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D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
1
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D. Dry Docking
Trim
Effect on G
Simpson Rules
Dry Docking
Simplified Stab
Statical Stab
Revisions Ex.
Inclining Test
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
2
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LEARNING OBJECTIVES
• To understand the virtual loss of GM and
the calculations.
• To calculate the maximum trim allowed
to maintain a minimum stated GM.
• To understand the safe requirements for
a ship prior enter into dry dock.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
3
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LEARNING OBJECTIVES
• To understand the critical period during
dry docking process.
• To calculate the ship’s drafts after the
water level has fallen and after the ship
has taken the block overall.
• Effect to stability when vessel has run
aground (single point).
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Anybody would like to share their experience during
dry docking….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Before enter into dry dock, vessel must have…
• Positive initial GM (GM fluid)
• Upright
• Trim - if possible even-keel or
slight trim by stern
• Double bottom tank kept either dry
or pressed up - reduced FSE
• If initial GM is small - D.B. tank to
be pressed up to increase GM
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock:
• The vessel will line-up with her
centerline vertically over the keel
blocks
• Dock gate will be closed and
commence pumping out water
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
7
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F
No effect on ship’s Initial Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock:
• The rate of pumping will be
reduced as the ship's sternpost
near the block.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Sueing Point
F
Commence touching the ground… ‘Sueing Point’
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock:
• Once the sternpost is touching the
block, the UP-THRUST forces start
to act against the sternpost.
• At this moment part of ship's
weight gets transferred to the keel
blocks.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
11
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P
F
P is the Upthrust Force acting at first point of
touching the ground. Commence Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Sueing Point
at ‘AP’
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
K
P is the Upthrust Force acting at first point of
touching the ground. Commence Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock:
• When ship's weight gets transferred to
the keel blocks, vessel will suffer loss on
her GM.
• The time interval between the sternpost
landing on the blocks and the ship taking
the blocks overall is referred to as the
CRITICAL PERIOD.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
F
P force is increasing gradually as the trim change
by Head…Vessel is still in Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock:
• Vessel must have positive effective
GM that to be maintained
throughout the critical period.
• If not vessel may heel over, slip off
the blocks when there is an
external force acting and heel the
ship.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
F
Vessel is fully rest on the blocks… End of Critical
Period
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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<>
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
19
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M
G1
G
Initial GM loss by GG1 after
completed the Critical
Period…
This is due to Upthrust
Force or ‘P’ Force…
B
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
20
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P
F
What is the total P Force during Critical Period __?___ tonnes
“How much weight to be discharged in order to bring the ship
from trim by stern to even-keel…”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
21
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CALCULATION
OF UPTHRUST FORCE
AT THE STERNPOST
- 'P' FORCE
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
22
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w
d
F
Weight discharged to even keel the draft…
Trimming Moment
D. Dry Docking
=
w x d t-m by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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F
After weight discharged…
T M By Head
D. Dry Docking
=
T M By Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
24
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P
d
F
P is the Upthrust Force or weight discharged to
the blocks…
T.M
D. Dry Docking
=
wxd
=
P x d t-m by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
25
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P
F
Vessel is fully rest on the blocks, Change of Trim
by Head and finally vessel at even keel drafts…
End of Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
26
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Change of Trim =
Trimming Moment (TM)
MCTC
Whereby TM
=
w x d
Change of Trim =
P x d
MCTC
P
D. Dry Docking
=
=
P x d
COT x MCTC tonnes
d
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
27
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Exercise in classroom
MV OneSuch, LBP 120m is going to dry dock
at the following condition in sea water
Draft forward is 3.5m and aft is 4.0m,
distance sueing point (AP) to F is 57.5m.
Her displacement is 4600 tonnes, MCTC is 86
t-m and TPC 15.45
Calculate
i. The amount of up-thrust force (P) at
the end of Critical Period?
ii. Final drafts forward and aft?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
28
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Calculation of P force…
P
P
D. Dry Docking
=
COT x MCTC
d
=
50 x 86
57.5
=
74.8 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
29
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Body rise
CODA =
=
D. Dry Docking
=
P
TPC
=
74.8
15.45
=
4.8cm
=
0.048m
57.5 x 50
120
24cm
COT
=
Pxd
MCTC
=
74.8 x 57.5
86
=
50cm
CODF =
=
50 – 24
26cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Forward
Aft
Initial draft
3.500m
4.000m
Body rise
0.048m -
0.048m –
COD
0.260m +
0.240m –
Final draft
3.712m
3.712m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
31
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P
d
F
P is the Upthrust Force or weight discharged to
the blocks…
T.M
D. Dry Docking
=
wxd
=
P x d t-m by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
32
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P
F
Vessel is fully rest on the blocks, Change of Trim
by Stern and finally vessel at even keel drafts…
End of Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
33
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Change of Trim
=
Trimming Moment (TM)
MCTC
Whereby
TM
=
w x d
Change of Trim
=
P x d
MCTC
=
COT x MCTC tonnes
d
P
D. Dry Docking
=
P x d
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
34
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Virtual Loss Of GM
During
Critical Period
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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D. Dry Docking
•
Method 1 – GG1
•
Method 2 – MM1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
36
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Method 1
• When the vessel comes in contact
with the blocks, it is assumed that
there is a transfer of weight 'P'
from the keel to the blocks.
• Hence there is a virtual rise of
ship's G (discharged of weight
below G)
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
37
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P
d
F
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
38
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P
d
F
Trimming Moment by… Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
39
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P
K
P is the Upthrust Force acting at first point of touching
the ground. Commence Critical Period…”weight
discharged from the ship”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
40
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M
G1
G
K
Reduction or Loss of GM = GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
41
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M
G1
G
Initial GM loss by GG1 during
the Critical Period…
This is due to Upthrust Force
or ‘P’ Force…
B
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
42
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Method 1
GG1
D. Dry Docking
=
P x KG
W-P
in metres
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
43
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P
M
G1
G
B
W
During Critical Period… part of ship body is still floating
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
44
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P
W-P
External Force
M

