DRY D O C K I N G.

DRY D O C K I N G

Trim 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).

SHARING EXPERIENCES... Graving Dock Floating Dock Marine Railway
Anybody would like to share their experience during dry docking….? Graving Dock Floating Dock Marine Railway

Positive initial GM (GM fluid) Upright
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

F No effect on ship’s Initial Stability…

The rate of pumping will be reduced as the ship's sternpost near the block.

Sueing Point F Commence touching the ground… ‘Sueing Point’

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.

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

Sueing Point at ‘AP’

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

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.

P F P force is increasing gradually as the trim change by Head…Vessel is still in Critical Period…

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.

P F Vessel is fully rest on the blocks… End of Critical Period

M G1 Initial GM loss by GG1 after completed the Critical Period… This is due to Upthrust Force or ‘P’ Force… G B

P F What is the total P Force during Critical Period __?___ tonnes
“How much weight to be discharged in order to bring the ship from trim by stern to even-keel…”

CALCULATION OF UPTHRUST FORCE AT THE STERNPOST - 'P' FORCE

Trimming Moment F w d Weight discharged to even keel the draft…
Trimming Moment = w x d t-m by Head

Even Keel Draft F After weight discharged… T M By Head = T M By Stern

Trimming Moment P d F P is the Upthrust Force or weight discharged to the blocks… T.M = w x d = P x d t-m by Head

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

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 MCTC tonnes 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?

Calculation of P force…
P = COT x MCTC d = 50 x 86 57.5 P = 74.8 tonnes

Body rise = P 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

Forward Aft Initial draft m m Body rise m m – COD m m – Final draft m m

P F d P is the Upthrust Force or weight discharged to the blocks…
T.M = w x d = P x d t-m by Stern

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

Virtual Loss Of GM During Critical Period

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)

Trimming Moment by… Head
P d F Trimming Moment by… Head

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

Reduction or Loss of GM = GG1
K Reduction or Loss of GM = GG1

M G1 Initial GM loss by GG1 during the Critical Period… This is due to Upthrust Force or ‘P’ Force… G B

Method 1 GG1 = P x KG in metres W - P

During Critical Period… part of ship body is still floating
M G1 G B W During Critical Period… part of ship body is still floating

Vessel is inclined to a small angle by an external force…
P W - P External Force M  G  B B1 W Vessel is inclined to a small angle by an external force…

Method 1 Discharged of weight, shift of GG1
P External Force M G1 G  K W - P Method 1 Discharged of weight, shift of GG1

P M X X = KG1 Sin  G1 G  K W - P Method 1 Discharged of weight, shift of GG1

P M Y G1 G  K W - P W Method 1 Discharged of weight, shift of GG1

Y G1 Y = GG1 Sin   G Method 1… Discharged of weight, shift of GG1…

P M Y G1 X G  K W - P W Method 1 Discharged of weight, shift of GG1

P X G1 Y G1 = G  G  K W W

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 GG1) = 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

Therefore the formula is…
GG1 = P x KG W - P

Righting Moment at small angle of heel…
W - P External Force M  G1 Z B B1 W - P Righting Moment at small angle of heel…

 M G1 Z W - P Righting Moment = W x GZ = W x GM Sin  In this case,
= (W – P) x G1M Sin   G1 Z W - P

Method 1 – GG1 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.

P d F Reduction in Buoyancy

P d Reduction in Buoyancy F

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

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

Reduction or Loss of GM = MM1
B B1 K Reduction or Loss of GM = MM1

Method 2 MM1 = P x KM in metres W

M Initial GM loss by MM1 after the Critical Period… This is due to Upthrust Force or ‘P’ Force… M1 G B

P M M1 G B W During Critical Period, part of ship body is still floating

P W - P External Force M  G  B B1 W Vessel is inclined to a small angle by an external force…

 P M Y M1 X G K W Method 2… Transferred of buoyancy, shift of MM1…
W - P W M Y M1 X G  K W Method 2… Transferred of buoyancy, shift of MM1…

W - P P W W X M M1  = G M1  Y K

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

P (KM1 + MM1) = W x MM1 P x KM = W x MM1 P x KM = MM1 W Therefore the formula is… MM1 = P x KM W

W M External Force M1  G Z B B1 W Righting Moment at small angle of heel…

 M1 G Z W Righting Moment = W x GZ = W x GM Sin  In this case,

SUMMARY… 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.

