No !
Unrollsrf will not check if a surface is developable.
below an example of a saddle-like-surface.
it is not developable.
_unrollSrf result in the middle
_squish result at right.
because of the symmetry (mirrored / rotated) of the initial surface, you would expect the same symmetry in the result. - but unrollSrf does not deliver it
squish does.
A surface is developable if its gaussian curvature is 0 at all surface points, see above Wikipedia link.
_curvatureAnalysis can somehow give this information, but the values might be hard to interpret.
_unrollSrf and squish will both report a deviation between 3d and flat version in the commandline.
in above case:
Command: UnrollSrf
One surface unrolled.
Area is 166.5231 sq millimeters (1.24 % ) bigger after unrolling
while squish will report:
Command: Squish
Area: expands 0.011% (3d area is larger)
Compression: average=0.72%, maximum=3.12% (in 3% of pattern)
Expansion: average=0.93%, maximum=3.83% (in 97% of pattern)
1.24% deviation is a failed unrollsrf.
you would expect very very little - maybe something like 0.. 0.01% deviation for a developable surface
curvature Analysis shows
regarding production / cutting-patterns the squish result is preferred and the command offers more control (compression / expansion)
their might be some rare cases where the unrollSrf result is useful - but i think, it s not the intention of the command.
hope this answers your question @Gurgen_Badiryan
kind regards -tom