INFINITY MARINER: Dry-docking and Grounding

Learning Objectives

After completing this topic you will be able to:

Dry Docking

State

Part of the weight is taken by the blocks as soon as the ship touches the blocks and reduces buoyancy force by the same amount

Define

Critical Instant
Critical Period
Declivity of the Dock

Explain

Up thrust P causes a virtual loss of GM
Up thrust P $= \frac{\text{MCTC} \times Tc}{\text{Distance from the Center of Flotation}}$
Why GM must remain positive until the critical instant
Vessel taking the blocks first at any point on the ships length

Calculate (Using Hindship Stability Tables)

Minimum ships GM to ensure stability until on the blocks overall
Maximum trim to ensure ship remains stable at the point of ship taking the blocks overall
Virtual loss of GM and Drafts after water level has dropped by a stated amount
Draft on Taking the Blocks overall
Loss in GM for fall in water level after taking the blocks overall

Grounding

Explain

How the stability of a ship that has grounded at one point on the centerline is reduced the same way as in dry dock.
How up thrust increase with fall in tide, increases the heeling moment and reduces the stability

Calculate (Using Hindship Stability Tables)

Virtual loss in GM and drafts of a ship after tide has fallen by a stated amount.
Point of grounding given initial drafts and drafts after grounding.

Dry docking of a vessel

Dry docking of a vessel

Vessels are required to dry dock for carrying out mandatory/ statutory surveys and bottom repairs. There are two types of dry dock

Graving

Graving Dock showing keel and bilge blocks

Graving Dock showing keel blocks and side shores

Up to the 60’s, when ships were much narrower, they used to dry dock on a single row of keel blocks, with the toppling controlled by side shores. Later with beam sizes increasing, they docked on a central row of keel blocks and one or more rows of blocks on either side.

Floating

Floating dock

A floating dry dock is literally another vessel which is ballasted so that the ship can move in afloat, after which the dock is deballasted until the vessel lined up on the blocks sits on them. It should be noted that since the dock is also a floating body, the stability of the dock with the ship on it requires to be considered.

The following media explains about the Dock, Bilge and Side Support Arms:

Declivity of docks

Dry docks are trimmed by the stern i.e. the bottom of the dry dock slopes towards the caisson. This slope or trim is called the declivity of the dock and causes residual water to flow into pits from which it is pumped out totally empting the dock.

Part of the weight is taken by blocks as soon as the ship touches the blocks and reduces buoyancy force by same amount

After the ship has been taken into the lock and the caisson gate (Graving dock) closed, the water/ dock tanks are pumped out. This causes the ship to sink within the dock, albeight with no change in her draft. Since the vessel enters with a stern trim the sole piece of the stern frame is the first to touch the keel block. Subsequent to this the aft draft falls and the ford draft increases as the water is pumped out.

With the ships keel touching the blocks, and the water dropping, the buoyancy is no longer able to support the displacement. The reduction in trim is an attempt to regain the buoyancy by increasing the volumetric buoyancy. In terms of trim this may be considered by considering the reaction at the stern block as a reduction in weight giving rise to a trimming moment by the head.

The following animation explains about the dry docking

Critical period

Upthrust P on stern touching the blocks

As the stern touches the block the resultant upthrust/ loss in buoyancy which causes a virtual loss in GM. This is critical since if the loss is great enough for the GM to turn negative the ship could topple of the block and smash into the sides of the dock. The danger of toppling disappears once the vessel is completely on the blocks/ the side shores have been firmed up. Therefore the period from first touching the block to the point where the vessel “takes to” the blocks overall is called the critical period.

Critical instant and and Why GM must remain positive until the critical instant

After the vessel initially touches the block and the critical period commences, the water is steadily pumped out, resulting in further loss of buoyancy/ increase in the upthrust. This steadily increase until the instant before the ship is on the blocks overall after which it is supported by the side shores/ bilge blocks. This instant is called the critical instant.

The increase in the loss of buoyancy from the commencement of the critical period causes a steadily increasing virtual loss of GM and it is extremely important that until the vessel is fully supported either by side shores or bottom blocks the initial GM is sufficient to accommodate this virtual loss in GM without becoming negative and turning unstable.

Up thrust P=(MCTC × )/Distance from the Center of Flotation and Resultant virtual loss of GM

As stated earlier, the loss in buoyancy/ upthrust at the sole piece acts as if a weight that has been discharged l mtrs from the tipping center F^1

. This creates a trimming moment.

(Change in trim once the critical period commences) = $\frac{(P \times l)}{MCTC}$

$P = \frac{\text{MCTC} \times T_c}{l}$

The virtual loss in GM may be considered from two viewpoints, as a loss of GM created by a virtual rise in G ( GG_1

) which we will call Type 1 and a reduction in the metacenter M ( MM_1

) which we will call Type 2

Diag 1

In our efforts to deduce the virtual loss in GM we will consider a vessel whose stern is resting on a single row of keel blocks and heel her by a small angle $\theta$ . The displacement of the vessel W acts downwards through G and the reaction P acts upward from K. B shifts to B_1

and the force of buoyancy now reduced by P i.e. W-P acts upward through it.

