三次Bezier反向用GetPoint公式：浮子矢量< = GT;矢量浮法 [英] Cubic Bezier reverse GetPoint equation: float for Vector <=> Vector for float

问题描述

是否有可能获得浮丁字裤给出的结果值和四点？
。如果是这样，怎么样？

Is it possible to get float t back given the resulting value and the four points? If so, how?

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;
}

的代码从本教程由碧玉弗里克

推荐答案

它是，并且涉及实施三度函数求根。这样做的一个直接的方法就是推行卡尔达诺的算法寻找根源三度的多项式 - 一个JavaScript实现，可以发现这里。根据曲线的参数，你会得到最多三个同样correcet答案，所以这取决于你试图找到 T 的价值，你就必须做更多。努力找出那些先进的三个值的需要

It is, and involves implementing root finding for third degree functions. One direct way of doing that is to implement Cardano's Algorithm for finding the roots for a polynomial of degree three - a JavaScript implementation of that can be found here. Depending on the curve's parameters, you will get up to three equally correcet answers, so depending on what you were trying to find the t value for, you'll have to do more work to find out which of those up-to-three values you need.

// 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];
  }
}

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