Stability – I : Chapter 5

A ship of 16000t displacement and TPC is 20 floating in SW at a draft of 8.0m . Find her draft in FW .

Solution:

Displacement (W) = 16000 t
TPC = 20 & SW draft = 8.0m

We know that:

Displacement when in SW = ( L X B x draft) x  1.025
Displacement when in FW  =( L X B X D) x 1

It is understood that displacement of ship will remain constant , as displacement is independent of change in density ,(is referred as MASS).

So, (L x B x 8) x 1.025 =  (L x B x draft) x 1
Hence draft  = ( 8 x 1.025)
= 8.2 m

2nd  Method :

As we know that

FWA = (W/40TPC)
= 16000/(40 x 20)
= 20cm

We can calculated:

Change in draft = 

(Change in RD )x(FWA)
             0.025

= (1.025 – 1) x 20 / 0.025
= 20c
=0.2m

So new draft of ship in FW = (8.0 +0.2)
= 8.2 m

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Graduated from M.E.R.I. Mumbai (Mumbai University), After a brief sailing founded this website with the idea to bring the maritime education online which must be free and available for all at all times and to find basic solutions that are of extreme importance to a seafarer by our innovative ideas.

3 Comments

  • A ship loads in fresh water to her salt water marks and proceeds along a
    river to a second port consuming 20 tonnes of bunkers. At the second port,
    where the density is 1016 kg per cu. m, after 120 tonnes of cargo have been
    loaded, the ship is again at the load salt water marks. Find the ship’s load
    displacement in salt water.

  • A mariner Enters port with 17,700 tons displacement,KG=28 feet, and GG=1.05 feet.A piece of Heavy machinery which weighs 180 tons is lifted by shore based cranes and placed on death so that its centre of gravity is 47 ft above the ship Khel and 19 feet two star board of centerline what will be the final mean draught gvm and angle of list when loading is complete?

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