1. Buoyancy & Subdivision Heel & Trim Stability
Hydrostatics
Buoyancy & Subdivision
Heel & Trim
Stability

2. Stuff Floats … why? G
A 30” pine 2x4 (240 cu. In.) weighs approximately 4.44 lbs (density ≈ lb/in3)
The downward force (weight) is considered to act through the Center of Gravity, G, of the object

3. Stuff Floats … why? G B
4.44 lbs of sea water (density ≈ .037 lb/in3) occupies only 120 cu.in.
Thus our pine 2x4 need only be half submerged to displace its weight
The upward Buoyant Force (equal to the weight displaced) is considered to act through the Center of Buoyancy, B.

4. Some stuff doesn’t … why?
A 2.5 cu.ft. steel anvil weighs 1200 lbs …
and would displace 2.5 cu. Ft. of water (161 lbs)
The upward Buoyant Force (equal to the weight displaced) would not support the weight

5. But some big, steel stuff does float …
1 sq. ft. of one-inch thick steel plate weighs about 40 lbs.
| 1 ft |
40 lbs.
Consider a box 500 ft long, 80 ft wide & 40 ft deep made from “40 lb” steel plate:
500 ft
40 ft
80 ft

6. But some big, steel stuff does float …
This box weighs:
[ 2(80x40) + 2(40x500) + (80x500) ] x 40 lbs
= 3,456,000 lbs = 1543 Long Tons
(1 Long Ton = 2240 lbs)
40 lb.
40 x 500
80 x 40
80 x 500

7. But some big, steel stuff does float …
35 cu. ft. of seawater weighs 1 Long Ton
Our “barge” displaces 1543 x 35
= 54,000 cu.ft of sea water
40 x 500
80 x 40
80 x 500

8. But some big, steel stuff does float …
54,000 cu. ft. of seawater would fill a hole
80 ft wide, 500 ft long
And only 1.35 ft deep !
The draft of our vessel is only 1.35 ft!
40 lb.

9. But some big, steel stuff does float …
To displace 80 x 500 x 20 ft (800,000 cu.ft)
We can carry another 746,000/35
= 21,314 Tons of cargo
( a total displacement of 22,857 Tons)
Draft, T
= 20 ft

10. Flooding If the hull were breached …
the internal volume would no longer be contributing to the displacement (lost buoyancy), only the steel shell itself ...
and the steel weight is greater than the volume (of water) it displaces.

11. Flooding
By partitioning the vessel into (longitudinally) water tight compartments ,,,
a breach of one compartment only loses the buoyancy due to the volume of that compartment
If there is sufficient reserve buoyancy from the remaining compartments the vessel will remain afloat (at a deeper draft).
lost buoyancy
regained
lost buoyancy
buoyancy

12. Flooding Flooding forward or aft is particularly critical
Buoyancy must be regained before a second becomes flooded “over the top” of the watertight partition
If not …
lost buoyancy
lost buoyancy

13. Flooding Flooding forward or aft is particularly critical
Buoyancy must be regained before a second becomes flooded “over the top” of the watertight partition
If not …
How can we prevent this?
lost buoyancy
lost buoyancy

14. Flooding Construct smaller subdivisions fore & aft
A breach at an extreme end then results in less lost buoyancy..
And the vessel remains afloat.
How do we determine the size (length) of watertight compartments?

15. The Floodable Length Curve
Floodable length is amount of ship length (centered at the ordinate) which may be flooded without the waterline rising above the margin line.
FL (ft)

16. Floodable Length Notes:
In the previous example, flooding of any single compartment will not cause the ship to founder
This is a single compartment vessel
If the floodable length everywhere spans two compartments, we have a two compartment vessel
Empire State VI is a single compartment vessel
The Titanic was a four compartment vessel.

17. Trim 0 trim (even keel) means same drafts fore & aft
B is directly under G (but not necessarily amidships)
If a weight is shifted longitudinally (say, aft) G moves proportionately and a trim moment is produced w G
LCG
Da= 30 ft B
LCB
Df= 30 ft

18. Trim
Trimming moment causes ship to submerge at the stern shifting B aft to equilibrium point under G
Note that trimming axis is not amidships but about a point on the original water plane called the center of floatation
The ship is now trimmed “2 ft by the stern” w G F
LCF
Da= 30.8 ft B B
Df= 28.8 ft
Da – Df = 30.8 – 28.8 = 2 ft

19. Trim notes:
The mean draft is the draft at F (still 30 ft.) not the average of Da & Df
The change in drafts (+.8 at the stern & -1.2 at the bow) are not equal but determined by the location of F
LCF, mean draft, and (0 trim) LCB are determined by the shape of the hull and change with displacement G F
Da= 30.8 ft
Df= 28.8 ft B
Da – Df = 30.8 – 28.8 = 2 ft

