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A ship listed 8degree to port , displaces 12000t and has KM 7.54m and KG 6.8m . Find how many tonnes of SW ballast must be transfered from NO 2 port DB tank to NO2 stbd DB tank , to up right and vessel , if the tank- centre are 10m apart .
Solution :
List = 8 degree (P) ,
Displacement (W) = 12000t,
KM = 7.54m &
KG = 6.8m
We know that:
GM = (KM – KG)
= (7.54 – 6.80)
= 0.74m
Again, listing moment (LM)= (W x GM x tanθ)
= (12000 x 0.74 x tan80)
= 1248 tm
We know that ,
Listing moment (LM) = (W x D)
1248 tm = (W x 10m)
So, W = 124.8 t
Here to upright the vessel the lighting moment caused on particle must be same at stbd side. So we have to transfer 124.8t of sea water to the stbd side of DB tank to upright the vessel.
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