BILGING &

PERMEABILITY
Flooding, damage stability calculations
Syllabus Requirements
• Bilging a single compartments in a box-shaped
vessels.
• Calculations include
• Bilging of an empty midships compartment
• Bilging of a midships compartment containing cargo
• Bilging of either of the above with a watertight flat
• Bilging of an empty side compartment
• Bilging of an empty end compartments with or without a
watertight flat
Definition
• Bilging
is the flooding of compartments which have been holed or
damaged.
• Permeability (μ)
of a space is the proportion of the immersed volume of that
space which can be occupied by water.
Methods
• Lost Buoyancy Method
• Flooded compartment dose not supply buoyancy
• Displacement dose not change
• Centre of gravity remains the same
• No free surface effect
• Added Weight Method
• Flooded compartment is part of the ship
• Displacement changes
• Centre of gravity changes
• Free surface effect accounted for
LOST BUOYANCY
METHOD
Bilging of an empty midships compartment without
watertight flat
Bilging Amidships Compartment
• When a vessel floats in still water it displaces its own
weight
B

G
W L

W
Bilging Amidships Compartment
• The buoyancy provided by the bilged compartment is lost.
• The draft has increased and the vessel now floats at the
waterline W1L1displacing its own weight
• The buoyancy is regained from intact compartments.

W1 L1
W y X z L

V
Bilging Amidships Compartment
• ‘X’ represents the increase in draft due to bilging.
• The volume of lost buoyancy (v) is regained by the intact
volumes ‘y’ and ‘z’.

W1 L1
W y X z L

V
 Bilging Amidships Compartment
• Let ‘A’ be the area of the water-plane before bilging, and
let ‘a’ be the area of the bilged compartment.

A
W L1
1
W y X z L

V
Bilging Amidships Compartment

W1 L1
W y X z L

V
Example
• A box shaped vessel 150m x 24m x 12m, even keel draft
of 5 m. GM 0.9m Bilges an amidships empty compartment
20m long. Find the new draft and new GM

20 x 24 x 5
Increase in draft = = 0.77 m
(150 – 20) x 24

New draft = 5 + 0.77 = 5.77 m

Example
• Although the position of G has not changed, the KM
certainly has:
• With the increase in draft, the KB has increased.
• With a loss in intact water plane area, the BM has
reduced
For ship shaped vessel For box shaped vessel

1
• KB= Simpson’s Rules = Draft
2
LB3 LBBB B2
• BM=
12V = 12 * L B d =
12d
Example M
W L
1 draft = 5/2 = 2.5 m G
• Old KB =
2 B
½d
B2 242 K
• Old BM = = = 9.6 m
12d 12 x 5

• Old KM = Old KB + Old BM = (2.5 + 9.6) = 12.1 m

1
• New KB = New draft = 5.77 / 2 = 2.89 m
2
a
LB3
• New BM =
12V
A
(150 – 20) x 243
• New BM = = 8.32 m
12 x 150 x 24 x5
Example

• New KM = New KB + New BM = (2.89 + 8.32) = 11.21 m

• Old KG = Old KM – Old GM = 12.1 – 0.9 = 11.2 m

• New GM = New KM – Old KG = 11.21 – 11.2 = 0.01 m

M
W L
G
B
½d
K
Example 2
• A box shaped vessel, length 90m, breadth 30m floats on
an even keel draught of 4.0m in saltwater, with a KG of
5m. An empty midship compartment 30m long is bilged.
Calculate the new draft, and the new GM
 L = 90 m, B = 30 m, d =4.0 m, KG =5 m
Example 2 compartment = 30 m

Volume lost = 30m x 30m x 4m = 3600 m³

Intact WPA = ( 90 m – 30 m ) x 30m = 1800 m²
Increase in draft = 3600 ÷ 1800 = 2m.
The new draft = 4 + 2 = 6m.
KM1 = KB1 + BM1 = 2m + ( B2 ÷ 12d )
KM1 = KB1 + BM1 = 2m + ( 30 x 30 ÷ 12 x 4 ) = 20.75m
GM1 =KM1~ KG = 20.75 ~ 5 = 15.75m
KM2 = KB2 + BM2 = 3m + ( LB³ ÷ 12V )
KM2 = 3m + (( 60 x 30 x 30 x 30) ÷ (12 x 90 x 30 x 4 ))=15.5m
GM2 = KM2 ~ KG = 15.5m ~ 5m = 10.5m
LOST BUOYANCY
METHOD
Bilging of an empty midships compartment with a
watertight flat Below Water Line
Example
• A box shaped barge is 90m long, 40m wide, she has three
equal sized holds with double bottoms, and is floating at a
mean draft of 5m. The empty midships double bottom
compartment 30m in length & 2m in depth is bilged.
• Calculate the new draft and the change in GM.

