Section 3.1
Archimedes Principle:
The Buoyant Force on an object is equal to the weight
of the volume of the water displaced by the object
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
Horizontal
components of
pressure force
are negated by
equal force on
opposite side of
barge.
G
B
FB
A
z
FB=PA;
P=rgz;
FB=rgzA;
V=zA;
FB=rgV
Static Equilibrium
F=0
Weight, Buoyancy, Drag and Lift forces all sum to
zero in each dimension
M=0
All forces in each dimension are colinear and cancel;
i.e. there are no separation of the action points of
forces such that couples or moments are generated.
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)
So far we’ve looked at ships that are in STATIC EQUILIBRIUM:
• SFx = 0
• SFy = 0
• SFz = 0
• SMp = 0
Go
Bo
K
BL
CL
Now let’s take a look at what happens when a weight is added
to disturb this equilibrium
A change in weight (either adding or removing it) will cause a change in
the location of G, the center of gravity of the ship
• A change in VCG (or KG)
• A change in the TCG
TCG
G1
Go
KGnew
It also causes a change in the
longitudinal CG (LCG), but
we’ll discuss that later...
K
CL
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
BL
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
K
CL
BL
In the case of a weight SHIFT, the CG first shifts AWAY from the
removed weight….
•g
G2
•g
G0
G1
K
BL
CL
…and the TOWARDS the relocated weight
Let’s first consider a weight added directly over the centerline
This will cause the location of the CG to move TOWARD the weight ...
•g
G1
G0
KGnew
KGold
K
BL
CL
… Causing a change in the VERTICAL distance, or KG
Use the concept of weighted averages to determine the new CG:
•g
G1
G0
K
CL
KGnew =
Dsold x KGold +
BL
wadd x Kg
Dsold + wadd
It’s the same deal for removing a weight, only this time the weight is negative (i.e.
removed):
•g
G0
G1
KGold
KGnew
K
CL
KGnew =
Kg
BL
Dsold x KGold + (-w) x Kg
Dsold + (-w)
In a relocation of a weight, look at it as SUBTRACTING one weight, and
ADDING another weight.
•g
G1
G0
KGnew
Kg2
•g
Kg1
KGold
K
CL
BL
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
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!
Ds
G0
G1
150
LT
19.5’
?
6’
K
CL
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
BL
Step 3: Write the GENERAL Equation
KGnew =
Dsold x KGold + Swi x Kgi
Dsnew
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
TCGo=0ft
Tcg#2RFT= -12ft (i.e. port of centerline);
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:
WL
D
FB
BL
G0
B0
G shifts:
D
WL
G1
FB B0
K
K
CL
CL
Ship responds:
D G1
WL
FB B1
F1
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:
Where the lines of action of the various
centers of buoyancy cross* is the Metacenter
M
F2(-) F1(+)
WL
G2
FB B2
F2
WL
G0 B1
B2
FB
BL
B0
K
CL
*Lines of action cross at a single point only for
“small” angles of inclination (<10º).
Shapes which impact KM:
M
M
B2
BL
B0
B1
WL
K
CL
Highly curved hull cross-section:
Little buoyant volume at large
lever arm: M is at/near center of
curvature
B2
BL
B0
B1
WL
K
CL
Very flat hull cross-section:
Large buoyant volume at large
lever arm: M is high
ML
Distance from B to MT =
Transverse Metacentric Radius
Distance from G to MT =
Metacentric Height =
Major player in stability
calculations (+ keeps
ship upright)
MT
GMT
KMT BMT
KB
TCG/TCB (-)
BL
WL
G
B
K
CL
KG
TCG/TCB (+)
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
G1: Support is outboard center of gravity
creating a couple which returns the chair
upright.
Support
G2
G2: Support is aligned with center of gravity
eliminating any couple. The chair maintains
position.
Support
G3: Support is inboard center of gravity creating
a couple which tips the chair over.
G3
Support
The center of curvature of the rocking chair’s
“skids” correspond to a ship’s metacenter.
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
G0
WL
F
B0
FB
D
Gf
Bf
CL
BL
gf
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….
Go
Bo
K
…MT, or the Transverse Metacenter
CL
BL
The Transverse Metacenter (MT) represents a convenient point of reference for
small changes in the angle of inclination, F, (less than 10o)
MT
Go
Bo
K
CL
BL
For small changes in inclination, the point MT is where the ship is assumed
to rotate.
MT
F
B1
...The MT is generally about 10-30ft above the keel
There is also a Longitudinal Metacenter, or ML...
ML
ϑ
…usually in the magnitude of 100 to 1000ft above the keel
When the ship reacts to an off-center load (which
will change the ship’s CG),...
MT
F
...the center of buoyancy will shift
until it is vertically aligned
with the new CG...
