d1mc semester 1 / ship stability / march 2007 /capt. mrd. dry docking 1 1 to be a world class...
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 11
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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 22
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Revisions Ex.Revisions Ex.
Simplified StabSimplified Stab
Simpson RulesSimpson Rules
TrimTrim Effect on GEffect on G
Dry DockingDry Docking
Statical StabStatical Stab
Inclining TestInclining Test
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 33
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LEARNING OBJECTIVESLEARNING OBJECTIVES
• To understand the virtual loss of GM and To understand the virtual loss of GM and the calculations. the calculations.
• To calculate the maximum trim allowed To calculate the maximum trim allowed to maintain a minimum stated GM. to maintain a minimum stated GM.
• To understand the safe requirements for To understand the safe requirements for a ship prior enter into dry dock. a ship prior enter into dry dock.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 44
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LEARNING OBJECTIVESLEARNING OBJECTIVES
• To understand the critical period during To understand the critical period during dry docking process. dry docking process.
• To calculate the ship’s drafts after the To calculate the ship’s drafts after the water level has fallen and after the ship water level has fallen and after the ship has taken the block overall. has taken the block overall.
• Effect to stability when vessel has run Effect to stability when vessel has run aground (single point). aground (single point).
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 55
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Anybody would like to share their experience Anybody would like to share their experience during dry docking….?during dry docking….?
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 66
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Before enter into dry dock, vessel must have…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
• 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
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 77
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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
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
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 88
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FF
No effect on ship’s Initial Stability…No effect on ship’s Initial Stability…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 99
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When coming into Dry Dock: • The rate of pumping will be
reduced as the ship's sternpost near the block.
When coming into Dry Dock: • The rate of pumping will be
reduced as the ship's sternpost near the block.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1010
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Commence touching the ground… ‘Commence touching the ground… ‘Sueing PointSueing Point’’
Sueing PointSueing Point
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1111
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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.
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.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1212
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PP
P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the ground. Commence touching the ground. Commence Critical PeriodCritical Period……
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1313
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Sueing PointSueing Point
at ‘at ‘APAP’’
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1414
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PP
KKKK
P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the ground. Commence touching the ground. Commence Critical PeriodCritical Period……
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1515
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When coming into Dry Dock:When coming into Dry Dock:
• When ship's weight gets transferred to When ship's weight gets transferred to the keel blocks, vessel will suffer loss on the keel blocks, vessel will suffer loss on her GM. her GM.
• The time interval between the sternpost The time interval between the sternpost landing on the blocks and the ship taking landing on the blocks and the ship taking the blocks overall is referred to as the the blocks overall is referred to as the CRITICAL PERIODCRITICAL PERIOD. .
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1616
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PP
P force is increasing gradually as the trim change P force is increasing gradually as the trim change by Head…Vessel is still in by Head…Vessel is still in Critical PeriodCritical Period……
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1717
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When coming into Dry Dock:When coming into Dry Dock:
• Vessel must have positive effective Vessel must have positive effective GM that to be maintained GM that to be maintained throughout the critical period. throughout the critical period.
• If not vessel may heel over, slip off If not vessel may heel over, slip off the blocks when there is an the blocks when there is an external force acting and heel the external force acting and heel the ship. ship.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1818
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P
Vessel is fully rest on the blocks… Vessel is fully rest on the blocks… End of Critical End of Critical PeriodPeriod
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1919
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<>
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2020
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M
G1
G
B
Initial GM loss by Initial GM loss by GGGG11 after after completed the Critical completed the Critical Period…Period…
This is due to Upthrust This is due to Upthrust Force or ‘P’ Force…Force or ‘P’ Force…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2121
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PP
What is the total P Force during Critical Period __What is the total P Force during Critical Period __??___ tonnes___ tonnes
““How much weight to be discharged in order to bring the ship How much weight to be discharged in order to bring the ship from trim by stern to even-keel…from trim by stern to even-keel…””
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2222
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CALCULATION CALCULATION
OF UPTHRUST FORCE OF UPTHRUST FORCE
AT THE STERNPOSTAT THE STERNPOST
- 'P' FORCE - 'P' FORCE
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2323
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ww
d
FF
Weight discharged to even keel the draft…
Trimming Moment = w x d t-m by Head
Weight discharged to even keel the draft…
Trimming Moment = w x d t-m by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2424
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FF
After weight discharged…
T M By Head = T M By Stern
After weight discharged…
T M By Head = T M By Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2525
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P
P is the Upthrust Force or weight dischargedP is the Upthrust Force or weight discharged to to the blocks…the blocks…
T.M T.M = = ww x d x d = = P P xx dd t-m by Head t-m by Head
d
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2626
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PP
Vessel is fully rest on the blocks, Change of Trim Vessel is fully rest on the blocks, Change of Trim by Head and finally vessel at even keel drafts… by Head and finally vessel at even keel drafts… End of End of Critical PeriodCritical Period……
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2727
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Change of TrimChange of Trim == Trimming Moment (TM)Trimming Moment (TM) MCTCMCTC
WherebyWhereby TM TM == w x dw x d
Change of TrimChange of Trim == PP x d x d MCTCMCTC
PP == COT x MCTCCOT x MCTC tonnestonnesdd
PP == COT x MCTCCOT x MCTC tonnestonnesdd
== PP x d x d
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2828
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Exercise in classroomExercise in classroom
MV OneSuch, LBP 120m is going to dry dock MV OneSuch, LBP 120m is going to dry dock at the following condition in sea waterat the following condition in sea water
Draft forward is 3.5m and aft is 4.0m, Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m.distance sueing point (AP) to F is 57.5m.
Her displacement is 4600 tonnes, MCTC is 86 Her displacement is 4600 tonnes, MCTC is 86 t-m and TPC 15.45t-m and TPC 15.45
Calculate Calculate i.i. The amount of up-thrust force (The amount of up-thrust force (PP) at ) at
the the end of Critical Period? end of Critical Period? ii.ii. Final drafts forward and aft?Final drafts forward and aft?
