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From a ship of W 8000t KM 8.6m KG 8.0m some deck cargo was washed overboard KG 10m, 8m from the centre line .if the resultant list is 3 degree, find the quantity of cargo lost.
Solution:
| Ship’s wt | KG | VM | LM |
| 8000 t | 8 | 64000 | |
| X t | 10 | (-) 10X |
Final W = (8000 – X) t Final VM =(64000 – 10X)tm
We know that:
Final KG = (Final VM)/(Final W)
= (64000 – 10X)/(8000 –X)
Again, Final GM = (KM – Final KG)
= 8.6 – (64000 – 10X )/(8000 – X)
= 8.6 (8000 -X ) – (64000 – 10X) /(8000 –X)
= (68800 – 8.6X – 64000 + 10X)/(8000 – X)
= (4800 + 1.4X)/(8000 – X)
We know that:
Tan θ = (Final LM)/ (Final W x Final GM)
= 8X / ((8000 – X ) x (4800 + 1.4X)/(8000 –X))
= 8X /(4800 + 1.4X)
8X = tan 30 x (4800 + 1.4X)
= (251.56 + 0.073X)
(8X – 0 .073X) = 251.56
Hence, X = (251.56/7.927)
= 31.73t
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