Set object density, volume and fluid to calculate buoyancy
Parameters
Results
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Buoyancy Fb (N)
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Weight W (N)
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Net force (N) (+ up)
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Fb / W ratio
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Status
Visualization
Theory & Key Formulas
Buoyancy: $F_b = \rho_{fluid} \cdot g \cdot V_{submerged}$
Weight: $W = \rho_{obj} \cdot g \cdot V_{obj}$
Net force: $F_{net} = F_b - W = g \cdot V(\rho_{fluid} - \rho_{obj})$ (fully submerged)
$F_{net} \gt 0$ → floats, $F_{net} \lt 0$ → sinks, $= 0$ → neutrally buoyant
Partially submerged buoyancy (ships, icebergs)
Equilibrium (floating): $\rho_{obj} \cdot V_{total} \cdot g = \rho_{fluid} \cdot V_{sub} \cdot g$
Submersion ratio: $\dfrac{V_{sub}}{V_{total}} = \dfrac{\rho_{obj}}{\rho_{fluid}}$
Ice: ρ_ice / ρ_water ≈ 0.917 → 91.7% submerged below sea level (9/10 of an iceberg is underwater)
FAQ
What is the Archimedes principle?
An object submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. It floats if its density is less than the fluid.
Why does a steel ship float?
The average density of the entire ship (hull + air) is less than water. The hollow hull displaces enough water to provide sufficient buoyancy.
Why do people float easily in the Dead Sea?
The Dead Sea has about 30% salinity, giving it a density of ~1240 kg/m³, higher than the human body (~1010 kg/m³). The extra buoyancy makes floating effortless.
How do submarines control depth?
Flooding ballast tanks with seawater increases average density to dive; blowing tanks with compressed air decreases density to surface. This is Archimedes principle in action.
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I can see the simulation updating, but what exactly is being calculated here?
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Great question! The simulator solves the governing equations in real time as you move the sliders. Each parameter you control directly affects the physical outcome you see in the graph. The key is to build an intuitive feel for how each variable influences the result — that's how engineers develop physical judgment.
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So when I increase this parameter, the curve shifts significantly. Is that a linear relationship?
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It depends on the model. Some relationships are linear, but many engineering phenomena are nonlinear. Try moving the sliders to extreme values and see if the output changes proportionally — if the graph shape changes, that's a sign of nonlinearity. This hands-on exploration is exactly what simulations are best for.
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Where is this kind of analysis actually used in practice?
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Constantly! Engineers run these calculations during the design phase to quickly screen parameters before investing in expensive physical tests or detailed finite element simulations. Getting comfortable with these simplified models is a real engineering skill.
Set fluid density in sRhoNum (kg/m³). Use 1000 for freshwater, 1025 for seawater, 870 for diesel oil.
Click Calculate to obtain buoyant force in Newtons using F = ρ_fluid × g × V_submerged.
Worked Example
A steel pontoon (ρ = 7850 kg/m³) with volume 2.5 m³ floats in seawater (ρ = 1025 kg/m³). If 60% submerged: F_b = 1025 × 9.81 × (2.5 × 0.60) = 15,074 N (15.07 kN). The weight supported equals this buoyant force. For a floating vessel at equilibrium, 100% submerged displaced volume generates maximum lift; partial submersion represents vessel draft during operation.
Practical Notes
For salvage operations on sunken steel structures (ρ = 7850 kg/m³), increase submersion incrementally to estimate required lift capacity before airbag deployment.
Composite hulls (ρ = 1600 kg/m³) in seawater require 92% submersion at equilibrium; monitor stability when sSub approaches 100%.
Temperature shifts alter sRhoNum significantly—fresh water expands 0.02% per 10°C, affecting buoyancy calculations by 0.2% in precision engineering.
Validate object density against published material specs; homogeneous assumption fails for hollow or air-filled structures without accounting for trapped volume.
