-
A vessel’s statutory freeboard is 2.0m. She is loading in DW of RDD 1.015 and her freeboard is 2.1m . TPC =24. FWA = 200mm. find the DWT available .
Solution:
Statutory freeboard is = 2.0m
DW RD = 1.015 & Freeboard = 2.1m
TPC = 24, FWA = 200mm = 20cm
As we know :
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.015 – 1.025 ) x 20 /0.025
= ( 0.010 x 20) /0 .025
= 8cm
Hence , actual freeboard available =( 2.1 + 0.08)m
= 2.18m
Sinkage available = ( 2.18 – 2.00)
= 0.18m
= 18cm
TPC = 24 ( Given)
24 = (A / 100) x(1.025)
A = (24 x 100)/ 1.025
A = (24 / 1.025)
= 23.414 m2
Now, TPC for RD 1.015 = (A/ 100) x(1.015)
= (23.414 x 1.015)
= 23.76 t/cm
DWT available = (sinkage x TPC)
= (18 x 23.76 )
= 427.68t.
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?