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If 200t cargo was shifted downwards by 10m and to starboard by 5m on a ship of W 10000t , KG 7.0m, KM 7.4m, find the list.
Solution :-
Given :
Cargo shifted (w) = 200t,
Distance = 10m,
d = 5m from CL to stbd
W = 10000 t ,
KG = 7.0m &
KM = 7.4m
| Ship’s wt | KG | VM |
| 10000 t | 7.0m | 70000 tm |
| 200t (shift) | 10m | (-) 2000 tm |
Final W =10000t Final VM= 68000 tm
We know that:
Final KG = (final VM)/ (Final W)
Final KG = (68000/10000)
= 6.8m
Final GM= (KM – KG)
= (7.4 – 6.8)
= 0.6m
Listing moment (LM) = (weight x distance )
= (200 x 5)
= 1000tm
tanθ = LM/(W x GM)
= 1000 /(10000 x 0.6)
= 9.46 degree
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