Stability fr mates

7/29/2019 Stability fr mates

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Stability

Displacement

Mass, Weight, Force and Gravity

Mass is the amount of matter that is contained within a body.

The S.I. units of Mass are:

Grammes

Kilogrammes = 1000 grammes

Metric ton = 1000 kilogrammes

Force is the product of Mass and acceleration

The S.I. Unit of force is:

Kilogramme m/ s2

or Newton (N)

Example: The car hit the tree with a great force. What would be the great force, this may be

calculated by applying the above.

However the car may not have been speeding or increasing the acceleration but may

have been traveling at a constant speed, in that case we come to Momentum

Momentum is the product of Mass and velocity.

So in case of nautical terms the constant velocity of a ship is of great importance.

If a ship bangs against a jetty with some velocity then there will be damage to the jetty but if the

same ship reduces her speed or velocity then the impact damage will be considerably less.

Coming back to Mass and Weight

Weight and Mass are often confused in everyday life.

Weight is actually the resultant force that acts on a body having some mass.

Weight is thus a product of the mass of the body and the acceleration due to the earths gravity.

So, the S.I. Units of Weight should actually be kg m/s2

or Newton (N)

Here since the acceleration due to gravity is known as 9.81m/s2

Therefore we may write:

a mass of 1 kg having a weight of 1kg 9.81 m/s2

as 9.81 kg m/s2

Or simply 1 kgf, which is saying 9.81 N

Or conveniently since 9.81 is constant on the surface of the earth


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We may write the weight to be:

1 kgf, this is the force that is being exerted on a mass of 1 kg.

But if have to express this in Newton then it would be:

9.81 N

However again since the gravity factor is common the unit of Weight is also expressed as kg.

Thus:

1 tonne = 1 metric ton force = 1000 kgf

Or 1 tonne is a measure of 1 metric ton weight.

Moment is the product of force and distance

The S.I. Units of Moment is the Newton-metre (Nm)

Since we have seen that force is expressed in kgf or N and the S.I. Unit of distance kg is the

metre

Thus, 9.81 N = 1 kgf

And, 9810 N = 1000 kgf or 1 tonne

So, the unit generally used for large moments is the tonnes-metre

Pressure is the force that acts on a body to cause it to change in some form.

If it does not change and there is room for it to move then it does so.

Pressure is thrust or force per unit area and is expressed as:

Kilogrammes-force units per square metre or

Kilogrammes-force units per square centimetre or for larger pressure in tonnes-metre (t/m2

)

Density is defined as mass per unit volume or is expressed as unit of mass per unit of volume

Or grammes/ cubic centimetre (gms/cc or gms/cm3)

Fresh water has a density of 1 gm/cm3

or 1000kg/m3

Both are correct since:

1 kg is 1000 gms and 1 metre is 100 cm, since we are talking of cubic quantity 1 cubic metre

would be 100x100x100 cubic cm

So to equate it would be 1000 kg/m3

Or 1 t/m3

Thus the density of FW may be expressed as 1gm/cm3

or 1t/m3

Relative density is a factor without any unit.

Relative density is expressed as the density of the substance divided by the density of FW

Thus the RD of FW would be 1/1 or 1


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And the RD of SW would be 1.025/1 or 1.025

So basically it is expressed as the same numerical value but without a unit.

Archimedes found that when a body is immersed in water then the volume of water that

overflowed as a result of this immersion was equal to the volume of the body.

However the weight of the body plays an important part in this.

Although the volume of the displaced water is the same as that of the body the weight may not

be the same.

Let us assume that a log of wood of dimension 1metre by 1metre by 12 metres is taken (thus the

volume is 12 metre3, or 12 cbm), let the weight of the log be 8 t. (assumed density of the log at

0.667 t/m3)

This log when it is fully immersed (using external force) in a tank full of water will make some

water overflow, the quantity of water that would overflow would be 12 metre3, or 12 cbm

But the weight of this water would be 12 t at the density of 1t/cbm

So we see that the weight of the water is more than the weight of the fully immersed log of wood

and so the log will float.

But at what level?

Now if we remove the force that was holding the log underwater the log will bounce back to the

surface and only a portion of the log will remain underwater.

This amount will depend on the volume that it displaces and the weight of that displaced water.

Both have to be equal.

If we assume that only 67% of the log is immersed (12cbm x 0.67) then the volume of the waterdisplaced would also be 8 cbm and its weight would be 8 t and that was the weight of the log.

So the log would float in a state of equilibrium

However the log would still be capable to taking extra load, and we can place weight on the log

up to a maximum of 4t, any weight beyond that, and the log would sink.

Let us work out the same example with a bar of iron of the same dimensions, thus the volume

would be 12 cbm and at a density of iron at 7.86 gm/m3

the weight of the bar would be 94t.

The volume of water that this bar would displace would be 12 cbm but the weight would be only

12 t.

This being a much lesser figure than the weight of the iron bar, the iron bar would sink.

Can we now make this bar of iron float?

Yes, we can but we then need to flatten it out to a sheet of iron.

We then need to bend the four edges so that the sheet is turned into a open cardboard box.


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This will give the iron sheet a much larger volume, the empty space on top of the sheet would

also contribute to the volume but without adding to the weight (assuming the weight of air to be

negligible)

The sheet + air combination however has the same weight.

Now it will float on the water at a level as determined by the weight of water that it woulddisplace at that level.

Centre of Gravity is the point of a body at which all the mass of the body may be assumed to be

concentrated.

The force of gravity acts vertically downwards from this point with a force equal to the weight of

the body.

Basically the body would balance around this point.

The Centre of Gravity of a homogeneous body is at its geometrical centre.

Buoyancy and Centre of Buoyancy

So what makes the log or the open box iron sheet float.

The fact that they are on the surface of the water is due to the earths gravity or the weight of the

body.

That it does not sink is due to Archimedes principle.

We may also say that a force is pushing up the box. This force is dependent on the volume of the

box within the water as well as its weight.

This force is termed as the force of Buoyancy.

It will act in case of a uniformly loaded box shaped vessel through the centre of gravity of the

underwater volume of the box.

However if the loading is not uniform, by which we mean that if say only the fore part is loaded

with some other weight then obviously the underwater volume of the box will change and the

centre of buoyancy will pass through centre of gravity of the new underwater volume of the box.

Centre of Buoyancy can be defined as the geometrical centre of the underwater volume and the

point through which the total force of buoyancy may be considered to act vertically upwards

with a force equal to the weight of the water displaced by the body.

Reserve Buoyancy

We have seen the condition of the sheet of iron, which was turned, into an open cardboard box,

which floated very nicely on the surface of the water.

What happens if you now decide to tilt the box, depending on how high the edges are the water

will enter the enclosed area and the combination of sheet+ air will become sheet + air + some

water.


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This may make the box much more heavier than the weight of the volume of water displaced and

the box would sink.

Thus we require to put a water tight cover on the open box. This would ensure that no water

would enter the open space within the box and the sheet+ air combination would remain intact

and the box would float perpetually.

Thus what we have created is Reserve buoyancy.

A ship in a sea way floats on water, which may be calm and also may be rough.

When in a rough seaway the ship rides the waves, the waves support sometimes the ends of the

ship and then at the midway mark.

In either of the case the ship would have a tendency to sink to a lower level since the weight of

the ship and that of the water that it displaces would be different.

