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

Q1. The TPC of a ship are as follows:

The displacement and KB at 2m draft are 3280t and 1.2m respectively. Find the displacement and KB at 6m draft in SW.

Solution –

Given,

Sum of product of moment = 524.04 ⨯ 1/3 ⨯ 100/1.025 ⨯ 1
= 17041.95m⁴

Total volume including appendage =?
Given displacement of appendage = 3280 tonnes at draft 2m
We know displacement = ( v/w volume X  density of water displaced)
Volume of appendage = 3280/1.025
= 3200m³

Total volume including appendage = (8357.398 + 3200) m³
= 11557.398 m³

COG of the structure without appendage
= volume/Moment
= 2.039m

COG from 2.0 mtr draft = 2.039 m
Now total moment = moment of appendage + moment of structure
11846.33 ⨯  X   = (3280 ⨯ 2) + (8566.33 ⨯ 4.039)

 X = 3.253m

Q2. The breadths of a ship’s water-plane 120m long, measured at equal intervals from aft, are: 1.2, 9.6, 13.2, 15.0, 15.3, 15.6, 15.6, 14.7, 12.9, 9.0 and 0.0m respectively. Find:

• The water plane area;

• TPC in sea water;

• FWA if displacement is 6690t.

Solution –

Water plane Area = ( H x SOP )/ 3
= 1483.2m²