C# 在两个顶点高度可定制的点之间沿弧线路径查找航路点&；距离

我有一个功能性脚本，可以在两个游戏对象（vector3）之间为cinemachine轨迹创建一条弧，我想添加一个变量，以便控制弧的高度，如图所示：

目前，我所使用的方法是基于两点之间距离的半径创建圆弧，我希望能够控制圆弧的高度，以便它计算出与图中绘制的线类似的点。以及控制最大高度出现在弧中与posA和posB之间的距离相关的位置

下面是我用来创建当前圆弧的代码，用于计算半径为两点之间距离一半的圆弧：

static void CalcWaypoints(CinemachineSmoothPath path, Vector3 posA, Vector3 posB)
{
    int greenRingPointsNum = 8;
    float metersPerWaypoint = 10;
    //Here we calculate how many segments will fit between the two points
    int segmentsToCreate = Mathf.RoundToInt(Vector3.Distance(posA, posB) / metersPerWaypoint);
    path.m_Waypoints = new CinemachineSmoothPath.Waypoint[segmentsToCreate + greenRingPointsNum];
    Debug.Log("Creating " + segmentsToCreate + " waypoints");
    
    
    // get circle center and radius
    var radius = Vector3.Distance(posA, posB) / 2f;
    var centerPos = (posA + posB) / 2f;

    // get a rotation that looks in the direction of the target gameobject
    var centerDirection = Quaternion.LookRotation((posA - posB).normalized);

    for (var i = 0; i < segmentsToCreate; i++)
    {
        
        var angle = Mathf.PI * (i) / (segmentsToCreate + 1f);
        var y = Mathf.Sin(angle) * radius;
        var z = Mathf.Cos(angle) * radius;
        var pos = new Vector3(0, y, z);
        // Rotate the pos vector according to the centerDirection
        pos = centerDirection * pos;
        path.m_Waypoints[i] = new CinemachineSmoothPath.Waypoint();
        path.m_Waypoints[i].position = centerPos + pos;
    }

    //create a circle of points around the target gameobject at a give radius
    float greenRadius = 20f;
    int waypointNum = segmentsToCreate;
    for (int i = 0; i < greenRingPointsNum; i++)
    {
        float angle = i * Mathf.PI * 2f / greenRingPointsNum;
        Vector3 newPos = posB + new Vector3(Mathf.Cos(angle) * greenRadius, posB.y, Mathf.Sin(angle) * greenRadius);
        path.m_Waypoints[waypointNum] = new CinemachineSmoothPath.Waypoint();
        path.m_Waypoints[waypointNum].position = newPos;
        waypointNum++;
    }

}

static void CalcWaypoints（路径、矢量3位置A、矢量3位置B）
{
int格林林点snum=8；
浮子流量计通过点=10；
//在这里，我们计算两点之间适合的线段数
int segmentsToCreate=Mathf.RoundToInt（矢量3.距离（位置A、位置B）/米/路径点）；
path.m_航路点=新航路点[segmentsToCreate+GreenringPointsUM]；
Debug.Log（“创建”+segmentsToCreate+“航路点”）；
//获取圆心和半径
var半径=矢量3.距离（位置A、位置B）/2f；
var centerPos=（posA+posB）/2f；
//获得朝向目标游戏对象方向的旋转
var centerDirection=Quaternion.LookRotation（（posA-posB）.normalized）；
对于（变量i=0；i

希望你们中的一位伟人能帮上忙：）

如下图所示的圆弧，我可以控制高度以及两点之间的距离：

---

最简单的方法可能是动态创建并遵循它们

首先，我们定义顶点的方向，并找到顶点：

float apexHeightFromA = 5f;  // apex is 5 world units above A
float apexDistanceFactor = 0.5f; // apex is halfway from A to B
Vector3 upDirection = Vector3.up; // direction apex is in relative to points

Vector3 aToB = posB - posA;
Vector3 flatAToB = Vector3.ProjectOnPlane(aToB, upDirection);

Vector3 posApex = posA 
        + flatAToB * apexDistanceFactor
        + apexHeightFromA * upDirection;

现在，可以定义Bézier曲线了。如果我们使用两条三次Bézier曲线，我们将需要两个控制点，一个在顶点的两侧。也就是说，一个朝向A点，一个朝向B点

Vector3 controlPointApexA;
Vector3 controlPointApexB;

如何定义这些是任意的。一个好的起点可能是水平移动到每个控制点朝向的终点的一半

Vector3 apexToA = posA - posApex;
Vector3 apexToB = posB - posApex;

float controlPointDistanceFactor = 0.5f;

controlPointA = posApex 
        + controlPointDistanceFactor * Vector3.ProjectOnPlane(apexToA, upDirection);
controlPointB = posApex 
        + controlPointDistanceFactor * Vector3.ProjectOnPlane(apexToB, upDirection);

不管我们如何定义控制点，我们都可以沿着曲线进行迭代

首先，我们需要确定顶点前后应该有多少个航路点。一个合理的假设是根据其在各点之间的位置来实现

float apexTravelFactor = apexDistanceFactor; 

然后，我们可以使用二次Bézier曲线的公式

。。。从A到顶点或从顶点到B取决于我们在曲线中的位置：

// cache for efficiency
Vector3 controlAToA = posA - controlPointA;
Vector3 controlBToB = posB - controlPointB;

