Chapter 2-Water Pressure Exercise 1 (Stability Calculation)

1. Find the thrust experienced by a flat keel plate 10m x 2m when the draft is 8m in SW.

Solution :

Area of the flat keel plate = (10 x 2) = 20m2

Pressure = (depth x density) =(8 x 1.025) =8.2t /m2

Thrust = (pressure x area) = 8.2 x 20 =164 t .

2. A box-shaped vessel 150m x20m x 12m is floating in a dock of RD 1.010 at an even keel draft of 10m.
Find the total water pressure experienced by the hull.

Solution:
The total water pressure exerted on the hull to the water = (Thrust of keel plate which acting horizontally +
Thrust of forward and aft which act vertically +Thrust of port and stbd side which act vertically) Thrust of Keel
plate = (pressure x area) Pressure = (depth x density) =10 x1.010 t/m2
Area = 150m x 20m = 3000m2 Thrust =(10 x1.010 x 3000) = 30, 300 t
Thrust of forward and Aft =(pressure x area)
Pressure = depth x density =( 5 x1.010)t/m2
Area =L X B =(20 x10) = 200m2
Thrust = (P x A) = (5×1.010 x200) = 1010t
Thrust acting both forward and aft = (1010 x2) = 2020 t
Thrust of port and stbd side = (pressure x area)
Pressure =( depth x density) =( 5×1.010)
Area = (L X B) = 150m x 10 = 1500m2
Thrust = ( P X A ) = 5 x 1.010 x1500 = 7575 t
Thrust acting on Both sides = 7575 x 2 =15150 t
Hence , Total pressure on the hull = (30300 +2020 + 15150) =47,470 tonnes.

3. A submarine has a surface area of 650m2 and can with stand a total water pressure of 1332500 t . Find
at what approximate depth in SW she would collapse.

Solution:
Area = 650m2 Thrust = 1332500t RD = 1.025 Depth =?
Thrust = ( P x A ) 1332500 = (P x 650) P = 2050 t/m2
Now, P =(depth x density) 2050 = (depth x 1.025)
Hence , Depth = 2000m.

4. A rectangular lock gate 40m wide and 20m high has water of RD 1.010 12m deep on one sideand water
of RD 1.020 11m deep on the other. Find the resultant thrust experienced and direction in which it acts.
Solution:
Water of RD =1.010 Depth =12 m C = (12/2) =6m
Pressure = (depth x density) =(6 x1.010) t/m2
Thrust =(P x A) =( 6 x 1.010 x40 x12) = 2908.8t
Water of RD = 1.020 Depth = 11m C = (11/2) =5.5m
Pressure =(depth x density) = (5.5 x 1.020 ) t/m2
Area =( L X B) =(40 x 11) = 440m2
Thrust =( P X A) = (5.5 x 1.020 x 440) = 2468.4 t
Thus we can Resultant thrust =(2908.8 – 2468.4) = 440.4t
5. A rectangular lock gate 36m wide and 20m high has FW on one side to a depth of 16m. Find what depth
of SW on the other side will equalize the thrust.
Solution :
Inside the lock gate area =( 36 x 16)m2
Pressure = depth x density =(16/2 x 1) = 8 t/m2
Thrust inside the lock gate = (pressure x area) Thrust =(8 x 36 x 16) = 4608 t
Let ‘X’ depth of salt water on other side will equalize the thrust .
Pressure = depth x density =( X/2 x 1.025) Area =(36 x X ) Thrust = (P x A)
4608 = (X/2 x 1.025 x 36X ) 4608 = (1.025 x 36X x X/2) 4608 =
(1.025 x 36X2/2) = ( 1.025 x 18X2) X2 = (4608 /1.025 x 18)
X = 15.8 m
Hence, Depth of other side of lockgate =15.8m

6. A collision bulkhead is triangular in shape. Its maximum breadth is 12m and its high 15m. Find the
thrust experienced by it if the fore peak tank is pressed up to a head of 3m of SW.
Solution :

Breadth of collision bulkhead = 12m Height = 15m Pressure = (depth x density) Thrust = (pressure x area) Breadth =
12m Height of the water inside the tank = 15m Water pressed up to ahead of = 3m ‘C’ inside the tank = (1/3 x15) = 5 m
Outside the tank = 3m so total =5 + 3 =8m
Now pressure =(depth x density) =( 8 x 1.025)
Area = (1/2 x 12 x15) = 90m2
Thrust experienced = (pressure x area) = (90 x 8 x 1.025) =738t.

