Section 3.1. Archimedes Principle: Forces on a body in water

Transcription

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 =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 Horizontal components of pressure force are negated by equal force on opposite side of barge. F B =PA; P=rgz; F B =rgza; G B F B A V=zA; F B =rgv z

3 Static Equilibrium F=0 Weight, Buoyancy, Drag and Lift forces all sum to zero in each dimension M=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 P hyd =rgz=64lb/ft³ 5ft [1ft²/144in²]=2.22psi F hyd =P hyd 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=t SW =10ft w=rg SW V SW =64lb/ft³ 50ft 100ft 10ft=3,200,000lb V FW =w/rg FW =3,200,000lb/62.4lb/ft³=51,280ft³ T FW =V FW /A WP =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 D s = Displacement of ship (LT) W = Magnitude of Gravitational Force/Weight of object (LT)

10 So far we ve looked at ships that are in STATIC EQUILIBRIUM: SF x = 0 SF y = 0 SF z = 0 SM p = 0 G o B o K B L C L 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 G o G 1 KG new It also causes a change in the longitudinal CG (LCG), but we ll discuss that later... K C L B L

12 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 B L

13 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 G 1 G 0 K B L C L

14 In the case of a weight SHIFT, the CG first shifts AWAY from the removed weight. g G 2 g G 0 G 1 K B L C L and the TOWARDS the relocated weight

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 G 1 G 0 KG new KG old K B L C L Causing a change in the VERTICAL distance, or KG

16 Use the concept of weighted averages to determine the new CG: g G 1 G 0 K C L B L KG new = Ds old x KG old + w add x Kg Ds old + w add

17 It s the same deal for removing a weight, only this time the weight is negative (i.e. removed): g G 0 G 1 Kg KG old KG new K C L B L KG new = Ds old x KG old + (-w) x Kg Ds old + (-w)

18 In a relocation of a weight, look at it as SUBTRACTING one weight, and ADDING another weight. g G 1 G 0 KG old KG new g Kg 1 Kg 2 K B L C L

19 In this unique case, Ds new and Ds old and are the SAME THING! w 1 and w 2 are also the same thing! The weight has only moved, not been removed So we can rearrange the formula: Ds G new G old = w g 2 g 1...This is ONLY for a single vertical weight shift!!

20 Ds G new G old = w g 2 g 1 Where: G new G old is the distance between the old and new CG s g 2 g 1 is the distance between the old and new Cg locations of the relocated weight...this relation will become important in the Inclining Experiment

21 We can generalize the formula for vertical changes in CG by the following: KG new = Ds old x KG old + Sw i x Kg i Ds old + Sw i

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 Step 1: Draw picture! Ds G LT G 1 K? B L C L 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 KG new = Ds old x KG old + Sw i x Kg i Ds new Step 4: Substitute in values into the general equation KG new = 4410LT x 19.5ft + 150LT x 6ft 4410LT + 150LT KG new = LT-ft + 900LT-ft 4560 LT

25 KG new = LT-ft 4560 LT KG new = 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: D o =6900LT Tcg #2RFT = -12ft (i.e. port of centerline); TCG o =0ft Capacity #2RFT =5000gal fw

27 Example Answer w #2RFT =rgv =5000gal [1ft³/7.4805gal] rg fw =668.4ft³ 62.4lb/ft³ [1LT/2240lb] =18.62LT D f = D 0 +Sw a -Sw r =6900LT-19LT=6881LT TCG f =(TCG 0 D 0 +STcg a w a -STcg r w r )/D f =(0ft 6900LT-[-12ft] 19LT)/6881LT =0.033ft (stbd of centerline) (Removed weight from port side)

28 Example Answer TCG f =(TCG 0 D 0 +STcg a w a -STcg r w r )/D f 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 Section 3.3: What happens when G leaves the Centerline? Initial Condition: G shifts: Ship responds: D G 0 WL D WL G 1 D G 1 WL F B B 0 F B B 0 F B B 1 B L K C L K C L F 1 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 Ship responds to opposite weight shift: G 2 WL Where the lines of action of the various centers of buoyancy cross* is the Metacenter M F 2 (-) F 1 (+) B 2 G 0 WL B 1 F B B 2 F 2 F B B 0 B L K C L *Lines of action cross at a single point only for small angles of inclination (<10º).