W1 L1

W L
 L = 90 m, B = 40 m, d =5.0 m,
Example compartment = 30 m x 2 m

30 x 40 x 2
• Increase in draft = = 0.666m
90 x 40

• New draft = 5 + 0.666 = 5.666 m

W1 L1
• BUT: the new KB is not half the new draft as the centroids
W ofbuoyancy are not uniform above the keel for all L
compartments.
• So we must find the new KB by taking volumetric
moments about the keel.
 L = 90 m, B = 40 m, d =5.0 m,
Example compartment = 30 m x 2 m

• KB (centroid V1) = 5.666 / 2 = 2.833 m

• KB (centroid V3) = 5.666 / 2 = 2.833 m
• Centroid V2 = (New draft – Tank depth) / 2
• Centroid V2 = (5.666 – 2) / 2 = 1.833
• KB (V2) = 1.833 + 2 = 3.833 m

W1 L1

W V2 L
B2
V3 B3 B1 V1
 L = 90 m, B = 40 m, d =5.0 m,
Example compartment = 30 m x 2 m

• New KB by taking volumetric moments about the keel.

• V1 = 30 x 40 x 5.666 = 6799.2 m³ KB = 2.833 m
• V3 = 30 x 40 x 5.666 = 6799.2 m³ KB = 2.833 m
• V2 = 30 x 40 x 3.666 = 4399.2 m³ KB = 3.833 m

W1 L1

W V2 W
V3 V1
Example
Volume KB Moment
V1 6799.2 m³ 2.833 m 19,262.13 m⁴
V2 6799.2 m³ 2.833 m 19,262.13 m⁴
V3 4399.2 m³ 3.833 m 16,862.13 m⁴
Total 17997.6 m³ 55386.39 m⁴
New KB = Total moments ÷ Total volume. M
M
New KB = 55386.39 ÷ 17997.6 = 3.077 m
G G
Full WPA intact – no change in BM B
B
change in GM = KB1 ~ KB2 K K
change in GM = 2.5m ~ 3.077m = 0.577 m INCREASE
LOST BUOYANCY
METHOD
Bilging of an empty midships compartment with a
watertight flat Above Water Line
Example
• A box shaped barge is 90m long, 30m wide, she has three
equal sized holds with double bottoms, and is floating at a
mean draft of 5m. The empty midships double bottom
compartment 30m in length & 6 m in depth is bilged.
• Calculate the new draft and the change in GM.

W1 L1

W L
 L= 90m, B=30m, d=5m,
Example Compartment= 30m x 6m

• When height of compartment is given and above water level calculate

sinkage by recoverable buoyancy method.
• Buoyancy still to be recover = Lost buoyancy – Recoverable Buoyancy
• Volume of Lost Buoyancy = l x b x draft
• Recoverable Buoyancy = ( L – l ) x B x ( Depth – Draft )

W1 L1

W L
 L= 90m, B=30m, d=5m,
Example Compartment= 30m x 6m

Volume of Lost Buoyancy = l x b x draft

Volume lost = 30 x 30 x 5 = 4500m³
Recoverable Buoyancy = ( L – l ) x B x ( Depth – Draft )
Volume recovered to the watertight flat = (90-30) x 30 x (6-5)= 1800m³
Buoyancy still to be recover = Lost buoyancy – Recoverable Buoyancy
Buoyancy still to be recover = 4500 – 1800= 2700 m³
Volume still to recover @ a Full waterplane area intact = 2700 m³

W1 L1
W L
 L= 90m, B=30m, d=5m,
Example Compartment= 30m x 6m

• Volume still to recover @ a Full waterplane area intact = 2700 m³

Sinkage ABOVE 6m draft = 2700 ÷ 90 x 30 = 1m
• To find the Final Draft, add the Sinkage to Tank’s height
• New mean draft = 6 + 1 = 7m
 L= 90m, B=30m, d=5m,
Example OR 1 Compartment= 30m x 6m

