Miscelleneous

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

Q1. The TPC of a ship are as follows:
Draft (m)
6.0 5.0 4.0 3.0 2.0
TPC
22.45 22.04 21.53 20.91 19.68
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,

Draft
TPC
Area (TPC= A/100 ⨯ density)
SM
Product
Lever
Moment
6.0 22.45 (100)⨯22.45/1.025 1 22.45x 4h 89.8xh
5.0 22.04 22.04x 4 88.16x 3h 264.48xh
4.0 21.53 21.53x 2 43.06x 2h 86.12xh
3.0 20.91 20.91x 4 83.64x 1h 83.64xh
2.0 19.68 19.68x 1 19.68x 0h 0

SSR 01


Sum of product of volume = 256.99x

Volume = 1/3 ⨯ 256.99 ⨯ 100/1.025
= 8357.398m3

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 –
Ordinate
SM
Product 
1.2 1 1.2
 9.6 4 38.4
13.2 2 26.4
15.0 4 60.0
15.3 2 30.6
15.6 4 62.4
15.6 2 31.2
14.7 4 58.8
12.9 2 25.8
9.0 4 36.0
0.0 1 0.0
    370.8

SSR 03

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

About the author

Manish Mayank

Graduated from M.E.R.I. (Mumbai). A cool, calm, composed and the brain behind the development of the database. The strong will to contribute in maritime education and to present it in completely different and innovative way is his source of inspiration.

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