Download - Dry-docking-All About to Know
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2
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Revisions Ex.
Simplified Stab
Simpson Rules
Trim
Effect on G
Dry Docking
Statical Stab
Inclining Test
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3
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LEARNING OBJECTIVES
To understand the virtual loss of GM andthe calculations.
To calculate the maximum trim allowedto maintain a minimum stated GM.
To understand the safe requirements fora ship prior enter into dry dock.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4
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LEARNING OBJECTIVES
To understand the critical period duringdry docking process.
To calculate the ships drafts after thewater level has fallen and after the shiphas taken the block overall.
Effect to stability when vessel has runaground (single point).
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5
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Anybody would like to share their experience duringdry docking.?
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6
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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 tobe pressed up to increase GM
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When coming into Dry Dock:
The vessel will line-up with hercenterline vertically over the keel
blocks
Dock gate will be closed andcommence pumping out water
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 8
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F
No effect on ships Initial Stability
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 9
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When coming into Dry Dock:
The rate of pumping will bereduced as the ship's sternpost
near the block.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 10
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Commence touching the ground Sueing Point
Sueing Point
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 11
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When coming into Dry Dock:
Once the sternpost is touching the
block, the UP-THRUSTforces startto act against the sternpost.
At this moment part of ship's
weight gets transferred to the keelblocks.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 12
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P
P is the Upthrust Forceacting at first point oftouching the ground. Commence Critical Period
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 13
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Sueing Point
at AP
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 14
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P
K
P is the Upthrust Forceacting at first point oftouching the ground. Commence Critical Period
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 15
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When coming into Dry Dock:
When ship's weight gets transferred tothe keel blocks, vessel will suffer loss on
her GM.
The time interval between the sternpostlanding on the blocks and the ship takingthe blocks overall is referred to as theCRITICAL PERIOD.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 16
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P
P force is increasing gradually as the trim changeby HeadVessel is still in Critical Period
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 17
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When coming into Dry Dock:
Vessel must have positive effectiveGM that to be maintainedthroughout the critical period.
If not vessel may heel over, slip off
the blocks when there is anexternal force acting and heel theship.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 18
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P
Vessel is fully rest on the blocks End of CriticalPeriod
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 19
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 20
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M
G1
G
B
Initial GM loss by GG1aftercompleted the Critical
Period
This is due to UpthrustForce or P Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 21
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P
What is the total P Force during Critical Period __?___ tonnes
How much weight to be discharged in order to bring the shipfrom trim by stern to even-keel
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 22
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CALCULATION
OF UPTHRUST FORCE
AT THE STERNPOST
- 'P' FORCE
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 24
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F
After weight discharged
T M By Head = T M By Stern
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 25
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P
P is the Upthrust Force or weight dischargedtothe blocks
T.M = wx d = P xd t-m by Head
d
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 26
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P
Vessel is fully rest on the blocks, Change of Trimby Head and finally vessel at even keel draftsEnd of Critical Period
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 27
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Change of Trim = Trimming Moment (TM)
MCTC
Whereby TM = w x d
Change of Trim = P x dMCTC
P = COT x MCTC tonnesd
= P x d
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 28
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Exercise in classroom
MV OneSuch, LBP 120m is going to dry dockat 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 86t-m and TPC 15.45
Calculatei. The amount of up-thrust force (P) at
the end of Critical Period?ii. Final drafts forward and aft?
