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问题是我对编程很陌生，现在我需要编写一个程序来计算许多球体的中心（最大 36，最小 3），每个球体有 4 个点 X、Y、Z。因为我的程序读取了一个 TXT 文件，其中包含我将其存储在列表中的点数据，其结构如下

bolas[n].xyz[row,element]

这意味着我对球体 1 的第一组点是这样的：

bolas[0] = 
 row0.  -> [0] [1] [2]
 row1.  -> [0] [1] [2]
 row2.  -> [0] [1] [2]
 row3.  -> [0] [1] [2]

因此，如果我想使用球体中第 1 行中的 X 值，我必须这样做：

bolas[0].xyz[0,0]

在网上搜索我发现有人转换了 java 代码并为 c# 实现它来计算球体的中心，他创建了一个类，但我很新，我不知道如何在他的类中使用元素，如何我应该把我的数据介绍给他的课吗？我如何得到结果；这是课程：

/// <summary>
/// Given four points in 3D space, solves for a sphere such that all four points
/// lie on the sphere's surface.
/// </summary>
/// <remarks>
/// Translated from Javascript on http://www.convertalot.com/sphere_solver.html, originally
/// linked to by http://stackoverflow.com/questions/13600739/calculate-centre-of-sphere-whose-surface-contains-4-points-c.
/// </remarks>
public class CircumcentreSolver
{
    private const float ZERO = 0;
    private double m_X0, m_Y0, m_Z0;
    private double m_Radius;
    private double[,] P = 
            {
                { ZERO, ZERO, ZERO },
                { ZERO, ZERO, ZERO },
                { ZERO, ZERO, ZERO },
                { ZERO, ZERO, ZERO }
            };

    /// <summary>
    /// The centre of the resulting sphere.
    /// </summary>
    public double[] Centre
    {
        get { return new double[] { this.m_X0, this.m_Y0, this.m_Z0 }; }
    }

    /// <summary>
    /// The radius of the resulting sphere.
    /// </summary>
    public double Radius
    {
        get { return this.m_Radius; }
    }

    /// <summary>
    /// Whether the result was a valid sphere.
    /// </summary>
    public bool Valid
    {
        get { return this.m_Radius != 0; }
    }

    /// <summary>
    /// Computes the centre of a sphere such that all four specified points in
    /// 3D space lie on the sphere's surface.
    /// </summary>
    /// <param name="a">The first point (array of 3 doubles for X, Y, Z).</param>
    /// <param name="b">The second point (array of 3 doubles for X, Y, Z).</param>
    /// <param name="c">The third point (array of 3 doubles for X, Y, Z).</param>
    /// <param name="d">The fourth point (array of 3 doubles for X, Y, Z).</param>
    public CircumcentreSolver(double[] a, double[] b, double[] c, double[] d)
    {
        this.Compute(a, b, c, d);
    }

    /// <summary>
    /// Evaluate the determinant.
    /// </summary>
    private void Compute(double[] a, double[] b, double[] c, double[] d)
    {
        P[0, 0] = a[0];
        P[0, 1] = a[1];
        P[0, 2] = a[2];
        P[1, 0] = b[0];
        P[1, 1] = b[1];
        P[1, 2] = b[2];
        P[2, 0] = c[0];
        P[2, 1] = c[1];
        P[2, 2] = c[2];
        P[3, 0] = d[0];
        P[3, 1] = d[1];
        P[3, 2] = d[2];

        // Compute result sphere.
        this.Sphere();
    }

    private void Sphere()
    {
        double r, m11, m12, m13, m14, m15;
        double[,] a =
                {
                    { ZERO, ZERO, ZERO, ZERO },
                    { ZERO, ZERO, ZERO, ZERO },
                    { ZERO, ZERO, ZERO, ZERO },
                    { ZERO, ZERO, ZERO, ZERO }
                };

        // Find minor 1, 1.
        for (int i = 0; i < 4; i++)
        {
            a[i, 0] = P[i, 0];
            a[i, 1] = P[i, 1];
            a[i, 2] = P[i, 2];
            a[i, 3] = 1;
        }
        m11 = this.Determinant(a, 4);

        // Find minor 1, 2.
        for (int i = 0; i < 4; i++)
        {
            a[i, 0] = P[i, 0] * P[i, 0] + P[i, 1] * P[i, 1] + P[i, 2] * P[i, 2];
            a[i, 1] = P[i, 1];
            a[i, 2] = P[i, 2];
            a[i, 3] = 1;
        }
        m12 = this.Determinant(a, 4);

        // Find minor 1, 3.
        for (int i = 0; i < 4; i++)
        {
            a[i, 0] = P[i, 0] * P[i, 0] + P[i, 1] * P[i, 1] + P[i, 2] * P[i, 2];
            a[i, 1] = P[i, 0];
            a[i, 2] = P[i, 2];
            a[i, 3] = 1;
        }
        m13 = this.Determinant(a, 4);

        // Find minor 1, 4.
        for (int i = 0; i < 4; i++)
        {
            a[i, 0] = P[i, 0] * P[i, 0] + P[i, 1] * P[i, 1] + P[i, 2] * P[i, 2];
            a[i, 1] = P[i, 0];
            a[i, 2] = P[i, 1];
            a[i, 3] = 1;
        }
        m14 = this.Determinant(a, 4);