7. A collision bulkhead is triangular shaped, having a breadth of 14m at the tank top and a height of 12m.
As a result of a collision, the forepeak tank gets ruptured and SW enters the tank to a sounding of 9m.
Calculate the thrust on the bulkhead.
Solution :
 B = 14m, H =12m As we know that : Pressure = (depth x density) Thrust = (pressure x area)
An right angle triangle PBC , PB =7m Since, Height divide AB equally into two parts. Triangle PBC and FEC are
similar So by law of similar angle triangle
PB/ FE = PC / FC 7 / FE =12 / 9
Hence, FE = (63 / 12) = 5.25m
In triangle DEC DF = FE DE = (2 X FE) = 2X 5.25 =10.5m
Area of triangle DEC = (1/2 x10.5×9) =47.25m2
Pressure = (depth x density) = (9 x 1/3 x 1.025) = 3.075 t/m2
Hence Thrust experienced = (P X A) = (3.075 x 47.25) =145. 29 t

8. A tank has a triangular bulkhead, apex upwards. Its base is 14m and its sides, 15m each. It has a
circular inspection hole of radius 0.5m the centre of a manhole 0.8m above the base and 1.6 from one
corner. Find the thrust on the manhole cover when the tank contain oil of RD 0.95 to a sounding of
10m.(Assume π to be 3.1416).
Solution:

Base = 14m , Sides = 15m , Depth of the oil = 10m

Pressure = (depth x density) = (10 – 0.8 x 0.95) = 8.74 t/m2
Area of the circle = π r2 = (3.1416 x 0.5 x 0.5) =0.7854m2
Thrust = (P X A) =( 8.74 x 0.7854) = 6.864 t.

9. A rectangular deep tank is 22m x 20m x10m. Above the crown of the tank is a rectangular trunkway
0.2m high, 5m long and 4m wide. Find the thrust on the tank lid when the tank is pressed up with SW to a
head of 2.64m above the crown of the tank .
Solution :
Rectangular deep tank = 22m x 20m x10m Trunk way = (5m x 4m x 0.2m) Pressure = (depth x density) = (2.64m – 0.2)
x1.025
Since , Depth is (2.64m – 0.2 ) and we have the calculate the thrust act on the trunkway and the water level
above the trunkway will act as thrust )
If this trunkway would have been placed inside the tank then the depth will be
(10m + 2.64m – 0.2/2m ).
Area of the trunkway = (L X B) = 4×5 =20 m2
Thrust experienced = (P X A) = (2.501 x 20 ) = 50.02tonnes.

10. A double bottom tank measures 25m x20m x 2m. Find the thrust on the tank top when pressed upto a
head 16m of SW. Also find the resultant thrust on the tank bottom, and the direction that it acts, if the
ship’s draft in SW is 10m.
Solution :

Volume of tank = (L X B X H) = (25m x 20m x 2m)

Area =(25 x20) = 500m2
Pressure = (depth x density) = (16 x 1.025) =16.4 t/m2
Thrust on tank top = (PX A) = (16.4 X 500) = 8200 t
Pressure on the tank bottom = (depth x density) = (18 x1.025)
= 8.45t/m2
Thrust act on the tank bottom = (P X A) = (500 x 18 .45) = 9225t
Pressure act on outside of the tank bottom = (depth x density) = (10 x 1.025) = 10.25t
Hence, Thrust acting = (P X A) = (10.25 X500) = 5125 tonnes.
Finally, Resultant thrust = (9225 -5125) = 4100 tonnes.

Chapter 2

1. A box-shaped vessel 120m long and 15m wide has alight draft of 4m and load draft of 9.8m in SW.
find her light displacement, load displacement and DWT.
Solution :
Area of box shape vessel = (L x B)
= 120m x 15m
Light draft = 4m

Light displacement = (u/w volume) x ( density)

= (L x b x d )x (1.025)
= (120 x 15 x 4) x (1.025)
= 7380 t
Load draft = 9.8m

So load displacement = (u/w volume) x (1.025)

= ( L x b x d) x (1.025)
= (120 x 15 x 9.8) x (1.025)
= 18081t

Hence, DWT = (Load displacement – Light displacement)

= ( 18081 – 7380 )
= 10, 701t

2. A box-shaped vessel 100m long and 14m wide is floating in SW at a draft of 7.6m. Her light draft
is 3.6 m and load draft 8.5m. Find her present displacement, DWT aboard and DWT available.