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 29
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P = COT x MCTCd
= 50 x 8657.5
P = 74.8 tonnes
Calculation of P force
ld Cl i i d
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 30
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CODF = 50 24
= 26cm
COT = P x dMCTC
= 74.8 x 57.586
= 50cm
CODA = 57.5 x 50120
= 24cm
Body rise = PTPC
= 74.815.45
= 4.8cm
= 0.048m
T B A W ld Cl M iti A d
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 31
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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
T B A W ld Cl M iti A d
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 32
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P
P is the Upthrust Force or weight dischargedtothe blocks
T.M = wx d = Px d t-m by Stern
d
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 33
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P
Vessel is fully rest on the blocks, Change of Trimby Sternand finally vessel at even keel draftsEnd of Critical Period
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 34
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Change of Trim = Trimming Moment (TM)
MCTC
Whereby TM = w x d = P x d
Change of Trim = P x dMCTC
P = COT x MCTC tonnesd
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 35
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Virtual Loss Of GM
During
Critical Period
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 36
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Method 1 GG1
Method 2 MM1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 37
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Method 1
When the vessel comes in contactwith 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 weightbelow G)
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 38
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P
d
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 39
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P
d
F
Trimming Moment by Head
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 40
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P
K
P is the Upthrust Force acting at first point of touchingthe ground. Commence Critical Periodweightdischarged from the ship
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 41
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K
G
G1
M
Reduction orLoss of GM= GG1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 42
o e o d C ass a t e cade y
M
G1
G
B
Initial GM loss by GG1duringthe Critical Period
This is due to UpthrustForceor P Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 43
y
Method 1
GG1 = P x KG in metresW - P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 44
y
During Critical Period part of ship body is still floating
P
B
M
W
G
G1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 45
y
Vessel is inclined to a small angle by anexternal force
P
B1
B
M
W - P
W
G
External Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 46
Method 1Discharged of weight, shift of GG1
P
G
M
G1
W - P
K
External Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 47
P
G
M
G1
W - P
X
K
X = KG1Sin
Method 1Discharged of weight, shift of GG1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 48
P
G
M
W
G1
W - P
Y
K
Method 1Discharged of weight, shift of GG1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 49
Method 1Discharged of weight, shift of GG1
G1
G
Y
Y = GG1 Sin
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 50
P
G
M
W
G1
W - P
X
Y
K
Method 1Discharged of weight, shift of GG1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 51
=
G1
P
K
XG
1
G
Y
G
WW
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 52
X= KG1x Sin Y= GG1Sin
PX = WY
P x KG1x Sin = W x GG1Sin
P x KG1 = W x GG1
P x (KG + GG1) = W x GG1
(P x KG) + (P x GG1) = W x GG1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 53
(P x KG) + (P x GG1) = W x GG1
P x KG = (W x GG1) (P x GG1)
P x KG = (W P) x GG1
P x KG = GG1W P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 54
Therefore the formula is
GG1 = P x KGW - P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 55
Righting Moment at small angle of heel
B1
B
G1
M
W - P
W - P
Z
External Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 56
G1 Z
M
W - P
W - P
Righting Moment= W x GZ= W x GM Sin
In this case,
Righting Moment= (W P) x G1M Sin
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 57
Method 1 GG1
Method 2 MM1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 58
Method 2
When the vessel comes in contactwith the blocks, it is assumed thatthere is a transfer of buoyancy'P'
to the keel blocks.
Hence there is a reduction in KM
while the weight and KG areremains constant.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 59
Reduction in Buoyancy
P
d
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 60
Reduction in Buoyancy
P
d
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 61
P
K
P is the Upthrust Force acting at first point oftouching the ground. Commence CriticalPeriodbuoyancy reduction from the ship
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 62
P
K
ReductionBuoyancy
P is the Upthrust Force acting at first point oftouching the ground. Commence CriticalPeriodbuoyancy reduction from the ship