…G1 can be assumed to move
PERPENDICULAR from the CL
Remember, this is only for listing
of 10o or less
G1
B1
FB
MT
Look at the right triangle formed by this
shifting…
F
The short leg is G0G1
The long leg is G0MT
The hypotenuse is G1MT
G1
SO….
tan F = G0G1
G0MT
B1
(tan F = opp/adj… remember?)
FB
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
MT
How?
• G0MT = KMT - KG0
Go
KG
• From the Curves of Form you can get KMT
• The Vertical Center of Gravity is KG0
KMT
• G0G1 is the change in the transverse Center of Gravity
Bo
K
CL
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)
D s1
KG1 = 3960LT x 20ft - (50LT x 15ft)
(3960 - 50)LT
KG1 = 78,450LT-ft
3910LT
KG1 = 20.06ft
Step 3: Find the Transverse CG (TCG)
TCG1 = Ds0 x TCG0 - (w x Tcg)
D s1
TCG1 = 3960LT x 0ft - (50LT x -11ft)
(minus because it’s to port)
(3960 - 50)LT
TCG1 = 0 - (-550LT-ft)
3910LT
TCG1 = 0.141ft
(shifts to starboard, away from
removed weight)
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
Step 5: (Almost there!) Find tan f:
MT
F
tan f = opposite
adjacent
tan f = G0G1
G0MT
tan f = .141ft
2.6ft
tan f = 0.0541
G1
atan 0.0541 = f
3.10o = f
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?
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:
Inclining Moment, wt (LT-ft)
Mast
dadj
Tangent of Inclining Angle (Tan[F])
dopp
F
tan(F)=dopp/dadj
Scale
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
G1
SO….
tan F = G0G1
G0MT
...And so we can find the angle of list
B1
FB
MT
F
Up to now, however, G0MT has
been given based upon a KG that
has been provided.
G1
We’ll now see how KG can be
found by determining G0MT
This is done by the Inclining Experiment
B1
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
G0MT tan F = G0G1
= wt
G0G1 = wt
Ds
G0MT tan F = wt
Ds
G0MT tan F = wt
Ds
That’s nice,… but not nice enough... One more rearrangement
and we’ll have what we really want, G0MT:
G0MT =
wt
tan F Ds
G 0 MT =
wt
tan F Ds
Let’s review what we know:
• “w” is a known weight that is relocated
• “t” is the distance the weight is moved
• “tan f” is the angle created by the weight shift
• “Ds” is the displacement of the ship
This will be the formula that governs the Inclining Experiment
G 0 MT =
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
G 0 MT =
wt
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
When you vary the distance t, and thus the moment, you’ll vary the inclination
angle. The result is plotted in an example here:
Moment v. tan f
moment (LT-ft)
1000
800
600
400
200
-0.06
-0.05
-0.04
-0.03
-0.02
0
-0.01 -200 0
-400
-600
0.01
0.02
0.03
0.04
0.05
0.06
-800
-1000
tan f
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
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 onlyno 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)
-50
-25
0
25
50
What is KGlight?
List Angle(°)
-12.8
-6.4
0
6.5
12.7
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
Inclining Moment (LT-ft)
15000
10000
5000
0
-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
-5000
-10000
-15000
-20000
-25000
Tangent of the Inclining Angle
0.1
0.15
0.2
0.25
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
Tfwd
Taft
F
Ap
daft
dfwd
Fp
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.
_
.
_
.
The ship will pivot about the center of flotation, F.
The change in draft will be evident in a change of draft
forward...
…and aft.
dTaft
dTrim
_
.
F
dTfwd
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:
AP
)(
O
FP
F
_
.
As weight is added, the the ship rotates about F:
)(
O
AP
F
FP
_
.
You can simulate this on your paper by turning the sheet in the direction that the
bow or stern would sink because of the added weight, then drawing a line to
represent the new position.
Now, rotate the sheet so that the line drawn becomes level and acts as
the new waterline:
AP
w
FP
_
.
The changes in draft can now be read directly…
w
dTaft
dTfwd
dTrim
_
.
dTaft is below the WL, so it’s subtracted. dTfwd is above the waterline, so it’s added to the
draft.
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
The value for dTrim will be for the entire length of the ship:
dTrim = wl
MT1”
l
w
dTfwd
dTrim
dTaft
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:
l
w
dTaft
daft
dfw
Lpp
d
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
dTfwd
dTrim
_
.
l
w
dTfwd
dTaft
daft
dTrim
_
.
dfw
Lpp
d
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!)
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 = 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!
Dfwd = 55.8
Daft = 45.9
19.95
_
.
dTaft
)(
O
F
101.7
10LT
dTfwd
dTrim
Dfwd = 55.8
Daft = 45.9
19.95
_
.
