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.
 A: 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 $n=4$.
A: Try this formula 
$$180[(n-2)k]°$$
$n$=number of sides
$k$=number of revolution of 360°