Solution:

Area of box shape vessel = (L x B)

= (100m x 14m)

Present draft = 7.6m

Present displacement = (Lx b x d) x (density)
= (100 x 14 x 7.6) x (1.025)
=10906 t
Light draft = 3.6m

Light displacement = (L x B x D) x(1.025)

= (100 x 14 x 3.6 ) x (1.025)
= 5166t

When Load draft is = 8.5m

Load displacement = ( L x b x d )x (1.025)

= (100 x 14 x 8.5) x (1.025)
= 12,197.5t

DWT aboard = (present displacement – Light displacement)

= (10906 t – 5166 t )
= 5740 t

DWT available = (Load displacement – present displacement)

= ( 12,197.5 t – 10,906 t )
= 1291.5t.

3. A ship is 200m long 20m wide at the waterline. If the coefficient of fineness of the water-plane is
0.8, find her TPC in SW, FW and DW of RD 1.015.

Solution :

Given : (L x B) = (200m x20) , Cw = 0.8

Cw = Area of water plane / ( L x B )
0.8 = Area of water plane / (200 x 20)
(0.8 x 200 x 20) = Water plane area

A = 3200m2

TPC in SW = (A/ 100 x 1.025)

= (3200 / 100 x 1.025)
= 32.8 t/cm

TPC in FW = (A / 100x 1 )
= (3200/100)
= 32 t/cm

TPC in RD of 1.015 =( A /100 x 1.015)

= (3200 /100 x 1.015)
= 32.48 t/cm.

Note : here “A” refers to water plane area.

4. A double bottom tank 20m x 10.5m x 1.0m has a block a coefficient of 0.82. Calculate how much
fuel oil of RD 0.9, it can hold .

Solution :

Given : – (L x B x H) of double bottom tank =( 20 x 10 .5 x 1)

Cb = 0.82 , RD = 0.95

Cb = (volume of tank) /( L x B x depth)

0.82 = (volume of tank) /(20 x 10.5×1)
Volume of tank = 0.82 x ( 20 x 10.5x 1)
= 172.2 m3

Weight of oil can be calculated as :- (volume of tank) x(density)

= (172.2 x 0.95)
= 163.59 t

5. A ship floating in SW at a draft of 8m is 110m long and 14m wide at the water line . If her block
coefficient is 0,72, find her displacement. If her load displacement is 12000 t, find the DWT
available .

Solution:

Given: – (L x B) = 110m x 14m,

Present draft = 8m
Cb = 0.72

Load displacement = 12000 t ,

Cb = (u/w volume ) / (Lx B x D)

0.72 = (u/w volume) / (110 x 14 x8)
u/w volume = 0.72 x ( 110 x 14 x 8 )
= 8870.4 t

DWT available = (Load displacement – present displacement)

= (12000 – 8870.4)
= 3129.6t.

6. A vessel of 14000t displacement is 160m long and 20m wide at the water line. If she is floating in
SW at a draft of 6.1m, find her block coefficient.

Solution:

Given : – W = 14000 t ,
(L x B) of waterline = (160m x 20m),
Present draft = 6.1m

Weight = (u/w volume )x (density of displaced water)

14000 = (u/w volume) x 1.025
u/w volume = 14000/ 1.025
= 13658.536 t

Cb = (u/w volume )/( Lx B x D)

Cb = 13658.536 /(160 x 20 x6.1)
= 0.699m
= 0.7m

7. A box-shaped vessel 18m x 5m x2m floats in SW at a draft of 1.4m. Calculate her RB %.

Solution :

Given :- (L x B x H) of box shape vessel =(18m x 5m x 2m), Present draft = 6.1mRB % = (Above water volume
)/(total volume ) x 100
Total volume can be calculated as = (L x B x D)
= 18 x 5x 2m
= 180m3

U/w volume = (L x B x present draft)

= 18 x 5 x 1.4
= 126m3

Hence, Above water volume = (180 – 126)

= 54m3

RB % = (Above water volume) / (total volume ) x 100

Hence, RB % = (543/180 x 100)
= 30%

8. A box –shaped of 2000t displacement is 50m x 10m x 7m. Calculate her RB% in FW

Solution :
Given :- W = 2000t,

(L x B x H) of box shaped vessel = (50m x 10m x 7m)

