bilging

Copyright 2007 Southampton Solent University 1

WARSASH M

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E ACADEMY SOUTHAMPTON SOLENT UNIVERSITY

WARSASH M

ARITIM

E ACADEMY S

OUTHAMPTON SOLENT UNIVERSITY

Warsash Maritime Academy

Bilging

Jonathan Ridley

Bilging

AIMS:

At the end of this section you should be able to:

Calculate the change in draught of a vessel at the LCF

is a compartment is bilged.

.


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Bilging

Bilging is the term used to describe a situation where a

compartment is open to the sea, so that water is free to

flow in and out of the compartment.

This is usually due to severe damage.

There are several methods of determining GM of a

vessel after bilging.

The standard IMO method is known as the lost

volume method.

For reasons that will be clearer later, this name is a bit

misleading

.

Bilging

All bilging problems are solved in these stages:

.

Start

Find parallel

sinkage

Find KB and TCB

or KB and LCB

Find BM or BML

Find GM or GML

Find list or trim


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Parallel Sinkage

Consider a box shaped vessel, as shown below, KG=3m,

with a watertight compartment at amidships:

100m

10m2m

20m

.

Parallel Sinkage

If the amidships compartment is bilged, the vessel will

sink lower into the water:

100m

10m2m

20m

.


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Parallel Sinkage

The compartment which has bilged is effectively no

longer part of the vessel we can now consider the

vessel to look like this:

.

Parallel Sinkage

The vessel has sunk as a result of losing buoyancy from

the bilged compartment. As the vessel sinks down, the

underwater volume, and hence buoyancy, increase

again, until the vessel is in equilibrium.

This happens when the volume gained is equal to the

volume lost.

100m

10m2m

20m

.


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Parallel Sinkage

100m

10m2m

20m

.

Parallel Sinkage

.


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Parallel Sinkage

.

100m

10m

2.5m

20m

100m

10m

2m

20m

Worked Example 1

A box shaped vessel has a length of 80 metres, a beam

of 9 metres and a draught of 4 metres. A full beam

amidships compartment, with a length of 10 metres, is

bilged. Determine the final draught of the vessel.

.


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Student Question 1




bilged. Determine the final draught of the vessel.

.

( ) ( )( ) ( )

3.529m0.5293 Draught Final

0.529m89860

389Sinkage

BLBL

TBLSinkage

tCompartmen BilgedtCompartmen BilgedVesselVessel

tCompartmen BilgedtCompartmen Bilged

=+=

=

=

=

Try this in your own time please ask for help if you need it

Student Question 2



amidships compartment, is bilged. The final draught is

2.857m. Determine the length of the compartment.

.

( ) ( )( ) ( )

( ) ( )15m L

0.857m10L500

20L

0.857m10L1050

210LSinkage

BLBL

TBLSinkage

tCompartmen Bilged

tCompartmen Bilged

tCompartmen Bilged

tCompartmen Bilged

tCompartmen Bilged



=

=

=

=

=



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Permeability

The theory so far assumes that the compartment

completely floods.

In practice, this may not happen, as solid objects such

as machinery, cargo and structure cannot flood.

This is modelled using a factor known as compartment

permeability.

This is a decimal or percentage value which tells us

how much of the compartment floods.

For example, a permeability of 0.6 means 60% of the

compartment can flood.

Permeability has the symbol .

.

Permeability

Permeability affects both the volume flooded and the

waterplane area lost.

This means that we have to adjust the parallel sinkage

formula:

.


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Worked Example 2


of 6 metres and a draught of 1 metre. A full beam


bilged. Determine the final draught of the vessel if the

permeability is 95%.

.

Worked Example 3




bilged. Determine the final draught of the vessel if the

permeability of the compartment is 0.82.

.


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Student Question 3



amidships compartment, 8 metres long, is bilged. The

final draught is 3.302m. Determine the permeability of

the compartment.

.