Diag 2

Type 1 (GG_1)

Diag - 3

Diag 3

Consider the two parallel but opposite forces W and P. Their resultant W-P acts downward through G_1

. This resultant is horizontally separated by the downward acting W by y mtrs, and from the upward acting P by x mtrs.

The heel $\theta$ results in a transverse righting moment $(W-P) \times G_1 M \ \sin \theta$

$W \times y = P \times x$

$W \times GG_1 \sin \theta = P \times K G_1 \sin \theta$

$W \times GG_1 = P \times KG_1 = P (KG + GG_1) = PKG + P GG_1$

$GG_1 (W-P) = P \times KG$

(Since GM, is the new Gm, GG_1

is the virtual loss in GM)

$GG_1 = \frac{(P \times KG)}{(W-P)}$

is the reduction in GM and G_1M

is the final metacentric height after taking into account the virtual loss.

Type 2 (MM_1)

Diag – 2

Of the three forces, we replace the two forces acting upwards W, and W-P, with the resultant W acting upwards from a point M_1

. This resultant is horizontally separated by the upward acting (W-P) by y mtrs, and from the upward acting P by x mtrs.

$(W-P) \times y = P \times x$

The heel $\theta$ results in a transverse righting moment $(W) \times GM_1 \sin \theta$

$(W-P)MM_1 \sin \theta =P \times KM_1 \sin \theta$

$(W-P)MM_1 =P \times KM_1$

$W \times MM_1 = P(KM_1 + MM_1) = P \times KM$

$MM_1 = \frac{(P \times KM)}{W}$

is the reduction in GM and GM_1

is the final metacentric height after taking into account the virtual loss.

Conditions involved in Dry Docking:

Stability status of the ship before entering dry dock shall be as follows

Ship’s displacement should be as light as possible considering that the ship is in operational condition. Cargo should not be there and ballast should be minimum.
There should not be any list.
Trim should be by stern based on following considerations

Minimum trim as prescribed by the dry dock authorities.Even keel condition or very small trim by stern is not acceptable to them because they will find it difficult to align the full length of the ship exactly in line with the keel blocks.
Maximum trim as decided by the Master.During dry-docking the ship suffers loss of GM, which is directly proportional to its trim. Hence, if the trim is too much, the loss of GM suffered by the ship will be large and it could make the ship unstable before it is secured in the dry dock.
Trim by head would bring the bows in contact with the keel blocks, and the load of the ship will come on it, which could damage the bows, as it is not strengthened for this purpose. Trim by stern will bring the stern frame in contact with the keel blocks, and it has been strengthened to take the load of the ship.

Maximum draft should be as prescribed by the dry dock authorities to ensure safe entry into the dry dock.
GM (F) should be adequate so that loss of GM during dry-docking will not make the ship unstable (Zero or –ve GM) before the ship is adequately secured.

The following media explains about the virtual loss of GM during dry docking and also explains how it occurs:

In modern dry docks the timber shores have been replaced by Bilge blocks, which are placed on both sides, below the ship, in line with the ship’s turn of the bilge. These blocks are adjustable so that once the ship’s bows comes to rest on the keel blocks, the bilge blocks are moved transversely and jacked-up to embrace the ship effectively in accordance with its underwater shape. This way the ship is actually resting on three lines of blocks.

In ideal conditions the ship may be undocked in the same as docking condition. However in practical operations it is not likely. Hence it is imperative that the ship’s officers should always calculate the stability of the ship before undocking by taking into account all changes, which may have taken place in the dry dock, with respect to various tanks and other weights on the ship. Primarily it should be ensured that the ship’s GM(F) at the critical instant, i.e. when the bows is about to lift off and the support of bilge blocks or side shores are not available, is not zero or –ve.

The following media explains about the Dry docking:

Modern Dry Docking

Vessel taking the blocks first at any point on the ships length

Other than grounding which is taken up in the next section, a vessel in damaged condition can enter a dock, trimmed by the head. In this case the critical period commences when the bow touches the block, and the critical instant is just before the stern takes to the blocks.

For purposes of calculating the virtual loss in GM, the same formula may be used, however if trim is required then the trimming moment will have to be taken from the first point of contact with the bow to the center of flotation.

$\text T_\text c = \frac{\text{P} \times l}{\text{MCTC}}$

Floating Dry Dock

Stages in Dry Docking in a Floating Dry Dock

Now we come to the other type of dry docks, the ones which float. Floating dry docks have their own stability issues stemming from the fact that they have to float and bear the load of the vessel under repair at the same time.

A floating dry dock is deballasted and trimmed as per the trim of the vessel to be dry docked. This results in a shorter Critical period. Once the ship is on the blocks over all the stability of the ship and dry dock has to be considered as one unit. The dock has to ensure that it maintains positive stability when lifting the ship clear of water. Hence the weight of the ship being dry docked becomes a factor. The KM of the vessel is now no longer governed only by the buoyancy of herself but it is a resultant of the ship dry dock unit.

