I would assume this would have to be based on coordinates of a view/sheet, but if it could grab the coordinates of the images and trace the borders, that’s exactly what I’m looking to do.
Another node works! Since each only gives two points, is there a node I don’t know of that could use just these two to draw a rectangle or do I need all four?
It gives you the full bounding box, it’s just that a bounding box only has the min and max points as properties. So you’ll have to get the perimeter yourself.
I’d recommend two ways to do that:
Build the points: Get the min and max point and build the other two missing points based on the X and Y coordinates. Because the image is a “flat” object, the Z coordinates shouldn’t matter.
Get the bounding box boundaries: Convert the BoundingBox to a Cuboid, then get the Edges of the cuboid topology. Again, because the image is flat, you should get just the one set of edge boundaries. The edges can then be converted directly into curves.
I have all of them coincident like so, and I have a feeling this would double up the line thickness for plotting. Is there a way to delete the overlapping lines after these are created? Or filter out duplicates of the overlapping horizontal and vertical edges beforehand? I was messing with a Clean.list script but didn’t really get it to work well…
If they really are coincident then they won’t be any thicker. Line weight is just line weight.
You can filter out duplicates but it’s a little tricky due to curve direction. Search the forum for topics on comparing lines/curves. There are quite a few that should at least get you started.
I just have the image import script to start the next one in the row at (x+width of bounding box). If you zoom in, the lines are technically not touching, but I’m not sure of another way to line the images up perfectly without aligning in Revit
That would only work if all your images are exactly the same width. Technically it would be half of the previous image’s width plus half of the new image’s width. width1 / 2 + width2 / 2;