04_longitudinal_stability_trim
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Subject
Mechanical Engineering
Date
Oct 30, 2023
Type
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22
Uploaded by CaptainCapybaraMaster915
NE224 Spring 2023
Instructor: An Wang
5
Trim and Longitudinal Stability
The analysis of longitudinal stability is similar to transverse stability. Trim is also similar
to heel but use different measurement.
5.1
Geometry
Figure 5.1: The geometry of trim.
Terminology and definitions:
Trimming moment:
Moments created by external forces that tend to cause the ship
to rotate about a transverse axis.
On an even keel:
T
a
=
T
f
, draft is equal along the ship.
(Have zero trim)
Trim by the stern
:
T
a
> T
f
, deeper drafts aft than forward.
Trim by the head or by the bow:
T
a
< T
f
, Deeper draft forward than aft.
The measure of the amount of trim is simply the difference between the aft and the
forward drafts.
L
:
length of ship between perpendiculars, assumed to be the same as between draft
marks.
F
: distance of center of floatation (point
F
) from the forward perpendicular (FP) or
mark.
A
: distance of center of flotation from the after perpendicular (AP) or mark.
LCF: distance of center of flotation from amidship.
T
A
: draft aft.
T
F
: draft forward.
1
NE224 Spring 2023
Instructor: An Wang
T
M
: mean draft of draft amidship.
T
M
= (
T
A
+
T
F
)
/
2
.
(5.1)
T
O
: draft at center of flotation, also called the corresponding even keel draft.
f
: forward difference in drafts, even keel to trimmed waterlines.
a
: aft difference in drafts, even keel to trimmed waterlines.
m
: amidship difference in drafts, even keel to trimmed waterlines.
t
: trim.
If trim by stern:
t
=
T
A
−
T
F
(5.2)
If trim by head:
t
=
T
F
−
T
A
(5.3)
θ
: angle of trim.
tan
θ
=
t
L
=
f
F
=
a
A
=
m
LCF
=
x
X
(5.4)
, where
X
is the longitudinal distance from the center of flotation to any point on the ship,
and
x
is the distance between the even keel and trimmed drafts at the point in question.
Convention for units:
Drafts
Trim and change in draft
Longitudinal measurements
(
T
A
,
T
F
,
T
O
,
T
M
)
(
t
,
a
,
f
,
m
)
(
L
,
A
,
F
, LCF)
U.S. units
ft-in
in, ft-in
ft (decimal)
SI units
meters
centimeters
meters
If use U.S. units (drafts in ft-in,
t
in inches):
tan
θ
=
t
12
L
=
f
12
F
=
a
12
A
=
m
12LCF
If use SI units (drafts in m,
t
in cm):
tan
θ
=
t
100
L
=
f
100
F
=
a
100
A
=
m
100LCF
Change of trim
:
Change of trim = Final trim
−
Initial trim
Need to pay attention to the direction of the change and the direction of the final and
initial trim.
2
NE224 Spring 2023
Instructor: An Wang
5.2
Longitudinal stability
5.2.1
Longitudinal mecacenter
The assessment of the longitudinal stability is similar to the transverse stability.
Figure 5.2: Longitudinal stability.
Assume the displacement does not change and the angle of trim is very small, the buoy-
ancy before and after the trim intersect at
longitudinal metacenter (
M
L
)
. The longitu-
dinal metacentric radius:
BM
L
=
I
L
∇
(5.5)
, where
I
L
: the longitudinal moment of inertia of the waterplane (about center of flotation)
∇
: volume of displacement
BM
L
: longitudinal metacentric radius
5.2.2
Axis of trim
For heeling: the immersed wedge and emerged wedge have the same shape due to sym-
metry (at small angle of heel).
3
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For trimming: The immersed wedge and emerged wedge are different in shape, but are
equal in volume.
Note:
The ship trims about a transverse axis passing through the center
of flotation of its waterplane.
Therefore, the axis of trim will likely be at different
longitudinal positions for each draft.
Proof:
(see Zubaly, p. 138)
Figure 5.3: Geometry of the immersed and emerged wedges.
Note:
The longitudinal metacentric radius is much greater than the transverse meta-
centric radius.
BM
L
BM
T
=
I
L
/
∇
I
T
/
∇
=
I
L
I
T
If assuming the waterplane is a rectangle,
I
L
/I
T
= (
BL
3
)
/
(
LB
3
) = (
L/B
)
2
.
The ship’s
length is much greater than beam. If
L
= 10
B
, then the
I
L
= 100
I
T
and
BM
L
= 100
BM
T
.
Therefore, a ship’s longitudinal stability is very large and it is quite impossible to render
it unstable longitudinally. The longitudinal metacenter of a typical surface ship is several
hundred feet above its center of gravity.
Therefore, typically there is no concern about the loss of stability in the longitudinal
direction.
