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 
Water plane Area = ( H x SOP )/ 3
= 1483.2m^{2}
We know that – TPC = A/100 ⨯ density
= 15.2 tonnes
FWA = W/40 ⨯ TPC
= 6690/40 ⨯ 15.2
= 11.00cm
Q3. A ship of length 300m floating at a draft of 20m has half ordinates of water plane as follows, commencing from the after perpendicular.
Station 
0 
1/2 
1 
2 
3 
4 
5 
5 1/2 
6 
½ ordinates 
0.1 
7.5 
10 
12 
12.3 
11.4 
8 
5.2 
1.0 
There is an appendage forward with an area of 2.8 m^{2}.
Find : (a) the area of the water Plane.
(b) The TPC in SW and
(c) The FWA at 20m draft.
Solution –
Half Ordinate 
SM 
Product 
0.1  0.5  0.05 
7.5  2  15 
10  1.5  15 
12  4  48 
12.3  2  24.6 
11.4  4  45.6 
8  1.5  12 
5.2  2  10.4 
1.0  0.5  0.5 
SOP = 171.15 
Area = 50/3 ⨯ 171.15
= 2852.5m^{2}
TCP = 29.23 t/cm
Q4. The area of a ship’s waterplanes, commencing from the load waterplane and spaced 1m apart, are 800, 760, 700, 600, 450 and 10sq.m. Midway between the lowest two waterplanes, the area is 180 sq.m. Find the load displacement in salt water and the height of centre of buoyancy above the keel.
Solution –
Area 
SM 
Product of volume 
Lever 
Product of moment 
800  1  800  0h  0 
760  4  3040  h  3040h 
700  2  1400  2h  2800 
600  4  2400  3h  7200h 
450  1.5  675  4h  2700h 
180  2  360  4.5h  1620h 
10  0.5  05  5h  25h 
SOP = 8680  SOM = 17385h 
Volume = 8680/3
= 2893.33
Displacement = 2965.67 tonnes
COG from Top = 2.00288m
COG from bottom = 2.99m
Q5. A wallsided vessel of constant waterplane area of length 105m has the following equally spaced halfordinates of water plane:
1, 7, 3.4, 5.6, 6.7, 6.7, 5.6, 3.4 and 1.7 mtrs. When floating at an even keel draft of 4.5m, an empty compartment with transverse end bulkheads at the 6.7m ordinates is bilged. Calculate the end draft at which she would settle.
Solution – Will be Uploaded Soon.
Q6. A wall sided double bottom tank of depth 1.5m has half ordinates of breadth as follows at 4m intervals commencing at the after bulkhead:
Station 
0 
1 
2 
3 
4 
5 
6 
½ ord mtr 
6.1 
6.0 
5.9 
5.7 
5.4 
4.9 
4.3 
The after bulkhead is 65m forward of the centre of floatation. Calculate the change of trim if the is filled with fuel oil density 0.95, MCTC 200 tm.
Solution – Will be Uploaded Soon.
Q7. The water plane areas of a ship at 1m intervals, commencing from the keel are as follows:
Draft (m) 
0 
1 
2 
3 
4 
5 
6 
Areas (sq.m) 
5850 
5885 
5900 
5915 
5943 
5975 
5995 
Calculate her KB and FWA at a draft of 6m.
Solution –
Draft 
Area 
SM 
Product 
Lever 
Product of moment 
0  5850  1  5850  0h  0h 
1  5885  4  23540  1h  23540h 
2  5900  2  11800  2h  23600h 
3  5915  4  23660  3h  70980h 
4  5943  2  11886  4h  47544h 
5  5975  4  23900  5h  119500h 
6  5995  1  5995  6h  35970h 
SOP = 106631  SOM = 321134h 
KB = 3.012m
Volume = 3554.667
W = 36432.258
TPC = A/100 ⨯ density of water
= 61.448
We can calculate – FWA = W/40 ⨯ TPC
= 14.82cm
Q8. The breadths of a bulkhead, at 3.0m intervals from the top, are : 19.2, 18.0, 17.1, 16.2, 14.4, 12.0, 9.3 and 6.0 m. Find the distance of its geometric cente from the top. (use Simpson’s I Rule for first five ordinates and II rule for remaining ordinates).
Solution –
Ordinate 
SM 
Product 
Lever 
moment 
19.2  1  19.2  0h  0 
18.0  4  72.0  3h  216 
17.1  2  34.2  6h  205.2 
16.2  4  64.8  9h  583.2 
14.4  1  14.4  12h  172.8 
SOP = 204.6  SOM = 1179.2h 
Moment = 3/3 ⨯ 1179.2
= 1179.2
Area = 3/3 ⨯ 204.6
= 204.6
Ordinate 
SM 
Product 
Lever 
Moment 
14.4  1  14.4  12h  172.8h 
12.0  3  36.0  15h  540h 
9.3  3  27.9  18h  502.2h 
6.0  1  6.0  21h  126h 
SOP=84.3  SOM = 1341h 
Moment = (3 ⨯ 3/8) ⨯ 1341
= 1508.625
Area = 9/8 ⨯ 84.3
= 94.83
Total moment = 2685.825
Area = 299.43
Therefore,
Geometric Centre = ( Total moment / Area)
= 8.97m from top.
Q9. A ship’s water plane is 216m in length. The half ordinates of the waterplane commencing from forward are as follows;
0.4, 4.4, 8.8, 11.0, 11.6, 11.8, 11.8, 11.6, 9.6, 7.0, 0.4 respectively.
The spacing between the first three and last three ordinates is half of the spacing between the other half ordinates. Calculate her waterplane area and the position of the CF with respect to the midlength.
Solution –
Ordinate 
SM 
Product 
Hours 
Moment 
0.4  0.5  0.2  0h  0h 
4.4  2  8.8  0.5h  4.4h 
8.8  1.5  13.2  h  13.2h 
11.0  4  44.0  3h  8.8h 
11.6  2  23.2  3h  69.6h 
11.8  24  47.2  4h  181.8h 
11.8  1.5  23.6  5h  118h 
11.6  4  46.4  6h  278h 
9.6  1.5  14.4  7h  100.8h 
7.0  2  14.0  7.5h  105h 
0.4  0.5  0.2  8h  1.6h 
SOP=235.2  SOM=967.8h 
Area = 2 ⨯ 27/3 ⨯235.2
= 4233.6m^{2}
COG from one end = 111.099
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]
= 130.375m^{3}
Weight = 131.94 tonnes
For moment = 0.75/12 [3 ⨯ 158 + 8 ⨯ 186 – 196]
= 50.203m^{4}
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
Semiordinate 
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 
Water Plane area = 25/3 ⨯ 116
= 966.67 ⨯ 2
= 1933.33m^{2}
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
= 966.67m^{2}
Volume between 9m & 9m = 966.67m^{2}
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 waterplane at 12m intervals from aft are: 0.0, 3.3, 4.5, 4.8, 4.5, 3.6, 2.7 and 1.5m. The halfbreadth, 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 24m^{2} and its GC 2m from forward extremity. Find the area of the waterplane 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
= 592.4m^{2}
Total area = 616.4 m^{2}
COG from aft = 41.33m
COG = (592.4 ⨯ 41.33) + (24 ⨯ 86.5)/616.4
= 43.08m from aft
Q13. Three semiordinate spaced at equal distance of 20m are 7.5, 11.8 and 15.8m. Calculate