Total volume = 3500m3
W = ( u/w volume) x (density)
2000 = (u/w volume) x (1.00)
u/w volume = 2000 m3

Above water volume = (Total volume) – (u/w volume )

= (3500 – 2000)
= 1500 m3

RB % = (Above water volume) / (total volume ) x 100

RB % = (1500 /3500)x 100
= 42.857%

9. The TPC of a ship in SW is 30. Calculate her TPC in FW and in DW of RD1.018.

Solution:

TPC = (A / 100) x (density)

30 = (A / 100) x (1.025)
A = (30 x 100)/ 1.025
A = 2926.83 m2

TPC in FW can be determined as = ( 2926.83/ 100)x(1)

=29.268 t/cm.

TPC in DW can be determined as = (2926.83 / 100) x(1.018)

= 29.79 t/cm.

10. A ship is a floating a draft of 8.2m in DW of RD 1.010. If her TPC in SW is 40, find how much cargo
she can load to bring her draft in DW to 8.4m.

Solution :

Given:-
RD of DW = 1.010,
Present Draft = 8.2m
Cargo can be load up to draft = 8.4m,
So, Sinkage = 0.2m = 20cm

TPC in SW = (A / 100) x (1.025)

40 = (40/100) x (1.025)
A = (40 x100)/ (1.025)
= 3902.44m2

Now, TPC in DW = (A/100) x (1.010)

= (3902.44 /100) x(1.010)
=39.41 t/cm.
Cargo can be load = (39.41 x 20)
= 788.2 t

Chapter 3

1. A ship of 16000t displacement and TPC is 20 floating in SW at a draft of 8.0m . Find her draft in
FW .
Solution:

Displacement (W) = 16000 t

TPC = 20 & SW draft = 8.0m

Displacement when in SW = ( L X B x draft) x 1.025

Displacement when in FW =( L X B X D) x 1

It is understood that displacement of ship will remain constant , as displacement is independent of change in
density ,(is referred as MASS).

So, (L x B x 8) x 1.025 = (L x B x draft) x 1

Hence draft = ( 8 x 1.025)
= 8.2 m

2nd Method :

FWA = (W/40TPC)
= 16000/(40 x 20)
= 20cm

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1) x 20 / 0.025
= 20c
=0.2m

So new draft of ship in FW = (8.0 +0.2)

= 8.2 m

2. A ship goes from water of RD 1.008 to SW. Find the change in draft , if her FWA is 180mm, and
state whether it would be sinkage or rise.
Solution :

FWA = 180mm = 18cm

Change in draft =

(Change in RD )x(FWA)
0.025
= (1.008 – 1.025 ) x 18 / 0.025
=(0.017 x 18)/0.025
= 12.24c
= 0.12m

Here , Change in draft is 0.122m and it will be rise.

3. A vessel goes from water of RD 1.010 to FW. If her FWA is 160mm, State whether she would sink
or rise and by how much.
Solution :

FWA = 160mm = 16cm

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.010 – 1)x 16 /0.025

= 0.01 x 16 /0.025
= 6.4 cm
= 0.064 m

Here change in draft would lead to sinkage.

4. A ship of FWA 175mm goes from water of RD 1.006 to water 0f RD 1.018 . Find the amount of
sinkage or rise.
Solution:

FWA = 175mm = 17.5cm

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.018 – 1.006) x 17.5 /(0 .025)

= (0.012 x 17.5) /(0 .025)
= 8.4cm

Here change in draft would cause rise of vessel.

5. A ship’s stability data book gives her load displacement to be 18000 t and TPC to be 25. If she is
now loading in DW of RD 1.018, by how much may her be loadline be immerse so that she would
not be over loaded.
Solution :

Load displacement = 18000t,

TPC =25.
FWA = W/(40 TPC)
= 18000/( 40 x 25 )
=18cm.

Change in draft =

(Change in RD )x(FWA)
0.025

=(1.025 – 1.018 ) x 18 /0.025

=5.04cm
= .05m

Since, Her load line should immersed to 0.05 m so that she will not be loaded.

