# Chapter 10 - Stability Calculation

### 1.** On a ship of 5000 t displacement , a tank is partly full of DO of RD 0.88. If the moment of inertia of the tank about its centreline is 242m4 , find the FSC .**

Solution:

W = 5000t, RD = 0.88, I = 242m4
** FSC = (i di /W )**
= (242 x 0.88 ) /5000
= 0.042m.

**2. If the tank in Question 1 was partly full of SW instead of DO, find the FSC**.

Solution:

RD = 1.025
** FSC = (i di/W )**
= (242 x1.025) /5000
= 0.049m.

**3. On a ship of W 6000 t, KM 7.4m, KG 6.6m, a double bottom tank of i 1200m4 is partly full of FW . Find the GM fluid .**

Solution:

W = 6000 t & KM = 7.4m, KG = 6.6m, i = 1200m4, RD = 1.00

**GM (solid) = (KM – KG)**
= (7.4 – 6.6)
= 0.8m

**FSC = (i di /W )**
= (1200 x 1) /6000
= 0.2m

Fluid GM = GM (solid) – FSC = (0.8 – 0.2) = 0.6m

**4. Given the following particulars, Find the GM fluid : W= 8800 t, tank of i = 1166 m4 is partly full of HFO of RD 0.95, KM 10.1m, KG 9.4m .**

Solution:

W = 8800 t & i = 1166m4, RD = 0.95, KM = 10.1 m & KG = 9.0m.

**GM (solid) = (KM – KG)**
= (10.1 – 9.0)
= 1.1mFSC = (i di /W )
= (1166 x 0.95)/ 8800
= 0.126m

GM fluid = GM (solid) – FSC = (1.1 – 0.126) = 0.974m.

**5. On a vessel of W 16000 t, NO.4 port DB tank 20m long and 8m wide is partly full of DW ballast of RD 1.010. Find the FSC.**

Solution:

W = 16000t, L = 20m & B = 8m,RD = 1.010

FSC = (i di / W)
** FSC =( LB3 X di )/( 12 x W)**
= (20 x 83 x 1.010)/(12 x 16000)
= 0.054m.

**6. A vessel has a deep tank on the starboard side 12m long 9m wide which is partly full of coconut oil of RD 0.72 . If W = 1200 t , KM = 9m and KG = 8.5m ,find the GM fluid .**

Solution:

L = 12m, B = 9m, RD = 0.72 W = 12000 t, KM = 9m & KG = 8.5m

GM (solid) = (KM – KG) = (9 – 8.5) = 0.5m

**FSC = (i di /W )**
= (LB3 x di ) / (12 x W )
= (12 x 93 x 0.72) / (12 x 1200)
= 0.044m

Fluid GM = GM (solid) -FSC = (0.5 – 0.044) = 0.456m.

**A vessel displacing 8000t , has a rectangular deep tank 10m long 8m wide and 9m deep full of SW . The KM is 7m and KG 6.2m. Find the GM when 1 / 3 of this tank as pumped out.**

Solution:

W = 8000t. L = 10m, B = 8m & D = 9m RD = 1.025, KM = 7m, KG = 6.2m

**Mass of the tank = (Volume x Density)**
= (L x B x D) x 1.025
= (10 x 8 x 9) x (1.025)
= 738t

According to question, 1/3 of water pumped out = (1/3 x738) = 246t

After pumping out the water the KG of the tank = (6 + 1.5) = 7.5m

Final W = 7754 t Final VM = 47755tm

**Final KG = (Final VM)/ (Final W)**
Final KG = (47755)/(7754)
= 6.159m

GM (solid) = (KM – KG) = (7.0 – 6.159) = 0.841m

FSC = (i di / W) = (LB3 x di)/ (12 x W) = (10 x 83 x 1.025)/(12 x 7754) = 0.05m

So, GM fluid = GM (solid) – FSC = (0.841 – 0.055) = 0.785m.