The following discussion will focus on the trim and the change of trim caused by shifting
weight and loading/discharging weight.
4
NE224 Spring 2023
Instructor: An Wang
5.3
Trimming moment, trim and drafts when shifting weight lon-
gitudinally
When a weight
w
is shifted along the longitudinal direction from
x
0
to
x
1
, the angle of
trim can be calculated by
tan
θ
=
GG
1
GM
L
=
w
(
x
1
−
x
0
)
∆
GM
L
(5.6)
, where
GG
1
is the shift in distance of the center of gravity of the ship and
w
(
x
1
−
x
0
) equals
to the trimming moment caused by the shift.
Equation 5.6 is only accurate at small angle of trim, which is almost always true for a
typical ship.
In practice, the change in trim or change in draft at forward/aft are often used, instead
of the angle of trim.
In U.S. units:
t
12
L
=
w
(
x
1
−
x
0
)
∆
GM
L
t
=
w
(
x
1
−
x
0
)
∆
GM
L
12
L
(
w
and ∆ in ton,
L
in feet and
t
in inches)
(5.7)
In SI units
t
=
w
(
x
1
−
x
0
)
∆
m
GM
L
100
L
(
w
and ∆
m
in MT,
L
is in m and
t
is in cm)
(5.8)
Resulting change in draft at bow:
f
=
F
L
t
(5.9)
Resulting change in draft at stern:
a
=
A
L
t
(5.10)
The change in draft at bow
f
and change in draft at stern
a
should satisfy:
a
+
f
=
t
(5.11)
5
NE224 Spring 2023
Instructor: An Wang
Example:
(example 5-1, Zubaly) A tanker 1,045 ft long with a displacement of 305,000 tons
floats in seawater at drafts of 62’0” forward and 65’0” aft.
KB
= 33
.
0 ft,
KG
= 45
.
0 ft,
KM
L
= 740 ft and the center of flotation is 15.5 ft aft of amidship. What final drafts will
result if 2,200 tons of cargo are shifted from cargo tanks #2 to # 5, a distance of 285 ft
toward aft?
Solution:
Need to use equation 5.7 to calculate change in trim. In equation 5.7, we first need to
find the Longitudinal metacentric height
GM
L
.
Longitudinal metacentric height:
GM
L
=
KM
L
−
KG
= 740 ft
−
45
.
0 ft = 695 ft
Use equation 5.7 to calculate the change of trim due to the shift:
t
=
w
(
x
1
−
x
0
)
∆
GM
L
12
L
=
2
,
200 ton
×
285 ft
305
,
000 ton
×
695 ft
12
×
1
,
045 ft
= 37
.
1 in
Weight is shifted toward the stern, so the change of trim is by the stern, increasing draft
aft and decreasing draft forward. In order to calculate the changes in draft at forward and
aft, we need to first determine the location of center of flotation relative to forward and aft,
F
and
A
, respectively. LCF =
−
15
.
5 ft (relative to amidship) is given, so
F
=
L
2
−
LCF =
1045 ft
2
−
(
−
15
.
5 ft) = 538 ft
A
=
L
2
+ LCF =
1045 ft
2
−
15
.
5 ft = 507 ft
(Or
A
=
L
−
F
= 507 ft.)
Therefore, the change in draft at forward and aft:
f
=
F
L
t
=
538 ft
1045 ft
37
.
1 in = 19
.
1 in = 1
′
7
.
1
′′
decrease of draft forward
a
=
A
L
t
=
507 ft
1045 ft
37
.
1 in = 18
.
0 in = 1
′
6
.
0
′′
increase of draft aft
(Or
a
=
t
−
f
= 18
.
0 in. )
Final draft at forward and aft:
T
F
=
T
F
0
−
f
= 62
′
0
′′
−
1
′
7
.
1
′′
= 60
′
4
.
9
′′
6
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T
A
=
T
A
0
−
a
= 65
′
0
′′
+ 1
′
6
.
0
′′
= 66
′
6
.
0
′′
Note:
The above example gives us the displacement at the trimmed condition. If we
know the ship’s trim at stern and bow, how to find out the displacement from the hydrostatics
curves (curves of form) or table?
The hydrostatic properties of a ship is typically determined at even keel condition. Once
the ship is trimmed, these properties are not exact but still acceptable.
However, if the
trim is so large or the center of flotation is especially distant from amidship, or both, the
hydrostatic properties need to be corrected for trimmed conditions.
5.4
Determine displacement from forward and aft draft*
* This subsection is not presented in lecture.
Recall: Hydrostatic curves (curves of form)
Previously, we discussed how to determine the displacement by summation of weights.
If we know the draft at stern and bow, there are three methods to determine the trim:
Method 1:
Determine the displacement (∆) from the hydrostatic curves or table at
mean draft,
T
M
= (
T
A
+
T
F
)
/
2. This method is least accurate but acceptable (especially
when center of flotation is near amidship).
