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Worked Example 1: Pool

• A box shaped vessel has a length of 50 metres, a beam

of 8 metres and a lightship draft of 2 metres. Thelightship KG is 3 metres.

•The vessel has a swimming pool, with the base of the

pool 4 metres above the keel. The pool is 15 metres

long, and 8 metres wide.

•Determine the effective GM of the vessel if the pool is

filled to a depth of 1 metre with fresh water.

.

Worked Example 1: Pool

• The lightship displacement can be found from the

geometry, draft and water density:

tonnes8201.0252850

∆=ρLBT

∆=ρ×

=×××

×

∇

.

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Worked Example 1: Pool

• To start with, we treat the fluid as a solid cargo. The

process of calculation is identical to loading a solidcargo:

tonnes120=1.000×1×8×15=MassCargo

metres5.4=2

1+4=KGCargo

2

DepthFluid+KeelAboveFluidOf Base=KGCargo

.

Worked Example 1: Pool

• A loading table can be used to find the solid KG of the

vessel:

24603820Lightship

3000940Totals

5404.5120Pool Water

MomentKGMassItem

m19.3=940

3000=

MassTotal

MomentTotal=KGSolid

.

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Worked Example 1: Pool

• The draft, BM,KB and GM can be found in the loaded

(solid) condition:

m2921.025850

940T

ρLB

∆T

.=××

=

×=

metres1512

292

2

T=KB .

.

== metres332292850

12

850I

=BM

3

.

.

=××

×

=∇

metres0.283.19-2.331.15GM

KG-BM+KB=GM

=+=

.

Worked Example 1: Pool

• The free surface effect can now be calculated:

metres68.0=1×940

1.000×12

8×15

=GMInLoss

n×∆

ρ×I=GMInLoss

2

3

2

F/SF/S

metres-0.40=0.68-0.28=GMFluid

GMInLoss-GMSolid=GMFluid

.

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Worked Example 1: Pool

• Clearly the vessel is dangerously unstable. Free surface

effect is strongly influenced by the beam of the freesurface. Reducing the beam by adding longitudinal

watertight baffles in the fluid. This is modelled by the n

value (not the tank length and beam).

metres17.0=2×940

1.000×12

8×15

=GMInLoss

n×∆

ρ×I=GMInLoss

2

3

2

F/SF/S

metres11.0=0.17-0.28=GMFluidGMInLoss-GMSolid=GMFluid

.