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Improve Curve#isFlatEnough() by finding the right threshold through testing, and remove other less precise solutions.

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Jürg Lehni 2011-11-12 01:23:41 +01:00
parent d43b54a531
commit a326b189b4
1 changed files with 2 additions and 48 deletions
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50
src/path/Curve.js
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@ -533,49 +533,6 @@ var Curve = this.Curve = Base.extend(/** @lends Curve# */{
},
isFlatEnough: function(v) {
// Code from Nearest Point-on-Curve Problem and by Philip J.
// Schneider from "Graphics Gems", Academic Press, 1990, adapted
// and optimised for cubic bezier curves.
// Derive the implicit equation for line connecting first and last
// control points
/*
var p1x = v[0], p1y = v[1],
c1x = v[2], c1y = v[3],
c2x = v[4], c2y = v[5],
p2x = v[6], p2y = v[7],
a = p1y - p2y,
b = p2x - p1x,
c = p1x * p2y - p2x * p1y,
// Find the largest distance from each of the points to the line
v1 = a * c1x + b * c1y + c,
v2 = a * c2x + b * c2y + c;
// Compute intercepts of bounding box
return Math.abs((v1 * v1 + v2 * v2) / (a * (a * a + b * b))) < 0.005;
*/
// Inspired by Skia, but to be tested:
// Calculate 1/3 (m1) and 2/3 (m2) along the line between start (p1)
// and end (p2), measure distance from there the control points and
// see if they are further away than 1/2.
// Seems all very inaccurate, especially since the distance
// measurement is just the bigger one of x / y...
// TODO: Find a more accurate and still fast way to determine this.
/*
var p1x = v[0], p1y = v[1],
c1x = v[2], c1y = v[3],
c2x = v[4], c2y = v[5],
p2x = v[6], p2y = v[7],
vx = (p2x - p1x) / 3,
vy = (p2y - p1y) / 3,
m1x = p1x + vx,
m1y = p1y + vy,
m2x = p2x - vx,
m2y = p2y - vy;
return Math.max(
Math.abs(m1x - c1x), Math.abs(m1y - c1y),
Math.abs(m2x - c1x), Math.abs(m1y - c1y)) < 1 / 2;
*/
// Thanks to Kaspar Fischer for the following:
// http://www.inf.ethz.ch/personal/fischerk/pubs/bez.pdf
var p1x = v[0], p1y = v[1],
@ -585,11 +542,8 @@ var Curve = this.Curve = Base.extend(/** @lends Curve# */{
ux = 3 * c1x - 2 * p1x - p2x,
uy = 3 * c1y - 2 * p1y - p2y,
vx = 3 * c2x - 2 * p2x - p1x,
vy = 3 * c2y - 2 * p2y - p1y,
tol = Numerical.TOLERNACE;
// Tolerance is 16 * tol ^ 2
return Math.max(ux * ux, vx * vx) + Math.max(uy * uy, vy * vy)
<= 16 * tol * tol;
vy = 3 * c2y - 2 * p2y - p1y;
return Math.max(ux * ux, vx * vx) + Math.max(uy * uy, vy * vy) < 1;
}
}
}, new function() { // Scope for methods that require numerical integration