Stability 1 – Chapter 6 Centre of Gravity
- In a vessel of 8800 tonnes displacement and KG 6.2m, 200 tonnes of cargo was loaded in the lower hold 1.7m above the keel. Find the final KG.
Solution:
Given:-
Weight = 8800t,
Original KG =6.2m,
Cargo loaded = 200t,
KG = 1.7m
| Ship’s weight | KG | VM |
| 8800 t | 6.2m | 54560tm |
| (+) 200 load | 1.7m | Final VM 54900 tm |
Final W – 9000 t Final VM 54900 tm
So, we can calculate
Final KG = (Final VM/ Final W)
Final KG = (54900 / 9000)
= 6.1m
Here ‘VM’ stands for vertical moments.
- 600 tonnes of cargo were discharged from a vessel from the upper 11m above the keel. In the original KG displacement were 6m and 12600 tonnes, calculate the KG.
Solution:-
Given:-
Weight =12600 t
Original KG =6m
Cargo discharged = 600 t,
Cargo discharge 11 m above the keel
| Ship’s weight | KG | VM |
| 12,600t | 6m | 75,600tm |
| ( -) 600t ( discharge) | 11m | (-) 6600tm |
Final Weight =12000 t Final VM =69000tm
We know that :
Final KG = (Final VM / Final Weight)
Final KG = ( 69,000 / 12000)
= 5.75m
- In a vessel of 9900 tonnes displacement and KG 4m, a heavy lift of 100 tonnes is loaded on the UD (KG 15m ). Find the final KG.
Solution:
| Ship’s weight | KG | VM |
| 12,600t | 4m | 39,600tm |
| (+)100 t load | 15m | ( +)1500tm |
Final W 10000t Final VM =41000tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = ( 41000 / 10000 )
= 4.11m.
- 500 tonnes of cargo of discharge from the lower the lower hold (KG 3m ) of a vessel whose displacement and KG before discharging were 11500 tonnes and 6.3m . Find the final KG.
Solution:
Given:-
Weight = 11500t
Initial KG = 6.3m
Cargo discharged = 500t, KG =3m
| Ship’s weight | KG | VM |
| 11500t | 4m | 39,600tm |
| (+)100 t load | 15m | ( +)1500tm |
Final W =11000t Final VM =70950tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (70,950/11000)
=6.45m
- 500 tonnes of cargo was shifted 15 meters vertically downwards in a vessel of 10000 tonnes displacement. Find the effect it has on the KG of the vessel and whether KG increases or decreases.
Solution:
Given:-
Weight =10000t,
Cargo shifted downward =500t,
We know that:-
GG1() = (w x d)/( total weight)
= (500 x 15)/ 10000
=0.75m
Since, The weight is shifted vertically downward, the KG of ship will also go down parallel to the KG of the weight
Hence, KG of the ship decrease by 0.75m
- In a vessel of 9000 tonnes displacement , KG 10.5m, 300 tonnes of cargo was shifted from the LH (KG 2.5m)to the UDC (KG 11.5m). Find the resultant KG of teh vessel.
Solution :
Weight =9000 t
Original KG = 10.5m ,
Cargo shifted = 300 t
Vertical Distance of cargo shifted
= (height of UD – height of LH)
= (11.5 -2.5)
= 9m upwards.
| Ship’s weight | KG | VM |
| 9000t | 10.5m | 94500tm |
| (+)300t | 9m | (+)2700tm |
Final W = 9000 Final VM =200t
We know that :
Final KG = (Final VM / Final Weight)
Final KG = (97200 / 9,000)
= 10.8m.
- In a vessel of 9009 tones displacement, KG 8.7m, how many tonnes of cargo can be loaded on the upper deck (KG 15m ) so that the final KG would be comes 9m.
Solution:-
Given:-
Weight = 9009 t,
Original KG =8.7m,
Let cargo loaded be X tonnes
Final KG =9m
| Ship’s weight | KG | VM |
| 9009 t | 8.7m | 94500tm |
| (+) X t (load) | 15m | (+) (15 X) tm |
Final W = (9009 + X) t Final VM =(78378.3 + 15X) tm
We know that:
Final KG = (Final VM / Final Weight)
Final KG = (78378.3 +15X ) /(9009 + X)
9.0 = (78378.3 + 15X)/ (9009+X)
9 x (9009+X) = (78378.3 + 15X)
(81081 + 9X) = (15X + 78378.3)
Hence, 6X = 78378 .3 – 81081
6X = 2702.7t
X = 450.45 t