Curve of Intact Statical Stability / “Righting Arm Curve” Assumes: – Quasi-static conditions – Given Displacement – Given KG Cross Curves of Stability – Since M T moves as a function of φ , Righting Arms are calculated for each φ at regular intervals – Assumes a value of KG
4.4 Measure of Overall Stability From the Curves of Intact Stability the following Measures of Overall Stability can be made: – Range of Stability – Maximum Righting Moment – Angle of Maximum Righting Moment – Dynamical Stability – Measure of Tenderness or Stiffness
Measure of Overall Stability Range of Stability – The range of angles for which there exists a positive righting moment. – The greater the range of stability, the less likely the ship will capsize. – If the ship is heeled to any angle in the range of stability, the ship will exhibit an internal righting moment that will right the ship if the external moment ceases.
Measure of Overall Stability Maximum Righting Moment – The largest Static Moment the ship can produce. – Calculated by multiplying the displacement of the vessel times the maximum Righting Arm. – The larger the Maximum Righting Moment, the less likely the vessel is to capsize.
Measure of Overall Stability Angle of Maximum Righting Arm – The angle of inclination where the maximum Righting Arm occurs. Beyond this angle, the Righting Arm decreases. – It is desirable to have a larger maximum angle so that at large angles of heel in a rolling ship the righting moment will continue to increase.
Measure of Overall Stability Dynamical Stability: – The work done by quasi-statically rolling the ship through its range of stability to the capsizing angle. D – Can be calculated by the equation: . This is GZ d s equal to the product of the ship’s displacement with the area under the Curve of Intact Statical Stability. – Not shown directly by the Curve of Intact Statical Stability. – Does not account for the actual dynamics, because it neglects the impact of waves and momentum.
Measure of Overall Stability Measure of “Tenderness” or “Stiffness” – The initial slope of the intact statical stability curve indicates the rate at which a righting arm is developed as the ship is heeled over. This slope is GM! – A steep initial slope indicates the rapid development of a righting arm and the vessel is said to be stiff. Stiff vessels have short roll periods and react strongly to external heeling moments. – A small initial slope indicates the slower development of a righting arm and the vessel is said to be tender. Tender vessel have longer roll periods and react sluggishly to external heeling moments.
Example: Plot the Intact Statical Stability Curve for an FFG-7 displacing 5000LT Step #1. From the Cross Curves of Form, find the 5000LT displacement value on the x-axis. Step #2. Record the righting arm value for each curve, from φ = 0 to 80 degrees Step #3. Draw the curve, using φ as x-axis, and GZ as y-axis
Intact Statical Stability Curve for FFG-7 @ s = 5000LT f GZ 0 0.00 5 2.00 Intact Statical Stability, FFG-7 25 10 3.80 15 5.80 20 7.75 20 Moment arm GZ 25 9.75 15 30 11.75 35 13.30 40 14.75 10 45 16.10 50 17.20 5 55 18.00 60 18.60 0 65 19.00 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 70 19.30 Heeling Angle 80 19.50 … But a correction must still be made!!
Example Problem The Statical Stability curve applies to a ship Curve of Intact Statical Stability with D=3600LT. 5 4 Righting Arm (GZ)(ft) 3 2 1 0 The ship is being -10 0 10 20 30 40 50 60 70 80 90 100 -1 pulled sideways -2 into a 10° list by a Angle of Heel (Degrees) tug attached to the ship 10ft above the Waterline. How much force is the tug applying to the tow line ?