Method 2:
Determine the displacement from the hydrostatic curves or table at mean
draft,
T
M
= (
T
A
+
T
F
)
/
2, plus the displacement of a correction layer.
This method has
better accuracy than Method 1.
Assumptions:
The LCFs of the waterline
W
O
L
O
and
W
M
L
M
are the same.
Figure 5.4: Displacement when trimmed: definitions.
In figure 5.4, we want to find the displacement at trimmed waterline
WL
.
— The displacement at trimmed waterline
WL
equals to the displacement at even-keel
waterline
W
O
L
O
.
7
NE224 Spring 2023
Instructor: An Wang
— The displacement at
W
O
L
O
equals to the displacement at waterline
W
M
L
M
(even keel
and at mean draft
T
M
of the trimmed condition), plus the displacement of the correction
layer of thickness
m
=
T
O
−
T
M
.
Procedures of method 2:
Step 1:
At mean draft,
T
M
= (
T
A
+
T
F
)
/
2, use the hydrostatic curves or table to
determine LCF, TPI or TPC and the uncorrected displacement, ∆
M
.
Step 2: Determine the thickness of the correction layer:
m
= (LCF
/L
)
t
.
Step 3: Determine the displacement of the correction layer:
In U.S. units
d
∆ =
m
×
TPI =
TPI
×
LCF
L
t
(m in inches and
d
∆ in tons)
In SI units:
d
∆
m
=
m
×
TPC =
TPC
×
LCF
L
t
(m in cm and
d
∆
m
in MT)
Step 4: Determine the displacement at trimmed condition:
Corrected displacement (∆) = uncorrected displacement (∆
M
) +
displacement of the correction layer (
d
∆ or
d
∆
m
)
Step 5: Determine the even keel draft (or draft at center of flotation at trimmed condi-
tion):
T
O
=
T
M
+
m
=
T
M
+
LCF
L
t
Question:
Can we find
T
O
in step 5 first, then find the displacement at
T
O
from the
hydrostatic curves?
Sign:
a) Trim by stern, LCF aft, correction must be added.
b) Trim by stern, LCF fwd, correction must be subtracted.
c) Trim by head, LCF aft, correction must be subtracted.
d) Trim by head, LCF fwd, correction must be added.
Example:
(example 5-2, Zubaly) A ship of LBP = 150 m floats in seawater at drafts
of 8.2 m forward and 9.8 m aft.
At the mean draft of 9.0 m, the curves of form give
∆
m
= 20
,
200 MT, TPC = 26
.
5 MT/cm and LCF = 4
.
3 m aft of amidship.
Determine
8
NE224 Spring 2023
Instructor: An Wang
the even-keel draft correction corresponding to the trimmed waterline and the displacement
corrected for trim. (Use the above method 2)
Solution:
Trim:
t
=
T
A
−
T
F
= 9
.
8 m
−
8
.
2 m = 1
.
6 m = 160 cm
by the stern
Corresponding even-keel draft:
T
0
=
T
M
+
LCF
L
t
= 9
.
0 m +
4
.
3 m
150 m
×
1
.
6 m = 9
.
04 m
In the above calculation, trim is by stern and LCF is aft of amidship, so the correction layer
should be added.
The displacement of the correction layer:
d
∆
m
=
TPC
×
LCF
L
t
=
26
.
5 MT
/
cm
×
4
.
3m
150 m
×
160 cm = 122 MT
Therefore, the displacement at the trimmed condition:
∆
m
= 20
,
200 MT + 122 MT = 20
,
322 MT
Method 3:
Determine the displacement of a trimmed ship by direct numerical inte-
gration the displaced volume of the ship floating at the trimmed waterline. The immersed
sectional area can be found on Bonjean curves (discussed previously). This method is the
most accurate.
5.5
Moment to change trim
In equation 5.7, if we let
t
= 1 inch we can define:
moment to change trim one inch:
MT1 =
∆
GM
L
12
L
(∆ in ton,
GM
L
and
L
in ft)
(5.12)
Therefore, if a weight is shifted longitudinally from
x
0
to
x
1
, the trim change can be
written as
t
=
w
(
x
1
−
x
0
)
MT1
(
w
in ton, (
x
1
−
x
0
) in ft)
(5.13)
or if using SI units, let
t
= 1 cm, we can define
9
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moment to change trim one cm:
MTcm =
∆
m
GM
L
100
L
(∆
m
in MT,
GM
L
and
L
in m)
(5.14)
if a weight is shifted longitudinally from
x
0
to
x
1
, the trim change can be written as
t
=
w
(
x
1
−
x
0
)
MTcm
(
w
in MT, (
x
1
−
x
0
) in m)
(5.15)
In the discussion of transverse stability, we need to know the center of gravity
G
to
determine
GM
T
(=
KB
+
BM
T
−
KG
). However, in the discussion of longitudinal stability,
since
BM
L
(on the order of ship length) is much greater than
KB
and
KG
,
GM
L
can be
approximated by
BM
L
, or
GM
L
≈
BM
L
(5.16)
Therefore, the value of MT1 and MTcm can be calculated before knowing the loading
condition of the ship.