G

B
B1
W
Vessel is inclined to a small angle by an
external force…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
45
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P
M
External Force
G1
G

K
W-P
Method 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
46
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P
M
X
X = KG1 Sin 
G1
G

K
W-P
Method 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
47
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P
M
Y
G1
G

K
W-P
W
Method 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
48
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Y

G1
Y = GG1 Sin 
G
Method 1… Discharged of weight, shift of GG1…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
49
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P
M
Y
G1
X
G

K
W-P
W
Method 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
50
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P
X
G

K
D. Dry Docking
W
G1
=
Y
G1

G
W
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
51
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X=
KG1 x Sin 
PX
Y=
GG1 Sin 
=
WY
=
W x GG1 Sin 
=
W x GG1
=
W x GG1
(P x KG) + (P x GG1) =
W x GG1
P x KG1 x Sin 
P x KG1
P x (KG + GG1)
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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(P x KG) + (P x GG1)
D. Dry Docking
=
W x GG1
P x KG
=
(W x GG1) – (P x GG1)
P x KG
=
(W – P) x GG1
P x KG
W–P
=
GG1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Therefore the formula is…
GG1
D. Dry Docking
=
P x KG
W-P
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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W-P
External Force
M

G1
Z
B
B1
W-P
Righting Moment at small angle of heel…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
55
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W-P
M

Righting Moment
= W x GZ
= W x GM Sin 
In this case,
Righting Moment
= (W – P) x G1M Sin 
G1
Z
W-P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
56
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D. Dry Docking
•
Method 1 – GG1
•
Method 2 – MM1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
57
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Method 2
• When the vessel comes in contact
with the blocks, it is assumed that
there is a transfer of buoyancy 'P'
to the keel blocks.
• Hence there is a reduction in KM
while the weight and KG are
remains constant.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
58
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P
d
F
Reduction in Buoyancy
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
d
Reduction in Buoyancy
F
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
60
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P
K
P is the Upthrust Force acting at first point of
touching the ground. Commence Critical
Period…”buoyancy reduction from the ship”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
61
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P
Buoyancy Reduction
K
P is the Upthrust Force acting at first point of
touching the ground. Commence Critical
Period…”buoyancy reduction from the ship”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
62
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M
M1
B
B1
K
Reduction or Loss of GM = MM1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Method 2
MM1
D. Dry Docking
=
P x KM in metres
W
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
64
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M
M1
G
Initial GM loss by MM1 after
the Critical Period…
This is due to Upthrust
Force or ‘P’ Force…
B
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
65
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P
M
M1
G
B
W
During Critical Period, part of ship body is still floating
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
66
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P
W-P
External Force
M

G

B
B1
W
Vessel is inclined to a small angle by an external force…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
67
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W-P
P
W
M
Y
M1
X
G

K
W
Method 2… Transferred of buoyancy, shift of MM1…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
68
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P
X
G

W
M1
W
W-P
M

=
M1
Y
K
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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X = KM1 x Sin 
D. Dry Docking
Y=
MM1 Sin 
PX
=
(W – P) x Y
P x KM1 x Sin 
=
(W – P) x MM1 Sin 
P x KM1
=
(W – P) x MM1
P x KM1
=
W x MM1 – P x MM1
P x KM1 + P x MM1 =
W x MM1
P (KM1 + MM1)
W x MM1
=
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
70
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P (KM1 + MM1) =
W x MM1
P x KM
=
W x MM1
P x KM
W
=
MM1
Therefore the formula is…
MM1
D. Dry Docking
=
P x KM
W
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
71
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W
M
External Force
M1

