| Scenario | Inputs | Output (approx.) |
|---|---|---|
| Archimedes in seawater | ρ = 1025 kg/m³, V = 0.015 m³, g = 9.80665 | ≈ 150.8 N (≈ 33.9 lbf) |
| Archimedes in freshwater | ρ = 1000 kg/m³, V = 0.002 m³, g = 9.80665 | ≈ 19.6 N (≈ 4.41 lbf) |
| Archimedes (US gal in water) | ρ = 1000 kg/m³, V = 2.0 US gal, g = 9.80665 | ≈ 74.2 N (≈ 16.7 lbf) |
| Shape volume (cylinder) | ρ = 1000 kg/m³, r = 5 cm, h = 20 cm, g = 9.80665 | ≈ 154.1 N (≈ 34.6 lbf) |
| Shape volume (sphere) | ρ = 1000 kg/m³, r = 6 cm, g = 9.80665 | ≈ 53.2 N (≈ 12.0 lbf) |
| Weight difference | m_air = 10 kg, m_sub = 8.5 kg, g = 9.80665 | ≈ 14.7 N (≈ 3.30 lbf) |
| Floating estimate (wood-like) | m = 1.2 kg, V = 0.0015 m³, ρ = 1000 kg/m³ | Fraction submerged ≈ 0.80 |
| Floating estimate (dense object) | m = 3.0 kg, V = 0.0010 m³, ρ = 1000 kg/m³ | Fraction submerged > 1 (sinks) |
- Archimedes: Fb = ρf · g · Vdisp
- Weight difference: Fb = Wair − Wsub, where W = m · g
- Floating estimate: ρobj = m / V, fraction ≈ ρobj/ρf, and at equilibrium Fb = W
- Select a calculation method that matches your data.
- Enter gravity and the fluid density in your preferred units.
- Provide volume (directly or using the shape tool).
- For weight difference, enter air and submerged readings.
- Click Calculate to view results and extra outputs.
- Use Download CSV or Download PDF for reporting.
1) Archimedes principle in one line
Buoyant force equals the weight of the displaced fluid: Fb=ρ·g·V. If you displace 0.010 m³ of seawater (ρ≈1025 kg/m³) at g=9.80665 m/s², the lift is about 100.6 N, which is roughly 22.6 lbf.
2) Typical fluid densities you can enter
Freshwater is commonly 1000 kg/m³, seawater about 1020–1030 kg/m³, and engine oil often 850–920 kg/m³. Mercury is around 13,534 kg/m³. Air near sea level is about 1.2 kg/m³, so buoyancy in air is small but measurable for large volumes and precision scales.
3) Choosing a volume method
If you know displaced volume from a tank or overflow measurement, use “Direct volume.” If you only know dimensions, use the shape tool. For a cylinder with r=5 cm and h=20 cm, volume is πr²h≈0.001571 m³, producing about 15.4 N of buoyancy in water. For irregular objects, measure displacement by water overflow, then enter the collected volume directly accurately.
4) Weight-difference measurements
With a spring scale, the difference between the reading in air and the submerged reading equals buoyant force. Example: 10.0 kg in air and 8.5 kg submerged implies Δm=1.5 kg, so Fb=Δm·g≈14.7 N. The optional air correction refines this for high accuracy when the object volume is large.
5) Floating and fraction submerged
For a floating object, buoyancy balances weight. The fraction submerged is approximately ρobj/ρfluid. A 1.2 kg object with 0.0015 m³ volume has ρ≈800 kg/m³, so it floats in water with about 0.80 of its volume below the surface.
6) Unit conversions and output checks
This calculator converts inputs to SI internally, then shows results in N, kN, and lbf. A quick sanity check: doubling displaced volume doubles buoyant force, and increasing fluid density from 1000 to 1025 kg/m³ increases buoyancy by 2.5% for the same volume. Remember 1 lbf ≈ 4.448 N.
7) Practical tips for real experiments
Ensure the object is fully submerged for the Archimedes method, remove trapped air bubbles, and measure dimensions carefully. For liquids with temperature variation, density can change noticeably; even a 1% density change alters buoyant force by 1%, which matters in calibration and design work.
1) What does buoyant force represent?
It is the upward force from a fluid on an object. Numerically, it equals the weight of the fluid displaced by the object’s submerged volume, so higher density fluids and larger displaced volumes produce more lift.
2) Does the object need to be fully submerged?
Only for the “displaced volume” method when you treat the whole object as displaced volume. If the object floats, only the submerged portion displaces fluid, and the calculator’s floating estimate uses density ratio to approximate that fraction.
3) Can I use this calculator for gases?
Yes. Enter gas density (for example, air ≈ 1.2 kg/m³) and the displaced volume. The resulting buoyant force is usually small, but it matters for balloons, large enclosures, and precision mass measurements.
4) Why can buoyancy be large, yet the object still sinks?
An object sinks when its weight exceeds the maximum buoyant force available at full submersion. In density terms, if ρobj > ρfluid, the required displaced volume would exceed the object’s own volume.
5) How do I estimate displaced volume for irregular shapes?
Use a displacement container. Submerge the object carefully and collect the overflow, then measure that collected volume with a graduated cylinder or scale. Enter the measured value in the “Direct volume” option for best results.
6) How accurate are the unit conversions and results?
Conversions use standard constants and SI internal math. Accuracy depends mostly on your inputs: density, gravity, and volume. Measure dimensions precisely and avoid trapped bubbles. For critical work, compare with a lab measurement or calibration mass.