chapter3 cog shift

8/12/2019 Chapter3 Cog Shift

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Overview

Chapter 3 - Buoyancy versus gravity =

stability

(see Chapter Objectives in text)

Builds on Chapters 1 and 2

6-week exam is Chapters 1-3!


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HYDROSTATICSReview (3.1)

Archimedes Principle:

An object partially or fully submerged in a fluidwill experience a resultant vertical force equal

in magnitude to the weight of the volume of

fluid displaced by the object.

This force is called the buoyant force or the

force of buoyancy(FB).


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HYDROSTATICSReview (3.1)

Mathematical Equation:

Where. . .

FB is the magnitude of the resultant buoyant force in lb,

is the density of the fluid in lb s2/ ft4,g is the magnitude of the acceleration of gravity normally

taken to be 32.17 ft / s2.

is the volume of fluid displaced by the object in ft3.

BF g weight


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Hydrostatics

The forceslead totranslations:

Heave

Surge Sway

The momentsleadto rotations:

Roll

Pitch Yaw

Vessel Degrees of Freedom

And Static Equilibrium


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HYDROSTATICSStatic Equilibrium : Forces and Moments

(3.1.2.1-2) Sum of the Resultant Forces:

Sum of the Moments about a reference point:

Static equilibrium must consist of both

conditions!

0 0 0Mx My Mz

BF g weight 0 ?

0 ?

0 ?

Fx

Fy

Fz


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HYDROSTATICSStatic Equilibrium : Stability

t

B

Is this boat in static equilibrium?

What are the component forces and moments?

Are they internal or external?


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HYDROSTATICSStatic Equilibrium (3.1.1.2)

G

B

LW

Port Starboard

DistibutedHydrostaticForces

ResultantWeight,s

Resultant VerticalBuoyant Force

FB

tmosphericPressure

HydrostaticPressure

What is the hydrostatic pressure? F=p*A

Wave?


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HYDROSTATICS

Static Equilibrium : Stability (3.2)

B

t


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HYDROSTATICS

Changes in the Center of Gravity (3.2) The Center of Gravity (G) is the point at which all ofthe mass of the ship can be considered to be located (for

most problems).

Terminology ! UPPERCASE for sh ip; lowercase for a

sm aller weight.

It is referenced vertically from the keel of the ship (KG

or VCG or Kg).

(1) Shifting, (2) adding, or (3) removing weight changes

the location of the Center of Gravity.


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HYDROSTATICSStatic Equilibrium : Stability

t

B

Where is the Center of Gravity?

The Center of Buoyancy?

Are they vertically aligned? Why/Why not?


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HYDROSTATICSChanges in the Center of Gravity (3.2.1.1)

C

B

L

L

WL

g

Gow Gf

K

When weight is added to

a ship, the CG will move in

a straight line from its

current position toward thecg of the weight being

added. G0to Gf. The

distance is a ratio of the

weight and disp.

What happens to the

Center of Buoyancy

(and the ship)?


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HYDROSTATICSChanges in the Center of Gravity (3.2.1.2)

When weight is

removed from a ship,

G will move

in a straight line fromits current position

away from the center

of gravity of the

weight being

removed. G0to Gf.

C

B

L

L

WL

g

Gow Gf

K


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Changes in the Center of Gravity (3.2.1.3)

When a small

weight is shifted (but

not added or removed, CG

will move parallel to theweight shift but a much

smaller distance because it

is only a small fraction of

the total weight of the ship.

C

B

L

L

W L

go

gf

GoGfw

w

K

StarboardPort

HYDROSTATICS


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HYDROSTATICS

Vertical Shift in the Center of Gravity (3.2.2.1)

Where: (note: some use the term initial for old andfinal for new

KGnew is the final vertical position of the center of gravity

of the ship as referenced from the keel. KGs are in feet.

KGold is the initial vertical position of the center of gravity

of the ship as referenced from the keel.

old old addedweight addedweight

new

old addedweight

KG Kg wKG

w


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HYDROSTATICS

Vertical Shift in the Center of Gravity (3.2.2.1)And,

s new is the final displacement of the ship in LT. In this

example, it is equal to the initialdisplacement plus or minus the weight added.

s old is the initial displacement of the ship in LT.

Kg added weight is the vertical position of the center of gravity of

the weight being added as referenced from the keel.

This line segment is a distance in feet.

w added weight is the weight of the weight to be added in LT.


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HYDROSTATICS

Vertical Shift in the Center of Gravity (3.2.2.1)

The first equation was for a weight addition or

removal. What do we do for a weight shift? What is

different

Re-examine our first vertical shift equation.

What changes?

old old addedweight addedweight

new

old addedweight

KG Kg wKG

w


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HYDROSTATICS

Vertical Shift in the Center of Gravity (3.2.2.3) So, the final equation for vertical shifts is:

( )old old new old new

old

KG w Kg Kg

KG w

Example: A 150 pound person climbs in a 10 pound canoe and sitsdown. How much has KG shifted? KGold=0.5 ft Kg=?


