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Some refactoring and added comments to @sapics code in Curve.getNearestParameter()

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Jürg Lehni 2015-10-24 11:14:08 +02:00
parent cbefa668f4
commit 4678697638
1 changed files with 21 additions and 8 deletions
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29
src/path/Curve.js
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@ -645,16 +645,29 @@ statics: {
},
getNearestParameter: function(v, point) {
if (!this.hasHandles(v)) {
if (!Curve.hasHandles(v)) {
var p1x = v[0], p1y = v[1],
p2x = v[6], p2y = v[7],
v12x = p2x - p1x, v12y = p2y - p1x,
t = ((point.x - p1x) * v12x + (point.y - p1y) * v12y) /
(v12x * v12x + v12y * v12y),
epsilon = /*#=*/Numerical.EPSILON;
if (t < epsilon) return 0;
if (t + epsilon > 1) return 1;
return 0.5 - Math.cos(Math.acos(2 * t - 1) / 3 + Math.PI * 4 / 3) || 0.5;
vx = p2x - p1x, vy = p2y - p1x;
// Line has zero length, avoid divisions by zero
if (vx === 0 && vy === 0)
return 0;
// Project the point onto the line and calculate its linear
// parameter u along the line:
// u = (point - p1).dot(v) / v.dot(v)
var u = ((point.x - p1x) * vx + (point.y - p1y) * vy) /
(vx * vx + vy * vy);
if (u < /*#=*/Numerical.EPSILON)
return 0;
if (u > /*#=*/(1 - Numerical.EPSILON))
return 1;
// Now translate the linear u to the curve-time t:
// B(t) = (1-t)^3 P0 + 3(1-t)^2 t P1 + 3(1-t)t^2 P2 + t^3 P3
// This case is linear, so B(t) = u, with: P0 = P1 = 0, P2 = P3 = 1
// -> u = 3(1-t)t^2 + t^3
// -> 2 * t^3 - 3 * t^2 + u = 0
// We can calculate t from u using the trigonometric method:
return 0.5 - Math.cos((Math.acos(2 * u - 1) + Math.PI * 4) / 3);
}
var count = 100,