( ) ( )( ) ( )

( ) ( )( )

80%or 0.80

24.16-211.4240

10810700.3023108

0.302m1081070

3108Sinkage

BLBL

TBLSinkage



=

=

=

=

=

=


Parallel Sinkage

Remember bilging calculations are undertaken in

three stages. The first is parallel sinkage the change in

draught at the LCF.

Even in cases where the vessel will list or trim, parallel

sinkage is still the first stage.

The process of calculation is exactly the same as the

previous cases.

.


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Worked Example 4

100m

10m2m

20m

.

Parallel Sinkage

The parallel sinkage can be complicated by the addition

of watertight flats within compartments.

These restrict flooding vertically, for example, in a

double bottom.

These problems have to be analysed logically look at

the actual lost volume, and look at the final waterplane

area.

.


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Worked Example 5

The vessel shown below is bilged in the amidships

double bottom compartment. Determine the parallel

sinkage.

100m

10mInitial draught = 2m

20m

.

1.5m

Worked Example 5

100m


20m

.

1.5m


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Worked Example 5

100m


20m

.

1.5m

The first step is to determine the lost volume. This is

found from the bilged compartment dimensions.

Worked Example 5

100m


20m

.

The next step is to determine the final waterplane

area. You need to carefully consider which

compartments are dry:

1.5m


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Worked Example 5

100m


20m

.

1.5m

Student Question 4



sinkage.

50m


10m

.

0.7m



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50m


10m

.

0.7m

0.140m850

0.7810Sinkage

AreaWaterplane FinalSinkage LOST

=

=

=

Student Question 4




sinkage if the compartment permeability is 95%.

80m


15m

.

1m

Student Question 5



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Student Question 5

80m


15m

.

1m

0.178m1280

0.9511215Sinkage


=

=

=


Parallel Sinkage

When bilging, particularly as a result of damage from

collisions, it is possible to be in a situation where the

vessel is bilged in a compartment above a water tight

flat.

In situations like these, the approach is the same, but

care must be taken with the waterplane area.

.


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Worked Example 6


compartment, above the double bottom. Determine the

parallel sinkage.

100m


20m

.

1.5m

Worked Example 6

100m


20m

.

1.5m


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Worked Example 6

.



100m


20m

1.5m

Worked Example 6

.




100m


20m

1.5m


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Worked Example 6

.

100m


20m

1.5m



parallel sinkage.

80m


22m

.

2m

Student Question 6



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80m


22m

.

2m


=

Student Question 6


.

3

LOST 198m1922 ==



80m


22m

2m

Student Question 6



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.

2522m9)(229)(80 AreaWaterplane Final ==




80m


22m

2m

Student Question 6


.

0.379m522

198Sinkage


==

=

80m


22m

2m

Student Question 6



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parallel sinkage if the permeability of the compartment

is 77%.

70m


25m

.

3m

Student Question 7


.


=

70m


25m

3m

Student Question 7



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.

3

LOST 308m0.772825 ==



70m


25m

3m

Student Question 7


.

2406m0.77)8(258)(70 AreaWaterplane Final ==




70m


25m

3m

Student Question 7



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.

0.759m406

308Sinkage


==

=

70m


25m

3m

Student Question 7


Bilging

This is the end of the section. You should be able to:

Understand the effects of bilging amidships, end, full

beam and side compartments with permeability.

Be able to calculate the metacentric height of a vessel

with a bilged compartment.

.


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Bilging: Centres Of Buoyancy

AIMS:


Calculate the longitudinal, vertical and transverse

position of the centre of buoyancy of a vessel after

bilging a compartment

.

Stability After Bilging

Once a compartment has been bilged, it is important to

be able to determine both GM and GML in order to assess

the stability of the vessel.

GM does not always reduce after bilging!

Finding GM involves finding KB, BM and KG.

.


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KB After Bilging

Finding KB after bilging requires a table of moments of

volume.

The vessel, in its bilged condition needs to be broken

down into a series of rectangular blocks, and the centre

of each block determined.

Each of these blocks is then put into a table in a similar

way to a loading table, except volume and KB are used

instead of mass and KG.

