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A ship of 14000t displacement , KM 9.0m , KG 7.8m has a total FSM of 2100tm and is listed 8degre to port . How many tonnes must be shifted transversely by 10m to upright the ship?
Solution :
W = 14000t ,
KM = 9.0m KG = 7.8m ,
Initial GM = 9 – 7.8 = 1.2
FSM = 2100 tm thit = 8degp,
d = 10m
FSC = 2100 / 1400
=0.15
Fluid GM = 1.2 – 0.15
= 1.05m
LM = W x GM tanthi
= 1400 x 1.05 x tan thit
= 2065.1tm
To upright the vessel we have to shift ballast from to which is to the same LM of on stbd
We know that
LM = W x D
2065.1 = W x 10
W = 2065.1 / 10 = 206.51t to ss
Q.7
Final listing moment is 700(p), not starboard
For case 2; since, both the tanks are slack, therefore, Total FSC = 0.3456m x 2 = 0.6912m.
Good luck
Ex -11 List Ques No. 18, Part 2 have a mistake in total Vertical movements…..it should be 87000 not 87200 which gives us a final answer as follows
List = 3* 51′ or 3.85 degrees
Q.7 listing moment is 700 port not stbd