Thus the requirement for a ship to have reserve buoyancy, to meet any eventual sea condition

where more sheet + air combinations would be required to be brought into use.

Coefficient of fineness of water-plane area (Cw):

A ship floats on water. If at the water line the ship were to be cut off then the area at the water

level is known as the ships water plane

If we now divide this area of water plane with an imaginary rectangle having the length similar

to the maximum length of the water plane and breadth similar to the maximum breadth of the

water plane then this ration is termed as the coefficient of fineness of water plane area or Cw

Cw = Area of water-plane/Area of rectangle ABCD

Similarly if we know the Cw at a particular draft then we may find the actual water plane area of

the ship by measuring the maximum length and the greatest breadth.

Area of the water-plane = L x B x CW

The block coefficient of fineness of displacement (Cb):

In exactly the same manner as we obtained the water plane area, if we were to measure the

volume of the underwater part of the ship and divide this with the volume of a box having its


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length as that of the ship at that particular draft and breadth of the box as the maximum breadth

of the underwater volume, then we would arrive at a ratio.

This ratio is termed as the block coefficient of fineness of displacement,

or Cb

Cb at any particular draft is the ratio of the volume of displacement at that draft tothe the volume of a rectangular block having the same overall length, breadth and depth.

Knowing how the Cb was arrived at, we understand that for a box shaped vessel the ratio of

Cb be 1.

Also finer the lines of a ship the lower would be the Cb.

Thus a VLCC would be tending towards 1, whereas a slender yacht or a warship would be closer

to 0.5.

Again this value of Cb would depend on the draft of that particular ship, since at the load draft a

ship, even a small one appears quite box shaped but as the light draft is approached the fine

curvature of the ship is apparent.

For merchant ship, this value (depending upon draft) will range from about 0.500 to 0.850, with

some typical values as shown below:

ULCC 0.850 General Cargo ships 0.700

Oil tankers 0.800 Passenger ships 0.625

Bulk carriers 0.750 Container / Ro-Ro 0.575

Tugs 0.500

Cb = Volume of displacement / L x B x draft

Therefore as in the case of Cw, the underwater volume of a ship may be found at that particular

draft by:

Volume of displacement = L x B x draft x Cb

The value of Cb is used to determine the carrying capacity of a Life Boat.

In figure, the shaded portion represents the volume of the ships displacement at the draft

concerned, enclosed in a rectangular block having the same dimensions.


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SHIPS LIFEBOAT BLOCK COEFFICIENT

The problem of loading/ declaring the number of persons that it can carry in a Life Boat is that

we do not have any load line marks to guide us.

And even if there was one it would be difficult to embark looking at the load line mark.

So how is the number of passengers determined for a life boat.

The block coefficient of the boat is taken, in this case there is no need to launch the boat in the

water and note the draft.

Say the If we accept that the Cb of wooden lifeboat is 0.6

Therefore, volume of the entire lifeboat would be given by

L x B x draft x 0.6 cubic metres

Now that the volume of the lifeboat has been found, the next step is to determine the number of

persons that it would safely carry.

To determine this the following is used and result is the closest whole number so obtained.

Volume of the boat / volume of each person

(both in cubic metres)

Here the size of the person is generally not taken into consideration but the volume is adjusted

with the length of the boat.

For lifeboat lengths:

7.3 m or more the volume of a person is taken as 0.283

4.9 m the volume of a person is taken as 0.396

For intermediate boat lengths the values are interpolated.

Effect of change of density on draft when the displacement is constant

It has been already explained that the body floats on water at a particular level/ draft, as long as

the weight of the body is equal to the weight of the volume of water that is displaced by the

underwater volume of the body.

Thus the volume of the water depends on the underwater volume of the body, and

The weight of this volume of water depends on the density of the water.

Thus when a ship moves from water of a higher density to a water of lesser density, the weight ofthe water volume will become less.

To compensate for this weight loss an additional volume of water has to be displaced, this is only

possible if the underwater volume of the body is increased.

So the body/ ship will sink lower in the water of a lesser density, or the draft will increase.


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For box shaped vessels since the shape is uniform all the way from the top to the bottom, the

walls being all vertical, it is easy to calculate the sinkage or the rising of the vessel with the

change in the density.

The resulting effect on box shaped vessels will be:

New mass of water displaced = Old mass of water displacedNew volume x New density = Old volume x Old density

New Volume = Old density

Old Volume New density

But volume = L x B x draft

L x B x New draft = Old density

L x B x Old draft New density


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New draft = Old density

Old draft New density

The resulting effect on ship shape vessels will be:

New displacement = Old displacement

New volume/Old volume = Old density/New density

Due to the fact that the ships underwater shape is not like a box shaped vessel, the underwater

volume does not linearly change.

To find the change in draft of a ship shape, the FWA must be known. This is the number of mm

that a ships draft changes when passing from SW to FW.

FWA (in mm) = Displacement/(4 x TPC).

When the density of the water lies between these two (SW & FW) then the value (in mm) that

the ships draft changes when she enters the SW is called the Dock Water Allowance.

DWA (in mm) = FWA (1025DW density)/25

Keeping the draft constant, in effect means that no load has been added or removed.

But if the draft remains unchanged, even when the density of the water has changed implies that

some change to the displacement has occurred.

Let us consider:

A ship floats in the water at a certain draft, therefore the underwater volume of the ship displaces

an equal volume of water. This volume of water when multiplied with the density of the water

gives us the weight of the water, which again is equal to the weight of the whole ship.

Now if the density of the water is reduced (travelling from SW to FW), the following would

happen:

The weight of the displaced water would become less. And consequently to compensate for this

loss in weight an additional volume of water would have to be displaced.

To get an additional volume of water displaced means that the unde5rwater volume of the ship

has to increase.

If we do not want the underwater volume of the ship to increase then we have to remove weights

fr0om the ship.

Thus we see that to keep the draft constant, in a changing density scenario we have to either

lighten the ship or we have add more weights to the ship.

However, since the draft has not changed, the volume of water displaced also has remained

unchanged.

New vol. of water displaced = Old vol. of water displaced

New displacement = Old displacement

New density Old density


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New displacement = New density

Old displacement Old density

Tonnes per centimetre immersion (TPC)

This is the mass that must be added/ or removed to a ship in order that the mean draft of a ship

changes by a value of ONE centimetre.The figures that are given are for SALT WATER only and corrections have to be applied for

obtaining the values in FW and in other dock waters.

The TPC is not constant for the ship in all states of loading. The TPC changes as the underwater

form changes, thus the TPCs are given against the drafts.

For every draft there is a different TPC, the most notable changes are between the light draft and

the half way load draft, close to the summer draft the values changes are very small..

The Tonnes per Centimetre is therefore dependent on the underwater form of the ship and this is

determined by the water plane at the surface of the water.

So to calculate the TPC the water plane is essential.

TPC = (water plane area x density of water) / 100

water plane area (WPA) is in m2

Density is in t/m3.

Now let the mass w tones be loaded such that the draft increases by 1 cm & the ship now floats

at new Waterline WL

Since the draft increase is by 1 cm the mass loaded is equal to TPC.

Also as the displaced water quantity increases by some amount, this weight of extra water

displaced equals to TPC as well.