Minimum ships GM to ensure stability until the ship sits on the blocks overall

All examples of dry-docking are solved by the following formulae ---

$\text T_\text C$ during dry-docking = $\text P\times\text {LCF}\div \text {MCTC}$
Loss of GM during dry-docking = $\text P\times \text {KG}\div\text (\text W-\text P)\text {OR}=\text P\times \text {KM}\div \text W$
Final GM(F) = KM – Initial KG – Loss of GM - FSC
Drop (cm.) = $(\text P\div\text {TPC})+\left\{(\text P\times \text {LCF}^2)\div (\text {MCTC}\times \text {LBP})\right\}$
P is loss of displacement.

Student should bear in mind that the last formula should only be used when the drop of water level is given and the ship has not reached the C.I.

Example 1

Fixed Data Type

A ship of 6500 mt displacement, KM 7.1 m, KG 6.5 m, MCTC 85 tm, and center of flotation 50 m from APP, with a trim of 0.25 mts by the stern enters a graving dock. Find her GM at the critical instant.

Trimming moment = 25 cm $\times$ 85 = 2125 tm

Upthrust $\text P=\frac{2125}{LCF}=\frac{2125}{50}=42.5\ mt$

Virtual loss using Type 1

$GG_1=\frac{(P \times KG)}{(W-P)}$

$=\frac{(42.5 \times 6.5)}{(6500-42.5)}$

= 0.043 m

ANS: Final GM at critical instant = (KM - KG) - GG_1 = 0.6 - 0.043 = 0.557 m

Virtual loss using Type 2

$MM_1=\frac{(P \times KM)}{W}$

$=\frac{(42.5 \times 7.1)}{6500}=0.046\ m$

ANS: Final GM at critical instant = (KM - KG) - GG_1 = 0.6 - 0.046 = 0.554 m

Maximum trim to ensure ship remains stable at the point of ship taking the blocks overall

Example 2

A ship displacing 3250 mt, 90 m long, KM 5.9 m, KG 5.3 m with center of flotation 42 m ford of APP, MCTC 38 tm. Find the maximum trim at which the vessel can enter the dry-dock if the GM at the critical instant is limited to 0.25,

Initial GM = 5.9 – 5.3 m = 0.6 m

Final GM required = 0.25

Allowable virtual loss = 0.35 m

Type 1

$GG_1=\frac{(P \times KG)}{(W-P)}$

$0.35=\frac{(P \times 5.3)}{(3250-P)}$

$5.65\ \text P = 3250 \times 0.35$

$P = 201.227\ mt$

$Tc=\frac{(P \times LCF)}{MCTC}$

$=\frac{(201.33 \times 42)}{38}$

= 222.52 cm

ANS: Maximum allowable trim = 2.225 m

Type 2

$MM_1=\frac{(P \times KM)}{W}$

$0.35 = \frac{(P \times 5.9)}{3250}$

$P=\frac{3250 \times 0.35}{5.9}= 192.8 \ mt$

$Tc=\frac{(P \times LCF)}{MCTC}$

$=\frac{192.8 \times 42}{38}= 213.09\ cm$

ANS: Maximum allowable trim = 2.131 m

Virtual loss of GM and Drafts after water level has dropped by a stated amount

Example 3

A ship displacing 6500 mt, 200 m long, KM 7.2 m, KG 6.5 m with center of flotation 102 m ford of APP, MCTC 150 tm, FSM 400 tm, draft F/5.20 m A/6.10 m. If the metacenter at the critical instant rises by 0.25 m calculate the virtual GM(F)at critical instant.

Initial Trim = 6.10 – 5.20 = 0.9 m

Upthrust P at critical instant = $T_c \times \frac{MCTC}{LCF}=\frac{(90 \times 150)}{102}= 132.353\ tons$

$FSC=\frac{FSM}{W-P}=\frac{400}{6367.65 \ m}= 0.063\ m$

GM(F) at C.I=KM + Rise in KM - KG - Loss of GM $(\text {GG}_1)$ - FSC

= 7.2 + 0.25 - 6.5 - 0.135 = 0.063 = 0.752m.

Type 1

$GG_1=\frac{(P \times KG)}{(W-P)}$

$=\frac{(132.35 \times 6.5)}{(6500-132.35)}$

= 0.135 m

ANS: Virtual GM = GM(F) at C.I=KM + Rise in KM - KG - Loss of GM $(\text {GG}_1)$ - FSC

= 7.2 + 0.25 - 6.5 - 0.135 = 0.063 = 0.752m

Type 2

$MM_1=\frac{P \times KM}{W}$

$=\frac{(132.35 \times 7.2 + 0.25)}{6500}$

= 0.152 m

ANS: Virtual GM = 0.887 – 0.152 = 0.74 m

Draft on Taking the Blocks overall

Example 4

Hindship Type

(To be solved with data obtained from “Trim & Stability Particular of M V Hindship published by Bhandarkar Publications

Hindship at a draft of F/3.82 A/5.38 m in SW is being dry docked. KG 8.3. m FSC 0.09 m. Calculate i) The GM after virtual loss at the critical instant. ii) Calculate the forward and aft drafts and the drop in water level when the GM(F) becomes zero.

