I was hoping someone could help me figure out a computationally inexpensive method for detecting kinks in a line drawn parallel to a Bezier curve as you can see here
What I would like to do is be able to determine the intersection of the kink, the segment with a starting point before the intersection and the first segment with an ending point after the kink. This way I can simply remove any unnecessary segments and adjust the first and last segments to meet at the intersection.
Apologies if I'm using the incorrect terms. But as far as I understand it the way I'm positioning these segments is by determining the unit vector of the segments for the Bezier curve (yellow) and multiplying it by the offset and finding the normal vector to create two new start and end points for the offset segment (white).
Mathematics isn't my strong suit so I'm hoping someone can give me a push in the right direction.
EDIT: The image has actually been resized by HTML so if you're having a hard time seeing what I'm talking about here's the direct link: http://i.stack.imgur.com/xtils.png
As a first approximation, compute the radius of curvature of your Bezier curve. If the offset is greater or equal to the radius of curvature, you should look for a kink.
Specifically, for a cubic Bezier with control points
P0, P1, P2, P3:
B(t) = P0 * (1-t)^3 + P1 * 3*t*(1-t)^2 + P2 * 3*t^2*(1-t) + P3 * t^3 -> B'(t) = (P1-P0) * 3*(1-t)^2 + (P2-P1) * 6*t*(1-t) + (P3-P2) * 3*t^2 -> B''(t) = (P2+P0-2*P1) * 6*(1-t) + (P3+P1-2*P2) * 6*t let: cross2d(p, q) = p.x*q.y - p.y*q.x then, radius of curvature = |B'(t)|^3 / cross2d(B'(t), B''(t))
I have left the radius of curvature in signed form; the sign should indicate the side of the curve on which you can expect the kink.
Note: you can have zero radius of curvature, or infinite radius of curvature; it may be better to compare
signed_offset * cross2d(B'(t), B''(t)) instead.
B''(t)should be 2-D vectors, just like
P3. Also, note that
|B'(t)|should take the length of the 2-D vector. It is probably best to use a
Vector2structure (or any existing structure you are currently using for your 2-D points), so you don't have to recompute the Bezier weights multiple times. However, you can write scalar functions and run them for each axis if you want. Just keep in mind that you will need to put them together again when computing the vector length and the cross2d() - comingstorm 2012-04-05 20:40