Lines Matching refs:cubic
2 http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi
6 Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2.
18 If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since
26 mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the
32 Compute the Tdiv as the root of (cubic) equation
34 if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a
37 0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point
39 Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation
57 static double calc_t_div(const SkDCubic& cubic, double precision, double start) {
62 cPtr = &cubic;
65 sub = cubic.subDivide(start, 1);
91 static bool add_simple_ts(const SkDCubic& cubic, double precision, SkTArray<double, true>* ts) {
92 double tDiv = calc_t_div(cubic, precision, 0);
103 static void addTs(const SkDCubic& cubic, double precision, double start, double end,
105 double tDiv = calc_t_div(cubic, precision, 0);
116 // FIXME: when called from recursive intersect 2, this could take the original cubic
118 // it would still take the prechopped cubic for reduce order and find cubic inflections