Example Answer F =10° F Tug 10ft F G B Water F B Resistance D RM=GZ Δ =1.2ft×3600LT=4320ft-LT Upsetting Moment from Tug=F Tug ×10ft=RM (in static equilibrium) F Tug =4320ft-LT/10ft=432LT
4.5 Effect of a Vertical Shift in the Center of Gravity on the Righting Arm In the Cross Curves of Stability, the data is presented assuming that: KG = 0 (on the keel) This is, of course, not realistic. It is done this way so that the curves may be generalized for all drafts. Once the curve data is recorded and plotted, a sine correction factor must be applied,shifting the KG to its correct position in order to get the TRUE MOMENT RIGHTING ARM VALUE . – Must Apply a Sine Correction if: • Using the Curve of Intact Statical Stability to correct for G not being located at K • Correcting the Curve of Intact Statical Stability for vertical movements of G
The external moment couple causes the creation of the internal moment couple to oppose it . M T s Wind B Water Resistance F B External Moment Couple = Internal Moment Couple
When the ship heels over, the center of buoyancy, B, shifts. The shift creates a distance or “moment arm”. M T f Z 0 B 1 G 0 Z 0 = Moment Arm
For values taken from the Cross Curves of Stability, G 0 is at the keel... M T f B 1 Z 0 G 0 Z 0 = Moment Arm This value is recorded as G 0 Z 0 , the Initial Moment Arm.
The KG value for the ship is given… this is the ACTUAL G position from the keel... M T f B 1 KG forms a similar triangle that gives the Z 0 value for the SINE correction
Sin f = opp hyp opp =correction factor hyp = K G Sine Correction factor = KG Sin f
M T f B 1 Z 0 Sin Correction = KG Sin f
M T f Z v B 1 Z 0 G v Z v = G 0 Z 0 - KG sin f
Effect of a Vertical Shift in the Center of Gravity on the Righting Arm As KG rises the righting arm (GZ) decreases. This change in GZ can be found from: G v Z v = G 0 Z 0 - G 0 G v sin F Where: – G v is the final vertical location of the center of gravity. – G 0 is the initial location of KG. – Typically, G 0 G V =KG final
Effect of a Vertical Shift in the Center of Gravity on the Righting Arm M T Sine Correction: D s F 0 Z 0 - G 0 G v sin F G v Z v = G G Z F G o , Z o =initial locations P G Z 0 0 W L G v , Z v =final positions B 0 B F b
Effect of Increased Displacement on the Righting Arm A higher displacement should increase the Righting Moment as RM= Displacement * RA But, if the added weight is high, then the KG increase could cause a reduction in GZ Weight added low down usually increases stability
Effect of a Vertical Shift in the Center of Gravity on the Righting Arm
4.6 Stability Change for Transverse Shift in CG So far we have only considered the case where the Center of Gravity is on the centerline (TCG=0). The center of gravity may be moved off the centerline by weight additions, removals, or shifts such as cargo loading, ordnance firing, and movement of crew.
Stability Change for Transverse Shift in the CG M T F G v G t cos G v F F G t Port Starboard D s W 1 L 1 F G v Z V Z t G t F B 1 F b
The red line indicates the COSINE Correction factor for a transverse change in G. f Z v G T Z T Note that G v G T is the TCG value and is the hypotenuse of this correction triangle... Cosine Corr = G v G T cos f
The final moment arm, G T Z T , is the correct moment arm f Z v G T Z T G T Z T = G v Z v - G v G T cos f
Stability Change for Transverse Shift in the CG The new righting arm created by a shift in TCG may be computed at each angle from the Cosine Correction: G t Z t = G v Z v - G v G t cos F ...Typically, G V G t =TCG final
Stability Change for Transverse Shift in the CG The new righting arm (G t Z t ) created due to the shift in the transverse center of gravity is either shorter or longer than the righting arm created if TCG=0. The range of stability has decreased on the side that the transverse center of gravity has shifted to but has increased on the side it shifted from.
Combining both the vertical and horizontal corrections by substituting for G v Z v you can get a final general formula for determining moment arms: G T Z T = G 0 Z 0 - KG sin f - G v G T cos f
Example Curves With Cosine Correction Statical Stability Curve and Corrections 15 Righting Arm from Cross Curves 10 With Sine Correction With Cosine Correction Righting Arm (GZ)(ft) 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 3 2 1 0 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 0 1 2 3 4 1 1 1 1 1 1 1 1 1 1 - - - - - - - - - - - - - - -5 -10 -15 Angle of Heel (degrees)
Various Righting Arm Conditions D D G G Z D B B Z G B F B F B F B F =30 F =0 F =RA Max D D G G B B F B F B F =Capsize Angle F = >Capsize Angle
Example Problem FFG-7, with draft 13.5ft , which would otherwise be on an even keel, is heeling 15° to starboard in a gale. KG is 17ft. What is the Righting Moment?
