Section 3.1 Archimedes Principle: Forces on a body in water

Section 3.1 Archimedes Principle: Forces on a body in water
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

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

Static Equilibrium åF=0 åM=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.

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?

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

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

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

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 2

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)

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

the location of G, the center of gravity of the ship
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 G1 A change in VCG (or KG) KGnew A change in the TCG TCG g Go B L It also causes a change in the longitudinal CG (LCG), but we’ll discuss that later... 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 C L K Go g G1 B 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 G0 G1 B L K C L

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

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

Dsold x KGold Dsold + wadd KGnew = + wadd x Kg
Use the concept of weighted averages to determine the new CG: g G1 G0 B K L C L KGnew = + wadd x Kg Dsold + wadd Dsold x KGold

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

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

In this unique case, Dsnew and Dsold and 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: Ds GnewGold = w g2g1 ...This is ONLY for a single vertical weight shift!!

Ds GnewGold = w g2g1 Where: GnewGold is 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

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

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

Ds Ds = 4410LT Ds = 147 x 30LT Step 1: Draw picture! B K C L G0 ?
150 LT 6’ 19.5’ ? Ds 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

Dsold x KGold Dsnew KGnew = + Swi x Kgi KGnew = KGnew =
Step 3: Write the GENERAL Equation KGnew = + Swi x Kgi Dsnew Dsold x KGold Step 4: Substitute in values into the general equation KGnew = 4410LT x 19.5ft LT x 6ft 4410LT + 150LT KGnew = LT-ft LT-ft 4560 LT

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

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

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)

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.

Section 3.3: What happens when “G” leaves the Centerline?
Initial Condition: G shifts: Ship responds: CL K WL WL D D WL D G1 G0 G1 B0 B0 FB FB FB B1 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.

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

Shapes which impact KM:
WL B2 B1 B0 WL B2 B1 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

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

Example Problem G3 G2 G1 Radius =3ft 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?

Example Answer 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. G1 Support G2 Support G3 Support

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

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 C L B Go Bo K One point of particular note remains…. …MT, or the Transverse Metacenter

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

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

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

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

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

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

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

Step 2: Find the new vertical CG (KG)
KG1 = Ds0 x KG0 - (w x Kg) Ds1 KG1 = 3960LT x 20ft - (50LT x 15ft) ( )LT KG1 = 78,450LT-ft 3910LT KG1 = ft

Step 3: Find the Transverse CG (TCG)
TCG1 = Ds0 x TCG0 - (w x Tcg) Ds1 TCG1 = 3960LT x 0ft - (50LT x -11ft) ( )LT (minus because it’s to port) TCG1 = (-550LT-ft) 3910LT TCG1 = ft (shifts to starboard, away from removed weight)

G0G1 is the change in the Transverse CG: G0 = 0 (on the centerline)
Step 4: Define lengths of G0G1 and G0MT 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

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

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/(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? 2

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) 3

Inclining Experiment Tools
-Plot: -Plumb Bob: Mast dadj F tan(F)=dopp/dadj Inclining Moment, wt (LT-ft) Scale dopp Tangent of Inclining Angle (Tan[F]) 4

So far we’ve established that the angle of list can be found using
MT So far we’ve established that the angle of list can be found using the right triangle identified here: F The short leg is G0G1 The long leg is G0MT The hypotenuse is G1MT Go G1 SO…. Bo tan F = G0G1 G0MT B1 C L FB ...And so we can find the angle of list

Up to now, however, G0MT has been given based upon a KG that
F Up to now, however, G0MT has been given based upon a KG that has been provided. Go G1 We’ll now see how KG can be found by determining G0MT Bo B1 C L This is done by the Inclining Experiment FB

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

In earlier discussions an equation was derived for a shift in of
a single weight: Ds G0G1 = w g0g1 …where g0g1 was the distance that the weight was shifted. Let’s call that distance “t”. Sooo,... Ds G0G1 = wt And re-look at the equation for the angle of list: tan F = G0G1 G0MT

