Case – 3
Ship’s wt |
KG |
VM |
LM |
| 10000 | 6.8 | 68000 | 437.44 P |
| 100 | 8 | – 800 | 600.00 S |
Final W= 10000t Final VM= 67200tm FLM= 162.56 S
We know that:
Final KG = (Final VM/Final W)
= (67200/10000)
= 6.72m
Now, Final GM = (KM – KG)
=(7.3 – 6.72)m
= 0.58m
Again, tanθ = Final LM/ (W x GM)
= 162.56 /(10000 x 0.58)
= 1.6 degree (S)
-
A ship of W 13000t, KM 8.75m, KG 8.0m, has the list of 6degree to starboard . A heavy-lift weighing 150t lying on the upper deck 9m above the keel and 5m stbd of the centre line , is to be discharge using the ship’s jumbo derrick whose head is 22m above the keel . Calculate
- The list as soon as the load is taken by the derrick.
- When the load the hanging over the port side of the ship with an outreach of 10m from the centre line .
- After discharging the heavy-lift.
Solution:
Given :
Displacement (W) = 13000t,
KM = 8.75m &
KG = 8.0m
Initial GM =0.75m
List = 6 degree stbd ,
w = 150t,
KG = 9m lying = 5m to stbd ofC
derrick = 22m
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