F 3.82 m

A 5.38 m

M 4.60 m

Trim 1.56 m by stern

LCF for mean draft of 4.60 m = 73.017 m

Aft draft correction = $\frac{(Trim \times LCF)}{LBP}=\frac{1.56 \times 73.017}{143.16}= 0.796\ m$

Hydrostatic draft = Aft draft – Correction = 5.38 – 0.796 = 4.584 m

At Hydrostatic draft

Displacement 8875.04 mt

MCTC 163.688 tm

LCF 73.020 m

KM 9.211 m

KG 8.300 m

FSC 0.090

GM (Fluid) 0.821 m

Drop in water level = Initial aft draft before dry-docking – Final aft draft after dry-docking.
In this example the student has to calculate the final drafts and drop in water level when GM(F) = 0. For this purpose it is necessary to ascertain whether the zero GM will occur before or after critical instant (C.I.). Hence first calculate the GM(F) at C.I. as follows ---
At C.I. ---
Trim change $(\text T_\text C)$ = Initial trim = $1.56\times 100=\text P\times \text {LCF}\div \text {MCTC}$
= $\text P\times 73.02\div163.688$
P = 349.703 tons ; New Displacement = 8875.04 – 349.703 = 8525.337 tons
Hydrostatic draft = 4.424m. ;KM = 9.363m.
$\text {FSC}=0.09 \times 8875.04\div8525.337=0.094\text m.$
$\text {Loss}\ \text {of}\ \text {GM}=\text P\times \text {KG}\div\text {New}\text W=349.703\times 8.3\div 8525.337=0.340\text m.$
$\text {GM}(\text F)@\text {C.I}=\text {KM}-\text {KG}-\text {Loss}\text\ {of} \text\ {GM}-\text {FSC}$
= 9.363 – 8.3 – 0.34 – 0.094 = 0.629m.
Hence GM(F) will become zero after the C.I. This means that the ship will be on even keel, side shores or bilge blocks will be in position, the ship will be fully secured, andso there will be no free surface effect.

Method 1

$\text {GM}(\text F)=0=\text {KM}\ (\text {at}\text\ {C.I})-\text {Initial KG}-\left\{(\text P\times \text {KG})\div(\text W-\text P)\right\}$
$=9.363-8.3-\left\{(\text P\times 8.3)\div (8857.04-\text P)\right\}$
P = 1005.557 tons ; New W = 8857.04 – 1005.557 = 7851.483 tons
Hyd. = 4.114m. = Ford. and aft drafts ; Drop = 5.38 – 4.114 = 1.266m.

Method 2

$\text {GM}(\text F)=0=\text {KM}\ (\text {at}\text {C.I})-\text {Initial KG}-\left\{(\text P\times \text {KM})\div \text W\right\}$
$=9.363-8.3-\left\{(\text P\times 9.363)\div8857.04\right\}$
P = 1005.557 tons ; New W = 8857.04 – 1005.557 = 7851.483 tons
Hyd. = 4.114m. = Ford. and aft drafts ; Drop = 5.38 – 4.114 = 1.266m.

Example 5

M.V. Hindship is floating at drafts F 3.82m., A 5.38m., KG 8.3m.and FSC 0.09m. If this ship has to be dry-docked, calculate the forward and aft drafts and the drop in water level when the GM(F) becomes 0.7m.

Data given in this example is same as that of Example 4. Hence, as calculated in previous example, initial ---Trim = 1.56m. ; Hydrostatic draft = 4.584m.
Displacement = 8875.04 tons ;MCTC = 163.688 tm. ; LCF = 73.02m. ; KM = 9.211m.
As calculated in the previous example, GM(F) at C.I. = 0.629m., which means the required GM(F) of 0.7m. will occur before the C.I. and the ship will not be resting fully on the keel blocks.
$\text {GM}(\text F)=0.7- \text {Initial KM}-\text {Initial KG}-\left\{(\text P\times \text {KG}\div (\text W-\text P)\right\}-\left\{\text {FSC}\times \text W\div (\text W-\text P)\right\}$
$=0.7=9.211-8.3-\left\{(\text P\times \text 8.3\div (8875.04-\text P)\right\}-\left\{0.09\times 8875.04\div (8875.04-\text P)\right\}$
$0.211=\left\{(\text P\times 8.3)+798.754\right\}\div(8875.04-\text P)$
$1872.633=(\text P\times 8.511)+798.754$
P = 126.175 tons ; New W = 8748.865 tons ; Hyd. Draft = 4.526m.
MCTC = 163.282 tm. ; LCF = 73.033m.
$\text T_\text C=\text P\times \text {LCF}\div\text {MCTC}=126.175\div73.033\div163.282=56.436\text{cm}=0.564\text m.$
Final trim = Initial trim - $\text T_\text C$ = 1.56 – 0.564 = 0.996m.
$\text T_\text A=\text T\times \text {LCF}\div\text {LBP}=0.996\times73.033\div143.16=0.508\text m.$
Aft draft = 4.526 + 0.508 = 5.034m. ; Ford. draft = 5.034 – 0.996 = 4.038m.
Drop in water level = 5.38 – 5.034 = 0.346m.

