Hey, I never had a problem like this, so I found it challenging and just started to expirement :P

This is what I came up with:


(Sorry for the poor drawing, but I only had openoffice installed and I m short on time right now)

- The red vectors are the outlines of the polygon.
- 2 blue vectors represent the unit-vectors of the outline vectors at each point
- The green vectors are the 2 blue unit-vectors added together
- The black is the orthogonal unit-vector of the green one (which can be multiplyed by the thickness to get thicker lines as this one points to the inside)

You can also get the black vector by inverting the first blue vector.

With that knowledge it shouldn't be too hard. There are probably other solutions for this, but this just came to my mind.
But you should remember that you won't get a line thickness in 3D (just incase you want to do this) because the inner vectors are always at the polygon-points-plane.

//Edit: You must also take the angle difference (between the outline and the vector pointing into the polygon) into account.

Quote:Original post by J-Fox
- The black is the orthogonal unit-vector of the green one (which can be multiplyed by the thickness to get thicker lines as this one points to the inside)

Shouldn't line thickness be perpendicular to the edge? The black vector isn't perpendicular to the edges.

I'm not saying that it can't be used, I'm just saying you can't use a constant thickness factor with it.
You have to use some more trigonometry to get its true length. If you know the length and angle of one side
of a triangle, you can get the remaining sides.

At every corner ABC, you want to find contour point that corresponds to B.
In code:
Vector Normal(Vector V){
return Vector(-V.y,V.x)
}
void Normalize(Vector &V){/// modifies V to make it have length 1.
l=1/sqrt(V.x*V.x + V.y*V.y)
V.x*=l;
V.y*=l;
}
...

N1=Normal(A-B);
N2=Normal(B-C);
Normalize(N1);
Normalize(N2);
D=(N1+N2);
s=D.x*D.x+D.y*D.y;
D=D*2*r/s;

where r is thickness. All capital letters are vectors.
How it works:
N1 and N2 are inward pointing normals. (or outward, depend to winding order)
let alpha= angle ABC
Sum of normals gives a vector which is pointing towards the contour point, and has length equal to 2*sin(alpha/2) [easy to see if you make a drawing]

Angle between D and AB is alpha/2
When contour point is at distance l along the line D , its distance to AB (and BC) is
l*sin(alpha/2) = r
l = r/sin(alpha/2)
Length of D is already equal to 2*sin(alpha/2) . So, when we multiply D by 2*r/[length of d squared] , we obtain length sin(alpha/2)*2*2*r/(2*sin(alpha/2))^2 = r/sin(alpha/2)

FYI Dmytry's solution is equivalent to explicitly building the lines and intersecting.

But if your current solution is working then it's probably good enough for now ;)