**Ship 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/s^{2} or Newton (N)

Here since the acceleration due to gravity is known as 9.81m/s^{2}

Therefore we may write:

a mass of 1 kg having a weight of 1kg 9.81 m/s^{2} as 9.81 kg m/s^{2}

Or simply 1 kgf, which is saying 9.81 N

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

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/m^{2})

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/cm^{3})

Fresh water has a density of 1 gm/cm3 or 1000kg/m^{3}

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/m^{3}

Or 1 t/m^{3}

Thus the density of FW may be expressed as 1gm/cm^{3} or 1t/m^{3}

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

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 metre^{3}, or 12 cbm), let the weight of the log be 8 t. (assumed density of the log at 0.667 t/m^{3})

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 metre^{3}, 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 water displaced 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.

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 would displace 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 earthâ€™s gravity or the weight of the body.

That it does not sink is due to Archimedesâ€™ 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.

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 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 to the 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 â€“ 0.850 General Cargo ships â€“ 0.700

Oil tankers â€“ 0.800 Passenger ships â€“ 0.625

Bulk carriers â€“ 0.750 Container / Ro-Ro â€“ 0.575

Tugs â€“ 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.

**SHIPâ€™S 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 of the 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.

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 displaced

New volume x New density = Old volume x Old density

New Volume / Old Volume = Old density / New density

But volume = L x B x draft

L x B x New draft / L x B x Old draft = Old density / New density

New draft / Old draft = Old density / 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 (1025 â€“ DW 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 underwater 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 from 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 / New density = Old displacement / Old density

New displacement / Old displacement = New density / 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 TPCâ€™s 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 m^{2}

Density is in t/m^{3}.

Now let the mass â€˜wâ€™ tones be loaded such that the draft increases by 1 cm & the ship now floats at new Waterline Wâ€™Lâ€™

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 area â€“ which determines the underwater volume of the ship

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

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 factors affect 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 TPCâ€™s 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 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.

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.

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.

So we can replace the word level by the nautical word â€˜draftâ€™

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.

As a ship moves from SW to FW, the weight of the displaced water reduces â€“ RD 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.

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 â€˜Vâ€™.

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

Let â€˜vâ€™ be 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 â€˜wâ€™ is the mass of SW in volume v and â€˜Wâ€™ 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

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 â€˜Wâ€™ = 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

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

Assume that â€˜wâ€™ is the mass of SW in volume v and â€˜Wâ€™ in volume V,

Then, replacing the factor as obtained above we get:

w = W/40

Displacement = FWA x ( 4 x TPC)

But w = TPC x (FWA/10)

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

Where â€˜Wâ€™ = Loaded SW displacement in tonnes.

**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.

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-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 dock â€“ this as a safety factor)

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

Written by Capt. Jayant RoyWritten by Capt. Jayant Roy

If Captain Roy is right then “Dock water is the water whose density is neither that of fresh water or salt water but in-between the two”.

In fact dock water is for example also:

a) water of density = 0.9954 T/m3 in Gatun Lake of Panama Canal

b) water of density = 1.030 T/m3.

If steel hydrometer has reading = density in FW of density = 1 T/m3 and SG4C/SG4C = 1.000027 then whenever such hydrometer has reading = 1 T/m3 = assumed unit density and in such water every vessel of no doubt true molded volume = Vm = L*B*T*Cb for S draft has:

a) allowed F draft = S draft + FWA;

b) allowed molded displacement = Dm = Vm * 1.025 T/m3.

In dock water of hydrometer reading = 0.996 T/m3 every vessel has:

a) allowed draft = S draft + DWA = S draft + 1.16 * FWA;

b) allowed Dm = Vm * 1.025 T/m3.

In water of the same hydrometer reading = assumed density = 1.030 T/m3 the same vessels have:

a) allowed draft = S draft + DWA = S draft – 0.2 * FWA;

b) allowed Dm = Vm * 1.025 T/m3.

Other way in SW of reading of hydrometer = assumed density = 1.025 T/m3 at various temperatures and in SW of density = 1.025 T/m3 at 4C = reading of hydrometers all vessels for S drafts have the same Dm = Vm * density of SW. In waters of other densities the vessels for S drafts have allowed the same Dm and allowed draft = S draft + DWA.

A. Bratek

bratek_a@op.pl

The fact is that pure water at 20C has (https://icllabs.com/mass-calibration-metrology-terms-explained-in-plain-english/):

a) density = 0.9982 T/m3 and SG = 0.9982 Tf/m3;

b) apparent density = conventional mass per unit volume = 0.99715 T/m3 and apparent SG = 0.99715 Tf/m3.

By the same international standard = OIML standard gold has apparent density > apparent density, while steel of density = 8 T/m3 has apparent density = 8 T/m3.

Consequently a steel lightship of displacement = 8 T for draft X has extreme mass = extreme apparent mass = extreme conventional mass = extreme displacement = 8 T for both technical and commercial purposes.

If pure water as above has apparent density = 0.9989481 * density, then SW of density = 1.025 T/m3 has apparent density = 1.0239218 T/m3 a. If in SW a lightship of draft = X has:

a) assigned molded volume = Vi =L*B*T*Cb + volume of shell = 7.7966914 m3 according to Rules of American Bureau of shipping;

b) assigned extreme volume = Ve = L*B*T*Cb + volume of shell + volume of appendages = 7.804878 m3 in the rest of the world,

then the same lightship for the same draft in SW has:

a) assigned molded displacement = molded mass = Di = Vi * density of SW = 7.9916086 T in imperial world = in world of ABS;

b) assigned extreme displacement = extreme mass = De = Ve * density of SW = 8 T in the rest of the world.

If a target of draft surveyors is extreme apparent mass = M, then:

a) M = Ve * apparent density of SW = 7.9915847

b) M = apparent density of SW * (De / density of SW) = 7.991584 T.

Other way in fact a target of draft surveyors is Di = 7.9916086 T as above i.e. molded mass = extreme apparent mass.

Where however a target of draft survey is extreme mass = conventional mass = De, there lightship as above has for draft survey purposes:

a) De = 1.00105 * assigned Di = (1.00105 * density of SW )* (Di / density of SW) = 1.0260762 T/m3 * (Di / density of SW) = assumed density of SW * (Di/density of SW) = 1.00105 * 7.9916086 T = 8 T in imperial world;

b) De = assigned displacement in metric world.

Other way to convert mechanically assigned Di of imperial vessels to De one needs a special purposes hydrometer of reading in SW = 1.0260762 T/m3. The closest is standard 60F hydrometer of reading in SW = SG60F = RD60F = 1.025/0.999 = 1.026026 known as load line hydrometer.