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

COG from m/ship = 3.099.

Q10. A double bottom tank is 1.5m deep. The horizontal Areas of the tank at equal intervals, commencing from the tank top are 192, 186 and 158 sq.m. respectively. The tank is ballasted to a sounding of 0.75m with water of R.D. 1.012. Calculate the weight of ballast and its KG.

Solution –

Using Simpson’s 3 rules

Volume = 0.75/1 [5 ⨯ 158 + 8 ⨯ 186 – 196]
= 130.375m³

Weight = 131.94 tonnes

For moment = 0.75/12 [3 ⨯ 158 + 8 ⨯ 186 – 196]
= 50.203m⁴

KG = 0.385m

Q11. A wall sided vessel, 150m in length of constant waterplane, the semiordinates of which at equal intervals are : 0, 5, 9, 10, 9, 5 and 0m, has a double bottom 1m deep extending from side to side between the 9m and 9m half ordinates given. Calculate the thrust on the tanktop if the tank is run up by opening the sea valve, if her original evenkeel draft was 5.2m.

Solution –

Original even Keel draft = 5.2m

Semi-ordinate	SM	Product area
0	1	0
5	4	20
9	2	18
10	4	40
9	2	18
5	4	20
0	1	0
		SOP = 116