Vector3 controlAToApex = posApex - controlPointA;
Vector3 controlBToApex = posApex - controlPointB;


for (var i = 0; i < segmentsToCreate; i++)
{
    float overallT = (float)i / segmentsToCreate;

    Vector3 control, controlToOrigin, controlToDest;
    float t;

    // are we going from a to apex or apex to b?
    if (overallT < apexTravelFactor)
    {
        // going from a to apex
        control = controlPointA;
        controlToOrigin = controlAToA;
        controlToDest = controlAToApex;

        t = overallT / apexTravelFactor;
    }
    else 
    {
        // going from apex to b
        control = controlPointB;
        controlToOrigin = controlBToApex;
        controlToDest = controlBToB;

        t = (overallT - apexTravelFactor) / (1f - apexTravelFactor);
    }

    Vector3 currentPos = control 
            + Mathf.Pow(1f - t, 2f) * controlToOrigin
            + Mathf.Pow(t, 2f) * controlToDest;
    
    path.m_Waypoints[i] = new CinemachineSmoothPath.Waypoint();
    path.m_Waypoints[i].position = currentPos;
}

---

最简单的方法可能是动态创建并遵循它们

首先，我们定义顶点的方向，并找到顶点：

float apexHeightFromA = 5f;  // apex is 5 world units above A
float apexDistanceFactor = 0.5f; // apex is halfway from A to B
Vector3 upDirection = Vector3.up; // direction apex is in relative to points

Vector3 aToB = posB - posA;
Vector3 flatAToB = Vector3.ProjectOnPlane(aToB, upDirection);

Vector3 posApex = posA 
        + flatAToB * apexDistanceFactor
        + apexHeightFromA * upDirection;

现在，可以定义Bézier曲线了。如果我们使用两条三次Bézier曲线，我们将需要两个控制点，一个在顶点的两侧。也就是说，一个朝向A点，一个朝向B点

Vector3 controlPointApexA;
Vector3 controlPointApexB;

如何定义这些是任意的。一个好的起点可能是水平移动到每个控制点朝向的终点的一半

Vector3 apexToA = posA - posApex;
Vector3 apexToB = posB - posApex;

float controlPointDistanceFactor = 0.5f;

controlPointA = posApex 
        + controlPointDistanceFactor * Vector3.ProjectOnPlane(apexToA, upDirection);
controlPointB = posApex 
        + controlPointDistanceFactor * Vector3.ProjectOnPlane(apexToB, upDirection);

不管我们如何定义控制点，我们都可以沿着曲线进行迭代

首先，我们需要确定顶点前后应该有多少个航路点。一个合理的假设是根据其在各点之间的位置来实现

float apexTravelFactor = apexDistanceFactor; 

然后，我们可以使用二次Bézier曲线的公式

。。。从A到顶点或从顶点到B取决于我们在曲线中的位置：

// cache for efficiency
Vector3 controlAToA = posA - controlPointA;
Vector3 controlBToB = posB - controlPointB;

Vector3 controlAToApex = posApex - controlPointA;
Vector3 controlBToApex = posApex - controlPointB;


for (var i = 0; i < segmentsToCreate; i++)
{
    float overallT = (float)i / segmentsToCreate;

    Vector3 control, controlToOrigin, controlToDest;
    float t;

    // are we going from a to apex or apex to b?
    if (overallT < apexTravelFactor)
    {
        // going from a to apex
        control = controlPointA;
        controlToOrigin = controlAToA;
        controlToDest = controlAToApex;

        t = overallT / apexTravelFactor;
    }
    else 
    {
        // going from apex to b
        control = controlPointB;
        controlToOrigin = controlBToApex;
        controlToDest = controlBToB;

        t = (overallT - apexTravelFactor) / (1f - apexTravelFactor);
    }

    Vector3 currentPos = control 
            + Mathf.Pow(1f - t, 2f) * controlToOrigin
            + Mathf.Pow(t, 2f) * controlToDest;
    
    path.m_Waypoints[i] = new CinemachineSmoothPath.Waypoint();
    path.m_Waypoints[i].position = currentPos;
}

---

不，不是抛物线，我只想在posA和PosB之间创建一个平滑的圆弧，但能够控制三个潜在圆弧的最大高度。双曲线，抛物线，双曲线（近似）再考虑一下。。。如果你想绕着圆走更长的路，那么圆弧可以达到任意高：p尽管如此，你还是怀疑这是你想要的！所以我更新了这个问题，是的，我想控制弧中的最大高度，与posA和posB之间的距离有关，感谢@Ruzihm的链接，但是公式让我困惑，我在学校做这种数学已经40年了，是的，40年了：）不，不是抛物线，我只想在posA和PosB之间创建一个平滑的圆弧，但能够控制三个潜在圆弧的最大高度。双曲线，抛物线，双曲线（近似）再考虑一下。。。如果你想绕着圆走更长的路，那么圆弧可以达到任意高：p尽管如此，你还是怀疑这是你想要的！所以我更新了这个问题，是的，我想控制弧线中的最大高度，相对于posA和posB之间的距离，感谢链接@Ruzihm，