Mass = Volume x Density

= Area x 1/100 x 1.025 tonnes

= 1.025A/100 tonnes

TPCSW = 1.025 A/100

TPCFW = A/100

TPCDW = (RDDW x TPCSW )/1.025

Note: TPC is always stated for Salt Water unless otherwise specifically mentioned.

Effect of draft and density on TPC

Since the TPC as has been seen is dependent on the 2 factors:

1. Water plane areawhich determines the underwater volume of the ship

2. And the density of the water on which the ship is floating


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Thus if any of these two factors change the TPC will be affected.

For box shaped vessels the 1st

factor is not applicable since the shape is uniform all the way from

the top to the bottom, the walls are all vertical. The 2nd

factor of density needs to be attended to.

As the density increases the TPC also increases.

However for most ships being ship shaped meaning not box shaped, means that both the factorsaffect the TPC. The water plane area would change as the ship sinks deeper into the water or is

lightened. Also the density affects the TPC in the same way as for a box shaped vessel.

TPC Curves

TPC is calculated for a range of drafts extending beyond the light and loaded drafts.

This calculated TPC is then tabulated or plotted in a graphical form and these graphs are called

the TPC curves.

On board a ship the TPCs are given in both a tabulated form alongside the drafts as well as in a

graphical form.

Displacement Curves

Displacement of the ship in SW (1.025) at various drafts is given in both a tabular form as well

as in a graphical form.

A displacement curve is one from which the displacement of the ship at any particular draft can

be found, and vice versa.

Fresh Water Allowance (FWA)

In the basic principle of why a ship floats it is understood that the weight of the volume of water

displaced by a ship is equal to weight of the entire ship.

The volume of the displaced water is again equal to the volume of the underwater volume of theship.

Now when the weight of this displaced water is calculated we take the product of the volume of

the water and the density of the water.

So, if the density of the water changes, then the weight of the displaced water changes, the

weight of the ship remaining unchanged.

Thus to keep the ship floating something has to be adjusted and adjustment is in the underwater

volume of the ship.

So a ship floating in waters of different densities will do so at different levels.


volume of the ship.


So we can replace the word levelby the nautical word draft

Thus we may now define Fresh Water Allowance as the amount in millimetres by which a ships

MEAN DRAFT changes when she moves between SALT WATER and FRESH WATER.


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As a ship moves from SW to FW, the weight of the displaced water reducesRD of SW at

1.025 and FW at 1.000, so additional volume of water is required to float the ship, this means

that the underwater volume of the ship has to increase so the ship sinks lower to compensate the

above. So the draft increases.

In the same way if a ship moves from FW to SW, the weight of the displaced water would be

more than the weight of the ship, so the weight of the water has to be reduced, this may bereduced if the volume of the water is reduced, this again depends on the underwater volume of

the ship, so the underwater volume of the ship is reduced.

And so the ship rises a little and the draft of the ship reduces.

FWA (in mm) = Displacement/ 4x ( (water plane area x density of water) / 100)

Or FWA = Displacement / ( 4 x TPC)


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Effect of draft on FWA:

For box shaped vessel, FWA is the same at all drafts.

For ship shaped vessels, FWA increases with draft. As the draft increases, both the displacement

and the TPC increase, but the rate of change of displacement is higher than that of the TPC.

Derivation of the FWA formula

Consider a ship floating in SW at load Summer draft at waterline WL.

Let volume of SW displaced at this draft be V.

Now let W1L1 be the waterline for the ship when displacing the same mass of fresh water.

Let vbe the extra volume of water displaced in FW.

Total volume of fresh water displaced will be V + v.

Mass = Volume x density

Mass of SW displaced = 1025V

Mass of fresh water displaced = 1000 (V + v)

But mass of FW displaced = Mass of SW displaced.

1000(V + v) = 1025V

v = V/40

Assume that w is the mass of SW in volume v and W in volume V,

Then, replacing the factor as obtained above we get:

w = W/40

But w is a factor that is a product of the FWA and the TPC

Now since the FWA is in mm and the TPC is in cm, they both have to be converted to metres


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Thus:

W = (((FWA mm x 100)cm X TPC cm) / 100 ) metres

Simplifying we have:

w = (FWA x 100 x TPC) / 100 = W / 40

Or (FWA x TPC) = W / 40

But w = TPC x (FWA/10)

Hence W/40 = TPC (FWA/10) or FWA = W/(4 x TPC).

Where W = Loaded SW displacement in tonnes.

Mass = Volume x density (22*100*12)/100=242=W/40

W=9680

9680/4/12=2420/12=242/1.2=22




1000(V + v) = 1025V

v = V/40 TPC = (water plane area x density of water) / 100



w = W/40

Displacement = FWA x ( 4 x TPC)




Dock Water Allowance (DWA)

As a ship sails the seas the SW density is assumed to be constant at 1.025 gms/cc, however the

density of the SW is never the same everywhere, especially in partially enclosed salt water

bodies, this does not make much difference since the depth of the water is very substantial.

However when a ship enters a river from the sea the density of the water changes from SW to

FW, gradually. The density of the river may never attain pure FW conditions and may be in

between.

Thus the need to calculate this intermediate correction for the new density.


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Docks (enclosed port areas containing jetties) have water that is intermediate between SW

and FW, the water is brackish and may have a density of 1.010 gms/ cc.

Thus Dock Water Allowance is similar to FWA and is the amount in millimetres by which the

ships mean draft changes when a vessel moves between a salt water and dock water.

Dock water is the water whose density is neither that of fresh water or salt water butin-between the two. RD between 1.000 and 1.025.

To get the correction in millimetres the formula that may be used is:

(Please note however that the DWA allowed for should be for the minimum density that will be

encountered by the ship while proceeding to the dockthis as a safety factor)

DWA = (FWA (1025density of dock water)) / 25

Buoyancy

The Laws Of Buoyancy

Floating objects possess the property of buoyancy.

A floating body displaces a volume of water equal in weight to the weight of the body.

A body immersed (or floating) in water is buoyed up by a force equal to the weight of the water

displaced.

Centre of buoyancy

C of B can be defined as the geometrical centre of the underwater volume and the point through

which the total force of buoyancy may be considered to act vertically upwards with a force equalto the weight of the water displaced by the body.

For the purposes of freeboard computation, ships are divided into type A and type B.

Type A ships

A type A ship is one which:

Is designed to carry only liquid cargoes in bulk;

Has a high integrity of the exposed deck with only small access openings to cargo compartments,

closed by watertight gasketed covers of steel or equivalent material; and

Has low permeability of loaded cargo compartments.


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Type B ships

All ships, which do not come within the provisions regarding type A ships in above

paragraphs, are considered as type B ships.

Type B ships, which have hatchways fitted with hatch covers, are assigned freeboards based

upon the values given in the rules.

Conditions of equilibrium

The condition of equilibrium after flooding shall be regarded as satisfactory provided:

The final waterline after flooding, taking into account sinkage, heel and trim, is below the lower

edge of any opening through which progressive down flooding may take place.

Such openings shall include air pipes, ventilators and openings which are closed by means of

weather tight doors or hatch covers, and may exclude those openings closed by means of

manhole covers and flush scuttles, cargo hatch covers, remotely operated sliding watertight

doors, and sidescuttles of the non-opening type.