.

Worked Example 7

Determine KB for the vessel shown below:

.

80m

9mBilged draught =

3.379m

22m

2m


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Worked Example 7

A good starting point is to draw the vessel from the

side, in the bilged condition:

.

Worked Example 7

Then label the intact compartments:

.

80mBilged T = 3.379m

22m

2m


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Worked Example 7

Calculate the volume (length x beam x depth) of each

of the intact compartments:

.

80mFinal T = 3.379m

22m

2mAFT

MIDFWD

Fwd

Mid

Aft

Volume (m3)DepthBreadthLengthCompartment

Worked Example 7

Calculate the position of the vertical centre of volume

of each of the intact compartments:

.

80mFinal T = 3.379m

22m

2mAFT

MIDFWD

Fwd

Mid

Aft

Centre Above Keel

Base Above Keel

Half DepthDepthCompartment


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Worked Example 7

Use the centres and volumes of each compartment to

find the overall centre:

.

Fwd

Totals

Mid

Aft

Moment (m4)

Centre Above Keel (m)

Volume (m3)Compartment

Determine KB for the vessel shown below:

.

90m

11mBilged draught = 5m

15m

3m

Student Question 8



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of the intact compartments:

.

AFTMID

FWD

2062.551137.5Fwd

49531115Mid

2062.551137.5Aft


90mFinal T = 5m

15m

3m

Student Question 8




.

AFTMID

FWD

2.502.55Fwd

1.501.53Mid

2.502.55Aft

Centre Above Keel

Base Above Keel


90mFinal T = 5m

15m

3m

Student Question 8



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.

5156.252.52062.5Fwd

110554620Totals

742.51.5495Mid

5156.252.52062.5Aft

Moment (m4)



2.393m4620

11055

VolumeTotal

MomentTotalKB ===

Student Question 8


KB After Bilging

If permeability is involved, then the calculation

becomes slightly more complex.

Permeability () is the measure of the amount of the

compartment which floods.

1- is the amount of buoyancy still remaining in the

compartment.

For example, if 95% ( =0.95) of the compartment

floods, 5% must still be providing buoyancy.

.


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Worked Example 8

Determine KB for the vessel shown below if the bilged

compartment has a permeability of 70%:

.

80m

9mFinal draught =

3.379m

22m

2m

Again, good starting point is to draw the vessel from

the side, in the bilged condition:

.

Worked Example 8


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Then label the intact and permeable compartments:

.

80mFinal T = 3.379m

22m

2m

Worked Example 8


of the intact and permeable compartments:

.

80mFinal T = 3.379m

22m

2mAFT

MIDFWD

Bilged

Fwd

Mid

Aft


Worked Example 8


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.

80mFinal T = 3.379m

22m

2mAFT

MIDFWD

Bilged

Fwd

Mid

Aft

Centre Above Keel

Base Above Keel


Worked Example 8



.

Bilged

Fwd

Totals

Mid

Aft

Moment (m4)



Worked Example 8


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.

90m

12mFinal draught = 4m

10m

1m

Student Question 9


Label the intact and permeable compartments:

.

AFTMID

FWD

90mFinal T = 4m

10m

1m

Student Question 9



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.

AFTMID

FWD

7231210Bilged

192041240Fwd

12011210Mid

192041240Aft


90mFinal T = 4m

10m

1m

Student Question 9




.

AFTMID

FWD

2.511.53Bilged

2024Fwd

0.500.51Mid

2024Aft

Centre Above Keel

Base Above Keel


90mFinal T = 4m

10m

1m

Student Question 9



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.

1802.572Bilged

384021920Fwd

79204032Totals

600.5120Mid

384021920Aft

Moment (m4)



1.964m4032

7920

VolumeTotal

MomentTotalKB ===

Student Question 9


KB After Bilging

If the compartment is a side or end compartment, the

process is exactly the same, but be careful with the

dimensions.

The fact that the vessel may trim or list as a result of

bilging is not important at this stage.

.


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Worked Example 9



.