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 63
K
B
M1
M
Reduction orLoss of GM= MM1
B1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 64
Method 2
MM1 = P x KM in metresW
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 65
M
M1G
B
Initial GM loss by MM1afterthe Critical Period
This is due to UpthrustForce or P Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 66
During Critical Period, part of ship body is still floating
P
B
M
W
G
M1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 67
Vessel is inclined to a small angle by an external force
P
B1
B
G
M
W - P
W
External Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 68
Method 2Transferred of buoyancy, shift of MM1
P
G
M
W
M1
W - P
X
Y
K
W
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 70
X = KM1x Sin Y= MM1Sin
PX = (W P) x Y
P x KM1x Sin = (W P) x MM1Sin
P x KM1 = (W P) x MM1
P x KM1 = W x MM1P x MM1
P x KM1+ P x MM1= W x MM1
P (KM1
+ MM1
) = W x MM1
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 71
P (KM1+ MM1) = W x MM1
P x KM = W x MM1
P x KM = MM1W
Therefore the formula is
MM1 = P x KMW
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 72
Righting Moment at small angle of heel
B1
B
G
M1
W
W
Z
M
External Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 73
G Z
M1
W
W
Righting Moment= W x GZ= W x GM Sin
In this case,
Righting Moment= W x GM1Sin
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 74
Method 1 GG1: Weight transferred
Method 2 MM1: Buoyancy transferred
SUMMARY
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 75
Exercise in classroom continued
MV OneSuch is going to dry dock at thefollowing 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 86t-m,
Calculate the amount of up-thrust force (P)during Critical Period and the virtual loss ofGMif KM is 8.0m and KG is 7.2m.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 76
P = COT x MCTCd
= 50 x 8657.5
P = 74.8 tonnes
Calculation of P force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 77
GG1 = P x KGW - P
= 74.8 x 7.24600 74.8
GG1 = 0.119m
Virtual loss of GM (GG1) method
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 78
MM1 = P x KMW
= 74.8 x 8.04600
MM1 = 0.130m
Virtual loss of GM (MM1) method
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 79
Comparison the Virtual loss of GM between
(MM1) and (GG1) method
Different is
= 0.130 0.119
= 0.011m
1cm
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 80
Effect ofTrim
In
Dry Docking
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 81
Change of Trim = Trimming Moment (TM)MCTC
Whereby TM = w x d = P x d
Change of Trim = P x dMCTC
P = COT x MCTC tonnes
d
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 82
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?
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 83
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 84
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Calculate P
P = MCTC x trim
d
Calculate P
P = MCTC x trim
d
Calculate P
P = MCTC x trim
d
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 85
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Calculate P
P = MCTC x trim
d
= 200 x 0
80
Calculate P
P = MCTC x trim
d
= 200 x 50
80
Calculate P
P = MCTC x trim
d
= 200 x 500
80
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 87
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
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 88
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Virtual Loss of GM
GG1 = P x KG
W P
Virtual Loss of GM
GG1 = P x KG
W P
Virtual Loss of GM
GG1 = P x KG
W - P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 89
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Virtual Loss of GM
GG1 = P x KG
W P
= 0x 6
5000 0
Virtual Loss of GM
GG1 = P x KG
W P
= 125x 6
5000 125
Virtual Loss of GM
GG1 = P x KG
W - P
= 1250x 6
5000 1250
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 90
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Virtual Loss of GM
GG1 = P x KG
W P
= 0 x 6
5000 0
GG1 = 0 m
Virtual Loss of GM
GG1 = P x KG
W P
= 125 x 6
5000 125
GG1 = 0.154 m
Virtual Loss of GM
GG1 = P x KG
W - P
= 1250 x 6
5000 1250
GG1 = 2.0 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 91
Case 1
0 m / Even keel
Case 2
0.5 m by Stern
Case 3
5 m by Stern
Old GM = 1.0m
New GM
= GM - GG1
= 1.0 0
= 1.0 m
Old GM = 1.0m
New GM
= GM - GG1
= 1.0 0.154
= 0.846 m
Old GM = 1.0m
New GM
= GM - GG1
= 1.0 2.0
= - 1.0 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 92
Residual GM
TRIM0
1.0
0.5
0.846
- 1.0
TRIM increasedGM decreasedMAX. TRIM?
5.0
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 93
Vessel displacement 5000 tonnes,distance sueing point to CF is 80m, MCTC 200 t-m, KM 7.0 m and
KG 6.0 m.
Maximum Trim is.?
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 94
P force is ? Initial GM 1.0m
Virtual Loss of GM = 1.0m
G
G1M
G
GG1is VirtualLoss of GM
During Critical Period
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 95
P force is ? Initial GM 1.0m
Virtual Loss of GM = 1.0m
GG1 = P x KG
W - P
1.0 = P x 65000 P
5000 - P = 6P
5000 = 7P
P = 714.28 tonnes
G
G1
M
GG1is VirtualLoss of GM
During Critical Period
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 96
P force is ? Initial GM 1.0m Maximum trim is ?
