1. You are loading a cargo of canned 1. You are loading a cargo of canned goods with a stowage factor of 65. If you goods with a stowage factor of 65. If you allow 15% for broken stowage, how many allow 15% for broken stowage, how many tons can be loaded in a space of 55,000 tons can be loaded in a space of 55,000 cubic feet?cubic feet?

Ans. 719 Tons

• 11. Solution;. Solution;

• Weight = Volume x Allowance

• Stowage Factor

• = 55,000 x 0.85

• 65

• Weight = 719.23 tonsWeight = 719.23 tons

2. What is the stowage factor of cargo of 2. What is the stowage factor of cargo of canned goods packed in cartons weighing canned goods packed in cartons weighing 340 pounds and 36 cubic feet each 340 pounds and 36 cubic feet each carton?carton?

Ans. 237 ft³/ton

2. Solution:2. Solution:

• SF = Volume

• Weight / 2240 lbs.

• SF = 36 ft³

• 340 tons / 2240 lbs

• SF = 237 ft³/tonSF = 237 ft³/ton

3. If the stowage factor of cotton is 56, 3. If the stowage factor of cotton is 56, allowing a stowage loss of 15%, it will allowing a stowage loss of 15%, it will stow at a factor of ___.stow at a factor of ___.

Ans. 65.9Ans. 65.9

• 3. Solution:3. Solution:

• S. Factor = Stowage Factor of Cotton

• Allowance

• S. Factor = 56

• 0.85

• S. Factor = 65.9S. Factor = 65.9

4. A ship of 11,000 tons displacement has 4. A ship of 11,000 tons displacement has a moment of statical stability of 500 tons a moment of statical stability of 500 tons meters when heeled 5°. Find the initial meters when heeled 5°. Find the initial metacentric height? metacentric height?

Ans. 0.522 m Ans. 0.522 m

• 4. Solution;4. Solution;

• MSS = ∆ x GM x Sin θ

• ∆ x GM x Sin θ = MSS

• GM = MSS

• ∆ x Sin θ

• = 500 tons meter

• 11,000 tons x Sin 5º

• GM = 0.52 mGM = 0.52 m

5. A vessel displacing 18,000 tons has a 5. A vessel displacing 18,000 tons has a KG of 50 feet. A crane is used to lift cargo KG of 50 feet. A crane is used to lift cargo weighing 20 tons from a supply vessel. weighing 20 tons from a supply vessel. When lifting, the head of the crane boom When lifting, the head of the crane boom is 150 feet above the keel. What is the is 150 feet above the keel. What is the change in KG?change in KG?

Ans. 0.11 footAns. 0.11 foot

• 5. Solution:5. Solution:

• GG’ = W x D

• ∆

• GG’ = 20 t x ( 150 ft – 50 ft )

• 18,000 t

• GG’ = 20 t x 100 ft.

• 18,000 t

• GG’ = 0.1111 footGG’ = 0.1111 foot

6. A weight of 500 tons is loaded into a 6. A weight of 500 tons is loaded into a ship so that its centre of gravity is 10 ship so that its centre of gravity is 10 meters from that of the ship. Find the shift meters from that of the ship. Find the shift of G if the ship's original displacement of G if the ship's original displacement was 3,000 tons.was 3,000 tons.

Ans. 1.428 mAns. 1.428 m

• 6. Solution:6. Solution:

• GG’ = GG’ = W x DW x D

• ∆∆

• GG’ = GG’ = 500 t x 10 m500 t x 10 m

• 3,500 tons3,500 tons

• GG’ = 1.428 mGG’ = 1.428 m

7. What will be the reduction of GM on a 7. What will be the reduction of GM on a 5,000 tons displacement ship with 6.3 KG 5,000 tons displacement ship with 6.3 KG in sea water after half of the sea water in sea water after half of the sea water ballast is discharged from the double ballast is discharged from the double bottom ballast tank 22 m x 10 m x 2 m with bottom ballast tank 22 m x 10 m x 2 m with 1.3 m KG?1.3 m KG?

Ans. 0.236 m.Ans. 0.236 m.

• 7. Solution:7. Solution:• Wt. of Ballast = L x B x D x R. Den.• = 22 x 10 x 1 x 1.025• Wt. of Ballast = 225.5 tonsWt. of Ballast = 225.5 tons • Old ∆ = 5,000 t• - 225.5 t• New New ∆ = 4774.5 tons∆ = 4774.5 tons• Dist = KG of Vsl.– KG of Tank• = 6.3 m – 1.3 m

• Dist = 5.0 mDist = 5.0 m

GG’ = W x D ∆ = 225.5 t x 5 m 4774.5 t GG’ = 0.236 mGG’ = 0.236 m

8. The original displacement of a ship 8. The original displacement of a ship was 4,285 tons and her KG was 6.00 was 4,285 tons and her KG was 6.00 meters. Find her new KG after she has meters. Find her new KG after she has loaded the following weights; 800 tons at loaded the following weights; 800 tons at 3.6 meters above the keel, 440 tons at 7.0 3.6 meters above the keel, 440 tons at 7.0 meters above the keel. 110 tons at 5.8 meters above the keel. 110 tons at 5.8 meters above the keel, 630 tons at 3.0 meters above the keel, 630 tons at 3.0 meters above the keel.meters above the keel.

Ans. 5.46 mAns. 5.46 m

• 8. Solution:8. Solution:• Weight KG MomentsWeight KG Moments• 4,285 6.0 m 25,710• 800 3.6 m 2,880• 440 7.0 m 3,080• 110 5.8 m 638• 630 3.0 m 1,890 • 6,265 34,198• New KG = Total Moments / Total Weights• = 34,198 / 6265• New KG = 5.46 mNew KG = 5.46 m

9. A vessel displacing 18,000 short tons 9. A vessel displacing 18,000 short tons and has a VCG of 70 ft. You dump 100 and has a VCG of 70 ft. You dump 100 short tons of water, which had a VCG of short tons of water, which had a VCG of 88 ft., what is the new KG?88 ft., what is the new KG?