20. Heel q Heel, or angle of heel, is the amount of tipping to one side
Heel is measured in degrees
There are many causes of heel

21. Heel A permanent heel results if G is not on the centerline
A heel due to an off-center G is called a list
A list may be corrected by shifting fuel, ballast or FW to the high side w G G B B

22. Heel When a ship rolls, it takes on various angles of heel
If a ship has a list, rolling is centered about the listing angle
Stability is concerned with the ship’s ability to return to its equilibrium position

23. Stability & Equilibrium
Equilibrium is a state of balance. In the upright position, with B vertically aligned with G, a vessel is in equilibrium. Equilibrium may be either stable or unstable
Stability is the tendency to return to equilibrium if disturbed
Positive stability is the tendency to return to the original (stable) equilibrium position after being disturbed
Negative stability is the tendency to move toward another equilibrium position when disturbed

24. Initial Stability
With G on the centerline, as a ship rolls B moves outboard & up
For small angles of heel, the locus of B is a circular arc with its center at M, the metacenter M G B B B

25. Initial Stability
KB & KM the location of B and M with respect to the keel K are functions of the underwater shape of the hull and may be calculated for any given displacement (draft)
KG depends on the arrangements of weights within the vessel and must be recalculated each time the vessel is loaded M G B K

26. Initial Stability BM KM GM KG KB
The Metacentric Radius, BM, fixed by the location of B and M with respect to the keel K, is also a function of the underwater shape of the hull and may be calculated for any given displacement (draft)
GM depends on the location of G (arrangements of weights within the vessel)
Thus shifting weights within the vessel affects G and GM, but not M and B M BM KM GM G KG B KB K

27. Initial Stability
As the ship heels, the upward buoyant force is aligned through B’ &M, but not G
The distance between the lines of action of B and the vessel weight D (through G) is the Righting Arm, GZ
This action of two opposite & equal forces, separated by distance GZ is the Righting moment D x GZ
Since GZ = GM sin q, the Righting Moment at any angle of heal is proportional to GM
The Righting Moment acts to return the ship to the vertical, equilibrium position M q Z G B’ B K

28. Initial Stability
In the unlikely (and unsafe) condition where G is located above M (negative GM) …
Any disturbance results in similar, parallel forces creating a moment, but …
In this case the moment tends to further heel the ship.
This is an example of negative stability, and the vessel will heel to some other, non-upright, equilibrium position—maybe even capsizing
GM then, is a measure of initial stability. G Z M B’ B K

29. Large Angle Stability: (Uncorrected) Static Stability Curve – GZ vs
Large Angle Stability: (Uncorrected) Static Stability Curve – GZ vs. Heel plotted for some assumed KG (must be “corrected” for actual KG)
GZ—Righting Arm
8 ft
Maximum Righting Arm
6 ft
Initial Slope = GM
4 ft
Range of Stability
GZ≈GM sinq
2 ft
15o o o o o o o Heel, q

30. Large Angle Stability:M
Raising G increases KG, reduces GM (less stable)
Lowering G, decreases KG, increases GM (more stable) G G GZ G
8 ft
larger GM K
6 ft
uncorr GM
Uncorrected KG
4 ft
smaller GM
2 ft
15o o o o o o o Heel, q

31. Large Angle Stability:M
As the GM increases,
so does the Maximum Righting Arm
..and so does the Range of Stability G G GZ G
8 ft K
6 ft
4 ft
2 ft
15o o o o o o o Heel, q

32. Additional GM notes:
Since GM is a measure of initial stability, there is a recommended minimum GM at each draft, but …
Rolling period is inversely proportional to GM
Thus a large GM, though more stable, results in short, quick rolls – the ship is said to be stiff
And a small GM yields longer, gentile rolls – the ship is said to be tender.

33. Roll Period, T ∝Beam /√GM

■■■■■■■■■■■■■■■■

Tankers tend to have a low center of gravity and large KM (large GM), little freeboard and are stiff.
Passenger ships, with large freeboard are designed to be tender for comfort.

34. Impaired Stability
Reduction of GM by addition of topside weight (e.g., icing)
Flooding
Solid Flooding (lost buoyancy
Partial Flooding (free surface effect)
Flooding “in communication with the sea” (worst case: lost buoyancy + free surface + communication effects)
Grounding
Upward grounding force reduces GM and stability
Structural damage from impact or severe bending stress can cause flooding

35. Impaired Stability D0 B P
Upon grounding, only part of the vessel’s weight is supported by the Buoyancy force
The rest is support by a Ground Force, P
This results in a virtual rise of G (to G’)
If G’ rises above the metacenter M, there is negative stability
The ship will certainly heel and maybe capsize

■■■■■■■■■ G M K

■■■■■■■■■ M G’ G’ P B D0
Initial Displ, D0
Grounded Displ, D1
Ground Force, P = D0 – D1