Virtual Loss of GM = 1.0m
GG1 = P x KG
W - P
1.0 = P x 6
5000 P
5000 - P = 6P
5000 = 7P
P = 714.28 tonnes
P = 714.28 tonnes
P = MCTC x trim
d
Trim = P x d
MCTC
= 714.28 x 80
200
Trim = 285.7 cms
Trim = 2.86 m by Stern
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 97
Residual GM
TRIM0
1.0
0.5
0.846
- 1.0
5.0
MAX. TRIM 2.86mMAX. TRIM ?
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 98
CONCLUSION:
The virtual loss of GM is NILas vesselhaving zero trim.
The loss is increased as the trim increased.
Maximum trim is depend upon the initialGM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 99
WORKED EXAMPLE 1
A ship of 140min length, displacement 5000tand upright is
to enter dry dock with drafts forward 3.84m, aft 4.60m. Giventhe following hydrostatic particulars:
TPC 20tonnesMCTC 150t- mCF 5mforward of amidshipsKM 9.75m
The blocks of the dry dock are horizontal.
i. Calculate the drafts of the vessel at the instants when she istaking 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 5.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 100
F
No effect on ships Initial Stability
4.60m
3.84mTrim 76 cm by Stern
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 101
F
P
P is the Upthrust Forceacting at first point oftouching the ground, commence Critical Period
4.60m3.84mTrim 76 cm by Stern
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 102
P
What is the total P Force during Critical Period?End of Critical Period
FEven keel draft
Change of Trim 76cms by Head
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 103
Ships trimmed = 4.60 3.84 = 0.76 m by Stern
i. P = MCTC x trim = 150 x 76d 75
P = 152 tonnes
a. Bodily rise = P = 152 = 7.6 cms = 0.076 mTPC 20
b. Change of Trim = 76 cms by Head
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 104
c. Change of draft aft due COT
= l x COT
L
= 75 x 76
140
= 40.7cm
= 0.407 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 105
d. Change of draft forward due COT
= COT Change of draft aft
= 76 40.7
= 35.3cm
= 0.353 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 106
e. Fwd Aft
Initial drafts 3.840 4.600Bodily rise 0.076 - 0.076 -
Change of drafts 0.353 + 0.407 -
Final drafts 4.117 m 4.117 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 107
F4.117m 4.117m
End of Critical Period, vessel is fully rested onblocks, draft is at even keel
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 108
i. ALTERNATIVE METHOD
a. Mean draft = 4.220 m.
b. True mean draft correction
= Dist. CF to amidships x trim
LBP
= 5 x 0.76140
= 0.027 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 109
i. ALTERNATIVE METHOD
c. True mean draft= Mean draft correction= 4.220 0.027= 4.193 m
d. Therefore:
True mean draft = 4.193 mBodily rise = 0.076 m -Final drafts = 4.117 m even keel
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 110
ii. GG1 = P x KG = 152 x 7.75
W P 5000 152
= 1178 = 0.243 m4848
Initial GM = KM KG = 9.75 m 7.75= 2.00 m
Effective GM = 2.00 0.243 = 1.757 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 111
MG1
G
B
Initial GM loss by GG1afterthe Critical Period
This is due to UpthrustForce or P Force
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 112
OR
MM1 = P x KM = 152 x 9.75W 5000
= 0.296 m
Effective GM = 2.00 0.296
= 1.704 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 113
M
M1
G
B
Initial GM loss by MM1afterthe Critical Period
This is due to UpthrustForce or P Force
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W P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 114
Righting Moment at small angle of heel
B1
B
G1
M
W - P
W - P
Z
External Force
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W P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 115
G1 Z
M
W - P
W - P
Righting Moment= W x GZ= W x GM Sin
In this case,
Righting Moment= (W P) x G1M Sin
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W
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 116
Righting Moment at small angle of heel
B1
B
G
M1
W
W
Z
MExternal Force
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W
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 117
G Z
M1
W
W
Righting Moment= W x GZ= W x GM Sin
In this case,
Righting Moment= W x GM1Sin
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 118
iii. RM = (W P) x G1M Sin
= (5000 152) x 1.757 x Sin 5
= 742.4 t-mOR
RM = W x GM1Sin
= 5000 x 1.704 x Sin 5
= 742.6 t-m
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WORKED EXAMPLE 2
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 119
WORKED EXAMPLE 2
A ship of length 165m, KG 7.30m is floating in a
graving dock with drafts forward 5.50m, aft 7.86minwater RD 1.025. At the aft perpendicular the keel is0.24mabove the top of the horizontal blocks. If thewater level has fallen in the dock by 1.22m, theshipsbecome unstable (GM = 0m).
