C L DUBEY – EXERCISE – 02 : Simpson’s Rule

Water Plane area = 25/3 ⨯ 116
= 966.67 ⨯ 2
= 1933.33m²

TPC = 1933.33/100 ⨯ 1.025
= 19.816

Volume between ordinate of 9m & 9m =?
Area between 9m & 9m = 25/12 ⨯ [45 + 80 – 9] ⨯2 ⨯ 2
= 966.67m²

Volume between 9m & 9m = 966.67m²
Weight of water = 990.83 tonnes
Sinkage = W/TPC
= 0.5m

Therefore, final draft = 5.7m = (5.2 + 0.5)
Thrust on tank top = (Thrust due to draft – Thrust due to tank water)

Thrust = Depth ⨯ Density ⨯ Area
= (5.7 ⨯ 1.025 ⨯ 977.07) – (1 ⨯ 1.025 ⨯ 966.07)
= 5647.769 – 990.83

= 4656.939 tonnes

Q12. The half breadths of a ship’s water-plane at 12m intervals from aft are: 0.0, 3.3, 4.5, 4.8, 4.5, 3.6, 2.7 and 1.5m. The half-breadth, midway between the first two from aft is 2m. At the fore end is an appendage by way of a bulbose bow 4.5m long. Its area is 24m² and its GC 2m from forward extremity. Find the area of the water-plane and the position of the COF. 

Solution –

Ordinate	SM	Product	Lever	POM
0.0	0.5	0	0	0
2.0	2	4	6	24
3.3	1.5	4.95	12	59.4
4.5	4	18	24	432
4.8	2	9.6	36	345.6
4.5	4	18	48	864
3.6	2	7.2	60	432
2.7	4	10.8	72	777.6
1.5	1	1.5	84	126
		74.05		3060.6

Area = (2 ⨯ 12/3) ⨯ 74.05
= 592.4m²