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 2929
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PP == COT x MCTCCOT x MCTC dd
== 50 x 8650 x 86 57.557.5
PP == 74.8 tonnes74.8 tonnes
Calculation of P force…Calculation of P force…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3030
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CODF CODF == 50 – 2450 – 24
== 26cm26cm
COT COT == P x dP x dMCTCMCTC
== 74.8 x 57.574.8 x 57.5 8686
== 50cm50cm
CODA CODA == 57.557.5 x 50 x 50120120
== 24cm24cm
Body rise Body rise == PPTPCTPC
== 74.874.815.4515.45
= = 4.8cm4.8cm
== 0.048m0.048m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3131
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ForwardForward AftAft
Initial draftInitial draft 3.500m3.500m 4.000m4.000m
Body riseBody rise 0.048m -0.048m - 0.048m –0.048m –
CODCOD 0.260m +0.260m + 0.240m –0.240m –
Final draftFinal draft 3.712m3.712m 3.712m3.712m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3232
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P
P is the Upthrust Force or weight dischargedP is the Upthrust Force or weight discharged to to the blocks…the blocks…
T.M T.M = = ww x d x d = = PP x d x d t-m by Stern t-m by Stern
d
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3333
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PP
Vessel is fully rest on the blocks, Change of Trim Vessel is fully rest on the blocks, Change of Trim by by SternStern and finally vessel at even keel drafts… and finally vessel at even keel drafts… End of Critical Period…End of Critical Period…
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3434
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Change of TrimChange of Trim == Trimming Moment (TM)Trimming Moment (TM) MCTCMCTC
WherebyWhereby TM TM == w x dw x d == PP x d x d
Change of TrimChange of Trim == PP x d x d MCTCMCTC
PP == COT x MCTCCOT x MCTC tonnestonnesdd
PP == COT x MCTCCOT x MCTC tonnestonnesdd
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3535
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Virtual Loss Of GM
During
Critical Period
Virtual Loss Of GM
During
Critical Period
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3636
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• Method 1 – GG1
• Method 2 – MM1
• Method 1 – GG1
• Method 2 – MM1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3737
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Method 1Method 1 • When the vessel comes in contact When the vessel comes in contact
with the blocks, it is assumed that with the blocks, it is assumed that there is a there is a transfer of weighttransfer of weight 'P' 'P' from the keel to the blocks.from the keel to the blocks.
• Hence there is a virtual rise of Hence there is a virtual rise of ship's G (ship's G (discharged of weight discharged of weight below Gbelow G))
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3838
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PPd
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 3939
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PP
dd
FF
Trimming Moment by… Trimming Moment by… HeadHead
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4040
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PP
KKKK
P is the Upthrust Force acting at first point of touching P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”the ground. Commence Critical Period…”weight weight discharged from the shipdischarged from the ship””
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4141
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KKKK
GGGG
GG11
MM
Reduction Reduction oror Loss of GM Loss of GM = GG = GG11
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4242
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M
G1
G
B
Initial GM loss by GGInitial GM loss by GG11 during during the Critical Period…the Critical Period…
This is due to This is due to UpthrustUpthrust Force Force or ‘P’ Force…or ‘P’ Force…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4343
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Method 1Method 1
GGGG11 == P x KGP x KG in metres in metres W - PW - P
GGGG11 == P x KGP x KG in metres in metres W - PW - P
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4444
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During Critical Period… part of ship body is still floating During Critical Period… part of ship body is still floating
PP
BBBB
MM
WW
GGGG
GG11
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4545
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Vessel is inclined to a small angle by an external force… Vessel is inclined to a small angle by an external force…
P
B1B1
BB
M
W - P
W
G
External Force
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4646
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Method 1 Discharged of weight, shift of GG1Method 1 Discharged of weight, shift of GG1
PP
GG
MM
GG11
W - PW - P
KK
External Force
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4747
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P
G
M
G1
W - PW - P
X
KK
X = KGX = KG11 Sin Sin
Method 1 Discharged of weight, shift of GG1Method 1 Discharged of weight, shift of GG1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4848
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P
G
M
W
G1
W - PW - P
Y
KK
Method 1 Discharged of weight, shift of GG1Method 1 Discharged of weight, shift of GG1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 4949
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Method 1… Discharged of weight, shift of GG1…Method 1… Discharged of weight, shift of GG1…
G1
G
Y
Y = GGY = GG1 1 Sin Sin
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5050
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P
G
M
W
G1
W - PW - PW - PW - P
X
Y
KK
Method 1 Discharged of weight, shift of GG1Method 1 Discharged of weight, shift of GG1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5151
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=
GG11
P
KK
X GG11
GG
Y
GG
WWW
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5252
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XX = = KGKG11 x Sin x Sin YY = = GGGG11 Sin Sin
PPXX = = WWYY
P x P x KGKG11 x Sin x Sin == W x W x GGGG11 Sin Sin
P x KGP x KG11 == W x GGW x GG11
P x (KG + GGP x (KG + GG11)) == W x GGW x GG11
(P x KG) + (P x GG(P x KG) + (P x GG11)) == W x GGW x GG11
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5353
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(P x KG) + (P x GG(P x KG) + (P x GG11)) == W x GGW x GG11
P x KGP x KG == (W x GG(W x GG11) – (P x GG) – (P x GG11))
P x KGP x KG == (W – P) x GG(W – P) x GG11
P x KGP x KG == GGGG11
W – PW – P
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5454
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Therefore the formula is… Therefore the formula is…
GGGG11 == P x KGP x KG W - PW - PGGGG11 == P x KGP x KG W - PW - P
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5555
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Righting Moment at small angle of heel… Righting Moment at small angle of heel…
B1B1
B
G1
M
W - P
W - P
Z
External Force
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5656
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G1 Z
MW - P
W - P
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = (W – P) x G(W – P) x G11M Sin M Sin
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = (W – P) x G(W – P) x G11M Sin M Sin
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5757
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• Method 1 – GG1
• Method 2 – MM1
• Method 1 – GG1
• Method 2 – MM1
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5858
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Method 2Method 2 • When the vessel comes in contact When the vessel comes in contact
with the blocks, it is assumed that with the blocks, it is assumed that there is a there is a transfer of buoyancytransfer of buoyancy 'P' 'P' to the keel blocks.to the keel blocks.
