1. STABILITY
PROBLEM 5

2. 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?

3. 1. Solution:
INC. IN DR = VOL. OF LOST BOUYANCY
AREA OF INTACT W.P.
= x 16 x 9
200 x x 16 =
INC. IN DR = m

4. 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?

5. 2. 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
= √
N Spd = Knots

6. 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.?

7. 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, nm

8. 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 m?

9. NEW BM = 8.989 m NEW KB = ½ x DR NEW KB = 4.737 m (+) = ½(9.474)
INC. IN DR = VOL. OF LOST BOUYANCY
AREA OF INTACT W.P.
= x 16 x 9
200 x x 16 =
INC. IN DR = m
OLD DR = m
NEW DR = m
NEW KB = ½ x DR
= ½(9.474)
NEW KB = m
NEW BM = m
NEW KB = m (+)
NEW KM = m

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

11. 5. Solution
Ch. of trim = W X D
MCTC
= x 10
878.79
Ch. Of Trim = cm

12. 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?

13. 6. Solution:
Weight Distance Moment
3,700 tons m ,780
380 tons m ( + )
4, ,008
NKG = Total Moment
Total Weight
= 35,008
4,080
NKG = m

14. 7. A ship of 45,000 displacement has a KG of 9. 49m, KM=12
7. A ship of 45,000 displacement has a KG of 9.49m, KM=12.53m, GGo = 0.18m. Find the value of GZ for a 20 degrees heel. (KN = 4.45m)

15. 7. Solution:
GZ = KN – KG X SIN Ф
= KN – ( KG + GG0 X SIN Ф )
= 4.45 – ( X SIN 20)
GZ = m

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

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

18. 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?

19. 9. Solution:
∆ = L x B x Dr x Dens.
= 25 x 6 x 1.10 x 1.025 =
(Pig Iron)
∆ =
∆ = L x B x Dr x Dens.
Dr = ∆
L x B x Dens. =
25 x 6 x 1.025
Dr = m

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

21. 10. Solution:
VIRTUAL LOSS
OF GM = LB³ x d1
12V d2
= 10(15)³ x .92
(12)(6,000)
Virtual Loss of GM = m

22. 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:

23. 11. Solution:
GG’ = W X D Δ
= 2,500 x 21.5
9,500
GG’ = 5.66 UPWARD

24. 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?

25. 12. Solution:
Force = Weight x ( % N.O.S )
Mechanical Advantage
Force = Weight
Force = lbs 2
Force = lbs

26. 13. On arrival at the discharging port, the displacement was 7,800 t
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?

27. 13. Solution:
WT DIST MOMENT
7800 – 3,200 = 4600
DISCH = 3,200 x = 18,560
FINAL DISPL= 4,600 x = 28,244 (+)
INITIAL DISPL = 7, ,804
KG = MOMENT / WEIGHT
= 46,804
7,800
OLD KG = m

28. 14. Your vessel tank measure 30 ft. long, 20 feet wide and 15 ft
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 Find the free surface constant if your vessel is floating to a density 1.024?

29. 14. Solution:
r =
1.024
r = .615
FSK = r x l x b³
420
FSK = x 30 x 20³
FSK =

30. 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.

31. 15. Solution:
FSC = FSK Δ
FSC = 4,272
12,500 t
FSC = ft.