1 A boxshaped vessel 200 m in length

• 1. A box-shaped vessel 200 m in length, 32 m breadth, floats in SW at an even keel draft of 9. 0 m. The KG is 10. 0 m. The vessel has a continuous center line bulkhead which is watertight. What is the bodily sinkage if an empty compartment 20. 0 m in length and symmetrical about amidships is bilged on one side?

• 1. Solution: 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 - 320 INC. IN DR = 0. 474 m

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

• • • 2. Solution: 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 knots. and consumed 326 T of fuel. If you have 210 T of usable fuel remaining, how 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 breadth, floats in SW at an even keel draft of 9. 0 m. The KG is 10. 0 m. It has a continuous centre line bulkhead which is watertight. An empty compartment 20. 0 m in length and symmetrical about amidships is bilged on one side. What is the KM 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 - 320 INC. IN DR = 0. 474 m OLD DR = 9. 000 m NEW DR = 9. 474 m NEW BM = 8. 989 m NEW KB = ½ x DR = ½(9. 474) NEW KB = 4. 737 m (+) NEW KB = 4. 737 m NEW KM = 13. 726 m

5. A box-shaped vessel 200 m in length, 30 m breadth, is floating in SW at even keel draft of 9. 0 m. What will be the change of trim if the forward end compartment 10. 0 m long is bilged with 2767. 5 tons SW assuming an MCTC = 878. 79 tons meters?

• • • 5. Solution Ch. of trim = W X D MCTC = 2767. 5 x 10 878. 79 Ch. Of Trim = 31. 492 cm

6. A double-bottom tank, when full, has its center of gravity at a height of 60 cm above the keel and can hold 380 tons of water. The KG of the ship is 9. 4 meters and her displacement is 3700 tons when the tank is empty. What will be her KG when the tank is filled?

• • • 6. Solution: Solution Weight Distance 3, 700 tons 9. 4 m 380 tons 0. 6 m 4, 080 NKG = Total Moment Total Weight = 35, 008 4, 080 NKG = 8. 56 m Moment 34, 780 228 ( + ) 35, 008

7. A ship of 45, 000 displacement has a KG of 9. 49 m, KM=12. 53 m, GGo = 0. 18 m. Find the value of GZ for a 20 degrees heel. (KN = 4. 45 m)

• 7. Solution: • GZ = KN – KG X SIN Ф • = KN – ( KG + GG 0 X SIN Ф ) • = 4. 45 – (9. 49 + 0. 18 X SIN 20) • GZ = 1. 1427 m

• 8. What is the bodily sinkage of a boxshaped vessel 80 m x 14 m floating at an even keel draft of 4 m if an empty midships DB tank is bilged 16 m x 14 m x 4. 2 m?

• • • 8. Solution: TPC = 1. 025 x A 100 = 1. 025 (80) (14) 100 TPC = 11. 48 BODILY SINKAGE = WT. TPC = (16)(14)(4. 2)(1. 025) 11. 48 BODILY SINKAGE = 84 cms

9. A box-shaped lighter is 25 meters long, 6 meters wide and floats at a draft of 1. 10 meters fore and aft. What will be her new draft after 30 tons of pig-iron have been spread evenly over the bottom?

• • • 9. Solution: ∆ = L x B x Dr x Dens. = 25 x 6 x 1. 10 x 1. 025 = 169. 125 + 30. 000 (Pig Iron) ∆ = L x B x Dr x Dens. ∆ = 199. 125 Dr = ∆ L x B x Dens. = 199. 125 25 x 6 x 1. 025 Dr = 1. 295 m

10. A ship of 6, 000 tonnes displacement is floating in fresh water and has a deep tank (10 m x 15 m x 6 m) which is undivided and is partly filled with nut oil of relative density 0. 92. Find the virtual loss of GM due to the free surface.

• 10. Solution: • VIRTUAL LOSS • OF GM = LB³ x d 1 • 12 V d 2 • = 10(15)³ x. 92 • (12)(6, 000) 1. 000 • Virtual Loss of GM = 0. 431 m

• 11. Your ship of 12, 000 tons displacement has a center of gravity of 21. 5 ft. above the keel. You run aground and estimate the weight aground is 2, 500 tons. The virtual rise in the center of gravity is:

• • • 11. Solution: GG’ = W X D Δ = 2, 500 x 21. 5 9, 500 GG’ = 5. 66 UPWARD

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

12. Solution: Force = Weight x ( 1 + 10% N. O. S ) Mechanical Advantage Force = Weight Mechanical Advantage Force = 800 lbs 2 Force = 400 lbs

• 13. On arrival at the discharging port, the displacement was 7, 800 t. After Discharging 3, 200 t of cargo with an average KG of 5. 8 m the new KG was found to be 6. 14 m. What was the vessel’s KG prior to discharge?

• • • 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, 800 OLD KG = 6. 00 m

• 14. Your vessel tank measure 30 ft. long, 20 feet wide and 15 ft. deep and the specific gravity of liquid in the tank is 0. 63. Find the free surface constant if 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 correction for vessel having a dimension of 45 ft. long, 36 ft. wide and 25 ft. deep, the free surface constant is 4, 272 and the vessel has displacement of 12, 500 T.

• 15. Solution: • FSC = FSK • Δ • FSC = 4, 272 • 12, 500 t • FSC = 0. 342 ft.