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Effect on G

Simpson Rules

Dry Docking

Simplified Stab

Statical Stab

Revisions Ex.

Inclining Test

- LEARNING OBJECTIVES
- To understand the virtual loss of GM and the calculations.
- To calculate the maximum trim allowed to maintain a minimum stated GM.
- To understand the safe requirements for a ship prior enter into dry dock.

- LEARNING OBJECTIVES
- To understand the critical period during dry docking process.
- To calculate the ship’s drafts after the water level has fallen and after the ship has taken the block overall.
- Effect to stability when vessel has run aground (single point).

Anybody would like to share their experience during dry docking….?

Graving Dock

Floating Dock

Marine Railway

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

- 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

No effect on ship’s Initial Stability…

- 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:
- 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.

at ‘AP’

- When coming into Dry Dock:
- When ship's weight gets transferred to the keel blocks, vessel will suffer loss on her GM.
- The time interval between the sternpost landing on the blocks and the ship taking the blocks overall is referred to as the CRITICAL PERIOD.

- When coming into Dry Dock:
- Vessel must have positive effective GM that to be maintained throughout the critical period.
- If not vessel may heel over, slip off the blocks when there is an external force acting and heel the ship.

G1

Initial GM loss by GG1 after completed the Critical Period…

This is due to Upthrust Force or ‘P’ Force…

G

B

F

What is the total P Force during Critical Period __?___ tonnes

“How much weight to be discharged in order to bring the ship from trim by stern to even-keel…”

P

d

F

P is the Upthrust Force or weight dischargedto the blocks…

T.M = w x d = P xd t-m by Head

F

Vessel is fully rest on the blocks, Change of Trim by Head and finally vessel at even keel drafts… End of Critical Period…

Change of Trim = Trimming Moment (TM)

MCTC

= P x d

Whereby TM = w x d

Change of Trim = P x d

MCTC

P = COT x MCTCtonnes

d

- Exercise in classroom
- MV OneSuch, LBP 120m is going to dry dock at the following condition in sea water
- Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m.
- Her displacement is 4600 tonnes, MCTC is 86 t-m and TPC 15.45
- Calculate
- The amount of up-thrust force (P) at the end of Critical Period?
- Final drafts forward and aft?

TPC

= 74.8

15.45

= 4.8cm

= 0.048m

COT = P x d

MCTC

= 74.8 x 57.5

86

= 50cm

CODA = 57.5 x 50

120

= 24cm

CODF = 50 – 24

= 26cm

Initial draft 3.500m 4.000m

Body rise 0.048m - 0.048m –

COD 0.260m + 0.240m –

Final draft 3.712m 3.712m

F

Vessel is fully rest on the blocks, Change of Trim by Stern and finally vessel at even keel drafts… End of Critical Period…

Change of Trim = Trimming Moment (TM)

MCTC

Whereby TM = w x d = P x d

Change of Trim = P x d

MCTC

P = COT x MCTCtonnes

d

- Method 1 – GG1
- Method 2 – MM1

- Method 1
- When the vessel comes in contact with the blocks, it is assumed that there is a transfer of weight 'P' from the keel to the blocks.
- Hence there is a virtual rise of ship's G (discharged of weight below G)

K

P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”weight discharged from the ship”

G1

Initial GM loss by GG1 during the Critical Period…

This is due to Upthrust Force or ‘P’ Force…

G

B

X = KG1 x Sin Y = GG1 Sin

PX = WY

P x KG1 x Sin = W x GG1 Sin

P x KG1 = W x GG1

P x (KG + GG1) = W x GG1

(P x KG) + (P x GG1) = W x GG1

(P x KG) + (P x GG 1) = W x GG1

P x KG = (W x GG1) – (P x GG1)

P x KG = (W – P) x GG1

P x KG = GG1

W – P

M

Righting Moment

= W x GZ

= W x GM Sin

In this case,

Righting Moment

= (W – P) x G1M Sin

G1

Z

W - P

- Method 1 – GG 1
- Method 2 – MM1

- Method 2
- When the vessel comes in contact with the blocks, it is assumed that there is a transfer of buoyancy 'P' to the keel blocks.
- Hence there is a reduction in KM while the weight and KG are remains constant.

