The immersed cross-sectional areas of a ship 120 m long, commencing from aft, are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate: (a) displacement (b) longitudinal position of the centre of buoyancy.

 The immersed cross-sectional areas of a ship 120 m long, commencing from aft, are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m². Calculate:

(a) displacement

(b) longitudinal position of the centre of buoyancy.

Reference: REED'S NAVAL ARCHITECTURE FOR MARINE ENGINEERS (4^th edition), Exercise – 3.11

 

Solution:

a)

	Cross-section	SM	f(vol)	levers	f(FM)
AFT	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	0
	104	2	208	+1	+208
	103	4	412	+2	+824
	97	2	194	+3	+582
	58	4	232	+4	+928
FOR	0	1	0	+5	0
			∑f(vol) = 2388		∑f(FM) = 412

 

Common interval, h = 120/10 = 12 m

Displacement  = h/3 * ∑f(vol) * 1.025

                        = 12/3 * 2388 * 1.025

                        = 9790.8 tons

b)

Center of buoyancy  = ∑f(FM)/ ∑f(vol) * h

                                  = 412/2388 * 12

                                     = 2.070 m FOR