6. A box-shaped vessel 20x 4 x2 m has mean draft of 1.05m in SW. Calculate her draft in DW of RD
1.012.
Solution:

Volume of box shape vessel = (Lx B x H)

= (20 X 4 X 2)

Mean draft = 1.05m

Displacement = (u/w volume )x density

Again, displacement can be calculated as (W)

=(L x B x D) x(density)
= (20 x 4 x 1.05) x( 1.025)
=86.1t

Let ‘X’ be the displacement at RD of 1.012

So, X = (L x B x D) x ( 1.012)

Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as
MASS.
(20 x 4 x 1.05) x(1.025) = (20 x 4 x d) x(1.012)
D = (1.05 x 1.025) / 1.012
= 1.06m

7. A box-shaped vessel 18 x5 x2 m floats in DW of RD1.000 at a draft 1.4m. Calculate her percentage

reserve buoyancy when she enters the SW .
Solution:

Volume of box shape vessel = (L x B x H )

= (18m x 5m x 2m)

RD of DW = 1.000 & Depth = 1.4m

Displacement at DW of RD 1.000 (W)

= (u/w volume)x (density)
=( Lx B x 1.4) x(1)

Displacement at SW (W1) = (u/w volume) x (density)

=(L x B x Draft) x (1.025)
Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as
MASS.

W = W1
(L x B x 1.4) x (1) = (L x B x Draft) x (1.025)

Hence, Draft = (1.4/1.025)

= 1.365m

Total volume of the ship = (L x B x H)

= (18 x 5 x 2)
= 180m3

Again, U/w volume of SW = (L x B x draft)

=(18 x 5 x 1.365)
= 122.85m3

Above water volume = (180 -122.85)

= 57.15m3

Hence, RB % = (Above water volume/ Total volume) x 100

= (57.15/ 180) x 100
= 31.75%.

8. The hydrostatic particular of a ship indicate that her displacement in SW at a draft of 5m is 3000t.
Find her displacement when floating at 5m draft in water of RD 1.018.
Solution:

Displacement(W) = 3000t,
Draft= 5m & RD = 1.018

W = (u/w volume) x (density)

= (L x B x draft) x(1.025)
Draft = W/ (L x B x 1.025)

Let W1 be the displacement at RD = 1.018

W1 = ( L x B x d) x (1.018 )
D1 = W1/( Lx B x 1.018)

But according to question ,

Draft = d1
W/ (L x B x1.025 )= W1 /(L x B x 1.018)
3000 / 1.025 = W1 / 1.018
W1 = 3000 x (1.018 / 1.025)
= 2979.51 t

9. A vessel displaces 4500 t of FW at a certain draft . Find her displacement at the same draft in
water of RD 1.020 .
Solution:

Displacement (W) = 4500t,

Displacement = (u/w volume ) x (density)

= (L x B x draft) x 1.025

Draft = W / (L x B x 1.025) m

Let W1 be the displacement at RD 1.020

So W1 =( L x B x d )x (1.020)
D = (W1 / (Lx B x 1.020)

According to question, Draft = d

W /( L x B x 1.025) = W1 /( L x B x 1.020)
W1 = (4500 x 1.020) / 1.025
= 4478.04t.

10. A ship 100m long and 20m wide, block coefficient 0.8m, floats in SW at a mean draft of 8.0 m.
Calculate the difference in displacement when floating at the same draft in FW .
Solution:

Area of Waterplane = (L X B)
= (100m x 20m)

Cb = 0.8, Depth = 8m

Displacement = (u/w volume ) x (density)

= ( Lx B x draft ) x (0.8)
= (100 x 20 x 8) x(0.8)
= 12800m3

Displacement (W) = (u/w volume)x (density)

= (12800 x 1.025)
= 13120t ,

Draft = W / ( L x B x Cb x 1.025)

Let W1 be the displacement in FW

= (u/w volume) x (density)
=(L x B x d1 x Cb ) x ( density)

D1 = (W1 /( L x B x 1 )x (Cb)

According to question, Draft = d1

W /( L x B x Cb x 1.025) = ( W1 /( L x B x Cb)
W1 = ( 13120 x 1) /( 1.025)
= 12800t
So, Difference is displacement = (13120 – 12800)
= 320 t .

11. A vessel displaces 14500 tonnes, if floating in SW up to her winter load- line. If she is in a dock of
RD 1.010 , with her winter load- line on the surface water , find how much cargo she can load, so
that she would floats at her winter load-line in SW .
Solution:

Displacement (W )= 14500t
RD = 1.025

Displacement (W) =( u/w volume ) x (density)

= (L x B x draft) x (1.025)

Draft = W /( L x B x 1.025)

Let W1 be the displacement at RD of 1.010

So W1 = (u/w volume )x( density)

= (L x B x d1 ) x (1.010)
Hence, d1 = W1 / (L x B x 1.010)

According to question, Draft = d1

So, W / ( L x B x 1.025) = W1 / (L x B x 1.010)

14500 / (L x B x 1.025) = W1 / (L x B x 1.010)
W1 = (14500 x 1.010) / 1.025
= 14287.8 t

Hence, Cargo to load = (14500 – 14287.8)

= 212.2t.

