BILGING

Type of Midships compartment Bilged

1) Compartment WITHOUT watertight Flat

WPA Top view

Side
View

- Here the volume of lost buoyancy is the volume between the keel and
the initial waterline.
- Height of lost buoyancy = Initial draft
- Then, Intact WPA = (A – a)
Intact L = (L – l)
- Vol of lost buoyancy (v) = a x Initial draft
- Sinkage = Lost volume(v) / Intact WPA
- New mean draft = Initial draft + Sinkage
- KB = ½ new mean draft

2) Compartment WITH watertight Flat - 3 type of compartments

i) Flat BELOW waterline - Bilged ABOVE flat

(Ht of flat from keel > Initial Draft)
 - Here the Lost Volume is the volume between the flat and the initial
waterline.
- Height of Lost Volume = Initial draft - Ht of flat from keel
- Then, Intact WPA = (A – a)
Intact L = (L – l)
- Vol of lost buoyancy (v) = a x Height of Lost Volume
- Sinkage = Lost volume(v) / Intact WPA
- New mean draft = Initial draft + Sinkage
- KB by moments method

ii) Flat BELOW waterline- Bilged BELOW flat

(Ht of flat from keel > Initial Draft)

- Here the Lost Volume is the volume of the (DB) or compartment

- Height of Lost Volume = ht of flat from keel or ht of DB tk or
compartment
- Then, Intact WPA = A
Intact L = Initial L
- Vol of lost buoyancy (v) = a x Height of Lost Volume
- Sinkage = Lost volume(v) / whole WPA
- New mean draft = Initial draft + Sinkage
- KB by moments method

iii) Flat CLOSE to waterline - Bilged below flat

(Ht of flat from keel > Initial Draft)

- Here the Lost Volume is the volume between the keel and the initial
waterline.
- Height of Lost Volume = Initial draft
- The shape of the WPA will depend on whether the final
waterline is above or below the level of the flat as shown below.
 To check if this is the case, compare the Lost Volume with the
Recoverable volume up to a draft equal to that of the flat.

Recoverable Volume to level upto flat = [ (A - a) x c) ]

OR = [ (A - µa) x c) ] If there is permeability (µ)

where, c = Ht of compartment – Initial draft

a) If, Lost Volume (v) > Recoverable Volume up to flat (v1), then the new
waterline will be above the flat.

- Then, whole WPA = A

L = Initial L

- Lost volume yet to be recovered (v2)

= Lost vol (v) – Recoverable vol upto flat(v1)
- Addnl sinkage = v2 / whole WPA
- New mean draft = Ht of flat above keel + Addnl sinkage

b) If, Lost Volume (v) < Recoverable volume up to level of flat, then the new
waterline will be below the flat.
- Then, Intact WPA = ( A – a )
Intact L = (L – l)
- Vol of lost buoyancy (v) = a x Initial draft
- Sinkage = Lost volume(v) / Intact WPA
- New mean draft = Initial draft + Sinkage
SINKAGE (No liquid in Compartment)
Sinkage = Lost volume = v (m)
Intact WPA (A-a)

A – Initial WPA = (L x B)
L – Initial Length of ship
B – Breadth of ship / compartment
v – Lost volume = l x b x ht
a – Lost WPA = ( l x b )
where, l – length of lost volume
b – breadth of lost volume
ht = Height of lost volume, depending upon the type of compartment bilged

Also,
Sinkage = w/TPC (cm)

PERMEABILITY
If the compartment has Permeability (µ),
i) When cargo broken stowage is given (BS)
Permeability (µ) = BS of cargo x 100 (percentage)
SF of cargo

ii) When RD of cargo (t/m³) is given

Solid SF of cargo = 1 / RD of cargo (m³/t)

Permeability (µ) = SF of cargo - Solid SF of cargo x 100 (percentage)

SF of cargo

Sinkage = Lost volume = µv (m)

Intact WPA (A - µ a)

NEW KB after bilging

TO FIND NEW KB
Find the following:
- X = Sinkage
- New mean draft = Initial mean draft + Sinkage
- New KB = ½ New mean draft