• Hence there is a reduction in KM Hence there is a reduction in KM while the weight and KG are while the weight and KG are remains constant. remains constant.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 5959
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Reduction in BuoyancyReduction in Buoyancy
PPd
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6060
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Reduction in BuoyancyReduction in Buoyancy
PPd
FF
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6161
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PP
KKKK
P is the Upthrust Force acting at first point of P is the Upthrust Force acting at first point of touching the ground. Commence Critical touching the ground. Commence Critical Period…”Period…”buoyancy reduction from the shipbuoyancy reduction from the ship””
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6262
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PP
KKKK
Reduction Reduction Buoyancy Buoyancy
P is the Upthrust Force acting at first point of P is the Upthrust Force acting at first point of touching the ground. Commence Critical touching the ground. Commence Critical Period…”Period…”buoyancy reduction from the shipbuoyancy reduction from the ship””
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6363
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KKKK
BB
MM11
MM
Reduction Reduction oror Loss of GM Loss of GM = MM = MM11
BB11BB11
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6464
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Method 2Method 2
MMMM11 == P x KMP x KM in metres in metres WW
MMMM11 == P x KMP x KM in metres in metres WW
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6565
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M
M1
G
B
Initial GM loss by MM1 after the Critical Period…
This is due to Upthrust Force or ‘P’ Force…
Initial GM loss by MM1 after the Critical Period…
This is due to Upthrust Force or ‘P’ Force…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6666
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During Critical Period, part of ship body is still floatingDuring Critical Period, part of ship body is still floating
PP
BB
MM
W
GG
MM11MM11
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6767
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Vessel is inclined to a small angle by an external Vessel is inclined to a small angle by an external force…force…
PP
B1B1
B
GG
MM
W - PW - P
W
External Force
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6868
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Method 2Method 2…… Transferred of buoyancy, shift of MM Transferred of buoyancy, shift of MM11……
P
G
M
W
M1
W - P
X
Y
KK
W
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 6969
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==
MM11
P
KK
X MM
MM11 Y
GGGG
W W W - P
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 7070
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X X = = KMKM11 x Sin x Sin YY = = MMMM11 Sin Sin
PPXX == (W – P) x (W – P) x YY
P x P x KMKM11 x Sin x Sin == (W – P) x (W – P) x MMMM11 Sin Sin
P x KMP x KM11 == (W – P) x MM(W – P) x MM11
P x KMP x KM11 == W x MMW x MM11 – P x MM – P x MM11
P x KMP x KM11 + P x MM + P x MM11 = = W x MMW x MM11
P (KMP (KM11 + MM + MM11)) == W x MMW x MM11
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 7171
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P (KMP (KM11 + MM + MM11)) == W x MMW x MM11
P x KMP x KM == W x MMW x MM11
P x KMP x KM == MMMM11 W W
Therefore the formula is…Therefore the formula is…
MMMM11 == P x KMP x KM WW
MMMM11 == P x KMP x KM WW
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 7272
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Righting Moment at small angle of heel… Righting Moment at small angle of heel…
BB11BB11
BB
G
M1
W
W
Z
MExternal Force
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 7373
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G Z
M1
W
W
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = W x GMW x GM11 Sin Sin
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = W x GMW x GM11 Sin Sin
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• Method 1 – GG1 : Weight transferred
• Method 2 – MM1 : Buoyancy transferred
• Method 1 – GG1 : Weight transferred
• Method 2 – MM1 : Buoyancy transferred
SUMMARY…SUMMARY…
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Exercise in classroom Exercise in classroom …continued…continued
MV OneSuch is going to dry dock at the MV OneSuch is going to dry dock at the following condition in sea waterfollowing condition in sea water
Draft forward is 3.5m and aft is 4.0m, Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m.distance sueing point (AP) to F is 57.5m.
Her displacement is 4600 tonnes, MCTC is 86 Her displacement is 4600 tonnes, MCTC is 86 t-m,t-m,
Calculate the amount of up-thrust force (P) Calculate the amount of up-thrust force (P) during Critical Period and the during Critical Period and the virtual loss of virtual loss of GMGM if KM is 8.0m and KG is 7.2m. if KM is 8.0m and KG is 7.2m.
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PP == COT x MCTCCOT x MCTC dd
== 50 x 8650 x 86 57.557.5
PP == 74.8 tonnes74.8 tonnes
Calculation of P force…Calculation of P force…
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GGGG11 == P x KGP x KG W - PW - P
== 74.8 x 7.274.8 x 7.24600 – 74.84600 – 74.8
GGGG11 == 0.119m0.119m
Virtual loss of GM (Virtual loss of GM (GGGG11) method…) method…
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MMMM11 == P x KMP x KM WW
== 74.8 x 8.074.8 x 8.0 46004600
MMMM11 == 0.130m0.130m
Virtual loss of GM (Virtual loss of GM (MMMM11) method…) method…
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Comparison the Virtual loss of GM between Comparison the Virtual loss of GM between ((MMMM11) and () and (GGGG11) method…) method…
Different is… Different is…
== 0.130 – 0.119 0.130 – 0.119
= = 0.011m0.011m
… … ±± 1cm1cm
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Effect of Trim
In
Dry Docking
Effect of Trim
In
Dry Docking
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Change of TrimChange of Trim == Trimming Moment (TM)Trimming Moment (TM) MCTCMCTC
WherebyWhereby TM TM == w x dw x d == PP x d x d
Change of TrimChange of Trim == PP x d x d MCTCMCTC
PP == COT x MCTCCOT x MCTC tonnestonnesdd
PP == COT x MCTCCOT x MCTC tonnestonnesdd
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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?
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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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Calculate ‘Calculate ‘PP’…’…
P = P = MCTC x trimMCTC x trim
dd
Calculate ‘Calculate ‘PP’…’…
P = P = MCTC x trimMCTC x trim
dd
Calculate ‘Calculate ‘PP’…’…
P = P = MCTC x trimMCTC x trim
dd
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Calculate ‘P’…Calculate ‘P’…
P = P = MCTC x trimMCTC x trim
dd
= = 200 x 200 x 00
8080
Calculate ‘P’…Calculate ‘P’…
P = P = MCTC x trimMCTC x trim
dd
= = 200 x 200 x 5050
8080
Calculate ‘P’…Calculate ‘P’…
P = P = MCTC x trimMCTC x trim
dd
= = 200 x 200 x 500500
8080
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Calculate ‘P’…Calculate ‘P’…
P = P = MCTC x trimMCTC x trim
dd
= = 200 x 200 x 00
8080
P = P = 0 tonne0 tonne
Calculate ‘P’…Calculate ‘P’…
P = P = MCTC x trimMCTC x trim
dd
= = 200 x 200 x 5050
8080
P = P = 125 tonnes125 tonnes
Calculate ‘P’…Calculate ‘P’…
P = P = MCTC x trimMCTC x trim
dd
= = 200 x 200 x 500500
8080
P = P = 1250 tonnes1250 tonnes
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Virtual Loss of GM…Virtual Loss of GM… Virtual Loss of GM…Virtual Loss of GM… Virtual Loss of GM…Virtual Loss of GM…
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W – PW – P
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W – PW – P
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W - PW - P
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W – PW – P
= = 00 x 6 x 6
5000 – 5000 – 00
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W – PW – P
= = 125125 x 6 x 6
5000 –5000 –125125
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W - PW - P
= = 12501250 x 6 x 6
5000 –5000 –12501250
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W – PW – P
= = 0 x 60 x 6
5000 – 05000 – 0
GGGG11 = = 0 m0 m
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W – PW – P
= = 125 x 6125 x 6
5000 –125 5000 –125
GGGG11 = = 0.154 m0.154 m
Virtual Loss of GM…Virtual Loss of GM…
GGGG11 = = P x KGP x KG
W - PW - P
= = 1250 x 61250 x 6
5000 –12505000 –1250
GGGG11 = = 2.0 m2.0 m
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Case 1 Case 1
0 m / Even keel0 m / Even keelCase 2Case 2
0.5 m by Stern0.5 m by SternCase 3 Case 3
5 m by Stern5 m by Stern
Old GM = 1.0mOld GM = 1.0m
New GM…New GM…
= GM - GG= GM - GG11
= 1.0 – 0= 1.0 – 0
= = 1.0 m 1.0 m
Old GM = 1.0mOld GM = 1.0m
New GM…New GM…
= GM - GG= GM - GG11
= 1.0 – 0.154= 1.0 – 0.154
= = 0.846 m 0.846 m
Old GM = 1.0mOld GM = 1.0m
New GM…New GM…
= GM - GG= GM - GG11
= 1.0 – 2.0= 1.0 – 2.0
= = - 1.0 m - 1.0 m
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Residual GMResidual GM
TRIMTRIM00
1.01.0
0.50.5
0.8460.846
- 1.0- 1.0
TRIM increasedTRIM increasedGM decreasedGM decreased
MAX. TRIM…?MAX. TRIM…?
