Cube distortion EDIT1:
A unit cube ABCDEFGH has rods pinned at its 8 vertices. It is distorted by bringing together opposite vertices A and G ( initial distance AG $=\sqrt3 $ now reduced) thereby increasing angle  between spread out  edges at A and G from a right angle $90^{\circ}$to $\theta\le 120^{\circ }.$ None of the 6 quadrilaterals (skewed rhombuses) are in a plane.
What are the other 18 angles in terms of distance $x=AG?$ How are they computed?
A rough sketch before deformation with 8 ball joints $x= \sqrt 3$, and,
when AG$\;\to 0$ after full flattening  regular hexagon in the plane. Except these two all rhombuses are skewed, not planar or flat.
Thanks!

 A: You can think of this distortion as a contraction scaling operation along $AG$, but combined with an expansion scaling in the other directions so that all the edges/rods remain the same length. A frame made from 12 rods and flexible joints has many more degrees of freedom, but provided that the arrangement of rods at $A$ look the same as those at $G$, and remain symmetrically arranged around the diagonal $AG$, then you will get a squashed cube with rhombus faces. In practice this is very hard to achieve since the slightest misalignment of any of the rods at $A$ or $G$ will cause the whole figure to become misshapen. These edges must not only have the same angles between each other, their angle to the $AG$ diagonal must also all be the same.
In a cube, the tetrahedron $ABDE$ has a height from vertex $A$ of $\frac{|AG|}3$. This will still be the case after the cube is squashed along $AG$. Assuming the edges remain of length $1$, you can use Pythagoras to find that the circumradius of triangle $BDE$ is $r = \sqrt{1-\frac{|AG|^2}9}$. The length of $BD$ is then $r\sqrt3$. This is also the long diagonal of a rhombus face, which makes the obtuse vertex angle equal to $\theta=2 \arcsin (\frac{r\sqrt{3}}2)$.
We can check that this calculation gives correct answers for the extreme cases.
$$|AG|=0 \implies r=1 \implies \theta=2 \arcsin (\frac{\sqrt{3}}2) = 2\cdot 60^\circ =120^\circ $$
$$|AG|=3 \implies r=0 \implies \theta=2 \arcsin 0 = 2\cdot 0^\circ =0^\circ $$
$$|AG|=\sqrt3 \implies r=\sqrt\frac23 \implies \theta=2 \arcsin (\frac{\sqrt{2}}2) = 2\cdot 45^\circ =90^\circ $$
