Ship’s wt |
KG |
VM |
LM |
| 10000 | 6.8 | 6800 | 437.44P |
| 100t | 18m | (+) 1800 | 0 |
Final W =10000t Final VM = 69800tm FLM=437.44 P
We know that:
Final KG = (Final VM/Final W)
= (74800/10000)
= 6.98m
Again , Final GM = ( KM – KG)
= (7.3 – 6.98)
= 0.32m
tanθ =LM/(W x GM)
= 437.44/(10000 x 0.32)
= 7.78 degree P
Case – 2
We know that :
List = ( 100 x 6)
= 600 tm (S)
Tan θ = (LM/ (W x GM)
= 600/(10000 x 0.32)
= 10.6 degree
Initial LM as we calculated = 437.44 tm(P)
LM caused = 600 tm (S)
Now resultant listing moment = (600 – 437.44)
= 162.56 tm S
Again,Tanθ = 162.56 /(10,000 x 0.32)
= 2.9 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