Note that the common term in both equations is G0G1. So
let’s isolate it in each equation: tan F = G0G1 G0MT Ds G0G1 = wt G0MT tan F = G0G1 G0G1 = wt Ds G0MT tan F = wt Ds

wt Ds tan F Ds G0MT = wt G0MT tan F =
That’s nice,… but not nice enough... One more rearrangement and we’ll have what we really want, G0MT: G0MT = wt tan F Ds

wt G0MT = 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

“w” and Ds will remain constant
G0MT = 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 G0MT

Ds Dy DWt Dx Dtan f wt (slope of wt vs tan F curve) G0MT = tan F Ds
Remember, slope is Dy/Dx: Or... = Dy Dx DWt Dtan f So... Average G0MT = (slope of wt vs tan F curve) Ds

Ds (slope of wt vs tan F curve)
When you vary the distance t, and thus the moment, you’ll vary the inclination angle. The result is plotted in an example here: The slope of the “best fit” line, Dy/Dx, when divided by the displacement, will give the average G0MT distance: Average G0MT = (slope of wt vs tan F curve) Ds

KG light = KG inclined x Ds old - Kg x w Ds new
Having found the Average G0MT, you can find the KG when the ship is loaded with the inclining weight: KG = KMT - G0MT 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 only- no crew, stores, fuel, etc.- is what we wanted!!

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 G0MT can be found Once G0MT is found, you can find KG of the light ship

Example Problem What is KGlight?
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. Dlightship=18700LT; KMT=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 KGlight?

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:

Example Answer Slope=(18750-[-18750]LT-ft)/(.225-[-.227]) =83000LT-ft
GMTincl=slope/Dincl=83000LT-ft/19100LT =4.35ft KGinc l=KMT-GMTincl=21ft-4.35ft=16.65ft KGlight =(KGinclDincl-Kgwtswwts)/Dlight =(16.65ft×19100LT-47ft×400LT)/18700LT =16ft

Section 3.6 Longitudinal Changes
Tm=(Taft+Tfwd)/2 Trim=Taft - Tfwd If ship is “trimmed by the stern”, Bow is up Taft> Tfwd Trim is (+) DWL WL Taft Tfwd F Ap daft dfwd Fp 2

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

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

The ship will pivot about the center of flotation, F.
dTfwd The change in draft will be evident in a change of draft forward... dTaft …and aft. w dTrim The difference between the fore and aft drafts is the change in trim: Trim = dTaft - dTfwd _ . F

Graphically, it looks like this
Graphically, it looks like this. First, the ship is represented with a line representing its initial state: FP AP O )( _ . F As weight is added, the the ship rotates about F: O )( AP FP _ . F w 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.

Now, rotate the sheet so that the line drawn becomes level and acts as
the new waterline: O )( AP w FP _ . F w O )( F _ . dTfwd dTaft dTrim The changes in draft can now be read directly… dTaft is below the WL, so it’s subtracted. dTfwd is above the waterline, so it’s added to the draft.

dTPS = w TPI dTrim = wl MT1”
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”: dTPS = 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”

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” dTPS = w TPI Tfwd new = Tfwd old +/- dTPS +/- dTrim AND Taft new = Taft old +/- dTPS +/- dTrim

Let’s consider change due to the parallel sinkage of the vessel first:
dTPS = 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

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

...Now we need to find how much of the trim is aft and how
The value for dTrim will be for the entire length of the ship: dTrim = wl MT1” l dTfwd O )( dTrim w _ . dTaft F Lpp ...Now we need to find how much of the trim is aft and how much is forward!

To find the trim distribution, consider the similar triangles formed below:
dTfwd O )( dTrim w _ . dTaft F daft dfwd Lpp The largest triangle shows the TOTAL change in trim, dTrim The hatched green triangle shows the forward trim dTfwd The hatched yellow area triangle shows the aft trim, dTaft

dTrim Lpp dTaft daft dTfwd dfwd = _ . O daft dfwd
)( dTrim w _ . dTaft F daft dfwd Lpp For these similar triangles there is a ratio aspect that relates to each: dTrim Lpp dTaft daft dTfwd dfwd = (The short leg divided by the long leg of the triangle!)

Tfwd new = Tfwd old +/- dTPS +/- dT
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: Tfwd new = Tfwd old +/- dTPS +/- dT AND Taft new = Taft old +/- dTPS +/- dT

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.