K

P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”buoyancy reduction from the ship”

Buoyancy

Reduction

K

P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”buoyancy reduction from the ship”

Initial GM loss by MM1 after the Critical Period…

This is due to Upthrust Force or ‘P’ Force…

M1

G

B

X = KM1 x Sin Y = MM1 Sin

PX = (W – P) x Y

P x KM1 x Sin = (W – P) x MM1 Sin

P x KM1 = (W – P) x MM1

P x KM1 = W x MM1 – P x MM1

P x KM1 + P x MM1 = W x MM1

P (KM1 + MM1) = W x MM1

M1

Righting Moment

= W x GZ

= W x GM Sin

In this case,

Righting Moment

= W x GM1 Sin

G

Z

W

- Method 1 – GG1 : Weight transferred
- Method 2 – MM1 : Buoyancy transferred

Exercise in classroom …continued

MV OneSuch is going to dry dock at the following condition in sea water

Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m.

Her displacement is 4600 tonnes, MCTC is 86 t-m,

Calculate the amount of up-thrust force (P) during Critical Period and the virtual loss of GM if KM is 8.0m and KG is 7.2m.

Comparison the Virtual loss of GM between ( MM1) and (GG1) method…

Different is…

= 0.130 – 0.119

= 0.011m

… ±1cm

Change of Trim = Trimming Moment (TM)

MCTC

Whereby TM = w x d = P x d

Change of Trim = P x d

MCTC

P = COT x MCTCtonnes

d

- 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?

- Vessel displacement 5000 tonnes, distance sueing point to CF is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.
- Maximum Trim is….?

During Critical Period… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

M

G1

GG1 is Virtual Loss of GM

G

G

During Critical Period… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

M

G1

GG1 is Virtual Loss of GM

G

Residual GM is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

1.0

MAX. TRIM ……?

MAX. TRIM 2.86m

0.846

TRIM

0

0.5

5.0

- 1.0

- CONCLUSION: is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.
- The virtual loss of GM is NIL as vessel having zero trim.
- The loss is increased as the trim increased.
- Maximum trim is depend upon the initial GM

- WORKED EXAMPLE 1 is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.
- A ship of 140m in length, displacement 5000t and upright is to enter dry dock with drafts forward 3.84m, aft 4.60m. Given the following hydrostatic particulars:
- TPC 20 tonnes
- MCTC 150 t- m
- CF 5m forward of amidships
- KM 9.75m
- The blocks of the dry dock are horizontal.
- Calculate the drafts of the vessel at the instants when she is taking the blocks forward and aft.
- The ship's effective GM at this moment if the KG is 7.75m
- The Righting Moment at this instant for an angle of heel 5º.

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

3.84m

Trim 76 cm by Stern

4.60m

No effect on ship’s Initial Stability…

P is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

F

3.84m

Trim 76 cm by Stern

4.60m

P is the Upthrust Force acting at first point of touching the ground, commence Critical Period…

P is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

F

Change of Trim 76cms by Head

Even keel draft

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

Ship’s trimmed = 4.60 – 3.84 = is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. 0.76 m by Stern

i. P = MCTC x trim = 150 x 76

d 75

P = 152 tonnes

a. Bodily rise = P = 152= 7.6 cms = 0.076 m TPC 20

b. Change of Trim = 76 cms by Head

- Change of draft aft due COT is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.
- = lx COT
- L
- = 75 x 76
- 140
- = 40.7cm
- = 0.407 m

d. Change of draft forward due COT is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

= COT – Change of draft aft

= 76 – 40.7

= 35.3cm

= 0.353 m

e. Fwd Aft is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

Initial drafts 3.840 4.600

Bodily rise 0.076 - 0.076 -

Change of drafts 0.353 +0.407 -

Final drafts 4.117 m4.117 m

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

4.117m

4.117m

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

i. ALTERNATIVE METHOD is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

a. Mean draft = 4.220 m.

b. True mean draft correction

= Dist. CF to amidships x trim

LBP

= 5 x 0.76

140

= 0.027 m

- i. ALTERNATIVE METHOD is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.
- True mean draft
- = Mean draft – correction
- = 4.220 – 0.027
- = 4.193 m
- d. Therefore:
- True mean draft = 4.193 m
- Bodily rise = 0.076 m -
- Final drafts = 4.117 m even keel

ii. GG is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. 1 = P x KG = 152 x 7.75