31 Shapes which impact KM: M B 2 B 1 B 0 B L K C L WL Highly curved hull cross-section: Little buoyant volume at large lever arm: M is at/near center of curvature B 2 M B 0 K B L B 1 C L WL Very flat hull cross-section: Large buoyant volume at large lever arm: M is high

32 M L Distance from B to M T = Transverse Metacentric Radius M T Distance from G to M T = Metacentric Height = Major player in stability calculations (+ keeps ship upright) KM T BM T G GM T WL KB B KG TCG/TCB (-) B L C L K TCG/TCB (+) Locations and Line Segments for Hydrostatic Calculations

33 Example Problem G 3 G 2 A rocking chair s skids have a radius of 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? G 1 Radius =3ft

34 Example Answer G 1 Support G 2 G 1 : Support is outboard center of gravity creating a couple which returns the chair upright. G 2 : Support is aligned with center of gravity eliminating any couple. The chair maintains position. Support G 3 G 3 : Support is inboard center of gravity creating a couple which tips the chair over. Support The center of curvature of the rocking chair s skids correspond to a ship s metacenter.

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? g 0 t WL G 0 M T F D G f g f B 0 F B B f C L B L

36 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. G o B o K B L M T, or the Transverse Metacenter C L

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

38 For small changes in inclination, the point M T is where the ship is assumed to rotate. M T F B 1...The MT is generally about 10-30ft above the keel

39 There is also a Longitudinal Metacenter, or M L... M L 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),... M T F...the center of buoyancy will shift until it is vertically aligned with the new CG... G 1 G 1 can be assumed to move PERPENDICULAR from the CL B 1 Remember, this is only for listing of 10 o or less F B

41 M T Look at the right triangle formed by this shifting F The short leg is G 0 G 1 The long leg is G 0 M T The hypotenuse is G 1 M T SO. G 1 tan F = G 0 G 1 G 0 M T B 1 (tan F = opp/adj remember?) F B

42 With that fact understood, we can now determine the ANGLE OF LIST of a vessel due to a change in loading. How? tan F = G 0 G 1 G 0 M T M T G 0 G 1 is the change in the transverse Center of Gravity From the Curves of Form you can get KM T KM T G o The Vertical Center of Gravity is KG 0 G 0 M T = KM T - KG 0 KG B o K C L

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 x 30LT = 3960LT

44 Step 2: Find the new vertical CG (KG) KG 1 = Ds 0 x KG 0 - (w x Kg) Ds 1 KG 1 = 3960LT x 20ft - (50LT x 15ft) ( )LT KG 1 = 78,450LT-ft 3910LT KG 1 = 20.06ft

45 Step 3: Find the Transverse CG (TCG) TCG 1 = Ds 0 x TCG 0 - (w x Tcg) Ds 1 TCG 1 = 3960LT x 0ft - (50LT x -11ft) ( )LT TCG 1 = 0 - (-550LT-ft) 3910LT TCG 1 = 0.141ft (minus because it s to port) (shifts to starboard, away from removed weight)

46 Step 4: Define lengths of G 0 G 1 and G 0 M T G 0 G 1 is the change in the Transverse CG: G 0 = 0 (on the centerline) G 1 =.141ft G 0 G 1 =.141ft G 0 M T = KM T - KG 0 KM T from curves is 113 x.2ft = 22.6ft KG 0 = 20ft G 0 M T = 2.6ft

47 Step 5: (Almost there!) Find tan f: tan f = opposite adjacent M T F tan f = G 0 G 1 G 0 M T tan f =.141ft 2.6ft tan f = G 1 atan = f 3.10 o = f

48 Section 3.5 The Inclining Experiment In the previous section, we derived the relationship between a shift in weight and the resultant list/trim angle: tan(f) = wt/(dg 0 M T ) w,t are the weight and distance moved usually known The location of M T 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 G 0?