G
Z
B
B1
W
Righting Moment at small angle of heel…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
72
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W
M1

Righting Moment
= W x GZ
= W x GM Sin 
In this case,
Righting Moment
= W x GM1 Sin 
G
Z
W
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
73
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SUMMARY…
•
Method 1 – GG1 : Weight transferred
•
Method 2 – MM1 : Buoyancy transferred
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
74
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Exercise in classroom …continued
MV OneSuch is going to dry dock at the
following condition in sea water
Draft forward is 3.5m and aft is 4.0m,
distance sueing point (AP) to F is 57.5m.
Her displacement is 4600 tonnes, MCTC is 86
t-m,
Calculate the amount of up-thrust force (P)
during Critical Period and the virtual loss of
GM if KM is 8.0m and KG is 7.2m.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
75
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Calculation of P force…
P
P
D. Dry Docking
=
COT x MCTC
d
=
50 x 86
57.5
=
74.8 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
76
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Virtual loss of GM (GG1) method…
GG1 =
=
GG1 =
D. Dry Docking
P x KG
W-P
74.8 x 7.2
4600 – 74.8
0.119m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
77
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Virtual loss of GM (MM1) method…
MM1 =
=
MM1 =
D. Dry Docking
P x KM
W
74.8 x 8.0
4600
0.130m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
78
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Comparison the Virtual loss of GM between
(MM1) and (GG1) method…
Different is…
=
0.130 – 0.119
=
0.011m
…
D. Dry Docking
± 1cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Effect of Trim
In
Dry Docking
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Change of Trim
=
Trimming Moment (TM)
MCTC
Whereby
TM
=
w x d
Change of Trim
=
P x d
MCTC
=
COT x MCTC tonnes
d
P
D. Dry Docking
=
P x d
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
81
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Example
• Vessel displacement 5000 tonnes, distance
sueing point to CF is 80 m, MCTC 200 t-m,
KM 7.0 m and KG 6.0 m.
What will be the maximum trim allowed?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
82
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Case 1
0 m / Even keel
D. Dry Docking
Case 2
0.5 m by Stern
Case 3
5 m by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
83
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Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Calculate ‘P’…
Calculate ‘P’…
Calculate ‘P’…
P
P
P
= MCTC x trim
d
D. Dry Docking
= MCTC x trim
d
= MCTC x trim
d
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
84
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Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Calculate ‘P’…
Calculate ‘P’…
Calculate ‘P’…
P
P
P
= MCTC x trim
d
= 200 x 0
80
D. Dry Docking
= MCTC x trim
d
= 200 x 50
80
= MCTC x trim
d
= 200 x 500
80
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
85
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Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Calculate ‘P’…
Calculate ‘P’…
Calculate ‘P’…
P
P
P
= MCTC x trim
d
= 200 x 0
80
P
= 0 tonne
D. Dry Docking
= MCTC x trim
d
= 200 x 50
80
P = 125 tonnes
= MCTC x trim
d
= 200 x 500
80
P = 1250 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
86
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Case 1
0 m / Even keel
Virtual Loss of GM…
D. Dry Docking
Case 2
0.5 m by Stern
Virtual Loss of GM…
Case 3
5 m by Stern
Virtual Loss of GM…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
87
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Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Virtual Loss of GM…
Virtual Loss of GM…
Virtual Loss of GM…
GG1
GG1
GG1
D. Dry Docking
=
P
x KG
W–P
=
P
x KG
W–P
=
P
x KG
W-P
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
88
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Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Virtual Loss of GM…
Virtual Loss of GM…
Virtual Loss of GM…
GG1
GG1
GG1
D. Dry Docking
=
P
x KG
W–P
=
0x6
5000 – 0
=
P
x KG
W–P
= 125 x 6
5000 –125
=
P
x KG
W-P
= 1250 x 6
5000 –1250
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
89
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Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Virtual Loss of GM…
Virtual Loss of GM…
Virtual Loss of GM…
GG1
GG1
GG1
GG1
D. Dry Docking
=
P
x KG
W–P
=
0x6
5000 – 0
= 0m
=
P
x KG
W–P
=
0.154 m
P
x KG
W-P
= 1250 x 6
5000 –1250
= 125 x 6
5000 –125
GG1
=
GG1
= 2.0 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
90
To Be A World Class Maritime Academy
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Old GM = 1.0m
Old GM = 1.0m
Old GM = 1.0m
New GM…
New GM…
New GM…
=
GM - GG1
=
GM - GG1
=
GM - GG1
=
1.0 – 0
=
1.0 – 0.154
=
1.0 – 2.0
=
1.0 m
=
=
- 1.0 m
D. Dry Docking
0.846 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
91
To Be A World Class Maritime Academy
Residual GM
MAX. TRIM…?
1.0
TRIM increased
GM decreased
0.846
0
0.5
5.0
TRIM
- 1.0
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
92
To Be A World Class Maritime Academy
