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A ship of W 18000 t, KG 7.75m discharge 1500t (6.0m above the keel and 3m port of the centre line ) and loads 500t (10m above the keel and 4m port of the centre line ).
- Cargo was then shifted as follows ;
- 500t upwards 2m and to starboard 4m
- 800 t downwards 2m and to port 3m.
If the final KM is 8.935 m, find the list .
Solution:
Ship’s wt |
KG |
VM |
D |
LM |
| 18000t | 7.73m | 139500 | ||
| (-) 1500t | 6.0m | (-) 9000 | 3P | 4500(S) |
| (+) 500t | 10m | (+) 5000 | 4P | 2000(P) |
| 500t | 2m | (+) 1000 | 4S | 2000(S) |
| 800t | 2m | (-) 1600 | 3P | 2400(P) |
Final W = 17000 Final VM = 134900 tm FLM =2100(S)
We know that:
Final KG = (Final VM) / (Final W)
Final KG = (134900/17000)
= 7.935m
Final GM = (KM –KG)
= (8.935 – 7.935)
= 1.0 m
We know that :
Tan θ = Final LM/(Final W x GM)
= 2100 /(17000 x 1)
= 7 degree 2.6min
Question no.19. Sub question 3
Kg will change when being shifted from derrick to lower hold 2m above kg so GM will also change
Gm= 0.58 and list will be 1.60degree to starboard
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