• 1. A box-shaped vessel 24m x 6m x 3m 1. A box-shaped vessel 24m x 6m x 3m displaces 150 tonnes of water. Find the displaces 150 tonnes of water. Find the draft when the vessel is floating in salt draft when the vessel is floating in salt water.water.

• Ans. 1.016 m

• 1. Solution:1. Solution:

• ∆ = L x B x Draft x Rel. Density

• 150 tons = 24m x 6m x Draft x 1.025

• Draft = 150

• 24m x 6m x 1.025

• Draft = 150

• 147.6

• Draft = 1.016 mDraft = 1.016 m

• 2. A box-ship 150m x 20m x 12m on an 2. A box-ship 150m x 20m x 12m on an even keel at 5m draft. A compartment even keel at 5m draft. A compartment amidships is 15m long and contains amidships is 15m long and contains timber of relative density 0.8 and a timber of relative density 0.8 and a stowage factor 1.5 cubic meters per ton. stowage factor 1.5 cubic meters per ton. Calculate the new draft if this Calculate the new draft if this compartment is now bilged.compartment is now bilged.

• Ans. 5.085 mAns. 5.085 m

• 2. Solution:2. Solution:• Space occupied = 1• By timber 0.8• = 1.25 m³• Stowage Factor = 1.50 m³• Space Occupied = 1.25 m³ ( - )• Broken Stowage = 0.25 m³• Permeability = Broken Stowage• Stowage Factor• = 0.25 / 1.50 x 100• Permeability = 16.67% or 0.1667Permeability = 16.67% or 0.1667

x 100

• Inc. in Draft = Vol. of Lost Buoyancy

• Area of Intact WP

• = 0.1667 x 15 x 20 x 5

• 150 x 20 – 01667 x 15 x 20

• 250 / 2950

• Increase in Draft = 0.085 m

• Old Draft = 5.000 m ( + )

• New Draft = 5.085 mNew Draft = 5.085 m

• 3. A box-ship draws 7.5 m in dock water 3. A box-ship draws 7.5 m in dock water of density 1.006 ton per cu.m. Find the of density 1.006 ton per cu.m. Find the draft when she is floating in sea water.draft when she is floating in sea water.

• Ans. 7.361 mAns. 7.361 m

• 3. Solution:3. Solution:

• New Draft Old Density

• Old Draft New Density

• New Draft 1.006 ton/m³

• 7.5 m 1.025 ton/m³

• New Draft = 7.5 m x 1006 ton/m³

• 1.025 ton/m³

• New Draft = 7.361 mNew Draft = 7.361 m

• 4. A box-ship 40 meters long, 6 meters 4. A box-ship 40 meters long, 6 meters beam, is floating at a draft of 2m. F and beam, is floating at a draft of 2m. F and A. She has an amidships compartment A. She has an amidships compartment 10m long which is empty. If the original 10m long which is empty. If the original GM is 0.6m, find the new GM if this GM is 0.6m, find the new GM if this compartment is bilged.compartment is bilged.

• Ans. 0.558 meterAns. 0.558 meter

• 4. Solution:4. Solution:

• A. Find the KGA. Find the KG

• KB = ½ x Draft

• KB = ½ x 2m

• KB = 1.00 mKB = 1.00 m

BM = I or LB³ V 12 x V = 40m x 6³m 12 x 40m x 6m x 2m = 8,640 / 5,760BM = 1.50mBM = 1.50mKB = 1.00m ( + )KM = 2.50mGM = 0.6m ( - )KG = 1.90mKG = 1.90m

• B. Find the New DraftB. Find the New Draft

• Inc. in Draft = Vol. of Lost Buoyancy

• Area of Intact Waterplane

• = 10m x 6m x 2m

• 40m x 6m – 10m x 6m

• = 120 / 240 – 60

• = 120 / 180

• Inc. in Draft = 0.67mInc. in Draft = 0.67m

• Old Draft = 2.00m ( + )

