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Section 3.1

Section 3.1. Archimedes Principle: The Buoyant Force on an object is equal to the weight of the volume of the water displaced by the object F B = r gV Forces on a body in water Distributed forces: Gravity : Distributed throughout volume of body based on mass density.

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Section 3.1

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  1. Section 3.1 Archimedes Principle: The Buoyant Force on an object is equal to the weight of the volume of the water displaced by the object FB=rgV Forces on a body in water Distributed forces: • Gravity: Distributed throughout volume of body based on mass density. • Buoyancy: Distributed over wetted surface of body based on hydrostatic pressure • Drag/Lift: Distributed over surface of body based on flow field when moving relative to medium

  2. Buoyant Forces Box Shaped Barge: Weight G Horizontal components of pressure force are negated by equal force on opposite side of barge. B A FB z FB=PA; P=rgz; FB=rgzA; V=zA; FB=rgV

  3. Static Equilibrium åF=0 Weight, Buoyancy, Drag and Lift forces all sum to zero in each dimension åM=0 All forces in each dimension are colinear and cancel; i.e. there are no separation of the action points of forces such that couples or moments are generated.

  4. Example Problem A boxed shaped barge 50ft wide, 100ft long, and 15ft deep has 10ft of freeboard in sea water: • What is its draft? • What is the hydrostatic pressure (psi) acting on the barge’s keel? • What is the magnitude (LT) of the total hydrostatic force acting on the barge’s keel? • What is the weight (LT) of the water displaced by the barge? • Assuming that the buoyant force acts through a single point, what is the location of that point in 3 dimensions? • Assuming minimum freeboard is 5ft, how many ft³ of coal at 50lb/ft³ can we load on the barge in seawater? • If we then take the barge into freshwater, what will the new draft be?

  5. Example Answer Draft=Depth-Freeboard=15ft-10ft=5ft Phyd=rgz=64lb/ft³×5ft×[1ft²/144in²]=2.22psi Fhyd=Phyd×A=2.22lb/in²×50ft×100ft×[144in²/ft²]×[1LT/2240lb]=714LT w=rgV=64lb/ft³×50ft×100ft×5ft×[1LT/2240lb]=714LT Center of Buoyancy=at amidships, on centerline, 2.5ft above keel

  6. Example Answer TPI=AWP[ft²]{LT/in}/420=50ft×100ft/420{LT/in}=11.9LT/in Change in draft=10ft-5ft=5ft×[12in/ft]=60in Change in weight=60in×11.9LT/in=714LT V=w/rg=714LT/(50lb/ft³)×2240lb/LT=32,000ft³

  7. Example Answer Current draft=TSW=10ft w=rgSWVSW=64lb/ft³×50ft×100ft×10ft=3,200,000lb VFW=w/rgFW=3,200,000lb/62.4lb/ft³=51,280ft³ TFW=VFW/AWP=51,280ft³/(50ft×100ft)=10.26ft Increased draft means reduced freeboard below minimum spec

  8. Section 3.2 Center of Mass/Gravity The weighted average over area or volume based on given distribution summed such that result is equivalent to the total force applied through a single point. What can change the Center of Gravity? • Add/subtract weight • Move weight/change distribution

  9. Notation: G=Location of Center of Gravity for ship g=Location of Center of Gravity for object Ds= Displacement of ship (LT) W = Magnitude of Gravitational Force/Weight of object (LT)

  10. Go Bo B L K C L • So far we’ve looked at ships that are in STATIC EQUILIBRIUM: • SFx = 0 • SFy = 0 • SFz = 0 • SMp = 0 Now let’s take a look at what happens when a weight is added to disturb this equilibrium

  11. A change in weight (either adding or removing it) will cause a change in the location of G, the center of gravity of the ship • A change in VCG (or KG) • A change in the TCG TCG G1 KGnew • g Go B L K It also causes a change in the longitudinal CG (LCG), but we’ll discuss that later... C L

  12. g G1 Go B L K C L When a weight is ADDED, the CG shifts TOWARD the added weight in line with the CG of the ship and the cg of the weight

  13. G1 G0 B L C L When a weight is REMOVED, the CG shifts AWAY from the added weight in line with the CG of the ship and the cg of the weight • g K

  14. In the case of a weight SHIFT, the CG first shifts AWAY from the removed weight…. • g G2 G0 G1 • g B L C L …and the TOWARDS the relocated weight K

  15. Let’s first consider a weight added directly over the centerline This will cause the location of the CG to move TOWARD the weight ... • g G1 G0 KGold KGnew B L C L … Causing a change in the VERTICAL distance, or KG K

  16. DsoldxKGold KGnew = + waddx Kg Dsold+ wadd Use the concept of weighted averages to determine the new CG: • g G1 G0 B L K C L

  17. Kg Dsoldx KGold KGnew = + (-w) x Kg Dsold+ (-w) It’s the same deal for removing a weight, only this time the weight is negative (i.e. removed): • g G0 G1 KGnew KGold B L K C L

  18. C L In a relocation of a weight, look at it as SUBTRACTING one weight, and ADDING another weight. • g G1 G0 Kg2 • g KGnew Kg1 KGold B L K

  19. Ds GnewGold = w g2g1 • In this unique case, Dsnew and Dsoldand are the SAME THING! • w1 and w2 are also the same thing! • The weight has only moved, not been removed • So we can rearrange the formula: ...This is ONLY for a single vertical weight shift!!

