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A ship of W 9000t, KG 8.3m loads 600t of cargo (KG 4.0m, 3m to port of the centre line ) and discharge 400t of cargo ( KG 9.0m, , from 5m to port of the centre line ). 200t of cargo is then shifted upwards by 5m and to starboard by 8m. 300t of cargo is then is then shifted 1m downwards and 4m port .Find the list if the final KM is 8.95m.
Solution :
Ship’s wt |
KG |
VM |
d |
LM |
| 9000t | 8.3 | 74700 | ||
| (+) 600t | 4.0 | (+) 2400 | 3(P) | 1800 |
| (-) 400t | 9.0 | (-) 3600 | 5(P) | 2000 |
| 200 t | 5 | (+) 1000 | 8 (S) | 1600 |
| 300 t | 1 | (-) 300 | 4(P) | 1200 |
Final W = 9200 t Final VM = 74200 tm FLM = 600(S)
Final KG = (Final VM)/(Final W)
Final KG = (74200)/(9200)
= 8.065m
Final GM = (KM – KG)
= (8.95 – 8.065)
= 0.885m
tanθ = (600 )/(9200 x 0.885)
= 4 degree ( S)
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