Calculatei. The drafts forward and aft at which it occursii. The original/initial GM
GivenDisplacement for a hydrostatic mean draft of 6.65mis 9151tonnes. TPC 24, MCTC 120t-m and CF 3.66mabaft amidships.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 120
F
No effect on ships Initial Stability,initial trim is2.36m by Stern
7.86m
5.50m
Clearance 24cm
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 121
No effect on ships Initial Stability
5.50m
Depth of water 7.86 + 0.24 = 8.10m
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 122
No effect on ships Initial Stability
7.86m
5.50m
Clearance 24cm
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 123
Drop of water level by 8cm. No effect on shipsInitial Stability.
7.86m 5.50m
Clearance 16cm
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 124
7.86m
5.50m
Clearance 12cm
F
Drop of water level by 12cm. No effect on shipsInitial Stability.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 125
7.86m
5.50m
Clearance 6cm
F
Drop of water level by 18cm. No effect on shipsInitial Stability.
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 126
5.50m7.86m
F
Drop of water level by 24cm stern post start totouch the block
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 127
5.50m
P
7.86m
P is the Upthrust Forceacting at first point oftouching the block. Commence Critical Period
F
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 128
P
Drop of water level 98cm, Vessel becomeunstable Zero GM. Vessel is still in CriticalPeriod
6.88m
F
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WL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 129
7.86m
WL
WL
6.88m
Reduction : 98cms
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P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 130
P
d
F
Body rise &Trimming Moment by Head
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WL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 131
7.86m
WL
WL
6.88m
A : Body rise
WL
7.86m -Br
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WL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 132
7.86m
WL
WL
6.88m
WL
7.86m -Br
B: Change of draft aft due to COT by Head
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WL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 133
7.86m
WL
WL
6.88m
WL
7.86m -Br
B: Change of draft aft due to COT by Head
A : Body rise Reduction : 98cm
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 135
Fallen of water level = A + B
where A Body RiseB Change of draft aft
due to COT
Fallen of water level
= Body rise + Change of draft aft
due to COT
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 136
Fallen WL = P + l x TMTPC L MCTC
Fallen WL = P + l x P x dTPC L MCTC
98 = P + 78.84 x P x 78.8424 165 120
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98 = P + [78.84 x Px 78.84 ]
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 137
[ ]24 165 120
98 = P + 0.313926545P24 1
98 = P + 7.534P24
2352 = 8.534P
P = 275.6 tonnes
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 138
P = 275.6 tonnes
If we calculate until vessel is FULLY REST,
P = MCTC x trim = 120 x 236d 78.84
P = 359.2 tonnes
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 139
To find the drafts forward and aft
i. Bodily rise = P = 275.6
TPC 24
= 11.5 cms
= 0.115 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 140
To find the drafts forward and aft
ii. COT = P x dMCTC
= 275.6 x 78.84120
= 181cm by Head
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 141
iii. Change of draft aft due COT
= lx COT = 78.84 x 181
L 165
= 86.5cm
= 0.865 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 142
iv. Change of draft Forward
= COT Change of draft aft
= 181 86.5
= 94.5cm
= 0.945 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 143
v. Fwd(m) Aft(m)
Initial drafts 5.500 7.860Bodily rise 0.115 - 0.115 -
Change of drafts 0.945 + 0.865 -
Final drafts 6.330 6.880
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To find the initial GM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 144
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 TrimLBP
= 3.66 x 2.36 = 0.052 m165
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To find the initial GM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 145
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
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To find the initial GM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 146
To find the initial GM
Therefore additional displacement
= 8.2 cm x TPC (24)= 196.8 t
Displacement for TMD 6.732 m
= 9151 + 196.8= 9347.8 t
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To find the initial GM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 147
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 = 275.5 x 7.3W P 9347.8 275.5
= 2011.15 = 0.222 m9072.3
Initial GM = 0.222 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 148
Then what will be the MAXIMUM TRIM
allowed, safely docked
if the initial GM is 0.222 m.?