49 How do we find the location of G 0? 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 D ship ), move to known distances port and starboard of centerline and measure tan(f) using plum bob. Measure & record D incl using draft and Curves of Form. 3. Plot wt vs. tan(f); divide slope by D incl to get G incl M T 4. Calculate KG incl = KM T (from Curve of Form) G incl M T 5. KG 0 =KG light =(KG incl D incl Kg inclwts w inclwts )/(D incl w inclwts )

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

51 So far we ve established that the angle of list can be found using the right triangle identified here: M T F The short leg is G 0 G 1 The long leg is G 0 M T The hypotenuse is G 1 M T SO. G 1 tan F = G 0 G 1 G 0 M T B 1...And so we can find the angle of list F B

52 M T F Up to now, however, G 0 M T has been given based upon a KG that has been provided. G 1 We ll now see how KG can be found by determining G 0 M T B 1 This is done by the Inclining Experiment F B

53 By using a known weight and placing it at a known distance an angle of list can be measured By repeating this process - port and starboard- we can graph the relationship between the moment created by the weight and the angle of inclination This will allow an average inclined KG to be determined, and from that a KG for the ship in an condition of no list or trim can be established

54 In earlier discussions an equation was derived for a shift in of a single weight: Ds G 0 G 1 = w g 0 g 1 where g 0 g 1 was the distance that the weight was shifted. Let s call that distance t. Sooo,... Ds G 0 G 1 = wt And re-look at the equation for the angle of list: tan F = G 0 G 1 G 0 M T

55 Note that the common term in both equations is G 0 G 1. So let s isolate it in each equation: tan F = G 0 G 1 G 0 M T Ds G 0 G 1 = wt G 0 M T tan F = G 0 G 1 G 0 G 1 = wt Ds G 0 M T tan F = wt Ds

56 G 0 M T tan F = wt Ds That s nice, but not nice enough... One more rearrangement and we ll have what we really want, G 0 M T : G 0 M T = wt tan F Ds

57 G 0 M T = wt tan F Ds Let s review what we know: w is a known weight that is relocated t is the distance the weight is moved tan f is the angle created by the weight shift Ds is the displacement of the ship This will be the formula that governs the Inclining Experiment

58 G 0 M T = wt tan F Ds In the Inclining Experiment: The distance t is varied, changing the angle of list, tan f w and Ds will remain constant By varying t, thus varying the created moment of wt, the angle of inclination will change By plotting wt versus tan f, you can determine the average G 0 M T

59 G 0 M T = wt tan F Ds Remember, slope is Dy/Dx: Or... Dy Dx = So... DWt Dtan f Average G 0 M T = (slope of wt vs tan F curve) Ds

60 When you vary the distance t, and thus the moment, you ll vary the inclination angle. The result is plotted in an example here: Moment v. tan f m o m e n t (L T-ft) tan f The slope of the best fit line, Dy/Dx, when divided by the displacement, will give the average G 0 M T distance: Average G 0 M T = (slope of wt vs tan F curve) Ds

61 Having found the Average G 0 M T, you can find the KG when the ship is loaded with the inclining weight: KG = KM T - G 0 M T The problem now degenerates to a simple change in vertical center of gravity, KG, equation: KG light = KG inclined x Ds old - Kg x w Ds new KG light, the KG of the ship with considering the ship s weight onlyno crew, stores, fuel, etc.- is what we wanted!!

62 In Summary: Using a known weight and a measured distance, a moment is created The moment creates a list that can be measured By repeating the process with the same weight over different distances and plotting the results, the average G 0 M T can be found Once G 0 M T is found, you can find KG of the light ship

63 Example Problem The USS OHIO has just completed her Overhaul and Conversion from an SSBN to an SSGN and Special Operations Forces platform. She is pierside performing a required Inclining Experiment. D lightship =18700LT; KM T =21ft. The inclining gear weighs 400LTs and is centered 47ft above the keel. 375LTs is moved to the following transverse distances resulting in the corresponding list angles. Distance to Starboard(ft) List Angle( ) What is KG light?