However, in the case of doors separating a main machinery space from a steering gear

compartment, watertight doors may be of a hinged, quick-acting type kept closed at sea, whilst

not in use, provided also that the lower sill of such doors is above the summer load waterline.

If pipes, ducts or tunnels are situated within the assumed extent of damage penetration,

arrangements shall be made so that progressive flooding cannot thereby extend to compartments

other than those assumed to be floodable in the calculation for each case of damage.

The angle of heel due to unsymmetrical flooding does not exceed 15deg. If no part of the deck is

immersed, an angle of heel of up to 17deg. may be accepted.

The metacentric height in the flooded condition is positive.

When any part of the deck outside the compartment assumed flooded in a particular case of

damage is immersed, or in any case where the margin of stability in the flooded condition may

be considered doubtful, the residual stability is to be investigated.

It may be regarded as sufficient if the righting lever curve has a minimum range of 20deg.

beyond the position of equilibrium with a maximum righting lever of at least 0.1 m within this

range. The area under the righting lever curve within this range shall be not less than 0.0175

m. rad.

The Administration shall give consideration to the potential hazard presented by protected or

unprotected openings, which may become temporarily immersed within the range of residual

stability.

The Administration is satisfied that the stability is sufficient during intermediate stages of

flooding.

EXAMPLE OF GRAVITY -VS- BUOYANCY


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1 ton of steel 1 ton of steel

If the cube of steel is placed in water it sinks. There is not enough displaced volume for the

forces ofbuoyancy to act upon. If the ships hull is placed in the water it will float. The larger

volume of the ships hull allows the forces of buoyancy to support the hulls weight.

The ships hull will sink to a draft where the forces of buoyancy and the forces of gravity are

equal.

Displacement

The weight of the volume of water that is displaced by the underwater portion of the hull is equal

to the weight of the ship. This is known as a ships displacement.

The unit of measurement for displacement is the Metric Tonne.

Gravity

The force of gravity acts vertically downward through the ships center of gravity. The

magnitude of the force depends on the ships total weight.

Units Of Measure

Force: A push or pull that tends to produce motion or a change in motion. Units: Newton, etc.

Parallel forces may be mathematically summed to produce one Net Force considered to act

through one point.

Weight: The force of gravity acting on a body. This force acts towards the center of the earth.

Units: kilograms, etc.

Moment: The tendency of a force to produce a rotation about a pivot point. This works like a

torque wrench acting on a bolt. Units: Newton meters, etc.

Moment = Weight x Lever Arm

Volume = Length x Breadth x Height

Volume: The number of cubic units in an object.

Units: cubic metres (cbm), etc. The volume of any compartment onboard a ship can be found

using the equation:

Salt Water = 1.025 gms/cc

Fresh Water = 1.00 gms/cc

Diesel Fuel = 0.92 gms/cc


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Calculating The Weight Of Flooding Water

A compartment has the following dimensions:

Length = 20 M

Breadth = 20 M

Height = 8 M

The compartment is now flooded with salt water to a depth of 6 M

1. First, calculate the volume of water that has been added to the compartment.

Volume = Length x Breadth x Depth of Flooding Water

= 20 M x 20 M x 6 M

= 2400 cbm

2. Second, multiply the volume of water by its specific gravity.


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Stability Reference Points

M - Metacentre

G - Center of Gravity

B - Center of BuoyancyK - Keel

K - Keel: The base line reference point from which all other reference point measurements are

compared.

B - Center ofBuoyancy: The geometric center of the ships underwater hull body. It is the point

at which all the forces of buoyancy may be considered to act in a vertically upward direction.


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The Center of Buoyancy will move as the shape of the underwater portion of the hull body

changes. When the ship rolls to starboard, B moves to starboard, and when the ship rolls to

port, B moves to port.

When the ships hull is made heavier, the drafts increase as the ship sits deeper in the water. B

will move up.


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When the ships hull is lightened, the drafts decrease as the ship sits shallower in the water. B

will move down.

The Center of Buoyancy moves in the same direction as the ships waterline.

G - Center of Gravity: The point at which all forces of gravity acting on the ship can be

considered to act. G is the center of mass of the vessel. The position of G is dependent uponthe distribution of weights within the ship. As the distribution of weights is altered, the position

of G will react as follows:


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1. G moves towards a weight addition

2. G moves away from a weight removal

3. G moves in the same direction as a weight shift

M - Metacenter: As the ship is inclined through small angles of heel, the lines of buoyant force

intersect at a point called the metacenter.

As the ship is inclined, the center of buoyancy moves in an arc as it continues to seek

the geometric center of the underwater hull body. This arc describes the metacentric

radius.

As the ship continues to heel in excess of 7-10 degrees, the metacenter will move as shown.


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The position of the metacenter is a function of the position of the center of buoyancy, thus a

function of the displacement of the ship.

The position of M moves as follows:

As the Center of Buoyancy moves up, the Metacenter moves down.

As the Center of Buoyancy moves down, the Metacenter move

Fresh Water Al lowance

Fresh Water Allowance (FWA)

In the basic principle of why a ship floats it is understood that the weight of the volume of water

displaced by a ship is equal to weight of the entire ship.

The volume of the displaced water is again equal to the volume of the underwater volume of the

ship.

Now when the weight of this displaced water is calculated we take the product of the volume of

the water and the density of the water.

So, if the density of the water changes, then the weight of the displaced water changes, the

weight of the ship remaining unchanged.


volume of the ship.



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Let us take the example of a ship with a weight of 10000 MT, let this ship float at a certain level

(assume the water level is at the mid level of the ship)

Then the underwater part of the ship would be displacing a volume of water that would be equal

to the volume of the underwater part of the ship.

Also the weight of this water would have to be equal to the entire weight of the ship.So we have,

Displaced water = underwater part (volume) of the ship

Weight of this displaced water = entire weight of ship

We know,

Weight of this displaced water = volume of displaced water x specific gravity of the water

So now if the specific gravity of the water changes, then to keep the weight of the water constant

the volume of the displaced water has to changeand this is the reason that the ship either sinks

lower or rises up when traversing from FW to SW and vice versa.


volume of the ship.


So we can replace the word level by the nautical word draft

Thus we may now define Fresh Water Allowance as the amount in millimetres by which a ships

MEAN DRAFT changes when she moves between SALT WATER and FRESH WATER and

vice versa

As a ship moves from SW to FW, the weight of the displaced water reducesRD of SW at

1.025 and FW at 1.000, so additional volume of water is required to float the ship, this means

that the underwater volume of the ship has to increase so the ship sinks lower to compensate the

above. So the draft increases.

In the same way if a ship moves from FW to SW, the weight of the displaced water would be

more than the weight of the ship, so the weight of the water has to be reduced, this may be

reduced if the volume of the water is reduced, this again depends on the underwater volume of

the ship, so the underwater volume of the ship is reduced.

And so the ship rises a little and the draft of the ship reduces.

FWA (in mm) = Displacement/ 4x ( (water plane area x density of water) / 100)

Or FWA = Displacement / ( 4 x TPC)

Effect of draft on FWA

For box shaped vessel, FWA is the same at all drafts.

For ship shaped vessels, FWA increases with draft. As the draft increases, both the displacement

and the TPC increase, but the rate of change of displacement is higher than that of the TPC.


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Derivation of the FWA formula

Consider a ship floating in SW at load Summer draft at waterline WL.

Let volume of SW displaced at this draft be V.

