Got my level…
(1) square as intended with corners A,B,C,D
(3) same as square (1) after skew printing with corners Ap,Bp,Cp,Dp
(2) same as square (1) rotated over 45 degrees with corners A’,B’,C’,D’
(4) square (2) after skew printing with corners A’p, B’p, C’p, D’p
Skew printing does not change the length of A-B,A-D etc. in (1) versus Ap-Bp,Ap-Dp etc. in (3) since these lengths are directly related to the timing belts of the printer.
However angle DAB is affected. And after printing the square has become a rhombus in (3).
Therefore, the length of diagonal A-C increase while length of diagonal B-D decreases by the same fraction. Both the diagonals A-C and B-D in (1) and Ap-Cp and Bp-Dp in (3) are perpendicular.
Now in drawing (2) and (4) the square in (1) is rotated over 45 degrees (2).
Basically the diagonal directions in (1) and (3) are swapped with the sides in (2) and (4) and the other way around.
Skew printing of (2) results in (4). Since the original directions A-C and B-D in (1) remain square in skew printing, A’-B’ and A’-D’ (2) versus A’p-B’p and A’p-D’p will remain square in skew printing.
However in (1) compared to A-C,B-D Ap-Cp has increased in length while Bp-Dp has decreased. Translating this to the rotated print from (2) to (4), the skewed A’p-B’p in (4) will increase with respect to the original A’-B’ in (1), while A’p-D’p in (4) will decrease (by the same fraction) with respect to A’-D’ in (2).
Bottom line, by rotating the original print object by 45 degrees a square becomes a rectangle (with perpendicular sides) instead of a rhombus. It should be possible to measure this back in those octagon prints.
As you can be seen from the drawings, this will only work for small skew angles. Which is hopefully the case.
Fixed typo in the description.