•
Vessel displacement 5000 tonnes,
distance sueing point to CF is 80
m, MCTC 200 t-m, KM 7.0 m and
KG 6.0 m.
Maximum Trim is….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
93
To Be A World Class Maritime Academy
During Critical Period…
P force is …? Initial GM 1.0m
M
G1
Virtual Loss of GM = 1.0m
GG1 is Virtual
Loss of GM
G
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
94
To Be A World Class Maritime Academy
During Critical Period…
P force is …? Initial GM 1.0m
M
G1
Virtual Loss of GM = 1.0m
GG1
1.0
=
x KG
W-P
= Px6
5000 – P
5000 - P
5000
P
D. Dry Docking
P
= 6P
= 7P
= 714.28 tonnes
GG1 is Virtual
Loss of GM
G
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
95
To Be A World Class Maritime Academy
P force is …? Initial GM 1.0m
Maximum trim is …?
Virtual Loss of GM = 1.0m
P
= 714.28 tonnes
GG1
P
Trim
= MCTC x trim
d
= Pxd
MCTC
= 714.28 x 80
200
= 285.7 cms
Trim
= 2.86 m by Stern
1.0
=
x KG
W-P
= Px6
5000 – P
5000 - P
5000
P
D. Dry Docking
P
= 6P
= 7P
= 714.28 tonnes
Trim
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
96
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Residual GM
1.0
MAX. TRIM ……?
2.86m
0.846
0
0.5
5.0
TRIM
- 1.0
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
97
To Be A World Class Maritime Academy
CONCLUSION:
• The virtual loss of GM is NIL as vessel
having zero trim.
• The loss is increased as the trim increased.
• Maximum trim is depend upon the initial
GM
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
98
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WORKED EXAMPLE 1
A ship of 140m in length, displacement 5000t and upright is
to enter dry dock with drafts forward 3.84m, aft 4.60m. Given
the following hydrostatic particulars:
TPC
20 tonnes
MCTC
150 t- m
CF
5m forward of amidships
KM
9.75m
The blocks of the dry dock are horizontal.
i.
Calculate the drafts of the vessel at the instants when she is
taking the blocks forward and aft.
ii.
The ship's effective GM at this moment if the KG is 7.75m
iii. The Righting Moment at this instant for an angle of heel 5º.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
99
To Be A World Class Maritime Academy
F
4.60m
Trim 76 cm by Stern
3.84m
No effect on ship’s Initial Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 100
To Be A World Class Maritime Academy
P
F
4.60m
Trim 76 cm by Stern
3.84m
P is the Upthrust Force acting at first point of
touching the ground, commence Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 101
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P
F
Change of Trim 76cms by Head
Even keel draft
What is the total P Force during Critical Period?
End of Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 102
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Ship’s trimmed
i.
=
P
=
MCTC x trim
d
P
=
152 tonnes
a. Bodily rise =
=
P
= 152
TPC
20
b. Change of Trim
D. Dry Docking
4.60 – 3.84 = 0.76 m by Stern
=
150 x 76
75
= 7.6 cms =
0.076 m
76 cms by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 103
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c. Change of draft aft due COT
= l x COT
L
= 75 x 76
140
= 40.7cm
= 0.407 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 104
To Be A World Class Maritime Academy
d. Change of draft forward due COT
= COT – Change of draft aft
= 76 – 40.7
= 35.3cm
= 0.353 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 105
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e.
Fwd
Aft
Initial drafts
Bodily rise
Change of drafts
3.840
0.076 0.353 +
4.600
0.076 0.407 -
Final drafts
4.117 m
4.117 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 106
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F
4.117m
4.117m
End of Critical Period, vessel is fully rested on
blocks, draft is at even keel
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 107
To Be A World Class Maritime Academy
i.
ALTERNATIVE METHOD
a.
Mean draft
b.
True mean draft correction
D. Dry Docking
=
4.220 m.
=
Dist. CF to amidships x trim
LBP
=
5 x 0.76
140
=
0.027 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 108
To Be A World Class Maritime Academy
i.
ALTERNATIVE METHOD
c. True mean draft
=
=
=
Mean draft – correction
4.220 – 0.027
4.193 m
=
=
=
4.193 m
0.076 m 4.117 m even keel
d. Therefore:
True mean draft
Bodily rise
Final drafts
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 109
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ii.
GG1
Initial GM
=
P x KG
W–P
=
152 x 7.75
5000 – 152
=
1178
4848
=
0.243 m
=
=
KM – KG
2.00 m
=
9.75 m – 7.75
Effective GM =
D. Dry Docking
2.00 – 0.243=
1.757 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 110
To Be A World Class Maritime Academy
M
G1
G
Initial GM loss by GG1 after
the Critical Period…
This is due to Upthrust
Force or ‘P’ Force…
B
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 111
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OR
MM1
=
P x KM
W
Effective GM
D. Dry Docking
=
152 x 9.75
5000
=
0.296 m
=
2.00 – 0.296
=
1.704 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 112
To Be A World Class Maritime Academy
M
M1
G
Initial GM loss by MM1 after
the Critical Period…
This is due to Upthrust
Force or ‘P’ Force…
B
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 113
To Be A World Class Maritime Academy
W-P
External Force
M