64 Example Answer Multiply transverse distances by 375LT to get inclining moment. Take tangent of list angle and plot the two derived sets of data against one another: Data From Inclining Experiment Inclining Moment (LT-ft) Tangent of the Inclining Angle

65 Example Answer Slope=(18750-[-18750]LT-ft)/(.225-[-.227]) =83000LT-ft GM Tincl =slope/d incl =83000LT-ft/19100LT =4.35ft KG inc l =KM T -GM Tincl =21ft-4.35ft=16.65ft KG light =(KG incl D incl -Kg wts w wts )/D light =(16.65ft 19100LT-47ft 400LT)/18700LT =16ft

66 Section 3.6 Longitudinal Changes T m =(T aft +T fwd )/2 Trim=T aft - T fwd If ship is trimmed by the stern, Bow is up T aft > T fwd Trim is (+) T aft Ap d aft d fwd F Fp DWL WL T fwd

67 Consider a ship floating on an even keel, that is, no list or trim... _.

68 When a weight, w, is added, it causes a change draft. _. _.

69 The ship will pivot about the center of flotation, F. The change in draft will be evident in a change of draft forward... and aft. dt aft dtrim F _. dt fwd The difference between the fore and aft drafts is the change in trim: Trim = dt aft - dt fwd

70 Graphically, it looks like this. First, the ship is represented with a line representing its initial state: AP F FP O)( _. As weight is added, the the ship rotates about F: AP F O)( FP _. You can simulate this on your paper by turning the sheet in the direction that the bow or stern would sink because of the added weight, then drawing a line to represent the new position.

71 Now, rotate the sheet so that the line drawn becomes level and acts as the new waterline: AP w FP _. The changes in draft can now be read directly dt fwd dtrim w _. dt aft dt aft is below the WL, so it s subtracted. dt fwd draft. is above the waterline, so it s added to the

72 There are two aspects of draft to consider when finding the change in draft: 1. Change in draft due to the parallel sinkage of the vessel due to the added weight, w : dt PS = w TPI 2. Change in draft due to the moment created by the added weight at a distance from F, or wl : dtrim = wl MT1

73 These two measurements- change due to parallel sinkage and change in trim due to moment- when added with the initial draft will give you the TOTAL draft, forward and aft: dtrim = wl MT1 dt PS = w TPI T fwd new = T fwd old +/- dt PS +/- dtrim AND T aft new = T aft old +/- dt PS +/- dtrim

74 Let s consider change due to the parallel sinkage of the vessel first: dt PS = w TPI TPI, Tons Per Inch Immersion is a geometric function of the vessel at a given draft and is taken from the Curves of Form The added weight, w, will cause the vessel to sink a small distance for the length of the entire vessel We assume that the weight is applied at F! This assures that the sinkage is uniform over the length of the ship

75 Now consider the change in trim due to the created moment of the added weight: dtrim = wl MT1 MT1, or the Moment to Trim 1, is also from the Curves of Form The weight, w, at a distance, l, from the center of flotation, F, creates a moment that causes the ship to rotate about F This rotation causes one end to sink and the other end to rise The degree of rise or fall depends on the location of F with regard to the entire length of the ship as given by Lpp

76 The value for dtrim will be for the entire length of the ship: dtrim = wl MT1 l dt fwd dtrim w _. dt aft Lpp...Now we need to find how much of the trim is aft and how much is forward!

77 To find the trim distribution, consider the similar triangles formed below: l dt aft dt fwd dtrim w _. d aft Lpp d fw d The largest triangle shows the TOTAL change in trim, dtrim The hatched green triangle shows the forward trim dt fwd The hatched yellow area triangle shows the aft trim, dt aft

78 l dt fwd dtrim w _. dt aft d aft Lpp d fw d For these similar triangles there is a ratio aspect that relates to each: dtrim L pp dt aft d aft = = dt fwd d fwd (The short leg divided by the long leg of the triangle!)

79 Knowing how to find the change in draft from both parallel sinkage and from the induced moment, you can now find the total draft change, fore and aft: T fwd new = T fwd old +/- dt PS +/- dt AND T aft new = T aft old +/- dt PS +/- dt

80 Calculating Draft Changes Procedure: Calculate impact of weight addition/removal to mean draft using TPI. Calculate impact of weight addition/removal to trim at given distance from center of floatation. Calculate trim effect on fwd and aft drafts separately. Separately add mean draft impact to trim effects to determine final drafts fwd and aft.

