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 objectFB=rgV

Forces on a body in waterDistributed 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

Box Shaped Barge:

FB=PA;P=rgz;FB=rgzA;

V=zA;FB=rgV

A

z

BG

Weight

FB

Horizontalcomponents ofpressure forceare negated byequal force onopposite side ofbarge.

Buoyant Forces

åF=0Weight, Buoyancy, Drag and Lift forces all sum tozero in each dimension

åM=0All forces in each dimension are colinear and cancel;i.e. there are no separation of the action points offorces such that couples or moments are generated.

Static Equilibrium

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

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)

CL

BL

Go

Bo

K

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

CL

BLK

Go

• A change in VCG (or KG)

KGnew

• g

A change in weight (either adding or removing it) will cause a change inthe location of G, the center of gravity of the ship

G1

• A change in the TCG

TCG

It also causes a change in thelongitudinal CG (LCG), butwe’ll discuss that later...

BL

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

CL

K

Go

• g

G1

CL

BLK

G1

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

CL

BLK

G0• g

In the case of a weight SHIFT, the CG first shifts AWAY from theremoved weight….

G1

…and the TOWARDS the relocated weight

• g

G2

CL

BLK

• g

Let’s first consider a weight added directly over the centerline

G0

… Causing a change in the VERTICAL distance, or KG

KGold

G1

This will cause the location of the CG to move TOWARD the weight ...

KGnew

CL

BLK

G0

G1

• g

KGnew = + wadd x Kg

Dsold + wadd

Dsold x KGold

Use the concept of weighted averages to determine the new CG:

CL

BLK

G1

G0

• g

KGnew = + (-w) x Kg

Dsold + (-w)

Dsold x KGold

KGoldKGnew

Kg

It’s the same deal for removing a weight, only this time the weight is negative (i.e. removed):

CL

BLK

• g

• g

KGold

KGnew

G0

G1

Kg1

Kg2

In a relocation of a weight, look at it as SUBTRACTING one weight, and ADDING another weight.

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:

...This is ONLY for a single vertical weight shift!!

Ds GnewGold =

w g2g1

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

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

Step 1: Draw picture!

CL

BLK

G1

G0

150LT

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

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 + 150LT x 6ft

4410LT + 150LT

KGnew = 86215.5 LT-ft + 900LT-ft

4560 LT

KGnew =87115.5 LT-ft

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

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?

WL

CL

KBL

G0

Initial Condition:

B0

WL

CL

K

G1

G shifts:

B0

WLG1

Ship responds:

B1

CL

K

As the ship lists/trims, the shape of thesubmerged volume changes movingB outboard until it slides under G.*Since the total weight of the ship hasnot changed, the total submerged volumeremains constant, but its shape changes.

FB

D D

FB FB

D

F1

CL

KCL

K

WL

Ship respondsto opposite weightshift:

B2

G2

WL

CL

K

Where the lines of action of the variouscenters of buoyancy cross* is the Metacenter

B0BL

B1B2

M

G0

*Lines of action cross at a single point only for“small” angles of inclination (<10º).

FBFB

F2

F1(+)F2(-)

Shapes which impact KM:

WL

CL

K

B0BL

B1B2

M

WL

CL

KBL

B0

B1B2

Highly curved hull cross-section:Little buoyant volume at largelever arm: M is at/near center ofcurvature

Very flat hull cross-section:Large buoyant volume at largelever arm: M is high

M

WL

CL

KBL

G

B

MT

KB

BMTKMT

KG

GMT

ML

Locations and Line Segments for Hydrostatic Calculations

Distance from G to MT = Metacentric Height =

Major player in stability calculations (+ keeps

ship upright)

Distance from B to MT = Transverse Metacentric Radius

TCG/TCB (+)TCG/TCB (-)

Example Problem

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?

Radius=3ft

G1

G2

G3

Example AnswerG1: 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

G2

G3

Support

Support

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?

WL

BL

CL

MT

G0 Gf

Bf

B0

g0

gf

FB

DF

t

CL

BL

Go

Bo

K

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

CL

BL

Go

Bo

K

The Transverse Metacenter (MT) represents a convenient point of reference for

small changes in the angle of inclination, F, (less than 10o)

MT

CL

BL

Go

Bo

K

For small changes in inclination, the point MT is where the ship is assumed to rotate.

