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 
Sum of product of volume = 256.99x
Volume = 1/3 ⨯ 256.99 ⨯ 100/1.025
= 8357.398m^{3}
Sum of product of moment = 524.04 ⨯ 1/3 ⨯ 100/1.025 ⨯ 1
= 17041.95m^{4}
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
= 3200m^{3}
Total volume including appendage = (8357.398 + 3200) m^{3}
= 11557.398 m^{3}
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 waterplane 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 