TO FIND NEW KB (By Moments Method)

- Whenever there is a flat and the compartment is bilged above or below the flat and the
new waterline is above the flat (i.e. Ht of flat from keel < New Mean Draft)

Reference to above Type of compartments are :

1. Midship compartment bilged

a) First find the Total Volume (V1) for the New Mean Draft & KB1 of V1
i. V1 = Lx B x new mean draft
ii. KB1 = ½ New mean draft

b) Find the bilged volume (v3) wrt new mean draft & its Kb
i) If compartment is bilged below flat
 - v3 = l x b x ht of flat above keel
- kb = ½ ht of flat

ii) If compartment is bilged above flat

- v3 = l x b x (New mean draft – ht of flat)
- kb = ht of flat + ½ (New mean draft – ht of flat)
Then,
Volume KB Moments
V1 KB1 V1 x KB1
. – v3 kb - (v1 x kb1)
V Moments

New KB = Moments / V

To find: I = Moment of Inertia (transverse)

First Draw and Find the shape of the WPA / Intact WPA

a) Bilged - Midships, Intermediate & End Compartments

- When Empty
ICL (Intact WPA) = (Intact L)B³/12 Intact L = L – l

b) Bilged - Midship compartments, Intermediate & End Compartments

- Has permeability (µ)
ICL (whole WPA) = LB³/12
ICL (lost WPA) = (lb³/12) µ
ICL (Intact WPA) = ICL (whole) - ICL (lost)
OR
ICL (Intact WPA) = [ (L – µl) B³ ] /12
To find Transverse Stability - (KM/GM or Loss of stability)
Find following :
- New KB
- New BM
- New KM = New KB + New BM
- New GM = New KM – (KG)
- Old KB = ½ old draft
- Old BM = L x B³
12 x V
- Old KM = Old KB + Old BM
- Loss of stability = Old KM – New KM

Midship compartment WITHOUT W/T flat and bilged

Q. A box shaped vessel 190m x 25m, floating at an even keel draft of 8m has a midship
compartment 15m in length extending from side to side, KG 6.861m. Find the GM after this
compartment gets bilged.
Solution:

WPA Top view

Side
View

Lost buoyancy = v = (15 x 25 x 8) = 3000m³

Intact WPA = A - a = [ (190 x 25) – (15 x 25) ] = 4375m²

Sinkage = Volume of lost buoyancy = 3000 = 0.686m

Area of intact waterplane 4375

New draught = 8.0 + 0.686 = 8.686m

New KB = ½ new draft = 4.343m

ICL (Intact WPA) = [ (L – l) B³ ]/12 = { [ 190 – 15 ] x 25³ } /12 = 227864.6 m⁴

BM = ICL (Intact WPA) / V = 227864.6 / (190 x 25 x 8) = 5.996m

New KM = New KB + New BM = 4.343 + 5.996 = 10.339m

New GM = New KM – KG = 10.339 – 6.861 = 3.478m
Midship compartment WITH W/T flat and bilged BELOW the flat
Q. A box shaped vessel 170m x 25m floats at an even keel draft of 7m. A midship
compartment 15m long has a watertight flat 4m above the keel. If that compartment is
bilged below the watertight flat, find her draft and KM.
Solution:

Lost buoyancy = v = (15 x 25 x 4) = 1500m³

Whole WPA = [ (170 x 25) ] = 4250m²

Sinkage = Volume of lost buoyancy = 1500 = 0.353m

Area of intact waterplane 4250

New draught = 7.0 + 0.353 = 7.353m

V1 = L x B x New draft = 170 x 25 x 7.353 = 31250.25 m3

KB1 = ½ new draft = ½ x 7.353 = 3.6765m
v1 = l x B x ht of flat = 15 x 25 x 4 = 1500 m3
kb1 = ½ ht of flat = ½ 4 = 2.0m

To find KB (Moments method)

