I'm looking for a way to get the coordinates of a polygon offset x pixels within a known polygon. This seems to be a non-trivial problem, but I thought of a trick - What if I have the original polygon drawn by Windows with a thick line, the inside edges of that thick line will describe a new polygon. Does anyone know a way to get the coordinates of the inside edges of this thickly drawn polygon ?

    - Jeff

I didn't know there's an inside and an outside edge. Assuming I comprehended your question correctly... This is a geometry related topic, whereas such imaginary laws (postulates and theorems) apply. Anyways enough rambling...

Suppose the polygon is a square. Given its borders are thick enough to produce a polygon-within-a-polygon illusion.

  • thickness of a line = th
  • length = x
  • width = x
  • Assuming this is a 2D square, and we're talking about enclosure of pixels, finding the Area of the square is necessary, as respresented as A.

Now the trick is to include from the outermost edges of the borders to retrieve the Area of the outside square. And the formula as follows A = (x + th)^2

What do you think the formula for the innermost polygon should be?

Edited 3 Months Ago by Stagnant: Text formatting and neatness.

Please elaborate.

Yes. More info is needed. I was thinking about this one and it's clear that Stagnant and I already interpreted this problem differently as far as how the "thick line" would be drawn. Regardless, I concur that this is a Geometry problem, not a problem of Windows drawing a line and doing the Geometry for you. I'm all for using someone else figuring out the Geometry for me and simply calling their function, but I'm wondering if the time you take investigating how that line is drawn and grabbing the points and making sure it matches your needs would take longer than simply working the Geometry yourself. I believe Stagnant was drawing the thick line so that the original square edges are in the MIDDLE of that thick line. I was going to draw that line so that no part of my thick pen ever strayed outside of the original square. One can also define "x pixels within a known polygon" is also vague and needs to be defined. That could refer to x pixels from the edge or x pixels from the vertex. Calculating it from the vertex might be easier, I think, but x would no longer be the thickness of the brush/pen and could be microscopically thin with very small interior angles. So the first thing to do is define exactly what you want.

It's times like these when I wish I knew how to use PowerPoint, Visio, Dia, etc., or actually knew how to draw. It's 2016 and I'm still paper/penciling things and scanning them in as a JPEG with a scanner, which is on the fritz, so I have no diagrams to post. Hopefully the words get the point across, but things are so much easier with drawings.

You tagged this with THREE languages. I have to lead with you know how to use the 3 shells (Demolition Man.)

So while that's a little misleading not one of those languages directly treats a polygon as an object so you have to step back and think about how you would do this as a Human. Now that you can do that on your own, you "port" from Human (some call that Psuedocode) to code in the computer language of your choice.

. PolygonOffsetProblem.jpg

I have tried to look into this from Geometry standpoint - the concept is called "Erosion of a polygon" and is not as simple as one would think. This is why I was trying to use the trick of having Windows draw the polygon with thick line and then trying to get the coordinates of vertices of the inner edges of the thickly drawn polygon.

As per "AssertNull 292" I am thinking no part of the thick line would be outside the boundry of the original polygon. So the original polygon defines outer boundry of the thickly drawn polygon, and inner boundry edges of this thickly drawn polygon defines the polygon I am looking for.

I am thinking of offset distance X being thickness of the drawn line - so perpandicular distance from the original polygon to the internal polygon.

Note - it is very easy for a rectangle, but gets complex if the polygon can have convex and concave areas.

I have added a picture here which I hope will help explain

So I got a little knowledge after watching this video. https://www.youtube.com/watch?v=ThrfCh_Y0YA

I figure you want the de-scalar of the polygon, so I applied the reference center point, the red dot. It didn't seem right to have the red dot outside of the polygon, but it looked a lot better placing within the polygon. The scaling part was tricky because connecting the first segment, point A to red dot was so wide... Scaling factor of 2 and -2 were nonsensical so then I realize it's the inverse of 2, 1/2, as depicted as midpoints of the all those segments. Connect the dots and there is the polygon "within". It looks like a prism!

Now to find the coordinates of your eroded polygon at each of its vertices, apply the midpoint formula. Welcome back to 8th grade Geometry.

Edited 3 Months Ago by Stagnant: Changed link label

Attachments calculating_descale_destination.png 38.68 KB half_scale_factor_complete.png 40.13 KB irregular_polygon.png 26.98 KB
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