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Blending Function/Bezier

开发者 https://www.devze.com 2023-03-01 13:45 出处:网络
Am I calculating the Bezier blend wrong? Any help would be appreciated. Thank you very much. double bezierBlend(int i, double u, int 开发者_如何学Cm) {

Am I calculating the Bezier blend wrong? Any help would be appreciated. Thank you very much.

double bezierBlend(int i, double u, int 开发者_如何学Cm) {       
   double blend = 1;
   blend = factorial(m) * pow(u, i) * pow(1 - u, (m - i)) / (factorial(i) * factorial(m - i));
   return blend;
}


Here's a sample to compute the Bezier blend function, following directly from the formulation:

double choose( long n, long k )
{
    long j;
    double a;

    a = 1;
    for (j = k + 1; j <= n; j++)
        a *= j;
    for (j = 1; j <= n - k; j++)
        a /= j;
    return a;
};

double bezierBlend( int i, double t, int n ) 
{
    return choose( n, i ) * pow(1 - t, n - i) * pow( t, i );
}

For most applications though, computing the powers and the binomial coefficients each time is absurdly inefficient. In typical applications, the degree of the curve is constant (e.g., 2 for quadratic or 3 for cubic), and you can compute the function much more efficiently by pre-expanding the formula. Here's an example for cubic curves:

double BezCoef(int i, double t)
{
  double tmp = 1-t;
  switch (i)
  {
  case 0: return tmp*tmp*tmp;

  case 1: return 3*tmp*tmp*t;

  case 2: return 3*tmp*t*t;

  case 3: return t*t*t;
  }
  return 0; // not reached
}
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