81 Example: The YP floats at a draft 10.5 ft aft and 10.1ft forward. A load of 10LT is placed 15ft forward of amidships. Find the final forward and aft drafts. GIVEN: Lpp = ft amidships = ft Draft = ( )/2 = 10.3ft Ds = 2LT x 205 = 410LT LCF = 55.8 ft from FP, or 4.95 ft aft of amidships DRAW A PICTURE! D aft = 45.9 D fwd = )( _ O 10LT dt fwd dtrim. dt aft F 101.7

82 D aft = 45.9 D fwd = )( _ O 10LT dt fwd dtrim. dt aft F Step 1: Find change due to parallel sinkage dt PS = w TPI dt PS = 10LT 235 x.02lt/in dt PS = 2.13in

83 D aft = 45.9 D fwd = )( _ O 10LT dt fwd dtrim. dt aft F Step 2: Find change due to moment dtrim = wl MT1 dtrim = 10LT x 19.95ft x.141 LT-ft/in dtrim = 5.60in

84 D aft = 45.9 D fwd = )( _ O 10LT dt fwd dtrim. dt aft F Step 3: Divide the dtrim based on similar triangles dtrim dt = aft = L pp d aft dt fwd d fwd dt aft 5.60in = 45.9ft x = 2.53 in 101.7ft 4.21in 101.7ft dt aft = = 45.9ft dt fwd 55.8ft dt fwd 5.60in = 55.8ft x = 3.07 in 101.7ft

85 D aft = 45.9 D fwd = )( _ O 10LT dt fwd dtrim. dt aft F Step 4: Sum the changes in draft fore and aft Forward: T fwd new = T fwd old +/- dt PS +/- dt moment T fwd new = 10.1ft + (2.13in in) x (1ft/12in) T fwd new = 10.1ft +.43ft T fwd new = 10.53ft

86 D aft = 45.9 D fwd = )( _ O 10LT dt fwd dtrim. dt aft F Step 4: Sum the changes in draft fore and aft Aft: T aft new = T aft old +/- dt PS +/- dt moment T aft new = 10.5ft + (2.13in in) x (1ft/12in) T aft new = 10.5ft -.033ft T aft new = ft

87 Background Lab 2 Lab Objectives Reinforce students understanding of Archimedes Principle Reinforce student s concept of static equilibrium Reinforce student s concept of the center of floatation

88 Background Lab 2 Concepts/Principles: Archimedes Principle Static Equilibrium Center of Floatation Simpson s First Rule Interpolation Hydrostatic Force TPI MT1

89 Terminology Displacement Buoyant Force Equations D=rg =F B Background Lab 2 General Safety Immediately clean up any water spilled to avoid fall hazard

90 Equipment Apparatus Floating bodies Tanks with weirs and spillways Buckets Scale Rulers 5 lb weights Procedures for taking measurements Record results measurements of models and weighing of buckets

91 Data Collection/Reduction Data to be collected & Expected results These should be equal Weight of model Weight of water Calculated water volume displaced Hydrostatic Force Longitudinal Center of Floatation (LCF) Sources of error Measurements Insufficient drip time

92 Data Collection/Reduction Calculations F B TPI MT1 Plots/sketches None

93 Section 3.8: Dry Docking How is the ship s weight shared between docking blocks and buoyant force? Requirements for Static Equilibrium still apply: SF=0; SM=0 SF V =(-)D+F B +F blocks =0 F B =rg S D =rg S + F blocks Since ship s weight remains constant, as hull comes out of water, submerged volume decreases, hence buoyant force decreases, and force from the blocks increases. (P= F blocks )

94 Dry-Docking If a list develops during docking, the increasing force from the blocks can work to capsize the ship WL D G M B WL D G M B F B F block =P=D-F B F B Solutions: Use side blocks to force a zero list Stop docking evolution and correct problem, if ship develops an increasing list P

95 Impact on Stability Consider force of blocks to be the same as a weight removal from the keel: What is the impact on KG and GM T? D f = D 0 -w r = D 0 -P Ship s weight/displacement is decreased KG f D f = KG 0 D 0 -Kg r w r, but Kg r =0; KG f D f = KG 0 D 0 ; Disturbance trying to roll the ship WL P=weight removed D 0 G 0 D f G f M B F B KG f = KG 0 D 0 /D f = KG 0 D 0 /(D 0 -P); Center of Gravity moves up due to keel weight removal GM T = KM T KG f Shorter distance between Center of Gravity and Metacenter gives less distance to develop a righting moment

