater volume of the barge = 3,750 m^{3} - Find the mass of the volume of water displaced:

3,750 m^{3}of salt water (1 m^{3}of salt water has a mass of 1,025 kg )

3,750 m^{3}× 1,025 kg / m^{3}= 3,843,750 kg = 3,843.75 tonnes - Find the mass of the barge:

The mass of the floating object = mass of water displaced = 3,843.75 tonnes

Δ = × ρ

Δ = 3,750 m^{3}× 1,025 kg / m^{3}= 3,843.75 tonnes

**3.2.2.a.(iii) Problem 3**

A barrel is 1 m in diameter and 1.5 m in height. Its mass when empty is 15 kg . It is filled with oil with a specific gravity of 0.864. What is the immersed height of the barrel when it is floating upright in water with a specific gravity of 1.015?

- Find the total volume of the barrel.

Volume of the barrel = (π×1^{2}) ÷ 4 × 1.5 = 1.178 m^{3} - Find the volume and mass of the oil in the barrel:

Volume of oil = volume of the barrel = 1.178 m^{3}

If the oil has a mass of 864 kg for 1 m^{3}, then for a volume of 1.178 m^{3}

1.178 m^{3}× 864 kg / m^{3}= 1,017.8 kg = 1.0178 tonne - Find the total mass of the barrel:

Total mass of the barrel = mass of the empty barrel + mass of the oil = 15 + 1,017.8 = 1,032.8 kg = 1.0328 tonne - The volume of water displaced = underwater volume of the barrel = 1.0175
m
^{3} - The barrel floats upright and displaces 1.0175
m
^{3}of water, therefore the underwater volume of the barrel is 1.0175 m^{3}. - Find the immersed height of the barrel:

Underwater volume = (π× D^{2}) ÷ 4 × height immersed

1.0175 m^{3}= (π×1^{2}) ÷ 4 × immersed height

Immersed height = 1.295 m

**3.2.2.a.(iv) Problem 4**

What is the weight (in Newtons) of a cast iron cube (
ρ
= 6,700
kg
/
m^{3}
) measuring 80
cm
along each edge when it is completely immersed in water with a specific gravity of 1.010?

- Find the volume and mass of the cube:

Volume = 0.8 m × 0.8 m × 0.8 m = 0.512 m^{3}

Mass = Volume of the block × ρ of the cast iron

0.512 m^{3}× 6,700 kg / m^{3}= 3,430.4 kg = 3.43 tonnes

The block of cast iron therefore has a mass of 3.43 tonnes - Find the mass of the water displaced by the block of cast iron:

Volume of water displaced = 0.512 m^{3}

Mass of water displaced = 0.512 m^{3}× 1,010 kg / m^{3}= 517.12 kg

This mass of water displaced creates upward buoyancy on the block of cast iron equivalent to 517.12 kg (according to Archimedes' law) - The cast iron cube has an equivalent mass of 3,430 kg – 517.12 kg = 2,912.88 kg
- The weight of the block of cast iron is found using
F
=
m
×
a

F (Newton) = 2,912.88 kg × 9.81 m/s^{2}= 28,575.3 Newtons =**28.575 kN**

### 3.2.3. Coefficients of form

#### 3.2.3.a. Coefficient of fineness

**Coefficient of fineness of the waterplane area** (
C_{w}
): This coefficient (variable according to the ship's draft) can be expressed as the ratio of the waterplane area to the area of a rectangle having the same length and breadth.

Waterplane area (
A_{W}
) =
L
×
B
×
C_{w}

C_{w}
= mean value of 0.6 to 0.8

Coefficient of fineness =

C_{w}
= Waterplane area (
A_{W}
) ÷ (
L
×
B
)

Coefficient of fineness of the waterplane area

#### 3.2.3.b. Block coefficient

**Block coefficient** (
C_{b}
): Coefficient (variable according to the ship's draft) that represents the ratio of the underwater volume of a ship to a rectangular block having the same length, breadth and depth.