Calculation of P force…
P = COT x MCTC d = 50 x 86 57.5 P = 74.8 tonnes

Virtual loss of GM (GG1) method…
GG1 = P x KG W - P = 74.8 x 7.2 4600 – 74.8 GG1 = 0.119m

Virtual loss of GM (MM1) method…
MM1 = P x KM W = 74.8 x 8.0 4600 MM1 = 0.130m

Comparison the Virtual loss of GM between (MM1) and (GG1) method…
Different is… = – 0.119 = m … ± 1cm

Effect of Trim In Dry Docking

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?

Case 1 0 m / Even keel Case 2 0.5 m by Stern Case 3 5 m by Stern

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

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 = 200 x 50 = 200 x 500

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 P = 0 tonne = 200 x 50 P = 125 tonnes = 200 x 500 P = 1250 tonnes

Case 1 0 m / Even keel Case 2 0.5 m by Stern Case 3 5 m by Stern Virtual Loss of GM…

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 W - P

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 = x 6 5000 – 0 = 125 x 6 5000 –125 W - P = x 6 5000 –1250

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 = x 6 5000 – 0 GG = 0 m = 125 x 6 5000 –125 GG = m W - P = x 6 5000 –1250 GG = 2.0 m

Old GM = 1.0m New GM… = GM - GG1 = 1.0 – 0 = 1.0 m = 1.0 – 0.154
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 = – 0 = m = – 0.154 = m = – 2.0 = m

TRIM increased GM decreased
Residual GM TRIM increased GM decreased 1.0 MAX. TRIM…? 0.846 TRIM 0.5 5.0 - 1.0

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…
P force is …? Initial GM 1.0m Virtual Loss of GM = 1.0m M G1 GG1 is Virtual Loss of GM G G

During Critical Period…
P force is …? Initial GM 1.0m Virtual Loss of GM = 1.0m GG1 = P x KG W - P = P x 6 5000 – P P = 6P = 7P P = tonnes M G1 GG1 is Virtual Loss of GM G

P force is …? Initial GM 1.0m Maximum trim is …? Virtual Loss of GM = 1.0m GG1 = P x KG W - P = P x 6 5000 – P P = 6P = 7P P = tonnes P = tonnes P = MCTC x trim d Trim = P x d MCTC = x 80 200 Trim = cms Trim = 2.86 m by Stern

Residual GM 1.0 MAX. TRIM ……? MAX. TRIM 2.86m 0.846 TRIM 0.5 5.0 - 1.0

CONCLUSION: The virtual loss of GM is NIL as vessel having zero trim. The loss is increased as the trim increased. Maximum trim is depend upon the initial GM

WORKED EXAMPLE 1 A ship of 140m in length, displacement 5000t and upright is to enter dry dock with drafts forward 3.84m, aft 4.60m. Given the following hydrostatic particulars: TPC 20 tonnes MCTC 150 t- m CF 5m forward of amidships KM m 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 No effect on ship’s Initial Stability… 3.84m Trim 76 cm by Stern

P 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 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 = 0.76 m by Stern
i. P = MCTC x trim = 150 x 76 d P = 152 tonnes a. Bodily rise = P = = 7.6 cms = m TPC 20 b. Change of Trim = 76 cms by Head

Change of draft aft due COT
= l x COT L = 75 x 76 140 = 40.7cm = m

d. Change of draft forward due COT
= COT – Change of draft aft = 76 – 40.7 = 35.3cm = m

e. Fwd Aft Initial drafts Bodily rise Change of drafts Final drafts m m

F 4.117m 4.117m End of Critical Period, vessel is fully rested on blocks, draft is at even keel

i. ALTERNATIVE METHOD a. Mean draft = m. b. True mean draft correction = Dist. CF to amidships x trim LBP = 5 x 0.76 140 = m

i. ALTERNATIVE METHOD True mean draft = Mean draft – correction = – 0.027 = m d. Therefore: True mean draft = m Bodily rise = m - Final drafts = m even keel

ii. GG1 = P x KG = 152 x 7.75 W – P – 152 = = m 4848 Initial GM = KM – KG = 9.75 m – 7.75 = m Effective GM = 2.00 – = m

M G1 Initial GM loss by GG1 after the Critical Period… This is due to Upthrust Force or ‘P’ Force… G B

OR MM1 = P x KM = 152 x W = m Effective GM = – 0.296 = m

M Initial GM loss by MM1 after the Critical Period… This is due to Upthrust Force or ‘P’ Force… M1 G B

