Cubic Bezier反向GetPoint方程：float for Vector Vector for float
鉴于结果值和四点，是否有可能获得浮动？ 如果是这样，怎么样？
public static Vector3 GetPoint (Vector3 p0, Vector3 p1, Vector3 p2, Vector3 p3, float t) { t = Mathf.Clamp01(t); float oneMinusT = 1f - t; return oneMinusT * oneMinusT * oneMinusT * p0 + 3f * oneMinusT * oneMinusT * t * p1 + 3f * oneMinusT * t * t * p2 + t * t * t * p3; } Jasper Flick撰写本教程的代码
它是，并涉及实现三阶函数的根查找。 一种直接的方法是实现Cardano的算法来查找三度多项式的根 – 可以在这里找到JavaScript的实现。 根据曲线的参数，您将获得最多三个相同的相关答案，因此根据您尝试查找t值的内容，您将需要做更多工作以找出哪些最多三个值你需要。
// Not in every toolbox, so: how to implement the cubic root // equivalent of the sqrt function (note that there are actually // three roots: one real, two complex, and we don't care about the latter): function crt(v) { if (v<0) return -pow(-v,1/3); return pow(v,1/3); } // Cardano's algorithm, based on // http://www.trans4mind.com/personal_development/mathematics/polynomials/cubicAlgebra.htm function cardano(curve, line) { // align curve with the intersecting line, translating/rotating // so that the first point becomes (0,0), and the last point // ends up lying on the line we're trying to use as root-intersect. var aligned = align(curve, line), // rewrite from [a(1-t)^3 + 3bt(1-t)^2 + 3c(1-t)t^2 + dt^3] form... pa = aligned[0].y, pb = aligned[1].y, pc = aligned[2].y, pd = aligned[3].y, // ...to [t^3 + at^2 + bt + c] form: d = ( -pa + 3*pb - 3*pc + pd), a = ( 3*pa - 6*pb + 3*pc) / d, b = (-3*pa + 3*pb) / d, c = pa / d, // then, determine p and q: p = (3*b - a*a)/3, p3 = p/3, q = (2*a*a*a - 9*a*b + 27*c)/27, q2 = q/2, // and determine the discriminant: discriminant = q2*q2 + p3*p3*p3, // and some reserved variables for later u1,v1,x1,x2,x3; // If the discriminant is negative, use polar coordinates // to get around square roots of negative numbers if (discriminant < 0) { var mp3 = -p/3, mp33 = mp3*mp3*mp3, r = sqrt( mp33 ), t = -q/(2*r), // deal with IEEE rounding yielding <-1 or >1 cosphi = t<-1 ? -1 : t>1 ? 1 : t, phi = acos(cosphi), crtr = crt(r), t1 = 2*crtr; x1 = t1 * cos(phi/3) - a/3; x2 = t1 * cos((phi+tau)/3) - a/3; x3 = t1 * cos((phi+2*tau)/3) - a/3; return [x1, x2, x3]; } else if(discriminant === 0) { u1 = q2 < 0 ? crt(-q2) : -crt(q2); x1 = 2*u1-a/3; x2 = -u1 - a/3; return [x1,x2]; } // one real root, and two imaginary roots else { var sd = sqrt(discriminant), tt = -q2+sd; u1 = crt(-q2+sd); v1 = crt(q2+sd); x1 = u1 - v1 - a/3; return [x1]; } }
求解下面的三次多项式应该揭示你原来的t：
(p3 - (3 * p2) + (3 * p1) - p0) * t^3 + ((3 * p2) - (6 * p1) + (3 * p0)) * t^2 + ((3 * p1) - (3 * p0)) * t + p0 = 0
我把它放在标准forms，这样你就可以很容易地从系数中得到根。