Angle Sum of Self-intersecting polygon

If a polygon is self intersecting, is there any way to calculate its angle sum of the interior angles perhaps in terms of the number of sides and the number of points of intersection?

Note we define interior angles to be the angles you get when you FOLLOW the polygon (the angle could be on the exterior of the shape). Exterior angles are simply the complement of the interior ones.

I have worked on this and found that it seems to rely on the number of times you do a 360 degree turn while following the edges of the polygon.

Do you think that this can be related to the number of points of intersection or something else?

I would really like to be enlightened on this issue.

The easy way to deal with this problem is to add up turning angles. By that I mean the angle you turn at a corner. If the turning angle at a corner is $\theta$ and the interior angle is $\phi$, then $\theta+\phi=\pi$ (or $180^\circ$ if you prefer) – I am assuming here that you count the left side of the edge as the “interior” side. Summing the turning angles over the whole polygon you must get an integer multiple of $2\pi$. You get exactly $\pm2\pi$ if the polygon is not self-intersecting, the sign depending on the orientation. But in general you get $2\pi k$ for some integer $k$ which is interpreted as the number of turns as you follow the edge.

So now, if there are $n$ corners, you add up the formula above for all corners to get $$\sum_{i=1}^n\theta_i+\sum_{i=1}^n\phi_i=n\pi,$$ or in other words $$\sum_{i=1}^n\phi_i=(n-2k)\pi$$ for the sum of “interior” angles.

The most familiar cases are of course $k=1$ while $n=3$ or $n=4$.

• I think this is only partially right: I think summing to $\pm 2 \pi$ is a necessary but not sufficient condition for not self-intersecting. I seem to be able to come up with polygons where the intersection occurs between edges separated by more than two vertices (so an edge and its neighbor's neighbor), and yet the angles sum to $2\pi$. – Jay Lemmon Apr 11 '15 at 2:17
• @JayLemmon I don't think I claimed that the summing to $\pm2\pi$ is sufficient for the polygon to be non-self-intersecting. It will be sufficient if all the turns have the same sign; then the polygon will be convex. But in general, counterexamples are quite easy to make, as you say. – Harald Hanche-Olsen Apr 11 '15 at 20:10
• Hmm, I think the wording of "you get exactly $\pm2\pi$ if the polygon is not self-intersecting" is maybe a little awkward if you didn't mean it as necessary and sufficient. Especially in the context of the OP's question: "Do you think that this can be related to the number of points of intersection or something else?". But reading it again, now, you don't explicitly say anything that would mean sufficient. But I had to read things very carefully not to draw that conclusion. – Jay Lemmon Apr 12 '15 at 8:33

Try this formula $$180[(n-2)k]°$$ $n$=number of sides $k$=number of revolution of 360°

• How did you get this ? – Shailesh May 12 '17 at 1:22