5.05.0
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• Vessel displacement 5000 tonnes, Vessel displacement 5000 tonnes, distance sueing point to CF is 80 distance sueing point to CF is 80 m, MCTC 200 t-m, KM 7.0 m and m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. KG 6.0 m.
Maximum Trim is….?Maximum Trim is….?
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P force is …? Initial GM 1.0mP force is …? Initial GM 1.0m
Virtual Loss of GM = Virtual Loss of GM = 1.0m1.0m
GG
GG11MM
GG
GGGG11 is Virtual is Virtual Loss of GMLoss of GM
During Critical Period…During Critical Period…
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P force is …? Initial GM 1.0mP force is …? Initial GM 1.0m
Virtual Loss of GM = Virtual Loss of GM = 1.0m1.0m
GGGG11 = = P x KGP x KG
W - PW - P
1.01.0 = = P x 6P x 6
5000 – P5000 – P
5000 - P5000 - P = 6P = 6P
5000 = 7P5000 = 7P
P = P = 714.28 tonnes714.28 tonnes GG
GG11MM
GGGG11 is Virtual is Virtual Loss of GMLoss of GM
During Critical Period…During Critical Period…
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P force is P force is …? Initial GM 1.0m…? Initial GM 1.0m Maximum trim is …?Maximum trim is …?
Virtual Loss of GM = 1.0mVirtual Loss of GM = 1.0m
GGGG11 = = P x KGP x KG
W - PW - P
1.0 = 1.0 = P x 6P x 6
5000 – P5000 – P
5000 - P5000 - P = 6P = 6P
5000 = 7P5000 = 7P
P = 714.28 tonnesP = 714.28 tonnes
P = P = 714.28 tonnes714.28 tonnes
P = P = MCTC x trimMCTC x trim
dd
Trim = Trim = P x dP x d
MCTCMCTC
= = 714.28 x 80714.28 x 80
200200
Trim = 285.7 cmsTrim = 285.7 cms
Trim = Trim = 2.86 m by Stern2.86 m by Stern
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Residual GM
TRIM00
1.01.0
0.50.5
0.8460.846
- 1.0- 1.0
5.0
MAX. TRIM MAX. TRIM 2.86m2.86mMAX. TRIM MAX. TRIM …………??
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CONCLUSION: CONCLUSION:
• The virtual loss of GM is The virtual loss of GM is NILNIL as vessel as vessel having zero trim. having zero trim.
• The loss is increased as the trim increased. The loss is increased as the trim increased.
• Maximum trim is depend upon the initial Maximum trim is depend upon the initial GMGM
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WORKED EXAMPLE 1WORKED EXAMPLE 1
A ship of A ship of 140m140m in length, displacement in length, displacement 5000t5000t and upright is and upright is to enter dry dock with drafts forward to enter dry dock with drafts forward 3.84m3.84m, aft , aft 4.60m4.60m. Given . Given the following hydrostatic particulars: the following hydrostatic particulars:
TPCTPC 2020 tonnes tonnesMCTCMCTC 150150 t- m t- mCFCF 5m5m forward of amidships forward of amidshipsKMKM 9.75m9.75m
The blocks of the dry dock are horizontal. The blocks of the dry dock are horizontal.
i.i. Calculate the drafts of the vessel at the instants when she is Calculate the drafts of the vessel at the instants when she is taking the blocks forward and aft.taking the blocks forward and aft.
ii.ii. The ship's effective GM at this moment if the KG is The ship's effective GM at this moment if the KG is 7.75m7.75m
iii.iii. The Righting Moment at this instant for an angle of heel The Righting Moment at this instant for an angle of heel 5º.5º.
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FF
No effect on ship’s No effect on ship’s Initial StabilityInitial Stability……
4.60m4.60m3.84m3.84m
Trim 76 cm by SternTrim 76 cm by Stern
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FF
PP
P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the ground, commence touching the ground, commence Critical Critical PeriodPeriod……
4.60m4.60m3.84m3.84mTrim 76 cm by SternTrim 76 cm by Stern
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PP
What is the total P Force during Critical What is the total P Force during Critical Period? End of Period? End of Critical PeriodCritical Period……
FFEven keel draftEven keel draftEven keel draftEven keel draft
Change of Trim 76cms by HeadChange of Trim 76cms by Head
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Ship’s trimmed Ship’s trimmed = = 4.60 – 3.84 = 4.60 – 3.84 = 0.76 m by Stern0.76 m by Stern
i.i. P P = = MCTC x trimMCTC x trim = = 150 x 76150 x 76 dd 7575
PP = = 152 tonnes152 tonnes
a.a. Bodily riseBodily rise= = P P = = 152152 = 7.6 cms = = 7.6 cms = 0.076 m0.076 m TPCTPC 20 20
b.b. Change of TrimChange of Trim == 76 cms by Head76 cms by Head
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c.c. Change of draft aft due COTChange of draft aft due COT
= = ll x COT x COT LL
= = 7575 x 76 x 76140140
= = 40.7cm 40.7cm
= = 0.407 m0.407 m
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d.d. Change of draft forward due COTChange of draft forward due COT
= = COT – Change of draft aft COT – Change of draft aft
= = 76 – 40.776 – 40.7
= = 35.3cm 35.3cm
= = 0.353 m0.353 m
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e.e. Fwd Fwd Aft Aft
Initial draftsInitial drafts 3.840 3.840 4.600 4.600 Bodily riseBodily rise 0.076 -0.076 - 0.076 -0.076 -Change of drafts Change of drafts 0.353 +0.353 + 0.407 -0.407 -
Final draftsFinal drafts 4.117 m4.117 m 4.117 m4.117 m
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FF4.117m4.117m 4.117m4.117m
End of Critical Period, vessel is fully rested on End of Critical Period, vessel is fully rested on blocks, draft is at even keelblocks, draft is at even keel
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i.i. ALTERNATIVE METHODALTERNATIVE METHOD
a.a. Mean draftMean draft = = 4.220 m.4.220 m.
