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A quantity of grain estimated to be 100t shifts transeversly by 12m and upwards by 1.5m, on a ship of W 12000t , GM 1.2m. Find the list caused.
Solution:
Grain shifted (w) = 100t,
Distance = 12m & transversely, d = 1.5( î)
W = 12000 t, GM = 1.2m
LM caused = (weight x distance)
= (100 x 12)
= 1200 tm
GG1 (î) = (distance x weight) /W
= (1.5 x 100) / 12000
= 0.0125 m
Final GM = (GM – GG1)
= (1.2 – 0.0125)m
= 1.1875m
Tanθ = LM /(W x GM )
= 1200/(12000 x 1.1875)
= 4 degree 8’.
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A ship displaces 4950t and has KG 4.85m, KM 5.79m. cargo weighing 50t is loaded 1.25m above the keel and 4m port of the centre line . Find the list
Solution:
W = 4950 t ,
KG = 4.85m
KM = 5.79m
Weight loaded (w) = 50 t
KG = 1.25m
d = 4 m to CL to port
LM caused = (weight x distance )
= (50 x 4)
= 200 tm to port .
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