Example Answer • RM=GZ Δ • G V Z V =G 0 Z 0 -G 0 G V sinF • From Curves of Form: Draft of 13.5ft-> Δ =100×30LT=3000LT • From Cross Curves for Δ =3000LT, φ =15°, G 0 Z 0 =6ft • G 0 G V =KG=17ft • G V Z V =6ft-17ft×sin(15°)=1.6ft • RM=1.6ft×3000LT=4800ft-LT
Example Problem • FFG-7, with draft 13.5ft, which would otherwise be on an even keel, is heeling 15° to port in a gale. KG is 17ft. While in this condition, 50LTs of unsecured stores shift from 20 ft starboard of centerline to 20ft port of centerline. • What is the Righting Moment? • What would the Righting Moment be if the weight had shifted the other way?
Example Answer RM=GZ Δ • G f Z f =G 0 Z 0 -G 0 G V sin φ -G V G t cos φ • • From Curves of Form: Draft of 13.5ft-> Δ =100×30LT=3000LT From Cross Curves for Δ =3000LT, φ =(-)15°, G 0 Z 0 =(-)6ft • • G 0 G V =KG=17ft G V G t = TCG f =(TCG 0 Δ 0 +STcg a w a -STcg r w r )/ Δ f • =(0ft×3000LT+(-)40ft×50LT)/3000LT=(-).67ft • G f Z f =(-)6ft-17ft×sin(-15°)-(-.67ft)×cos(-15°)=(-).953ft • RM=(-).953ft×3000LT=(-)2860ft-LT vice (-)4800ft-LT in the case without the weight shift (~40% reduction) • G f Z f =(-)6ft-17ft×sin(-15°)-(+.67ft)×cos(-15°)=(-)2.247ft • RM=(-)2.247ft×3000LT=(-)6741ft-LT vice (-)4800ft-LT in the case without the weight shift (~40% improvement)
4.7 How Does a Ship Sink? • 3 of the 6 Degrees of Freedom can sink a ship • Foundering(heave): Ship fills up with water from the bottom up and simply sinks on a relatively even keel (Loss of Buoyancy) • Overwhelm and Capsize: Ship still floats and is stable but has insufficient Righting Arm and Dynamical Stability for weather and sea conditions, which eventually roll the ship past range of stability • Plunging: Pitch angle becomes excessive causing the ship to sink bow or stern first • Progressive Flooding: Excessive list and/or trim angle and/or failure of bulkheads adjacent to flooded compartments resulting in one or more of the loss modes above.
Damage Stability “Flooding” - Water ingression such that the vessel has sinkage and trim but no list. May be intentional. “Damage” - Water ingression such that the vessel has sinkage, trim and list.
When a vessel is damaged, creating a a gap or hole in the hull, water will breech the ship. This results in: • Increase in draft • Change in trim • Permanent angle of list The result of this flooding can be determined two ways: • Lost Buoyancy Method • Added Weight Method
Consider a vessel that has been damaged such that a portion of the bottom is now open to the sea... The vessel ’ s draft will increase because an amount of the buoyancy was lost...
Lost buoyancy considers the amount of buoyancy “lost” as a result of the hole, and determines the value based upon the change in parallel sinkage that results. Original draft The change in the draft reflects the amount of buoyancy lost. The ship sinks until the available submerged volume is again equal to the ship’s displacement
Lost Buoyancy Method Analyzes damage by changes in buoyancy versus changes in the Center of Gravity. Premise is that the ship’s CG does not move. Since weight does not change, total buoyant volume must also be constant. Therefore, the ship makes up any lost buoyancy volume from damage by listing, trimming, and draft changes.