W – P 5000 – 152

= 1178 = 0.243 m

4848

Initial GM = KM – KG = 9.75 m – 7.75

= 2.00 m

Effective GM = 2.00 – 0.243 = 1.757 m

M is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

G1

Initial GM loss by GG1 after the Critical Period…

This is due to Upthrust Force or ‘P’ Force…

G

B

- OR is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.
- MM1 = P x KM = 152 x 9.75 W 5000
- = 0.296 m
- Effective GM = 2.00 – 0.296

- = 1.704 m

M is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

Initial GM loss by MM1 after the Critical Period…

This is due to Upthrust Force or ‘P’ Force…

M1

G

B

W - P is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

External Force

M

G1

Z

B

B1

W - P

Righting Moment at small angle of heel…

W - P is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

M

Righting Moment

= W x GZ

= W x GM Sin

In this case,

Righting Moment

= (W – P) x G1M Sin

G1

Z

W - P

W is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

M

External Force

M1

G

Z

B

B1

W

Righting Moment at small angle of heel…

W is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

M1

Righting Moment

= W x GZ

= W x GM Sin

In this case,

Righting Moment

= W x GM1 Sin

G

Z

W

iii. RM = (W – P) x G is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. 1M Sin

= (5000 – 152) x 1.757 x Sin 5

= 742.4 t-m

OR

RM = W x GM1 Sin

= 5000 x 1.704 x Sin 5

= 742.6 t-m

WORKED EXAMPLE 2 is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

A ship of length 165m, KG 7.30m is floating in a graving dock with drafts forward 5.50m, aft 7.86m in water RD 1.025. At the aft perpendicular the keel is 0.24m above the top of the horizontal blocks. If the water level has fallen in the dock by 1.22m, the ship’s become unstable (GM = 0m).

Calculate

i. The drafts forward and aft at which it occurs

ii. The original/initial GM

Given

Displacement for a hydrostatic mean draft of 6.65m is 9151 tonnes. TPC 24, MCTC 120 t-m and CF 3.66 m abaft amidships.

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

5.50m

7.86m

Clearance 24cm

No effect on ship’s Initial Stability,initial trim is 2.36m by Stern

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

5.50m

Depth of water 7.86 + 0.24 = 8.10m

No effect on ship’s Initial Stability…

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

5.50m

7.86m

Clearance 24cm

No effect on ship’s Initial Stability…

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

5.50m

7.86m

Clearance 16cm

Drop of water level by 8cm. No effect on ship’s Initial Stability.

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

5.50m

7.86m

Clearance 12cm

Drop of water level by 12cm. No effect on ship’s Initial Stability.

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

5.50m

7.86m

Clearance 6cm

Drop of water level by 18cm. No effect on ship’s Initial Stability.

F is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

5.50m

7.86m

Drop of water level by 24cm… stern post start to touch the block…

P is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

F

5.50m

7.86m

P is the Upthrust Force acting at first point of touching the block. Commence Critical Period…

P is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

F

6.88m

Drop of water level 98cm, Vessel become unstable… Zero GM. Vessel is still in Critical Period…

WL is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

7.86m

WL

7.86m -Br

B : Change of draft aft due to COT by Head

WL

6.88m

WL is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

7.86m

A : Body rise

Reduction : 98cm

WL

7.86m -Br

B : Change of draft aft due to COT by Head

WL

6.88m

Reduction = A is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. + B

where A Body Rise

B Change of draft aft

due to COT

REDUCTION

= Body rise + Change of draft aft due to COT

Fallen of water level = A is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. + B

where A Body Rise

B Change of draft aft

due to COT

Fallen of water level

= Body rise + Change of draft aft due to COT

Fallen WL = is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. P+l x TM TPC L MCTC

Fallen WL = P+l x P x d TPC L MCTC

98 = P+78.84 x P x 78.84

24 165 120

- 98 = is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. P + [78.84 x P x 78.84] 24 165 120
- = P + 0.313926545P
- 24 1
- = P + 7.534P
- 24
- 2352 = 8.534P
- P = 275.6 tonnes

P is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. = 275.6 tonnes

If we calculate until vessel is FULLY REST,

P = MCTC x trim = 120 x 236

d 78.84

P = 359.2 tonnes

What is it mean?

