Tangent for 5th order Bezier curve

Question

I'm using Bezier curves for gameobject placement. Mostly it's enough to use cubic curve but in some cases I need more precision. I'm not strong in math, so I found 5th order equation in wiki and the curve looks fine. Now I just need set angles. Can anyone help me with tangent equation for 5th order Bezier curve? [1]:

 //Cubic
 public Vector3 GetPosition(float t)
         {
             return (1 - t) * (1 - t) * (1 - t) * _p0 
                 + 3 * (1 - t) * (1 - t) * t * _p1
                 + 3 * (1 - t) * t * t * _p2 
                 + t * t * t * _p3;
         }
 
         public Vector3 GetTangent(float t)
         {
             float u = 1 - t;
             float uu = u * u;
             float tu = t * u;
             float tt = t * t;
 
             Vector3 P = _p0 * 3 * uu * (-1.0f);
             P += _p1 * 3 * (uu - 2 * tu);
             P += _p2 * 3 * (2 * tu - tt);
             P += _p3 * 3 * tt;
 
             return P.normalized;
         }
 
 //n5
 public Vector3 GetPosition(float t)
         {
             float u = 1 - t;
             float uu = u * u;
             float uuu = uu * u;
             float tt = t * t;
             float ttt = t * tt;
 
             Vector3 P = _p0 * uu * uuu;
             P += _p1 * 5 * uu * uu * t;
             P += _p2 * 10 * uuu * tt;
             P += _p3 * 10 * ttt * uu;
             P += _p4 * 5 * tt * tt * u;
             P += _p5 * ttt * tt;
 
             return P;
         }

Answer 1

Since I like math I just derived the tangent function myself (actually two times in two different ways):

 public Vector3 GetTangent(float t)
 {
     float u = 1f - t;
     float uu = u * u;
     float uuu = uu * u;
     float tt = t * t;
     float ttt = t * tt;
     Vector3 T = _p0 * (uuu * u)
         - _p1 * (uuu * (5 * t - 1))
         + _p2 * (2 * t * uu * (5 * t - 2))
         - _p3 * (2 * tt * u * (5 * t - 3))
         + _p4 * (ttt * (5 * t - 4))
         - _p5 * (ttt * t);
     return T;
 }

I have no time to test it, so feel free to try it out. Note I would recomment you also put brackets around your factors in your GetPosition method. Otherwise a lot performance would be wasted. A multiplication is carried out left to right by the compiler. So in your code you calculate _p0 * uu which returns a new vector3 and then you multiply that result by uuu. So those would be two Vector*float multiplciations. When you group the float factors in brackets they are calculated first and you only do a single Vector - float multiplication per control point. So something like this:

 public Vector3 GetPosition(float t)
 {
     float u = 1f - t;
     float uu = u * u;
     float uuu = uu * u;
     float tt = t * t;
     float ttt = t * tt;

     Vector3 P = _p0 * (uu * uuu)
         + _p1 * (5 * u * uuu * t)
         + _p2 * (10 * uuu * tt)
         + _p3 * (10 * ttt * uu)
         + _p4 * (5 * t * ttt * u)
         + _p5 * (tt * ttt);
     return P;
 }

Answer 2

Answer by spirit_2 · May 26 at 09:48 AM

Suddenly it hit me

         public Vector3 GetTangent(float t)
         {
             Vector3 p1 = GetPosition(t - 0.01f);
             Vector3 p2 = GetPosition(t + 0.01f);
 
             return (p2-p1).normalized;
         }

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Tangent for 5th order Bezier curve

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