**Q1. The TPC of a ship are as follows:**

**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.**

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

**Solution –**

**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**

**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:**

**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 –**

**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**

**= 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.**

**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}.

**There is an appendage forward with an area of 2.8 m**

^{2}.**Find : (a) the area of the water Plane.**

**Find : (a) the area of the water Plane.**

** (b) The TPC in SW and**

**(b) The TPC in SW and**

** (c) The FWA at 20m draft.**

**(c) The FWA at 20m draft.**

**Solution –**

**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**

**TCP = 29.23 t/cm**

**Q4. The area of a ship’s water-planes, commencing from the load water-plane and spaced 1m apart, are 800, 760, 700, 600, 450 and 10sq.m. Midway between the lowest two water-planes, the area is 180 sq.m. Find the load displacement in salt water and the height of centre of buoyancy above the keel.**

**Q4. The area of a ship’s water-planes, commencing from the load water-plane and spaced 1m apart, are 800, 760, 700, 600, 450 and 10sq.m. Midway between the lowest two water-planes, the area is 180 sq.m. Find the load displacement in salt water and the height of centre of buoyancy above the keel.**

**Solution –**

**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**

**Displacement = 2965.67 tonnes**

**COG from Top = 2.00288m**

** COG from bottom = 2.99m**

**Q5. A wall-sided vessel of constant water-plane area of length 105m has the following equally spaced half-ordinates 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.**

**Q5. A wall-sided vessel of constant water-plane area of length 105m has the following equally spaced half-ordinates 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:**

**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.**

**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:**

**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.**

**Calculate her KB and FWA at a draft of 6m.**

**Solution –**

**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

KB = 3.012m

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**

**= 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).**

**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 –**

**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.**

**= 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.**

**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 –**

**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.**

**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.**

**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 –**

**Solution –**

*Using Simpson’s 3 rules*

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

* = 130.375m ^{3}*

*Weight = 131.94 tonnes*

*Weight = 131.94 tonnes*

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

* = 50.203m ^{4}*

*KG = 0.385m*

*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.**

**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 –**

**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 |

**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**

**= 4656.939 tonnes**

**Q12. The half breadths of a ship’s water-plane at 12m intervals from aft are: 0.0, 3.3, 4.5, 4.8, 4.5, 3.6, 2.7 and 1.5m. The half-breadth, 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 water-plane and the position of the COF.

**Q12. The half breadths of a ship’s water-plane at 12m intervals from aft are: 0.0, 3.3, 4.5, 4.8, 4.5, 3.6, 2.7 and 1.5m. The half-breadth, 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 water-plane and the position of the COF.**Solution –**

**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}

**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**

**= 43.08m from aft**

**Q13. Three semi-ordinate spaced at equal distance of 20m are 7.5, 11.8 and 15.8m. Calculate**

**Q13. Three semi-ordinate 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 –**

**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.**

**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 –**

**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**

**Displacement = (v/w volume ⨯ density)**

** = 13291.9 tonnes**

**Displacement = (v/w volume ⨯ density)**

**= 13291.9 tonnes**