Platonic Solid Volume Calculator

Volume of regular polyhedra (Platonic solids)

There are exactly five Platonic solids. They are the convex polyhedra built from congruent regular polygons, with the same arrangement of faces around each vertex. Once you know the edge length a, every volume is fixed: the tetrahedron is V = a³/(6√2) ≈ 0.1178·a³, the cube is V = a³, the octahedron is V = a³·√2/3 ≈ 0.4714·a³, the dodecahedron is V ≈ 7.6631·a³ and the icosahedron is V ≈ 2.1817·a³. Take a = 5 and you get: cube → 125; tetrahedron → ≈ 14.73; octahedron → ≈ 58.93; dodecahedron → ≈ 957.89; icosahedron → ≈ 272.71.

Applications

You run into these shapes more often than you'd expect. In chemistry, methane CH₄ is tetrahedral, SF₆ is octahedral, and C₂₀H₂₀ is a dodecahedrane. Quantum chemistry and crystallography rely on them too, through close-packed lattices. Tabletop RPG dice are the everyday example (d4 tetrahedron, d6 cube, d8 octahedron, d12 dodecahedron, d20 icosahedron), and they roll fairly because symmetry gives every face the same chance. Virologists see icosahedral capsids in many viruses, and architects know them from Buckminster Fuller's geodesic domes.

FAQ

Why are there only five Platonic solids? Two conditions box you in. Euler's polyhedron formula V − E + F = 2 has to hold, and at least three regular polygons must meet at each vertex with their angles summing to less than 360°. Work through the cases and only five arrangements survive.

Which has the largest volume per edge? The dodecahedron wins easily at ≈ 7.66·a³. Its 12 pentagonal faces wrap around a lot more space than the triangular faces of the smaller solids.

Are RPG dice really fair? In theory, yes. The rotational symmetry group acts transitively on the faces, so every face comes up with the same probability on a roll. In practice, manufacturing tolerances can nudge things slightly off.