MT

...The MT is generally about 10-30ft above the keel

F

B1

There is also a Longitudinal Metacenter, or ML...

(O)

ML

…usually in the magnitude of 100 to 1000ft above the keel

ϑ

CL

BL

Go

Bo

MT

F

When the ship reacts to an off-center load (whichwill change the ship’s CG),...

FB

B1

...the center of buoyancy will shift until it is vertically aligned with the new CG...

Remember, this is only for listingof 10o or less

G1

…G1 can be assumed to movePERPENDICULAR from the CL

CL

Go

Bo

MT

F

FB

B1

G1

Look at the right triangle formed by this shifting…

The long leg is G0MT

The hypotenuse is G1MT

tan F = G0G1

G0MT

SO….

The short leg is G0G1

(tan F = opp/adj… remember?)

With that fact understood, we can now determine the ANGLE OF LISTof a vessel due to a change in loading.

tan F = G0G1

G0MT

• G0G1 is the change in the transverse Center of Gravity

CL

Go

Bo

K

MT

• From the Curves of Form you can get KMT

• The Vertical Center of Gravity is KG0

• G0MT = KMT - KG0

KM

T

KG

How?

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

KG1 = Ds0 x KG0 - (w x Kg)

Ds1

KG1 = 3960LT x 20ft - (50LT x 15ft)

(3960 - 50)LT

KG1 = 78,450LT-ft

3910LT

KG1 = 20.06ft

Step 2: Find the new vertical CG (KG)

TCG1 = Ds0 x TCG0 - (w x Tcg)

Ds1

TCG1 = 3960LT x 0ft - (50LT x -11ft)

(3960 - 50)LT

TCG1 = 0 - (-550LT-ft)

3910LT

TCG1 = 0.141ft

(minus because it’s to port)

(shifts to starboard, away from removed weight)

Step 3: Find the Transverse CG (TCG)

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 = 20ftG0MT = 2.6ft

Step 5: (Almost there!) Find tan :f

tan f = opposite adjacent

tan f = G0G1

G0MT

tan f = 0.0541

atan 0.0541 = f

3.10o = fCL

Go

MT

F

G1

tan f = .141ft2.6ft

2.6ft

.14ft

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?

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)

Inclining Experiment Tools

-Plot: -Plumb Bob:

Tangent of Inclining Angle (Tan[F])

Incl

inin

g M

omen

t, w

t (LT

-ft)

Fdadj

doppScale

Mast

tan(F)=dopp/dadj

CL

Go

Bo

MT

F

FB

B1

G1

The short leg is G0G1

The long leg is G0MT

The hypotenuse is G1MT

tan F = G0G1

G0MT

SO….

So far we’ve established that the angle of list can be found usingthe right triangle identified here:

...And so we can find the angle of list

CL

Go

Bo

MT

F

FB

B1

G1

Up to now, however, G0MT hasbeen given based upon a KG thathas been provided.

We’ll now see how KG can be found by determining G0MT

This is done by the Inclining Experiment

By using a known weight and placing it at a known distancean 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

Ds G0G1 = w g0g1

In earlier discussions an equation was derived for a shift in of a single weight:

…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 = wtDs

G0MT tan F = wtDs

That’s nice,… but not nice enough... One more rearrangementand we’ll have what we really want, G0MT:

G0MT = wttan F Ds

G0MT = wttan 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

G0MT = wttan 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

G0MT = wttan F Ds

Remember, slope is Dy/Dx:

Or...

=Dy

Dx

DWt

Dtan f

So...

Average G0MT = (slope of wt vs tan F curve)

Ds

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

Moment v. tan f

-1000

-800

-600

-400

-200

0

200

400

600

800

1000

-0.06 -0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05 0.06

tan f

mom

ent (

LT-ft

)

Having found the Average G0MT, you can find the KG when theship is loaded with the inclining weight:

KG = KMT - G0MT

The problem now degenerates to a simple “’change in verticalcenter 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• 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(°)-50 -12.8-25 -6.40 025 6.550 12.7

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:

Data From Inclining Experiment

-25000

-20000

-15000

-10000

-5000

0

5000

10000

15000

20000

25000

-0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25

Tangent of the Inclining Angle

Incl

inin

g M

om

ent

(LT

-ft)

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 (+)

WLTfwdTaft

DWL

FFpAp daft dfwd

_.

