细分Bezier曲线

Question

我想实现我可以细分具有给定距离的贝塞尔曲线。现在它可以工作，如果贝塞尔是一条直线，但如果我改变一个控制点（B＆C），使贝塞尔得到弯曲，计算点之间的差距就不再像给定的距离！

我通过网络浏览，但没有遇到类似的问题。

float t = Distance between subdividedParts / bezier length;

//A,B,C,D = ControllPoints of Bezier

GetPoint(A,B,C,D,t);

---

//GetPoint equation:
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;
     }

Answer 1

我现在已经设法得到非常准确的方法来分割Bezier并获得Positions =>但是它的性能消耗正在增加，因为它更准确。所以这可以在这段代码中得到改进：

 //if accuracy is 0.001 = good performance | if 0.000001 laggy performance
    public Vector3[] GetPoints (float gap,float accuracy){
     SimpsonVec sv = SV_Setup(0);
     Vector3 last_spawn = Bezier.GetPoint(sv.A,sv.B,sv.C,sv.D,0);

     List<Vector3> allPoints = new List<Vector3>();
     allPoints.Add(last_spawn);

     for(float t = accuracy;t <= 1.0f; t +=accuracy){
         Vector3 trial = Bezier.GetPoint(sv.A,sv.B,sv.C,sv.D,t);
         if(Vector3.Distance(trial,last_spawn) >= gap){
             last_spawn = trial;
             allPoints.Add(trial);
         }
     }
     return allPoints.ToArray();
 }

为了更好的表现，我现在做了这个：

Vector3 []数组输出的代码

public Vector3[] GetAllPoints(float gap,float acc){

    SimpsonVector = SV_SETUP_ALL();
    BezierPoints bp = new BezierPoints();
    bp.bp_vector3 = new List<Vector3>();
    bp.bp_lastSpawn = new List<Vector3>();

    for(int i = 0; i<points.Length / 3;i++){

        Vector3 ls = new Vector3();
        if(i == 0){
            ls = Bezier.GetPoint(SimpsonVector[0].A,SimpsonVector[0].B,SimpsonVector[0].C,SimpsonVector[0].D,0);
        }if (i > 0){
            ls = bp.bp_lastSpawn[i-1];
        }
        BezierPoints bp_temp = GetSegmentPoints(gap,acc,i,ls);
        bp.bp_lastSpawn.Add(bp_temp.bp_lastSpawn[0]);
        bp.bp_vector3.AddRange(bp_temp.bp_vector3);
        SimpsonVector_TEMP = SimpsonVector;
    }


    return bp.bp_vector3.ToArray();
}

BezierPoints GetSegmentPoints (float gap,float acc,int index, Vector3 ls)
{
    SimpsonVec sv = SimpsonVector[index];
    Vector3 last_spawn = ls;

    BezierPoints bp = new BezierPoints();
    bp.bp_vector3 = new List<Vector3>();
    bp.bp_lastSpawn = new List<Vector3>();

    float step = 0.1f;
    float t = step;
    float lastT = new float();

    while (t >= 0 && t <= 1f)
    {
        while (t < 1f && Vector3.Distance(Bezier.GetPoint(sv.A,sv.B,sv.C,sv.D,t), last_spawn) < gap){
            t += step;}
        step /= acc;
        while (t > lastT && Vector3.Distance(Bezier.GetPoint(sv.A,sv.B,sv.C,sv.D,t), last_spawn) > gap){
            t -= step;}
        step /= acc;
        if (t > 1f || t < lastT){
            break;}
        if(step < 0.000001f){
            last_spawn = Bezier.GetPoint(sv.A,sv.B,sv.C,sv.D,t);
            bp.bp_vector3.Add(last_spawn + transform.position);
            lastT = t;
            step = 0.1f;
        }
    }
    bp.bp_lastSpawn.Add(last_spawn);
    return bp;
}

结构：

public struct SimpsonVec{
    [SerializeField] public Vector3 A;
    [SerializeField] public Vector3 B;
    [SerializeField] public Vector3 C;
    [SerializeField] public Vector3 D;
}

public struct  BezierPoints
{
    [SerializeField] public List<Vector3> bp_vector3;
    [SerializeField] public List<Vector3> bp_lastSpawn; 
}

帮手方法：

public SimpsonVec SV_Setup(int index){
     SimpsonVec sv;
     sv.A = points[index];
     sv.B = points[index+1];
     sv.C = points[index+2];
     sv.D = points[index+3];
     return sv;

 }
 public SimpsonVec[] SV_SETUP_ALL(){
     SimpsonVec[] sv = new SimpsonVec[points.Length / 3];
     for(int i = 0; i<points.Length / 3;i++){
         sv[i] = SV_Setup(i*3);
     }
     return sv;
 }
 public 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;
 }

Answer 2

您无论如何都需要绘制曲线，因此请保持曲线的查找表并预先计算每个条目的距离，或者通过选择t值，计算距离，然后移动t值的一半来二进制搜索您的胜利之路/如果你关闭了，请重复，直到你达到所需的精度。而且因为二进制搜索是疯狂有效的，所以你在那里尝试的次数可以忽略不计。

请参阅https://pomax.github.io/bezierinfo/#tracing的基本原理，https://pomax.github.io/bezierinfo/#arclength用于计算曲线的长度（https://pomax.github.io/bezierinfo/#splitting是获取确定曲线长度t所需的值的明显部分），以及“所有https://pomax.github.io/bezierinfo获取更多信息”，真的。