scratchfoundation/paper.js
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Further simplify code.

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This commit is contained in:
Jürg Lehni 2015-10-24 11:29:15 +02:00
parent 4678697638
commit b9a0f5f659
1 changed files with 4 additions and 6 deletions
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10
src/path/Curve.js
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@ -649,7 +649,7 @@ statics: {
var p1x = v[0], p1y = v[1], var p1x = v[0], p1y = v[1],
p2x = v[6], p2y = v[7], p2x = v[6], p2y = v[7],
vx = p2x - p1x, vy = p2y - p1x; vx = p2x - p1x, vy = p2y - p1x;
// Line has zero length, avoid divisions by zero // Line has zero length, avoid divisions by zero.
if (vx === 0 && vy === 0) if (vx === 0 && vy === 0)
return 0; return 0;
// Project the point onto the line and calculate its linear // Project the point onto the line and calculate its linear
@ -657,17 +657,15 @@ statics: {
// u = (point - p1).dot(v) / v.dot(v) // u = (point - p1).dot(v) / v.dot(v)
var u = ((point.x - p1x) * vx + (point.y - p1y) * vy) / var u = ((point.x - p1x) * vx + (point.y - p1y) * vy) /
(vx * vx + vy * 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: // 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 // 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 // This case is linear, so B(t) = u, with: P0 = P1 = 0, P2 = P3 = 1
// -> u = 3(1-t)t^2 + t^3 // -> u = 3(1-t)t^2 + t^3
// -> 2 * t^3 - 3 * t^2 + u = 0 // -> 2 * t^3 - 3 * t^2 + u = 0
// We can calculate t from u using the trigonometric method: // We can calculate t from u using the trigonometric method:
return 0.5 - Math.cos((Math.acos(2 * u - 1) + Math.PI * 4) / 3); return u < /*#=*/Numerical.EPSILON ? 0
: u > /*#=*/(1 - Numerical.EPSILON) ? 1
: 0.5 - Math.cos((Math.acos(2 * u - 1) + Math.PI * 4) / 3);
} }
var count = 100, var count = 100,