What is Displacement of a Ship ?

Understanding the term Displacement 

Under normal circumstances, water is an incompressible fluid, meaning its density and volume do not change with applied pressure or force. When a finite body is placed in water, two things happen:

1. The body exerts or imparts a force on the water medium by virtue of its weight. 
2. The body tends to displace a certain volume of water with which it is directly interacting. 

These two phenomena are the basis for all our explanations.

When the first happens, according to Newton’s 3rd law (every action has an equal and opposite reaction), the fluid or water medium tends to impart a reaction force equal to the body’s weight acting as a downward force.  

However, like solids, if point 1) were solely the governing law of liquids, then a tiny coin or a screw weighing just a few grams would have easily been balanced by a large water body. But by the nature of liquids, point 1) is dependent on point 2).  

What does point 2) more technically mean? When a body is placed in water, it occupies some volume space in the fluid domain. Since the total volume of the system should remain constant from first principles, this locally lost or ‘displaced’ volume or supplanted space is adjusted by a uniform equivalent global increment in the volume of the surrounding water body.  

Now, note that we have used vague terms like ‘some volume’ or a ‘certain volume’ while discussing displacement in water. How do we definitively quantify this replaced volume? The answer is it depends.

This is because the volume displaced upon placement of a particular body in water or any other fluid is dependent on the following factors: 

• The weight of the body 
• The geometry and physical form of the body 
• The physical nature of the fluid (density) 

While we have talked about the mass of the body, manifested in the physical world as weight, in the realm of fluids, the geometry and form of the body are more crucial.  

Imagine placing a wooden box weighing 5 kgs of dimensions 2 metres X 2 metres X 2 metres in a small tank containing freshwater (density= 1 kg/ cubic. Metres). When this 5 kg mass, acting as Weight: 5 X acceleration due to gravity, g, is placed in water, it starts to sink slowly and becomes afloat soon. 

This is because it has reached a state of mechanical equilibrium and the reaction force offered by the fluid medium is exactly equal to the weight of the body. Now, since water is incompressible and highly elastic, how is this reaction force generated, or more accurately, from where does this come from? 

This is derived from the volume space occupied by the box, or the volume of fluid displaced away, multiplied by density (mass) times the acceleration due to gravity. 

Now, this volume space is what? Length X Breadth X Depth. As water is an ideal fluid, it takes any form based on the body it is interacting with. Thus, when the box is placed, the length and breadth of the volume of the water domain displaced is also the same, that is 2 X 2. How much is the depth? This is essentially the level till which the box has sunk, and the crux of all our discussion (variable ‘t’) The volume of the water displaced, or inversely, the space in the fluid domain that the box has occupied, is then: 2 X 2 X t. 

For mechanical equilibrium, owing to the equivalence of forces, the force imparted by the box by its weight, 5 times g, should be equal to the reaction force offered by the water, the volume displaced X density X g. 

Solving for equivalence, the level till which the box has sunken, t, comes out to be 1.25 meters. This is essentially known as the draft of the body. Hence, the net volume occupied by the box in a fluid medium, or conversely, the water volume substituted or replaced by the fluid is 2 X 2 X 1.25 = 5 cubic metres. 

Since density is 1, the reaction force offered by the water is 5 times g, the same weight that was being imposed by the box, and thanks to this balance, the body floats in water. 

This is the famous Archimedes’ Principle. A body floating in the water, partially or fully, is acted upon by an upward-acting force known as buoyancy that is mathematically the same as the weight of the water volume that the body displaces as well as the body’s own physical weight. 

From another perspective, the column or volume of water that gets replaced by the occupied body and equivalently distributed elsewhere weighs the same as the physical weight of the body, and as a reaction, the water domain offers an equal and opposite force that balances the system.  

This column or volume of water is known as displacement, or more specifically, volume displacement. When speaking on a mass scale, the weight of this displaced water volume equal to the physical weight of the body (measured quantitatively as mass) is known as mass displacement. 

Thus, in the above example, we can say that the mass displacement of the box is 5 kg and the volume displacement is 5 cubic meters. 

