dry-docking-all about to know

8/11/2019 Dry-docking-All About to Know

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D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 1

To Be A World Class Maritime Academy


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Revisions Ex.

Simplified Stab

Simpson Rules

Trim

Effect on G

Dry Docking

Statical Stab

Inclining Test


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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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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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Anybody would like to share their experience duringdry docking.?


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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


7/200D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 7


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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F

No effect on ships Initial Stability


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The rate of pumping will bereduced as the ship's sternpost

near the block.


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Commence touching the ground Sueing Point

Sueing Point

F


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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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P

P is the Upthrust Forceacting at first point oftouching the ground. Commence Critical Period

F


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Sueing Point

at AP


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P

K

P is the Upthrust Forceacting at first point oftouching the ground. Commence Critical Period


15/200




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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P

P force is increasing gradually as the trim changeby HeadVessel is still in Critical Period

F


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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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P

Vessel is fully rest on the blocks End of CriticalPeriod

F


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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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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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CALCULATION

OF UPTHRUST FORCE

AT THE STERNPOST

- 'P' FORCE


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F

After weight discharged

T M By Head = T M By Stern


25/200



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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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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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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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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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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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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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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P

P is the Upthrust Force or weight dischargedtothe blocks

T.M = wx d = Px d t-m by Stern

d

F

To Be A World Class Maritime Academ


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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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Change of Trim = Trimming Moment (TM)

MCTC

Whereby TM = w x d = P x d


P = COT x MCTC tonnesd



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Virtual Loss Of GM

During

Critical Period



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Method 1 GG1

Method 2 MM1



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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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P

d

F



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P

d

F

Trimming Moment by Head



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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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K

G

G1

M

Reduction orLoss of GM= GG1



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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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y

Method 1

GG1 = P x KG in metresW - P



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y

During Critical Period part of ship body is still floating

P

B

M

W

G

G1



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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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Method 1Discharged of weight, shift of GG1

P

G

M

G1

W - P

K

External Force



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P

G

M

G1

W - P

X

K

X = KG1Sin




48/200


P

G

M

W

G1

W - P

Y

K




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G1

G

Y

Y = GG1 Sin



50/200


P

G

M

W

G1

W - P

X

Y

K




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=

G1

P

K

XG

1

G

Y

G

WW



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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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(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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Therefore the formula is

GG1 = P x KGW - P



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Righting Moment at small angle of heel

B1

B

G1

M

W - P

W - P

Z

External Force



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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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Method 1 GG1

Method 2 MM1



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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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Reduction in Buoyancy

P

d

F



60/200


Reduction in Buoyancy

P

d

F



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P

K

P is the Upthrust Force acting at first point oftouching the ground. Commence CriticalPeriodbuoyancy reduction from the ship



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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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K

B

M1

M

Reduction orLoss of GM= MM1

B1



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Method 2

MM1 = P x KM in metresW



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M

M1G

B

Initial GM loss by MM1afterthe Critical Period




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During Critical Period, part of ship body is still floating

P

B

M

W

G

M1



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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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Method 2Transferred of buoyancy, shift of MM1

P

G

M

W

M1

W - P

X

Y

K

W


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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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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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B1

B

G

M1

W

W

Z

M

External Force



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G Z

M1

W

W


In this case,

Righting Moment= W x GM1Sin



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Method 1 GG1: Weight transferred

Method 2 MM1: Buoyancy transferred

SUMMARY



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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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P = COT x MCTCd

= 50 x 8657.5

P = 74.8 tonnes

Calculation of P force



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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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MM1 = P x KMW

= 74.8 x 8.04600

MM1 = 0.130m

Virtual loss of GM (MM1) method



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Comparison the Virtual loss of GM between

(MM1) and (GG1) method

Different is

= 0.130 0.119

= 0.011m

1cm



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Effect ofTrim

In

Dry Docking



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Change of Trim = Trimming Moment (TM)MCTC

Whereby TM = w x d = P x d


P = COT x MCTC tonnes

d



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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?



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Case 1

0 m / Even keel

Case 2

0.5 m by Stern

Case 3

5 m by Stern



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Case 1

0 m / Even keel

Case 2

0.5 m by Stern

Case 3

5 m by Stern

Calculate P

P = MCTC x trim

d

Calculate P

P = MCTC x trim

d

Calculate P

P = MCTC x trim

d



85/200


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


86/200



87/200


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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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



89/200


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



90/200


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



91/200


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



92/200


Residual GM

TRIM0

1.0

0.5

0.846

- 1.0

TRIM increasedGM decreasedMAX. TRIM?