Now let W1L1 be the waterline for the ship when displacing the same mass of fresh water.

Let v be the extra volume of water displaced in FW.

Total volume of fresh water displaced will be V + v.

Mass = Volume x density




1000(V + v) = 1025V

v = V/40



w = W/40

But w is a factor that is a product of the FWA and the TPC

Now since the FWA is in mm and the TPC is in cm, they both have to be converted to metres

Thus:

W = (((FWA mm x 100) cm X TPC cm) / 100) metres

Simplifying we have:

w = (FWA x 100 x TPC) / 100 = W / 40

Or (FWA x TPC) = W / 40




Dock Water Allowance (DWA)

As a ship sails the seas the SW density is assumed to be constant at 1.025 gms/cc, however thedensity of the SW is never the same everywhere, especially in partially enclosed salt water

bodies, this does not make much difference since the depth of the water is very substantial.

However when a ship enters a river from the sea the density of the water changes from SW to

FW, gradually. The density of the river may never attain pure FW conditions and may be in

between.

Thus the need to calculate this intermediate correction for the new density.


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Docks (enclosed port areas containing jetties) have water that is intermediate between SW

and FW, the water is brackish and may have a density of 1.010 gms/ cc.

Thus Dock Water Allowance is similar to FWA and is the amount in millimetres by which the

ships mean draft changes when a vessel moves between a salt water and dock water.

Dock water is the water whose density is neither that of fresh water or salt water but in-betweenthe two. RD between 1.000 and 1.025.

To get the correction in millimetres the formula that may be used is:

(Please note however that the DWA allowed for should be for the minimum density that will be

encountered by the ship while proceeding to the dockthis as a safety factor)

DWA = FWA (1025density of dock water)

25

StabilityMovement of the Centre of Gravi ty

Centre of gravity

It is the point of a body at which all the mass of the body may be assumed to be concentrated.

The force of gravity acts vertically downwards from this point with a force equal to the weight of

the body.

Basically the body would balance around this point.

The COG of a homogeneous body is at its geometrical centre.

Effect of removing or discharging mass

Consider a rectangular plank as shown. The effects of adding or removing weights would be as

shown:

Now cut the length of plank of mass w kg whose CGis d mtrs away from CG of the plank.

Note that a resultant moment of w x d kg m has been created in an anti-clockwise direction

about G.


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The CG of the new plank shifts from G to G1.

The new mass (W-w) kg now creates a tilting moment of (W-w) x GG1 about G.

Since both are referring to the same moment,

(W-w) x GG1 = w x d

GG1 = (w x d)/(W-w)

CONCLUSION: When a weight is removed from a body, the CG shifts directly away from the

CG of the mass removed, and the distance it moves is given by:

GG1 = (w x d)/Final mass metres

Where, GG1 is the shift of CG

w is the mass removed

d is the distance between the CG of the mass removed and the CG of the body.

Effect of adding or loading massEquating the tilting moments created due to the added weight, which must again be equal:

(W + w) x GG1 = w x d

GG1 = (w x d)/(W + w)

GG1 = (w x d)/ (Final mass) metres

Application to ships

DISCHARGING WEIGHTS:

GG1 = (w x d) metres

(Final displacement)

LOADING WEIGHTS


(Final displacement)


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Shifting Weights


(Displacement)

Vertical Weight Shifts

Shifting weight vertically, no matter where onboard it is, will always cause the ships center of

gravity to move in the same direction as the weight shift.

To calculate the height of the ships center of gravity after a vertical weight shift, the following

equation is used:

KG1 = ((W0 x KG0) +/- (w x kg)) / F

KGO = The original height of the ships center of gravity (M)

o = The ships displacement prior to shifting weight (MT)

w = The amount of weight shifted (MT)

kg = The vertical distance the weight was shifted (M)

F = The ships displacement after shifting the weight (MT)

(+) When the weight is shifted up use (+)

(-) When the weight is shifted down use (-)

Example Problem

10 MT of cargo is shifted up 3 M. O is 3500 MT and KGo is 6 M. What is the new height of the

ships center of gravity (KG1)?

KG1= ((o x KGo) +/- (w x kg)) / F

KG1 = ((3500 x 6) + (10 x 3)) / 3500

KG1 = 6.009 M


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Vertical Weight Additions/Removals

When weight is added or removed to/from a ship, the vertical shift in the center of gravity is

found using the same equation.

KG1= ((o x KGo) +/- (w x kg)) / F

KGO = The original height of the ships center of gravity (M)

O= Ships displacement prior to adding/removing weight (MT)

w = The amount of weight added or removed (MT)

kg = The height of the center of gravity of the added/removed weight above the keel (M)

F = The ships displacement after adding/removing the weight

(+) When the weight is added use (+)

(-) When the weight is removed use (-)

Example Problem

A 30 MT crate is added 10 M above the keel. o is 3500 MT and KG0 is 6 M. What is the new

height of the ships center of gravity (KG1)?

KG1= ((o x KGo) +/- (w x kg)) / F

KG1 = ((3500 x 6) + (30 x 10)) / 3530KG1 = 6.034 M

Horizontal Weight Shifts

Shifting weight horizontally, no matter where onboard it is, will always cause the ships center of

gravity to move in the same direction as the weight shift.


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NOTE: A weight shift causing the ships center of gravity to move off centerline will always

reduce the stability of the ship.

To calculate the horizontal movement of the ships center of gravity, the following equation is

used:

GG2= (w x d) / F

w= The amount of weight shifted (MT)

d = The horizontal distance the weight is shifted (M)

F = The ships displacement after the weight is shifted (MT)

Example Problem

A 50 MT weight is shifted 10 M to starboard. O is 32000 MT.

What is the change in the center of gravity (GG2)?

GG2= (w x d) / F

GG2 = (50 x 10) / 32000

GG2 = 0.01562 M

Horizontal Weight Additions/Removals

When an off-center weight is added or removed to/from a ship, the ships center of gravity will

move off centerline, the ship will develop a list.


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To calculate the horizontal movement of the ships center of gravity after adding/removing an

off-center weight, the same equation is used:

GG2= (w x d) / F

w= The amount of weight added/removed (MT)

d = The distance from the center of gravity of the weight to the ships centerline (M)

F= the ships displacement after the weight is shifted (MT)

Example Problem

50 MT of cargo is loaded onto the Tween deck, 10 M from centerline. O is 48000 MT. What is

the change in the center of gravity (GG2)?

GG2= (w x d) / F

GG2 = (50 x 10) / 48000

GG2 = 0.0104 M


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Effect of suspended weights

The CG of a body is the point through which the force of gravity may be considered to act

vertically downwards.

For a suspended weight, whether the vessel is upright or inclined, the point through which the

force a gravity may be considered to act vertically downwards is g1, the POINT OFSUSPENSION.


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Conclusions

The CG of a body will move directly TOWARDS the CG of any weight ADDED.

The CG of a body will move directly AWAY from the CG of any weight DISCHARGED.

The CG of a body will move PARALLEL to the shift of the CG of any weight MOVED within

the body.

The shift of the CG of the body in each case is given by the following formula:

GG1 = w x d metres

W

where w = weight added, removed or shifted.

W = final mass of the body

d = distance between the CG if weight added or removed, or the distance by which the

weight is shifted.

