2Stuff Floats … why?GA 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
3Stuff Floats … why?GB4.44 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 weightThe upward Buoyant Force (equal to the weight displaced) is considered to act through the Center of Buoyancy, B.
4Some 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
5But 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 ft40 ft80 ft
6But 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 50080 x 4080 x 500
7But some big, steel stuff does float … 35 cu. ft. of seawater weighs 1 Long TonOur “barge” displaces 1543 x 35= 54,000 cu.ft of sea water40 x 50080 x 4080 x 500
8But some big, steel stuff does float … 54,000 cu. ft. of seawater would fill a hole80 ft wide, 500 ft longAnd only 1.35 ft deep !The draft of our vessel is only 1.35 ft!40 lb.
9But 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
10Flooding 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.
11FloodingBy partitioning the vessel into (longitudinally) water tight compartments ,,,a breach of one compartment only loses the buoyancy due to the volume of that compartmentIf there is sufficient reserve buoyancy from the remaining compartments the vessel will remain afloat (at a deeper draft).lost buoyancyregainedlost buoyancybuoyancy
12Flooding Flooding forward or aft is particularly critical Buoyancy must be regained before a second becomes flooded “over the top” of the watertight partitionIf not …lost buoyancylost buoyancy
13Flooding Flooding forward or aft is particularly critical Buoyancy must be regained before a second becomes flooded “over the top” of the watertight partitionIf not …How can we prevent this?lost buoyancylost buoyancy
14Flooding 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?
15The 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)
16Floodable Length Notes: In the previous example, flooding of any single compartment will not cause the ship to founderThis is a single compartment vesselIf the floodable length everywhere spans two compartments, we have a two compartment vesselEmpire State VI is a single compartment vesselThe Titanic was a four compartment vessel.
17Trim 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 producedwGLCGDa= 30 ftBLCBDf= 30 ft
18TrimTrimming moment causes ship to submerge at the stern shifting B aft to equilibrium point under GNote that trimming axis is not amidships but about a point on the original water plane called the center of floatationThe ship is now trimmed “2 ft by the stern”wGFLCFDa= 30.8 ftBBDf= 28.8 ftDa – Df = 30.8 – 28.8 = 2 ft
19Trim notes:The mean draft is the draft at F (still 30 ft.) not the average of Da & DfThe change in drafts (+.8 at the stern & -1.2 at the bow) are not equal but determined by the location of FLCF, mean draft, and (0 trim) LCB are determined by the shape of the hull and change with displacementGFDa= 30.8 ftDf= 28.8 ftBDa – Df = 30.8 – 28.8 = 2 ft
20Heel q Heel, or angle of heel, is the amount of tipping to one side Heel is measured in degreesThere are many causes of heel
21Heel A permanent heel results if G is not on the centerline A heel due to an off-center G is called a listA list may be corrected by shifting fuel, ballast or FW to the high sidewGGBB
22Heel When a ship rolls, it takes on various angles of heel If a ship has a list, rolling is centered about the listing angleStability is concerned with the ship’s ability to return to its equilibrium position
23Stability & 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 unstableStability is the tendency to return to equilibrium if disturbedPositive stability is the tendency to return to the original (stable) equilibrium position after being disturbedNegative stability is the tendency to move toward another equilibrium position when disturbed
24Initial StabilityWith G on the centerline, as a ship rolls B moves outboard & upFor small angles of heel, the locus of B is a circular arc with its center at M, the metacenterMGBBB
25Initial StabilityKB & 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 loadedMGBK
26Initial 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 BMBMKMGMGKGBKBK
27Initial StabilityAs the ship heels, the upward buoyant force is aligned through B’ &M, but not GThe distance between the lines of action of B and the vessel weight D (through G) is the Righting Arm, GZThis action of two opposite & equal forces, separated by distance GZ is the Righting moment D x GZSince GZ = GM sin q, the Righting Moment at any angle of heal is proportional to GMThe Righting Moment acts to return the ship to the vertical, equilibrium positionMqZGB’BK
28Initial StabilityIn 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 capsizingGM then, is a measure of initial stability.GZMB’BK
29Large 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 Arm8 ftMaximum Righting Arm6 ftInitial Slope = GM4 ftRange of StabilityGZ≈GM sinq2 ft15o o o o o o o Heel, q
30Large Angle Stability: MRaising G increases KG, reduces GM (less stable)Lowering G, decreases KG, increases GM (more stable)GGGZG8 ftlarger GMK6 ftuncorr GMUncorrected KG4 ftsmaller GM2 ft15o o o o o o o Heel, q
31Large Angle Stability: MAs the GM increases,so does the Maximum Righting Arm..and so does the Range of StabilityGGGZG8 ftK6 ft4 ft2 ft15o o o o o o o Heel, q
32Additional 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 GMThus a large GM, though more stable, results in short, quick rolls – the ship is said to be stiffAnd a small GM yields longer, gentile rolls – the ship is said to be tender.
33Roll 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.
34Impaired StabilityReduction of GM by addition of topside weight (e.g., icing)FloodingSolid Flooding (lost buoyancyPartial Flooding (free surface effect)Flooding “in communication with the sea” (worst case: lost buoyancy + free surface + communication effects)GroundingUpward grounding force reduces GM and stabilityStructural damage from impact or severe bending stress can cause flooding
35Impaired Stability D0 B P Upon grounding, only part of the vessel’s weight is supported by the Buoyancy forceThe rest is support by a Ground Force, PThis results in a virtual rise of G (to G’)If G’ rises above the metacenter M, there is negative stabilityThe ship will certainly heel and maybe capsize■■■■■■■■■GMK■■■■■■■■■MG’G’PBD0Initial Displ, D0Grounded Displ, D1Ground Force, P = D0 – D1