Now, imagine a metal bar of weight 4 kg is placed inside the box. What happens? The mass of the whole system becomes 9 kg. Going by the above equation, the value of the unknown variable t becomes 2.25 meters. 

However, this is more than the maximum vertical extent of the box, 2 meters. This essentially means that the water level shoots past the uppermost limit of the box, and yet requires another 0.25 meters to attain the condition of equilibrium. 

The water immediately ingresses into the box and the entire system instantly sinks. Thus, when the extra 4 kg mass was applied, the entire system was not essentially designed to cater to the condition of mechanical equilibrium or the laws of floatation.

Now, suppose, you take another heavier 8 kg box of the same material and design but having dimensions as 3 X 3 X 2, and load the 4 kg mass.  

By the above equation, (8+4) x g = (3 X 3 X t) X 1 X g. 

 The draft value, t, becomes equal to approximately 1.33 m, and that means it comfortably floats with a considerable margin of about another 0.67 m. 

The increase of the dimensions by a meter led to essentially two things: 

• An improved mass distribution or density, that is a lower mass per unit volume in the system.  
• The displaced underwater volume or displacement is such that its weight (and thus buoyancy) is enough to balance the imposed weight of the body. 

Thus, the mass of an object is immaterial while considering floatation, and the condition of equilibrium only holds good if the form and design of the body cater to the balance of forces. This is the very reason that metal ships weighing thousands of tons stay afloat, whereas a small screw or a metal bar sinks instantly. 

Displacement in Ships 

When we say that a vessel has a displacement of 1000 Tones, it essentially means that it:

• Physically weighs 1000 tons in the ground 
• While afloat, this weight is manifested as displacement of the water domain such that this displaced volume also weighs 1000 tons and acts equally in an opposite direction in what is known as buoyancy. 

Lightweight and Deadweight 

The displacement of a vessel comprises two components: Lightweight and deadweight. 

A ship’s lightweight is the vessel’s weight under empty conditions. This includes the weight of the hull structure, machinery, fittings, and everything permanent and not expendable. 

The deadweight of the vessel is the sum of the weight of all expendables, including cargo, fuel, ballast water, crew and passengers, stores, and any other form of consumables. This is essentially the carrying capacity of a vessel. 

A fully loaded ship is when the vessel is filled up to its maximum capacity. In this case, due to higher weight, the vessel tends to sink more and displace a greater fraction of the underwater volume that provides the necessary buoyancy to maintain equilibrium. 

Thus, the displacement is higher, and as a result, the draft of the vessel is also higher. This is known as the full-load or deep-sea condition of the vessel. 

A lightship condition is when the vessel is loaded up to a bare minimum, excluding cargo and most of the other consumables. Since the weight is lesser, a lower volume of water weighing the same as the vessel can cater to the necessary buoyancy force (equal to the weight) for keeping the vessel upright. 

Density 

Density also has a very important role in deciding the flotation mechanics of a vessel. Note that in the above equation, the density was equal to 1 since it was freshwater. Now, if we consider seawater that has a value of 1.025 kg/ cubic metres, the equation becomes: 

5 X g = (2 X 2 X t) X 1.025 X g

Now, the value of t is roughly 1.21, less than 1.25, which was in the case of freshwater. This essentially means that in salt water, the same body has sunk less than it did in freshwater.

Further to our discussions, owing to the density, this means that for the same weight of a floating body, a greater volume of water needs to be displaced in a freshwater environment as compared to saline water for maintaining the necessary state of equilibrium. Thus, the higher the density, the lesser the volume displacement of a vessel, and vice-versa.  

You might also like to read-

• What are Semi-Displacement Vessels?
• Intact Statical Transverse Stability Of Displacement Vessels
• Understanding Stability of Cruise Ships

About Author

Subhodeep is a Naval Architecture and Ocean Engineering graduate. Interested in the intricacies of marine structures and goal-based design aspects, he is dedicated to sharing and propagation of common technical knowledge within this sector, which, at this very moment, requires a turnabout to flourish back to its old glory.

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