Curves of Stat. Stability | List | Free Surface Effetcs | Trim |

**Stability**

*Movement of the Centre of Gravity*

Centre of gravity

It is the
point of a body at which all the mass of the body may be assumed to be
concentrated.

The force
of gravity acts vertically downwards from this point with a force equal to the
weight of the body.

Basically
the body would balance around this point.

The COG of
a homogeneous body is at its geometrical centre.

Effect of removing or discharging mass

Consider a
rectangular plank as shown. The effects of adding or removing weights would be
as shown:

Now cut the
length of plank of mass ‘w’ kg whose CG is ‘d’ mtrs away from CG of the plank.

Note that a
resultant moment of ‘w x d’ kg m has been created in an anti-clockwise
direction about ‘G’.

The CG of
the new plank shifts from ‘G’ to ‘G1’.

The new
mass (W-w) kg now creates a tilting moment of (W-w) x GG1 about G.

Since both
are referring to the same moment,

(W-w) x GG1 = w x d

GG1 = (w x d)/(W-w)

CONCLUSION:
When a weight is removed from a body, the CG shifts directly away from the CG
of the mass removed, and the distance it moves is given by:

GG1 = (w x d)/Final mass metres

Where, GG1
is the shift of CG

w is the
mass removed

d is the
distance between the CG of the mass removed and the CG of the body.

Effect of adding or loading mass

Equating
the tilting moments created due to the added weight, which must again be equal:

(W + w) x GG1 = w x d

GG1 = (w x d)/(W
+ w)

GG1 = (w x d)/ (Final mass) metres

Application to ships

DISCHARGING
WEIGHTS:

GG1 = __(w x d)__ metres

(Final
displacement)

LOADING
WEIGHTS

GG1 = __(w x d) __ metres

(Final
displacement)

**Shifting Weights**

GG2 = __(w x d) __ metres

(Displacement)

Vertical Weight Shifts

Shifting weight vertically, no matter where onboard it
is, will always cause the ship’s center of gravity to
move in the same direction as the weight shift.

To calculate the height of the ship’s center of
gravity after a vertical weight shift, the following equation is used:

KG_{1} = ((W_{0} x KG_{0}) +/-
(w x kg)) / Δ_{F}

KG_{O }= The original
height of the ship’s center of gravity (M)

Δo = The ship’s displacement prior to shifting weight (MT)

w = The amount of weight
shifted (MT)

kg = The
vertical distance the weight was shifted (M)

Δ_{F} = The
ship’s displacement after shifting the weight (MT)

(+) When the weight is shifted up use (+)

(-) When the weight is shifted down use (-)

__Example
Problem__

10 MT of cargo is shifted up 3 M. Δ_{O}
is 3500 MT and KG_{o} is 6 M. What is the new
height of the ship’s center of gravity (KG_{1})?

KG_{1} = ((Δo x
KGo) +/- (w x kg)) / Δ_{F}

KG_{1} = ((3500 x 6) + (10 x 3)) / 3500

KG_{1} = 6.009 M

Vertical Weight Additions/Removals

When weight is added or removed to/from a ship, the
vertical shift in the center of gravity is found using the same equation.

KG_{1} = ((Δo x
KGo) +/- (w x kg)) / Δ_{F}

KG_{O }= The original
height of the ship’s center of gravity (M)

Δ_{O} = Ship’s displacement prior to
adding/removing weight (MT)

w = The amount of weight
added or removed (MT)

kg = The
height of the center of gravity of the added/removed weight above the keel (M)

Δ_{F} = The
ship’s displacement after adding/removing the weight

(+) When the weight is added use (+)

(-) When the weight is removed use (-)

__Example
Problem__

A 30 MT crate is added 10 M above the keel. Δo is 3500 MT and KG_{0} is 6 M. What is the
new height of the ship’s center of gravity (KG_{1})?

KG_{1} = ((Δo x
KGo) +/- (w x kg)) / Δ_{F}

KG_{1} = ((3500 x 6) + (30 x 10)) / 3530

KG_{1} = 6.034 M

Horizontal Weight Shifts

Shifting weight horizontally, no matter where onboard
it is, will always cause the ship’s center of gravity
to move in the same direction as the weight shift.

NOTE: A weight shift causing the ship’s center of
gravity to move off centerline will always reduce the stability of the ship.

To calculate the horizontal movement of the ship’s
center of gravity, the following equation is used:

GG_{2} = (w x d) / Δ_{F}

w_{ }= The amount of
weight shifted (MT)

d = The horizontal distance
the weight is shifted (M)

Δ_{F} = The
ship’s displacement after the weight is shifted (MT)

Example Problem

A 50 MT weight is shifted 10 M to starboard. Δ_{O}
is 32000 MT.

What is the change in the center of gravity (GG_{2})?

GG_{2} = (w x d) / Δ_{F}

GG_{2} = (50 x 10) / 32000

GG_{2} = 0.01562 M

Horizontal Weight Additions/Removals

When an off-center weight is added or removed to/from
a ship, the ship’s center of gravity will move off centerline, the ship will
develop a list.

To calculate the horizontal movement of the ship’s
center of gravity after adding/removing an off-center weight, the same equation
is used:

GG_{2} = (w x d) / Δ_{F}

w_{ }= The amount of
weight added/removed (MT)

d = The distance from the
center of gravity of the weight to the ship’s centerline (M)

Δ_{F} = the ship’s displacement after the
weight is shifted (MT)

__Example
Problem__

50 MT of cargo is loaded onto the Tween deck, 10 M
from centerline. Δ_{O} is 48000 MT. What is the change in the
center of gravity (GG_{2})?

GG_{2} = (w x d) / Δ_{F}

GG_{2} = (50 x 10) / 48000

GG_{2} = 0.0104 M

Effect of suspended weights

The CG of a
body is the point through which the force of gravity may be considered to act
vertically downwards.

For a
suspended weight, whether the vessel is upright or inclined, the point through
which the force a gravity may be considered to act
vertically downwards is g1, the POINT OF SUSPENSION.

**Conclusions**

The CG of a
body will move directly TOWARDS the CG of any weight ADDED.

The CG of a
body will move directly AWAY from the CG of any weight DISCHARGED.

The CG of a
body will move PARALLEL to the shift of the CG of any weight MOVED within the
body.

The shift
of the CG of the body in each case is given by the following formula:

GG1 = __w x d __ metres

W

where w =
weight added, removed or shifted.

W = final mass of the body

d = distance between the CG if
weight added or removed, or the distance by which the weight is shifted.

When a
weight is SUSPENDED, its CG is considered to be at the POINT OF SUSPENSION.