No thoughts off-hand, but only because I'm kinda swamped with some contract work right now - sorry.

Just a quick thought... you might need to get the inverse of that weight above ( weight = 1 - weight ), so that if threshold distance is (arbitrary number) 20 and distance is 20, you'd get 1.0, (full weight), but you want zero weight for that one (it's at the maximum distance) - however that works into the averaging of multiple vertices.

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