MT1
≈
∆
BM
L
12
L
(5.17)
Since
BM
L
=
I
L
/
∇
and ∆ =
ρg
∇
, the above equation can be simplified to
MT1
≈
ρgI
L
12
L
(5.18)
In SI units:
MTcm
≈
∆
m
BM
L
100
L
(5.19)
MTcm
≈
ρI
L
100
L
(5.20)
Therefore, MT1 and MTcm are only functions of the longitudinal moment of inertia of
the waterplane.
Note:
1) Strictly speaking, the “12” in equation 5.12 should be “12 in
/
ft”. Therefore, the units
of MT1 should be ton
·
ft
/
in. Similarly, the “100” in equation 5.14 should be “100 cm
/
m”.
Therefore, the units of MTcm should be MT
·
m
/
cm.
2) As shown in figure
??
, the MT1 or MTcm at various draft are usually plotted in the
hydrostatics curves or given in table of hydrostatic properties.
10
NE224 Spring 2023
Instructor: An Wang
Example:
(example 5-3, Zubaly) Using the data given for the tanker of Example 5-1,
calculate the exact and approximate MT1. Use the approximate MT1 to determine the final
drafts of the ship and compare them to the drafts calculated in example 5-1.
Example 5-1: A tanker 1,045 ft long with a displacement of 305,000 tons floats in seawater
at drafts of 62’0” forward and 65’0” aft.
KB
= 33
.
0 ft,
KG
= 45
.
0 ft,
KM
L
= 740 ft and
the center of flotation is 15.5 ft aft of amidship. What final drafts will result if 2,200 tons of
cargo are shifted from cargo tanks #2 to # 5, a distance of 285 ft toward aft?
Solution:
5.6
Loading and discharging small weights
Small weights method:
Assumption: when a small weight is loaded/discharged, the change in the displacement
of the ship is small. Under this assumption, the properties of the waterplane (such as TPI,
LCF, MT1) can be considered unchanged.
To determine the trim caused by loading/discharging small weight
w
at a given longitu-
dinal position, we can consider the process as two steps.
Step 1: Parallel sinkage or parallel rise caused by loading or discharging the weight at
the center of flotation (LCF). There is no change of trim in step 1. The change in draft (in
inches) at all longitudinal position of the ship due to parallel sinkage/rise is
t
p
=
w
TPI
Step 2: Shifting the weight from LCF to its final location (by a distance
d
from LCF, see
figure 5.5 below). First find MT1 from the hydrostatic properties at the original condition
(before loading/discharging). Then use equation 5.13 or 5.15 to calculate the change of trim,
t
=
wd/
(MT1). At a given longitudinal position
X
(
X
is the distance from LCF), the draft
change at this location,
t
x
, can be calculated by
t
x
=
X
L
t
=
X
L
wd
MT1
Combine the two steps, the total draft change at location
x
(distance
X
from LCF) is
δT
x
=
w
TPI
±
X
L
wd
MT1
(
w
in ton,
X
,
L
,
d
in ft,
δT
x
in inch)
(5.21)
11
NE224 Spring 2023
Instructor: An Wang
Figure 5.5: Draft change at any point
x
, as a result of loading/discharging.
Note:
The sign in equation 5.21 should be determined by whether the location of the
weight and the position of interest are on the same side of the center of flotation. If yes, use
plus “+”; if not, use “-”. For loading:
w
is positive; for discharging,
w
is negative.
Example:
(example 5-6) A Mariner class ship is floating in seawater at drafts of 25’6”
forward and 26’6” aft. The following loading then takes place:
Discharging 280 tons of cargo from hold #2, 2nd deck;
Discharging 440 tons of cargo from hold #3, 2nd deck;
Load 615 tons of cargo from hold #4, 3rd deck;
Load 385 tons of cargo from hold #4, 2nd deck;
Load 235 tons of cargo from hold #5, 2nd deck;
Load 220 tons of fuel in double bottom tank #3, centerline.
Determine the final drafts, using the small weights method.
12
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Figure 5.6: Example of loading table of Mariner class.
Solution:
The initial mean draft:
T
M
0
=
T
A
0
+
T
F
0
2
=
26
′
6
′′
+ 25
′
6
′′
2
= 26
′
Read the hydrostatic properties of the ship at the initial mean draft 26’ in figure
??
, we
find:
TPI = 67
.
3 tons
/
in
MT1 = 1
,
720 tons
·
ft
/
in
LCF = 277
.