80m

9mFinal draught =

3.379m

22m

2m



.

Worked Example 9


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.

80mFinal T = 3.379m

22m

2m

Worked Example 9



.

Bilged

Fwd

Aft


80mFinal T = 3.379m

22m

2mAFT

FWD

Worked Example 9


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.

Bilged

Fwd

Aft

Centre Above Keel

Base Above Keel


80mFinal T = 3.379m

22m

2mAFT

FWD

Worked Example 9



.

Bilged

Fwd

Totals

Aft

Moment (m4)



Worked Example 9


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KB After Bilging

If the bilged compartment runs all of the way from the

keel to the final waterline (ie, there are no watertight

flats) then there is a shortcut to finding KB for box

shaped vessels.

In these scenarios, KB is equal to half of the final

draught of the vessel.

.

LCB After Bilging

Finding the LCB after bilging is very similar to finding

KB, however the centres of each compartment are

measured from the aft perpendicular.

.


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Worked Example 10

Determine the LCB for the vessel shown below if the

bilged compartment has a permeability of 70%:

.

80m

9mFinal draught =

3.379m

22m

2m



.

Worked Example 10


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.

80mFinal T = 3.379m

22m

2mAFT

FWD

Worked Example 10



.

Bilged

Fwd

Aft


80mFinal T = 3.379m

22m

2mAFT

FWD

Worked Example 10


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Calculate the position of the longitudinal centre of

volume of each of the intact compartments:

.

Bilged

Fwd

Aft

Centre (m FOAP)

Aft End (m FOAP)

Half LengthLengthCompartment

80mFinal T = 3.379m

22m

2mAFT

FWD

Worked Example 10



.

Bilged

Fwd

Totals

Aft

Moment (m4)

Centre (m FOAP)Volume (m3)Compartment

Worked Example 10


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LCB After Bilging



flats), and is at the extreme end of the vessel, and there

is no permeability, then there is a shortcut to finding

LCB for boxed shaped vessels.

In these scenarios, the LCB is equal to half of the final

waterplane length of the vessel.

.

TCB After Bilging

Finding the TCB after bilging is very similar to finding

KB, however the centres of each compartment are

measured from the centreline of the vessel.

Distances to port are treated as positive, distances to

starboard are treated as negative.

.


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In the bilged condition, the draught of the vessel

below is 4 metres. Determine the TCB if the

permeability is 70%:

.

100m

10m

8m

2m

Worked Example 11

The first stage is to sketch the vessel from above, and

break the vessel down into simple rectangular sections:

.

Worked Example 11


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The volumes of each section are then found:

.

100m

10m

8m

2m

AFT

MID

FWD

Bilged

Width (m)

Fwd

Mid

Aft

Volume (m3)Draught (m)Length (m)Section

Worked Example 11

The centre of each section from the damaged edge is then found:

.

100m

10m

8m

2m

AFT

MID

FWD

Width (m)Width (m)

Bilged

Fwd

Mid

Aft

Centre From Damaged Edge (m)

Worked Example 11


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.

The overall centre of volume is found in a moment

table, in a similar way to KB and KG:

Bilged

Fwd

Volume (m3)

Totals

Mid

Aft

Moment (m4)Centre From Damaged Edge (m)

Section

Worked Example 11

.

This value is from the damaged edge of the vessel. TCB

should be quoted from the centreline:

Worked Example 11


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TCB After Bilging



flats) then there is a shortcut to finding TCB for box

shaped vessels.

In these scenarios, TCB is equal to the centre of area of

the waterplane, which will be found later as part of the

BM calculations.

.

Bilging

AIMS:

This is the end of this section. You should be able to:

Calculate the longitudinal, vertical and transverse

position of the centre of buoyancy of a vessel after

bilging a compartment

.


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Bilging: BM and BML

AIMS:


Calculate the BM and BML for a bilged vessel.

Combine sections of the theory to calculate the GM and GML for a bilged vessel.

.

BM After Bilging

To determine GM, BM is required.