• New Draft = 2.67mNew Draft = 2.67m

• C. Find the New GM

• KB = ½ x New Draft

• = ½ x 2.67m

• KB = 1.335m

• BM = I or LB³

• V 12 x V

• = 30m x 6³m

• 12 x 40m x 6m x 2m

• = 6480 /5760

• BM = 1.125mBM = 1.125m

New BM = 1.125mNew KB = 1.335m ( + )New KM = 2.460mNew KM = 2.460mNew KG = 1.900m ( - )New GM = 0.56mNew GM = 0.56m

• 5. A box-ship 80m x 10m x 6m is floating 5. A box-ship 80m x 10m x 6m is floating upright in salt water on an even keel at 4 upright in salt water on an even keel at 4 meters draft. She has an amidships meters draft. She has an amidships compartment 15m long which is filled compartment 15m long which is filled with timber (SF=1.5 cu.m. of ton). One with timber (SF=1.5 cu.m. of ton). One ton of solid timber would occupy 1.25 ton of solid timber would occupy 1.25 cu.m. of space. What would be the cu.m. of space. What would be the increase in draft if this compartment is increase in draft if this compartment is now bilged.now bilged.

• Ans. 0.129 meterAns. 0.129 meter

• 5. Solution:5. Solution:

• Stowage Factor = 1.50 m³

• Space Occupied = 1.25 m³ ( - )

• Broken Stowage = 0.25 m³

• Permeability = Broken Stowage

• Stowage Factor

• = 0.25 / 1.50 x 100

• Permeability = 16.67% or 0.1667Permeability = 16.67% or 0.1667

x 100

• Inc. in Draft = Vol. of Lost Buoyancy

• Area of Intact WP

• = 0.1667 x 15 x 10 x 4

• 80 x 10 – 0.1667 x 15 x 10

• 100.02 / 774.95

• Increase in Draft = 0.129 mIncrease in Draft = 0.129 m

• 6. A box-ship 75m long x 10m wide x 6m 6. A box-ship 75m long x 10m wide x 6m deep is floating in salt water on an even deep is floating in salt water on an even keel at a draft of 4.5 meters. Find the new keel at a draft of 4.5 meters. Find the new mean draft if a forward compartment 5 mean draft if a forward compartment 5 meters long is bilged.meters long is bilged.

• Ans. 4.821 mAns. 4.821 m

• 6. Solution:6. Solution:

• W = X x B x d x R. Density

• WW = Trimming Moment

• XX = Length of the Bilged Compartment

• W = 5m x 10m x 4.5m x 1.025 ton/m³

• W = 230.625 tonsW = 230.625 tons

• TPC = WPA / 97.56TPC = WPA / 97.56

• TPC = TPC = 70 x 1070 x 10

• 97.5697.56

• TPC = 7.175TPC = 7.175

• 6. Solution:6. Solution:

• Inc. in Draft = W / TPCInc. in Draft = W / TPC

• = 230.625 tons / 7.175= 230.625 tons / 7.175

• Inc. in Draft = 0.3214 m or 32.14 cmInc. in Draft = 0.3214 m or 32.14 cm

• Old Draft = 4.5 m ( + )Old Draft = 4.5 m ( + )

• New Mean Draft= 4.821 mNew Mean Draft= 4.821 m

• 7. A ship displaces 7500 cu.m. of water 7. A ship displaces 7500 cu.m. of water density 1,000 kgs per cu.m. Find the density 1,000 kgs per cu.m. Find the displacement in tons when the ship is displacement in tons when the ship is floating at the same draft in water floating at the same draft in water density 1,015 kgs per cu.m.density 1,015 kgs per cu.m.