3 ft from FP
Therefore, the distance from center of flotation to FP and AP:
F
= 277
.
3 ft
A
=
L
−
F
= 528 ft
−
277
.
3 ft = 250
.
7 ft
13
NE224 Spring 2023
Instructor: An Wang
There are more than one weights loaded/discharged, so we need to find out the total
weight and total moment created by all the weights. To do that, we need to find out the
location of the center of gravity of each weight relative a reference point, for example, FP.
The location of each cargo hold of Mariner class is shown in figure 5.6 above and in Appendix
A in Zubaly.
Then we use the following table for the calculation:
Center of gravity relative to FP of all the loaded/discharged weights:
240
,
871
/
735 = 327
.
7 ft
The loading is equivalent to loading a weight of 735 tons at 327.7 ft from FP. The distance
of the center of gravity of the weight loaded to the center of flotation:
d
= 327
.
7 ft
−
277
.
3 ft = 50
.
4 ft, aft of center of flotation
Now, we can use equation 5.21 to calculate the draft change at FP and AP.
FP and the loaded weight are at the opposite side of center of flotation, so we should
use “-” in equation 5.21. The change of draft at FP:
δT
F
=
w
TPI
−
F
L
wd
MT1
=
735 ton
67
.
3 tons
/
in
−
277
.
3 ft
528 ft
735 ton
×
50
.
4 ft
1
,
720 tons
·
ft
/
in
= 10
.
9 in
−
11
.
3 in =
−
0
.
4 in
AP and the loaded weight are at the same side of center of flotation, so we should use
“+” in equation 5.21. The change of draft at AP:
δT
A
=
w
TPI
+
A
L
wd
MT1
=
735 ton
67
.
3 tons
/
in
−
250
.
7 ft
528 ft
735 ton
×
50
.
4 ft
1
,
720 tons
·
ft
/
in
= 10
.
9 in + 10
.
2 in = +21
.
1 in
14
NE224 Spring 2023
Instructor: An Wang
The final draft at FP:
T
F
=
T
F
0
+
δT
F
= 25
′
6
′′
−
0
.
4
′′
= 25
′
5
.
6
′′
The final draft at AP:
T
A
=
T
A
0
+
δT
A
= 26
′
6
′′
+ 21
.
1
′′
= 28
′
3
.
1
′′
Question:
recall in transverse stabiliy, we derived the change in the transverse meta-
centric height
GM
and the angle of heel after a small weight is loaded/discharged at an
off-center transverse location.
Similarly, one can follow the same procedure to derive the
same equation for loading/discharging small weight at a longitudinal position different from
the center of flotation (only the transverse metacentric height will be replaced with longi-
tudinal metacentric height). Is the the approach discussed in the current subsection ( 5.6)
different?
— Recall equation for
G
1
M
1
(the new transverse metacentric height) after loading a
small weight. Similarly, the new longitudinal metacentric height after loading a small weight
above/below/at the center of flotation and at height
z
above baseline can be calculated by
G
1
M
L
1
=
GM
L
+
w
w
+ ∆
T
+
δT
2
−
z
−
GM
L
.
Because
T
+
δT/
2
−
z
is very small compared with
GM
L
, the above equation can be
approximated by
G
1
M
L
1
≈
GM
L
−
w
w
+ ∆
GM
L
=
∆
w
+ ∆
GM
L
Now move the weight longitudinally away from the center of flotation by distance
d
, the
resulting angle of trim:
tan
θ
=
wd
(
w
+ ∆)
G
1
M
L
1
≈
wd
(
w
+ ∆)
∆
w
+ ∆
GM
L
=
wd
∆
GM
L
Since
tan
θ
=
t
12
L
=
t
x
12
X
and
MT1 =
∆
GM
L
12
L
15
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NE224 Spring 2023
Instructor: An Wang
, we have
tan
θ
=
t
12
L
=
t
x
12
X
=
wd
12
L
×
MT1
Therefore
t
x
=
X
L
wd
MT1
Total change in draft at location
x
due to parallel sinkage and change in trim:
δT
x
=
t
p
±
t
x
=
w
TPI
±
X
L
wd
MT1
By this approach, we derived the same result as in equation 5.21. Therefore, for load-
ing/discharging small weight at a longitudinal position away from center of flotation, the
two methods are indeed equivalent for calculating the change of draft, if certain assumptions
are made.
5.7
Trim table
Figure 5.7: Mariner class trim table.
Note:
16
NE224 Spring 2023
Instructor: An Wang
1) Each column represent a longitudinal location (aligned with the ship profile drawing
above). The longitudinal positions are equally spaced (for example, in the above able, the
spacing is 10 ft).
2) The correction of draft at forward and aft in each column is for each 100 tons of
cargo loaded to its corresponding location. If loading a weight
w
different from 100 ton, the
numbers in the tables should be multiplied by
w/
100.
