**Final KG (Derrett Stability)**

**A ship of 3500 tonnes light displacement and light KG 6.4 m has to load 9600 tonnes of cargo. The KG of the lower hold is 4.5 m, and that of the tween deck is 9 m. The load KM is 6.2 m and, when loading is completed, the righting moment at 6 degrees of the heel is required to be 425 tonnes m. Calculate the amount of cargo to be loaded into the lower hold and tween deck, respectively. (Righting moment = W * GM * sin heel.)**

Let x tonne cargo need to load in Lower hold

So, (9600 – x) tonne cargo need to load in Twin deck

weight x * KG 4.5 = moment 4.5x

weight (9600 – x) * KG 9 = moment (9600 – x)9 = 86400 – 9x

weight 3500 * KG 6.4 = moment 22400

Total weight = x + 9600 – x + 3500 = 13100

Total moment = 4.5x + 86400 – 9x + 22400 = 108800 – 5x

Righting moment = W * GM * sin@

GM = 425 / (13100 * sin 6) = 0.3104 m

So, final KG = KM – GM = 6.2 – 0.3104 = 5.89 m

We know, final KG = final moment / final displacement

5.89 = (108800 – 5x) / (13100)

So, x = 7031.33 tonne cargo in Lower hold

And, 9600 – 7031.33 = 2568.7 tonne cargo in Twin deck

**Final KG**

**A ship of 5500 tonnes displacement has KG 5 m, and she proceeds to load the following cargo:**

**1000 tonne = KG 6 m700 tonne = KG 4 m300 tonne = KG 5 m,**

**She then discharges 200 tonnes of ballast, KG = 0.5 m. Find how much deck cargo (KG = 10 m) can be loaded so that the ship may sail with a positive GM of 0.3 meters. The load KM is 6.3 m.**

Let x tonnes deck cargo need to load

Load KM = 6.3 m

Final GM after loaded = 0.3 m

So, KG = 6.0 m

We know, Weight * KG = moment

1000 * 6 = 6000,

700 * 4 = 2800

300 * 5 = 1500

5500 * 5 = 27500

200 * 0.5 = 100

x * 10 = 10x

Total weight = 5500 + 1000 + 700 + 300 + x – 200 = 7300 + x

Total moment = 6000 + 2800 + 1500 + 27500 + 10x – 100 = 37700 + 10x

We know, Final KG = Final moment / Final displacement

6 = (37700 + 10x) / (7300 + x)

43800 + 6x = 37700 + 10x

4x = 6100

x = 1525 tonnes deck cargo need to load to bring the ship with a positive GM of 0.3 m. (Ans.)

**Derrett Stability**

**A ship has a displacement of 3200 tonnes (KG = 3 m and KM = 5.5 m). She then loads 5200 tonnes of cargo (KG = 5.2 m). Find how much deck cargo having a KG = 10 m may now be loaded if the ship is to complete loading with a positive GM of 0.3 m.**

KM = 5.5 m

GM = 0.3 m

Final KG after loaded condition = 5.2 m

Let x tonne deck cargo need to load

We know, weight * KG = moment

3200 * 3 = 9600

5200 * 5.2 = 27040

x * 10 = 10x

Final weight = 3200 + 5200 + x = 8400 + x

Final moment = 9600 + 27040 + 10x = 36640 + 10x

We know, Final KG = Final moment / Final displacement

5.2 = (36640 + 10x) / (8400 +x)

x = 1466.67 tonne. (Ans.)