        // Find minor 1, 5.
        for (int i = 0; i < 4; i++)
        {
            a[i, 0] = P[i, 0] * P[i, 0] + P[i, 1] * P[i, 1] + P[i, 2] * P[i, 2];
            a[i, 1] = P[i, 0];
            a[i, 2] = P[i, 1];
            a[i, 3] = P[i, 2];
        }
        m15 = this.Determinant(a, 4);

        // Calculate result.
        if (m11 == 0)
        {
            this.m_X0 = 0;
            this.m_Y0 = 0;
            this.m_Z0 = 0;
            this.m_Radius = 0;
        }
        else
        {
            this.m_X0 = 0.5 * m12 / m11;
            this.m_Y0 = -0.5 * m13 / m11;
            this.m_Z0 = 0.5 * m14 / m11;
            this.m_Radius = System.Math.Sqrt(this.m_X0 * this.m_X0 + this.m_Y0 * this.m_Y0 + this.m_Z0 * this.m_Z0 - m15 / m11);
        }
    }

    /// <summary>
    /// Recursive definition of determinate using expansion by minors.
    /// </summary>
    private double Determinant(double[,] a, double n)
    {
        int i, j, j1, j2;
        double d = 0;
        double[,] m = 
                {
                    { ZERO, ZERO, ZERO, ZERO },
                    { ZERO, ZERO, ZERO, ZERO },
                    { ZERO, ZERO, ZERO, ZERO },
                    { ZERO, ZERO, ZERO, ZERO }
                };

        if (n == 2)
        {
            // Terminate recursion.
            d = a[0, 0] * a[1, 1] - a[1, 0] * a[0, 1];
        }
        else
        {
            d = 0;
            for (j1 = 0; j1 < n; j1++) // Do each column.
            {
                for (i = 1; i < n; i++) // Create minor.
                {
                    j2 = 0;
                    for (j = 0; j < n; j++)
                    {
                        if (j == j1) continue;
                        m[i - 1, j2] = a[i, j];
                        j2++;
                    }
                }

                // Sum (+/-)cofactor * minor.
                d = d + System.Math.Pow(-1.0, j1) * a[0, j1] * this.Determinant(m, n - 1);
            }
        }

        return d;
    }
}

我怎么说我的数据球体编号可能会有所不同，但我最多有 36 个球体，每个球体有 4 个点 x、y、z。如果我可以将结果中心存储在另一个列表中，这将非常有用，可能类似于：

ballCent[0]=
center-> [0][1][2] //center of the sphere[x][y][z].
radius-> [0]       //radius of the sphere.

我希望我解释得足够清楚，我不是以英语为母语的人，我非常感谢社区的帮助。PS。我个人用我的数据尝试了该程序的 java 版本，它对我来说非常完美。这是链接：http://www.convertalot.com/sphere_solver.html

4

2 回答 2

0

给定四个点a、b、c和d，您可以通过将以下行列式设置为零并求解它来找到中心：

| (x^2  + y^2  + z^2)    x   y   z  1 |
| (ax^2 + ay^2 + az^2)  ax  ay  az  1 |
| (bx^2 + by^2 + bz^2)  bx  by  bz  1 | = 0.
| (cx^2 + cy^2 + cz^2)  cx  cy  cz  1 |
| (dx^2 + dy^2 + dz^2)  dx  dy  dz  1 |

数学很粗糙，但以下 C++ 代码实现了解决方案：

class Point {
  public:
    double x;
    double y;
    double z;
    Point() { x = 0; y = 0; z = 0; }
    Point(double x_, double y_, double z_) { x = x_; y = y_; z = z_; }
};
    
class Sphere {
  public:
    Point center;
    double radius;
    Sphere(Point center_, double radius_) {
        center = Point(center_.x, center_.y, center_.z);
        radius = radius_;
    }
};
    
Sphere
sphereFromFourPoints(Point a, Point b, Point c, Point d)
{
#define U(a,b,c,d,e,f,g,h) (a.z - b.z)*(c.x*d.y - d.x*c.y) - (e.z - f.z)*(g.x*h.y - h.x*g.y)
#define D(x,y,a,b,c) (a.x*(b.y-c.y) + b.x*(c.y-a.y) + c.x*(a.y-b.y))
#define E(x,y) ((ra*D(x,y,b,c,d) - rb*D(x,y,c,d,a) + rc*D(x,y,d,a,b) - rd*D(x,y,a,b,c)) / uvw)
    double u = U(a,b,c,d,b,c,d,a);
    double v = U(c,d,a,b,d,a,b,c);
    double w = U(a,c,d,b,b,d,a,c);
    double uvw = 2 * (u + v + w);
    if (uvw == 0.0) {
        // Oops.  The points are coplanar.
    }
    auto sq = [] (Point p) { return p.x*p.x + p.y*p.y + p.z*p.z; };
    double ra = sq(a);
    double rb = sq(b);
    double rc = sq(c);
    double rd = sq(d);
    double x0 = E(y,z);
    double y0 = E(z,x);
    double z0 = E(x,y);
    double radius = sqrt(sq(Point(a.x - x0, a.y - y0, a.z - z0)));
    return Sphere(Point(x0, y0, z0), radius);
}

我编写了这段代码，特此将其置于公共领域。你可以用它做任何你想做的事情。

于 2022-01-25T09:51:05.107 回答
-1

使用前两点定义一条线。对后两个做同样的事情。找到两条线之间的 POI。

于 2016-05-25T23:09:38.340 回答