b.b. True mean draft correction True mean draft correction
== Dist. CF to amidships x trimDist. CF to amidships x trimLBPLBP
= = 5 x 0.765 x 0.76 140140
= = 0.027 m0.027 m
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i.i. ALTERNATIVE METHODALTERNATIVE METHOD
c.c. True mean draft True mean draft = = Mean draft – Mean draft –
correction correction = = 4.220 – 0.027 4.220 – 0.027 = = 4.193 m4.193 m
d.d. Therefore:Therefore:
True mean draftTrue mean draft = = 4.193 m4.193 mBodily riseBodily rise = = 0.076 m -0.076 m -Final drafts Final drafts == 4.117 m4.117 m even keel even keel
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ii.ii. GGGG11 = = P x KGP x KG = = 152 x 7.75152 x 7.75 W – P W – P 5000 – 1525000 – 152
= = 11781178 = = 0.243 m0.243 m 48484848
Initial GM Initial GM == KM – KG KM – KG == 9.75 m – 7.75 9.75 m – 7.75 = = 2.00 m2.00 m
Effective GM =Effective GM = 2.00 – 0.2432.00 – 0.243 == 1.757 m1.757 m
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M
G1
G
B
Initial GM loss by GGInitial GM loss by GG11 after after the Critical Period…the Critical Period…
This is due to Upthrust This is due to Upthrust Force or ‘P’ Force…Force or ‘P’ Force…
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OROR
MMMM11 = = P x KMP x KM = = 152 x 9.75152 x 9.75 W W 5000 5000
= = 0.296 m0.296 m
Effective GM Effective GM = = 2.00 – 0.296 2.00 – 0.296
= = 1.704 m1.704 m
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MM
M1
GG
BB
Initial GM loss by MM1 after the Critical Period…
This is due to Upthrust Force or ‘P’ Force…
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Righting Moment at small angle of heel…Righting Moment at small angle of heel…
B1B1
BB
G1
M
W - PW - P
W - P
Z
External Force
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G1 Z
MW - P
W - P
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = (W – P) x G(W – P) x G11M Sin M Sin
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = (W – P) x G(W – P) x G11M Sin M Sin
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Righting Moment at small angle of heel…Righting Moment at small angle of heel…
B1B1
BB
G
M1
W
W
Z
MExternal Force
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G Z
M1
W
W
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = W x GMW x GM11 Sin Sin
Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin
In this case,In this case, Righting MomentRighting Moment
= = W x GMW x GM11 Sin Sin
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iii.iii. RM RM = = (W – P) x G(W – P) x G11M Sin M Sin
= = (5000 – 152) x 1.757 x Sin 5(5000 – 152) x 1.757 x Sin 5
= = 742.4 t-m742.4 t-mOROR
RM RM = = W x GMW x GM11 Sin Sin
= = 5000 x 1.704 x Sin 55000 x 1.704 x Sin 5
= = 742.6 t-m742.6 t-m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 119119
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WORKED EXAMPLE 2WORKED EXAMPLE 2 A ship of length A ship of length 165m165m, KG , KG 7.30m7.30m is floating in a is floating in a graving dock with drafts forward graving dock with drafts forward 5.50m5.50m, aft , aft 7.86m7.86m in in water RD 1.025. At the aft perpendicular the keel is water RD 1.025. At the aft perpendicular the keel is 0.24m0.24m above the top of the horizontal blocks. If the above the top of the horizontal blocks. If the water level has fallen in the dock by water level has fallen in the dock by 1.22m1.22m, the , the ship’s become unstable (ship’s become unstable (GM = 0mGM = 0m). ).
Calculate Calculate i.i. The drafts forward and aft at which it occursThe drafts forward and aft at which it occursii.ii. The original/initial GM The original/initial GM
Given Given Displacement for a hydrostatic mean draft of Displacement for a hydrostatic mean draft of 6.65m6.65m is is 91519151 tonnes. TPC tonnes. TPC 2424, MCTC , MCTC 120120 t-m and CF t-m and CF 3.663.66 m m abaft amidships.abaft amidships.
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F
No effect on ship’s Initial Stability,No effect on ship’s Initial Stability, initial trim is initial trim is 2.36m by Stern2.36m by Stern
7.86m7.86m5.50m5.50m
Clearance 24cmClearance 24cm
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No effect on ship’s No effect on ship’s Initial StabilityInitial Stability……
5.50m5.50m
Depth of water 7.86 + 0.24 = 8.10mDepth of water 7.86 + 0.24 = 8.10m
F
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No effect on ship’s No effect on ship’s Initial StabilityInitial Stability……
7.86m7.86m5.50m5.50m
Clearance 24cm Clearance 24cm
F
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Drop of water level by Drop of water level by 8cm8cm. No effect on ship’s . No effect on ship’s Initial Stability.Initial Stability.
7.86m7.86m 5.50m5.50m
Clearance 16cm Clearance 16cm
F
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 124124
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7.86m7.86m5.50m5.50m
Clearance 12cm Clearance 12cm
F
Drop of water level by Drop of water level by 12cm12cm. No effect on ship’s . No effect on ship’s Initial Stability.Initial Stability.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 125125
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7.86m7.86m5.50m5.50m
Clearance 6cm Clearance 6cm Clearance 6cm Clearance 6cm
F
Drop of water level by Drop of water level by 18cm18cm. No effect on ship’s . No effect on ship’s Initial Stability.Initial Stability.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 126126
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5.50m5.50m7.86m7.86m
F
Drop of water level by Drop of water level by 24cm24cm… stern post start … stern post start to touch the block…to touch the block…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 127127
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5.50m5.50m
PP
7.86m7.86m
P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the block. Commence touching the block. Commence Critical PeriodCritical Period……
F
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 128128
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P
Drop of water level Drop of water level 98cm, 98cm, Vessel become Vessel become unstable… unstable… Zero GM. Zero GM. Vessel is still in Critical Vessel is still in Critical Period…Period…
6.88m6.88m
F
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 129129
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7.86m7.86m
WLWL
WLWL
6.88m6.88m
Reduction : 98cmsReduction : 98cms
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 130130
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PP
dd
FF
Body rise & Trimming Moment by… Head Body rise & Trimming Moment by… Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 131131
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7.86m7.86m
WLWL
WLWL
6.88m6.88m
A : A : Body riseBody rise
WLWL7.86m 7.86m -- BrBr
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 132132
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7.86m7.86m
WLWL
WLWL
6.88m6.88m
WLWL7.86m 7.86m -- BrBr
BB : : Change of draft aft due to COT by Change of draft aft due to COT by HeadHead
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 133133
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7.86m7.86m
WLWL
WLWL
6.88m6.88m
WLWL7.86m 7.86m -- BrBr
BB : : Change of draft aft due to COT by Change of draft aft due to COT by HeadHead
