Here's an odd thought.
For parallel quads, is there a relationship between any of the following
points that could be exploited?
original control point
left_curve and right_curve control points
center point on the original (and rt and lt) curve
perpendicular of center point on the curve(s)
center of line between endpoints
I'm just basing this on a "gut feel" of how quad curves behave, not on
any mathematical intuition, though. It also makes sense given that you
are computing the intersection of parallel versions of the lines. I'm
guessing that the intersections of all parallel lines equidistant from
the original pair form a line themselves and that line might be related
to other points we have access to.
But, it might be some food for thought on how to make parallel quads go
a lot faster...
...jim
On 4/15/2011 2:20 PM, Jim Graham wrote:
Hi Denis,
The strategy I took to work around this was to simply check for a zero
denominator in the Miter method and return the average of the endpoints
instead of the intersection points. That makes a straight line via a
quad that has colinear points, but it greatly simplifies the impact of
the fix.
To be safe (performance-wise), I made a separate copy of the method
called "safecomputeMiter()" and put the test only in that copy (which is
only used from the quad function) until I have time to do some more
exhaustive performance tests. To be honest, though, I don't imagine that
single test could affect performance (given that "den" is already
computed as the difference of two values, the subtract operation already
sets condition codes so it is simply a matter of how fast the processor
can take the branch or not, and this is likely a case where branch
prediction would pay off a lot...)
...jim
On 4/15/2011 10:38 AM, Denis Lila wrote:
Hi.
Jim Graham pointed out this bug to me.
A fix is here:
http://icedtea.classpath.org/~dlila/webrevs/7036754/webrev/
It just checks for inf/nan and just emits a line joining
the endpoints in that case.
The stroking is no longer symmetric with respect to right
and left polynomial degrees. This is a bit more general.
I have a question:
The "curve is a straight line" check I use is this:
737 float dotsq = (dx1 * dx3 + dy1 * dy3);
738 dotsq = dotsq * dotsq;
739 float l1sq = dx1 * dx1 + dy1 * dy1, l3sq = dx3 * dx3 + dy3 * dy3;
740 if (Helpers.within(dotsq, l1sq * l3sq, 4 * Math.ulp(dotsq))) {
However, this isn't making much sense mathematically
right now. I would like to avoid redoing the math
so if someone can quickly confirm/deny that it works
that would be nice.
Thank you,
Denis.