12. A vessel of 12000 t displacement arrives at the mouth of a river , drawing 10.0 m in SW. how much
cargo must she discharge so that her draft in an up river port of RD 1.012 would be 10m .
Solution:

Displacement (W)= 12000t ,

Depth = 10m & RD = 1.025

Let W1 be the displacement at RD 1.012

Now, W =( L x B x draft ) x( 1.025)

W1 = (L x B x d1) x (1.012)

According to question ,

Draft = d1
W/( L x B x 1.025) = W1 / (L x B x 1.012)
W/ (L x B x 1.025) = W1/ ( L x B x 1.025)
W1 = (12000 x 1.012) / 1.025
W1 = 11847.8 t

Hence, Cargo she has to discharge = (12000 – 11847.8) t

= 152.2 t

13. A vessel floating in DW of RD 1.005 has the upper edge of her summer loadline in the water line
to starboard and 50mm above the waterline to port . If her FWA is 180mm and TPC is 24 , find the
amount of cargo which the vessel can load to bring her to her permissible draft .
Solution:

RD = 1.0
Starboard side = 0cm above by below summer draft
Port side = 50mm (5cm above the water line to summer draft)

Mean draft = (0+5 ) / 2

= 5/ 2
= 2.5cm above the water line

FWA = 180mm
= 18cm,

TPC = 24

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1.005 ) x 18 /0 .025

= (0.2 x 18) /0 .025
= 14.4cm

Total sinkage = (14.4 + 2.5) cm

= 16.9cm

TPC is SW = 24

TPC = (A / 100) x 1.025

= (24 x100)/ 1.025
A = 23.41 t/cm

Again ,

TPC in RD 1.005 = ( A / 100) x 1.005

= (23.41 x 1.005)
= 23.53t/cm

Cargo to load = sinkage x TPC

= 23.53 x 16.9
= 397.657 t
 14. A vessel is floating at 7.8m draft in DW of RD 1.010 . TPC is 18 and FWA is 250mm. the maximum
permissible draft .
Solution:

Present draft = 7.8m,

RD of DW = 1.010 & TPC = 18 t/cm
FWA = 250mm = 25cm
Maximum permissible draft = 8.0m

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.010 – 1.020 ) x FWA /0 .025

=(0.015 x 25) /0 .025
= 15cm

When vessel enters SW, the draft will rise by 15 cm,

Hence ,actual draft in SW = (7.80 -0. 15)

=7.65m

Sinkage available = (8.00 – 7.65)

=0.35m

TPC = 18 ( given)

TPC = (A/100) x ( density)

18 = (A/100) x 1.025
So, A = (18 x 1.025)/ 100
= 17.56 m2

Now , TPC at RD 1.010 = A / 100 x 1.010

= 17.56 x 1.010
= 17.73 t/cm.

DWT available =( sinkage x TPC)

= (35 x 17.73)
= 620.5t.

15. A vessel’s statutory freeboard is 2.0m. She is loading in DW of RDD 1.015 and her freeboard is
2.1m . TPC =24. FWA = 200mm. find the DWT available .
Solution:

Statutory freeboard is = 2.0m

DW RD = 1.015 & Freeboard = 2.1m
TPC = 24, FWA = 200mm = 20cm

Change in draft =
(Change in RD )x(FWA)
0.025

= (1.015 – 1.025 ) x 20 /0.025

= ( 0.010 x 20) /0 .025
= 8cm

Hence , actual freeboard available =( 2.1 + 0.08)m

= 2.18m

Sinkage available = ( 2.18 – 2.00)

= 0.18m
= 18cm

TPC = 24 ( Given)

24 = (A / 100) x(1.025)
A = (24 x 100)/ 1.025
A = (24 / 1.025)
= 23.414 m2

Now, TPC for RD 1.015 = (A/ 100) x(1.015)

= (23.414 x 1.015)
= 23.76 t/cm

DWT available = (sinkage x TPC)

= (18 x 23.76 )
= 427.68t.

