Solution Naval Numerical 19

Numerical 19: A ship of 15000 tonne displacement has righting levers of 0, 0.38, 1.0, 1.41 and 1.2 m at angles of hell of 0°, 15°, 30°, 45° and 60° respectively and an assumed KG of 7.0 m. The vessel is loaded to this displacement but the KG is found to be 6.80m and GM 1.5m – (i) Draw the amended stability curve; (ii) Estimate the dynamic stability at 60°.

Solution:

 

θ	sinθ	GG1sinθ	GZ	G1Z	SM	product
0	0	0	0	0	1	0
15	0.259	0.0518	0.38	0.43	4	1.72
30	0.500	0.100	1.0	1.1	2	2.2
45	0.707	0.1414	1.41	1.55	4	6.2
60	0.866	0.1732	1.2	1.37	1	1.37
						∑a=11.49

 

Dynamic stability is the area under the curve of statical stability;

To determine the area upto 60 degrees we already have the values

Common interval =h= 15 degrees 

or 15/57.3 radians

=0.26178 radians

Area under the curve=

=0.26178/3 x 11.49 

=1.0026178 unit 2

Dynamic stability = △ x g x area under the curve.

=15000 x 9.8 x 1.0026178

=147384.817 kN-m