Question

我已经看过无数的三维球体交叉问题，不幸的是，它们要么远远超出我的理解能力，要么不能满足我的要求。

这是在Unity游戏引擎中并使用c＃

我设法让这段代码正常工作：

public void calculatePoints_H(Vector3 c1p, Vector3 c2p, float c1r, float c2r, out Vector3 startLine, out Vector3 endLine)
{
    //c1p = circle one position
    //c1r = circle one radius

    Vector3 P0 = c1p;
    Vector3 P1 = c2p;

    float d,a,h;


    d = Vector3.Distance(P0,P1);

    a = (c1r*c1r - c2r*c2r + d*d)/(2*d);

    h = Mathf.Sqrt(c1r*c1r - a*a);


    Vector3 P2 = (P1 - P0);
            P2 = (P2 * (a/d));
            P2 = (P2 + P0);

    float x3,y3,x4,y4 = 0;

    x3 = P2.x + h*(P1.y - P0.y)/d;
    y3 = P2.y - h*(P1.x - P0.x)/d;

    x4 = P2.x - h*(P1.y - P0.y)/d;
    y4 = P2.y + h*(P1.x - P0.x)/d;;

    //draw visual to screen (unity 3D engine)
    Debug.DrawLine(new Vector3(x3,0,y3), new Vector3(x4,0,y4),Color.green);

    //out parameters for a line renderer
    startLine = new Vector3(x3,0,y3);
    endLine = new Vector3(x4,0,y4);


}

目前这段代码允许我计算两个球体相交的x和z轴上的两个点，然后绘制一条线。

我想要实现的是xyz交点，所以我也可以在方法中添加高度（y向量3值），这样我就可以从任何方向/高度得到一个与另一个相交的球体

有人可以帮我理解这样做的方法吗，我的脑子有点油炸，我担心这是一个我想念的简单解决方案？

Answer 1

球体很好，因为你知道交叉点，如果它们是触摸的，那么沿着 AB 的矢量 SphereA 中心到 SphereB 中心。该距离是球半径的函数：

float dA = SphereA.radius / AB.magnitude // % distance along AB starting from SphereA

从那里计算您的交叉点的 AB 的距离：

Vector3 p = SphereA.position + AB * dA; // eq: p`= p + direction * time

示例（使用Unity的内置球体预制）：

bool Intersect(out Vector3 ip, float threshold=0.1f){
    // vector from sphere 1 -> sphere 2
    Vector3 ab = Sphere2.transform.position - Sphere1.transform.position;

    // Calculate radius from Unity built-in sphere.
    // Unity spheres are unit spheres (diameter = 1)
    // So diameter = scale, thus radius = scale / 2.0f.
    // **Presumes uniform scaling.
    float r1 = Sphere1.transform.localScale.x / 2.0f;
    float r2 = Sphere2.transform.localScale.x / 2.0f;

    // When spheres are too close or too far apart, ignore intersection.
    float diff = Mathf.Abs(r2 + r1 - ab.magnitude);
    if( diff >= threshold) {
        ip = Vector3.zero;
        return false;
    }
    // Intersection is the distance along the vector between
    // the 2 spheres as a function of the sphere's radius.
    ip = Sphere1.transform.position + ab * r1/ab.magnitude;
    return true;
}

示例用法：

void FixedUpdate(){
    Vector3 p; //will hold intersection point if one is found
    if(Intersect(out p,0.1f)){
        IntersectionPoint.transform.position = p;
        IntersectionPoint.renderer.enabled = true;
    } else {
        IntersectionPoint.renderer.enabled = false;
    }
}

这只会返回一个交叉点。多个交叉点，如球体重叠时，是一个不同的问题需要解决。

球体 - 球体交点c＃，碰撞点的3D坐标

1 个答案: