Learning objectives
By the end of this chapter you will be able to:
- define mass, volume, density and relative density, and use consistent units for each;
- convert between density in kilograms per cubic metre, density in tonnes per cubic metre and relative density;
- state the Law of Archimedes and use it to calculate the upthrust on a submerged or partly submerged body;
- state the Law of Flotation and explain why a steel ship floats;
- define displacement, volume of displacement, light displacement and deadweight, and write them in the symbols of the examination formula sheet;
- calculate the draught of a box shaped vessel in water of any density;
- extract displacement and volume of displacement from the MV Ninja hydrostatic table and adjust them for water density.
Everything in ship stability rests on a handful of physical ideas: what density means, why a liquid pushes upward on anything placed in it, and the beautifully simple law, discovered by Archimedes of Syracuse more than two thousand years ago, that tells us exactly how hard it pushes. Master this chapter thoroughly. Every calculation you will ever perform as a deck officer, from a simple draught survey to a full damage stability assessment, begins here.
1.1 Mass, volume and density
The mass of a body is the quantity of matter it contains. At sea we measure mass in tonnes (t), where one tonne is 1000 kilograms. The volume of a body is the space it occupies, measured in cubic metres (m³).
Density connects the two. It is the mass of one unit of volume of a substance:
Fresh water has a density of 1.000 t/m³ (1000 kg/m³). Clean salt water at sea is denser because of the dissolved salts: the standard figure used throughout this series is 1.025 t/m³. Water in docks, rivers and estuaries usually lies somewhere between the two, and we call it dock water. Steel, by contrast, has a density of about 7.85 t/m³, nearly eight times that of water, which makes the fact that a steel ship floats at all seem rather remarkable. By the end of this chapter you will see that there is no mystery in it.
A double bottom tank on MV Ninja contains 250 m³ of fuel oil, and the mass of the oil is found to be 230 tonnes. Find the density of the oil.
density = mass ÷ volume = 230 ÷ 250 = 0.920 t/m³
Notice that the oil is less dense than fresh water, which is why oil spilt at sea floats on the surface.
A ballast tank of capacity 1800 m³ is pressed full with dock water of density 1.012 t/m³. Find the mass of water in the tank.
mass = volume × density = 1800 × 1.012 = 1821.6 tonnes
The same tank pressed full of salt water would hold 1800 × 1.025 = 1845 tonnes. The tank always holds the same volume; the mass depends on what is in it.
1.2 Relative density
The relative density (RD) of a substance, sometimes called specific gravity in older books, is the ratio of its density to the density of fresh water:
Because it is a ratio of two densities, relative density has no units. Fresh water has an RD of exactly 1.000, standard salt water 1.025, and the fuel oil of Worked example 1.1 has an RD of 0.920. The convenience of the tonne and the cubic metre is that the RD of a substance and its density in t/m³ are numerically identical, so converting between the two is effortless: an RD of 1.012 means a density of 1.012 t/m³, which is 1012 kg/m³.
On board ship, the density of the water the vessel floats in is measured with a hydrometer, a weighted glass float with a graduated stem. The instrument sinks until it has displaced its own weight of liquid, so in a denser liquid a smaller volume, and therefore less of the stem, is submerged. The reading is taken at the surface of the liquid.
Key point
Take the dock water density at the ship, at the time of sailing, and at several points around the ship if there is any doubt. River water layered over salt water can give misleading single readings, and a wrong density feeds a wrong answer into every calculation that follows.1.3 Pressure in a liquid and the origin of upthrust
The pressure at any point in a liquid at rest acts equally in all directions and increases with depth. At a depth h in a liquid of density ρ, the pressure due to the liquid is ρgh. Now think of a block held under water. The pressure on its bottom face, which is deeper, is greater than the pressure on its top face. The horizontal pressures on the sides cancel out, but the vertical pressures do not: the result is a net upward force on the block. This force is the upthrust, or force of buoyancy, and it exists on every immersed body, whether it floats or sinks.
1.4 The Law of Archimedes
Archimedes put a precise value on the upthrust:
The Law of Archimedes
When a body is wholly or partly immersed in a liquid, it experiences an upthrust equal to the weight of the liquid it displaces.The classic demonstration uses an overflow can. A steel block of volume 0.5 m³ and mass 3.9 tonnes hangs from a spring balance. As it is lowered into a brimful can of fresh water, exactly 0.5 m³ of water, weighing 0.5 tonnes, spills into a beaker on a weighing scale, and the spring balance reading falls from 3.9 t to 3.4 t. The loss of weight of the block equals the weight of the water displaced, exactly as the law states.
The same steel block (volume 0.5 m³, mass 3.9 t) is lowered fully below the surface, first in fresh water and then in salt water. Find its apparent weight in each case.
Fresh water: upthrust = 0.5 × 1.000 = 0.500 t; apparent weight = 3.9 − 0.500 = 3.400 t
Salt water: upthrust = 0.5 × 1.025 = 0.5125 t; apparent weight = 3.9 − 0.5125 = 3.3875 t
The upthrust depends only on the volume immersed and the density of the liquid, never on what the body is made of. This is why heavy lift calculations always account for the change in effective weight when a load enters or leaves the water.
Interactive: the Archimedes laboratory
Drag the slider to lower the steel block into the fresh water. Watch the spring balance, the water spilling into the beaker, and the scale beneath it.
