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A box-shaped vessel 18 x5 x2 m floats in DW of RD1.000 at a draft 1.4m. Calculate her percentage reserve buoyancy when she enters the SW .
Solution:
Volume of box shape vessel = (L x B x H )
= (18m x 5m x 2m)
RD of DW = 1.000 & Depth = 1.4m
Displacement at DW of RD 1.000 (W)
= (u/w volume)x (density)
=( Lx B x 1.4) x(1)
Displacement at SW (W1) = (u/w volume) x (density)
=(L x B x Draft) x (1.025)
Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as MASS.
W = W1
(L x B x 1.4) x (1) = (L x B x Draft) x (1.025)
Hence, Draft = (1.4/1.025)
= 1.365m
Total volume of the ship = (L x B x H)
= (18 x 5 x 2)
= 180m3
Again, U/w volume of SW = (L x B x draft)
=(18 x 5 x 1.365)
= 122.85m3
Above water volume = (180 -122.85)
= 57.15m3
Hence, RB % = (Above water volume/ Total volume) x 100
= (57.15/ 180) x 100
= 31.75%.
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?