Buoyancy Calculator (Archimedes)
Find the buoyant force on anything in a fluid, and whether it floats. Solve F = ρVg for force, volume or fluid density, see how much of a floating object sits under the surface, and work out what a raft or boat can carry.
How to Use
- Pick what to find: the buoyant force, the displaced volume, the fluid density, whether an object floats, or a boat’s load.
- Choose a fluid (fresh water, seawater, ethanol, mercury or air) or type your own density in any unit.
- Enter the known values, each with its own unit: litres or cubic feet, kilograms or pounds, newtons or pound-force.
- Read the answer, the drawing of the object at its waterline and the table of the result in other units.
- Press a preset to load an example, and check Show Work for every step and unit conversion.
Worked Example
A 1-litre block held under fresh water. V = 1 L = 0.001 m³ and fresh water at 25 °C is 997 kg/m³, so F = ρVg = 997 × 0.001 × 9.80665 = 9.78 N. That is the weight of the 0.997 kg of water the block pushes aside, and it is the same for a block of cork, oak or lead.
An iceberg in the sea. A floating object sinks until it displaces its own weight, so the fraction under the surface is ρobject ÷ ρfluid = 917 ÷ 1,025 = 0.8946. 89.46% of the ice is under water and 10.54% shows above it.
The common mistake: using the object’s density instead of the fluid’s. For a 1 L steel block (7.85 kg) under water, ρ in F = ρVg is the water’s 997 kg/m³, giving 9.78 N. Putting in steel’s 7,850 kg/m³ gives 76.98 N, which is simply the block’s own weight and would wrongly say the block hangs weightless. The scale really reads 76.98 − 9.78 = 67.20 N.
Show Work
Formulas
Archimedes, the Crown and the Load Line
Archimedes of Syracuse (c. 287–212 BC) set out the principle in On Floating Bodies: a body in a fluid is lifted by the weight of the fluid it displaces, and a floating body displaces its own weight. The famous story of King Hiero II’s gold crown, and of Archimedes running through the streets shouting “Eureka”, comes from the Roman architect Vitruvius, writing about two centuries later in De Architectura. In 1586 the young Galileo argued in La Bilancetta (“The Little Balance”) that Archimedes would really have weighed the crown against gold under water, which is exactly the apparent-weight method above.
The same sum keeps ships safe. After many overloaded ships were lost, the British Merchant Shipping Act of 1876, pushed through by the MP Samuel Plimsoll, made a load line painted on the hull compulsory. Modern load-line marks have separate levels for fresh and salt water, because a ship floats deeper in fresh water (997 kg/m³) than in the sea (about 1,025 kg/m³).
Densities used here: water and mercury at 25 °C and ethanol at 20 °C from the CRC Handbook of Chemistry and Physics; dry air at 20 °C and 1 atm; and 1,025 kg/m³ as a typical value for surface seawater, which ranges from about 1,020 to 1,030 kg/m³ with temperature and salinity.
About This Calculator
This calculator solves Archimedes’ principle for the buoyant force, the displaced volume or the fluid density, tells you whether an object floats and how deep it sits, gives its apparent weight when it sinks, and finds the load a box-shaped hull can carry with its draft and freeboard. Every value takes its own unit, so you can mix litres, cubic feet, pounds and newtons, and the conversions are shown in the working. The drawing shows the object at its real waterline with the weight and buoyant force to scale.
Everything runs in your browser; nothing you enter is sent anywhere.
Related tools: Density Calculator, Specific Heat Calculator, and Unit Converter.
Frequently Asked Questions
What is Archimedes’ principle?
An object in a fluid is pushed up by a force equal to the weight of the fluid it displaces: F = ρ × V × g. A 1-litre block held under fresh water (997 kg/m³) displaces 0.997 kg of water, so it is pushed up with 0.997 × 9.80665 = 9.78 N, whatever the block is made of.
How much of an iceberg is under water?
The fraction under the surface is the object’s density divided by the fluid’s. Ice at 917 kg/m³ in seawater at 1,025 kg/m³ sits 917 ÷ 1,025 = 89.5% under, with 10.5% showing. In fresh water (997 kg/m³) it sits 92.0% under.
Why does a steel ship float when steel sinks?
Steel is about 7,850 kg/m³, almost eight times as dense as water, but a ship is a steel shell full of air. What matters is the mass of the whole ship against the water its hull can displace. A 3 m × 1.2 m × 0.4 m box hull can displace 997 × 1.44 = 1,435.68 kg of fresh water, so with a 120 kg hull it carries 1,315.68 kg before the water reaches the rim.
Why is it easier to float in the sea?
Seawater is denser: about 1,025 kg/m³ against 997 kg/m³ for fresh water at 25 °C. The same body displaces the same volume, so the sea pushes up 1,025 ÷ 997 = 1.028 times as hard, 2.8% more.
What is apparent weight?
It is the weight minus the buoyant force: what a spring scale reads when the object hangs in the fluid. A 7.85 kg steel block of 1 L weighs 76.98 N in air; under fresh water the water takes 9.78 N, so the scale shows 67.20 N, as if it had a mass of 6.853 kg.
How do I use the Buoyancy Calculator (Archimedes)?
Simply type your numbers and read the result, which refreshes the instant you change something. There is nothing to submit and nothing to wait for.
Does it cost anything or need an account?
No. The tool is completely free, there is no account to create, and it keeps working offline after the page first loads.
Is anything I type uploaded?
No. The tool works entirely on your device, so the values you enter never leave your browser.
Common Use Cases
Rafts and small boats
A 3 × 1.2 × 0.4 m punt weighing 120 kg carries 1,315.68 kg in fresh water before it swamps; with 300 kg aboard it sits 11.7 cm deep with 28.3 cm of freeboard.
Lift bags and salvage
To lift 100 kg off the seabed, a bag must displace 100 ÷ 1,025 m³ = 97.56 L of seawater (plus whatever the bag itself weighs).
Measuring a liquid’s density
A 0.5 L sinker that loses 3.87 N when lowered into a liquid gives ρ = 3.87 ÷ (0.0005 × 9.80665) = 789 kg/m³, which is ethanol.
Balloons in air
One cubic metre of air at 20 °C weighs 1.204 kg, so 1 m³ of anything is pushed up with 11.81 N. Helium at 0.166 kg/m³ leaves a net lift of about 1.04 kg per cubic metre.
Physics homework
Steel (7,850 kg/m³) sinks in water but floats on mercury (13,534 kg/m³) with 58.0% of its volume under the surface.
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