1. STABILITY
PROBLEM 4

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

3. 1. Solution:
∆ = L x B x Draft x Rel. Density
150 tons = 24m x 6m x Draft x 1.025
Draft =
24m x 6m x 1.025
Draft =
147.6
Draft = m

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

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

6. Inc. in Draft = Vol. of Lost Buoyancy
Area of Intact WP
= x 15 x 20 x 5
150 x 20 – x 15 x 20
250 / 2950
Increase in Draft = m
Old Draft = m ( + )
New Draft = m

7. 3. A box-ship draws 7. 5 m in dock water of density 1. 006 ton per cu
3. A box-ship draws 7.5 m in dock water of density ton per cu.m. Find the draft when she is floating in sea water.
Ans m

8. 3. Solution:
New Draft Old Density
Old Draft New Density
New Draft ton/m³
7.5 m ton/m³
New Draft = 7.5 m x 1006 ton/m³
1.025 ton/m³
New Draft = m

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

10. 4. Solution:
A. Find the KG
KB = ½ x Draft
KB = ½ x 2m
KB = 1.00 m
BM = I or LB³
V x V
= m x 6³m
12 x 40m x 6m x 2m
= ,640 / 5,760
BM = m
KB = m ( + )
KM = m
GM = m ( - )
KG = m

11. B. Find the New Draft
Inc. in Draft = Vol. of Lost Buoyancy
Area of Intact Waterplane
= m x 6m x 2m
40m x 6m – 10m x 6m
= / 240 – 60
= / 180
Inc. in Draft = m
Old Draft = m ( + )
New Draft = m

12. C. Find the New GM KB = ½ x New Draft = ½ x 2.67m KB = 1.335m
BM = I or LB³
V x V
= m x 6³m
12 x 40m x 6m x 2m
= /5760
BM = m
New BM = m
New KB = m ( + )
New KM = m
New KG = m ( - )
New GM = 0.56m

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

14. Permeability = Broken Stowage Stowage Factor = 0.25 / 1.50 x 100
5. Solution:
Stowage Factor = m³
Space Occupied = m³ ( - )
Broken Stowage = m³
Permeability = Broken Stowage
Stowage Factor
= / 1.50 x 100
Permeability = % or
x 100

15. Inc. in Draft = Vol. of Lost Buoyancy
Area of Intact WP
= x 15 x 10 x 4
80 x 10 – x 15 x 10 /
Increase in Draft = m

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

17. 6. Solution:
W = X x B x d x R. Density
W = Trimming Moment
X = Length of the Bilged Compartment
W = 5m x 10m x 4.5m x ton/m³
W = tons
TPC = WPA / 97.56
TPC = 70 x 10
97.56
TPC =

18. 6. Solution:
Inc. in Draft = W / TPC
= tons / 7.175
Inc. in Draft = m or cm
Old Draft = m ( + )
New Mean Draft= m

19. 7. A ship displaces 7500 cu. m. of water density 1,000 kgs per cu. m
7. A ship displaces 7500 cu.m. of water density 1,000 kgs per cu.m. Find the displacement in tons when the ship is floating at the same draft in water density 1,015 kgs per cu.m.
Ans. 7,612.5 tons

20. 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 tons

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

23. 9. A vessel of 10,000 tonnes displacement has a KG of 8
9. A vessel of 10,000 tonnes displacement has a KG of 8.29 meters and KM of Find GM if weight of 20 tons is added 10 meters vertically above the base line.
Ans m

25. 10. A vessel of 8,000 tonnes displacement has a KG of 7
10. A vessel of 8,000 tonnes displacement has a KG of 7.00 meters and KM of Find GM if weight of 20 tons is added 10 meters vertically above the base line.
Ans m

27. 11. A vessel of 8,000 tons displacement has a KG of 7
11. A vessel of 8,000 tons displacement has a KG of 7.00 meters and KM of Find GM if weight of 20 tons is removed 10 meters vertically above the base line.
Ans m

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

31. 13. A box-ship of 75m long x 10m wide x 6m deep is floating in salt water on an even keel at a draft of 4.5 meters. Find the new drafts (final drafts) if a forward compartment 5 meters long is now bilged.
Ans. A= meters; F=6.002 meters

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

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