Case – 1
Ship’s wt |
KG |
VM |
LM |
| 13000 | 8.0 | 104,000 | 1024.76 |
| 150 | 13.0 | + 1950 | 0 |
Final W= 13000t Final VM= 105950tm FLM =1024.76
We know that:
Final KG = (Final VM/Final W)
= (105950/13,000)
Final KG = 8.15m
Again we can calculate
Final GM = (KM- KG)
= (8.75 – 8.15)
= 0.6m
Now, Tan =FLM /(W x GM)
= 1024.76/(13000 x 0.6 )
= 7degree 48minute (S)
Case – 2
When the load is hanging over the portside of the ship with on outreach of 10m from CL.
In this case the GM will be same but listing moment will change.
Listing moment caused =( W x d)
= (150 x 15)
= 2250 tm (P)
Hence, Final LM = (2250 – 1024.76)
= 1225.24 P
Tanθ = 1225.24/(13000 x 0.6)
= 8.9 degree P
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