Volume KB Moment
31250.25 3.6765 114891.5
– 1500 2.0 – 3000
29750.25 111891.5

New KB = 111891.5 = 3.761m

29750.25
ICL (Iwhole WPA) = (L x B³ ]/12 = [ 170 x 25³ } /12 = 221354.2m⁴

BMT = ICL (whole WPA) / V = 221654.2 / (170 x 25 x 7) = 7.440m

New KM = New KB + New BMT = 3.761 + 7.440 = 11.201m

Midship compartment WITH W/T flat and bilged ABOVE the flat
Q. A box shaped vessel 170m x 25m at an even keel draft of 7m has a midship compartment
15m in length, extending from side to side, with a watertight flat 4m above the keel. The KG
of the vessel is 5.8m. if the compartment is bilged above the watertight flat, calculate (i) the
draft at which she settles and (ii) her GM after bilging.
Solution:

Lost buoyancy = v = (15 x 25 x 3) = 1125m³

Intact WPA = A - a = [ (170 x 25) – (15 x 25) ] = 3875m²

Sinkage = Volume of lost buoyancy = 1125 = 0.290m

Area of intact waterplane 3875

New draught = 7.0 + 0.290 = 7.290m

V1 = L x B x New draft = 170 x 25 x 7.290 = 30982.5 m3

KB1 = ½ new draft = ½ x 7.29 = 3.645m
v1 = l x B x (New draft – ht of flat) = 15 x 25 x (7.29-4) = 1233.75 m3
kb1 = Ht of flat + ½ (New draft – ht of flat) = 4 + ½ (7.29 – 4) = 5.645m

To find KB (Moments method)

Volume KB Moment
30982.5 3.645 112931.2
– 1233.75 5.645 – 6964.5
29748.75 105966.7

New KB = 105966.7 = 3.562m

29748.75
ICL (Intact WPA) = [ (L – l) x B³ ]/12 = { [ 170 – (15) ] x 25³ } /12 = 201822.9m⁴
BMT = ICL (Intact WPA) / V = 201822.9 / (170 x 25 x 7) = 6.784m

New KM = New KB + New BMT = 3.562 + 6.784 = 10.346m

New GM = New KM – KG = 10.346 – 5.80 = 4.546m
Midship compartment WITH W/T flat CLOSE to the WL and bilged BELOW the W/T flat.
Q. A box shaped vessel 210m long and 25m wide floats in salt water at draft 10m F & A. KG
10.6m. An empty amidships compartment 30m long and 25m wide has a watertight flat
10.5m above the keel. Find the GM if this compartment is bilged below the watertight flat.

Midship compartment WITHOUT W/T flat bilged (With permeability)

A box shaped vessel is 60m long, 10m wide and floats at a draft of 4m F & A. A compartment
amidships is 15m long contains cargo of permeability 40%. Find the new drafts and new KM,
if the compartment is bilged.
Ans: Sinkage = 0.444m, New draft = 4.444m, KB = 2.222, BM = 1.875, KM = 4.097m

Midship compartment WITH W/T flat and bilged BELOW the flat (With permeability)
A box shaped vessel 80m x 16m at an even keel draft of 4.5m, KG 5.8m, has a midship
compartment 22m in length with a watertight flat 3m above the keel. If that compartment
with permeability of 65% is bilged below the flat, calculate her drafts and GM.
Ans: F & A draft = 5.036m, KB = 2.639m BM=4.741m GM = 1.579m

Midship compartment WITH W/T flat and bilged ABOVE the flat (With permeability)
Q. A box shaped vessel of length 80m, breadth 15m at an even keel draft of 4m has a
midship compartment 12m in length with a watertight flat 3m above the keel. The midship
compartment with a permeability of 70% is bilged above the watertight flat. Calculate the
drafts F & A and GM after bilging if KG in initial condition is 5.1m.
Solution:

Lost buoyancy = µv = 0.7 (12 x 15 x 1) = 126m³

Intact WPA = A - µa = [ (80 x 15) – (0.7 x 12 x 15) ] = 1074m²

Sinkage = Volume of lost buoyancy = 126 = 0.117m

Area of intact waterplane 1074

New draught = 4.0 + 0.117 = 4.117m

To find KB
Volume KB Moment
80 x 15 x 4.117 = 4940.4 2.0585 10169.8
– 12 x 15 x 1.117 x 0.7 = – 140.74 3.5585 – 500.8
4799.66 9669.0

KB = 9669.0 = 2.0145 m
4799.66

ICL (Intact WPA) = [ (L – µl) B³ ]/12 = { [ 80 – (0.7 x 12) ] x 15³ } /12 = 20137.5

BM = ICL (Intact WPA) / V = 20137.5 / (80 x 15 x 4) = 4.195m

New KM = New KB + New BM = 2.0145 + 4.195 = 6.2095m

New GM = New KM – KG = 6.2095 – 5.10 = 1.1095m

Midship compartment WITH W/T flat CLOSE to the WL and bilged BELOW the W/T flat.
Q. A box shaped vessel 210m long and 25m wide floats in salt water at draft 10m F & A. KG
10.6m. An empty amidships compartment 30m long and 25m wide has a watertight flat
10.5m above the keel. Find the GM if this compartment is bilged below the watertight flat.
Solution:

Lost buoyancy = v = (30 x 25 x 10) = 7500 m³

Intact WPA = A - a = [ (210 x 25) – (30 x 25) ] = 4500 m²

c = 10.5 – 10.0 = 0.5m
Recoverable volume up to level of flat (v1) = [ (A - a) x c ] = 4500 x 0.5 = 2250 m³
v > v1
So WL is above the flat

Buoyancy still to recover (v2) = 7500 – 2250 = 5250 m 3

Additional Sinkage = Volume of buoyancy still to recover = 5250 = 5250 = 1.0 m
whole WPA (210 x 25) 5250
New draught = Ht of flat + Addnl sinkage = 10.5 + 1.0 = 11.50m

To find KB
Volume KB Moment
210 x 25 x 11.5 = 60375 5.75 347156.25
– 30 x 25 x 10.5 = – 7875 5.25 – 41343.75
52500 305812.5

KB = 305812.5 = 5.825 m
52500

BM = LB³ = 210 x 25³ = 5.208m

12V 12(210 x 25 x 10)

KM = KB + BM = 5.825 + 5.208 = 11.033m

GM = KM – KG = 11.033 – 10.600 = 0.433m

Midship compartment WITH W/T flat CLOSE to the WL and bilged BELOW the W/T flat.
( with permeability)
Q. A box shaped vessel 210m long and 25m wide floats in salt water at draft 10m F & A. KG
9.90m. An amidships compartment 30m long and 25m wide has a watertight flat 10.55m
above the keel. Find the GM if this compartment which has a permeability of 30% is bilged
below the watertight flat.
Solution:

Lost buoyancy(v) = µv = 0.3 (10 x 30 x 25) = 2250 m³

Intact WPA = A - µa = [ (210 x 25) – (0.3 x 30 x 25) ] = 5025 m²
c = 10.55 – 10.0 = 0.55m
Recoverable volume up to level of flat (v1) = [ (A - µa) x c ] = 5025 x 0.55 = 2763.75 m³
v1 > v
So the WL is below the flat.

Sinkage = Volume of lost buoyancy = 2250 = 0.448m

Area of intact waterplane 5025

New draught = 10.0 + 0.448 = 10.448m

New KB = ½ new draft = ½ x 10.448 = 5.224m

ICL (whole WPA) = LB³/12 = (210 x 25³) / 12 = 273437.5

ICL (lost WPA) = (lb³/12) µ = [ (30 x 25³)/12] x 0.3 = 11718.75

ICL (Intact WPA) = ICL (whole) - ICL (lost) = 273437.5 – 11718.75 = 261718.75m

BM = ICL (Intact WPA) / V = 261718.75 / (210 x 25 x 10) = 4.985m

New KM = New KB + New BM = 5.224 + 4.985 = 10.209m

New GM = New KM – KG = 10.209 – 9.90 = 0.309m

Midship compartment WITH W/T flat CLOSE the WL and bilged BELOW the W/T flat.
(With permeability)
Q.A box shaped vessel, KG 5.8m, is 82m long, 14m wide and 7m deep is floating at an
even keel draught of 4.2m. There is a deep tank midships 15m long and 4.36m deep
extending over the full breadth of the vessel and tightly stowed with cargo of relative
density 1.04 and stowage factor 1.40m3/t.
Calculate the new draught and GM if the tank is bilged.
Solution:

Solid Factor = 1 = 1 = 0.96 m3/t.