Loss in GM for fall in water level after taking the blocks overall

Example 6

M.V. Hindship of displacement 9788 tons, trim 80 cm. by the stern, KG 7.9m. and FSC 0.15m., is to be dry-docked. Calculate

GM(F) before entering the dock.
GM(F) and forward and aft drafts when the drop in water level is 1.0m
Hyd. draft = 5.0m. ; MCTC = 166.8 tm. ; LCF = 72.917m. ; KM = 8.89m. ; T = 0.8m.
Corr. = 72.917 $\times$ 0.8 $\div$ 143.16 = 0.413m.
Aft draft = 5.0 + 0.413 = 5.413m. ; Forward draft = 5.413 – 0.8 = 4.213m.
GM(F) = 8.89 – 7.9 – 0.15 = 0.84m.
In this example the student has to calculate the final drafts and GM(F) when drop in water level is 1.0m. For this purpose it is necessary to ascertain whether the required drop will occur before or after the C.I. Hence first calculatedrop at C.I. as follows ---
At C.I. ---
Trim change $(\text T_\text C)$ = Initial trim = 0.8 $\times$ 100 = P $\times$ LCF $\div$ MCTC
= P $\times$ 73.917 $\div$ 166.8
P = 180.527tons ; New Displacement = 9788 – 180.527 = 9607.473 tons
Hydrostatic draft = 4.918m. = F & A drafts ; Drop @ C.I. = 5.413 – 4.918 = 0.495m.
Hence the required drop of 1.0m. will occur after the C.I. This means that the ship will be on even keel, side shores or bilge blocks will be in position, the ship will be fully secured, and so there will be no free surface effect.
Hence final Aft draft = 5.413 – 1.0 = 4.413 = Forward draft = Hyd. draft
Final displacement = 8501.405 tons; KM = 9.38m.;P = 9788 – 8501.405 = 1286.595 tons
Loss of GM = (P $\times$ KG) $\div$ (W – P) = (1286.595 $\times$ 7.9) $\div$ (9788 - 1286.595) = 1.196m.
Final GM(F) = 9.38 – 7.9 –1.196 = 0.284m.

Example 7

M.V. Hindship of displacement 9788 tons, trim 80 cm. by the stern, KG 7.9m. and FSC 0.15m., is to be dry-docked. Calculate GM(F) and forward and aft drafts when the drop in water level is 0.4 m.

Data given in this example is same as that of Example 6. Hence, as calculated in the previous example, initial --- Hyd. draft = 5.0m. ; MCTC = 166.8 tm. ; LCF = 72.917m. ; T = 0.8m. ; Aft draft = 5.413m. ; Forward draft = 4.213m. ; TPC = 22.08 t / cm.
As calculated in the previous example, drop at C.I. = 0.495m., which means the required drop of 0.4m. will occur before C.I. and ship will not be resting fully on the keel blocks.
This requires a new formula as follows ---
Drop (cm.) = (P $\div$ TPC) + {(P $\times$ LCF2) $\div$ (MCTC $\times$ LBP)}
Note --- Student should bear in mind that this formula should only be used when the drop of water level is given and the ship has not reached the C.I.
$0.4\times 100=(\text P\div22.08)+\{(\text P\times 72.917^2)\div(166.8\times 143.16)\}$
$40\times 22.08\times 166.8\times 143.16=(\text P\times 166.8\times 143.16)+(\text P\times 72.917^2\times22.08)$
21090010.52 = (P $\times$ 23879.088) + (P $\times$ 117396.907) = P $\times$ 141275.995
P = 149.282tons ; New W = 9788 - 149.282 = 9638.718 tons ; Hyd. = 4.932m.
MCTC = 166.324 tm. ; LCF = 72.936m. ; KM = 8.938m.
$\text T_\text C$ = P $\times$ LCF $\div$ MCTC = 149.282 $\times$ 72.936 $\div$ 166.324 = 65.463 cm. = 0.655m.
Final trim = 0.8 – 0.655 = 0.145m. ; $\text T_\text A$ = 0.145 $\times$ 72.936 $\div$ 143.16 = 0.074m.
Aft draft = 4.932 + 0.074 = 5.006m. ; Forward draft = 5.006 – 0.145 = 4.861m.
FSC = 0.15 $\times$ 9788 $\div$ 9638.718 = 0.152m.
Loss of GM = 149.282 $\times$ 7.9 $\div$ 9638.718 = 0.122m.
Final GM(F) = KM – KG – Loss of GM – FSC
= 8.938 – 7.9 – 0.122 – 0.152 = 0.764m.

Example 8

M.V. Hindship of displacement 19051 tons, KG 7.81m. and FSC 0.122m., is to be dry-docked. Calculate the maximum permissible trim so that the final GM(F) at critical instant is 0.3m. Also calculate initial drafts, and final drafts and drop at C.I.

Hyd. = 9.0m. ; MCTC = 211.8 tm. ; LCF = 69.768m. ; KM = 8.388m.

Initial GM(F) = 8.388 – 7.81 – 0.122 = 0.456m. ; Loss of GM @ C.I. = 0.456 – 0.3 = 0.156m.

