title: Intersection between two Arcs? (arc = distance between pair of angles)
body: <p>I'm trying to find a way to calculate the intersection between two arcs.
  I need to use this to determine how much of an Arc is visually on the right half of a circle, and how much on the left.
  I though about creating an arc of the right half, and intersect that with the actual arc.
  But it takes me wayyy to much time to solve this, so I thought about asking here - someone must have done it before.</p>
  
  <p>Edit:
  I'm sorry the previous illustration was provided when my head was too heavy after crunching angles. I'll try to explain again:</p>
  
  <p>In <a href="http://i.stack.imgur.com/QSu2D.png" rel="nofollow">this link</a> you can see that I cut the arc in the middle to two halves, the right part of the Arc contains 135 degrees, and the left part has 90.</p>
  
  <p>This Arc starts at -180 and ends at 45. (or starts at 180 and ends at 405 if normalized).</p>
  
  <p>I have managed to create this code in order to calculate the amount of arc degrees contained in the right part, and in the left part:</p>
  
  <pre><code>f1 = (angle2&gt;270.0f?270.0f:angle2) - (angle1&lt;90.0f?90.0f:angle1);
  if (f1 &lt; 0.0f) f1 = 0.0f;
  f2 = (angle2&gt;640.0f?640.0f:angle2) - (angle1&lt;450.0f?450.0f:angle1);
  if (f2 &lt; 0.0f) f2 = 0.0f;
  f3 = (angle2&gt;90.0f?90.0f:angle2) - angle1;
  if (f3&lt;0.0f) f3=0.0f;
  f4 = (angle2&gt;450.0f?450.0f:angle2) - (angle1&lt;270.0f?270.0f:angle1); 
  if (f4&lt;0.0f) f4=0.0f;
  </code></pre>
  
  <p>It works great after normalizing the angles to be non-negative, but starting below 360 of course.
  Then f1 + f2 gives me the sum of the left half, and f3 + f4 gives me the sum of the right half.
  It also does not consider a case when the arc is defined as more than 360, which may be an "error" case.</p>
  
  <p>BUT, this seems like more of a "workaround", and not a correct mathematical solution.
  I'm looking for a more elegant solution, which should be based on "intersection" between two arc (because math has no "sides", its not visual";</p>
  
  <p>Thanks!!</p>