96 Comparison to Grounding: Same stability concerns for both evolutions although grounding is obviously not planned or controlled. Since re-floating after grounding is generally not on level sea bed with a zero list, it should only be done at highest available tide to maximize buoyant force and righting moment and avoid capsizing. Pulling the ship directly off the shoal. WL F ground =P=D-F B D G M B F B

97 Floating the Ship Undocking has the same concerns as docking plus: The Center of Gravity may have been shifted by the work done in dock. All holes in the ship below the waterline need to be confirmed properly closed. Recovery from grounding concerns: The Center of Gravity may have been changed by flooded or damaged compartments. When ship floats again, damage previously held above the water could be submerged resulting in further damage.

98 Example Problem DD963 is preparing to enter drydock. It is currently moored pier side on an even keel and a draft of 18.5 feet. To ensure that the sonar dome rests properly on the blocks, the forward draft of the ship must be T f =17.5 feet. How much ballast must be removed from a tank located 100 feet forward of amidship? Give the answer in gallons of saltwater. Lpp=465 feet TPI=50LT/in MT1 =1400ft-LT/in LCF=25 feet aft of amidships

99 Example Answer T final fwd =T initial fwd ±dtps±dtfwd dtps=w/tpi dtfwd=dtrim Dfwd/Lpp dtrim=wl/mt1 AP Daft=207.5ft 100ft 25ft w 232.5ft F amidship l=125ft FP Lpp=465ft Dfwd=257.5ft T final fwd = T initial fwd ± w/tpi ± wl/mt1 Dfwd/Lpp = w/tpi ± wl/mt1 Dfwd/Lpp =(17.5ft-18.5ft) 12in/ft= -12 in =-w/(50lt/in) w(125ft)/(1400ft-lt/in) 257.5ft/465ft= (-)12in -12 in = -w/(50lt/in) w/(20.23lt/in) = -w/(14.4lt/in) w= -12in (-14.4LT/in)=172.8LT V=w/(rg)=172.8LT/[(64lb/ft³) 2240lb/LT gal/ft³]=45,243gal This is just another application of moments!

100 Example Problem An FFG-7 is in the process of undocking when the evolution is halted at 10ft of water on the hull. If D=3600LT, how much weight is being supported by the blocks? If the water level is raised 1in, how much additional weight is removed from the blocks?

101 Example Answer At T=10ft, F B = 62 30LT = 1860LT; P=D-F B =3600LT-1860LT = 1740LT At T=10ft, TPI= LT/in = 25.6LT/in; Raising water level 1in removes an additional 25.6LT from the blocks

102 Lab Objectives Background Lab 3 Reinforce students understanding of the theory behind inclining experiments Provide students with practical experience in conducting an inclining experiment Determine the KG of the 27-B-1 model for future laboratories

103 Background Lab 3 Concepts/Principles KG TCG M T Inclining Experiment

104 Background Lab 3 Terminology Light-ship condition Inclined ship condition Plum bob Equations G incl M T = wt/tan(f) 1/D KG incl = KM T (from Curve of Form) G incl M T KG 0 =KG light =(KG incl D incl Kg inclwts w inclwts )/(D incl w inclwts )

105 Apparatus General Safety Minimize water on the floor Equipment 27-B-1 Models Weights Plum bobs Procedures for taking measurements Record measurements

106 Data Collection/Reduction Data to be collected & Expected results 27-B-1 Model Numbers Weight of Models Drafts Model dimensions Water temperature tan(f) Where do you expect KG to be? Sources of error Measurement error Round off

107 Data Collection/Reduction Calculations Use equations Plots/sketches w t vs. tan(f)

108 Review of Chapters 1-3 for Six Week Exam Chapter 1: Engineering Fundamentals Chapter 2: Hull Form and Geometry Chapter 3: Hydrostatics Review Equation & Conversion Sheet

109 Chapter 1: Engineering Fundamentals Drawings, sketches, graphs Dependent/independent variables Region under and slope of a curve Unit analysis Significant figures Linear interpolation Forces, moments, couples, static equilibrium, hydrostatic pressure, mathematical moments Six degrees of freedom Bernoulli s Equation

