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A vessel displaces 14500 tonnes, if floating in SW up to her winter load- line. If she is in a dock of RD 1.010 , with her winter load- line on the surface water , find how much cargo she can load, so that she would floats at her winter load-line in SW .
Solution:
Displacement (W )= 14500t
RD = 1.025
Displacement (W) =( u/w volume ) x (density)
= (L x B x draft) x (1.025)
Draft = W /( L x B x 1.025)
Let W1 be the displacement at RD of 1.010
So W1 = (u/w volume )x( density)
= (L x B x d1 ) x (1.010)
Hence, d1 = W1 / (L x B x 1.010)
According to question, Draft = d1
So, W / ( L x B x 1.025) = W1 / (L x B x 1.010)
14500 / (L x B x 1.025) = W1 / (L x B x 1.010)
W1 = (14500 x 1.010) / 1.025
= 14287.8 t
Hence, Cargo to load = (14500 – 14287.8)
= 212.2t.
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.
Joe Bee. Can you please send me the solved question of the above question you posted. please
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?