Volume of Lost Buoyancy = Volume of Gained Buoyancy

Volume lost = compartment Volume up to the draft
Volume lost = 30 x 30 x 5 = 4500m³
Volume Gained = {L x B x increase in Draft (X)} – {l x b x(6-5) depth-draft}
Volume Gained = 90 x 30 x (X) – 30 x 30 x1
Volume Gained = 2700 (X) – 900
Volume of Lost Buoyancy = Volume of Gained Buoyancy
4500 = 2700 (X) – 900
increase in Draft (X) = 4500 + 900 / 2700 = 2 m
New mean draft = 5 + 2 = 7m
W1 L1
W L
 L= 90m, B=30m, d=5m,
Example OR 2 Compartment= 30m x 6m

Extra mass of water

• Increase in draught =
TPC
A 1.025
• TPC = = 90 x 30 x 1.025 /100 = 27.675 t
100

• Extra mass of water = 30 x 30 x 6 x 1.025 = 5535 t

• Increase in draught = 5535 / 27.675 = 200 cm
• New mean draft = 7 m
 L= 90m, B=30m, d=5m,
Compartment= 30m x 6m
Example New draft = 7m
New KB by taking volumetric moments about the keel.
• KB (centroid V1) = 7 / 2 = 3.5 m
• KB (centroid V3) = 7 / 2 = 3.5 m
• Centroid V2 = (New draft – Tank depth) / 2 =
• Centroid V2 = (7 – 6) / 2 = 0.5
• KB (V2) = 0.5 + 6 = 6.5 m

W1 L1

V2 B2
W W
V3 B3 B1 V1
 L= 90m, B=30m, d=5m,
Compartment= 30m x 6m
Example New draft = 7m

• new KB by taking volumetric moments about the keel.

• V1 = 30 x 40 x 7 = 8,400 m³ KB = 3.5 m
• V3 = 30 x 40 x 7 = 8,400 m³ KB = 3.5 m
• V2 = 30 x 40 x 1 = 1,200 m³ KB = 6.5 m

W1 L1

V2
W W
V3 V1
Example
Volume KB Moment
V1 8,400 m³ 3.5 m 29,400 m⁴
V2 8,400 m³ 3.5 m 29,400 m⁴
V3 1,200 m³ 6.5 m 7,800 m⁴
Total 18,000 m³ 66,600 m⁴
New KB = Total moments ÷ Total volume. M
M
New KB = 66,600 ÷ 18,000 = 3.7 m
G G
Full WPA intact – no change in BM B
B
change in GM = KB1 ~ KB2 K K
change in GM = 2.5m ~ 3.7m = 1.2 m INCREASE
PERMEABILITY ( μ )
Permeability ( μ )

Permeability (μ) of a space is the proportion of the

immersed volume of that space which can be occupied by
water.
Permeability ( μ )

• After calculating ‘Effective length’ always use this length

for tank’s length.
Example
• A box shaped vessel, length 60m, breadth 20m, is floating
at an even keel draught of 4m. A compartment midships is
20m long and contains cargo with a permeability of 30%.
Calculate the new draft and change in GM if this
compartment is bilged.
 L=60 m, B=20 m, d=4 m
Example Compartment = 20 m µ 30%

• Effective length of compartment = 20 x 0.3 = 6m.

• Effective length of waterplane = 60 – 6 = 54m.

• Sinkage = Lost volume ÷ IWPA

• Sinkage = ( 6 x 20 x 4) ÷ ( 54 x 20 ) = 480 ÷ 1080 = 0.4m
• New draft = 4 + 0.4 = 4.4m
• change in GM = KM1 ~ KM2
As both the KB and the BM have changed
Example 2
• A box shaped barge is 90m x 30m with an initial mean
draft of 4m,she has a midships compartment 30m in
length bilged.
• The compartment is tightly stowed with grain.
• R.D. of grain = 0.95 t/m³.
• The stowage factor of the cargo = 1.46 m³/tonne.
Calculate the new draft.
 L=90 m, B=30 m, d=4 m
Example 2 Compartment = 30 m
R.D= 0.95 t/m³, SF =1.46 m³/tonne

• Solid S.F = 1 ÷ 0.95 = 1.05 m³/t.