3) For discharging weight, the numbers in the table will reverse sign.
4) If the actual mean draft is different from the corresponding draft for each table (20 ft
and 30 ft), one can choose the table whose corresponding draft is the closest to the actual
mean draft. Or one can interpolate the results from two adjacent tables whose corresponding
drafts are closest to the actual mean draft.
Example:
Use the table of trim of Mariner class to determine: the draft at forward and
aft after loading a cargo of 250 tons to 120 ft forward of amidship. The initial draft of the
ship is 19’6” forward and 20’6” aft.
Solution
Mean draft: (19
′
6
′′
+ 20
′
6
′′
)
/
2 = 20
′
.
Use the 20’-0” table.
Count 120 ft forward, the correction is +6.2’ forward and -2.7”
aft (for 100 tons of cargo).
Therefore, when loading 250 tons of cargo, the correction is
+6
.
2
′
×
2
.
5 = 15
.
5
′
forward and
−
2
.
7
×
2
.
5 =
−
6
.
75
′
aft.
Final draft: 19
′
6
′′
+ 15
′
6
′′
= 35
′
forward and 20
′
6
′′
−
6
′
9
′′
= 13
′
9
′′
aft.
5.8
Loading and discharging large weights
When loading/discharging large weight, the small weight assumption is no longer valid,
i.e., the waterplane properties (such as
A
w
, LCF, TPI) changes significantly. Therefore, we
cannot use TPI to determine the parallel sinkage/rise.
Large weights method:
(also called LCG method or whole ship method)
Step 1:
Determine the displacement and LCG from the draft at forward and aft before
loading/discharging.
a) Determine the original displacement ∆
0
from the draft at forward and aft,
T
F
and
T
A
:
see the three methods discussed in
5.4 (usually method 1 can be used, i.e., use the mean
draft for determining the displacement).
Find the hydrostatic properties, such as LCB
0
,
MT1 at the original draft ∆
0
from hydrostatic curves or table.
17
NE224 Spring 2023
Instructor: An Wang
b) Determine the LCG at the trimmed condition before loading/discharging. See below.
The trim is first calculated by
t
=
T
A
−
T
F
(5.22)
According to equation 5.13, the trimming moment (which causes the ship to trim from
even-keel to the trimmed condition) is
˜
w
˜
d
=
t
×
MT1
, where MT1 is found in the hydrostatic curves of the ship at the mean draft before load-
ing/discharging, ˜
w
and
˜
d
are virtual weight and virtual distance that create the initial trim
from the even-keel condition.
Then, the longitudinal distance from the center of gravity at the initial trimmed condition
(before loading), to the center of gravity at even-keel condition (before loading),
GG
0
can
be calculated by
GG
0
=
˜
w
˜
d
∆
0
=
t
×
MT1
∆
0
(5.23)
Then, the longitudinal position of center of gravity at the initial trimmed condition
(before loading) can be determined by (see figure 5.8 below)
LCG = LCG
0
+
GG
0
= LCB
0
+
GG
0
(5.24)
, where LCB
0
can be found in the hydrostatic curves of the ship at the mean draft before
loading/discharging.
Note:
In equation 5.24, LCG is measured from FP (consistent with the hydrostatic
table shown in figure
??
). If trim by stern,
GG
0
is positive and
G
is aft of
B
0
; if trim by
head,
GG
0
is negative and
G
is forward of
B
0
.
Figure 5.8: Shift of center of gravity from
G
0
(at even-keel) to
G
(at the trimmed condition).
18
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NE224 Spring 2023
Instructor: An Wang
Example:
(example 5-8) A Mariner class ship floats in seawater at drafts of 19’2”
forward and 23’8” aft. Determine the ship’s LCG in ft from the forward perpendicular.
Solution:
Drafts at aft and forward are:
T
A
0
= 23
′
8
′′
T
F
0
= 19
′
2
′′
First calculate the mean draft and the trim (before any weight is loaded). The mean draft:
T
M
0
=
T
A
0
+
T
F
0
2
=
23
′
8
′′
+ 19
′
2
′′
2
= 21
′
5
′′
the trim
t
0
=
T
A
0
−
T
F
0
= 23
′
8
′′
−
19
′
2
′′
= 4
′
6
′′
= 54 in by the stern
At the mean draft
T
M
0
= 21
′
5
′′
, read the hydrostatic properties from figure
??
.
Displacement: ∆
0
= 14
,
330 ton
Longitudinal position of center of buoyancy at even-keel: LCB
0
= 265
.
45 ft from FP
Moment to change trim by 1 in: MT1 = 1
,
530 ton
·
ft/in
The longitudinal shift of center of gravity from even-keel condition to the trimmed con-
dition (equation 5.23,
GG
0
is positive because the ship is trimmed by stern):
GG
0
=
t
0
×
MT1
∆
0
=
54 in
×
1
,
530 ton
·
ft
/
in
14
,
330 ton
= 5
.
