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From the following details, Calculate the DWT available :- present free board: port 3.0m, starboard 2.9m in water of RD 1.020 FWA 200mm. TPC 30. Statutory summer free board 2.8m.
Solution :
For, Port Side = (3 -2.8) = 0.2m
= 20cm (above )
For , Starboard side =( 2.9 – 2.8) =0. 1m
= 10cm above
Hence ,Mean freeboard can be calculated as
= (20+ 10)/2
= 30 /2 cm
= 15cm available
As we know:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.020 ) x 20 /0 .025
= 4cm
Now, Total sinkage = (15 + 4)cm
= 19cm
TPC = 30( given)
TPC = (A/ 100) x density
30 = (A/100) x 1.025
A = 30/ 1.025
= 29.268m2
Again for calculating ,TPC at RD 1.020
= (A/100) x ( 1.020)
= 29.85 t/cm
Now , Cargo can be load =( TPC x sinkage)
= (29.85 x 19)
= 567.15 t
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?