Again, this needs consideration as to what the final

waterplane area looks like.

If the bilged compartment runs right across the vessel,

then determining BM is straight forward.

.


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BM After Bilging

.

Length

Beam

Compartment Length

Bilged

Intact

What is the difference in the underwater volume

before and after bilging?

The amidships compartment on the following vessel is

bilged. Determine BM after bilging.

100m

10m2m

20m

.

Worked Example 12


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After bilging, the final waterplane area will be as

shown:

.

Worked Example 12


bilged. Determine BM after bilging if the permeability is

70%.

80m

12m3m

10m

.

Worked Example 13


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shown:

Worked Example 13

BM After Bilging

This process is identical if the full beam compartment

is amidships, at the ends or anywhere along the length

of the vessel.

For side compartments, the process is a bit different.

When a side compartment is bilged, the final

waterplane is asymmetric about the centreline of the

ship.

The ship rolls around an axis which runs through the

transverse centre of the waterplane area.

.


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BM After Bilging

.

Bilged

Intact

BM After Bilging

We cannot directly calculate the transverse inertia

through the new roll axis, as the shape of the

waterplane is irregular.

To solve this we need to calculate the inertia about the

edge of the waterplane, and then use parallel axes

theory to determine the inertia through the centre of

the waterplane (which is the new roll axis).

This is best shown by example

.


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.

100m

10m

20m

Bilged

Intact

4m

A vessel with an initial draught of 2 metres is bilged in

a side compartment as shown. Find BM:

Worked Example 14

.

Bilged

Intact

The first stage is to determine the position of the

centre of the new waterplane area. This is done by

breaking the waterplane down into simple rectangular

sections:

Worked Example 14


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.

100m

10m

20m The areas of each section are then found:

4m

AFT

MID

FWD

Fwd

Mid

Aft

Area (m2)Width (m)Length (m)Section

Worked Example 14

.

100m

10m

20m The centre of each section from the damaged edge is then found:

4m

AFT

MID

FWD

Half Width (m)

Fwd

Mid

Aft


Distance From Damaged Edge (m)

Width (m)

Worked Example 14


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.

The overall centre of area is found in a moment table,

in a similar way to KB and KG:

Fwd

Area (m2)

Totals

Mid

Aft


Section

Worked Example 14

.

100m

10m

20m

Bilged

Intact

4m

This means that the actual centre of the final

waterplane, and hence the roll axis, is 5.261m from the

damaged edge:

5m

Worked Example 14


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.

The next stage is to determine the inertia through the

new centre. We cannot do this directly, but we can find

it about the edge, and use the parallel axes theory to

convert it.

To do this, we need to know the inertia of the

waterplane about one edge - this will be a remote

inertia.

For a rectangle, this can be found by formula:

Worked Example 14

.

100m

10m

20m

4m

The inertia about the damaged edge for the whole

vessel can be found:

Worked Example 14


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.

The parallel axes theory can now be used to convert

this remote value to be the centre value:

Worked Example 14

.

This finally allows BM to be found:

Worked Example 14


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a side compartment as shown. Find BM:

.

80m

8m

10m

Bilged

Intact

2m

Student Question 10


The areas of each section are found:

.

80m

8m

10m

2m

AFT

MID

FWD

280835Fwd

60610Mid

280835Aft


Student Question 10



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.

80m

8m

10m

2m

AFT

MID

FWD

4

3

4

Half Width (m)

408Fwd

526Mid

408Aft



Width (m)

Student Question 10


.



11204280Fwd

620

60

280

Area (m2)

2540Totals

3005Mid

11204Aft


Section

4.097m620

2540

AreaTotal

MomentTotalCentre Overall ===

Student Question 10



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.

80m

8m

10m

2m

433

EDGE

33

EDGE

13626.667m3

2101

3

880Inertia

3

Beam tCompartmenLength tCompartmen

3

BeamLengthInertia

=

=

=

Student Question 10


.