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HYDROSTATICS

Vertical Shift in the Center of Gravity (3.2.2.4) Last Comments:

The general equation covers all cases for a

change in KG. This is the equation you should

apply to the exams!


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HYDROSTATICSTransverse Shift in the Center of Gravity

(3.2.3)

Shifts side to side of the Center of Gravity.

Starboard is positive and port is negative!

As in Vertical case, the Transverse movement of

G may be caused by either (1) addition, (2)

removal, or (3) shifting of weights.


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HYDROSTATICSTransverse Shift in Center of Gravity (3.2.3)

Results in a List on the Vessel.

List occurs when a vessel is in static

equilibrium and down by either the port orstarboard side. No external forces are required to

maintain this condition and it is permanent unless

the Center of Gravity changes.

List is different from heeling. Heeling

occurs because an external couple is acting on

the vessel. Heeling is a more temporary

condition.


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Example (Listing or Heeling?)


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The Transverse Center of Gravity is

referenced in the transverse

(athwartships) direction from thecenterline of the ship and is labeled TCG.

The equation used for a transverse shift in

the Center of Gravity is the same as wasused for the vertical shift! (With some

changes in the notation.)


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HYDROSTATICSTransverse Shift in Center of Gravity (3.2.3.4)

Remember a weight shift is just like removing a

weight from its original location and adding it

to its final location. So for just a weight shift,

the generalized equation simplifies to:


old

TCG w TCg TCg TCG

w

Example: Your 100 LT ship is initially upright. You pump 5LT of water from a point 15 ft starboard of centerline to 10 ft

port of centerline. What is the new TCG? (We will use that

answer later to find the angle of heel.


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HYDROSTATICS


old

TCG w TCg TCg TCG

w

Vertical and Transverse Changes in G

The Key Equations!


old

KG w Kg KgKG

w

When faced with a change in weight (add, sub ormove), first sketch it, then solve KG, then solve

TCG!


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HYDROSTATICSMetacenter (3.3)

A reference point for hydrostatic calculations for smallangles of roll (less than 10 degrees) or pitch (less than five

degrees).

Defined as the intersection of the buoyancy forces and the

ship centerline.

Bo B1

S

O

FB

O

T


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HYDROSTATICSMetacenter (3.3)

The higher the metacenter, the more stable the ship

is!

There is a different metacenter for ship pitching in

the longitudinal direction and ships rolling in

the transverse direction.

BMTis for roll, BMLfor pitch. Which is higher?

If the subscript is omitted, it means BMT.


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HYDROSTATICSMetacentric Radius (3.3.1.1)

The distance from the Metacenter to the

Center of Buoyancy is defined as the

Metacentric Radius (BM).

Zero pt.

B

MT

K

TIBM

32

3TI y dx


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Quick Review Finding KMTfrom the Curves of Form

For a draft of 10 ft

Curve 8

Genl Scale =192

192*0.06 ft

KMT=11.5 ft

HYDROSTATICS


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HYDROSTATICSMetacentric Height (3.3.1.2)

The distance between the Center of Gravity (G)and the Metacenter (M) is defined as the

Metacentric Height (GM).

Zero pt.

G

MT

HYDROSTATICS


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HYDROSTATICSMetacentric Height (3.3.1.2)

Why is GM important?

If G is below M, then GM is said to be positive.

The ship does not want to capsize. This is

GOOD!

If G coincides with M, then GM is said to be

zero. A vessel would stay heeled. This is not

very good.

If G is above M, the GM is said to be negative.

The ship will tip over. This is REALLY BAD!


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B d ( ti ) GM!


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Bad (negative) GM!

The ship wants to roll over. G is

either too high or M is too low!

Zeropt.

HYDROSTATICS


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B and M are functions of the hull shape and aregenerally constant over the life of the ship. G is

based on the weights and changes constantly.

To be safe at sea, we need to find the ships KG to

make sure it is sufficiently below M!

KG = KM - GM

Where, KM is shown on the Curves of Form.

GM is found from both calculations and by an

Inclining Experiement


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KM=KB + BM where:

KB is found by numerical integration but

for most vessels is between 40-50% of the

draft

BM is found by:

ITis the Transverse Moment of Area of theWaterplane and has the units of ft4

For a box-shaped barge it simplifies to:

TT

IBM

3

12T

LBI

32

3TI y dx


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Example stability check

You have just bought a 30-foot long

floating dock, made some modifications

and will now put it in the water. It is 6 ftwide and 2.5 feet deep. KG=2 ft and it

has a 1 ft draft. Will it be stable? (eg

Find GM and determine if it is positive!)

HYDROSTATICS


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HYDROSTATICSCalculating Angle of List (3.4)

As a weight shifts across the deck of a vessel, thevessel lists (or inclines. How can we predict

the angle of inclination (list)?

Derivation of Equation

Draw two vessels, one upright and one listing.Show a weight moving, along with the CG and B.

