A ship’s water-plane is 80 metres long. The breadths commencing from forward are as follows: 0, 3.05, 7.1, 9.4, 10.2, 10.36, 10.3, 10.0, 8.84, 5.75 and 0m, respectively. The space between the first three and the last three ordinates is half of that between the other ordinates. Calculate the area of the water-plane, and the position of the centre of flotation.

A ship’s water-plane is 80 metres long. The breadths commencing from forward are as follows:

0, 3.05, 7.1, 9.4, 10.2, 10.36, 10.3, 10.0, 8.84, 5.75 and 0m, respectively.

 The space between the first three and the last three ordinates is half of that between the other ordinates. Calculate the area of the water-plane, and the position of the centre of flotation.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 10.8

 

Solution:

	Breadth	SM	f(A)	Levers(x)	f(m)
FOR	0	0.5	0	-3.5	0
	3.05	2	6.1	-3	-18.3
	7.1	1.5	10.65	-2.5	-26.625
	9.4	4	37.6	-2	-75.2
	10.2	2	20.4	-1	-20.4
	10.36	4	41.44	0	0
	10.3	2	20.6	+1	+20.6
	10.0	4	40	+2	+80
	8.84	1.5	13.26	+2.5	+33.15
	5.75	2	11.5	+3	+34.5
AFT	0	0.5	0	+3.5	0
			∑ = 201.55		∑ = 27.725

 

Common interval, h = 80/8 = 10 m

Waterplane area = h/3 * ∑f(A)

                           = 10/3 * 201.55

                           = 671.83 m²

 

Centre of flotation = ∑f(m)/∑f(A) * h

                               = 27.725/201.55 * 10

                               = 1.375 m