( )( )

4

CENTRE

2

CENTRE

2

CENTREREMOTE

3219.713mInertia

4.097620Inertia13626.667

DistanceAreaInertiaInertia

=

+=

+=

Student Question 10



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1.677m3880

3219.713InertiaBM Centre =

=

=

Student Question 10



a side compartment as shown. Find BM if the

compartment permeability is 82%:

.

80m

8m

10m

Bilged

Intact

2m

Student Question 10



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.

80m

8m

10m

2m

AFT

MID

FWD

3.6210Bilged

280835Fwd

60610Mid

280835Aft


Student Question 10



.

8m

10m

2m

AFT

MID

FWD

1012Bilged

4

3

4

Half Width (m)

408Fwd

526Mid

408Aft



Width (m)

Student Question 10



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3.613.6Bilged

11204280Fwd

623.6

60

280

Area (m2)

2543.6Totals

3005Mid

11204Aft


Section

4.079m623.6

2543.6

AreaTotal


Student Question 10




.

80m

8m

10m

2m

433

EDGE

33

EDGE

13631.467m3

2100.82

3

880Inertia

3

Beam tCompartmenLength tCompartmen

3

BeamLengthInertia

=

=

=

Student Question 10



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( )( )

4

CENTRE

2

CENTRE

2

CENTREREMOTE

3255.860mInertia

4.079623.6Inertia13631.467

DistanceAreaInertiaInertia

=

+=

+=

Student Question 10


.


1.272m4880

3255.860InertiaBM Centre =

=

=

Student Question 10



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BM After Bilging

If the bilged compartment runs all the way vertically

from the keel to the final waterline, then for a box

shaped vessel the TCB will be the same as the centre of

the waterplane.

This can be used to simplify the calculations for the

centre of the waterplane.

.

BML After Bilging

This process of finding BML is a little similar to finding

BM for the side compartment scenario.

BML requires the longitudinal inertia to be known at the

LCF.

This cannot be directly calculated, and so parallel axes

theory must be used to determine the value based on

remote values that can be calculated.

Again, this is best shown by example

.


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bilged. Determine BML after bilging.

100m

10m2m

20m

.

20m

Worked Example 15

The first stage is to determine the shape of the final

waterplane and the position of the LCF. This is similar to

the transverse case, but distances are found relative to

the AP of the vessel:

.

Bilged

Intact

Worked Example 15


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.

Fwd

Aft


100m

10m

20m 20m

Aft Fwd

Worked Example 15

The centre of each section from the AP is then found:

.

Half Length (m)

Fwd

Aft

Centre From AP (m)

Aft Bulkhead From AP (m)

Length (m)

10m

20m 20m

Aft Fwd

100m

Worked Example 15


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Fwd

Area (m2)

Totals

Aft

Moment (m3)Centre From AP (m)

Section

Worked Example 15

When the LCF is known, the distance from the LCF to

the centre of each section must be found:

.

20m 20m60m LCF

Aft Fwd

Worked Example 15


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.

The next stage is to find the longitudinal inertia

through the centre of each section. These values are the

inertia through the centre of each section:

Fwd

Beam (m)

Aft

InertiaL CENTRE (m4)

Length (m)Section

Worked Example 15

.

Then the parallel axes theory can be found to find the

inertia of each section through the overall LCF of the

vessel. These are remote inertia values:

Area (m2)

Fwd

InertiaCENTRE(m4)

Total

Aft

InertiaREMOTE(m4)

Distance (m to LCF from section

centre)

Section

Worked Example 15


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.

The total of the remote values is the total longitudinal

inertia measured at the LCF. This can finally be used to

find BML:

Worked Example 15


bilged. Determine BML after bilging.

100m

10m3m

20m

.

30m

Student Question 11



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.

100m

10m

20m

Bilged

Intact

30m

Aft Fwd

Student Question 11



.

3001030Fwd

5001050Aft


100m

10m

20m 30m

Aft Fwd

Student Question 11



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.

15

25

Half Length (m)

857030Fwd

25050Aft

Centre From AP (m)


Length (m)

10m

20m 30m

Aft Fwd

100m

Student Question 11


.