• Broken stowage = 1.46 ~ 1.05 = 0.41 m³/t

• μ ( permeability ) = 0.41 ÷ 1.46 = 28%

 L=90 m, B=30 m, d=4 m
Compartment = 30 m
Example 2 R.D= 0.95 t/m³, SF =1.46 m³/tonne

• μ ( permeability ) = 0.41 ÷ 1.46 = 28%

• Effective length of compartment = 30 x 0.28 = 8.4m

• Effective length of waterplane = 90 – 8.4 = 81.6m
• Sinkage = lost volume ÷ IWPA
• Sinkage = ( 8.4 x 30 x 4 ) ÷ ( 81.6 x 30 ) = 1008 ÷ 2448 = 0.41m
• New draft = 4.41m.
BILGING OF AN EMPTY
SIDE COMPARTMENT
Introduction
• When a compartment in a
ship is bilged the buoyancy
provided by that
compartment is lost.
• This causes the centre of
buoyancy of the ship to
move directly away from the
centre of the lost buoyancy
and,
• unless the centre of gravity
of the compartment is on the
ship's centre line, a listing
moment will be created,
 Introduction
• the ship will sink to the waterline W1L1.
That is, the lost buoyancy is made good by
the layer between WL and W1L1.
• The centre of buoyancy will move from B to
B1, directly away from the centre of gravity
of the lost buoyancy, and
• the distance BB1 is equal to w * d / W
• where w represents the lost buoyancy and
d represents the distance between the
ship's centre of buoyancy and the centre of
the lost buoyancy.
• The shift in the centre of buoyancy
produces a listing moment.

• Let  be the resultant list.

Tan  = GX / XM
= BB1 / XM

where XM represents the initial metacentric

height for the bilged condition.
Second moments of area
Moments of inertia
• The tendency of a body to resist acceleration; the
tendency of a body at rest to remain at rest or of a body in
straight line motion to stay in motion in a straight line
unless acted on by an outside force.

• To find the second moment of a rectangle about one of its

sides.
LB3
• I side=
3
• To find the second moment of a rectangle about an axis
parallel to one of its sides and passing through the
centroid.
LB3 I LB3
• I centroid = BM = = 12 V
12 V
The parallel axes theorem
the moment of inertia about any axis passing through the
centroid of a water plane area is equal to the moment of
inertia of the same water plane area about any parallel
axis, minus the area of the water plane multiplied by the
distance between the axes squared.

I B I
F
B
d 2
X X
The parallel axes theorem

I B I
F
B
d 2
X X

The moment of inertia about the axis II (which passes through

the centre of flotation)
is equal to the moment of inertia about the axis XX minus the
product of the water plane area and the distance between the
axes squared
III = Ixx - (A X d2)
The parallel axes theorem
A box-shaped vessel has length 20 m and breadth 6 m.

LB3
III = = 20 * 6 * 6 * 6 / 12 = 360 m4
12
LB3
Ixx= 3 = 20 * 6 * 6 * 6 / 3 = 1440 m4
OR III = Ixx - (A X d2)
III = 1440 - (20 * 6 * 3 * 3) = 360 m4
L

I B I
F
B
d 2
X X
bilged side compartment
When an amidships side compartment extending upwards
the full depth of the vessel becomes bilged the centre of
flotation (F) will move off the centre line (F 1) , its new
position with reference to side XX (axis XX) may be
calculated by taking moments of area about one side. This
will give the distance, d, between the two axes to be
considered.
bilged side compartment
When an amidships side compartment extending upwards the
full depth of the vessel becomes bilged
The centre of flotation (F) will move off the centre line (F 1) its
new position with reference to side XX (axis XX) may be
calculated by taking moments of area about one side.
This will give the distance, d, between the two axes to be
considered.
L
I F1 I
C F B L

d b
X X

l
bilged side compartment
In order to calculate the BM in the bilged condition it will be
necessary to calculate the new value of the transverse
moment of inertia of the remaining intact water plane area
about the new longitudinal axis of rotation passing through
the new position of the centre of flotation (about axis II).
This is achieved by using the parallel axis theorem.