  20. Ds GnewGold = w g2g1 Where: • GnewGoldis the distance between the old and new CG’s • g2g1 is the distance between the old and new Cg locations • of the relocated weight ...This relation will become important in the Inclining Experiment

  21. Dsoldx KGold + Swix Kgi KGnew = Dsold +Swi We can generalize the formula for vertical changes in CG by the following:

  22. Example: • Given • USS CURTS (FFG-38) floats on an even keel at a draft of 17ft • KG = 19.5ft • Lpp = 408ft • It takes on 150LT of fresh water in a tank 6ft above the keel on the CL • Find • New vertical center of gravity (KG) after taking on water

  23. Ds G1 ? 150 LT 19.5’ 6’ G0 B L K C L Step 1: Draw picture! Step 2: Find Ds when floating at 17ft draft • Go to curves of form for FFG in appendix • Using curve 1, find the intersection w/ 17ft Ds = 147 x 30LT Ds = 4410LT

  24. Step 3: Write the GENERAL Equation Dsoldx KGold KGnew = 4410LT x 19.5ft + 150LT x 6ft 4410LT + 150LT KGnew = 86215.5 LT-ft + 900LT-ft KGnew = + Swix Kgi 4560 LT Dsnew Step 4: Substitute in values into the general equation

  25. 87115.5 LT-ft KGnew = 4560 LT KGnew = 19.10 ft CHECK: Does this answer make sense? YES! The CG shifts toward the added weight, lower than the original CG

  26. Example Problem A 688 Class Submarine is in port, pier side undergoing a maintenance period. The tender will be pulling periscopes tomorrow which requires the ship to maintain zero list, i.e. TCG=0ft. The Engineering Dept needs to pump #2RFT dry to perform a tank inspection. What impact will this have on the sub’s TCG? The sub has 10 MK 48 ADCAP torpedoes at 2LT each. How far and in which direction should these torpedoes be shifted to restore the sub’s TCG to zero? Data: Do=6900LT Tcg#2RFT= -12ft (i.e. port of centerline); TCGo=0ft Capacity#2RFT=5000galfw

  27. Example Answer w#2RFT=rgV =5000gal×[1ft³/7.4805gal]×rgfw =668.4ft³×62.4lb/ft³×[1LT/2240lb] =18.62LT Df= D0+Swa-Swr=6900LT-19LT=6881LT TCGf =(TCG0D0+STcgawa-STcgrwr)/Df=(0ft×6900LT-[-12ft]×19LT)/6881LT =0.033ft (stbd of centerline) (Removed weight from port side)

  28. Example Answer TCGf=(TCG0D0+STcgawa-STcgrwr)/Df 0ft=(0.033ft×6881LT+dTcg×10torps×2LT/ torp)/6881LT dTcg = -(0.033ft×6881LT)/20LT = -11.4ft (to port) Shift 10 torpedoes each 11.4ft to port to compensate for the loss of weight on the port side of the sub.

  29. K CL Section 3.3:What happens when “G” leaves the Centerline? Initial Condition: G shifts: Ship responds: WL WL WL D D D G1 G0 G1 B0 B0 FB FB B1 FB BL F1 K K CL CL As the ship lists/trims, the shape of the submerged volume changes moving B outboard until it slides under G. *Since the total weight of the ship has not changed, the total submerged volume remains constant, but its shape changes.

  30. K K CL CL Ship responds to opposite weight shift: Where the lines of action of the various centers of buoyancy cross* is the Metacenter M F2(-) F1(+) WL G2 WL G0 B2 B1 FB B2 FB B0 BL F2 K CL *Lines of action cross at a single point only for “small” angles of inclination (<10º).

  31. Shapes which impact KM: M M WL B2 B1 B0 WL B1 B2 BL B0 K BL CL K CL Highly curved hull cross-section: Little buoyant volume at large lever arm: M is at/near center of curvature Very flat hull cross-section: Large buoyant volume at large lever arm: M is high

  32. Distance from G to MT = Metacentric Height = Major player in stability calculations (+ keeps ship upright) Distance from B to MT = Transverse Metacentric Radius ML MT GMT WL BMT G KMT B KG KB BL TCG/TCB (-) TCG/TCB (+) K CL Locations and Line Segments for Hydrostatic Calculations

  33. G3 Example Problem G2 G1 Radius =3ft A rocking chair’s “skids” have a radiusof curvature of 3ft. The chair’s initial center of gravity is 2.5ft above the skids. A box is put on the seat which raises the combined center of gravity to 3ft above the skids. Another box is put on top of the first which raises the combined center of gravity to 3.5ft above the skids. • For each of these conditions, when the chair is tipped 45°, show how the forces of gravity and support are spatially related and predict how the chair will react when released. • What point in this scenario is analogous to a ship’s metacenter?