0.156 = (P $\times$ 7.81) $\div$ (19051 – P) ; P = 373.08 tons

W @ C.I. = 19051 – 373.08 = 18677.92 tons ; Hyd. = 8.846m. ; MCTC = 209.644 tm.

LCF = 69.878m.

Forward and aft drafts @ C.I. = Hyd. = 8.846m.

$\text T_\text C$ = Maximum initial trim = 373.08 $\times$ 69.878 $\div$ 209.644 = 124.354 cm. = 1.244m.

Initial forward and aft drafts should be calculated using initial LCF.

$\text T_\text A$ = 1.244 $\times$ 69.768 $\div$ 143.16 = 0.606m.

Initial aft draft = 9.0 + 0.606 = 9.606m. ; forward draft = 9.606 – 1.244 = 8.362m.

Drop of W/L @ C.I. = 9.606 – 8.846 = 0.76m.

How the stability of a ship that has grounded at one point on the centerline is reduced the same way as in dry dock

A grounded vessel may be treated as a vessel being dry docked with minor deviations in the manner in which these cases are treated. The main difference is that a vessel may ground overall or at any single or multiple points between the two perpendiculars.

The pumping out of the dry dock is equated with the rise and fall in tide around the grounded vessel and the resultant upthrust caused by the loss in buoyancy of the vessel. The upthrust (P) is calculated in the same manner as a dry-docked vessel, by either treating the virtual rise in in GM to be caused by a rise in G or a drop in M.

A point of divergence is when the change in trim is considered. In a dry docked vessel the trimming moment is calculated as the product of the upthrust (P) and the LCF from the APP (And in the rare case of dry docking a vessel that is down by the head the FPP). If the vessel aground is grounded at a single point between the APP and the FPP, the trimming moment will be the product of the upthrust (P) and the distance of the grounding point from the APP. Similarly the change in trim is also similarly considered.

The most powerful force available for floating a grounded vessel is the additional buoyancy created by the rise in tide. Needless to say, if a vessel grounds at the "top of the tide", she is in extreme danger since the tide has a long way to fall, and there will hardly be any additional buoyancy to assist the ships refloating.

An inspection of the drafts all round the vessel gives us sufficient data to find the point of grounding by considering the change in trim as the tide rises or falls. Example 7 considers these aspects and a close study of it will help the student to have a fuller understanding of a grounded vessel.

In this section we will consider single point groundings on the centerline, and the list of vessels that ground at points to port or stbd of the centerline.

How up thrust increase with fall in tide, increases the heeling moment and reduces the stability

As discussed in the previous topic, a drop in tidal levels leads to a increase in upthrust (P) and with it a rise in the virtual loss of GM. If the point of grounding is off the centerline, this upthrust (P) also causes a heeling moment that increases with the fall in water level and heels the vessel and if the fall is high enough the vessel can capsize.

Grounded & Heeled

Virtual loss in GM and drafts of a ship after tide has fallen by a stated amount

Please refer to the topic 'Point of grounding given initial drafts and drafts after grounding'.

Point of grounding given initial drafts and drafts after grounding

All examples of grounding are solved by the following formulae

TC during grounding = P $\times$ D $\div$ MCTC
Loss of GM during dry-docking = P $\times$ KG $\div$ (W – P) OR = P $\times$ KM $\div$ W
Final GM(F) = KM – Initial KG – Loss of GM - FSC
Fall of tide (cm.) = (P $\div$ TPC) + { $(\text{P} \times \text{D}^2)$ $\div$ (MCTC $\times$ LBP)}
P is loss of displacement and D is the distance of point of grounding from the COF.

Student should bear in mind that the last formula should only be used in case of beach grounding, when the fall of tide is given and the ship has not reached the rest position i.e. it is not fully resting on the beach.

Example 1

M.V. Hindship in condition No. 10 runs aground on a rock. After fall of tide the drafts are Ford. 6.0m. and Aft 9.2m. Calculate the position of point of grounding, final GM(F) and the rise of tide required to refloat the ship.

Calculate initial data of the ship

W = 16133 tons ; LCG = 70.289m. ; KG = 7.57m. ; FSM = 1372 tm. ;

Hyd. = 7.788m. ; MCTC = 194.432 tm. ; LCB = 72.644m. ; LCF = 70.792m.

BG = LCB – LCG = 72.644 – 70.289 = 2.355m.

T = W $\times$ BG $\div$ MCTC = 16133 $\times$ 2.355 $\div$ 194.432 = 195.406 cm. = 1.954m. (Stern)

Calculate final data of the ship after fall of tide

T = 9.2 – 6.0 = 3.2m. ; Mean draft = (6.0 + 9.2) $\div$ 2 = 7.6m. ; LCF for mean draft = 70.979m.

Corr. = T $\times$ LCF $\div$ LBP = 3.2 $\times$ 70.979 $\div$ 143.16 = 1.587m.

Hyd. = Aft draft – Corr. = 9.2 – 1.587 = 7.613m.

$\text{W}_1$ = 15723.42 tons ; MCTC = 191.982 tm. ; LCF = 70.966m. ; KM = 8.238m.