Block coefficient =

C_{b}
=

÷ ( L × B × d )

Mean of
C_{b}
= 0.75

Fast ships = 0.50

Slow ships = 0.80

**Block coefficient**

#### 3.2.3.c. Solved problems

**3.2.3.c.(i) Example 1**

Find the waterplane area of a rectangular barge with a length of 50 m and breadth of 18 m .

A_{w}
= 50
m
× 18
m
= 900
m^{2}

Therefore,
C_{w}
= 900
m^{2}
÷ (50
m
× 18
m
) = 1

3.2.3.c.(ii) Example 2

A ship has a length of 100
m
and breadth of 20
m
. The coefficient of fineness of the waterplane area (
C_{w}
) is 0.85. Find the ship's waterplane area.

C_{w}
= Waterplane area ÷ (
L
×
B
) = Waterplane area ÷ (100
m
× 20
m
) = 0.85

Waterplane area = 0.85 × 100
m
× 20
m
= 1,700
m^{2}

3.2.3.c.(iii) Example 3

A ship with a length of 200 m and breadth of 22 m has a draft of 8 m. If the ship's block coefficient at this draft is 0.75, find the underwater volume of the hull (

).

C_{b}
=

÷ ( L × B × d ) = 0.75 =

÷ (200 × 22 × 8)

= 0.75 × (200 × 22 × 8) = 26,400
m^{3}

**3.2.3.c.(iv) Example 4**

What is the block coefficient (
C_{b}
) of a hull having a length of 150
m
, breadth of 20
m
and a displacement volume of 16,170
m^{3}
at a draft of 7
m
?

C_{b}
= 16,170
m^{3}
÷ (150
m
×20
m
×7
m
) = 0.77

### 3.2.4. Tonnes per centimetre

Tonnes per centimetre (
TPC
): This is the mass required to increase or decrease a ship's mean draft by 1
cm
. This value varies only according to the waterplane area (
A_{w}
), and the waterplane area can vary according to the ship's draft. Therefore, the
TPC
can vary according to the ship's draft.

TPC
= Tonnes per centimetre immersion

TPI
= Tonnes per inch immersion

#### 3.2.4.a. Solved problems

**3.2.4.a.(i) Example 1**

A rectangular barge having a length of 20
m
and breadth of 12
m
is floating in salt water (
ρ
= 1,025
kg
/
m^{3}
).

- What is its waterplane area?
- What is its TPC ?

- Find the waterplane area:

A_{w}= (length × breadth) at the waterline = 20 m × 12 m = 240 m^{2} - Find the
TPC
:

TPC = ( A_{w}× ρ ) ÷ 100 = (240 m^{2}× 1.025) ÷ 100 = 2.46 tonnes

This means that loading 2.46 tonnes on the ship will increase the draft by 1 cm . You can also say that unloading 2.46 tonnes from the ship will decrease the draft by 1 cm .

### 3.2.5. Effects on a ship's draft of changes in the specific gravity of water

If the specific gravity of the water in which the ship is floating changes without any changes to the ship's displacement, the ship's draft will change. The draft will change because the ship must displace the same mass of water, which no longer has the same density.

Δ
=

× ρ

If the displacement ( Δ ) remains constant and the density ( ρ ) changes, the underwater volume (

) must change; therefore, the ship's draft will change automatically.

If a ship goes from fresh water to salt water, buoyancy will increase and the draft will decrease. Inversely, a ship that goes from salt water to fresh water will see its draft increase.

A ship that loads to its marks (summer load lines S ) in salt water, will see its draft increase as it goes into fresh water and will exceed its marks. This situation is accepted because the absence of heavy weather on bodies of fresh water compensates for the ship's increased draft. This situation has even been made official by adding an additional mark ( F ) to the load lines.

