hi guys im trying to find the rectangular outer boundary of a irregular shape hope this image explains more clearly. help me out
thanks in advance
hi guys im trying to find the rectangular outer boundary of a irregular shape hope this image explains more clearly. help me out
not sure if this is efficient but here is some thoughts:
group adjacent points into pairs (points from edges).
construct right triangles with them, fill them as faces
filter out faces which intersect original irregular geometry
legs on the rest of triangles are what you want.
(But the result might not be what you want exactly if original â€śirregularâ€ť shape is concave.)
Hi @dineshsubramani,
It has been a while since i last used it, but the Springs package has a good node for that.
@DynamoTeam: This also could be a very good base for a non axis aligned boundingbox
just curious, is there any chance to apply new Transform to elementâ€™s bounding box? I wanted to construct non-axis-aligned bounding box for an element before, by assigning new Transform to bounding box but it always stick to x-Axis & y-Axis in project space. @MJB-online
i have already tried springs minimum area rectangle and its a greate node
what iâ€™m trying is to get enclosed boundary which has only 90 degree as angle
maybe you can take a look my post, see if it works.
please can you explain little more clear
points are from start points and end points on all edges, order them clockwise/counter-clockwise, group them in to pairs
construct right triangles base on each pair and the edge between them (edge as hypotenuse), fill triangles as faces
thereâ€™re faces that intersects the original irregular geometry and those do not are perhaps the ones you want
retrieve triangles from them, the legs of those triangle are perhaps the final result.
I think filtering is required since each pair creates two triangles if you follow vector (0, +1/-1, 0) and (+1/-1, 0, 0) to create legs.
sorry this is not what i need
i understand i will try and reply
I think that @jshial is on the right track with the triangles suggestion. I came up with a pretty rough graph that tries to brute-force the solution:
Give it a try on your own and if you canâ€™t crack it, Iâ€™ll share my current approach. Iâ€™m certain you can come up with something more elegant.