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Fix regression issues with Curve-Line intersection code

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hkrish 2013-10-11 20:48:34 +02:00
parent dfa1687a90
commit f0434548c6
1 changed files with 36 additions and 30 deletions
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66
src/path/Curve.js
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@ -1356,10 +1356,8 @@ new function() { // Scope for methods that require numerical integration
/*#*/ } // options.fatline
/**
* Intersections between curve and line becomes rather simple here mostly
* because of Numerical class. We can rotate the curve and line so that the
* line is on the X axis, and solve the implicit equations for the X axis
* and the curve.
* Intersections between curve and line based on the algebraic method
* described at http://www.particleincell.com/blog/2013/cubic-line-intersection/
*/
function addCurveLineIntersections(v1, v2, curve1, curve2, locations) {
var flip = Curve.isLinear(v1),
@ -1367,38 +1365,46 @@ new function() { // Scope for methods that require numerical integration
vl = flip ? v1 : v2,
l1x = vl[0], l1y = vl[1],
l2x = vl[6], l2y = vl[7],
// Rotate both curve and line around l1 so that line is on x axis
lvx = l2x - l1x,
lvy = l2y - l1y,
// Angle with x axis (1, 0)
angle = Math.atan2(-lvy, lvx),
sin = Math.sin(angle),
cos = Math.cos(angle),
// (rl1x, rl1y) = (0, 0)
rl2x = lvx * cos - lvy * sin,
vcr = [];
vc0 = vc[0], vc1 = vc[1], vc2 = vc[2], vc3 = vc[3],
vc4 = vc[4], vc5 = vc[5],
// Equation of the line Ax + By + C = 0
A = l2y-l1y, //A=y2-y1
B = l1x-l2x, //B=x1-x2
C = l1x*(l1y-l2y) + l1y*(l2x-l1x), //C=x1*(y1-y2)+y1*(x2-x1)
// Bernstein coefficients for the curve
bx0 = -vc0 + 3*vc2 + -3*vc4 + vc[6],
bx1 = 3*vc0 - 6*vc2 + 3*vc4,
bx2 = -3*vc0 + 3*vc2,
bx3 = vc0,
by0 = -vc1 + 3*vc3 + -3*vc5 + vc[7],
by1 = 3*vc1 - 6*vc3 + 3*vc5,
by2 = -3*vc1 + 3*vc3,
by3 = vc1,
// Form the cubic equation
// a*t^3 + b*t^2 + c*t + d = 0
a = A*bx0 + B*by0, /*t^3*/
b = A*bx1 + B*by1, /*t^2*/
c = A*bx2 + B*by2, /*t*/
d = A*bx3 + B*by3 + C, /*1*/
roots = [], count, x0, x1, t, tl;
for(var i = 0; i < 8; i += 2) {
var x = vc[i] - l1x,
y = vc[i + 1] - l1y;
vcr.push(
x * cos - y * sin,
y * cos + x * sin);
}
var roots = [],
count = Curve.solveCubic(vcr, 1, 0, roots);
// Solve the cubic equation
count = Numerical.solveCubic(a, b, c, d, roots);
// NOTE: count could be -1 for inifnite solutions, but that should only
// happen with lines, in which case we should not be here.
for (var i = 0; i < count; i++) {
var t = roots[i];
if (t >= 0 && t <= 1) {
var point = Curve.evaluate(vcr, t, 0);
for (var i=0;i<count;i++) {
t = roots[i];
if(t >= 0 && t <= 1.0){
x0 = bx0*t*t*t + bx1*t*t + bx2*t + bx3;
x1 = by0*t*t*t + by1*t*t + by2*t + by3;
// We do have a point on the infinite line. Check if it falls on
// the line *segment*.
if (point.x >= 0 && point.x <= rl2x) {
// tl is the parameter of the intersection point in line segment.
// First check if line is vertical
tl = (l2x-l1x) != 0 ? (x0-l1x)/(l2x-l1x) : (x1-l1y)/(l2y-l1y);
if(tl >= 0 && tl <= 1.0){
// Interpolate the parameter for the intersection on line.
var tl = point.x / rl2x,
t1 = flip ? tl : t,
var t1 = flip ? tl : t,
t2 = flip ? t : tl;
addLocation(locations,
curve1, t1, Curve.evaluate(v1, t1, 0),