#### 3.2.5.a. Fresh Water Allowance

FWA : Inversely, a ship that loads in fresh water can load up to its " F " line, so that when it is in salt water it will float at its regular marks.

This allowed increase in draft is called the "Fresh Water Allowance".

The FWA is therefore the change in draft when a ship goes from salt water to fresh water.

**Fresh Water Allowance**

### 3.2.6. Problems to solve

#### 3.2.6.a. Problem 1

A cube of wood with a specific gravity of 0.82 has sides of 35 cm . If a 3 kg mass is placed on the block of wood, what is its draft in salt water (specific gravity = 1.025)?

Answer: 28 cm

#### 3.2.6.b. Problem 2

A rectangular barge with a length of 18
m
, breadth of 5
m
and height of 2.25
m
is floating with a draft of 1.5
m
in water with a density of 1,013
kg
/
m^{3}
. Find the vessel’s displacement in tonnes.

Answer: 136.75 tonnes

#### 3.2.6.c. Problem 3

An empty barrel 1 m long and 50 cm in diameter has a mass of 20 kg . What mass should be placed in the barrel so that it floats in fresh water with half of its volume immersed?

Answer: 78.17 kg

#### 3.2.6.d. Problem 4

A rectangular wood beam 3 m long, 40 cm wide and 30 cm high is floating in fresh water with a draft of 20 cm . Find the beam's mass and density.

Answer: 240
kg
; 666.6
kg
/
m^{3}

#### 3.2.6.e. Problem 5

A ship has a length of 130
m
and breadth of 21
m
, with a coefficient of fineness of the waterplane area (
C_{w}
) of 0.85. Find the
TPC
in salt water at this draft.

Answer: 23.78 tonnes

#### 3.2.6.f. Problem 6

A ship with a length of 65
m
and a breadth of 10
m
is floating with a draft of 4.5
m
in salt water. The ship has a block coefficient of fineness of displacement (
C_{b}
) of 0.75. Find the ship's displacement at this draft.

Answer: 2,248.6 tonnes

## 3.3. Transverse statical stability

Transverse statical stability is the ship's stability at small angles of inclination. The rules of statical stability are considered to apply to angles of inclination less than 15°.

To be able to calculate stability, first the fixed and mobile points on a ship's cross section must be identified.

### 3.3.1. Reference point ( K )

**Reference point
K
** is applied to the ship's lowest point, which is the keel. This is a fixed point.

### 3.3.2. Centre of gravity ( G )

Transverse centre of gravity G

Before applying this concept to a ship, a few short definitions of a general nature are necessary.

The centre of gravity of a body can be defined as:

- The point at which the force of gravity is exerted vertically downwards;
- The point where a pivot can be placed that will keep the body balanced;
- The geometric centre of a uniform body.

For a ship, the centre of gravity is the point at which the force generated by the ship's mass is exerted vertically downwards. The position of the centre of gravity changes according to the ship's loading conditions.

#### 3.3.2.a. Effects of changing load

Effects of changing load: The height of the centre of gravity depends on the vertical distribution of the ship's mobile masses such as cargo, fuel and ballast). The height of the centre of gravity is measured from reference point K . The height of the centre of gravity is identified as a ship's KG .

For a ship to float without an angle of list, point G must be on the same vertical axis as K . As soon as G leaves this vertical axis, an angle of list is produced and the ship will no longer float upright.

The centre of gravity can thus be shifted vertically and horizontally by transferring, adding or removing mobile masses.

If mass is added to a ship, the centre of gravity shifts towards the position of the added mass. For example, fuel added to a ship's double bottom tanks will lower the centre of gravity. Deck cargo generally raises a ship's G . A concentration of cargo on a ship's port side will shift G to port, which will cause the ship to list to port.