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

w_.

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

_.

w_.

The ship will pivot about the center of flotation, F.

F

dTfwd

The change in draft will be evident in a change of draft forward...

dTaft

…and aft.

dTrim

The difference between the fore and aft drafts is the change in trim:

Trim = dTaft - dTfwd

Graphically, it looks like this. First, the ship is represented with a line representing its initial state:

O)( _.F

FPAP

You can simulate this on your paper by turning the sheet in the direction that thebow or stern would sink because of the added weight, then drawing a line to represent the new position.

As weight is added, the the ship rotates about F:

w

O)(F

_.AP FP

wO)(F

_.

Now, rotate the sheet so that the line drawn becomes level and acts as the new waterline:

wO)(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.

FPAP

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”

Tfwd new = Tfwd old +/- dTPS +/- dTrim

AND

Taft new = Taft old +/- dTPS +/- dTrim

These two measurements- change due to parallel sinkage andchange 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

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 sinkageis 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

The value for dTrim will be for the entire length of the ship:

dTrim = wl MT1”

...Now we need to find how much of the trim is aft and how much is forward!

wO)(F

_.

dTfwd

dTaft

dTrim

l

Lpp

To find the trim distribution, consider the similar triangles formed below:

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

wO)(F

_.

dTfwd

dTaft

dTrim

l

Lpp

daft dfwd

For these similar triangles there is a ratio aspect that relates to each:

dTrimLpp

dTaft

daft

dTfwd

dfwd

= =

(The short leg divided by the long leg of the triangle!)

wO)(F

_.

dTfwd

dTaft

dTrim

l

Lpp

daft dfwd

Tfwd new = Tfwd old +/- dTPS +/- dT

AND

Taft new = Taft 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:

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.

GIVEN:

Lpp = 101.7 ft Draft = (10.5 + 10.1)/2 = 10.3ft

amidships = 50.85 ft Ds = 2LT x 205 = 410LT

LCF = 55.8 ft from FP, or 4.95 ft aft of amidships

DRAW A PICTURE!

Example:

The YP floats at a draft 10.5 ft aft and 10.1ft forward. A load of 10LT is placed15ft forward of amidships. Find the final forward and aft drafts.

101.7

Daft = 45.9 Dfwd = 55.8

)(_. dTaft

dTfwd dTrimF

O 10LT

19.95

Step 1: Find change due to parallel sinkage

dTPS = w TPI

dTPS = 10LT 235 x .02LT/in

dTPS = 2.13in

101.7

Daft = 45.9 Dfwd = 55.8

)(_. dTaft

dTfwd dTrimF

O 10LT

19.95

Step 2: Find change due to moment

dTrim = wl MT1”

dTrim = 10LT x 19.95ft 252.5 x .141 LT-ft/in

dTrim = 5.60in

101.7

Daft = 45.9 Dfwd = 55.8

)(_. dTaft

dTfwd dTrimF

O 10LT

19.95

Step 3: Divide the dTrim based on similar triangles

dTrimLpp

dTaft

daft

dTfwd

dfwd

= =

4.21in

101.7ft

dTaft

45.9ft

dTfwd

55.8ft= =

5.60in

101.7ftdTaft 45.9ft x= 2.53 in=

5.60in

101.7ftdTfwd 55.8ft x= 3.07 in=

101.7

Daft = 45.9 Dfwd = 55.8

)(_. dTaft

dTfwd dTrimF

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 + 3.07in) x (1ft/12in)

Tfwd new = 10.1ft + .43ft

Tfwd new = 10.53ft

101.7

Daft = 45.9 Dfwd = 55.8

)(_. dTaft

dTfwd dTrimF

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 - 2.53in) x (1ft/12in)

Taft new = 10.5ft - .033ft

Taft new = 10.467ft

101.7

Daft = 45.9 Dfwd = 55.8

)(_. dTaft

dTfwd dTrimF

O 10LT

19.95

Background Lab 2

Lab ObjectivesReinforce students’ understanding of

Archimedes PrincipleReinforce student’s concept of static

equilibriumReinforce student’s concept of the center of

floatation

Concepts/Principles:Archimedes PrincipleStatic EquilibriumCenter of FloatationSimpson’s First RuleInterpolationHydrostatic ForceTPIMT1”