• Ans. 7,612.5 tonsAns. 7,612.5 tons

• 7. Solution:7. Solution:

• New ∆ New Density

• Old ∆ Old Density

• New ∆ 1,015 kgs./m³

• 7,500t 1,000 kgs./m³

• New ∆ = 7,500t x 1,015 kgs./m³

• 1,000 kgs/m³

• New ∆ = 7,612.5 tonsNew ∆ = 7,612.5 tons

• 8. A vessel of 10,000 tonnes 8. A vessel of 10,000 tonnes displacement has a KG of 8.00 meters displacement has a KG of 8.00 meters and KM of 9.78. Find GM if weight of 20 and KM of 9.78. Find GM if weight of 20 tonnes is removed 10 meters vertically tonnes is removed 10 meters vertically above the base line.above the base line.

• Ans. 1.784 mAns. 1.784 m

• 9. A vessel of 10,000 tonnes 9. A vessel of 10,000 tonnes displacement has a KG of 8.29 meters displacement has a KG of 8.29 meters and KM of 9.78. Find GM if weight of 20 and KM of 9.78. Find GM if weight of 20 tons is added 10 meters vertically above tons is added 10 meters vertically above the base line.the base line.

• Ans. 1.487 mAns. 1.487 m

• 10. A vessel of 8,000 tonnes 10. A vessel of 8,000 tonnes displacement has a KG of 7.00 meters displacement has a KG of 7.00 meters and KM of 9.78. Find GM if weight of 20 and KM of 9.78. Find GM if weight of 20 tons is added 10 meters vertically above tons is added 10 meters vertically above the base line.the base line.

• Ans. 2.773 mAns. 2.773 m

• 11. A vessel of 8,000 tons displacement 11. A vessel of 8,000 tons displacement has a KG of 7.00 meters and KM of 9.78. has a KG of 7.00 meters and KM of 9.78. Find GM if weight of 20 tons is removed Find GM if weight of 20 tons is removed 10 meters vertically above the base line.10 meters vertically above the base line.

• Ans. 2.805 mAns. 2.805 m

• 12. A ship is floating in salt water on an 12. A ship is floating in salt water on an even keel at 6 meters draft. TPC is 20 even keel at 6 meters draft. TPC is 20 tonnes. A rectangular-shaped tonnes. A rectangular-shaped compartment amidships is 20 meters compartment amidships is 20 meters long, 10 meters wide, and 4 meters deep. long, 10 meters wide, and 4 meters deep. The compartment contains cargo with The compartment contains cargo with permeability 25 percent. Find the new permeability 25 percent. Find the new draft if this compartment is bilged.draft if this compartment is bilged.

• Ans. 6.1025 metersAns. 6.1025 meters

• 13. A box-ship of 75m long x 10m wide x 13. A box-ship of 75m long x 10m wide x 6m deep is floating in salt water on an 6m deep is floating in salt water on an even keel at a draft of 4.5 meters. Find even keel at a draft of 4.5 meters. Find the new drafts (final drafts) if a forward the new drafts (final drafts) if a forward compartment 5 meters long is now compartment 5 meters long is now bilged.bilged.

• Ans. A= 3.788 meters; F=6.002 metersAns. A= 3.788 meters; F=6.002 meters

• 14. A box-ship of 75m long x 10m wide x 14. A box-ship of 75m long x 10m wide x 6m deep is floating in salt water on an 6m deep is floating in salt water on an even keel at a draft of 4.5 meters. Find even keel at a draft of 4.5 meters. Find the Change of Trim if a forward the Change of Trim if a forward compartment 5 meters long is now compartment 5 meters long is now bilged.bilged.

• Ans. 221.4 cm by the headAns. 221.4 cm by the head

• 15. A box-ship floats upright on an even 15. A box-ship floats upright on an even keel in a fresh water and the center of keel in a fresh water and the center of buoyancy is 0.50 meter above the keel. buoyancy is 0.50 meter above the keel. Find the KB when she floating in salt Find the KB when she floating in salt water.water.

• Ans. 0.488 mAns. 0.488 m