### 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 shipâ€™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 â€“ 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 â€˜Xâ€™ 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 vesselâ€™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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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 â€“ 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

### 20. From the following information , calculate the DWT available up to the tropical loadline SW :-

Persent freeboards : port 1.6m, starboard 1.79m inRD 1.017. Tropical SW freeboard : 1.63m Tropical of SW draft : 9.6m FWA 150mm, TPC 20.4 .

Solution :

Present freeboard port = 1.68cm Stbd side = 1.79m Water of RD = 1.017, Tropical SW freeboard = 1.63c Tropical SW draft = 9.6m FWA = 150mm TPC = 20.4

From the following information calculate the quantity of cargo to load so that the vessel will be floating at the tropical marks in salt water:

Present freeboards: port 1.68 m, starboard 1.79 m.

Relative density of dock water 1.017.

Tropical draft: 9.70 m.

Tropical freeboard: 1.63 m.

FWA: 145 mm.

TPC: 19.2.

after solving your preliminary loading plan for iron ore at 1.025 density .displacement. is 68,800 tons. tpc is 62.8 find out your new draft if the specific gravity of dock water is fresh water.

Here Q no. 14 is incomplete and I think it's wrong because tpc =18 in SW and max.. permissible draft in SW =8m so after calculation we have total 35 cm sinkage available in SW

So

DWA = 35cm Ã—TPC (which is 18 for SW given in Question) 35Ã—18=630t

But in book answer is 620.8t

On this. Web page answer came 620.5t nearest to book answer

So

May be I am wrong

Hey viewers reply your opinion and correct me

A vessel displaces 16,000 T at her summer load draft in SW. If she is now floating in DW of relative density 1.015 with her summer loadline on the water, calculate how much Deadweight is available.

Help please :(

A vessel displaces 16,000 T at her summer load draft in SW. If she is now floating in DW of relative density 1.015 with her summer loadline on the water, calculate how much Deadweight is available. help me :(