The Added Weight Method considers the resulting flooding as though it was a weight added to the ship. This is the method that will be used in this course. Original draft A flooded compartment does not fill completely with water, however. Compartments contain equipment, furniture, structural components, and cargo. A correction factor must be added to the volume of the compartment to accurately reflect conditions...
This correction factor is called: PERMEABILITY = AVAILABLE VOLUME TOTAL VOLUME Some typical factors are: Watertight compartment (warship) 97% Watertight compartment (merchant) 95% Accommodation spaces 95% Machinery spaces 85% Dry cargo spaces 70% Bunkers, stores, cargo holds 60%
Added Weight Method “The One We Will Use” Damaged Ship Modeled as Undamaged But with Water- Filled Spaces. Average Distances of Space from Keel, Midships, and Centerline Known & Water Density Known. Therefore Can Solve for Shifts in “G” as a Weight Addition Problem
Added Weight Method Independently Solve for Damaged Condition – KG – TCG – Draft and Trim Must know compartment contents to find Total of (Water Weight) Added. This involves a “Permeability” factor.
Permeability Compartments are rarely 100% flooded during damage, due to trapped air, equipment, etc. – Ratio of volume occupied by water to the total gross volume is defined as “permeability”. Permeability = Volume Available for Flooding Total Gross Volume – Permeability is always < or = to 100%!
Damage Stability Design Criteria Guiding rules for vessel design. Note that criteria used in static analysis will neglect the impact of dynamic forces such as wind and waves.
Damage Stability Design Criteria Three Main Criteria • “MARGIN LINE” • “LIST” • “EXTENT OF DAMAGE TO HULL”
MARGIN LINE LIMIT • Highest permissible location of any damaged waterplane. • Must be at least 3 inches (0.075 m) below top of the bulkhead deck at the side.
LIST LIMIT Heel by damage 20 degrees. • • Naval machinery to operate indefinitely at a permanent list 15 degrees (most will function up to ~25 degrees for a few hours). • Assumes personnel can continue damage control efforts effectively at a permanent list of 20 degrees. • Ship must possess adequate stability against weather to be towed when at 20 degree list.
EXTENT OF DAMAGE TO THE HULL LIMIT 100 ft LOA: must withstand flooding in one space. 100 - 300 ft LOA: flooding in two adjacent compartments. Warships, troop transports and hospital ships over 300 ft LOA: hull opening up to 15 % of Lpp. Others 300 ft: hull opening up to 12.5% of Lpp.
Foundering and Plunging A vessel as result of “damage” or other events can be lost several ways: Insufficient transverse stability. It rolls over. – (Could be static or dynamic.) Insufficient longitudinal stability. “Plunging” If insufficient buoyancy. It sinks. “Foundering”
Example Problem An FFG-7 with a draft of 13.5ft and a KG of 19ft on an even keel inport sails into the North Atlantic during Winter . – While there, topside becomes coated with a 6in thick coating of ice of density of 55lb/ft³. The topside area covered is 20,500ft² and has a Kg of 40ft. – In this condition, a space heater in CIC shorts generating a fire which is only extinguished by completely filling the 97% permeable 40ft×40ft×10ft space with firefighting (sea) water. The space is centered 45ft above the keel and 2.5ft port of centerline. What is the Righting Moment for a 15° port list and how could the resulting problem have been prevented?