When a weight is SUSPENDED, its CG is considered to be at the POINT OF SUSPENSION.

Angle of Loll

Angle of loll

Consider the following vessel in unstable equilibrium condition.


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As the angle of heel increases, the CB moves out further until it is directly under G. The

capsizing moment disappears now and this angle of heel at which this condition occurs is called

the angle of loll.


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The ship now moves around the angle of loll, but if the CB does not move out far enough to

move directly under G, then the vessel will capsize.

If the heel increases beyond the angle of loll, the CB moves out further to the low side and the

ship now moves around this angle.


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The angle of loll can be on either side depending upon the external inclining force, such as thewind and the waves.

However, there is always the threat of the G rising above the M and this will create a situation of

unstable equilibrium, thereby capsizing the ship.

List Caused By Negative Gm

When a ships center of gravity moves vertically upwards and slightly above the Metacenter, the

ship will develop a list (or possibly capsize.) The vessel may also flop over, developing the

same list to the other side.

Possible Causes

1. Removal of low weight


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2. Addition of high weight (ice)

3. Moving weight upward

4. Free Surface Effect

5. Free Flow Effect (if present)

How to Recognize

1. Vessel will not remain upright and will assume a list to either port or starboard.

2. Vessel flops to port or starboard.

3. Vessel will have a very long, slow roll period about the angle of list.

4. A small GM is known to exist plus any of the above.

Corrective Measures

1. Eliminate Free Surface and Free Flow Effects (if present)

2. Add low weight symmetrically about centerline.

3. Remove high weight symmetrically.

4. Shift weight down symmetrically.

List Caused By Off-Center Weight And Negative Gm

The vessels stability is reduced by both an increase in the height of the center of gravity and

movement from centerline. A negative GM condition exists, represented by the uncorrected

curve. An off-center weight, represented by the cosine curve, is added and a larger list develops.

Possible Causes

1. A combination of the previous causes of list.


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How to Recognize

1. Vessel will assume a permanent list either port or starboard (vessel will not flop).

2. Very slow roll period about this permanent list.

3. The known off-center weight isnt proportional to the ships list.

Corrective Measures

Correct Negative GM first.

a. Eliminate Free Surface and Free Flow Effects (if present)

b. Shift weight down, add weight low, or jettison weight high.

Correct for Gravity Off Centerline

a. Add weight to higher side

b. Remove weight from lower side

c. Shift weight to higher side

*** ALWAYS correct Negative GM prior to shifting weights transversely ***

Curves of Statical Stabil i ty

Load Line requirements for minimum stability conditions

The area under the GZ curve shall not be less than

0.55 m-rad up to an angle of 30

0.09 m-rad up to an angle of either 40 or the lesser angle at which the lower edges of any

openings which can not be closed weather-tight are immersed

0.03 m-rad between the angles of heel of 30 and 40 or such lesser angle as mentioned above

The Righting Lever (GZ) shall be at least 0.20m at an angle of heel equal to or greater than 30

The maximum GZ shall occur at an angle of heel of not less than 30

Initial transverse metacentric height shall not be less than 0.15m. For ship carrying timber deckcargo complying with (a), this may be reduced to not less than 0.05 metres.

Curve Of Statical Stability

Graph where GZ is plotted against the angle of heel.

Drawn for each voyage condition by the ships officer.

This curve is for a particular displacement and KG.


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From this curve it is possible to ascertain the following:

Initial metacentric heightpoint of intersection of the tangent drawn to the curve at the initial

point and a vertical through the angle of heel of 57.3 (1 radian).

Angle of contraflexurethe angle of heel up to which the rate of increase of GZ with heel is

increasing. Though the GZ may increase further, the rate of increase of GZ begins to decrease atthis angle.

The range of stabilitywhere all GZ values are positive.

The maximum GZ lever & the angle at which it occurs.

The angle of vanishing stabilitybeyond which the vessel will capsize.

The area of negative stability

The moment of statical stability at any given angle of heel (GZ x Displacement of the ship).

The moment of dynamical stabilitywork done in heeling the ship to a particular angle.

Dynamical stability at = W x A (in t-m-rad)

W = Displacement (in tonnes)

A = area between the curve and the baseline up to the given angle of heel (in metre-

radians).


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GZ Cross Curves of Stability

To draw the curve of statical stability, we need GZ values for various angles of heel.

For this we use the GZ cross curves of stability.

These curves are provided for an assumed KG, tabulating GZ values for various displacements

and angles of list.

Called cross curves because the various curves actually cross each other.

Since the curves are plotted for an assumed KG, if the actual KG differs from this a correction

(GG1Sine) needs to be applied.

This correction is positive if the actual KG is less than the assumed KG and vice-versa.

After obtaining the GZ values at various angles, the curve of statical stability is prepared.

KN Cross Curves of Stability

Same as the GZ cross curves and also used to get the GZ values for making the curve

of statical stability.

The only difference being that here the KG is assumed to be ZERO.

This solves the problem of a sometimes positive and sometimes negative correction, as now the

correction is always subtracted.

GZ = KNKG Sine

List

Definitions

Roll: The action of a vessel involving a recurrent motion, usually caused by wave

action.

Heel: Semi-permanent angle of inclination caused by external forces, such as high-speed turns,

beam winds, and seas.

List: Permanent angle of inclination, caused by:

1. Ships Center of Gravity transversely shifted from centerline.

2. Negative Metacentric Height (-GM)

3. Combination of Gravity off-centerline andGM

Moment To Heel 1o

Equation


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When a ship experiences an Inclining Moment (IM) the vessel will list or heel until the Righting

Moment (RM) is equal to the Inclining Moment (RM = IM). The Inclining Moment is simply a

force acting through some distance.

IM = w x d

This is only true when the ship has a negligible heel or list.

As the vessel inclines, the distance between the forces changes.

A relationship can be developed to solve for the distance between forces for all angles of heel.

Using an expanded drawing of the triangle from the above diagram:


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Using the cosine equation to solve for the distance X:

X = d x cos

Therefore:

IM = w x d x cos

A Righting Moment is created by the ship to keep itself upright. In this case, the force is equal to

the ships displacement (WF) and the distance is the ships righting arm (GZ) at each particular

angle of heel.

RM = WF x GZ

The Righting Arm (GZ) changes with inclination of the ship. Using the relationship derived for

small angles of heel:

GZ = GM x sin

NOTE: This relationship holds true for angles less than 7-10

Therefore:

RM = GM x WF x sin

The initial premise was that RM = IM:

W x d x cos = GM x WF x sin

Transferring cosine to the right:

(sin / cos ) = tan

w x d = GM x WF x tan

Choosing a specific angle, the moment (w x d) required to create that list or heel can be found.

Using 1o:

tan 1o

= 0.01746

Therefore:

MH 1 = GM x WF x 0.01746

This formula is valid for angles less than 10o

due to movement of the metacenter. To check this

formula for all inclinations less than 10o, a comparison between the MH10

oand 10 times

MH1o

is made.

MH 10 = GM x WF x tan 10 -vs- 10 x (MH 1 = GM x WF x 0.01746)


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MH 10 = GM x WF x (0.01746)

And

10 x (MH 1) = 10 x GM x WF x (0.01746)

There is a 0.0017 difference over the 10range. This error is negligible. The list equation cannow be used.