**Shifting of G resulting from movement of loads**

The inverse is true when mobile masses are removed. For example, fuel consumption reduces the mass in the storage tanks, and if they are double bottom tanks, the loss of mass shifts G upwards. If deck cargo is unloaded, G is shifted downwards. Unloading cargo on the starboard side will shift G to port, which will result in a list to port.

If a mass already on board is moved, the position of the ship's centre of gravity will shift in the same direction as the mass. For example, shifting port ballast to starboard will shift G to starboard, which will tend to cause a list to starboard.

### 3.3.3. Centre of buoyancy ( B )

**Transverse centre of buoyancy
B
**

A ship's centre of buoyancy can be defined as the point through which the force of buoyancy acts vertically upwards. You can also say that the centre of buoyancy is the geometric centre of the ship's underwater volume. The height of the centre of buoyancy is measured from reference point K and is thus KB .

**Position of point
B
**

### 3.3.4. Ship's inclination

If a ship is floating upright, its symmetrical construction will place point B on the same axis as K and G . The only way to shift the transverse centre of buoyancy is to change the ship's inclination. The underwater volume will be a different shape and B will shift to reach the new geometric centre of this underwater volume. Since the position of B depends solely on the geometry of the cross section, if the hull shape is known, it is easy to identify B based on the load and list conditions.

If a ship is floating upright, points
K
,
G
and
B
will all be on the same vertical axis. If the ship is inclined by external forces (wind, waves, tight mooring at the wharf),
G
should not change position (no mass has shifted), but
B
will shift to the geometric centre of the new underwater volume. Points
B
and
G
will no longer be on the same vertical axis, consequently the ship's weight will act vertically downwards through
G
and the force of buoyancy will push upwards from
B_{1}
.

#### 3.3.4.a. Righting moment

When a ship is inclined, these two forces are no longer on the same vertical axis and a righting moment is created. The righting moment tends to bring the ship back to an upright position. This moment is equal to a force multiplied by a distance. The value of the force is the same for the upwards and downwards vectors, and is equal to the ship's displacement.

**Forces of gravity and buoyancy**

#### 3.3.4.b. Righting lever ( GZ )

The distance between the two vectors is called GZ and represents the righting lever. The larger the righting lever, the higher the righting moment. The size of the righting lever increases with the ship's inclination. In other words, up to a certain angle of inclination (usually between 40° and 60°), the more the ship lists, the greater its tendency to return to an upright position. If the maximum righting angle is exceeded, the righting lever decreases and the ship's ability to right itself also decreases until it reaches an angle where the righting lever is zero and the ship is in serious danger of capsizing.

Inversely, if G is located high on the centreline, the righting lever will be smaller so the righting moment will be weaker. The ship will right itself more slowly.

The value of the righting moment (also called the moment of statical stability,
MSS
) is calculated by the formula

MSS
=
Δ
×
GZ

To find the value of
GZ
at small angles of inclination, the following trigonometric equation is used:

GZ
=
GM
sin
Θ
.

Θ
being the ship's angle of inclination.

**Righting moment**

### 3.3.5. Metacentre ( M )

Looking at the inclination diagram, you can see that a point M has appeared. Point M is located at the intersection of the buoyancy vector and the centreline and is called the metacentre. For small angles of inclination (less than 15°), M is considered to be fixed. The presence of M allows us to introduce a new concept that actually controls stability at small angles of inclination.

### 3.3.6. Metacentric height ( GM )

This is the distance between G and M , which is identified as distance GM , also called the metacentric height.

**Righting moment with a reduced
GM
**

The position of G in relation to M is crucial in a ship's ability to right itself. Under normal conditions, G should always be below M . The GM is then said to be positive. The greater the distance between these two points, the higher the positive GM . As stated in the previous paragraph, the larger the GM , the larger the righting lever. If G approaches M , the righting lever decreases and the righting moment is weak.

If GM is zero, meaning that G coincides with M , the righting lever is non-existent. If an external force then makes the ship heel to a small angle, the ship will remain heeled at this angle because there is no righting moment.