G1
Z
B
B1
W-P
Righting Moment at small angle of heel…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 114
To Be A World Class Maritime Academy
W-P
M
Righting Moment
= W x GZ
= W x GM Sin 

In this case,
Righting Moment
= (W – P) x G1M Sin 
G1
Z
W-P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 115
To Be A World Class Maritime Academy
W
M
External Force
M1

G
Z
B
B1
W
Righting Moment at small angle of heel…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 116
To Be A World Class Maritime Academy
W
M1
Righting Moment
= W x GZ
= W x GM Sin 

In this case,
Righting Moment
= W x GM1 Sin 
G
Z
W
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 117
To Be A World Class Maritime Academy
iii.
RM
OR
RM
D. Dry Docking
=
(W – P) x G1M Sin 
=
(5000 – 152) x 1.757 x Sin 5
=
742.4 t-m
=
W x GM1 Sin 
=
5000 x 1.704 x Sin 5
=
742.6 t-m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 118
To Be A World Class Maritime Academy
WORKED EXAMPLE 2
A ship of length 165m, KG 7.30m is floating in a
graving dock with drafts forward 5.50m, aft 7.86m in
water RD 1.025. At the aft perpendicular the keel is
0.24m above the top of the horizontal blocks. If the
water level has fallen in the dock by 1.22m, the
ship’s become unstable (GM = 0m).
Calculate
i. The drafts forward and aft at which it occurs
ii. The original/initial GM
Given
Displacement for a hydrostatic mean draft of 6.65m
is 9151 tonnes. TPC 24, MCTC 120 t-m and CF 3.66 m
abaft amidships.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 119
To Be A World Class Maritime Academy
F
5.50m
7.86m
Clearance 24cm
No effect on ship’s Initial Stability, initial trim is
2.36m by Stern
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 120
To Be A World Class Maritime Academy
F
5.50m
Depth of water 7.86 + 0.24 = 8.10m
No effect on ship’s Initial Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 121
To Be A World Class Maritime Academy
F
5.50m
7.86m
Clearance 24cm
No effect on ship’s Initial Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 122
To Be A World Class Maritime Academy
F
7.86m
5.50m
Clearance 16cm
Drop of water level by 8cm. No effect on ship’s
Initial Stability.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 123
To Be A World Class Maritime Academy
F
7.86m
5.50m
Clearance 12cm
Drop of water level by 12cm. No effect on ship’s
Initial Stability.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 124
To Be A World Class Maritime Academy
F
7.86m
5.50m
Clearance 6cm
Drop of water level by 18cm. No effect on ship’s
Initial Stability.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 125
To Be A World Class Maritime Academy
F
7.86m
5.50m
Drop of water level by 24cm… stern post start to
touch the block…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 126
To Be A World Class Maritime Academy
P
F
7.86m
5.50m
P is the Upthrust Force acting at first point of
touching the block. Commence Critical Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 127
To Be A World Class Maritime Academy
P
F
6.88m
Drop of water level 98cm, Vessel become
unstable… Zero GM. Vessel is still in Critical
Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 128
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WL
7.86m
Reduction : 98cms
6.88m
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 129
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P
d
F
Body rise & Trimming Moment by… Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 130
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7.86m
A:
Body rise
7.86m - Br
6.88m
D. Dry Docking
WL
WL
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 131
To Be A World Class Maritime Academy
WL
7.86m
WL
7.86m - Br
B:
Change of draft aft due to COT by Head
6.88m
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 132
To Be A World Class Maritime Academy
WL
7.86m
A:
Body rise
Reduction : 98cm
WL
7.86m - Br
B:
Change of draft aft due to COT by Head
6.88m
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 133
To Be A World Class Maritime Academy
Reduction = A
+
where
Body Rise
Change of draft aft
due to COT
A
B
B
REDUCTION
=
D. Dry Docking
Body rise + Change of draft aft
due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 134
To Be A World Class Maritime Academy
Fallen of water level =
where
A
B
A
+
B
Body Rise
Change of draft aft
due to COT
Fallen of water level
=
D. Dry Docking
Body rise + Change of draft aft
due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 135
To Be A World Class Maritime Academy
Fallen WL
=
P
TPC
+
l
L
x
Fallen WL
=
P
TPC
+
l
L
x Pxd
MCTC
=
P
24
+
78.84
