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A ship of FWA 175mm goes from water of RD 1.006 to water 0f RD 1.018 . Find the amount of sinkage or rise.
Solution:
FWA = 175mm = 17.5cm
We can calculate:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.018 – 1.006) x 17.5 /(0 .025)
= (0.012 x 17.5) /(0 .025)
= 8.4cm
Here change in draft would cause rise of vessel.
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A ship’s stability data book gives her load displacement to be 18000 t and TPC to be 25. If she is now loading in DW of RD 1.018, by how much may her be loadline be immerse so that she would not be over loaded.
Solution :
Load displacement = 18000t,
TPC =25.
We know that:
FWA = W/(40 TPC)
= 18000/( 40 x 25 )
=18cm.
We can calculate:
Change in draft =
(Change in RD )x(FWA)
0.025
=(1.025 – 1.018 ) x 18 /0.025
=5.04cm
= .05m
Since, Her load line should immersed to 0.05 m so that she will not be loaded.
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?