To find the drafts forward and aft… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

i. Bodily rise = P = 275.6

TPC 24

= 11.5 cms

= 0.115 m

To find the drafts forward and aft… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

ii. COT = P x d

MCTC

= 275.6 x 78.84

120

= 181cm by Head

iii. is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. Change of draft aft due COT

= lx COT = 78.84 x 181

L 165

= 86.5cm

= 0.865 m

iv. is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. Change of draft Forward

= COT – Change of draft aft

= 181 – 86.5

= 94.5cm

= 0.945 m

v. is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. Fwd(m) Aft(m)

Initial drafts 5.500 7.860

Bodily rise 0.115 - 0.115 -

Change of drafts 0.945 + 0.865 -

Final drafts6.3306.880

To find the initial GM… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

Mean draft = 6.680m. Trim = 2.36m by stern

CF is 3.66m abaft amidships.

TMD Correction

= Dist. CF to Amidships x Trim

LBP

= 3.66 x 2.36 = 0.052 m

165

To find the initial GM… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

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

To find the initial GM… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

Therefore additional displacement

= 8.2 cm x TPC (24)

= 196.8 t

Displacement for TMD 6.732 m

= 9151 + 196.8

= 9347.8 t

To find the initial GM… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

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 7.3

W – P 9347.8 – 275.5

= 2011.15 = 0.222 m

9072.3

Initial GM = 0.222 m

Then… what will be the is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. MAXIMUM TRIM

allowed, safely docked

if the initial GM is 0.222 m….?

Then… what will be the is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. final drafts…

if the initial GM is 0.222 m and trim now is

1.81m by stern….?

To find the drafts forward and aft… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

i. Bodily rise = P = 275.5

TPC 24

= 11.5 cm

= 0.115 m

To find the drafts forward and aft… is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m.

ii. COT = P x d = 275.5 x 78.84

MCTC 120

= 181cm by Head

iii. is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. Change of draft aft due COT

= l x COT = 78.84 x 181

L 165

= 86.5cm = 0.865 m

iv. is 80 m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. Change of draft Forward

= COT – Change of draft aft

= 181 – 86.5 = 94.5cm

= 0.945 m

Assuming aft draft maintain at 7.86m, new trim is 1.81m by astern, therefore forward draft now is 6.05m…

v. Fwd(m) Aft(m)

Initial drafts 6.050 7.860

Bodily rise 0.115 - 0.115 -

Change of drafts 0.945 + 0.865 -

Final drafts6.8806.880

Worked Example 3 astern, therefore forward draft now is 6.05m…

Your vessel is going to dry dock with the following conditions:

Draft forward 8.00 m and aft 9.00 m. Her displacement is 30 000 tonnes. KM is 11.50 m, KG 10.90 m. MCTC 400 t-m. TPC 38. LCF is 1.5 m abaft the amidships and LBP is 160 m.

The depth of water in the dock is initially 9.50m.

i. Find the effective GM and her new draft after water level has fallen by 95cm in the dock.

ii. How much will be the further drop of water level so that vessel will take the blocks overall?

F astern, therefore forward draft now is 6.05m…

9.0m

9.5m

Clearance 50cm

No effect on ship’s Initial Stability…

P astern, therefore forward draft now is 6.05m…

F

50cm drop of water level

9.0m

9.0m

Drop of water level by 50cm, No effect on ship’s Initial Stability…

P astern, therefore forward draft now is 6.05m…

F

45cm drop of water level

8.55m

Drop of water level by 45cm, effect on ship’s Initial Stability…

WL astern, therefore forward draft now is 6.05m…

9.00m

WL

9.00m -Br

B : Change of draft aft due to COT by Head

WL

8.55m

WL astern, therefore forward draft now is 6.05m…

9.00m

A : Body rise

Reduction : 45cm

WL

9.00m -Br

B : Change of draft aft due to COT by Head

WL

8.55m

Reduction = A astern, therefore forward draft now is 6.05m…+ B

where A Body Rise

B Change of draft aft

due to COT

REDUCTION

= Body rise + Change of draft aft due to COT

Fallen of water level = A astern, therefore forward draft now is 6.05m…+ B

where A Body Rise

B Change of draft aft

due to COT

Fallen of water level

= Body rise + Change of draft aft due to COT

Fallen WL = astern, therefore forward draft now is 6.05m…P+l x TM TPC L MCTC

Fallen WL = P+l x P x d TPC L MCTC

45 = P+78.5 x P x 78.5

38 160 400

45 = astern, therefore forward draft now is 6.05m…P + [ 78.5 x P x 78.5] 38 160 400