R.D. 1.04
Permeability = SF of Cargo – Solid Factor = 1.40 – 0.96 = 0.314
SF of Cargo 1.40
Lost buoyancy = µv = 0.314 (4.2 x 14 x 15) = 276.9 m³
A - µa = [ (82 x 14) – (0.314 x 15 x 14) ] =
c = 4.36 – 4.2 = 0.16m

Recoverable volume up to level of flat = [ (A - µa) x c ] = 1082.1 x 0.16 = 173.1 m 3

Buoyancy still to recover = 276.9 – 173.1 = 103.8 m3
Sinkage = Volume of lost buoyancy = 103.8 = 0.090m
Area of intact waterplane 82 x 14

Final draught = 4.360 + 0.090 = 4.450m

To find KB
Volume KB Moment
4.45 x 14 x 82 = 5108.6 2.225 11366.6
– 4.36 x 14 x 4.71 = – 287.5 2.180 – 626.75
4821.1 10739.85

KB = 10739.85 = 2.228m
4821.1

BM = LB3 = 82 x 14 x 14 x14 = 3.889m

12V 12 x 82 x 14 x 4.2

KM = KB + BM = 2.228 + 3.889 = 6.117m

GM = KM – KG = 6.117 – 5.800 = 0.317m

Midship Centre, P & S Compartments WITHOUT W/T flat and bilged *******
Q. A box shaped vessel 190m x 25m @ an E.K draft of 8m, KG 6.823m has a midships
compartment 15m in length divided in port, centre & stbd compartments, P & S
compartments are 6m wide and centre 13m wide.
Calculate GM of vessel (a) If only centre compartment is bilged.
(b) If only P & S compartments are bilged.
Midship compartment WITHOUT W/T flat and bilged and Ordinates of WPA are given
(Use Simpson’s rule) ********
Q. A wall-sided vessel of constant water-plane area of length 105m has the following equally
spaced half ordinates of water plane:
1.7, 3.4, 5.6, 6.7, 6.7, 5.6, 3.4, and 1.7m. When floating at an even keel draft of 4.5m, an
empty compartment with transverse end bulkheads at the 6.7m ordinates is bilged.
Calculate the end drafts at which she would settle.
Solution:

Solution:
h = 105/7 = 15m
½ ord SM Product of semi area ½ ord SM Product of semi area
1.7 1 1.7 6.7 1 6.7
3.4 3 13.6 5.6 3 16.8
5.6 3 11.2 3.4 3 10.2
6.7 1 6.7 1.7 1 1.7
SOP = 35.4 SOP = 35.4

Area 1 = 2 x 3h/8 x SOP Area 2 = 2 x 3h/8 x SOP

= 2 x 3 x 15/8 x 35.4 = 398.25m² = 2 x 3 x 15/8 x 35.4 = 398.25m²

Intact WPA = 398.5 x 2 = 796.5m²

½ ord SM POA
5.6 -1 -5.6
6.7 8 53.6
6.7 5 33.5
SOP = 81.5

Lost WPA = Area between 6.7 & 6.7 ords = 2 x h/12 x SOP = 2 x 15/12 x 81.5 = 203.75m²

Lost buoyancy = lost WPA x draft = 203.75 x 4.5 = 916.9m³

Sinkage = Lost buoyancy / Intact WPA = 916.9 / 796.5 = 1.151m

New draft = Old draft + sinkage = 4.5 + 1.151 = 5.651m