2550085300Fwd

800

500

Area (m2)

38000Totals

1250025Aft


Section

FOAP 47.5m800

38000

AreaTotal


Student Question 11



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.

20m 30m50m LCF

47.5m

Aft Fwd

25m

85m

22.5m 37.5m

Student Question 11


.




225003010Fwd

10

Beam (m)

104166.6750Aft


Length (m)Section

12

BLInertia

3

L =

Student Question 11



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300

500

Area (m2)

44437537.522500Fwd

104166.67

InertiaCENTRE(m4)

801666.67Total

357291.6722.5Aft

InertiaREMOTE(m4)


centre)

Section

( )2CENTREREMOTE DistanceAreaInertiaInertia +=

Student Question 11


.



find BML:

267.222m310100

801666.67InertiaBM LL =

=

=

Student Question 11



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bilged. Determine BML after bilging, if the compartment

permeability is 70%.

100m

10m3m

20m

.

30m

Worked Example 16





.

Worked Example 16


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.

Bilged

Fwd

Aft


100m

10m

20m 30m

Aft Fwd

Worked Example 16


.

Bilged

Half Length (m)

Fwd

Aft

Centre From AP (m)


Length (m)

10m

20m 30m

Aft Fwd

100m

Worked Example 16


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Bilged

Fwd

Area (m2)

Totals

Aft


Section

Worked Example 16



.

20m 30m50m LCF

Aft Fwd

Worked Example 16


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Bilged

Fwd

Beam (m)

Aft


Length (m)Section

Worked Example 16

.




Bilged

Area (m2)

Fwd

InertiaCENTRE(m4)

Total

Aft

InertiaREMOTE(m4)


centre)

Section

Worked Example 16


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find BML:

Worked Example 16

BML After Bilging

If the bilged compartment is at the end of the vessel,

and there is no permeability, then a shortcut can be

used to find BML.

.


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KG After Bilging

Finding KG after bilging is extremely simple.

The lost buoyancy method works by changing the

distribution of buoyancy of the vessel.

This in turn changes the stability of the vessel.

We have seen that overall the total displacement is

constant.

We can make the assumption that the distribution of

mass has not changed the water in the bilged

compartment is not technically adding any mass to the

vessel.

Although there may be a small change due to damage,

we can assume that KG REMAINS CONSTANT..

GM and GML After Bilging

As has been seen, for a variety of situations, KB, BM

and BML can be found, and as KG is constant, GM and

GML can be found.

These can be used to determine the transverse stability

and list, as well as longitudinal stability and trim, after

bilging.

.


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Bilging

AIMS:

This is the end of this section. You should be able to:

Calculate the BM and BML for a bilged vessel.

Combine sections of the theory to calculate the GM

and GML for a bilged vessel.

.

Bilging: List and Trim

AIMS:


Calculate the list and trim of a vessel after bilging.

.


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List After Bilging

When a compartment is bilged, the centre of gravity

remains constant.

The centre of buoyancy, which is the centre of the

underwater volume, moves transversely away from the

side compartment if a side compartment is bilged.

.

List After Bilging

The mass acts down from the centre of gravity, and the buoyancy acts up through the centre of buoyancy.

B

G

B

.


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List After Bilging

The vessel heels under the action of the misaligned forces:

G

B

.

List After Bilging

The vessel heels until the centres are again aligned:

G

B

.


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List After Bilging

The angle of list can be calculated:

.

The vessel shown below bilges the compartment shown.

Determine the transverse stability and the list if the permeability is 80%, and KG is 4 metres.

100m

10m2m

20m

.

3m

Worked Example 17


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.

100m

10m2m

20m

3m

Worked Example 17

.

As the bilging runs all the way from the keel to the

final waterline, the KB in the final condition will be half

draught in the final condition.

Worked Example 17


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The first stage is to sketch the vessel from above, and

break the vessel down into simple rectangular sections:

.

Worked Example 17

The volumes of each section are then found:

.

