• 1. A box-shaped vessel 200 m in length, 1. A box-shaped vessel 200 m in length, 32 m breadth, floats in SW at an even 32 m breadth, floats in SW at an even keel draft of 9.0 m. The KG is 10.0 m. The keel draft of 9.0 m. The KG is 10.0 m. The vessel has a continuous center line vessel has a continuous center line bulkhead which is watertight. What is bulkhead which is watertight. What is the the bodily sinkage bodily sinkage if an empty if an empty compartment 20.0 m in length and compartment 20.0 m in length and symmetrical about amidships is bilged symmetrical about amidships is bilged on one side?on one side?

• 1. Solution1. Solution:

INC. IN DR = VOL. OF LOST BOUYANCY AREA OF INTACT W.P. = 20 x 16 x 9 200 x 32 - 20 x 16

= 2880 6400 - 320INC. IN DR = 0.474 mINC. IN DR = 0.474 m

2. Your vessel arrives in port with 2. Your vessel arrives in port with sufficient fuel to steam 550 miles at 13 kts. sufficient fuel to steam 550 miles at 13 kts. If you are unable to load bunkers, at what If you are unable to load bunkers, at what speed must you proceed to reach your speed must you proceed to reach your next port which is 683 miles away?next port which is 683 miles away?

• 2. Solution2. Solution:

• N Cons = N Spd² x N Dist

• O Cons O Spd² x O Dist

• N Spd = √ N Cons x O Spd² x O Dist

• O Cons x N Dist

• N Spd = √ x (x) 13² x 550 nm

• x (x) 683 nm

• = √ 136.09077

• N Spd = 11.66 Knots

• 3. You have steamed 1,124 miles at 21 3. You have steamed 1,124 miles at 21 knots. and consumed 326 T of fuel. If you knots. and consumed 326 T of fuel. If you have 210 T of usable fuel remaining, how have 210 T of usable fuel remaining, how far can you steam at 17 knots.?far can you steam at 17 knots.?

• 3. Solution:

• N Cons = N Spd² x N Dist

• O Cons O Spd² x O Dist.

• N Dist = N Cons x O Spd² x O Dist

O Cons x N Spd²

• N Dist = 210 tons x 21² x 1,124 nm

326 tons x 17²

• N Dist = 1,104.86 nm

• 4. A box-shaped vessel 200 m length, 32 m 4. A box-shaped vessel 200 m length, 32 m breadth, floats in SW at an even keel draft breadth, floats in SW at an even keel draft of 9.0 m. The KG is 10.0 m. It has a of 9.0 m. The KG is 10.0 m. It has a continuous centre line bulkheadcontinuous centre line bulkhead which is which is watertight. An empty compartment 20.0 m watertight. An empty compartment 20.0 m in length and symmetrical about in length and symmetrical about amidships is bilged on one side. amidships is bilged on one side. What is What is the KMthe KM if the BM is 8.989 m? if the BM is 8.989 m?

INC. IN DR = VOL. OF LOST BOUYANCY AREA OF INTACT W.P. = 20 x 16 x 9 200 x 32 - 20 x 16

= 2880 6400 - 320INC. IN DR = 0.474 m OLD DR = 9.000 m NEW DR = 9.474 mNEW DR = 9.474 m

NEW KB = ½ x DR = ½(9.474)NEW KB = 4.737 m NEW KB = 4.737 m

NEW BM = 8.989 mNEW KB = 4.737 m (+)NEW KM = 13.726 m

5. A box-shaped vessel 200m in length, 5. A box-shaped vessel 200m in length, 30m breadth, is floating in SW at even 30m breadth, is floating in SW at even keel draft of 9.0m. What will be the keel draft of 9.0m. What will be the change of trim if the forward end change of trim if the forward end compartment 10.0m long is bilged with compartment 10.0m long is bilged with 2767.5 tons SW assuming an MCTC = 2767.5 tons SW assuming an MCTC = 878.79 tons meters?878.79 tons meters?

• 5. Solution5. Solution

• Ch. of trim = W X D

• MCTC

• = 2767.5 x 10

• 878.79

• Ch. Of Trim = 31.492 cmCh. Of Trim = 31.492 cm

6. A double-bottom tank, when full, has 6. A double-bottom tank, when full, has its center of gravity at a height of 60 cm its center of gravity at a height of 60 cm above the keel and can hold 380 tons of above the keel and can hold 380 tons of water. The KG of the ship is 9.4 meters water. The KG of the ship is 9.4 meters and her displacement is 3700 tons when and her displacement is 3700 tons when the tank is empty. What will be her KG the tank is empty. What will be her KG when the tank is filled?when the tank is filled?

• 6. Solution6. Solution:

• Weight Distance MomentWeight Distance Moment

• 3,700 tons 9.4 m 34,780

• 380 tons 0.6 m 228 ( + )

• 4,080 35,0084,080 35,008

• NKG = Total Moment

• Total Weight

• = 35,008

• 4,080

• NKG = 8.56 mNKG = 8.56 m

7. A ship of 45,000 displacement has a 7. A ship of 45,000 displacement has a KG of 9.49m, KM=12.53m, GGo = 0.18m. KG of 9.49m, KM=12.53m, GGo = 0.18m. Find the value of GZ for a 20 degrees Find the value of GZ for a 20 degrees heel. (KN = 4.45m)heel. (KN = 4.45m)

• 7. Solution:7. Solution:

• GZ = KN – KG X SIN Ф

• = KN – ( KG + GG0 X SIN Ф )

• = 4.45 – (9.49 + 0.18 X SIN 20)

• GZ = 1.1427 mGZ = 1.1427 m

• 8. What is the bodily sinkage of a box-8. What is the bodily sinkage of a box-shaped vessel 80m x 14m floating at an shaped vessel 80m x 14m floating at an even keel draft of 4m if an empty even keel draft of 4m if an empty midships DB tank is bilged 16m x 14m x midships DB tank is bilged 16m x 14m x 4.2m?4.2m?

