I've found an implementation that uses whether bc is a 'right turn'
from ab as its sole corvergence criterion (where a b and c are sites
in neighboring regions). The code it uses is:
if ((b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y) 0)
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You
I just tested it . . .right turn detection did the trick. so there's
your answer.
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Better answer for #1 that I'm pretty sure will work:
calculate the center and radius of the circle. if the radius is not
NaN (i.e the sites are not collinear) and the center of the circle is
between the current locations of the breakpoints, they will converge.
On Nov 21, 6:36 am, bordaigorl
Also, in response to your idea, it doesn't make sense. . .you calculate
the center of the circle BY calculating where the bisectors intersect.
Or vice versa. The two values are always equivalent. Also, your fear
about numerical instability is unfounded. Calculating the orthocenter
by bisector