W - P External Force M  G1 Z B B1 W - P Righting Moment at small angle of heel…

 M G1 Z W - P Righting Moment = W x GZ = W x GM Sin  In this case,
= (W – P) x G1M Sin   G1 Z W - P

W M External Force M1  G Z B B1 W Righting Moment at small angle of heel…

M1  G Z W Righting Moment = W x GZ = W x GM Sin  In this case,

iii. RM = (W – P) x G1M Sin  = (5000 – 152) x x Sin 5 = t-m OR RM = W x GM1 Sin  = x x Sin 5 = t-m

WORKED EXAMPLE 2 A ship of length 165m, KG 7.30m is floating in a graving dock with drafts forward 5.50m, aft 7.86m in water RD 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 5.50m 7.86m Clearance 24cm No effect on ship’s Initial Stability, initial trim is 2.36m by Stern

F No effect on ship’s Initial Stability… 5.50m
Depth of water = 8.10m No effect on ship’s Initial Stability…

F 5.50m 7.86m Clearance 24cm No effect on ship’s Initial Stability…

F Drop of water level by 8cm. No effect on ship’s Initial Stability.
Clearance 16cm Drop of water level by 8cm. No effect on ship’s Initial Stability.

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

P F 7.86m 5.50m P is the Upthrust Force acting at first point of touching the block. Commence Critical Period…

P F 6.88m Drop of water level 98cm, Vessel become unstable… Zero GM. Vessel is still in Critical Period…

WL 7.86m Reduction : 98cms WL 6.88m

Body rise & Trimming Moment by… Head
P d F Body rise & Trimming Moment by… Head

WL 7.86m A : Body rise WL 7.86m - Br WL 6.88m

WL 7.86m WL 7.86m - Br B : Change of draft aft due to COT by Head WL 6.88m

WL 7.86m A : Body rise WL 7.86m - Br WL 6.88m Reduction : 98cm
B : Change of draft aft due to COT by Head WL 6.88m

Reduction = A + 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 + 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 = P + l x TM TPC L MCTC
Fallen WL = P + l x P x d TPC L MCTC 98 = P x P x 78.84

98 = P + [ x P x ] = P P 24 1 = P P 24 2352 = 8.534P P = tonnes

What is it mean? P = 275.6 tonnes
If we calculate until vessel is FULLY REST, P = MCTC x trim = 120 x 236 d P = tonnes What is it mean?

To find the drafts forward and aft…
i. Bodily rise = P = TPC = cms = m

ii. COT = P x d MCTC = x 78.84 120 = 181cm by Head

iii. Change of draft aft due COT
= l x COT = x 181 L = 86.5cm = m

iv. Change of draft Forward
= COT – Change of draft aft = 181 – 86.5 = cm = m

v Fwd(m) Aft(m) Initial drafts Bodily rise Change of drafts Final drafts

To find the initial GM… Mean draft = 6.680m. Trim = 2.36m by stern CF is 3.66m abaft amidships. TMD Correction = Dist. CF to Amidships x Trim LBP = x = m 165

To find the initial GM… True Mean Draft (TMD) = Mean draft + TMD Correction = = m Diff of TMD = – 6.650 = m = 8.2 cm

To find the initial GM… Therefore additional displacement = 8.2 cm x TPC (24) = t Displacement for TMD m = = t

To find the initial GM… When the ship become unstable, the GM = 0 m, therefore loss of GM must be equal to initial GM. GG1 = P x KG = x 7.3 W – P – 275.5 = = m 9072.3 Initial GM = m

Then… what will be the MAXIMUM TRIM
allowed, safely docked if the initial GM is m….?

GM 0.222 MAX. TRIM…? 2.36m TRIM -ve

Initial GM 0.222m P force is …? Maximum trim is …? Virtual Loss of GM = 0.222m P = tonnes

Initial GM 0.222m P force is …? Maximum trim is …? Virtual Loss of GM = 0.222m P = tonnes P = tonnes P = MCTC x trim d Trim = P x d MCTC = x 78.84 120 Trim = 181cm Trim = 1.81m by Stern

GM MAX. TRIM 1.81m 0.222 2.36m TRIM -ve

Trim 1.81m by Stern Virtual loss of GM…? P = MCTC x trim d P = x 181 78.84 P = tonnes GG = P x KG W – P = x 7.3 – 275.5 GG = Residual GM = – 0.222 Residual GM =