1.5 The Law of Flotation
Now let the body float freely instead of hanging from a balance. A floating body sinks into the liquid until the upthrust exactly equals its weight, and there it rests. Combining this with the Law of Archimedes gives us:
The Law of Flotation
A floating body displaces its own weight of the liquid in which it floats.Two forces act on a ship floating freely at rest: her weight, acting vertically downward through her centre of gravity G, and the force of buoyancy, equal to her weight, acting vertically upward through the centre of buoyancy B, which is the geometric centre of the underwater volume. For the ship to rest in equilibrium the two forces must be equal, opposite and in the same vertical line.
Here also is the answer to the steel ship paradox. A solid block of steel sinks because its weight exceeds the weight of water its volume can displace. But shape the same steel into a hollow hull and the ship can now displace a far greater volume of water than the volume of the steel itself. She settles at the draught where the displaced water weighs exactly as much as she does, and floats with capacity to spare for cargo.
1.6 Displacement, and meeting MV Ninja
The word displacement (symbol Δ, capital delta, in tonnes) means the mass of the ship and everything on board, which by the Law of Flotation equals the mass of water displaced. The volume of displacement (symbol ∇, called nabla, in m³) is the volume of that water, the volume of the hull below the waterline. These are the symbols of the examination formula sheet. The two quantities are connected through the density of the water:
Read this in both directions, because the two directions answer different questions. If the ship's mass is known, her displacement is fixed whatever water she floats in, and it is the volume ∇ that adjusts to the density: she sinks deeper in fresh water than in salt water. If instead the draught is known, it is the volume ∇ that is fixed, and the displacement Δ adjusts to the density: the same draught means a smaller mass afloat in fresh water than in salt water.
Two related terms are needed. The light displacement is the mass of the empty ship as built, with no cargo, fuel, water, stores or crew. The deadweight is everything she can carry: the difference between her displacement at any draught and her light displacement. Throughout this series our worked examples use the training vessel MV Ninja, whose full particulars are in the data booklet that accompanies every volume.
| MV Ninja, summer condition (from the data booklet) | |
|---|---|
| Summer draught | 9.600 m |
| Summer displacement (salt water) | 30456 t |
| Volume of displacement | 30456 ÷ 1.025 = 29713 m³ |
| Light displacement | 4950 t |
| Summer deadweight | 30456 − 4950 = 25506 t |
MV Ninja floats on an even keel at a draught of 6.00 m in salt water. Using the hydrostatic table, find (a) her displacement, (b) her volume of displacement, and (c) the displacement she would have at the same draught in fresh water.
(a) From the table at 6.00 m: Δ = 18064 t in salt water.
(b) ∇ = Δ ÷ ρ = 18064 ÷ 1.025 = 17623 m³
(c) At the same draught the underwater volume is the same, because the hull has not changed shape and the waterline is at the same height on it, so in fresh water Δ = ∇ × 1.000 = 17623 t. The fresh water column of the table prints 17624 t at 6.00 m; the one tonne difference is rounding, since each column of the table is rounded to the nearest tonne separately. The table's two displacement columns are simply the same volume multiplied by two different densities.
MV Ninja lies at a draught of 6.00 m in dock water of RD 1.010. Find her displacement.
The draught fixes the underwater volume. From Worked example 1.4, carrying one more figure than the table prints, ∇ = 18064 ÷ 1.025 = 17623.4 m³.
Δ = ∇ × ρ = 17623.4 × 1.010 = 17800 t
A quicker route for any intermediate density goes straight from the salt water column, since ∇ is common to both: Δ = SW displacement × (dock water RD ÷ 1.025) = 18064 × 1.010 ÷ 1.025 = 17800 t. Had we used the rounded volume, 17623 × 1.010 = 17799 t; the tonne is lost in the intermediate rounding, and the direct ratio, which avoids it, is the better habit.
Interactive: the MV Ninja hydrostatic lookup
Enter any even keel draught between 2.60 m and 10.40 m and a water density. The tool interpolates the booklet table exactly as you will learn to do by hand in Volume Two.
1.7 The effect of density on draught
For a box shaped vessel, wall sided and rectangular in plan, the underwater volume is simply length × breadth × draught, which makes it the perfect training ground before we tackle real hull forms:
A box shaped barge is 40 m long and 8 m wide, and her displacement is 984 tonnes. Find her draught (a) in salt water and (b) in fresh water.
(a) d = Δ ÷ (L × B × ρ) = 984 ÷ (40 × 8 × 1.025) = 984 ÷ 328 = 3.000 m
(b) d = 984 ÷ (40 × 8 × 1.000) = 984 ÷ 320 = 3.075 m
Passing from salt water into fresh water, her mass unchanged, the barge sinks 75 mm deeper. A ship behaves the same way: for MV Ninja the corresponding change of draught between salt and fresh water at the summer displacement is her fresh water allowance of 216 mm, a subject we take up properly in Chapter 5.
Interactive: the floating barge laboratory
The barge from Worked example 1.6 (40 m × 8 m, Δ = 984 t) floats in water whose density you control. Slide the density from fresh to salt and watch the draught respond. The vertical movement is exaggerated so that the 75 mm change can be seen.
Chapter summary
- Density is mass per unit volume; RD is density relative to fresh water and has no units. In t/m³, density and RD are numerically the same.
- Law of Archimedes: upthrust equals the weight of liquid displaced.
- Law of Flotation: a floating body displaces its own weight of the liquid it floats in.
- Δ = ∇ × ρ. A known mass fixes Δ, and ∇ adjusts to the density; a known draught fixes ∇, and Δ adjusts to the density.
- Deadweight = displacement minus light displacement.
- For a box shaped vessel, d = Δ ÷ (L × B × ρ). Less dense water means a deeper draught for the same mass.
Self test questions
Answer each question, then check yourself. Your running score appears in the bar at the foot of the page. Work every question with pencil and paper first; the options are close enough to punish guessing.