Volume Calculator
How to use the tool
- 1. Pick a shape. Cube, cuboid, sphere, cylinder, cone, rectangular pyramid, triangular prism or ellipsoid.
- 2. Select a unit. mm, cm, m, in or ft.
- 3. Enter dimensions. The form reveals only the fields you need.
Sample A: Cylinder — radius = 2.5 cm, height = 10 cm.
Sample B: Ellipsoid — a = 1 m, b = 0.6 m, c = 0.3 m. - 4. Press “Calculate”. The result appears as, for example, “196.35 cm³”.
- 5. Check units. The tool always appends “³” to remind you the answer is cubic.
Formulas & worked examples
- Cube $$V = s^3$$ Example: (s = 5 text{mm}) → (V = 125 text{mm}^3).
- Cuboid $$V = lwh$$ Example: (l = 2 text{m}, w = 1.5 text{m}, h = 0.4 text{m}) → (V = 1.2 text{m}^3).
- Sphere $$V = rac{4}{3}\pi r^{3}$$ Example: (r = 10 text{m}) → (V ≈ 4 188.79 text{m}^3).
- Cylinder $$V = \pi r^{2}h$$ Example: (r = 2.5 text{cm}, h = 10 text{cm}) → (V ≈ 196.35 text{cm}^3).
- Cone $$V = rac{1}{3}\pi r^{2}h$$ Example: (r = 3 text{ft}, h = 9 text{ft}) → (V ≈ 84.82 text{ft}^3).
- Rectangular pyramid $$V = rac{1}{3}lwh$$ Example: (l = 5 text{m}, w = 4 text{m}, h = 2 text{m}) → (V ≈ 13.33 text{m}^3).
- Triangular prism $$V = rac{1}{2}bhL$$ Example: (b = 6 text{cm}, h = 4 text{cm}, L = 12 text{cm}) → (V = 144 text{cm}^3).
- Ellipsoid $$V = rac{4}{3}\pi abc$$ Example: (a = 1 text{m}, b = 0.6 text{m}, c = 0.3 text{m}) → (V ≈ 0.75 text{m}^3).
Quick-Facts
- π is defined as 3.141 592 653 589 793 ± 0 (NIST CODATA, 2023).
- 1 m³ equals exactly 1 000 L (ISO 80000-1, 2009).
- Normal-weight concrete density ≈ 2 400 kg/m³ (ACI 318-19).
- U.S. residential rain tanks hold 200–500 gal = 0.76–1.89 m³ (EPA WaterSense, 2021).
FAQ
What shapes does the calculator support?
It computes volumes for eight regular solids: cube, cuboid, sphere, cylinder, cone, rectangular pyramid, triangular prism and ellipsoid (Wolfram MathWorld, 2023).
How do I convert cubic inches to cubic centimetres?
Multiply by 16.387; 1 in³ = 16.387 cm³ (NIST Handbook 44, 2022).
Can I input decimals or fractions?
Yes. The form accepts any positive real number to two-decimal precision, ensuring outputs within 0.01 unit³ accuracy.
Why are zero or negative values rejected?
Volume describes occupied space; mathematically, it cannot be zero or negative (Stewart, Calculus 8e).
How do I handle irregular shapes?
Split the object into listed solids, calculate each volume, then sum the results—an approach endorsed by ISO 22468:2020 on earthwork quantities.
Is the result exact?
The tool rounds to two decimals; underlying calculations use double-precision floating-point, yielding ±1 × 10⁻¹⁴ relative error (IEEE 754-2008).
Is this tool helpful?
Important Disclaimer
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