A : A : Body riseBody rise Reduction : 98cmReduction : 98cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 134134
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Reduction = Reduction = AA ++ BB
wherewhere AA Body RiseBody RiseB B Change of draft aftChange of draft aft
due to COTdue to COT
REDUCTIONREDUCTION
== Body rise Body rise ++ Change of draft aft Change of draft aft due to COT due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 135135
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Fallen of water level = Fallen of water level = AA ++ BB
wherewhere AA Body RiseBody RiseB B Change of draft aftChange of draft aft
due to COTdue to COT
Fallen of water level Fallen of water level
== Body rise Body rise + + Change of draft aft Change of draft aft due to COT due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 136136
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Fallen WLFallen WL = = PP ++ ll x TM x TM TPCTPC L MCTCL MCTC
Fallen WL Fallen WL == PP ++ ll x P x d x P x d TPCTPC L MCTCL MCTC
9898 == PP ++ 78.8478.84 x x P x 78.84P x 78.842424 165165 120120
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 137137
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98 98 == PP ++ [[78.8478.84 x x PP x x 78.8478.84 ]] 2424 165165 120 120
9898 == PP ++ 0.3139265450.313926545PP
2424 11
9898 == PP ++ 7.5347.534PP2424
23522352 == 8.5348.534PP
P P = = 275.6 tonnes275.6 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 138138
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P P = = 275.6 tonnes275.6 tonnes If we calculate until vessel is If we calculate until vessel is FULLY RESTFULLY REST,,
PP == MCTC x trimMCTC x trim == 120 x 120 x 236236dd 78.84 78.84
PP == 359.2 tonnes359.2 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 139139
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To find the drafts forward and aft…To find the drafts forward and aft…
i.i. Bodily riseBodily rise= = P P = = 275.6275.6 TPCTPC 24 24
= = 11.5 cms 11.5 cms
= = 0.115 m0.115 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 140140
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To find the drafts forward and aft…To find the drafts forward and aft…
ii.ii. COT COT = = P x dP x d MCTCMCTC
= = 275.6 x 78.84275.6 x 78.84120120
= = 181cm by Head181cm by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 141141
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iii.iii. Change of draft aft due COTChange of draft aft due COT
= = ll x COT x COT = = 78.8478.84 x 181 x 181 LL 165 165
= = 86.5cm 86.5cm
= = 0.865 m0.865 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 142142
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iv.iv. Change of draft ForwardChange of draft Forward
= = COTCOT – – Change of draft aftChange of draft aft
== 181 181 – – 86.586.5
= = 94.5cm94.5cm
= = 0.945 m0.945 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 143143
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v. v. Fwd(m)Fwd(m) Aft(m)Aft(m)
Initial draftsInitial drafts 5.500 5.500 7.860 7.860 Bodily riseBodily rise 0.115 -0.115 - 0.115 -0.115 -Change of draftsChange of drafts 0.9450.945 + + 0.8650.865 - -
Final draftsFinal drafts 6.3306.330 6.8806.880
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 144144
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To find the initial GM…To find the initial GM…
Mean draft Mean draft = 6.680m. = 6.680m. Trim =Trim = 2.36m by stern2.36m by stern
CF is 3.66m abaft amidships.CF is 3.66m abaft amidships.
TMD CorrectionTMD Correction = = Dist. CF to Amidships x TrimDist. CF to Amidships x Trim
LBPLBP = = 3.66 x 2.363.66 x 2.36 = = 0.052 m0.052 m
165 165
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 145145
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To find the initial GM…To find the initial GM…
True Mean Draft (TMD)True Mean Draft (TMD)
= = Mean draft + TMD Correction Mean draft + TMD Correction = = 6.680 + 0.052 6.680 + 0.052 = = 6.732 m6.732 m
Diff of TMD Diff of TMD = = 6.732 – 6.650 6.732 – 6.650 = = 0.082 m 0.082 m = = 8.2 cm8.2 cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 146146
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To find the initial GM…To find the initial GM…
Therefore additional displacementTherefore additional displacement
= = 8.2 cm x TPC (24) 8.2 cm x TPC (24) = = 196.8 t196.8 t
Displacement for TMD 6.732 mDisplacement for TMD 6.732 m
= = 9151 + 196.8 9151 + 196.8 = = 9347.8 t9347.8 t
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 147147
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To find the initial GM…To find the initial GM…
When the ship become unstable, the GM = 0 m, When the ship become unstable, the GM = 0 m, therefore loss of GM must be equal to initial GM.therefore loss of GM must be equal to initial GM.
GGGG11 == P x KGP x KG = = 275.5 x 7.3275.5 x 7.3
W – PW – P 9347.8 – 275.59347.8 – 275.5
= = 2011.152011.15 = = 0.222 m0.222 m 9072.39072.3
Initial GM Initial GM = = 0.222 m0.222 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 148148
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Then… what will be the Then… what will be the MAXIMUM TRIMMAXIMUM TRIM
allowed, safely dockedallowed, safely docked
if the initial GM is if the initial GM is 0.222 m0.222 m….?….?
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 149149
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TRIMTRIM
MAX. TRIM…MAX. TRIM…??
00
0.2220.222
GMGM
2.36m2.36m
-ve-ve
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 150150
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Initial GM 0.222mInitial GM 0.222m
P force is …? P force is …? Maximum trim is …?Maximum trim is …?
Virtual Loss of GM = 0.222mVirtual Loss of GM = 0.222m
P = 275.5 tonnesP = 275.5 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 151151
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Initial GM 0.222mInitial GM 0.222m
P force is …? P force is …? Maximum trim is …?Maximum trim is …?
Virtual Loss of GM = 0.222mVirtual Loss of GM = 0.222m
P = 275.5 tonnesP = 275.5 tonnes
P = P = 275.5 tonnes275.5 tonnes
P = P = MCTC x trimMCTC x trim
dd
Trim = Trim = P x dP x d
MCTCMCTC
= = 275.5 x 78.84275.5 x 78.84
120120
Trim = 181cmTrim = 181cm
Trim = Trim = 1.81m by Stern1.81m by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 152152
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TRIMTRIM
MAX. TRIM MAX. TRIM 1.81m1.81m
00
0.2220.222
GMGM
2.36m2.36m
-ve-ve
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 153153
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Trim 1.81m by SternTrim 1.81m by Stern Virtual loss of GM…?Virtual loss of GM…?
P = P = MCTC x trimMCTC x trim
dd
P = P = 120 x 181120 x 181
78.8478.84
P = P = 275.5 tonnes275.5 tonnes
GGGG11 = = P x KGP x KG
W – PW – P
= = 275.5 x 7.3275.5 x 7.3
9347.8 – 275.59347.8 – 275.5
GGGG11 = = 0.2220.222
Residual GM = 0.222 – 0.222Residual GM = 0.222 – 0.222
Residual GM = Residual GM = 0.0000.000
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 154154
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Then… what will be the Then… what will be the final draftsfinal drafts……
if the initial GM is if the initial GM is 0.222 m 0.222 m and trim now isand trim now is
1.81m 1.81m by stern….?by stern….?
