Regular Tetrahedron Volume

Also known as volume of a tetrahedron · triangular pyramid volume

V = \frac{s^3}{6\sqrt{2}}

Worked example: edge 2 m → 2 sqrt(2)/3 m3 (0.9428090 m3) — press Try an example to run it live, then adjust anything.

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Constant used — built into this formula, no need to enter

\sqrt{2} = 1.4142135623730951

Square Root of Two · exact

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Regular Tetrahedron Volume explained

The regular tetrahedron is the simplest of the five Platonic solids: four vertices, six equal edges, four equilateral faces, and no way to make anything simpler in three dimensions. Its volume from the edge alone is $V = s^3/(6\sqrt{2})$ , equivalently $s^3\sqrt{2}/12$ , which works out to only about 11.8 percent of the cube built on the same edge. It is a startlingly empty shape for its footprint.

A neat way to see where the constant comes from: take a cube of side $a$ and connect four alternating corners. The result is a regular tetrahedron of edge $a\sqrt{2}$ , and the four corner pieces you cut away each have volume $a^3/6$ , leaving $a^3/3$ behind. Substituting $s = a\sqrt{2}$ reproduces the formula exactly.

Tetrahedra show up wherever rigidity matters, because a triangle cannot be deformed without changing a side length and a tetrahedron inherits that stubbornness in three dimensions. Space frames, geodesic structures, and the carbon bonds in a diamond lattice all lean on it.

Regular Tetrahedron Volume formula

V = \frac{s^3}{6\sqrt{2}}

Where

$V$ = Volume (L)
$s$ = Edge length (m)

Missing one of these? Work it out first, then come back

Volume — Cone Frustum Volume (Truncated Cone), Torus Volume
Edge length — Regular Tetrahedron Surface Area, Cube Face Diagonal