The angle "alpha" through which the 4 cubes must rotate in order to define a regular Dodecahedron is given by the equation:

sin(alpha/2) = sqrt(3/2) / (2 * p)

where "p" is the Goldern ratio

p = (1 + sqrt(5)) / 2

alpha approximately equals 44.47751219 degrees.

One half of this value is approximately 22.2387561 degrees.

To derive angle alpha, I considered 3 vertices of the 120 Polyhedron.

- v1 = (0, p, p^3)
- v2 = (p^3, 0, P)
- v3 = (p^2, -p^2, p^2)

These vectors all have the same magnitude given by

mag = mag(v3) = sqrt(p^4 + p^4 + p^4) = sqrt(3)p^2

The dot product v2 and v1:

v1.v2 = mag(v1)mag(v2)cos(theta) = p^4

(sqrt(3)p^2)(sqrt(3)p^2)cos(theta) = p^4

cos(theta) = 1/3

The projection of v2 onto v1:

A = mag(v2)cos(theta) = (sqrt(3)p^2)/3 = p^2 / sqrt(3)

The vector A defined to be in the direction of v1 and of
length A can be defined as

A = (0, p/3, (p^3/3))

Then the distance from A to v2 is

d = 2 sqrt(2/3) p^2

The distance from v2 to v3 is

mag(v2 - v1) = mag(p^2 - p^3, -p^2, p^2 - p)

which reduces to mag = 2p.

So we have a triangle with 2 sides equal to

d = 2 sqrt(2/3) p^2

and one side equal to

mag = 2p

With this information, we find the angle alpha to be

2 sqrt(2/3) p^2 sin(alpha / 2) = p

Solving for sin(alpha/2) we get

sin(alpha/2) = sqrt(3/2) / (2 * p)