165
98
D. Dry Docking
TM
MCTC
x P x 78.84
120
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 136
To Be A World Class Maritime Academy
98
=
P
24
+
[78.84
165
98
=
P
24
+
0.313926545P
1
98
=
P
+
24
7.534P
2352 =
8.534P
P
275.6 tonnes
D. Dry Docking
=
x
P x 78.84 ]
120
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 137
To Be A World Class Maritime Academy
P
=
275.6 tonnes
If we calculate until vessel is FULLY REST,
P
=
MCTC x trim
d
P
=
359.2 tonnes
D. Dry Docking
=
120 x 236
78.84
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 138
To Be A World Class Maritime Academy
To find the drafts forward and aft…
i.
D. Dry Docking
Bodily rise=
P
=
TPC
=
11.5 cms
=
0.115 m
275.6
24
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 139
To Be A World Class Maritime Academy
To find the drafts forward and aft…
ii.
D. Dry Docking
COT =
Pxd
MCTC
=
275.6 x 78.84
120
=
181cm by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 140
To Be A World Class Maritime Academy
iii.
D. Dry Docking
Change of draft aft due COT
=
l x COT
L
=
86.5cm
=
0.865 m
=
78.84
165
x 181
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 141
To Be A World Class Maritime Academy
iv.
D. Dry Docking
Change of draft Forward
=
COT –
Change of draft aft
=
181 –
86.5
=
94.5cm
=
0.945 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 142
To Be A World Class Maritime Academy
v.
D. Dry Docking
Fwd(m)
Aft(m)
Initial drafts
Bodily rise
Change of drafts
5.500
0.115 0.945 +
7.860
0.115 0.865 -
Final drafts
6.330
6.880
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 143
To Be A World Class Maritime Academy
To find the initial GM…
Mean draft
= 6.680m. Trim = 2.36m by stern
CF is 3.66m abaft amidships.
TMD Correction
=
Dist. CF to Amidships x Trim
LBP
=
3.66 x 2.36 =
165
D. Dry Docking
0.052 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 144
To Be A World Class Maritime Academy
To find the initial GM…
True Mean Draft (TMD)
=
=
=
Mean draft + TMD Correction
6.680 + 0.052
6.732 m
Diff of TMD
=
6.732 – 6.650
=
0.082 m
=
8.2 cm
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 145
To Be A World Class Maritime Academy
To find the initial GM…
Therefore additional displacement
=
=
8.2 cm x TPC (24)
196.8 t
Displacement for TMD 6.732 m
=
=
D. Dry Docking
9151 + 196.8
9347.8 t
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 146
To Be A World Class Maritime Academy
To find the initial GM…
When the ship become unstable, the GM = 0 m,
therefore loss of GM must be equal to initial GM.
GG1 =
=
P x KG
W–P
=
275.5 x 7.3
9347.8 – 275.5
2011.15
9072.3
=
0.222 m
=
0.222 m
Initial GM
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 147
To Be A World Class Maritime Academy
Then… what will be the MAXIMUM TRIM
allowed, safely docked
if the initial GM is 0.222 m….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 148
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GM
0.222
MAX. TRIM…?
2.36m
0
TRIM
-ve
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 149
To Be A World Class Maritime Academy
Initial GM 0.222m
P force is …?
Maximum trim is …?
Virtual Loss of GM = 0.222m
P = 275.5 tonnes
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 150
To Be A World Class Maritime Academy
Initial GM 0.222m
P force is …?
Virtual Loss of GM = 0.222m
Maximum trim is …?
P
= 275.5 tonnes
P
= MCTC x trim
d
= Pxd
MCTC
Trim
P = 275.5 tonnes
=
D. Dry Docking
Trim
275.5 x 78.84
120
= 181cm
Trim
= 1.81m by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 151
To Be A World Class Maritime Academy
GM
0.222
MAX. TRIM 1.81m
2.36m
0
TRIM
-ve
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 152
To Be A World Class Maritime Academy
Trim 1.81m by Stern
Virtual loss of GM…?
P = MCTC x trim
d
GG1
P = 120 x 181
78.84
P = 275.5 tonnes
= P x KG
W–P
=
GG1
275.5 x 7.3
9347.8 – 275.5
= 0.222
Residual GM = 0.222 – 0.222
Residual GM = 0.000
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 153
To Be A World Class Maritime Academy
Then… what will be the final drafts…
if the initial GM is 0.222 m and trim now is
1.81m by stern….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 154
To Be A World Class Maritime Academy
To find the drafts forward and aft…
i.
D. Dry Docking
Bodily rise=
P
=
TPC
=
11.5 cm
=
0.115 m
275.5
24
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 155
To Be A World Class Maritime Academy
To find the drafts forward and aft…
ii.
D. Dry Docking
COT
=
Pxd =
MCTC
275.5 x 78.84
120
=
181cm by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 156
To Be A World Class Maritime Academy
iii.
D. Dry Docking
