Regular Tetrahedron Volume Calculator
Compute regular tetrahedron volume from a single edge length.
For 4-sided dice (d4), pyramid tea bags, and crystal structure modeling.
A regular tetrahedron is one of the five Platonic solids. All four faces are equilateral triangles, all six edges are equal, and all four vertices are equivalent.
V = s³ / (6√2) = s³ × √2 / 12 ≈ 0.1178 × s³
Where s is the edge length (the same for all six edges).
Worked example — D4 die (4-sided gaming die): A standard 16 mm tabletop D4 (Dungeons & Dragons gaming die) has edge length s = 16 mm. V = 16³ × 0.1178 ≈ 482.4 mm³ ≈ 0.48 cm³.
At plastic density 1.2 g/cm³: 0.58 g per die. A bag of 10 dice weighs ~6 g — light enough to ship in regular mail with no extra postage.
Worked example — Tetrahedral box of milk (Tetra Classic): A 1950s Tetra Pak milk carton was a regular tetrahedron with s = 130 mm (for a 250 mL pack). V = 130³ × 0.1178 ≈ 259,000 mm³ = 259 mL. Close to the labeled 250 mL with a little headroom for thermal expansion.
The reason Tetra Pak invented the tetrahedron packaging in 1952: a tetrahedral shape can be made from a single rectangular sheet of paper with just two glued seams — incredibly cheap to manufacture. Later they switched to brick-shaped (Tetra Brik) cartons for stacking efficiency.
Where regular tetrahedra appear in real measurements:
- D4 dice. Standard 4-sided gaming dice — tabletop role-playing games, math classes.
- Tetrahedral kites. Multi-cell kites made from many small tetrahedra (Alexander Graham Bell pioneered these around 1900).
- Crystallography models. Methane (CH₄), silicon dioxide (SiO₂), and many other compounds have tetrahedral molecular geometry. The model usually has a central atom at the centroid with bonds pointing to the four vertices.
- Diamond crystal structure. Carbon atoms in diamond bond tetrahedrally — each carbon connected to four others at the vertices of a regular tetrahedron.
- Tetrahedral packaging (Tetra Pak Classic). Vintage milk and juice cartons from the 1950s-70s.
- Tetrahedron Platonic-solid puzzles. Tactile geometry teaching aids.
Useful tetrahedron measurements (all derived from s):
| Quantity | Formula | Value for s = 1 |
|---|---|---|
| Edge length | s | 1 |
| Face area (equilateral triangle) | (√3 / 4) × s² | 0.433 |
| Face perimeter | 3s | 3 |
| Height (apex to opposite face) | s × √(2/3) | 0.816 |
| Total surface area | √3 × s² | 1.732 |
| Volume | √2 × s³ / 12 | 0.118 |
| Inscribed sphere radius (insphere) | s / (2√6) | 0.204 |
| Circumscribed sphere radius (circumsphere) | s × √6 / 4 | 0.612 |
Volume vs. cube comparison:
A tetrahedron with edge s has volume ~0.118s³. A cube with edge s has volume s³. The tetrahedron holds only 11.8% of the cube’s volume for the same edge length.
For the SAME bounding sphere: a tetrahedron inscribed in a sphere of radius r has edge s = r × 4/√6, so V = (8/(9√3)) × r³ ≈ 0.513 × r³. A cube inscribed in the same sphere has edge 2r/√3 and volume 8r³/(3√3) ≈ 1.54 × r³. The cube holds 3× more than the tetrahedron in the same sphere.
Sanity check:
- s = 0: V = 0. ✓
- s = 1: V = √2 / 12 ≈ 0.118. ✓
How we build and check this calculator
This calculator runs entirely in your browser, so the numbers you enter stay on your device. The math behind it is written by hand and tested against worked examples and standard references before the page goes live.
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