Chapter3 Cog Shift
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Transcript of 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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HYDROSTATICSTransverse Shift in Center of Gravity (3.2.3)
Example (Listing or Heeling?)
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HYDROSTATICSTransverse Shift in Center of Gravity (3.2.3)
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HYDROSTATICSTransverse Shift in Center of Gravity (3.2.3)
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 old new old new
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 old new old new
old
TCG w TCg TCg TCG
w
Vertical and Transverse Changes in G
The Key Equations!
( )old old new old new
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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HYDROSTATICSMetacentric Radius (3.3.2.2)
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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HYDROSTATICSMetacentric Radius (3.3.2.2)
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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HYDROSTATICSCalculating Angle of List (3.4.3)
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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HYDROSTATICSCalculating Angle of List (3.4.3)
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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HYDROSTATICSInclining Experiment (3.5.1)
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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HYDROSTATICSInclining Experiment (3.5.1)
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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HYDROSTATICSInclining Experiment (3.5.3)
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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HYDROSTATICSLongitudinal Changes in the Center of Gravity
(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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HYDROSTATICSLongitudinal Changes in the Center of Gravity
(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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HYDROSTATICSLongitudinal Changes in the Center of Gravity
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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HYDROSTATICSLongitudinal Changes in the Center of Gravity
(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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HYDROSTATICSLongitudinal Changes in the Center of Gravity
(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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HYDROSTATICSLongitudinal Changes in the Center of Gravity
(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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HYDROSTATICSLongitudinal Changes in the Center of Gravity
(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