L
I F1 I
C F B L

d b
X X

l
bilged side compartment
III = Ixx - (A X d2)
LB3 lb3
Ixx= -
3 3

A = LB – lb

LB3 lb3
III ={ 3 - 3 } - {(LB – lb) * d2}
L
I F1 I
C F B L

d b
X X

l
Example
• A box shaped vessel has length 96 m and breadth 18 m
and floats at an even keel draught of 4.6 m and KG 5.80
m in salt water. An amidships side compartment of length
24 m extending in from the side 6 m is bilged. Calculate:
• (a) the initial BM value;
• (b) the BM in the damaged condition.
• (c) the New draft
• (d) the angle of list when this compartment becomes
bilged.
L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
the initial BM value
L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list

I LB3
• In the intact condition BM = =
V 12 V

• BM = (96 * 18 * 18 * 18) / (12 * 96 * 4.6 )

• BM initial = 5.870 m
BM in the damaged Condition
L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list

LB3
LB – lb) * d2}
lb 3
III ={ - } - {(
3 3
Calculate the distance of the new CF from one side of the
WPA (XX)
Take moments of area about side XX

L
I F1 I
C F B L

d b
X X

l
 BM in the damaged Condition
L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
LB3 lb3
III ={ 3 - 3 } - {(LB – lb) * d2}
Take moments of area about side XX
Area m2 Dist. from XX m Moment
Total Area 96 * 18 9 15552

- Bilged Area 24 * 6 3 432

Total 1584 9.545 15120
L
I F1 I
C F B L

d b
X X

l
BM in the damaged Condition
L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
d (XX-II) =9.545m

LB3 lb3
III ={ 3
- 3
} - {(LB – lb) * d2}
96 * 183 24 * 63
III ={ 3
- 3
} - {( 96 *18 – 24*6) *9.545 2}
III = 40582 m4
L
I F1 I
C F B L

d b
X X

l
 BM in the damaged Condition
L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
d (XX-II) =9.545m, III=40582m4
I
BM = V
40582
BMdamaged = 96 * 18 * 4.6
= 5.105 m

Using the lost buoyancy (constant displacement) method the volume of

displacement remains constant
L
I F1 I
C F B L

d b
X X

l
the New draft
L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
d (XX-II) =9.545m, III=40582m4 BMdamaged=5.105 m

• Volume of buoyancy lost = volume of buoyancy gained

• Let x = sinkage
• 24 * 6 * 4.6 = {(96 * 18) – (24 * 6) }x
• 662.4 = 1584 x
• X = 0.418 m
• New draft = 4.6 + .418 = 5.018 m
 L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
angle of list d (XX-II) =9.545m, III=40582m4 BMdamaged=5.105 m
new draft = 5.018 m
• Because of the symmetry of a box-shaped vessel BBH is
equal to the movement of the centre of flotation off the
centre line (FF1) that is found by taking moments of area
of the water plane area about one edge
D

Vertical Q
distance
G to M

BH B

D
 L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
angle of list d (XX-II) =9.545m, III=40582m4 BMdamaged=5.105 m
new draft = 5.018 m

• Find GM Bilged

• KB bilged = draft bilged / 2

• KB bilged = 5.018 / 2 = 2.509 m
• KM Bilged = KB bilged + BM bilged
• KM Bilged = 2.509 + 5.105 = 7.614 m
• GM Bilged = KM Bilged – KG (constant)
• GM Bilged = 7.614 – 5.8 = 1.814 m
 L= 96m, B = 18m, draft=4.6m, KG=5.8m, l= 24m, b=6m
BM initial and damaged, New draft, Angle of list
angle of list d (XX-II) =9.545m, III=40582m4 BMdamaged=5.105 m
new draft = 5.018 m, GM Bilged =1.814 m

• Since the CF has moved of the centre line by an amount

equal to: 9.545 -9.0 = 0.545 m
• BBH is also equal to 0.545 m.
• Tan  list =0.545 / 1.814 = 0.30044
•  list = 16.7º
L
I F1 I
C F B L

d b
X X

l
SQA – July 2006
• A box shaped vessel floating upright on an even keel in
salt water has the following particulars:
• LengthBP: 150.00m Breadth: 28.00m Even keel draught:
8.60m KG: 9.20m
• The vessel has two longitudinal bulkheads each 9.00m
from the side of the vessel.
• Calculate the angle of heel if an amidship side
compartment 24.00m is bilged.

