Solution Naval Numerical 14

Numerical 14: The immersed cross-sectional area of a ship 120m long, commencing from aft are 2,40,79,100,103,104,104,103,97, 58 and 0 m2 calculate – (i) Displacement; (ii) Longitudinal position of the centre of buoyancy.

Solution: 

Cross section area	SM	Product for Volume	Lever	Product for first moment
2	1	2	-5	-10
40	4	160	-4	-640
79	2	158	-3	-474
100	4	400	-2	-800
103	2	206	-1	-206
104	4	416	0	∑mA = -2130
104	2	208	1	208
103	4	412	2	824
97	2	194	3	582
58	4	232	4	928
0	1	0	5	0
		∑⛛ = 2388		∑mF =2542

Common interval (h) =120/10 = 12m
(i) Displacement = ρ x (h/3) x ∑⛛ = 1.025 x (12/3) x 2388 = 9790.8 tonne.
(ii) Centre of buoyancy from midships = {h x (∑mA + ∑mF)}/∑⛛
=12 x (-2130+2542)/2388
=2.07m forward of midships.