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A vessel floating in DW of RD 1.005 has the upper edge of her summer loadline in the water line to starboard and 50mm above the waterline to port . If her FWA is 180mm and TPC is 24 , find the amount of cargo which the vessel can load to bring her to her permissible draft .
Solution:
RD = 1.0
Starboard side = 0cm above by below summer draft
Port side = 50mm (5cm above the water line to summer draft)
Mean draft = (0+5 ) / 2
= 5/ 2
= 2.5cm above the water line
FWA = 180mm
= 18cm,
TPC = 24
As we know :
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.005 ) x 18 /0 .025
= (0.2 x 18) /0 .025
= 14.4cm
Total sinkage = (14.4 + 2.5) cm
= 16.9cm
TPC is SW = 24
TPC = (A / 100) x 1.025
= (24 x100)/ 1.025
A = 23.41 t/cm
Again ,
TPC in RD 1.005 = ( A / 100) x 1.005
= (23.41 x 1.005)
= 23.53t/cm
Cargo to load = sinkage x TPC
= 23.53 x 16.9
= 397.657 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?