77 ft
The longitudinal position of the center of gravity at the trimmed condition (equation 5.24):
LCG = LCB
0
+
GG
0
= 265
.
45 ft + 5
.
77 ft = 271
.
22 ft from FP
Step 2:
Determine the final displacement, ∆
′
, and center of gravity, LCG
′
after load-
ing/discharging. In the following, we use prime to denote the variables after the loading.
∆
′
= ∆ +
X
w
i
(5.25)
LCG
′
=
∆
×
LCG +
∑
w
i
×
x
i
∆
′
(5.26)
, where
x
i
is the distance of each loaded/discharged weight from the reference point.
Example:
(example 5-9) The Mariner ship in example 5-8 is to be loaded as follows:
19
NE224 Spring 2023
Instructor: An Wang
Load 320 tons in hold #2 on the tanktop.
Load 355 tons in hold #2 on the 3rd deck.
Load 640 tons in hold #3 on the 3rd deck.
Load 368 tons in hold #3 on the 2nd deck.
Load 652 tons in hold #4 on the 3rd deck.
Load 448 tons in hold #5 on the 2nd deck.
Load 685 tons in hold #6 on the 3rd deck.
Load 422 tons in hold #6 on the 2nd deck.
Load 355 tons in hold #7 on the 3rd deck.
Determine the displacement and the LCG of the ship when the loading is completed.
Solution:
Use the following table to calculate the total weight of the ship and the LCG
after loading is completed.
The total weight of the ship after loading: ∆
′
= 18
,
593 ton
The LCG of the ship after loading: LCG
′
= 271
.
88 ft from FP.
Step 3:
Determine the new drafts from the displacement and LCG after loading/discharging
(from step 2).
Use the new displacement ∆
′
determined in step 2, from the hydrostatic properties of
the ship, read out the following:
— New mean draft
T
′
0
— New center of buoyancy: LCB
′
0
at even keel condition at mean draft
T
′
0
.
— New MT1
′
— New LCF
′
20
NE224 Spring 2023
Instructor: An Wang
Then the procedure in step 1 is reversed to find the final draft. See below:
The shift of center of gravity from the even keel condition to the trimmed condition.
G
′
G
′
0
= LCG
′
−
LCB
′
0
(5.27)
The new trim (after loading/discharging) can be found by
t
′
=
∆
′
G
′
G
′
0
MT1
′
(5.28)
With the new LCF
′
, the new distance of FP and AP to the new center of flotation can
be found:
F
′
=
x
F
−
LCF
′
and
A
′
=
x
A
−
LCF
′
Then, the change of draft at forward and aft:
f
′
=
F
′
L
t
′
and
a
′
=
A
′
L
t
′
The final draft at forward and aft:
T
′
F
=
T
′
0
±
f
′
T
′
A
=
T
′
0
±
a
′
Example:
(example 5-10) Determine the drafts forward and aft of the Mariner ship
after loading as described in Example 5-9.
Solution:
Read the hydrostatic properties of the Mariner class (figure
??
), at the new displacement
after the loading is completed ∆ = 18
,
593 ton (calculated previously in example 5-9):
Mean draft at even-keel:
T
′
0
= 26
′
9
′′
.
Longitudinal position of the center of buoyancy at even-keel: LCB
′
0
= 267
.
70 ft from FP
Moment to change trim by 1 in: MT1
′
= 1
,
758 ton
·
ft/in
Longitudinal position of center of flotation: LCF
′
= 278
.
15 ft from FP
21
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NE224 Spring 2023
Instructor: An Wang
Calculate the shift of the center of gravity of the ship from the even-keel condition to
the trimmed condition (equation 5.27):
G
′
G
′
0
= LCG
′
−
LCB
′
0
= 271
.
88 ft
−
267
.
70 Ft = 4
.
18 ft ,
G
′
is aft of
B
′
0
The trim of the ship after the loading is completed (equation 5.28):
t
′
=
∆
′
G
′
G
′
0
MT1
′
=
18
,
593 ton
×
4
.
18 ft
1
,
758 ton
·
ft
/
in
= 44
.
2 in
The draft is by stern, because
G
′
is aft of
B
′
0
(LCG
′
>
LCB
′
0
). Therefore, the draft at
FP should decrease from the even-keel condition and the draft at AP should increase from
the even-keel condition.
Change of draft at FP:
f
′
=
F
′
L
t
′
=
LCF
′
L
t
′
=
278
.
15 ft
528 ft
×
44
.
2 in = 23
.
3 in = 1
′
11
.
3
′′
(subtract)
Change of draft at AP:
a
′
=
A
′
L
t
′
=
L
−
LCF
′
L
t
′
=
249
.
85 ft
528 ft
×
44
.
2 in = 20
.