Calculating Angle of List (3 4)


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Calculating Angle of List (3.4)

LB

L

W GoGt

BoBf

S

MT O

FB

O

StarboardPort

HYDROSTATICS


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HYDROSTATICSCalculating Angle of List (3.4.2)

The weight is shifted causing a shift in the Center

of Gravity.

A moment is created causing the vessel to incline.

The underwater shape of the hull changes

causing the Center of Buoyancy (B) to move

until it is in line with the Center of Gravity (G)

and the vessel is back in static equilibrium.

HYDROSTATICS


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From the geometry and then some substitution,we get:

Zeropt.

tan

tan

O F OG G G M

GM w t

W

t

G

M

B

HYDROSTATICS


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This equation only works for small angles

because it assumes that the Metacenter does

not move!

Note that for small angles, tan = sin! So you can

calculate GM from either along the old or new

inclined axis.

Example: You move a 1 LT weight 25 feet to

starboard on your 100 LT ship and it lists 2 degrees.

What is GM? How would you find KG?

HYDROSTATICS


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HYDROSTATICSInclining Experiment (3.5)

Uses small-angle hydrostatics to find the

vertical center of gravity (KG) of a ship.

Process: A weight is moved a transverse

distance, causing a shift in the TCG, and

resulting in measurable inclination (list).

HYDROSTATICS


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HYDROSTATICSInclining Experiment (3.5)

Navy 44 Incline Experiment


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HYDROSTATICSInclining Experiment (3.5.1)

Solving the Angle of Heel equation for themetacentric height (GM), we find:

The easiest way to do this experiment is to use

one set of weights at one distance off centerline.Alas, this would have significant experimental

errors, so we measure the inclination with different

weights and different positions.

tan

w t

GM


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We then plot the data on a graph where the y-

axis is the Inclining Moment (wt) and the x-

axis is the Tangent of the inclining angle (Tan ).

The average value of GM can be found from the

slope of the line. We can see that:

1 1 1

tan

w t rise yGM slope

run x

HYDROSTATICS


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Recall: We want to find the Center of Gravitywhich can be found by the equation:

KG=KM-GM

KM is found from the Curves of Form

GM is found from the Inclining Experiment

HYDROSTATICS


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HYDROSTATICS

Inclining Experiment (3.5.2)

Removing the Inclining Apparatus we must

recalculate KG. This is done as a weight

removal problem:( )old old new old

new

old

KG w Kg KgKG

w

HYDROSTATICS


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Shipboard Considerations: No initial list.

Minimum trim.

Dry bilges.

Liquid fuel and oil to be in accordance with the

Shipyard Memo.

Sluice valves closed.

All consumables are to be inventoried.

Minimum number of personnel remain onboard.

See the example in your text!

HYDROSTATICS


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HYDROSTATICSLongitudinal Changes in the Center of Gravity

(3.6) A longitudinal shift in the CG will result in the

vessel having some trim.

Trim is the difference between the forward and aftdrafts, Tfand Ta. It may be calculated by:

Trim = Taft

- Tfwd

Tm = ()(Ta+ Tf )

The Mean Draft is:

Ex. A ship has a draft of 15

fwd and 16 aft. Trim = 1 ft

HYDROSTATICS


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(3.6)

A vessel is trimmed by the bow when the bow has

a deeper draft. This is indicated by a negative

trim.

A vessel is trimmed by the stern when the stern

has a deeper draft. This is indicated by a

positive trim.

What is the point which the vessel trims about?

HYDROSTATICS


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(3.6) What happens when a weight is shifted forward or

aft?

The vessel goes down by the bow or stern

depending on the direction of the weight shift.

Note that the change in trim is independent ofthe original location of the weight. (i.e. It only

matters whether the weight moves forward or aft)

HYDROSTATICS


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The Trim Problem (3.6) Draw a picture of what is happening when a

vessel trims due to a weight shift:

dAFT dFWD

LPPAP

lFFP

w

HYDROSTATICS


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(3.6) As the weight shifts forward, a new operating waterline is

created and the draft decreases aft and increases

forward.

dAFT dFWD

LPPAP

lF

FPw

dTaft

dTfwd

HYDROSTATICS


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(3.6) We now have two similar triangles and will draw a

third which represents the change in trim.

Recall: Trim = Taft-TfwdSo the total Trim with a change in trim is:

And with no initial trim, then the change in TRIM is:

aft aft fwd fwd TRIM T T T T

aft fwd TRIM T T

HYDROSTATICS


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(3.6) To calculate the final drafts we will need to find:

Where MT1 is from the Curves of Form (2.10)

We use similar triangles (ratios) to find the change

in draft due to the weight shift.

1 aft fwd w l

TRIM T T MT

aft fwd

aft fwd

T T TRIM

d d LPP

HYDROSTATICS


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(3.6) Example: You have a 1000x 200 x 90 foot tanker

(100,000 LT) with F at Stn 6. It

has zero TRIM. You move

1000 LT of oil 450 ft aft. What

is the new draft at the stern?

MT1~ IL/420L=150,000 FTLT/in

chapter3 cog shift

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