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. ✓