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 Draft = ( )/2 = 10.3ft amidships = ft Ds = 2LT x 205 = 410LT LCF = 55.8 ft from FP, or 4.95 ft aft of amidships DRAW A PICTURE! 101.7 Daft = 45.9 Dfwd = 55.8 )( _ . dTaft dTfwd dTrim F O 10LT 19.95

_ . dTPS = w TPI dTPS = 10LT 235 x .02LT/in dTPS = 2.13in O
101.7 Daft = 45.9 Dfwd = 55.8 )( _ . dTaft dTfwd dTrim F O 10LT 19.95 Step 1: Find change due to parallel sinkage dTPS = w TPI dTPS = LT 235 x .02LT/in dTPS = in

dTrim = 10LT x 19.95ft 252.5 x .141 LT-ft/in
101.7 Daft = 45.9 Dfwd = 55.8 )( _ . dTaft dTfwd dTrim F O 10LT 19.95 Step 2: Find change due to moment dTrim = wl MT1” dTrim = LT x 19.95ft x .141 LT-ft/in dTrim = in

_ . dTrim Lpp dTaft daft dTfwd dfwd = O
101.7 Daft = 45.9 Dfwd = 55.8 )( _ . dTaft dTfwd dTrim F O 10LT 19.95 Step 3: Divide the dTrim based on similar triangles dTrim Lpp dTaft daft dTfwd dfwd = 4.21in 101.7ft 45.9ft 55.8ft 5.60in dTaft = 45.9ft x = 2.53 in 101.7ft 5.60in dTfwd = 55.8ft x = 3.07 in 101.7ft

Tfwd new = Tfwd old +/- dTPS +/- dTmoment
101.7 Daft = 45.9 Dfwd = 55.8 )( _ . dTaft dTfwd dTrim F O 10LT 19.95 Step 4: Sum the changes in draft fore and aft Forward: Tfwd new = Tfwd old +/- dTPS +/- dTmoment Tfwd new = 10.1ft + (2.13in in) x (1ft/12in) Tfwd new = 10.1ft + .43ft Tfwd new = ft

Taft new = Taft old +/- dTPS +/- dTmoment
101.7 Daft = 45.9 Dfwd = 55.8 )( _ . dTaft dTfwd dTrim F O 10LT 19.95 Step 4: Sum the changes in draft fore and aft Aft: Taft new = Taft old +/- dTPS +/- dTmoment Taft new = 10.5ft + (2.13in in) x (1ft/12in) Taft new = 10.5ft ft Taft new = ft

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

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

Background Lab 2 Terminology Equations General Safety Displacement
Buoyant Force Equations D=rgÑ=FB General Safety Immediately clean up any water spilled to avoid fall hazard

Apparatus Equipment Procedures for taking measurements 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

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

Calculations FB TPI MT1” Plots/sketches None

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 SFV=(-)D+FB+Fblocks=0 FB=rgÑS D =rgÑS+ Fblocks 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= Fblocks) 2

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

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 GMT? Df= D0-wr= D0-P Ship’s weight/displacement is decreased KGfDf= KG0D0-Kgrwr, but Kgr=0; KGfDf= KG0D0; KGf= KG0D0/Df= KG0D0/(D0-P); Center of Gravity moves up due to keel weight removal GMT= KMT – KGf Shorter distance between Center of Gravity and Metacenter gives less distance to develop a righting moment WL G0 M B FB P=weight removed D0 Gf Df Disturbance trying to roll the ship

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. D M G WL B FB Fground=P=D-FB

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.

Example Problem Lpp=465 feet TPI=50LT/in MT1”=1400ft-LT/in
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 Tf=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

Example Answer Tfinal fwd=Tinitial fwd±dTps±dTfwd dTps=w/TPI dTfwd=dTrim×Dfwd/Lpp dTrim=wl/MT1” Tfinal fwd= Tinitial 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 100ft 25ft w 232.5ft l=125ft F amidship FP AP Lpp=465ft Daft=207.5ft Dfwd=257.5ft V=w/(rg)=172.8LT/[(64lb/ft³)×2240lb/LT×7.4805gal/ft³]=45,243gal This is just another application of moments!