Loss of P = 16133 - 15723.42 = 409.58 tons

$\text{T}_\text{C}$ =Final trim – Initial trim = 3.2 – 1.954 = 1.246m. = 124.6 cm. (By stern)

Because the trim has increased, the point of contact will be forward of Center of Floatation.

$\text{T}_\text{C}$ = 124.6 = P $\times$ $\text{D}_1$ $\div$ MCTC = 409.58 $\times$ $\text{D}_1$ $\div$ 191.982

$\text{D}_1$ = 58.404m. = Distance of the point of contact from the COF, after grounding.

Distance of the point of contact from A.P. = 58.404 + 70.966m. = 129.37m.

Calculate draft at the point of contact in initial condition

Distance of point of contact from COF in initial condition = D = 129.37 – 70.792 = 58.578m.

Corr. = T $\times$ D $\div$ LBP = 1.954 $\times$ 58.578 $\div$ 143.26 = 0.8m.

Draft at point of contact = Hyd. – Corr. = 7.788 – 0.8 = 6.988m.

Note: Sign is –ve because the trim is by stern and point of contact is forward of COF.

Calculate draft at the point of contact after grounding

Corr. = T (Final) $\times$ $\text{D}_1$ $\div$ LBP = 3.2 $\times$ 58.404 $\div$ 143.26 = 1.305m.

Draft at the point of contact = Hyd. – Corr. = 7.613 - 1.305 = 6.308m.

Rise of tide required to refloat the ship = 6.988 – 6.308 = 0.68m.

Loss of GM during grounding = P $\times$ KG $\div$ W1 = 409.58 $\times$ 7.57 $\div$ 15723.42 = 0.197m.

Final FSC = FSM $\div$ $\text{W}_1$ = 1372 $\div$ 15723.42 = 0.087m.

GM(F) = Final KM – KG – Loss of GM – Final FSC

= 8.238 – 7.57 – 0.197 – 0.087 = 0.384m.

Example 2

M.V. Hindship floating at drafts F 7.9m., A 9.8m., KG 7.7m. and FSC 0.092m., runs aground at the forward end on a beach having a slope of 3m. by the stern. If the tide falls by 0.5m., what will be the final GM(F) and drafts?

T = 1.9m. (Stern) ; Mean draft = 8.85m. ; LCF for mean draft = 69.875m.

Corr. = T $\times$ LCF $\div$ LBP = 1.9 $\times$ 69.875 $\div$ 143.16 = 0.927m. ; Hyd. = 9.8 – 0.927 = 8.873m.

W = 18742.39 tons ; TPC = 24.064 T/cm. ; MCTC = 210.022 tm. ; LCF = 69.859m.

D = Distance of COF from F.P. = LBP – LCF = 143.16 – 69.859 = 73.301m.

For a fall of tide of 0.5m., the ship may not be fully resting on the beach or it may be so resting. To resolve this ambiguity, calculate the fall of tide when the ship just reaches the rest position.

TC till the rest position = 3.0 – 1.9 = 1.1m. = 110 cm. = P $\times$ D $\div$ MCTC

= P $\times$ 73.301 $\div$ 210.022

P (Till rest position) = 315.172 tons ; W (At rest) = 18742.39 – 315 172 = 18427.218 tons

Hyd. at rest = 8.743m. ; LCF at rest = 69.955m.

$\text{T}_\text{A}$ = T at rest (Slope of beach) $\times$ LCF $\div$ LBP = 3.0 $\times$ 69.955 $\div$ 143.16 = 1.466m.

Aft draft = 8.743 + 1.466 = 10.209m. ; Ford. draft = 10.209 – 3.0 = 7.209m.

Fall of tide = Initial Ford. draft – Final Ford. draft = 7.9 – 7.209 = 0.691m.

As the fall of tide given in the example is less than the fall of tide required to reach the rest position, hence the ship has not reached the rest position.

Fall of tide = (P $\div$ Initial TPC) + {(P $\times$ Initial $\text{D}^2$ ) $\div$ (Initial MCTC $\times$ LBP)}

Student should note that this formula should be used only when the fall of tide is given and the ship has not reached the rest position.

0.5 $\times$ 100 = 50 cm. = (P $\div$ 24.064) + { $(\text{P} \times 73.301^2)$ $\div$ (210.022 $\times$ 143.16)}

50 $\times$ 24.064 $\times$ 210.022 $\times$ 143.16 = (P $\times$ 210.022 $\times$ 143.16) + ( $\text{P} \times 73.301^2$ $\times$ 24.064)

36176313.02 = (P $\times$ 30066.750) + (P $\times$ 129296.753) = P $\times$ 159363.503

P = 227.005 tons ; $\text{W}_1$ = 18515.385 tons ; Hyd. = 8.779m. ; MCTC = 208.685 tm.

LCF = 69.927m. ; $\text{D}_1$ = LBP – LCF = 143.16 – 69.927 = 73.233m. ; KM = 8.348m.

$\text{T}_\text{C}$ = P $\times$ $\text{D}_1$ $\div$ MCTC = 227.005 $\times$ 73.233 $\div$ 208.685 = 79.662 cm. = 0.797m. (By stern)

Final trim = Initial trim + $\text{T}_\text{C}$ = 1.9 + 0.797 = 2.697m.