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GM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 149
TRIM
MAX. TRIM?
0
0.222
2.36m
-ve
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Initial GM 0 222m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 150
Initial GM 0.222m
P force is ?Maximum trim is ?
Virtual Loss of GM = 0.222m
P = 275.5 tonnes
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Initial GM 0 222m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 151
Initial GM 0.222m
P force is ?Maximum trim is ?
Virtual Loss of GM = 0.222m
P = 275.5 tonnes
P = 275.5 tonnes
P = MCTC x trim
d
Trim = P x dMCTC
= 275.5 x 78.84
120
Trim = 181cm
Trim = 1.81m by Stern
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GM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 152
TRIM
MAX. TRIM 1.81m
0
0.222
2.36m
-ve
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 153
Trim 1.81m by Stern Virtual loss of GM?
P = MCTC x trim
d
P = 120 x 181
78.84
P = 275.5 tonnes
GG1 = P x KG
WP
= 275.5 x 7.3
9347.8275.5
GG1 = 0.222
Residual GM = 0.2220.222
Residual GM = 0.000
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 154
Then what will be the final drafts
if the initial GM is 0.222 m and trim now is
1.81m by stern.?
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 155
To find the drafts forward and aft
i. Bodily rise = P = 275.5
TPC 24
= 11.5 cm
= 0.115 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 156
To find the drafts forward and aft
ii. COT = P x d = 275.5 x 78.84MCTC 120
= 181cm by Head
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 157
iii. Change of draft aft due COT
= l x COT = 78.84 x 181L 165
= 86.5cm = 0.865 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 158
iv. Change of draft Forward
= COT Change of draft aft
= 181 86.5 = 94.5cm
= 0.945 m
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Assuming aft draft maintain at 7.86m, new trim is 1.81m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 159
v. Fwd(m) Aft(m)
Initial drafts 6.050 7.860
Bodily rise 0.115 - 0.115 -Change of drafts 0.945 + 0.865 -
Final drafts 6.880 6.880
by astern, therefore forward draft now is 6.05m
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Worked Example 3
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 160
Your vessel is going to dry dock with the following
conditions:
Draft forward 8.00 m and aft 9.00 m. Her displacement is30 000 tonnes. KM is 11.50 m, KG 10.90 m. MCTC 400 t-m. TPC 38. LCF is 1.5 m abaft the amidships and LBP is160 m.
The depth of water in the dock is initially 9.50m.
i. Find the effective GM and her new draft after waterlevel has fallen by 95cmin the dock.
ii. How much will be the further drop of water level sothat vessel will take the blocks overall?
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 161
No effect on ships Initial Stability
Clearance 50cm
9.0m
F
9.5m
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P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 162
F
Drop of water level by 50cm, No effect on shipsInitial Stability
9.0m
50cm drop of water level
9.0m
P
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P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 163
8.55m
F 45cm drop of water level
Drop of water level by 45cm, effect on shipsInitial Stability
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9 00m WL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 165
9.00m
WL8.55m
A : Body rise
WL
9.00m -Br
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9.00m WL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 166
9.00m
WL8.55m
WL
9.00m -Br
B: Change of draft aft due to COT by Head
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9.00m WL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 167
9.00m
WL8.55m
WL
9.00m -Br
B: Change of draft aft due to COT by Head
A : Body rise Reduction : 45cm
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Reduction = A + B
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 168
where A Body RiseB Change of draft aft
due to COT
REDUCTION
= Body rise +Change of draft aft
due to COT
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Fallen of water level = A + B
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 169
where A Body RiseB Change of draft aft
due to COT
Fallen of water level
= Body rise + Change of draft aft
due to COT
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Fallen WL = P + l x TM
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 170
TPC L MCTC
Fallen WL = P + l x P x dTPC L MCTC
45 = P + 78.5 x P x 78.538 160 400
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45 = P + [ 78.5 x Px 78.5 ]38 160 400
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 171
45 = P + 0.096285156P38 1
45 = P + 3.659P38
1710 = 4.659P
P = 367.0 tonnes
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GG P KG 367 05 10 9