If GM is negative, meaning that G is above M , not only is the righting lever non-existent, but it also becomes a capsizing moment. If the ship is then subjected to a light external force, it will incline sharply and, depending on the shape of the hull, may even capsize completely. In any case, a negative GM is a situation that must absolutely be avoided.

**Neutral equilibrium when
GM
= 0**

**Capsizing moment with a negative
GM
**

### 3.3.7. Abrupt shifting of G

Two situations have a radical effect on the position of G . In both situations, an abrupt rise in G occurs, which in some extreme cases can lead to a situation where GM becomes negative. Both situations are a result of the free surface effect and the effect of suspended weight.

#### 3.3.7.a. Suspended weight

When cargo is handled using cranes or cargo booms mounted on a ship, the centre of gravity of the mass being handled is considered to be at the point of suspension, which is the end of the crane arm or cargo boom. For example, if a crane lifts a mass of 5 tonnes from the bottom of a hold, as soon as the mass leaves the surface it was resting on, the centre of gravity of these 5 tonnes is instantly transferred from the bottom of the hold to the head of the crane arm. This causes an instant and sometimes significant rise in the ship's G . If the GM was already small, this change in position can result in a negative GM .

**Effect of a suspended weight**

#### 3.3.7.b. Free surface effect

The other situation is the occurrence of the free surface effect. If a ship's tank is partially filled and the ship rolls, the mass of liquid in the tank moves uncontrollably. The centre of gravity of the liquid mass shifts from side to side, and the change in "shape" of the liquid can also cause the G of the moving mass to rise radically. In addition, the inertia of the liquid mass moving around affects the ship's transverse stability and the position of its G . The effect of the inertia of the moving liquid is applied by making a virtual change to the position of G . This change in the height of G of the liquid mass can have a radical effect on the ship's height of G , which can result in a negative GM . To reduce the free surface effect, anti-rolling devices are placed in the tanks.

**Free surface effect**

A combination of these two situations can occur when a ship is loading or unloading. Cargo handling is often combined with ballast handling. While in port, fuel or storage transfers can be done. Free surfaces can appear in ballast as well as fuel tanks. When this situation occurs during cargo handling with cranes or cargo booms, a negative GM can easily be created.

As an engineer, in some situations during a layover in port, you must check with the officer in charge of the ship's stability before transferring liquid masses.

All the values for the above terms for a ship can be found in the ship's stability book.

All the concepts covered in this chapter make it possible to maintain a ship's intact stability.

A ship's intact stability is defined as the stability of an undamaged ship that meets
IMO
requirements as set out in the *Code on Intact Stability for all Types of Ships Covered by
IMO
Instruments*.

## 3.4. Simple longitudinal stability

The principles of transverse stability apply partly to create longitudinal stability.

Points
K
,
G
,
B
and
M
from transverse stability are used with a ship's longitudinal section. They become points
G_{L}
,
B_{L}
and
M_{L}
.

The longitudinal metacentre (
M_{L}
) is found the same way as for transverse stability. It will be located at the intersection of the vertical lines passing through points
B_{L}
and
B_{L1}
when the trim is adjusted.

**Longitudinal stability**

In longitudinal stability, trim can be considered as the equivalent of list in transverse stability. Trim represents the longitudinal inclination of the ship and instead of being expressed in degrees, it is given as a difference between the forward and aft drafts.

When aft draft is greater than forward draft (usual situation), the trim is positive. When forward draft is greater, the trim is negative.

For example: a ship with a forward draft of 5 m and an aft draft of 5.75 m will have a positive trim of 75 cm .

Shifting, loading and unloading of a mass will affect a ship's trim. The change in trim when handling masses is measured using the concept of MCTC and MCTI .

MCTC is the Moment to Change Trim 1 centimetre.

MCTI is the Moment to Change Trim 1 inch.

All the values for the above terms for a ship can be found in the ship's stability book.