45 = P + 0.096285156P

38 1

45 = P + 3.659P

38

1710 = 4.659P

P = 367.0 tonnes

GG astern, therefore forward draft now is 6.05m…1 = P x KG = 367.05 x 10.9

W – P 30 000 – 367.0

GG1 = 0.135 m

Initial GM = 0.600 m

Effective GM = 0.600 – 0.135

= 0.465 m

OR astern, therefore forward draft now is 6.05m…

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

MM1= 0.141 m

Effective GM = 0.600 – 0.141

= 0.459 m

To find the drafts forward and aft… astern, therefore forward draft now is 6.05m…

i. Bodily rise = P = 367.0

TPC 38

= 9.66cm

= 0.097 m

To find the drafts forward and aft… astern, therefore forward draft now is 6.05m…

ii. COT = P x d = 367.0 x 78.5

MCTC 400

= 72cm by Head

iii. astern, therefore forward draft now is 6.05m…Change of draft aft due COT

= l x COT = 78.5 x 72

L 160

= 35.3cm

= 0.353 m

iv. astern, therefore forward draft now is 6.05m…Change of draft Forward

= COT – Change of draft aft

= 72.0 – 35.3

= 0.367 m

v. astern, therefore forward draft now is 6.05m…Fwd(m) Aft(m)

Initial drafts 8.000 9.000

Bodily rise 0.097 - 0.097 -

Change of drafts 0.367 + 0.353 -

Final drafts8.2708.550

New Trim = 8.55 – 8.27 = 0.28m by Stern assuming ‘F’ constant

P = MCTC x T = 400 x 28

d 78.5

P = 142.7 tonnes

Further drop = ‘F’ constantP + l x P x d vessel fully rest TPC L MCTC

= 142.7 + 78.5 x 142.7 x 78.5 38 160 400

= 17.5cm

SINGLE POINT ‘F’ constant

GROUNDING

SINGLE POINT GROUNDING ‘F’ constant

A vessel floating at drafts forward 8.70 m, aft 9.40 m grounds at a point 30 m aft of the forward perpendicular.

Estimate the drafts of the vessel and the GM after the tide has fallen by 70cm.

MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40 m, LBP 162 m. LCF 78 m forward of Aft Perpendicular and displacement is 29 000 tonnes.

WL ‘F’ constant

Draft at P

WL

Draft at P -Br

B : Change of draft at P due to COT by Stern

WL

New draft at P

WL ‘F’ constant

Draft at P

A : Body rise

Fallen of tide by 70cm

WL

Draft at P -Br

B : Change of draft at P due to COT by Stern

WL

New draft at P

Fallen of tide = A ‘F’ constant+ B

where A Body Rise

B Change of draft at P

due to COT by Stern

Fallen of tide

= Body rise + Change of draft at P due to COT Stern

Fallen of tide = ‘F’ constantP+l x TM TPC L MCTC

Fallen of tide = P+l x P x d TPC L MCTC

70 = P+54 x P x 54

28 162 340

70 = ‘F’ constantP + [ 54 x P x 54] 28 162 340

70 = P + 0.052941176P

28 1

70 = P + 1.482P

28

1960 = 2.482P

P = 789.7 tonnes

To find the drafts forward and aft… ‘F’ constant

ii. COT = P x d = 789.7 x 54

MCTC 340

= 125.4cm by Stern

v. ‘F’ constantFwd(m) Aft(m)

Initial drafts 8.700 9.400

Bodily rise 0.280 - 0.280 -

Change of drafts 0.650 -0.604 +

Final drafts7.770 m9.724 m

ii. Estimated GM ‘F’ constant

GG1 = P x KG = 789.7 x 7.60

W – P 29000 – 789.7

= 0.213 m

Initial GM = 8.40 m – 7.60 = 0.80 m

Effective GM = 0.800 – 0.213 = 0.587 m

ii. Estimated GM ‘F’ constant

MM1 = P x KM = 789.7 x 8.40 W 29000

= 0.229 m

Effective GM = 0.80 – 0.229

= 0.571 m

Thank you… ‘F’ constant

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