• List = 12.5°
Wallsided Formula
SMALL HEEL ANGLES ( < 6º )
M

G Z

B1

GZ = GM Sinθ
 At large angles of heel?
GZ = (GM + MM1) Sinθ or : GZ = Sinθ (GM + MM1)
and : MM1 = ½ BM x Tan² θ
So : GZ = ? M1
M

G Z

B1
 Wallsided Formula

GZ = Sinθ ( GM + MM1 ) and : MM1 = ½ BM x Tan² θ

So : GZ = Sinθ ( GM + ½ BM x Tan² θ )

M1
M

G Z

B1
Righting Moment

Righting moment = GZ Sine 

Use of the Wallsided Formula

• Used when angle of heel is > 6 degrees

• Only to be used on wallsided vessels with sides

that are parallel

• Only valid up to the Angle of deck edge Immersion

BILGING OF AN EMPTY END
COMPARTMENTS WITH OR
WITHOUT A WATERTIGHT
FLAT
Bilging of an End Compartment
• The loss of buoyancy at the end of the vessel causes:
• The draft will increase
• B will move vertically up to BV
• but also longitudinally to BH
• LCF will Move to the centre of the waterplan area

L
W1 F1 G L1

W F L
BV
BH B

l
Bilging of an End Compartment
• G and B are no longer in a vertical line causing a trimming
lever BBH
• trimming moment = W x BBH

L
W1 F1 G L1

W F L
BV
BH B

l
Bilging of an End Compartment
MCTC Δ
WL2
G
F
Vol gained
WL2 BV B1 Vol Lost
BH B b

Δ x GML
MCTC = Δ
100 x LBP
• GML is for the bilged condition.
• L is the full length of the vessel as the vessel
trims and sinks over the whole length.
• If the KG is not given the question then:
• Because B1 and G are very close to each other
then BML may be used instead of GML
• This assumption must be stated
 Δ
6m
Example 30 m
WL2

F
Vol gained
WL2 BV G Vol Lost

4m
BH B b

3m

A box shaped vessel Length 30 m, Breadth 6 m, is floating on

an even keel draft of 4m in salt water. KG 2.00 m.
Calculate the drafts F and A if an empty forward end
compartment of length 3 m and full beam is bilged.
 Δ

30 m 6m

L=30m, B=6m, d=4m, KG=2m, l=3m G

WL2

F
Vol gained
WL2 BV Vol Lost

4m
BH B b

3m

Volume L x B x d
30 x 6 x 4 =720 m3
Initial Δ (constant) 30 x 6 x 4 x 1.025 =738 t

Sinkage = Vol Bilged Comp x μ

IWPA

Sinkage = 3 x 6 x 4 x 1 0.444 m
[(30 – 3) x 6]
 Δ

30 m 6m
L=30m, B=6m, d=4m, KG=2m, l=3m WL2
G
TMDBilged= 4.444m, BBH = 1.5m F

Δ = 738 t Vol gained

WL BV Vol Lost
2

4m
BH B b

3m

Initial draft 4.000

Sinkage 0.444
Bilged TMD 4.444

Horizontal shift of B (BBH) = L comp 3 =1.5 m aft

2 2
 Δ

30 m 6m
WL2

L=30m, B=6m, d=4m, KG=2m, l=3m F

G

TMDBilged= 4.444m, BBH = 1.5m Vol gained

BV Vol Lost
Δ = 738 t, Trimming Moment =1107 tm WL 2

4m
BH B b

3m

Trimming Moment = w x s

Trimming Moment = Δ x BBH

738 x 1.5 1107 tm by head

MCTC Δ x GML
100 x L
 Δ

30 m 6m

L=30m, B=6m, d=4m, KG=2m, l=3m G

WL2

TMDBilged= 4.444m, BBH = 1.5m Vol gained

F

Δ = 738 t, Trimming Moment =1107 tm WL 2

BV Vol Lost

4m
BH B b
Bilged KB = 2.222 m,
Bilged BML13.669 m 3m

Bilged KB = d 4.444 2.222 m

2 2

Bilged BML IL L 3B 273 x 6 13.669 m

V 12 V (12 x 720)
 Δ
6m
L=30m, B=6m, d=4m, KG=2m, l=3m 30 m
WL2

TMDBilged= 4.444m, BBH = 1.5m F

G

Δ = 738 t, Trimming Moment =1107 tm

Vol gained
BV
WL Vol Lost
Bilged KB = 2.222 m, 2

4m
BH B b

Bilged BML13.669 m 3m
Bilged GML13.891 m Δ

Bilged KB 2.222
Bilged BML 13.669
Bilged KML 15.891
KG 2.000
Bilged GML 13.891
 Δ
L=30m, B=6m, d=4m, KG=2m, l=3m 30 m 6m