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 172
GG1 = P x KG = 367.05 x 10.9
W P 30 000 367.0
GG1 = 0.135 m
Initial GM = 0.600 m
Effective GM = 0.600 0.135
= 0.465 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 173
OR
MM1 = P x KM = 367.05 x 11.5W 30 000
MM1 = 0.141 m
Effective GM = 0.600 0.141
= 0.459 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 174
To find the drafts forward and aft
i. Bodily rise = P = 367.0TPC 38
= 9.66cm
= 0.097 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 175
To find the drafts forward and aft
ii. COT = P x d = 367.0 x 78.5MCTC 400
= 72cm by Head
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 176
iii. Change of draft aft due COT
= l x COT = 78.5 x 72L 160
= 35.3cm
= 0.353 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 177
iv. Change of draft Forward
= COT Change of draft aft
= 72.0 35.3
= 0.367 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 178
v. Fwd(m) Aft(m)
Initial drafts 8.000 9.000Bodily rise 0.097 - 0.097 -Change of drafts 0.367 + 0.353 -
Final drafts 8.270 8.550
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 179
New Trim = 8.55 8.27 = 0.28m by Sternassuming F constant
P = MCTC x T = 400 x 28d 78.5
P = 142.7 tonnes
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 180
Further drop = P + l x P x dvessel fully rest TPC L MCTC
= 142.7 + 78.5 x 142.7 x 78.538 160 400
= 17.5cm
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 181
SINGLE POINT
GROUNDING
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SINGLE POINT GROUNDING
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 182
A vessel floating at drafts forward 8.70 m, aft9.40 m grounds at a point 30 m aft of theforward perpendicular.
Estimate the drafts of the vessel and the GMafter 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 AftPerpendicular and displacement is 29 000tonnes.
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P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 183
F
9.40m 8.70m
Rock
30m
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P= ?Tide fallen by 70cms
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 184
Rock
F
Aft?
Fwd?
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Draft at PWL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 185
WLNew draft at P
Fallen of tide by 70cm
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Draft at PWL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 186
WLNew draft at P
A : Body rise
Draft at P -Br
WL
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Draft at PWL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 187
WLNew draft at P
Draft at P -Br
WL
B: Change of draft at Pdue to COT by Stern
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Draft at PWL
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 188
Fallen of tide by 70cm
WLNew draft at P
Draft at P -Br
WL
B: Change of draft at Pdue to COT by Stern
A : Body rise
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Fallen of tide = A + B
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 189
where A Body RiseB Change of draft at P
due to COT by Stern
Fallen of tide
= Body rise + Change of draft at Pdue to COT Stern
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Fallen of tide = P + l x TMTPC L MCTC
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 190
TPC L MCTC
Fallen of tide = P + l x P x d
TPC L MCTC
70 = P + 54 x P x 5428 162 340
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P
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 191
F
9.40m 8.70m
Rock
30m54m
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70 = P + [ 54 x Px 54 ]28 162 340
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 192
70 = P + 0.052941176P28 1
70 = P + 1.482P28
1960 = 2.482P
P = 789.7 tonnes
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 193
To find the drafts forward and aft
i. Bodily rise = P = 789.7TPC 28
= 28cm
= 0.280 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 194
To find the drafts forward and aft
ii. COT = P x d = 789.7 x 54MCTC 340
= 125.4cm by Stern
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 195
iii. Change of draft aft due COT
= l x COT = 78 x 125.4L 162
= 60.4 cm
= 0.604 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 196
iv. Change of draft Forward
= COT Change of draft aft
= 125.4 60.4
= 65cm
= 0.650 m
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 197
v. Fwd(m) Aft(m)
Initial drafts 8.700 9.400
Bodily rise 0.280 - 0.280 -
Change of drafts 0.650 - 0.604 +
Final drafts 7.770 m 9.724 m
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ii i d
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 199
ii. Estimated GM
MM1 = P x KM = 789.7 x 8.40W 29000
= 0.229 m
Effective GM = 0.80 0.229
= 0.571 m
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Thank you