Background Lab 2

Terminology– Displacement– Buoyant Force

Equations– D=rgÑ=FB

General Safety– Immediately clean up any water spilled to avoid fall

hazard

Background Lab 2

ApparatusEquipment

– 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/ReductionData to be collected & Expected results

– These should be equalWeight of modelWeight of waterCalculated water volume displacedHydrostatic Force

– Longitudinal Center of Floatation (LCF)

Sources of error– Measurements– Insufficient drip time

Data Collection/Reduction

CalculationsFB

TPIMT1”

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)

Dry-DockingIf 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

WL

GM

B

FBFblock=P=D-FB

D

WLG

M

B

FB

P

D

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

FBP=weightremoved

D0

Gf

Df

Disturbancetrying to rollthe 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.

WL

GM

B

FBFground=P=D-FB

D

Pulling the ship directly off the shoal.

Floating the ShipUndocking 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 ProblemDD963 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 AnswerTfinal fwd=Tinitial fwd±dTps±dTfwd

dTps=w/TPIdTfwd=dTrim×Dfwd/LppdTrim=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

w

AP

100ft25ft

232.5ftl=125ftF amidship FP

Lpp=465ft

Dfwd=257.5ftDaft=207.5ft

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 3Lab 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

Concepts/Principles– KG– TCG– MT

– Inclining Experiment

Background Lab 3

Terminology– 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)

Background Lab 3

Apparatus

General Safety– Minimize water on the floor

Equipment– 27-B-1 Models– Weights– Plum bobs

Procedures for taking measurements– Record measurements

Data Collection/ReductionData 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

Data Collection/Reduction

Calculations– Use equations

Plots/sketches– w×t vs. tan(F)

Review of Chapters 1-3for

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

Force × distanceEqual and opposite forces applied with an offset

distance to produce a rotation åF=0; åM=0P= rgzMx=òydA

Translational: heave, surge, swayRotational: roll, pitch, yawList, trim, heelp/r+V²/2+gz=constant

Chapter 1: Engineering Fundamentals

Chapter 2: Hull Form and Geometry

Categorizing shipsWays to represent the hull formTable of OffsetsHull form characteristicsCentroidsCenter of Flotation, Center of BuoyancySimpson’s RuleCurves of Form

PlansBody: 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

Chapter 2: Hull Form and Geometry

Simpson IntegralsSee your “Equations and Conversions” Sheet

Waterplane AreaAWP=2òydx; where integral is half breadths by station

Sectional AreaAsect=2òydz; where integral is half breadths by waterline

X

Y

Half-Breadths(feet)

Stations

y(x)

dx=Station Spacing

0

Z

YHalf-Breadths (feet)0

Waterlines

y(z)

dz=Waterline Spacing

(Body Plan)

(Half-Breadth Plan)

Simpson IntegralsSee your “Equations and Conversions” Sheet

Submerged Volume– ÑS=òAsectdx; where integral is sectional areas by station

Longitudinal Center of FloatationLCF=(2/AWP)*òxydx; where integral is product of distance

from FP & half breadths by station

X

Asect

SectionalAreas(feet²)

Stations

A(x)

0

dx=Station Spacing

X

Y

Half-Breadths(feet)

Stations

y(x)

dx=Station Spacing

0

(Half-Breadth Plan)

x

Chapter 3: HydrostaticsArchimedes 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

WL

CL

KBL

G0g0

Gi

gf Gf

G moves parallel toweight shift

Chapter 3: Hydrostatics

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)

M

WL

CL

K

B0BL

B1B2G0

FB

F1(+)F2(-)

WL

CL

KBL

G

B

MT

KB

BMTKMT

KG

GMT

ML

TCG/TCB (+)TCG/TCB (-)

Chapter 3: Hydrostatics

Trim Equations:– dTPS=w/TPI

– dTrim=wl/MT1”– dTfwd/aft/dfwd/aft =dTrim/Lpp

– Tfinal fwd/aft=Tinitial fwd/aft±dTPS±dTfwd/aft

w

dTPSFl

Ap Fpdaft dfwd

WeightAdded

dTfwddTaft

dTrim

Lpp

q

Tfinal fwd

Tfinal aft

Chapter 3: Hydrostatics

General Problem Solving TechniqueWrite 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 SheetAssigned homework problemsAdditional homework problemsExample problems worked in classExample Problems worked in text