Example Answer • w ice = ρ gV=55lb/ft³×.5ft×20,500ft²×1LT/2240lb =252LT(@Kg=40ft) • w ffwater = ρ gV=64lb/ft³×40ft×40ft×10ft×.97×1LT/2240lb =443LT(@Kg=45ft) • Δ (Curves of Form[T=13.5ft])=100×30LT=3000LT • KG f =(KG 0 Δ 0 +Kg a w a -Kg r w r )/ Δ f • KG f =(19ft×3000LT+40ft×252LT+45ft×443LT) /(3000LT+252LT+443LT) • KG f =23.5ft • TCG f =(TCG 0 Δ 0 +Tcg a w a -Tcg r w r )/ Δ f • TCG f =(0ft×3000LT+0ft×252LT+(-)2.5ft×443LT) /(3695LT) • TCG f =(-)0.3ft • G 0 Z 0 (Cross Curves[ Δ =3695LT; φ =(-)15°])=(-)6ft • G f Z f =G 0 Z 0 -KG f sin φ -TCG f cosF • G f Z f =(-)6ft-23.5ft×sin(-15°)-(-.3ft)×cos(-15°)=(+).372ft
Example Answer • R.M.= Δ ×G f Z f =3695LT×0.372ft=(+)1375ft-LT to port for a port list: The ship capsizes! • KM t (Curves of Form[ Δ =3695LT;T=15.25ft])=112*.2ft=22.4ft (GM t =KM t -KG f =22.4ft-23.5ft=(-)1.1ft; Stable?) • Center of Gravity is above Metacenter; ship rolls to port due to offset of flooded compartment and capsizes. • Prevent by keeping topside clear of ice and dewatering fire spaces as soon as possible .
4.8 Free Surface Correction (Small Angles of Heel) Free Surface - A “fluid” that moves freely. Fluid Shift is a weight and causes the CG to shift in both the vertical and horizontal directions. – Vertical shift is small for small angles and is usually ignored. – Horizontal shift always causes a reduction in the righting arm (GZ).
Free Surface Correction Free Surface Correction (FSC) The distance the center of gravity would have to rise to cause a reduction in the righting arm equivalent to that caused by the actual transverse shift . "Virtual" center of gravity (G v ) The effective position of this new VCG. Effective Metacentric Height (GM eff ) The distance from the virtual center of gravity (G v ) to the metacenter. Note: Dynamic effects are neglected.
Free Surface Effect Static effects for small angles ( F <=7°) – Effective “g” for tank is above tank g analogous to relationship between M and B M T g eff B 0 g 0 B f g f
Free Surface Correction The Big Picture M T GM eff F G v FSC F WL 1 Z F G Z 1 G 1 WL B B 1 . g g 1 K
Free Surface Effect The new, effective VCG is G v , so a sine correction is applied to get the statical stability curve G 1 Z 1 = G t Z t - GG v sin f or G 1 Z 1 = G t Z t - FSC sin f
Free Surface Correction The free surface correction to GM for small angle hydrostatics is: t i t FSC = s s where: t is the density of the fluid in the tank in lb s 2 /ft 4 s is the density of the water the ship is floating in lb s 2 /ft 4 i t is the transverse moment of area of the tank's free surface area in ft 4 . s is the underwater volume of the ship in ft 3 .
Free Surface Correction i t is calculated for a rectangular tank as: The dimensions are for the free surface! Tank C L i t = ( L ) ( B ) 3 X 12 L Y B
Effect on Ship “G” and Stability GZ eff =G 0 Z 0 -G 0 G v sin φ -G v G t cos φ -FSCsin φ – Calculation of KG, etc. is already accounted for in this equation – Free Surface Correction (FSC) already accounts for size of ship. GM eff =GM-FSC=KM-KG-FSC – A large FSC has exactly the same effects on list and stability as a higher KG.
How do we minimize adverse effects of free surface effect? • Compartmentalization • Pocketing (Keep tanks >95% full) • Empty Tanks • Compensated Fuel Oil Tanks • Dewater quickly after a casualty - flooding or fire
4.9 Metacentric Height Recall that Overall Stability is measured by: • Range of Stability • Dynamical Stability • Maximum righting moment • The angle at which the maximum righting moment occurs.
Initial Slope of the Curve of Intact Stability At small angles, a right triangle is formed between G, Z, and M. The righting arm may be computed: __ __ GZ = GM sin As 0, if the angle is given in radians the equation becomes: __ __ = GM GZ
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