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 155155
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To find the drafts forward and aft…To find the drafts forward and aft…
i.i. Bodily riseBodily rise= = P P = = 275.5275.5 TPCTPC 24 24
= = 11.5 cm 11.5 cm
= = 0.115 m0.115 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 156156
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To find the drafts forward and aft…To find the drafts forward and aft…
ii.ii. COT COT = = P x dP x d = = 275.5 x 78.84275.5 x 78.84MCTCMCTC 120 120
= = 181cm by Head181cm by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 157157
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iii.iii. Change of draft aft due COTChange of draft aft due COT
= = l l x COT x COT = = 78.8478.84 x 181 x 181 LL 165 165
= = 86.5cm 86.5cm = = 0.865 m0.865 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 158158
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iv.iv. Change of draft ForwardChange of draft Forward
= = COTCOT – – Change of draft aftChange of draft aft
== 181 – 181 – 86.586.5 = 94.5cm= 94.5cm
= = 0.945 m0.945 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 159159
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v. v. Fwd(m)Fwd(m) Aft(m)Aft(m)
Initial draftsInitial drafts 6.0506.050 7.860 7.860 Bodily riseBodily rise 0.115 -0.115 - 0.115 -0.115 -Change of draftsChange of drafts 0.9450.945 + + 0.8650.865 - -
Final draftsFinal drafts 6.8806.880 6.8806.880
Assuming aft draft maintain at 7.86m, new trim is 1.81m Assuming aft draft maintain at 7.86m, new trim is 1.81m by astern, therefore forward draft now is 6.05m…by astern, therefore forward draft now is 6.05m…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 160160
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Worked Example 3Worked Example 3
Your vessel is going to dry dock with the following Your vessel is going to dry dock with the following conditions:conditions:
Draft forward 8.00 m and aft 9.00 m. Her displacement is 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 t-30 000 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 is m. TPC 38. LCF is 1.5 m abaft the amidships and LBP is 160 m.160 m.
The depth of water in the dock is initially The depth of water in the dock is initially 9.50m9.50m..
i.i. Find the effective GM and her new draft after water Find the effective GM and her new draft after water level level has fallen by has fallen by 95cm95cm in the dock. in the dock.
ii.ii. How much will be the further drop of water level so How much will be the further drop of water level so that vessel will take the blocks overall?that vessel will take the blocks overall?
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 161161
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No effect on ship’s No effect on ship’s Initial StabilityInitial Stability……
Clearance 50cm Clearance 50cm
9.0m9.0m
FF
9.5m9.5m9.5m9.5m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 162162
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FF
Drop of water level by Drop of water level by 50cm50cm, No effect on ship’s , No effect on ship’s Initial Stability…Initial Stability…
9.0m9.0m9.0m9.0m
50cm drop of water level50cm drop of water level
9.0m9.0m9.0m9.0m
PP
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 163163
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8.55m8.55m
FF 45cm drop of water level45cm drop of water level
PP
Drop of water level by Drop of water level by 45cm45cm, effect on ship’s , effect on ship’s Initial Stability…Initial Stability…
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 164164
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9.00m9.00m
WLWL
WLWL
8.55m8.55m
Reduction : 45cmReduction : 45cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 165165
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9.00m9.00m
WLWL
WLWL
8.55m8.55m
A : A : Body riseBody rise
WLWL9.00m 9.00m -- BrBr
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 166166
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9.00m9.00m
WLWL
WLWL
8.55m8.55m
WLWL9.00m 9.00m -- BrBr
BB : : Change of draft aft due to COT by Change of draft aft due to COT by HeadHead
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 167167
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9.00m9.00m
WLWL
WLWL
8.55m8.55m
WLWL9.00m 9.00m -- BrBr
BB : : Change of draft aft due to COT by Change of draft aft due to COT by HeadHead
A : A : Body riseBody rise Reduction : 45cmReduction : 45cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 168168
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Reduction = Reduction = AA ++ BB
wherewhere AA Body RiseBody RiseB B Change of draft aftChange of draft aft
due to COTdue to COT
REDUCTIONREDUCTION
== Body rise Body rise ++ Change of draft aft Change of draft aft due to COT due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 169169
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Fallen of water level = Fallen of water level = AA ++ BB
wherewhere AA Body RiseBody RiseB B Change of draft aftChange of draft aft
due to COTdue to COT
Fallen of water level Fallen of water level
== Body rise Body rise + + Change of draft aft Change of draft aft due to COT due to COT
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 170170
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Fallen WLFallen WL = = PP ++ ll x TM x TM TPCTPC L MCTCL MCTC
Fallen WL Fallen WL == PP ++ ll x P x d x P x d TPCTPC L MCTCL MCTC
4545 == PP ++ 78.578.5 x x P x 78.5P x 78.53838 160160 400 400
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 171171
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45 45 == PP ++ [ [ 78.578.5 x x PP x x 78.578.5 ]] 3838 160160 400 400
4545 == PP ++ 0.0962851560.096285156PP
3838 11
4545 == PP ++ 3.6593.659PP3838
17101710 == 4.6594.659PP
P P = = 367.0 tonnes367.0 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 172172
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GGGG11 = = P x KGP x KG = = 367.05 x 10.9367.05 x 10.9 W – P W – P 30 000 – 367.030 000 – 367.0
GGGG11 = = 0.135 m0.135 m
Initial GM Initial GM == 0.600 m0.600 m
Effective GM Effective GM == 0.600 – 0.1350.600 – 0.135
== 0.465 m0.465 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 173173
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OROR
MMMM11 = = P x KMP x KM = = 367.05 x 11.5367.05 x 11.5 W W 30 000 30 000
MMMM11 = = 0.141 m0.141 m
Effective GM Effective GM = = 0.600 – 0.141 0.600 – 0.141
= = 0.459 m0.459 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 174174
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To find the drafts forward and aft…To find the drafts forward and aft…
i.i. Bodily riseBodily rise = = P P = = 367.0367.0 TPCTPC 38 38
= = 9.66cm9.66cm
= = 0.097 m0.097 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 175175
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To find the drafts forward and aft…To find the drafts forward and aft…
ii.ii. COT COT = = P x dP x d = = 367.0 x 78.5367.0 x 78.5MCTCMCTC 400 400
= = 72cm by Head72cm by Head
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 176176
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iii.iii. Change of draft aft due COTChange of draft aft due COT
= = l l x COT x COT = = 78.578.5 x 72 x 72 LL 160160
= = 35.3cm 35.3cm
= = 0.353 m0.353 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 177177
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iv.iv. Change of draft ForwardChange of draft Forward
= = COTCOT – – Change of draft aftChange of draft aft
== 72.0 – 72.0 – 35.335.3
= = 0.367 m0.367 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 178178
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v. v. Fwd(m)Fwd(m) Aft(m)Aft(m)
Initial draftsInitial drafts 8.000 8.000 9.000 9.000 Bodily riseBodily rise 0.097 -0.097 - 0.097 -0.097 -Change of draftsChange of drafts 0.3670.367 + + 0.3530.353 - -
Final draftsFinal drafts 8.2708.270 8.5508.550
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 179179
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New Trim = 8.55 – 8.27New Trim = 8.55 – 8.27 == 0.28m by Stern 0.28m by Stern assuming ‘F’ constantassuming ‘F’ constant
PP == MCTC x TMCTC x T == 400 x 28400 x 28 dd 78.5 78.5
PP == 142.7 tonnes142.7 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 180180
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Further drop Further drop = = PP + + ll x x P x dP x d vessel fully restvessel fully rest TPC TPC L L MCTC MCTC
== 142.7142.7 + + 78.578.5 x x 142.7 x 78.5 142.7 x 78.5
38 38 160160 400 400
== 17.5cm17.5cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 181181
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SINGLE POINTSINGLE POINT
GROUNDING GROUNDING
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 182182
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SINGLE POINT GROUNDINGSINGLE POINT GROUNDING
A vessel floating at drafts forward 8.70 m, A vessel floating at drafts forward 8.70 m, aft 9.40 m grounds at a point 30 m aft of the aft 9.40 m grounds at a point 30 m aft of the forward perpendicular. forward perpendicular.