110 Chapter 1: Engineering Fundamentals Force distance Equal and opposite forces applied with an offset distance to produce a rotation F=0; M=0 P= rgz M x = yda Translational: heave, surge, sway Rotational: roll, pitch, yaw List, trim, heel p/r+v²/2+gz=constant

111 Chapter 2: Hull Form and Geometry Categorizing ships Ways to represent the hull form Table of Offsets Hull form characteristics Centroids Center of Flotation, Center of Buoyancy Simpson s Rule Curves of Form

112 Plans Chapter 2: Hull Form and Geometry Body: Section Lines Sheer: Buttock Lines Half-Breadth: Waterlines Depth(D), draft(t), beam(b), freeboard Centroid (location): LCF=(2/A WP )* xda Center of waterplane area Center of submerged volume ydx=dx/3*[1y 0 +4y 1 +2y 2 +4y y n-2 +4y n-1 +1y n ] D, LCB, KB, TPI, A WP, LCF, MT1, KM L, KM T Draft->proper curve, proper axis, proper multiple/units

113 Simpson Integrals See your Equations and Conversions Sheet Y Half- Breadths (feet) (Half-Breadth Plan) y(x) dx=station Spacing Waterplane Area A WP =2 ydx; where integral is half breadths by station 0 Stations X Sectional Area A sect =2 ydz; where integral is half breadths by waterline Water lines Z (Body Plan) dz=waterline Spacing y(z) 0 Half-Breadths (feet) Y

114 Simpson Integrals See your Equations and Conversions Sheet Submerged Volume S = A sect dx; where integral is sectional areas by station Sectional Areas (feet²) A sect 0 A(x) dx=station Spacing X Stations Longitudinal Center of Floatation LCF=(2/A WP )* xydx; where integral is product of distance from FP & half breadths by station Y Half- Breadths (feet) 0 (Half-Breadth Plan) y(x) dx=station Spacing x Stations X

115 Chapter 3: Hydrostatics Archimedes Principle/Static Equilibrium Impact to G of weight addition, removal, movement Metacenter Angle of list Inclining Experiment Trim calculations Drydocking

116 Chapter 3: Hydrostatics The Buoyant Force on an object is equal to the weight of the volume of the water displaced by the object: F B =rgv For box shaped barge, F B = rgv = P A wp = rgza wp F=0; M=0 Center of Gravity (G) g f Gf WL D f = D 0 +Sw a -Sw r KG f D f = KG 0 D 0 +SKg a w a -SKg r w r g 0 G i G 0 G moves parallel to weight shift TCG f D f = TCG 0 D 0 +STcg a w a -STcg r w r B L K C L

117 M F 2 (-) F 1 (+) B G 0 2 B 1 Chapter 3: Hydrostatics WL KM T BM T M L M T G GM T WL B L F B B 0 K C L tan(f) = wt/(dg 0 M T ) B L KB TCG/TCB (-) B C L K KG TCG/TCB (+) To find KG: Plot wt vs. tan(f); divide slope by D incl to get G incl M T KG incl = KM T (from Curve of Form) G incl M T KG 0 =KG light =(KG incl D incl Kg inclwts w inclwts )/(D incl w inclwts )

118 Chapter 3: Hydrostatics Trim Equations: dt PS =w/tpi dtrim=wl/mt1 dt fwd/aft /d fwd/aft =dtrim/l pp T final fwd/aft =T initial fwd/aft ±dt PS ±dt fwd/aft dtrim dtaft Weight Added w l F dt PS q dt fwd Ap d aft L pp d fwd Fp T final fwd T final aft

119 General Problem Solving Technique Write down applicable reference equation which contains the desired answer variable. Solve the reference equation for the answer variable. Write down additional reference equations and solve for unknown variables in the answer variable equation, if needed. Draw a quick sketch to show what information is given and needed and identify variables, if applicable. Rewrite answer variable equation, substituting numeric values with units for variables. Simplify this expanded equation, including units, to arrive at the final answer. Check the answer: Do units match answer? Is the answer on the right order of magnitude?

120 Summary Equation Sheet Assigned homework problems Additional homework problems Example problems worked in class Example Problems worked in text