Geometric centre of the area between first two and last two ordinates.

Amount of cargo that can be loaded on the deck area between first two ordinates, if the load density of the deck is 10t/m^{2 }.
Solution –
Area between first two ordinate
= 2 ⨯ (20/12) ⨯ [5 ⨯ 7.5 + 8 ⨯ 11.8 – 15.8]
= 387m^{2}
Area between last two ordinate
= 2 ⨯ (20/12) ⨯ [5 ⨯ 15.8 + 8 ⨯ 11.8 – 7.5]
= 553m^{2}
Moment between first two ordinate
= 2 ⨯ (20 ⨯ 20/24) [3 ⨯ 7.5 + 10 ⨯ 11.8 – 15.8]
= 4156.67 tm
Moment between first last two ordinate from aft
= 2 ⨯ (20 ⨯ 20/24) [3 ⨯ 15.8 + 10 ⨯ 11.8 – 7.5]
= 5263.335 tm

Geometric centre between first two = 4156.67/387
= 10.74m

Geometric centre between last two = 5263.335/553
= 9.518m
Q14. The TPC values for a ship at 1.2 meters intervals of draught commencing at the keel, are 8.2, 16.5, 18.7, 19.4, 20.0, 20.5, 21.1 respectively. Calculate her displacement at 7.2 metres draught.
Solution –
TPC 
Ordinate 
SM 
Product of volume 
8.2  8.2x  1  8.2x 
16.5  16.5x  4  66x 
18.7  18.7x  2  37.4x 
19.4  19.4x  4  77.6x 
20.0  20.0x  2  40.0x 
20.5  20.5x  4  82.0x 
21.1  21.1x  1  21.1x 
332.2x 
TPC = (A/100) x density
A = TPC ⨯ (100/density)
X = 100/1.025
V/W volume = (1.2/3) ⨯ 332.3 ⨯ 100/1.025
= 12967.804