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How to modify just the inside or outside edge of a polygon?

I need to adjust the thickness of a polygon and was wondering if there is any function besides Partial to edit the sides of the polygon one segment at a time. For example say I have a ring. I want to keep the outer dimension the same and reduce the inner dimension by 300 nm. So then total thickness of the ring would go from 400 nm to 100nm. Is this possible and if so how do I do that? Please let me know if this is possible because with complex shapes it would be very helpful to do this.


  • What you mean by "thickness of a polygon" is not very clear.
    If the polygon is made of a hull with holes inside, you could create these two objects separately, then "size" the holes by the desired amount, and subtract the holes to the hull.

  • It is not a hull with holes inside. It is a complex design pattern imported from another program called intellisuite. If there is no functionality that is similar in Klayout I may just have to use intellisuite make the changes and import them. By thickness of a polygon I mean by moving the entire edge of it closer or further from its opposite edge. I can do this using the function Partial on each individual point but for complex shapes it would take way to long. I just wanted to know if there was a way to select an entire edge of a polygon or is it only points on the polygon?

  • Hi ocasta,

    you can select more than one edge in "Partial" mode by dragging a selection rectangle around the edges you want to modify. You can also use Shift+drag to extend the selection.

    To move the selected partials by a given distance, you can use Edit/Selection/Move By.

    For coded manipulations you can use the DRC feature which can manipulate complex polygons using the basic instructions of the DRC language and write the result to a new layer. But this feature operates layer-wise, not on selections.

    Does this help?


  • edited June 10

    Hi ocasta,

    Note: You are probably saying "Hi" to the wrong person. This is "opticsdude001" who is asking the question, not me.

  • @ocasta - Oh sorry ... so my answer is to @opticsdude001 and btw thanks for answering too :-)


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