Example 7.1. A ship displacing 10000 t in salt water has the following righting arm values

0 / 0 / 15 / 30 / 45 / 60 / 75 / 90
GZ (m) / 0 / 0.275 / 0.515 / 0.495 / 0.330 / 0.120 / -0.100

Find the new values of GZ following a rise in the centre of gravity by 0.25 m

Solution

 / GZ / sin / GG1sin / G1Z
0 / 0 / 0 / 0 / 0
15 / 0.275 / 0.259 / 0.065 / 0.210
30 / 0.515 / 0.500 / 0.125 / 0.390
45 / 0.495 / 0.707 / 0.177 / 0.318
60 / 0.330 / 0.866 / 0.217 / 0.113
75 / 0.120 / 0.966 / 0.241 / -0.121
90 / -0.100 / 1.000 / 0.250 / -0.350

The range of stability for the original loading condition

The range of stability for the new loading condition

The reduction in the range of stability is 83-67=16 degrees.

Example 7.2. In awarship displacing 10000 t a pallet of ammunition weighing 15 t was moved 8 m upwards and 5 m to starboard. Find the resulting change in the righting arm.

Solution

Change in the centre of gravity due to the vertical shift of weight

Change in the centre of gravity due to the horizontal shift of weight

Change in the righting arm

Example 7.3. Aship displacing 35000 t has the following righting arm values at a loading condition corresponding to KG=9.0 m.

0 / 0 / 15 / 30 / 45 / 60 / 75 / 90
GZ (m) / 0 / 0.86 / 2.07 / 2.45 / 1.85 / 0.76 / -0.50

Find the new values of GZ following a shift of 500 t weight 3.5 m up and 5 m across the deck.

Solution

Change in the centre of gravity due to the vertical shift of weight

Change in the centre of gravity due to the horizontal shift of weight

Change in the righting arm

/ GZ / / / G2Z
0 / 0 / 0 / 0.071 / -0.071
15 / 0.86 / 0.013 / 0.069 / 0.778
30 / 2.07 / 0.025 / 0.061 / 1.984
45 / 2.45 / 0.035 / 0.050 / 2.365
60 / 1.85 / 0.043 / 0.036 / 1.771
75 / 0.76 / 0.048 / 0.018 / 0.694
90 / -0.50 / 0.050 / 0 / -0.550

Example 7.4. Arectangular barge L=100 m long, B=12.5 m beam, and D=12 m deep floats at adraught of T=3.5 m. The height of the centre of gravity is given as KG=4 m. Find the righting arm value at a heel angle of 30 degrees.

Solution

The area of port and starboard sides

The moment with respect to centreline

Then the distance of the centre of buoyancy from centreline is

The moment with respect to keel

Then the distance of the centre of buoyancy from keel is

It can be shown that

The righting arm at 30 degrees of heel is

Example 7.5. Arectangular barge L=100 m long, B=12 m beam, and D=10 m deep floats at adraught of T=7.5 m. The height of the centre of gravity is given as KG=5 m. Find the righting arm value at a heel angle of 30 degrees.

Solution


The submerged area is the sum of 1, 2, and 3

The moment with respect to centreline

Then the distance of the centre of buoyancy from centreline is

The moment with respect to keel

Then the distance of the centre of buoyancy from keel is

The righting arm at 30 degrees of heel is

Example 7.6. Abarge with triangular sections L=100 m long, B=12 m beam, and D=10 m deep floats at adraught of T=6.5 m. The height of the centre of gravity is given as KG=3.75 m. Find the righting arm value at a heel angle of 30 degrees.

Solution

The inclined waterline can be represented by the following linear equation

The side of the barge can also be represented by another linear equation

Therefore the coordinates of A and B are as follows

The submerged area is

The moment with respect to centreline

Then the distance of the centre of buoyancy from centreline is

The moment with respect to keel

Then the distance of the centre of buoyancy from keel is

It can be shown that

The righting arm at 30 degrees of heel is

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