100m

10m

20m

3m

AFT

MID

FWD

Bilged

Width (m)

Fwd

Mid

Aft

Volume (m3)Draught (m)Length (m)Section

Worked Example 17


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.

100m

10m

20m

3m

AFT

MID

FWD

Width (m)Width (m)

Bilged

Fwd

Mid

Aft

Centre From Edge (m)

Worked Example 17

.

The overall centre of volume is found in a moment

table, in a similar way to KB and KG:

Bilged

Fwd

Volume (m3)

Totals

Mid

Aft


Section

Worked Example 17


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.

This value is from the damaged edge of the vessel. TCB

should be quoted from the centreline:

As the bilging runs all the way from the keel to the

final waterline, the centre of the waterplane in the final

condition will be the same as the TCB in the final

condition.

The area of the waterplane still needs to be found.

Worked Example 17


.

100m

10m

20m

3m

AFT

MID

FWD

Fwd

Bilged

Width (m)

Totals

Mid

Aft

Area (m2)Length (m)Section

Worked Example 17


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.

100m

10m

20m

3m

Worked Example 17

.



Worked Example 17


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.


Worked Example 17

.

Worked Example 17


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The angle of list can be found:

.

Worked Example 17

Trim After Bilging

When a compartment is bilged, the centre of gravity

remains constant.

The centre of buoyancy, which is the centre of the

underwater volume, moves longitudinally away from the

original position, which creates a longitudinal imbalance

between buoyancy and gravity.

The vessel trims as a result of this.

.


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Trim After Bilging

The trim can be found using:

.

The MCTC can be found using:

Worked Example 18

The vessel shown below bilges the compartment shown.

Determine the transverse stability and the end draughts

if the permeability is 90%, and KG is 1.4 metres.

100m

10m2m

20m

.

5m

1m


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100m

10m2m

20m

.

5m

1m

Worked Example 18



.

Bilged

Fwd

Mid

Aft


Worked Example 18


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.

Bilged

Fwd

Mid

Aft

Centre Above Keel

Base Above Keel


100mFinal T = 2.220m

20m

AFTMID

FWD1m

Worked Example 18



.

Bilged

Fwd

Totals

Mid

Aft

Moment (m4)



Worked Example 18


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.

Bilged

Fwd

Mid

Aft

Centre FOAP

Aft end FOAP

Half LengthLengthCompartment

100mFinal T = 2.220m

20m

AFTMID

FWD1m

Worked Example 18



.

Bilged

Fwd

Totals

Mid

Aft

Moment (m4)

Centre FOAP (m)Volume (m3)Compartment

Worked Example 18


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shown:

.

100m

10m2.22m

20m 5m

1m

Worked Example 18

.

Worked Example 18


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The areas of each waterplane section are found:

.

Bilged

Fwd

Aft


100m

10m

20m 5m

Aft FwdBilged

Worked Example 18


.

Bilged

Half Length (m)

Fwd

Aft

Centre From AP (m)


Length (m)

100m

10m

20m 5m

Aft FwdBilged

Worked Example 18


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.



Bilged

Fwd

Area (m2)

Totals

Aft


Section

Worked Example 18



.

75m LCF

10m

20m 5m

Aft FwdBilged

Worked Example 18


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.




Bilged

Fwd

Beam (m)

Aft


Length (m)Section

Worked Example 18

.




Bilged

Area (m2)

Fwd

InertiaCENTRE(m4)

Total

Aft

InertiaREMOTE(m4)


centre)

Section

Worked Example 18


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.


inertia measured at the LCF. This can be used to find

BML:

Worked Example 18

.

Worked Example 18


WARSASH M

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.

The values found so far can be used to determine the

MCTC:

Worked Example 18

.

The values found so far can be used to determine the

trim:

Worked Example 18


WARSASH M

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.

Worked Example 18

bilging

Documents

bilged compartment

box shaped vessel

parallel sinkagethe

watertight compartment

final draught

draught final0

lost volume method

underwater volume