A ship of 12000 tonnes displacement is 150 metres long at the waterline. The half-ordinates of the waterplane at equidistant intervals from forward are as follows: 0, 4, 8.5, 11.6, 12.2, 12.5, 12.5, 11.6, 5.2, 2.4 and 0.3m, respectively Calculate the longitudinal and transverse BM’

A ship of 12000 tonnes displacement is 150 metres long at the waterline. The half-ordinates of the waterplane at equidistant intervals from forward are as follows:

0, 4, 8.5, 11.6, 12.2, 12.5, 12.5, 11.6, 5.2, 2.4 and 0.3m, respectively

Calculate the longitudinal and transverse BM’

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 11.7

Solution:

				Transverse			Longitudinal
	½ ord (y)	SM (dx)	f(A)	Levers(x)	f(FM) xy*dx	f(SM) x2y*dx	Y2	Y3	f(FM) Y2dx	f(SM) Y3dx
AFT	0	1	0	-5	0	0	0	0	0	0
	4	4	16	-4	-64	256	16	64	64	256
	8.5	2	17	-3	-51	153	72.25	614.12	144.5	1228.25
	11.6	4	46.4	-2	-92.8	185.6	134.56	1560.89	538.24	6243.58
	12.2	2	24.4	-1	-24.4	24.4	148.84	1815.84	297.68	3631.69
	12.5	4	50	0	0	0	156.25	1953.12	625	7812.5
	12.5	2	25	+1	+2.5	2.5	156.25	1560.89	312.5	39.6.25
	11.6	4	46.4	+2	+92.8	185.6	134.56	1560.89	538.24	6243.58
	5.2	2	10..4	+3	+31.2	93.6	27.04	140.60	54.08	281.21
	2.4	4	9.6	+4	+38.4	153.6	5.76	13.824	23.04	55.29
FOR	0.3	1	0.3	+5	+1.5	7.5	0.009	0.027	0.09	0.027
			∑ = 245.5		∑ = -43.3	∑ = 1084.3			∑ = 2597.37	∑ = 29658.4

 

Volume = 12000/1.025 = 11707.32 m³

Common interval, h = 150/10 = 15 m

I_CL = 2 * h/3 ∑ Y³dx

       = 2 * 15/3 * 29658.4

       = 98861.34 m⁴

BM_T = I_CL/Volume

          = 98861.34/11707.32

          = 8.45 m

LCF_midship = ∑ xy*dx / ∑f(A) * h

                     = -43.3/245.5 * 15

                     = 2.645 m FOR

Waterplane area = 2 * h/3 * ∑f(A)

                             = 2 * 15/3 * 245.5

                             = 2455 m²

I_midship = 2 * h³/3 * ∑ x²y*dx

            = 2 * 15³/3 * 1084.3

            = 2439675 m⁴

BM_L = I_CF/volume

         = I_midship - AX²/volume

         = 2439675 – 2455*2.645²

         = 206.92 m