)(
O
dTaft
10LT
F
101.7
Step 1: Find change due to parallel sinkage
dTPS =
dTPS =
dTPS =
w
TPI
10LT
235 x .02LT/in
2.13in
dTfwd
dTrim
Dfwd = 55.8
Daft = 45.9
19.95
_
.
)(
O
dTaft
10LT
F
101.7
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
dTfwd
dTrim
Dfwd = 55.8
Daft = 45.9
19.95
_
.
)(
O
dTaft
dTfwd
10LT
dTrim
F
101.7
Step 3: Divide the dTrim based on similar triangles
dTrim dTaft dTfwd
=
=
Lpp
daft
dfwd
4.21in
=
101.7ft
dTaft
45.9ft
=
dTfwd
55.8ft
dTaft = 45.9ft x 5.60in = 2.53 in
101.7ft
dTfwd = 55.8ft x
5.60in
101.7ft
= 3.07 in
Dfwd = 55.8
Daft = 45.9
19.95
_
.
dTaft
)(
O
10LT
dTfwd
F
101.7
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
dTrim
Dfwd = 55.8
Daft = 45.9
19.95
_
.
dTaft
)(
O
10LT
dTfwd
F
101.7
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
dTrim
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
– Displacement
– Buoyant Force
Equations
– D=rg=FB
General Safety
– Immediately clean up any water spilled to avoid fall
hazard
Apparatus
Equipment
–
–
–
–
–
–
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
Data Collection/Reduction
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=rgS
 D =rgS+ 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-Docking
If a list develops during docking, the increasing force
from the blocks can work to capsize the ship
D
G
WL
M
G
B
Fblock=P=D-FB
D
M
WL
B
FB
FB
P
Solutions:
– Use side blocks to force a zero list
– Stop docking evolution and correct problem, if ship
develops an increasing list
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;
Df
Disturbance
trying to roll
the ship
WL
P=weight
removed
– 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
D0
G0
M
Gf
B
FB
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
G
WL
Fground=P=D-FB
M
B
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
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”
25ft
F
100ft
w
232.5ft
amidship l=125ft
FP
Lpp=465ft
AP
Daft=207.5ft
Dfwd=257.5ft
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
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
– 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
– Minimize water on the floor
Equipment
– 27-B-1 Models
– Weights
– Plum bobs
Procedures for taking measurements
– Record measurements
Data Collection/Reduction
Data to be collected & Expected results
–
–
–
–
–
–
–
27-B-1 Model Numbers
Weight of Models
Drafts
Model dimensions
Water temperature
tan(F)
Where do you expect KG to be?
Sources of error
– Measurement error
– Round off
Data Collection/Reduction
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
 F=0; M=0
 P= rgz
 Mx=ydA
 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)*xdA
Center of waterplane area
Center of submerged volume
ydx=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
See your “Equations and Conversions” Sheet
Y
(Half-Breadth Plan)
y(x)
HalfBreadths
(feet)
dx=Station Spacing
0
X
Stations
Waterplane Area
AWP=2ydx; where integral is half breadths by station
Sectional Area
Z
Asect=2ydz; where integral is half breadths by waterline
Water
lines
(Body Plan)
dz=Waterline Spacing
y(z)
0 Half-Breadths (feet) Y
Simpson Integrals
See your “Equations and Conversions” Sheet
Asect
Submerged Volume
 S=Asectdx; where integral is sectional areas
by station
A(x)
Sectional
Areas
(feet²)
dx=Station Spacing
0
(Half-Breadth Plan)
y(x)
Y
Longitudinal Center of Floatation
HalfBreadths
(feet)
LCF=(2/AWP)*xydx; where integral is product of distance
from FP & half breadths by station
0
X
Stations
dx=Station Spacing
x
Stations
X
Chapter 3: Hydrostatics
 Archimedes Principle/Static Equilibrium
 Impact to G of weight addition, removal, movement
 Metacenter
 Angle of list
 Inclining Experiment
 Trim calculations
 Drydocking
Chapter 3: Hydrostatics
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
F=0; M=0
gf
Center of Gravity (G)
Df= D0+Swa-Swr
Gi
g0
KGfDf= KG0D0+SKgawa-SKgrwr
TCGfDf= TCG0D0+STcgawa-STcgrwr
Gf WL
BL
G0
G moves parallel to
weight shift
K
CL
Chapter 3: Hydrostatics
M
F2(-) F1(+)
ML
MT
WL
G0
B2
FB
BL
B1
GMT
KMT BMT
B0
K
CL
tan(F) = wt/(DG0MT)
KB
BL
TCG/TCB (-)
WL
G
B
K
CL
KG
TCG/TCB (+)
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)
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
Weight
Added
dTrim
w
dTaft
Ap
Tfinal aft
l
daft
F
Lpp
dTPS
q
dfwd
dTfwd
Tfinal fwd
Fp
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