A unit cube ABCDEFGH has rods/edges connected at its 8 vertices by means of flexible ball-joints enabling rotation with two degrees of freedom . It is distorted by bringing together opposite vertices (B,D) of top face simultaneously pulling apart opposite vertices (F,H) of bottom face so all four vertices move in plane BDHF.
EDIT1:
Distance BD decreases and distance HF increases from the initial distance $=\sqrt2 $ by the same amount, say $\delta$. Angles at B,D increase and those at H,F decrease by the same amount away from $90^{\circ}$. None of the 6 quadrilaterals (skewed rhombuses) are in a plane after deformation.
Center of cube O in 3d space remains the same.
In the extreme deformation the cube shrinks to a radial tetrahedral unit vector bundle of three unit rod lengths at angle $ \cos^{-1}({-1}/{3})$ to each other about the cube center. So the (origin centered) Cartesian coordinates of vector tips are:
$$\{(0,0,1),(-\sqrt 2/3, \sqrt6/3,-1/3), (-\sqrt 2/3,- \sqrt6/3,-1/3),(2\sqrt 2/3, 0,-1/3)\};$$
What are the other new 18 angles in terms of distance BD or HF? How can they be computed? A rough sketch of pre-deformation situation is:
The question is similar to Cube Distortion.
Thanks for all thoughts on it!
If all six skewed "faces" are kept congruent among them, during the deformation, then $BDEG$ are the vertices of a regular tetrahedron (green in figure below). If $x=BD$ is the length of its edges, then $x=\sqrt2\,l$ at the beginning of the distortion ($l$ is the edge of the cube) and $x=0$ at the end.
Points $ACFH$ are then the vertices of four equal pyramids, whose bases are the faces of the tetrahedron. It's easy to compute the height $h$ of these pyramids, because lateral edges have length $l$: $$ h=\sqrt{l^2-{1\over3}x^2} $$
$ACFH$ are also the vertices of a regular tetrahedron (not shown in the figure), whose edge $FH$ can be computed (with some geometry) as $$ FH={x+2\sqrt{6l^2-2x^2}\over3}. $$
The initial value is $FH=\sqrt2\,l$ and the final value is ${2\over3}\sqrt6\,l$, but it reaches a maximum of $FH=\sqrt3\,l$ for $x=l/\sqrt3$. Hence distance $FH$ doesn't keep increasing throughout the transformation.
EDIT.
Here's an animated picture of the transformation.