Change of draft aft due COT
=
l x COT
L
=
78.84
165
x 181
=
86.5cm
=
0.865 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 157
To Be A World Class Maritime Academy
iv.
D. Dry Docking
Change of draft Forward
=
COT
–
=
181 – 86.5
=
0.945 m
Change of draft aft
=
94.5cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 158
To Be A World Class Maritime Academy
Assuming aft draft maintain at 7.86m, new trim is 1.81m
by astern, therefore forward draft now is 6.05m…
v.
D. Dry Docking
Fwd(m)
Aft(m)
Initial drafts
Bodily rise
Change of drafts
6.050
0.115 0.945 +
7.860
0.115 0.865 -
Final drafts
6.880
6.880
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 159
To Be A World Class Maritime Academy
Worked Example 3
Your vessel is going to dry dock with the following
conditions:
Draft forward 8.00 m and aft 9.00 m. Her displacement is
30 000 tonnes. KM is 11.50 m, KG 10.90 m. MCTC 400 tm. TPC 38. LCF is 1.5 m abaft the amidships and LBP is
160 m.
The depth of water in the dock is initially 9.50m.
i.
Find the effective GM and her new draft after water
level has fallen by 95cm in the dock.
ii. How much will be the further drop of water level so
that vessel will take the blocks overall?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 160
To Be A World Class Maritime Academy
F
9.0m
9.5m
Clearance 50cm
No effect on ship’s Initial Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 161
To Be A World Class Maritime Academy
P
F
9.0m
50cm drop of water level
9.0m
Drop of water level by 50cm, No effect on ship’s
Initial Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 162
To Be A World Class Maritime Academy
P
F
45cm drop of water level
8.55m
Drop of water level by 45cm, effect on ship’s
Initial Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 163
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WL
9.00m
Reduction : 45cm
8.55m
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 164
To Be A World Class Maritime Academy
9.00m
A:
Body rise
9.00m - Br
8.55m
D. Dry Docking
WL
WL
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 165
To Be A World Class Maritime Academy
WL
9.00m
WL
9.00m - Br
B:
Change of draft aft due to COT by Head
8.55m
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 166
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WL
9.00m
A:
Body rise
Reduction : 45cm
WL
9.00m - Br
B:
Change of draft aft due to COT by Head
8.55m
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 167
To Be A World Class Maritime Academy
Reduction = A
+
where
Body Rise
Change of draft aft
due to COT
A
B
B
REDUCTION
=
D. Dry Docking
Body rise + Change of draft aft
due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 168
To Be A World Class Maritime Academy
Fallen of water level =
where
A
B
A
+
B
Body Rise
Change of draft aft
due to COT
Fallen of water level
=
D. Dry Docking
Body rise + Change of draft aft
due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 169
To Be A World Class Maritime Academy
Fallen WL
=
P
TPC
+
l
L
x
Fallen WL
=
P
TPC
+
l
L
x Pxd
MCTC
=
P
38
+
78.5
160
45
D. Dry Docking
TM
MCTC
x P x 78.5
400
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 170
To Be A World Class Maritime Academy
45
=
P
38
+
[ 78.5
160
45
=
P
38
+
0.096285156P
1
45
=
P
+
38
3.659P
1710 =
4.659P
P
367.0 tonnes
D. Dry Docking
=
x
P x 78.5 ]
400
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 171
To Be A World Class Maritime Academy
GG1
=
P x KG
W–P
GG1
=
0.135 m
=
367.05 x 10.9
30 000 – 367.0
Initial GM
=
0.600 m
Effective GM
=
0.600 – 0.135
=
0.465 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 172
To Be A World Class Maritime Academy
OR
MM1
=
P x KM
W
MM1
Effective GM
D. Dry Docking
=
367.05 x 11.5
30 000
=
0.141 m
=
0.600 – 0.141
=
0.459 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 173
To Be A World Class Maritime Academy
To find the drafts forward and aft…
i.
D. Dry Docking
Bodily rise
=
P
TPC
=
367.0
38
=
9.66cm
=
0.097 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 174
To Be A World Class Maritime Academy
To find the drafts forward and aft…
ii.
COT
=
Pxd =
MCTC
=
D. Dry Docking
367.0 x 78.5
400
72cm by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 175
To Be A World Class Maritime Academy
iii.
Change of draft aft due COT
=
D. Dry Docking
l x COT
L
=
78.5
160
x 72
=
35.3cm
=
0.353 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 176
To Be A World Class Maritime Academy
iv.
D. Dry Docking
Change of draft Forward
=
COT
–
Change of draft aft
=
72.0
–
35.3
=
0.367 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 177
To Be A World Class Maritime Academy