$\text{T}_\text{A}$ = 2.697 $\times$ 69.927 $\div$ 143.16 = 1.317m.

Aft draft = 8.779 + 1.317 = 10.096m. ; Ford. draft = 10.096 – 2.697 = 7.399m.

Loss of GM = P $\times$ KG $\div$ W1 = 227.005 $\times$ 7.7 $\div$ 18515.385 = 0.094m.

New FSC = 0.092 $\times$ 18742.39 $\div$ 18515.385 = 0.093m.

Final GM(F) = KM – KG – Loss of GM – New FSC = 8.348 – 7.7 – 0.094 – 0.093 = 0.461m.

Example 3

M.V. Hindship floating in SW at drafts F 6.0m., A 9.0m., KG 7.5m. and FSM 1000 tm., runs aground on a sloping beach. Soundings taken immediately after grounding showed that the depth of water at forward end was 7m. and at aft end it was 9m. Calculate the fall of tide and the drafts when the GM(F) becomes 0.6m.

T = 3.0m. ; Mean draft = 7.5m. ; LCF for mean draft = 71.086m.

Corr. = 3 $\times$ 71.086 $\div$ 143.16 = 1.49m. ; Hyd. = 9 – 1.49 = 7.51m. ; W = 15482.85 tons

MCTC = 190.54 tm. ; LCF = D = 71.075m. ; KM = 8.238m.

From the given soundings it is clear that the ships bows are in deep waters while the stern has just touched the ground i.e. the point of contact is the A.P. and slope of the beach is 2m. by the stern. Hence in the formulae for grounding D = LCF.

For the GM(F) to become 0.6m., the ship may not be fully resting on the beach or it may be so resting. To resolve this ambiguity, calculate the GM(F) when the ship just reaches the rest position.

TC till the rest position = 3.0 – 2.0 = 1m. = 100 cm. (By head) = P $\times$ D $\div$ MCTC

= P $\times$ 71.075 $\div$ 190.54

P (Till rest position) = 268.083 tons ; W (At rest) = 15482.85 – 268.083 = 15214.767 tons

Hyd. at rest = 7.395m. ; KM at rest = 8.238m.

Loss of GM = P $\times$ KG $\div$ W(at rest) = 268.083 $\times$ 7.5 $\div$ 15214.767 = 0.132m.

FSC = 1000 $\div$ 15214.767 = 0.066m. ; GM(F) at rest = 8.238 – 7.5 – 0.132 – 0.066 = 0.54m.

As the GM(F) given in the example is more than the GM(F) at the rest position, hence the ship has not reached the rest position.

GM(F) = Initial KM – KG – Loss of GM – FSC

= Initial KM – KG – {(P $\times$ KG) $\div$ (W – P)} – {FSM $\div$ (W –P)}

0.6 = 8.238 – 7.5 – {(P $\times$ 7.5) $\div$ (15482.85 – P} – {1000 $\div$ (15482.85 – P)}

0.138 = {(P $\times$ 7.5) + 1000} $\div$ (15482.85 – P)

P $\times$ 7.638 = 1136.633 ; P = 148.813 tons

$\text{W}_1$ = 15334.037 tons ; Hyd. = 7.446m. ; MCTC = 189.644 tm. ; LCF = D = 71.144m.

$\text{T}_\text{C}$ = P $\times$ D $\div$ MCTC = 148.813 $\times$ 71.144 $\div$ 189.644 = 55.826 cm. = 0.558m. (By head)

Final trim = 3.0 – 0.558 = 2.442m. (Stern)

$\text{T}_\text{A}$ = T $\times$ LCF $\div$ LBP = 2.442 $\times$ 71.144 $\div$ 143.16 = 1.214m.

Aft draft = 7.446 + 1.214 = 8.66m. ; Ford. draft = 8.66 – 2.442 = 6.218m.

Fall of tide = Initial aft draft – final aft draft = 9.0 – 8.66 = 0.34m.

INFINITY MARINER

Saturday, June 8, 2024

Dry-docking and Grounding

Dry docking of a vessel

Declivity of docks

Part of the weight is taken by blocks as soon as the ship touches the blocks and reduces buoyancy force by same amount

Critical period

Critical instant and and Why GM must remain positive until the critical instant

Up thrust P=(MCTC × )/Distance from the Center of Flotation and Resultant virtual loss of GM

Vessel taking the blocks first at any point on the ships length

Floating Dry Dock

Minimum ships GM to ensure stability until the ship sits on the blocks overall

Maximum trim to ensure ship remains stable at the point of ship taking the blocks overall

Virtual loss of GM and Drafts after water level has dropped by a stated amount

Draft on Taking the Blocks overall

Loss in GM for fall in water level after taking the blocks overall

How the stability of a ship that has grounded at one point on the centerline is reduced the same way as in dry dock

How up thrust increase with fall in tide, increases the heeling moment and reduces the stability

Virtual loss in GM and drafts of a ship after tide has fallen by a stated amount

Point of grounding given initial drafts and drafts after grounding

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