5.0



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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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P force is ? Initial GM 1.0m

Virtual Loss of GM = 1.0m

G

G1M

G

GG1is VirtualLoss of GM

During Critical Period



95/200


P force is ? Initial 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



96/200


P force is ? Initial GM 1.0m Maximum trim is ?


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



97/200


Residual GM

TRIM0

1.0

0.5

0.846

- 1.0

5.0

MAX. TRIM 2.86mMAX. TRIM ?



98/200


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



99/200


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.



100/200


F


4.60m

3.84mTrim 76 cm by Stern



101/200


F

P

P is the Upthrust Forceacting at first point oftouching the ground, commence Critical Period

4.60m3.84mTrim 76 cm by Stern



102/200


P

What is the total P Force during Critical Period?End of Critical Period

FEven keel draft

Change of Trim 76cms by Head



103/200


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



104/200


c. Change of draft aft due COT

= l x COT

L

= 75 x 76

140

= 40.7cm

= 0.407 m



105/200


d. Change of draft forward due COT

= COT Change of draft aft

= 76 40.7

= 35.3cm

= 0.353 m



106/200


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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F4.117m 4.117m

End of Critical Period, vessel is fully rested onblocks, draft is at even keel



108/200


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



109/200


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



110/200


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



111/200


MG1

G

B

Initial GM loss by GG1afterthe Critical Period




112/200


OR

MM1 = P x KM = 152 x 9.75W 5000

= 0.296 m

Effective GM = 2.00 0.296

= 1.704 m



113/200


M

M1

G

B

Initial GM loss by MM1afterthe Critical Period



W P


114/200



B1

B

G1

M

W - P

W - P

Z

External Force


W P


115/200


G1 Z

M

W - P

W - P


In this case,

Righting Moment= (W P) x G1M Sin


W


116/200



B1

B

G

M1

W

W

Z

MExternal Force


W


117/200


G Z

M1

W

W


In this case,

Righting Moment= W x GM1Sin



118/200


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


WORKED EXAMPLE 2


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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.



120/200


F

No effect on ships Initial Stability,initial trim is2.36m by Stern

7.86m

5.50m

Clearance 24cm



121/200



5.50m

Depth of water 7.86 + 0.24 = 8.10m

F



122/200



7.86m

5.50m

Clearance 24cm

F



123/200


Drop of water level by 8cm. No effect on shipsInitial Stability.

7.86m 5.50m

Clearance 16cm

F



124/200


7.86m

5.50m

Clearance 12cm

F




125/200


7.86m

5.50m

Clearance 6cm

F




126/200


5.50m7.86m

F

Drop of water level by 24cm stern post start totouch the block



127/200


5.50m

P

7.86m

P is the Upthrust Forceacting at first point oftouching the block. Commence Critical Period

F



128/200


P

Drop of water level 98cm, Vessel becomeunstable Zero GM. Vessel is still in CriticalPeriod

6.88m

F


WL


129/200


7.86m

WL

WL

6.88m

Reduction : 98cms


P


130/200


P

d

F

Body rise &Trimming Moment by Head


WL


131/200


7.86m

WL

WL

6.88m

A : Body rise

WL

7.86m -Br


WL


132/200


7.86m

WL

WL

6.88m

WL

7.86m -Br

B: Change of draft aft due to COT by Head


WL


133/200


7.86m

WL

WL

6.88m

WL

7.86m -Br


A : Body rise Reduction : 98cm


134/200



135/200


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



136/200


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


98 = P + [78.84 x Px 78.84 ]


137/200


[ ]24 165 120

98 = P + 0.313926545P24 1

98 = P + 7.534P24

2352 = 8.534P

P = 275.6 tonnes



138/200


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



139/200


To find the drafts forward and aft

i. Bodily rise = P = 275.6

TPC 24

= 11.5 cms

= 0.115 m



140/200



ii. COT = P x dMCTC

= 275.6 x 78.84120

= 181cm by Head



141/200


iii. Change of draft aft due COT

= lx COT = 78.84 x 181

L 165

= 86.5cm

= 0.865 m



142/200


iv. Change of draft Forward


= 181 86.5

= 94.5cm

= 0.945 m



143/200


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


To find the initial GM


144/200



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




145/200



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




146/200



Therefore additional displacement

= 8.2 cm x TPC (24)= 196.8 t

Displacement for TMD 6.732 m

= 9151 + 196.8= 9347.8 t




147/200



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



148/200


Then what will be the MAXIMUM TRIM

allowed, safely docked

if the initial GM is 0.222 m.?


GM


149/200


TRIM

MAX. TRIM?

