Case – 3
After discharging the heavy lift.
Ship’s wt |
KG |
VM |
LM |
13000 | 8.0 | 104000 | 1024.76(S) |
(-) 150t | 9.0 | (-) 1350 | 750 (P) |
Final W= 12850 Final VM = 102650 tm FLM = 274.76(S)
We know that:
Final KG = (Final VM/ final W)
Final KG = (102650/12850)
Final KG = 7.988m
Now,Final GM =( KM – KG)
=(8.75 – 7.988)
= 0.762m
We know that :
Tanθ = FLM/( W x GM)
= 274.76/(12850 x 0.762)
= 1.6 degree (S).
-
A ship of 10000t displacement is floating in SW and has KM of 10.8m and KG of 9.0m .she is listing 10degre to stbd. she has two rectangular deep tanks, one on either side , each 12m long ,12m wide and 9m deep . The stbd tank is full of FW while the port one is empty . If FW is to be transferred from the stbd tank to the port one , find
- The quantity of FW to transfer to bring the ship upright.
- The list if one third of the original FW in the stbd tank is transferred to the port tank.
NOTE : Fluid GM should be used here.
Solution :
Displacement(W) = 10000t,
KM = 10.8m, KG = 9.0M & GM = 1.8M
List = 10 degree stbd
Volume of tank = (L x B x H)
= (12 x 12 x 9 )m3
Initial LM = (W x GM x tan 100 )
= (10000 x 1.8 x tan 100)
= 3173.88tm
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