LIST = (w x d) / MH 1


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Example

Your ship has a 1.5o

list to starboard. There are 50 MT of cargo placed on the starboard side. The

stevedores want to know how far to transfer the cargo to correct the list.

Step 1: Calculate MH1o:

MH 1 = GM x WF x (0.01746)

MH 1 = 0.8 M x 3500 MT x (0.01746)

MH 1 = 48.8 M

Step 2: Use the list equation to solve for distance:

List = (w x d) / MH 1

Or, 1.5 = (50 MT x d) / 48.8 M

d = (1.5 x 48.8) / 50 = 1.464 M


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Example

Your ship has a 2 list to port. The CO wants it corrected. There are 15 cbm of fuel in the port

wing tank (sp.gr. 0.94). The starboard wing tank is empty. Correct the list using the fuel and a set

of 5 cargo pallets (8 MT each). The cargo pallets may only be moved 5 M to starboard before

hitting the bulkhead. How long will it take to correct the list? Pump capacity is 40 cbm per hour.

WO = 12500 MT

KM = 7.1 M

KG = 6.02 M

Step 1: Calculate MH1:

MH 1 = GM x WF x (0.01746)

MH 1 = (7.1 6.02) x 12500 x (0.01746)

MH 1 = 235.7


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Step 2: Calculate the amount of list corrected by shifting fuel:

Weight of fuel = 15 x 0.94 = 14.1 MT

List = (w x d) / MH 1

Or,

List = (14.1 x 11) / 235.7 = 0.66

Step 3: So far, we have corrected 0.66o

of the 2o

list. Using the pallets, we will correct for the

remaining 1.34o

list.

List = (w x d) / MH 1 or

d = (1.34 x 235.7) / 40 = 7.9 M

Step 4: Finally, calculate how long it takes to transfer 15 cbm of fuel when the pump capacity is

40cbm/ hour.

Time = (15 cbm / 40 cbm/h) = 0.375 hr x 60 = 22.5 minutes

Assuming it takes less than 22.5 minutes to move the 5 pallets, this is the time required to correct

the list.

Important:

1. When attempting problems on List, first find out the GM of the vessel (if the KG has to

be calculated then do so) if it has not been stated.

2. If there are more than one shifting/ loading/ discharging involved then tabulate the

moments and get the final moment (w x d) to either port or to starboard.


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Example:

A ship of 8000 tonnes displacement has KM = 8.7 m, and KG = 7.6 m. The following weights

are then loaded and discharged:

a. Load 250 tonnes cargo KG 6.1m and centre of gravity 7.6m, to starboard of the centre

line.b. Load 300 tonnes fuel oil KG 0.6m, and centre of gravity 6.1m, to port of the centre line.

c. Discharge 50 tonnes of ballast KG 1.2m, and centre of gravity 4.6m, to port of the centre

line.

Find the final list.

Weight KG Moment about Keel (V-M)

Orig. Disp. 8000 7.6 60800

Load 250 6.1 1525

Load 300 0.6 180

Total 8550 62505

Disch. -50 1.2 -60

Final Disp. 8500 62445

Final KG = Final Moment / Final displacement = 62445 / 8500

KG = 7.34

KM = 8.7

Therefore, GM = 1.36

w dListingmoment

Port Stbd

250 7.6 1900

50 4.6 230

300 6.1 1830

From above we have Port: 1830 and Stbd: 2130

Therefore the final listing moment (w x d) = 300 to stbd.

Now, MH 1 = GM x WF x 0.01746 = 1.36 x 8500 x 0.01746 = 201.8376

List = (w x d) / MH 1 = 300 / 201.8376 = 1.49 to stbd.


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Inclining Experiment

The inclining experiment is completed upon commissioning of the vessel.

It is performed to obtain accurately the vertical height of the ships center of gravity above the

keel (KG).

Procedures:

The shipyard at which the inclining experiment is to be performed will issue a memorandum to

the ship outlining the necessary work to be done by ships force and by the yard to prepare the

ship for inclining.

1. Liquid load will be in accordance with the memorandum.

2. Inventory of all consumables to be made by ships crew and inclining party.

3. Inclining weights are placed on centerline.

4. Freeboard is measured, and a photo of the drafts is taken.

5. Salinity of saltwater is measured.

8. Pendulums set up forward, midships, and aft.

9. Weights are moved off-centerline.

10.Inclination of the ship measured.

Measurements are taken for several weight movements both port and starboard. The Naval

Architect then uses the following equation:

Where:

w = Inclining Weights (LT)


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d = Athwartships Distance Weights Were Moved (FT)

WF = Displacement of Ship (LT, with Inclining Weights)

tan = Movement of Pendulum Length of Pendulum

The inclining experiment measures GM accurately, and since the ships drafts are known, KM

can be found, KG is then found using KG = KM - GM.

Free Surface Effect

Liquid that only partially fills a compartment is said to have a free surface that tends to remain

horizontal (parallel to the waterline). When the ship is inclined, the liquid flows to the lower side

(in the direction of inclination), increasing the inclining moment.

Background:

If the tank contains a solid weight, and the ship is inclined, the center of buoyancy shifts in the

direction of the inclination and righting arms (GZ) are formed.


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Replacing the solid with a liquid of the same weight, when the ship is inclined, the surface of the

liquid remains horizontal. This results in a transfer of a wedge of water, which is equivalent to

a horizontal shift of weight, causing gravity to shift from G0 to G2.

The wedge of water transferred increases as the angle of inclination increases, therefore, the

center of gravity shifts a different amount for each inclination.


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Due to the horizontal shift of the center of gravity, the righting arm is now G2Z2. To determine

the effect on stability, a vertical line is projected upward through G2 (see below). Where this line

crosses the ships centerline is labeled G3. The righting arm G3Z3 is the same length as the

righting arm G2Z2. Therefore, moving the ships center of gravity to position G2 or G3 yields the

same effect on stability. Movement from G0 to G3 is referred to as a Virtual Rise of the center

gravity.

To calculate the virtual rise in the center of gravity due to the Free Surface Effect, use the

following equation:


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B = The breadth (width) of the compartment

L = The length of the compartment

WF = The ships final displacement (after flooding water added)

Factors Effecting Free Surface Effect

Pocketing

Free Surface Effect can be reduced, to some extent, by creating pocketing. Pocketing occurs

when the surface of the liquid contacts the top or bottom of the tank, reducing the breadth (B) of

the free surface area.

Since the effects of pocketing can not be calculated, it is an indeterminate safety factor. The Free

Surface correction will therefore indicate less overall stability than actually exists.


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Surface Permeability

Impermeable objects (engines, pumps, piping systems, etc) inside a flooded space project

through and above the liquid surface. These objects inhibit the moving water and the shifting of

the wedge may or may not be complete, thus reducing Free Surface Effect. The impermeable

objects also occupy volume, reducing the amount of flooding water (movable weight) that can

fill the space.


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Swash Bulkheads (Baffle Plates)

In addition to some structural support, these bulkheads are designed to reduce Free Surface

Effect. They are longitudinal bulkheads that hinder, but do not prevent, the flow of liquid from

side to side as the ship rolls or heels. They are found in tanks, voids, double bottoms, bilges, etc.

Sluice Valves

Free flow (Sluice) valves on tankers allow opposing tanks to be cross-connected. When large,

partially filled tanks are connected, Free Surface Effect increases, and the vessel becomes less

stable.