0

0.222

2.36m

-ve


Initial GM 0 222m


150/200


Initial GM 0.222m

P force is ?Maximum trim is ?


P = 275.5 tonnes


Initial GM 0 222m


151/200


Initial GM 0.222m

P force is ?Maximum trim is ?


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


GM


152/200


TRIM

MAX. TRIM 1.81m

0

0.222

2.36m

-ve



153/200


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



154/200


Then what will be the final drafts

if the initial GM is 0.222 m and trim now is

1.81m by stern.?



155/200



i. Bodily rise = P = 275.5

TPC 24

= 11.5 cm

= 0.115 m



156/200



ii. COT = P x d = 275.5 x 78.84MCTC 120

= 181cm by Head



157/200



= l x COT = 78.84 x 181L 165

= 86.5cm = 0.865 m



158/200




= 181 86.5 = 94.5cm

= 0.945 m


Assuming aft draft maintain at 7.86m, new trim is 1.81m


159/200


v. Fwd(m) Aft(m)

Initial drafts 6.050 7.860

Bodily rise 0.115 - 0.115 -Change of drafts 0.945 + 0.865 -


by astern, therefore forward draft now is 6.05m


Worked Example 3


160/200


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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Clearance 50cm

9.0m

F

9.5m


P


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F

Drop of water level by 50cm, No effect on shipsInitial Stability

9.0m

50cm drop of water level

9.0m

P


P


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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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9.00m

WL8.55m

A : Body rise

WL

9.00m -Br


9.00m WL


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9.00m

WL8.55m

WL

9.00m -Br



9.00m WL


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9.00m

WL8.55m

WL

9.00m -Br


A : Body rise Reduction : 45cm


Reduction = A + B


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due to COT

REDUCTION

= Body rise +Change of draft aft

due to COT


Fallen of water level = A + B


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due to COT

Fallen of water level

= Body rise + Change of draft aft

due to COT


Fallen WL = P + l x TM


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TPC L MCTC

Fallen WL = P + l x P x dTPC L MCTC

45 = P + 78.5 x P x 78.538 160 400


45 = P + [ 78.5 x Px 78.5 ]38 160 400


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45 = P + 0.096285156P38 1

45 = P + 3.659P38

1710 = 4.659P

P = 367.0 tonnes


GG P KG 367 05 10 9


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GG1 = P x KG = 367.05 x 10.9

W P 30 000 367.0

GG1 = 0.135 m

Initial GM = 0.600 m


= 0.465 m



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OR

MM1 = P x KM = 367.05 x 11.5W 30 000

MM1 = 0.141 m


= 0.459 m



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i. Bodily rise = P = 367.0TPC 38

= 9.66cm

= 0.097 m



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ii. COT = P x d = 367.0 x 78.5MCTC 400

= 72cm by Head



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= l x COT = 78.5 x 72L 160

= 35.3cm

= 0.353 m



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= 72.0 35.3

= 0.367 m



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v. Fwd(m) Aft(m)

Initial drafts 8.000 9.000Bodily rise 0.097 - 0.097 -Change of drafts 0.367 + 0.353 -




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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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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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SINGLE POINT

GROUNDING


SINGLE POINT GROUNDING


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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.


P


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F

9.40m 8.70m

Rock

30m


P= ?Tide fallen by 70cms


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Rock

F

Aft?

Fwd?


Draft at PWL


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WLNew draft at P

Fallen of tide by 70cm


Draft at PWL


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WLNew draft at P

A : Body rise

Draft at P -Br

WL


Draft at PWL


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WLNew draft at P

Draft at P -Br

WL

B: Change of draft at Pdue to COT by Stern


Draft at PWL


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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


Fallen of tide = A + B


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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


Fallen of tide = P + l x TMTPC L MCTC


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TPC L MCTC

Fallen of tide = P + l x P x d

TPC L MCTC

70 = P + 54 x P x 5428 162 340


P


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F

9.40m 8.70m

Rock

30m54m


70 = P + [ 54 x Px 54 ]28 162 340


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70 = P + 0.052941176P28 1

70 = P + 1.482P28

1960 = 2.482P

P = 789.7 tonnes



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i. Bodily rise = P = 789.7TPC 28

= 28cm

= 0.280 m



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ii. COT = P x d = 789.7 x 54MCTC 340

= 125.4cm by Stern



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= l x COT = 78 x 125.4L 162

= 60.4 cm

= 0.604 m



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= 125.4 60.4

= 65cm

= 0.650 m



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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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ii. Estimated GM

MM1 = P x KM = 789.7 x 8.40W 29000

= 0.229 m


= 0.571 m



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Thank you

dry-docking-all about to know

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