9 in = 1
′
8
.
9
′′
(add)
The final draft after the loading is completed:
At FP:
T
′
F
=
T
′
0
−
f
′
= 26
′
9
′′
−
1
′
11
.
3
′′
= 24
′
9
.
7
′′
At AP:
T
′
A
=
T
′
0
−
a
′
= 26
′
9
′′
−
1
′
8
.
9
′′
= 28
′
5
.
9
′′
22
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- The right answer should be F_lift= 105 kip and F_thrust= 415 kip. I tried many times with acceleration in horizontal but couldn’t get these correct numbers.arrow_forwardHello, please solve part b accurate and exact please.arrow_forwardItem 1 (no subje X C n My Drive X ▼ Part A ko = Value Submit Amazon.c μA Provide Feedback X Request Answer ProblemID=15630894 Al LO hp idle game Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O. Express your answer to three significant figures and include the appropriate units. Units X SHE ? ME Assign W Mastering h 4:57 PM 11/29/2022arrow_forward
- 6. A punch punches a 1-in. diameter hole in a steel plate % inch thick every 10 sec. The actual punching takes 1 sec. The ultimate shear strength of the plate is 60,000 psi. The flywheel of the punch press has a mass moment of inertia of 500 in-lb sec2and rotates at a mean speed of 150 rpm. What is the total speed fluctuation of the flywheel in rpm?arrow_forwardHi, I have an engineering dynamics question.arrow_forwardGiven: The plane accelerates in its current trajectory with a= 100 m/s^2 Farag Angle theta= 5° W=105 kips F_drag= 80 kips m= 1000 lbs Find: F_thrust, F_lift Please include the KD. Fthrust Futel t Fueight 000 BY NC SA 2013 Michael Swanbomarrow_forward
- A rectangular tube of outer width w = 7.5 m, outer height h = 5 m and thickness t = 0.17 m experiences bending through the x axis. What is the moment of inertia of the cross-sectional area In? y -- X MATLAB input variables: format shortEng W = 7.5; h = 5; t = 0.17; moment of inertia I, = m4arrow_forward1. Given: N Specific Weight = 25 m3 %3D • Mass = 2.2 Ibs m • Acceleration = 32.2 s2 Find: Diameter of Cylinder in inches • Remarks: Show your complete solution (with UOM) and conversion.arrow_forwardThe plot below of load vs. extension was obtained using a specimen (shown in the following figure) of an alloy remarkably similar to the aluminum-killed steel found in automotive fenders, hoods, etc. The crosshead speed, v, was 3.3x104 inch/second. The extension was measured using a 2" extensometer as shown (G). Eight points on the plastic part of the curve have been digitized for you. Use these points to help answer the following questions. 1. 900 800 (0.10, 630 ) 700 (0.50, 745) (0.30, 729) 600 (0.20, 699) (0.40, 741.5) 500- (0.004, 458) (0.80, 440 ) - 400 (0.0018, 405) 300 D= 3.3 " 0.03" 200 G=2.0" 100 - 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Extension, inches a. Determine the following quamtities. Do not neglect to include proper units in your answer. Young's Modulus Total elongation Yield stress Ultimate tensile strength Uniform elongation Engineering strain rate Post-uniform elongation b. Construct a table with the following headings, left-to-right: Extension, load, engineering strain,…arrow_forward
- 1. The figure below shows a one-dimensional, horizontal beam of uniform-density in static equilibrium. The beam is supported by a frictionless support at a point 1/3 of the way down its length (L). A rope running through a frictionless pulley is attached to the far end of the beam. The mass of the beam per-unit-meter (m) is known, and the magnitude of the acceleration due to gravity is g. y | 9 13 a) What is the total force due to the weight of the beam and where does it act? b) Draw the free body diagram (FBD) of this beam. c) Derive expressions for the vertical reaction force at the support (R) and the force applied to the end of the rope (F). Your expressions should be in terms of L, m,, and g; that is, you should find Ry = f(L, mr, g) and F = f(L, mz.g).arrow_forward5.Design a system to measure the bending moment (up and down) and torque in the tail boom of the human-powered aircraft shown below. You will need two Wheatstone bridges. Your system should measure the bending and torque independently, but not be sensitive to any other internal forces, like axial force, transverse shear, or bending moment about the vertical axis (back and forth). (If relevant) A clearly labeled diagram (or diagrams) about your analysis with a coordinate system and relevant labels. Final answer with appropriate units and significant figures. You can use the fprintf() command in MATLAB to format numerical results A 2-3 sentence reflection on your answer. Does it make sense? Why or why not? What are some implications? (a) Make a few clear, labeled sketches showing approximately where you would place the strain gauges. Consider the location (along the length of the boom), positioning (around the circumference of the boom), and orientation of the gauges. Show clearly…arrow_forwardPlease see attached photo for question information. Thanksarrow_forward
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