Then… what will be the final drafts…
if the initial GM is m and trim now is 1.81m by stern….?

i. Bodily rise = P = TPC = cm = m

ii. COT = P x d = x 78.84 MCTC = 181cm by Head

= l x COT = x 181 L = 86.5cm = m

= COT – Change of draft aft = 181 – = cm = m

Assuming aft draft maintain at 7. 86m, new trim is 1
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 Bodily rise Change of drafts Final drafts

Worked Example 3 Your vessel is going to dry dock with the following conditions: Draft forward 8.00 m and aft 9.00 m. Her displacement is tonnes. KM is m, KG 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 9.0m 9.5m Clearance 50cm No effect on ship’s Initial Stability…

F Drop of water level by 50cm, No effect on ship’s Initial Stability…
50cm drop of water level 9.0m 9.0m Drop of water level by 50cm, No effect on ship’s Initial Stability…

P F Drop of water level by 45cm, effect on ship’s Initial Stability…
45cm drop of water level 8.55m Drop of water level by 45cm, effect on ship’s Initial Stability…

WL 9.00m Reduction : 45cm WL 8.55m

WL 9.00m A : Body rise WL 9.00m - Br WL 8.55m

WL 9.00m WL 9.00m - Br B : Change of draft aft due to COT by Head WL 8.55m

WL 9.00m A : Body rise WL 9.00m - Br WL 8.55m Reduction : 45cm
B : Change of draft aft due to COT by Head WL 8.55m

Reduction = A + 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 + 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 = P + l x TM TPC L MCTC
Fallen WL = P + l x P x d TPC L MCTC 45 = P x P x 78.5

45 = P + [ x P x 78.5 ] 45 = P P 38 1 45 = P P 38 1710 = 4.659P P = tonnes

GG1 = P x KG = x 10.9 W – P – 367.0 GG1 = m Initial GM = m Effective GM = – 0.135 = m

OR MM1 = P x KM = x W MM1 = m Effective GM = – 0.141 = m

i. Bodily rise = P = TPC = cm = m

ii. COT = P x d = x 78.5 MCTC = 72cm by Head

= l x COT = x 72 L = 35.3cm = m

= COT – Change of draft aft = – 35.3 = m

v. Fwd(m) Aft(m) Initial drafts Bodily rise Change of drafts Final drafts

New Trim = 8.55 – 8.27 = 0.28m by Stern assuming ‘F’ constant
P = MCTC x T = 400 x 28 d P = tonnes

Further drop = P + l x P x d vessel fully rest TPC L MCTC
= x x = 17.5cm

SINGLE POINT GROUNDING

SINGLE POINT GROUNDING
A vessel floating at drafts forward 8.70 m, aft 9.40 m grounds at a point 30 m aft of the forward perpendicular. Estimate the drafts of the vessel and the GM after the tide has fallen by 70cm. MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40 m, LBP 162 m. LCF 78 m forward of Aft Perpendicular and displacement is tonnes.

P 30m F 8.70m 9.40m Rock

Tide fallen by 70cms… P = ? F Fwd…? Aft…? Rock

WL Draft at P Fallen of tide by 70cm WL New draft at P

WL Draft at P A : Body rise WL Draft at P - Br WL New draft at P

WL WL WL Draft at P Draft at P - Br
B : Change of draft at P due to COT by Stern WL New draft at P

WL A : Body rise WL WL Draft at P Fallen of tide by 70cm
Draft at P - Br B : Change of draft at P due to COT by Stern WL New draft at P

Fallen of tide = A + 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 = P + l x TM TPC L MCTC
Fallen of tide = P + l x P x d TPC L MCTC 70 = P x P x 54

P 30m 54m F 8.70m 9.40m Rock

70 = P + [ x P x 54 ] 70 = P P 28 1 70 = P P 28 1960 = 2.482P P = tonnes

i. Bodily rise = P = TPC = 28cm = m

ii. COT = P x d = x 54 MCTC = cm by Stern

= l x COT = x L = cm = m

= COT – Change of draft aft = – 60.4 = 65cm = m

v. Fwd(m) Aft(m) Initial drafts Bodily rise Change of drafts Final drafts m m

ii. Estimated GM GG1 = P x KG = x 7.60 W – P – 789.7 = m Initial GM = m – 7.60 = 0.80 m Effective GM = – = m

ii. Estimated GM MM1 = P x KM = x W = m Effective GM = – 0.229 = m

Thank you…

DRY D O C K I N G.

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