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?

Example Answer At T=10ft, FB= 62×30LT = 1860LT;
P=D-FB=3600LT-1860LT = 1740LT At T=10ft, TPI=128×0.2LT/in = 25.6LT/in; Raising water level 1in removes an additional 25.6LT from the blocks

Background Lab 3 Lab Objectives
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

Background Lab 3 Concepts/Principles KG TCG MT Inclining Experiment

Background Lab 3 Terminology Equations Light-ship condition
Inclined ship condition Plum bob Equations GinclMT= wt/tan(F)×1/D KGincl = KMT(from Curve of Form)–GinclMT KG0=KGlight=(KGinclDincl–Kginclwtswinclwts)/(Dincl–winclwts)

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

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

Calculations Use equations Plots/sketches w×t vs. tan(F)

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

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

Chapter 1: Engineering Fundamentals
Force × distance Equal and opposite forces applied with an offset distance to produce a rotation åF=0; åM=0 P= rgz Mx=òydA Translational: heave, surge, sway Rotational: roll, pitch, yaw List, trim, heel p/r+V²/2+gz=constant

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

Chapter 2: Hull Form and Geometry
Plans Body: Section Lines Sheer: Buttock Lines Half-Breadth: Waterlines Depth(D), draft(T), beam(B), freeboard Centroid (location): LCF=(2/AWP)*òxdA Center of waterplane area Center of submerged volume òydx=Dx/3*[1y0+4y1+2y2+4y3+…+2yn-2+4yn-1+1yn] D, LCB, KB, TPI, AWP, LCF, MT1”, KML, KMT Draft->proper curve, proper axis, proper multiple/units

Simpson Integrals Waterplane Area Sectional Area
(Half-Breadth Plan) Y y(x) Half- Breadths (feet) See your “Equations and Conversions” Sheet Waterplane Area AWP=2òydx; where integral is half breadths by station Sectional Area Asect=2òydz; where integral is half breadths by waterline dx=Station Spacing X Stations Z (Body Plan) dz=Waterline Spacing Water lines y(z) Y Half-Breadths (feet) 8

Simpson Integrals Submerged Volume Longitudinal Center of Floatation
See your “Equations and Conversions” Sheet Submerged Volume ÑS=òAsectdx; where integral is sectional areas by station Longitudinal Center of Floatation LCF=(2/AWP)*òxydx; where integral is product of distance from FP & half breadths by station Asect A(x) Sectional Areas (feet²) dx=Station Spacing X Stations (Half-Breadth Plan) Y y(x) Half- Breadths (feet) dx=Station Spacing x X Stations 8

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

The Buoyant Force on an object is equal to the weight of the volume of the water displaced by the object: FB=rgV For box shaped barge, FB= rgV = P×Awp= rgzAwp åF=0; åM=0 Center of Gravity (G) Df= D0+Swa-Swr KGfDf= KG0D0+SKgawa-SKgrwr TCGfDf= TCG0D0+STcgawa-STcgrwr gf Gf WL Gi G0 g0 G moves parallel to weight shift BL K CL

M WL CL K BL G B MT KB BMT KMT KG GMT ML TCG/TCB (+) TCG/TCB (-) WL CL K B0 BL B1 B2 G0 FB F1(+) F2(-) tan(F) = wt/(DG0MT) To find KG: Plot wt vs. tan(F); divide slope by Dincl to get GinclMT KGincl = KMT(from Curve of Form)–GinclMT KG0=KGlight=(KGinclDincl–Kginclwtswinclwts)/(Dincl–winclwts)

Trim Equations: dTPS=w/TPI dTrim=wl/MT1” dTfwd/aft/dfwd/aft =dTrim/Lpp Tfinal fwd/aft=Tinitial fwd/aft±dTPS±dTfwd/aft Weight Added dTPS F l w dTfwd dTrim q dTaft Tfinal fwd daft dfwd Ap Lpp Fp Tfinal aft

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?

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

Section 3.1 Archimedes Principle: Forces on a body in water

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Section 3.1 Archimedes Principle: Forces on a body in water

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