  34. Example Answer G1 G1: Support is outboard center of gravity creating a couple which returns the chair upright. G2: Support is aligned with center of gravity eliminating any couple. The chair maintains position. G3: Support is inboard center of gravity creating a couple which tips the chair over. The center of curvature of the rocking chair’s “skids” correspond to a ship’s metacenter. Support G2 Support G3 Support

  35. Section 3.4:Angle of List for Small Angles after Transverse Weight Shift For a given transverse weight shift, what is the corresponding change in list angle? g0 t MT gf F G0 D WL Gf B0 Bf FB CL BL

  36. Go Bo B L K C L • Up to now we’ve considered ship’s floating on an even keel • …(no list or trim). • The following points are noted: • K, keel • B, center of buoyancy • G, center of gravity One point of particular note remains…. …MT, or the Transverse Metacenter

  37. Go Bo B L K C L The Transverse Metacenter (MT) represents a convenient point of reference for small changes in the angle of inclination, F, (less than 10o) MT

  38. B1 B L C L For small changes in inclination, the point MT is where the ship is assumed to rotate. MT F Go Bo K ...The MT is generally about 10-30ft above the keel

  39. ML O ) ( There is also a Longitudinal Metacenter, or ML... ϑ …usually in the magnitude of 100 to 1000ft above the keel

  40. When the ship reacts to an off-center load (which will change the ship’s CG),... ...the center of buoyancy will shift until it is vertically aligned with the new CG... G1 …G1 can be assumed to move PERPENDICULAR from the CL B1 B L C L FB MT F Go Bo Remember, this is only for listing of 10o or less

  41. MT The short leg is G0G1 The long leg is G0MT The hypotenuse is G1MT tan F = G0G1 G0MT Look at the right triangle formed by this shifting… F Go G1 SO…. Bo B1 C L (tan F = opp/adj… remember?) FB

  42. tan F = G0G1 G0MT MT Go KMT • From the Curves of Form you can get KMT • The Vertical Center of Gravity is KG0 • G0MT =KMT - KG0 Bo KG C L K With that fact understood, we can now determine the ANGLE OF LIST of a vessel due to a change in loading. How? • G0G1is the change in the transverse Center of Gravity

  43. Example: The USS Simpson (FFG-56) floats on an even keel at a 16ft draft. The KG is 20ft above the keel. After 1 week, 50LT of fuel has been used from a tank 11ft to port and 15ft above the keel. Find the angle of list after the fuel has been used. Step 1: Find the ship’s displacement From the curves of form, curve #1, 16ft draft crosses at 132 132 x 30LT = 3960LT

  44. KG1 = Ds0x KG0 - (w x Kg) Ds1 KG1 = 3960LT x 20ft - (50LT x 15ft) (3960 - 50)LT KG1 = 78,450LT-ft 3910LT Step 2: Find the new vertical CG (KG) KG1 = 20.06ft

  45. TCG1 = Ds0x TCG0 - (w x Tcg) Ds1 TCG1 = 3960LT x 0ft - (50LT x -11ft) (3960 - 50)LT TCG1 = 0 - (-550LT-ft) 3910LT Step 3: Find the Transverse CG (TCG) (minus because it’s to port) TCG1 = 0.141ft (shifts to starboard, away from removed weight)

  46. G0G1 is the change in the Transverse CG: • G0 = 0 (on the centerline) • G1 = .141ft G0G1 = .141ft • G0MT =KMT - KG0 • KMT from curves is 113 x .2ft = 22.6ft • KG0 = 20ft G0MT = 2.6ft Step 4: Define lengths of G0G1 and G0MT

  47. MT tan f = opposite adjacent 2.6ft tan f = G0G1 G0MT tan f = .141ft 2.6ft .14ft Step 5: (Almost there!) Find tan f: F Go G1 tan f = 0.0541 atan 0.0541 = f 3.10o = f C L

  48. Section 3.5The Inclining Experiment In the previous section, we derived the relationship between a shift in weight and the resultant list/trim angle: tan(F) = wt/(DG0MT) • w,t are the weight and distance moved – usually known • The location of MT and the magnitude of D are properties of the hull shape read from the Curves of Form for the appropriate draft (T). • How do we find the location of G0?

  49. How do we find the location of G0? We determine it experimentally after new construction for a class or any major permanent complex weight redistributions for a given ship (alteration/conversion). Inclining Experiment Procedure: 1. Configure the ship in a “light” condition 2. Bring on large weights (~2% of Dship), move to known distances port and starboard of centerline and measure tan(F) using “plum bob”. Measure & record Dincl using draft and Curves of Form. 3. Plot wt vs. tan(F); divide slope by Dincl to get GinclMT 4. Calculate KGincl = KMT(from Curve of Form)–GinclMT 5. KG0=KGlight=(KGinclDincl–Kginclwtswinclwts)/(Dincl–winclwts)

  50. Inclining Experiment Tools -Plot: -Plumb Bob: Mast dadj tan(F)=dopp/dadj F Inclining Moment, wt (LT-ft) Scale dopp Tangent of Inclining Angle (Tan[F])

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