Conditions of Free Surface Effect

1. FSE increases with increased length and width of compartment

2. FSE increases when displacement decreases (de-ballasting)

3. FSE is independent of the depth of the liquid


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Example Problem

The firemain ruptures, flooding a compartment with 0.91 metre of saltwater. Displacement prior

to flooding was 4485 MT. The dimensions of the space are: L=9.14m B=12.8m

Calculate the weight added by the flooding water:

2. Calculate the new displacement:

3. Calculate the virtual rise in G due to Free Surface Effect:

Free Flow Effect

Free Flow Effect occurs when the ships hull is ruptured, allowing sea water to flow in and out asthe ship rolls. This continuous weight addition and removal causes a horizontal shift in the center

of gravity, which then equates to another virtual rise in the center gravity.

Three conditions must exist for Free Flow Effect:

The compartment must be open to the sea.

The compartment must be partially flooded.

The compartment must be off centerline or asymmetrical about centerline.

When the vessel below is inclined, it experiences a horizontal weight shift due to the Free

Surface Effect. The center of gravity shifts from G0 to G2. The center of gravity is shifted furtherfrom centerline due to the flooding weight addition/removal as the ship rolls. This reduces the

righting arm from G2Z2 to G4Z4. By extending the line of gravitational force up to the centerline,

position G5 is found. This increase from G3 to G5 is the virtual rise of gravity due to the Free

Flow Effect.


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The virtual rise in the center of gravity due to the Free Flow Effect (G3G5) is found using the

equation:

B = Breadth (width) of the compartment

L = Length of the compartment

Y = The distance from the center of gravity of the compartment to the Centerline of the ship

WF = The ships displacement following damage

The factors which minimize Free Surface Effect (pocketing, surface permeability, swash

bulkheads, etc) will also minimize Free Flow Effect. There is one additional factor associated

with Free Flow: the size of the hole in the ship.

How the size of the hole affects Free Flow is not something that can be calculated. The FCE

equation does not account for the hole. Basically, if the hole is small, less water will

be added/removed to/from the ship. The larger the hole, the closer Free Flow Effect is

to its calculated value.

Example Problem

A vessel has a hole in the starboard side of a compartment. Displacement prior to damage was

3700 MT. Flooding depth is 1.52 m. Calculate the total virtual rise in the center of gravity (FSE

+ FCE). Compartment length is 9.14 and the breadth is 8.23m. The compartment extends from

the Starboard shipside to a distance of 2.74 m beyond the centre line on the port side.


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1. Calculate the weight added due to flooding water:

2.

Calculate the ships final displacement:

3. Calculate the virtual rise in G due to Free Surface Effect:

4. Determine the distance Y forcalculating the Free Flow Effect:

The center of the compartment is 4.11 m from the inboard bulkhead, and the ships centerline is

2.74 m from the inboard bulkhead.

5. Calculate the virtual rise in G due to Free Flow Effect:

6. Calculate the total virtual rise in the center of gravity:

GG (virtual) = FSE + FCE = 0.11 + 0.038 = 0.148 m


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StabilityTrim

Trim

For a rectangular box shaped vessel, when a weight is added on to one side the vessel would list

to that side.

If however the weight is added either behind or ahead of the of the midship area but within the

centre line partition of the ship then the vessel would get tilted either forward or aft.

This tilting is known as TRIM

Thus trim is the longitudinal equivalent of list.

However there is a fundamental difference in the way the List and Trim are noted. List is as weknow expressed in degrees, trim may be measured also in degrees but the expression is in Feet or

Metres.

Thus Trim may be defined as the difference between the draft at the fore perpendicular and the

draft at the aft perpendicular.

Unlike list which is stated as Port or Starboard, Trim is stated as Positive or Negativemore

usually as Trimmeaning trimmed by stern, taken as positive. And Trim by headmeaning

negative and that the draft ahead is more than the draft astern.

Moment to Change Trim 1 cm (MCTC)

Now we have seen that to change the Trim we need to move weights in the fore and aft line ofthe ship.

This then brings about a moment, and the moment required to change the trim by 1 cm is given

by:

MCTC = (W x GMl) / 100 x L

Where W is the displacement of the vessel in tonnes

GMl is the longitudinal metacentric height (m)

L is the Length between perpendiculars (m)

Centre of Floatation

This is the imaginary point where the ship pivots. It is the centre of gravity of the water plane

area. The centre of Floatation is also referred to as the Tipping Centre

A box shaped vessel with a rectangular water plane area would have its centre of floatation

amidships, whereas on a ship shaped vessel the centre of floatation would be either slightly

forward or abaft of amidships.


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Remember all trimming moments are taken about the centre of floatation, since it is around this

point that the vessel pivots.

Change of Trim

This is the difference between a earlier trim and the latest trim. For example the trim that the

vessel had on departure and the proposed trim that the vessel would have on arrival at thedestination port.

Longitudinal Metacentre (ML)

In the manner of the Metacentre, the Longitudinal Metacentre is the point of intersection

between the verticals passing through the centre of buoyancy when the vessel is on an even keel

and when the vessel is trimmed.

Longitudinal Metacentric Height (GML)

This is the vertical distance between the Centre of gravity of the vessel and the longitudinal

Metacentre

In the above figure we see that

GG1 = (w x d) / W

Or W x GG1 = w x d

Trimming moment = W x GG1 = w x d


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The vessel trims until G and B come in the same vertical line again

Also take note that since the distance BG is very small as compared to BML, sometimes BML

may be substituted for GML in calculations, without any appreciable error


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Tan = trim / LBP = t /L where, trim in cm and LBP is in metres

Tan = GG1/ GML = (w x d) (W x GML) because

GG1 = (w x d)/ W

T/ 100L = (w x d) / (W x GML)

T = (w x d) x 100L (W x GML)

T = (w x d) / MCTC = Trimming Moment / MCTC

Where Trim obtained will be in cm.


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Trim = t / 100L

Where L = LBP

Ttrim in cms

To find the change of draft forward and aft due to change of trim

Change of trim = Trimming Moment / MCTC

Change of draft aft (cm) = (l x change of trim) / L

Where:

L is the distance of the centre of floatation from the aft perpendicular (m)

L is the LBP (m)

Change of draft forward (cm) = change of trimchange of draft aft

Or

Change of draft aft (m) = (L-l) / L x Change of Trim


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Effect of loading, discharging or shifting weights

Loading / discharging at the centre of floatation will produce no change of trim but the draft will

only change

Only if the weight is shifted to either forward d or aft will we get a trimming effect.

Shifting a weight will on the other hand give only a change of trim but not of draft

So, loading can be considered as loading at the centre of floatation and then shifting to the

desired place

Similarly discharging can be considered as shifting to centre of floatation first and then taking

the load off the ship

Effect of loading, discharging or shifting weights

So the two components to be calculated are:

a. Change of draft

b. Change of trim

Then we go on to calculate the draft forward and aft

Hence calculate these problems as follows:

Bodily sinkage = W / TPC

Then calculate the change of trim

Change of trim (cm) = Trimming Moment / MCTC

3. Then calculate the change of aft draftchange of aft draft (cm) = l / L x COT

4. Then calculate the change of draft forwardchange of draft forward (cm) = COTchange of

draft aft

OR

(L-l) / L x COT

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