The 1/2 ordinates of a waterplane 120 m long are as follows: Section AP 1/2 1 1.5 2 3 4 5 6 7 8 8.5 9 9.5 FP ½ ord 1.2 3.5 5.3 6.8 8.0 8.3 8.5 8.5 8.5 8.4 8.2 7.9 6.2 3.5 0 Calculate: (a) waterplane area (b) distance of centroid from midships.

 The 1/2 ordinates of a waterplane 120 m long are as follows:

Section	AP	1/2	1	1.5	2	3	4	5	6	7	8	8.5	9	9.5	FP
½ ord	1.2	3.5	5.3	6.8	8.0	8.3	8.5	8.5	8.5	8.4	8.2	7.9	6.2	3.5	0

Calculate:

(a) waterplane area

(b) distance of centroid from midships.

Reference: REED'S NAVAL ARCHITECTURE FOR MARINE ENGINEERS (4^th edition), Exercise – 3.4

 

Solution:

a)

	½ ord	SM	f(A)	Levers	f(m)
AP	1.2	½	0.6	-5	-3
½	3.5	2	7	-4.5	-31.5
1	5.3	1	5.3	-4	-21.2
1.5	6.8	2	13.6	-3.5	-47.6
2	8.0	1.5	12	-3	-36
3	8.3	4	33.2	-2	-66.4
4	8.5	2	17	-1	-17
5	8.5	4	34	0	0
6	8.5	2	17	+1	+17
7	8.4	4	33.6	+2	+67.2
8	8.2	1.5	12.3	+3	+36.9
8.5	7.9	2	15.8	+3.5	+55.3
9	6.2	1	6.2	+4	+24.8
9.5	3.5	2	7	+4.5	+31.5
FP	0	1	0	+5	0
			∑f(A) = 214.6		∑f(m) = +10

 

Common interval, h = 120/10 = 12 m

Waterplane area, A_w = 2 * h/3 * ∑f(A)

                                    = 2 * 12/3 * 214.6

                                    = 1716.8 m²

b)

Distance of centroid from midship, LCF = ∑f(m)/ ∑f(A) * h

                                                                 = 10/214.6 * 12

                                                                 = 0.559 m FOR