Ans. 69.9 feetAns. 69.9 feet

• 9. Solution:9. Solution:

• Weight VCG MomentsWeight VCG Moments

• 18,000 70 ft 1,260,000

• ( - ) 100 88 ft 8,800 ( - )

• 17,900 1,251,200

• New KG = Total Moments

• Total Weights

• New KG = 1,251,000 / 17,900

• New KG = 69.9 ftNew KG = 69.9 ft

10. A rectangular shaped vessel 200 m in 10. A rectangular shaped vessel 200 m in length and 32 m in breadth, floats at an length and 32 m in breadth, floats at an even keel draft of 9.0 m. The KG is 10.0 m, even keel draft of 9.0 m. The KG is 10.0 m, KB is 4.737, BM is 8.989 m. What is her KB is 4.737, BM is 8.989 m. What is her GM?GM?

Ans. 3.726 mAns. 3.726 m

• 10. Solution:10. Solution:

• KB = 4.737 m

• BM = 8.989 m ( + )

• KM = 13.726 m

• KG = 10.00 m ( - )

• GM = 3.726 mGM = 3.726 m

11. At a given draft, a ship of 120 meters 11. At a given draft, a ship of 120 meters length and 15 meters beam has a length and 15 meters beam has a coefficient of fineness of the coefficient of fineness of the waterplane of 0.770. What is her TPC waterplane of 0.770. What is her TPC at this draft?at this draft?

Ans. 14.20 tonsAns. 14.20 tons

• 11. Solution:11. Solution:

• TPC = L x B x Cw x Rel. Density

• 100

• TPC = 120 m x 15 m x 0.770 x 1.025

• 100

• TPC = 14.20 tons TPC = 14.20 tons

12. What is the TPC of a box-shaped 12. What is the TPC of a box-shaped vessel 120 m x 15 m x 10 m floating at vessel 120 m x 15 m x 10 m floating at even keel draft of 4.5 m?even keel draft of 4.5 m?

Ans. 18.45 tonsAns. 18.45 tons

• 12. Solution:12. Solution:

• TPC = 1.025 x Area of Waterplane

• 100

• TPC = 1.025 x 120 m x 15 m

• 100

• TPC = 1845

• 100

• TPC = 18.45 tons TPC = 18.45 tons

13. A rectangular shaped vessel 100 m in 13. A rectangular shaped vessel 100 m in length, 20 m breadth is floating in salt length, 20 m breadth is floating in salt water at an even keel draft of 6.0 m. A water at an even keel draft of 6.0 m. A forward hold 10.0 m long was bilged. What forward hold 10.0 m long was bilged. What is the TPC?is the TPC?

Ans. 18.45 tons Ans. 18.45 tons

13. Solution:13. Solution:

TPC = 1.025 x Area

100

TPC = 1.025 x ( 100m – 10m ) x 20 m

100

TPC = 1.025 x 90m x 20m

100

TPC = 18.45 tonsTPC = 18.45 tons

14. A drum which is 1.5 m long and 60 cm. 14. A drum which is 1.5 m long and 60 cm. in diameter has mass of 20 kgs when in diameter has mass of 20 kgs when empty. Find what will be her draft in water empty. Find what will be her draft in water of density 1,024 kg/m³ If it contains 200 of density 1,024 kg/m³ If it contains 200 liters of paraffin of relative density 0.6.liters of paraffin of relative density 0.6.

Ans. 0.484 mAns. 0.484 m

14. Solution:14. Solution:

Den. of Paraffin = RD of Paraffin x FW Density

= 0.6 x 1,000 kgs./m³

Den. Of Paraffin = 600 kgs./m³Den. Of Paraffin = 600 kgs./m³

Mass of the Paraffin = Volume x Density

= 0.2 m³ x 600 kgs./m³

Mass of the Paraffin = 120 kgs.Mass of the Paraffin = 120 kgs.

Mass of the Drum = 20 kgs. ( + )

Total Mass = 140 kgs.Total Mass = 140 kgs.

• 14. Solution:14. Solution:

Vol. of water displaced = Mass / Density

• =140 kgs. / 1,024 kgs./m³

• Vol. of water displaced = 0.137 m³

• Vol. of water displaced (V) = π²d

• d = V / π²

• d = 0.137 m³

• 3.1416 x .09

• Draft = 0.484 mDraft = 0.484 m

15. A box-shaped barge 16m x 6m x 5 m 15. A box-shaped barge 16m x 6m x 5 m is floating alongside a ship in fresh water is floating alongside a ship in fresh water at a mean draft of 3.5 m. The barge is to be at a mean draft of 3.5 m. The barge is to be lifted out of the water and loaded on to the lifted out of the water and loaded on to the ship with a heavy lift derrick. Find the load ship with a heavy lift derrick. Find the load in tons borne by the purchase when the in tons borne by the purchase when the draft of the barge is reduced to 2 m?draft of the barge is reduced to 2 m?

Ans. 144 tonsAns. 144 tons

• 15. Solution:15. Solution:

• Mass of the Barge = L x W x Dr x Rel. Den.

• = 16 x 6 x 3.5 x 1.0

• Mass of the Barge = 336 tons

• Mass of water = 16 x 6 x 2 x 1.0

• Displaced at 2m = 192 tons

•

• Mass of the Barge = 336 tons

• Mass of Barge at 2 m = 192 tons ( - )

• Load Borne by the = 144 tonsLoad Borne by the = 144 tons

• Purchase Purchase