• 8. Solution:8. Solution:• TPC = 1.025 x A• 100• = 1.025 (80) (14)• 100• TPC = 11.48TPC = 11.48• BODILY SINKAGE = WT.• TPC• = (16)(14)(4.2)(1.025)• 11.48 BODILY SINKAGE = 84 cmsBODILY SINKAGE = 84 cms

9. A box-shaped lighter is 25 meters long, 9. A box-shaped lighter is 25 meters long, 6 meters wide and floats at a draft of 1.10 6 meters wide and floats at a draft of 1.10 meters fore and aft. What will be her new meters fore and aft. What will be her new draft after 30 tons of pig-iron have been draft after 30 tons of pig-iron have been spread evenly over the bottom?spread evenly over the bottom?

• 9. Solution:9. Solution:

• ∆ = L x B x Dr x Dens.

• = 25 x 6 x 1.10 x 1.025

• = 169.125

• + 30.000 (Pig Iron)

• ∆ ∆ = 199.125= 199.125 ∆ = L x B x Dr x Dens. Dr = ∆ L x B x Dens. = 199.125 25 x 6 x 1.025 Dr = 1.295 mDr = 1.295 m

10. A ship of 6,000 tonnes displacement is 10. A ship of 6,000 tonnes displacement is floating in fresh water and has a deep floating in fresh water and has a deep tank (10m x 15m x 6m) which is undivided tank (10m x 15m x 6m) which is undivided and is partly filled with nut oil of relative and is partly filled with nut oil of relative density 0.92. Find the virtual loss of GM density 0.92. Find the virtual loss of GM due to the free surface.due to the free surface.

• 10. Solution:10. Solution:

• VIRTUAL LOSS

• OF GM = LB³ x d1

• 12V d2

• = 10(15)³ x .92

• (12)(6,000) 1.000

• Virtual Loss of GM = 0.431 mVirtual Loss of GM = 0.431 m

• 11. Your ship of 12,000 tons 11. Your ship of 12,000 tons displacement has a center of gravity of displacement has a center of gravity of 21.5 ft. above the keel. You run aground 21.5 ft. above the keel. You run aground and estimate the weight aground is 2,500 and estimate the weight aground is 2,500 tons. The virtual rise in the center of tons. The virtual rise in the center of gravity is:gravity is:

• 11. Solution:11. Solution:

• GG’ = W X D

• Δ

• = 2,500 x 21.5

• 9,500

• GG’ = 5.66 UPWARD GG’ = 5.66 UPWARD

• 12. When a weight of 800 lbs. is 12. When a weight of 800 lbs. is suspended, what is the stress on the suspended, what is the stress on the hauling part when using a gun tackle hauling part when using a gun tackle rove to least advantage?rove to least advantage?

12. Solution:12. Solution:

Force = Weight x ( 1 + 10% N.O.S )

Mechanical Advantage

Force = Weight

Mechanical Advantage

Force = 800 lbs

2

Force = 400 lbsForce = 400 lbs

• 13. On arrival at the discharging port, the 13. On arrival at the discharging port, the displacement was 7,800 t. After displacement was 7,800 t. After Discharging 3,200 t of cargo with an Discharging 3,200 t of cargo with an average KG of 5.8 m the new KG was average KG of 5.8 m the new KG was found to be 6.14 m. What was the found to be 6.14 m. What was the vessel’s KG prior to discharge?vessel’s KG prior to discharge?

• 13. Solution:13. Solution:

• WT DIST MOMENT

• 7800 – 3,200 = 4600

• DISCH = 3,200 x 5.8 = 18,560

• FINAL DISPL= 4,600 x 6.14 = 28,244 (+)

• INITIAL DISPL = 7,800 46,804

KG = MOMENT / WEIGHT = 46,804 7,800OLD KG = 6.00 mOLD KG = 6.00 m

• 14. Your vessel tank measure 30 ft. long, 14. Your vessel tank measure 30 ft. long, 20 feet wide and 15 ft. deep and the 20 feet wide and 15 ft. deep and the specific gravity of liquid in the tank is specific gravity of liquid in the tank is 0.63. Find the free surface constant if 0.63. Find the free surface constant if your vessel is floating to a density 1.024?your vessel is floating to a density 1.024?

• 14. Solution:

• r = .63

• 1.024

• r = .615

FSK = r x l x b³

• 420

• FSK = .615 x 30 x 20³

• 420

• FSK = 351.42

• 15. Compute for the free surface 15. Compute for the free surface correction for vessel having a dimension correction for vessel having a dimension of 45 ft. long, 36 ft. wide and 25 ft. deep, of 45 ft. long, 36 ft. wide and 25 ft. deep, the free surface constant is 4,272 and the the free surface constant is 4,272 and the vessel has displacement of 12,500 T.vessel has displacement of 12,500 T.

• 15. Solution:

• FSC = FSK

• Δ

• FSC = 4,272

• 12,500 t

• FSC = 0.342 ft.