16. A vessel is lying in a river berth of density 1.010 tonnes per m3 , with her summer loadline 20mm
above the water on the starboard side and 50mm above the water on the port side . Find how
much cargo she can load to bring her to her to her summer loadline in SW, if her summer
displacement is 15000 tonnes and TPC is 25.
Solution:

RD of river = 1.010
Summer load line on starboard side = 20mm above
Port side = 50mm above

Mean draft = ( 20 + 50) / 2

= (70/2)
= 3.5cm above

W = 15000t,

FWA = (W/TPC)
= 15000/(40 x 25)
= 15cm

Change in draft =

(Change in RD )x(FWA)
0.025
= (1.025 – 1.010 ) x 15 / 0.025
= 9cm

Total sinkage = (9 + 3.5 )

= 12.5cm

TPC = 25( Given)

TPC = (A/100) x 1.025

= (25 / 1.025)
= 24.39 m2

Now, TPC for RD 1.010 = (A /100) x( 1.010)

= (24.39 x 1.010)
= 24.63t/cm

Hence ,Cargo can be load = (sinkage x TPC)

= (12.5 x 24.63)
= 307.875 t

17. A vessel is floating in dockwater of RD 1.005with her starboard WNA mark 30mm below and her
port WNA mark 60mm below the waterline . If her summer SW draught is 8.4m, TPC is 30 and FWA
is 160 mm, calculate how much cargo can be loaded to bring the vessel to her vessel draught in
SW.
Solution:

Dockwater RD = 1.005,
Stbd WNA mark = 30mm below
Port WNA mark = 60mm below
Mean draft = (30 + 60) / 2
= 45mm
=4.5cm,below the water line.

Distance from WNA to water(W) = 50mm

= 5cm

So, sinkage available = (50 – 45)

= 5mm
= 0.5cm
Water (W) is 1/48 of summer draft
= (1/48 x 8.4)
= 0.175m
= 17.5cm

Sinkage = (17.5 + 0.5 )

= 18cm

Total sinkage = (18 + 12.8)

= 30.8cm

Change in draft =

(Change in RD )x(FWA)
 0.025

= (1.025 – 1.005 )x 16 / 0.025

= 12.8cm

Again, TPC = (A/100) x density

30 = (A/100) x 1.025
A = 30/ 1.025
A = 29.268 m2

Now , TPC for density of 1.005

TPC = (A/100) x 1.005
= (29.268 x 1.005)
= 29.41 t/cm.

Hence ,cargo can be loaded = (30.8 x 29.41)

= 905.82 t

18. A vessel is loading in a SW dock and is lying with her starboard winter loadline 60mm above and
her port winter loadline 20mm below the surface of water. If her summer draught in SW is 7.2m
and TPC is 20,, find how many tonnes of cargo the vessel can load to bring to her down to her
tropical loadline in SW.
Solution:

Starboard winter load line = 60mm above

Port winter load line = 20mm below
Mean draft = ( 60 – 20)/2
= (40/2)
= 20mm
= 2cm

Summer SW draft = 7.2m( given)

Waterline (W) = (1/48 of summer draft)

=(1/48 x 7.2)
= 0.15m
= 15cm

Total sinkage available to bring the vessel to her tropical load line = ( 2+ 15 + 15)
= 32cm

TPC =20( given)

So, Cargo can be load = (TPC x sinkage)
=(20 x 32)
= 640t.

19. From the following details, Calculate the DWT available :- present free board: port 3.0m, starboard
2.9m in water of RD 1.020 FWA 200mm. TPC 30. Statutory summer free board 2.8m.
Solution :
For, Port Side = (3 -2.8) = 0.2m
= 20cm (above )

For , Starboard side =( 2.9 – 2.8) =0. 1m

= 10cm above

Hence ,Mean freeboard can be calculated as

= (20+ 10)/2
= 30 /2 cm
= 15cm available

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1.020 ) x 20 /0 .025

= 4cm

Now, Total sinkage = (15 + 4)cm

= 19cm

TPC = 30( given)

TPC = (A/ 100) x density
30 = (A/100) x 1.025
A = 30/ 1.025
= 29.268m2

Again for calculating ,TPC at RD 1.020

= (A/100) x ( 1.020)
= 29.85 t/cm

Now , Cargo can be load =( TPC x sinkage)

= (29.85 x 19)
= 567.15 t