Lines Matching refs:bisector
191 // a,b are >90 degrees apart. Find the bisector of their interior normals instead. (Above 90
197 // a,b are <-90 degrees apart. Find the bisector of their interior normals instead. (Below
214 // midtangent. The bisector of tan0 and -tan1 is orthogonal to the midtangent:
220 SkVector bisector = SkFindBisector(tan0, -tan1);
222 // The midtangent can be found where (F' dot bisector) = 0:
224 // 0 = (F'(T) dot bisector) = |2*T 1| * |p0 - 2*p1 + p2| * |bisector.x|
225 // |-2*p0 + 2*p1 | |bisector.y|
230 // = 2*T * ((tan1 - tan0) dot bisector) + (2*tan0 dot bisector)
232 // T = (tan0 dot bisector) / ((tan0 - tan1) dot bisector)
233 float T = sk_ieee_float_divide(tan0.dot(bisector), (tan0 - tan1).dot(bisector));
608 // midtangent. The bisector of tan0 and -tan1 is orthogonal to the midtangent:
610 // bisector dot midtangent == 0
614 SkVector bisector = SkFindBisector(tan0, -tan1);
618 // midtangent dot bisector == 0, or using a tangent matrix C' in power basis form:
621 // |T^2 T 1| * |. . | * |bisector.x| == 0
622 // |. . | |bisector.y|
624 // The coeffs for the quadratic equation we need to solve are therefore: C' * bisector
634 Sk4f coeffs = C_x * bisector.x() + C_y * bisector.y();
1513 // midtangent. The bisector of tan0 and -tan1 is orthogonal to the midtangent:
1515 // bisector dot midtangent = 0
1519 SkVector bisector = SkFindBisector(tan0, -tan1);
1536 // Now solve for "bisector dot midtangent = 0":
1539 // bisector * |A B C| * |T | = 0
1542 float a = bisector.dot(A);
1543 float b = bisector.dot(B);
1544 float c = bisector.dot(C);
1698 // compute the bisector vector, and then rescale to be the off-curve point.