Estimate the drafts of the vessel and the GM Estimate the drafts of the vessel and the GM after the tide has fallen by 70cm.after the tide has fallen by 70cm.
MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40 MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40 m, LBP 162 m. LCF 78 m forward of Aft m, LBP 162 m. LCF 78 m forward of Aft Perpendicular and displacement is 29 000 Perpendicular and displacement is 29 000 tonnes. tonnes.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 183183
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FF
9.40m9.40m 8.70m8.70m
PP
RockRock
30m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 184184
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PP = = ??
RockRock
FF
Tide fallen by 70cms…Tide fallen by 70cms…
Aft…?Aft…?Fwd…?Fwd…?
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 185185
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Draft at PDraft at P
WLWL
WLWL
New draft at PNew draft at P
Fallen of tide by 70cmFallen of tide by 70cm
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 186186
To Be A World Class Maritime Academy
To Be A World Class Maritime Academy
Draft at PDraft at P
WLWL
WLWL
New draft at PNew draft at P
A : A : Body riseBody rise
Draft at P Draft at P -- BrBrWLWL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 187187
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To Be A World Class Maritime Academy
Draft at PDraft at P
WLWL
WLWL
New draft at PNew draft at P
Draft at P Draft at P -- BrBrWLWL
BB : : Change of draft at Change of draft at PP due to COT by Stern due to COT by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 188188
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To Be A World Class Maritime Academy
Fallen of tide by 70cmFallen of tide by 70cm
Draft at PDraft at P
WLWL
WLWL
New draft at PNew draft at P
Draft at P Draft at P -- BrBrWLWL
BB : : Change of draft at Change of draft at PP due to COT by Stern due to COT by Stern
A : A : Body riseBody rise
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 189189
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To Be A World Class Maritime Academy
Fallen of tide = Fallen of tide = AA ++ BB
wherewhere AA Body RiseBody RiseB B Change of draft at Change of draft at PP
due to COT by Sterndue to COT by Stern
Fallen of tide Fallen of tide
== Body rise Body rise + + Change of draft at Change of draft at PP due to COT Stern due to COT Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 190190
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Fallen of tideFallen of tide = = PP ++ ll x TM x TM TPCTPC L MCTCL MCTC
Fallen of tideFallen of tide == PP ++ ll x P x d x P x d TPCTPC L MCTCL MCTC
7070 == PP ++ 5454 x x P x 54P x 542828 162162 340 340
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 191191
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To Be A World Class Maritime Academy
FF
9.40m9.40m 8.70m8.70m
PP
RockRock
30m54m54m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 192192
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To Be A World Class Maritime Academy
70 70 == PP ++ [ [ 5454 x x PP x x 5454 ]] 2828 162162 340 340
7070 == PP ++ 0.0529411760.052941176PP
2828 11
7070 == PP ++ 1.4821.482PP2828
19601960 == 2.4822.482PP
P P = = 789.7 tonnes789.7 tonnes
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 193193
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To find the drafts forward and aft…To find the drafts forward and aft…
i.i. Bodily riseBodily rise = = P P = = 789.7 789.7 TPCTPC 28 28
= = 28cm28cm
= = 0.280 m 0.280 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 194194
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To Be A World Class Maritime Academy
To find the drafts forward and aft…To find the drafts forward and aft…
ii.ii. COT COT = = P x dP x d = = 789.7 x 54 789.7 x 54 MCTCMCTC 340340
= = 125.4cm by Stern 125.4cm by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 195195
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iii.iii. Change of draft aft due COTChange of draft aft due COT
= = l l x COT x COT = = 7878 x 125.4 x 125.4 LL 116262
= = 60.4 cm 60.4 cm
= = 0.604 m 0.604 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 196196
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To Be A World Class Maritime Academy
iv.iv. Change of draft ForwardChange of draft Forward
= = COTCOT – – Change of draft aftChange of draft aft
== 125.4 – 60.4125.4 – 60.4
= = 65cm 65cm
= = 0.650 m 0.650 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 197197
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To Be A World Class Maritime Academy
v. v. Fwd(m)Fwd(m) Aft(m)Aft(m)
Initial draftsInitial drafts 8.700 8.700 9.400 9.400
Bodily riseBodily rise 0.280 -0.280 - 0.280 -0.280 -
Change of draftsChange of drafts 0.650 -0.650 - 0.604 +0.604 +
Final draftsFinal drafts 7.770 m7.770 m 9.724 m9.724 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 198198
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ii.ii. Estimated GMEstimated GM
GGGG11 = = P x KGP x KG = = 789.7 x 7.60789.7 x 7.60 W – P W – P 29000 – 789.729000 – 789.7
= = 0.213 m0.213 m
Initial GM Initial GM = = 8.40 m – 7.608.40 m – 7.60 == 0.80 m0.80 m
Effective GM = Effective GM = 0.800 – 0.213 0.800 – 0.213 = = 0.587 m0.587 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 199199
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ii.ii. Estimated GMEstimated GM
MMMM11 = = P x KMP x KM = = 789.7 x 8.40789.7 x 8.40 W W 29000 29000
= = 0.229 m0.229 m
Effective GM Effective GM = = 0.80 – 0.229 0.80 – 0.229
= = 0.571 m0.571 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 200200
To Be A World Class Maritime Academy
To Be A World Class Maritime Academy
Thank you…Thank you…