Given two circle segments of the same circle: A=[a1, a2] and B=[b1, b2], with:
- a1, a2, b1, b2 values in degree between -inf and +inf
- a1 <= a2 ; b1 <= b2
- a2-a1<=360; b2-b1<=360
How can I find out if these two circle segments overlap? (i.E. if they intersect or touch in at least one point)
Examples:
A=[ -45°, 45°]; B=[ 10°, 20°] ==> overlap
A=[ -45°, 45°]; B=[ 90°, 180°] ==> no overlap
A=[ -45°, 45°]; B=[ 180°, 360°] ==> overlap
A=[ -405°, -315°]; B=[ 180°, 360°] ==> overlap
A=[-3600°, -3601°]; B=[ 3601°, 3602°] ==> overlap (touching counts as overlap)
A=[ 3600°, 3601°]; B=[-3601°,-3602°] ==> overlap (touching counts as overlap)
A=[ -1°, 1°]; B=[ 3602°, 3603°] ==> no overlap
This looks like a deceptively simple problem but I cannot wrap my head around it. I currently have a basic idea for a solution which involves splitting each segment into two if it crosses 0°, but I am not sure if that covers all cases, and I was wondering if there is an elegant formula.
How about normalizing each degree value to 0-360:
Then you just check for intersection:
As @admaoldak mentioned, normalize the degrees first:
Now to check if b1 is within (a1,a2),
Final answer is:
For an interval [i.X , i.Y] , let's define the normalization i_norm = normalize(i) so that :
then we define another operation i_slide = slide(i) so that :
we can prove that, for your input A and B , A circle-overlaps with B if and only if :
interval-overlaps is defined in the same way as "intersection" in adamoldak's post.
and both operations normalize() and slide() are easy to be implemented.
take your example:
A=[-45°,45°]; B=[10°,20°]
, we haveand [315,405] interval-overlaps with [370,380]