v.
D. Dry Docking
Fwd(m)
Aft(m)
Initial drafts
Bodily rise
Change of drafts
8.000
0.097 0.367 +
9.000
0.097 0.353 -
Final drafts
8.270
8.550
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 178
To Be A World Class Maritime Academy
New Trim = 8.55 – 8.27
=
0.28m by Stern
assuming ‘F’ constant
P
=
400 x 28
78.5
=
142.7 tonnes
=
MCTC x T
d
P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 179
To Be A World Class Maritime Academy
Further drop
vessel fully rest
D. Dry Docking
=
P
TPC
+
l
L
x
Pxd
MCTC
=
142.7
38
+
78.5 x 142.7 x 78.5
160
400
=
17.5cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 180
To Be A World Class Maritime Academy
SINGLE POINT
GROUNDING
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 181
To Be A World Class Maritime Academy
SINGLE POINT GROUNDING
A vessel floating at drafts forward 8.70 m, aft
9.40 m grounds at a point 30 m aft of the
forward perpendicular.
Estimate the drafts of the vessel and the GM
after the tide has fallen by 70cm.
MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40
m, LBP 162 m. LCF 78 m forward of Aft
Perpendicular and displacement is 29 000
tonnes.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 182
To Be A World Class Maritime Academy
P
30m
F
8.70m
9.40m
Rock
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 183
To Be A World Class Maritime Academy
Tide fallen by 70cms…
P=?
F
Fwd…?
Aft…?
Rock
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 184
To Be A World Class Maritime Academy
WL
Draft at P
Fallen of tide by 70cm
New draft at P
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 185
To Be A World Class Maritime Academy
Draft at P
A:
Body rise
Draft at P - Br
New draft at P
D. Dry Docking
WL
WL
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 186
To Be A World Class Maritime Academy
WL
Draft at P
WL
Draft at P - Br
B:
Change of draft at P due to COT by Stern
New draft at P
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 187
To Be A World Class Maritime Academy
WL
Draft at P
A:
Body rise
Fallen of tide by 70cm
WL
Draft at P - Br
B:
Change of draft at P due to COT by Stern
New draft at P
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 188
To Be A World Class Maritime Academy
Fallen of tide
where
=
A
B
A
+
B
Body Rise
Change of draft at P
due to COT by Stern
Fallen of tide
=
D. Dry Docking
Body rise + Change of draft at P
due to COT Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 189
To Be A World Class Maritime Academy
Fallen of tide
=
P
TPC
+
l x TM
L
MCTC
Fallen of tide
=
P
TPC
+
l
L
=
P
28
+
54 x P x 54
162
340
70
D. Dry Docking
x Pxd
MCTC
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 190
To Be A World Class Maritime Academy
P
30m
54m
F
8.70m
9.40m
Rock
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 191
To Be A World Class Maritime Academy
70
=
P
28
+
[
70
=
P
28
+
0.052941176P
1
70
=
P
+
28
1.482P
1960 =
2.482P
P
789.7 tonnes
D. Dry Docking
=
54 x
162
P x 54 ]
340
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 192
To Be A World Class Maritime Academy
To find the drafts forward and aft…
i.
D. Dry Docking
Bodily rise
=
P
TPC
=
789.7
28
=
28cm
=
0.280 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 193
To Be A World Class Maritime Academy
To find the drafts forward and aft…
ii.
COT
=
Pxd =
MCTC
=
D. Dry Docking
789.7 x 54
340
125.4cm by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 194
To Be A World Class Maritime Academy
iii.
Change of draft aft due COT
=
D. Dry Docking
l x COT
L
=
78 x
162
125.4
=
60.4 cm
=
0.604 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 195
To Be A World Class Maritime Academy
iv.
D. Dry Docking
Change of draft Forward
=
COT
–
=
125.4 – 60.4
=
65cm
=
0.650 m
Change of draft aft
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 196
To Be A World Class Maritime Academy
v.
D. Dry Docking
Fwd(m)
Aft(m)
Initial drafts
Bodily rise
Change of drafts
8.700
0.280 0.650 -
9.400
0.280 0.604 +
Final drafts
7.770 m
9.724 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 197
To Be A World Class Maritime Academy
ii.
Estimated GM
GG1
Initial GM
=
P x KG
W–P
=
789.7 x 7.60
29000 – 789.7
=
0.213 m
=
8.40 m – 7.60
=
0.80 m
Effective GM =
0.800 – 0.213
=
0.587 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 198
To Be A World Class Maritime Academy
ii.
Estimated GM
MM1
Effective GM
D. Dry Docking
=
P x KM
W
=
0.80 – 0.229
=
0.571 m
=
789.7 x 8.40
29000
=
0.229 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 199
To Be A World Class Maritime Academy
Thank you…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 200