TMDBilged= 4.444m, BBH = 1.5m G

WL2

Δ = 738 t, Trimming Moment =1107 tm

Vol gained
F

Bilged KB = 2.222 m, WL 2
BV Vol Lost

4m
BH B b
Bilged BML13.669 m
Bilged GML13.891 m 3m

CoT = 324.0 cm Δ

MCTC = Δ x GML
100 x LBP

738 x 13.891 = 3.417

(100 x 30)

Trim Mom = 1107 = 324.0 cm by head

CoT=
MCTC 3.417
 Δ
L=30m, B=6m, d=4m, KG=2m, l=3m 30 m 6m

TMDBilged= 4.444m, BBH = 1.5m G

WL2

Δ = 738 t, Trimming Moment =1107 tm

Vol gained
F

Bilged KB = 2.222 m, WL 2
BV Vol Lost

4m
BH B b
Bilged BML13.669 m
Bilged GML13.891 m 3m

CoT = 324.0 cm Δ

Proportion trim LCF not midships

LCF = 15 – 1.5 = 13.5 m foap

CoTA = CoT x LCF = 3.24 x 13.5 = 1.458

LBP 30

CoTF = 3.240 – 1.458 = 1.782

 Δ

30 m 6m
WL2
G
F
Vol gained
WL2 BV Vol Lost

4m
BH B b

3m

Find drafts F & A

Fwd Aft
Bilged TMD 4.444 4.444
CoTF/A + 1.782 - 1.458
Bilged Draft 6.226 2.986
March 2010
• A box shaped vessel floating on even keel in dock water
of RD 1.015 has the following particulars:
• Length 130.00m Breadth 21.00m
• Draught 8.000m MCTC (salt water) 300
• There is an empty watertight forward end compartment,
length 10.00m, height 6.70m, extending the full width of
the vessel.
• Calculate the draughts forward and aft, if this
compartment is bilged.
March 2010
• Volume before bilging = 130 x 21 x 8 = 21840 m3
• Displacement before bilging = 21840 x 1.015 = 22167.6 t
• Permeability of the bilged compartment = 1.00
• Volume of the compartment = 10 x 21 x 6.70 = 1407 m3
• Intact water plane area = 130 x 21 = 2730 m2

• Sinkage caused to bilging = 1407 / 2730 = 0.515m

• Bilged TMD = 8.000 + 0.515 = 8.515m
March 2010

What If MCTC and KG NOT GIVEN?

• Intact water plan area & BML same

Δ x GML
MCTC =
100 x LBP
• BML may be used instead of GML
• This assumption must be stated
March 2010

What If MCTC NOT GIVEN But KG is given?

• KB = take the moment about the keel

• Intact water plan area & BML same

Δ x GML
MCTC =
100 x LBP
• Calculate and use GML
 March 2010
• Bilged TMD = 8.000 + 0.515 = 8.515m

Volume (m3) Distance from AP Moments (m4)

(m)
130 x 21 x 8.515 65.0 1510986.75
(–) 1407 125.0 (–) 175875
21838.95 61.134 1335111.75

• BBH = 65 – 61.134 = 3.866m

• Trimming Moment = 22167.6 x 3.866 x 1.025= 85699.94 tm by
forward
 March 2010
𝑇𝑟𝑖𝑚𝑚𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡 22167.6 ×1.025
• COT = = = 288.5cm
𝑀𝐶𝑇𝐶 300 ×1.015

• Intact water plan area

• height of compartment = 6.70m
• Bilged TMD = 8.515m
• LCF unchanged amidship
• Ta=Tf= 288.5 / 2 = 144.3 cm
• Draught Fwd = 8. 515 +1.443 = 9.958 m
• Draught Aft = 8. 515 - 1.443 = 7.072 m
Counter measures in the event of flooding
• Close all water tight doors.
• Use of ship’s pumps to remove water from the flooded
compartment.
• Take measures to restrict ingress of water (other vessel to
stay in place initially and / or use of patch)
• Cross flooding – ballasting the other side of the vessel to
bring the ship upright (movement of weights may also be
considered).
• Ballasting or movement of weight to raise the damaged
area of the hull above the water line.
Counter measures in the event of flooding
• Removal of weight, particularly from the upper parts of the
vessel (empty swimming pool).
• consider beaching.
• Throughout the above, reference should be made to the
stability data on-board providing guidance for such
circumstances.
• In addition the SMS should be brought into operation. This
usually involves informing ship owners of the situation and
gaining access to advice from experts associated with
Classification society and / or salvage association.