c# - 找到两个 3D 线段之间的最短距离

Question

我有两条线段，在它们的起点/终点处用 3D 点表示。

线路:

class Line
{
    public string Name { get; set; }
    public Point3D Start { get; set; } = new Point3D();
    public Point3D End { get; set; } = new Point3D();
}

3D 点只是坐标 X、Y 和 Z 的 3 个 double 值。

3D 点:

class Point3D
{
    public double X { get; set; }
    public double Y { get; set; }
    public double Z { get; set; }
}

问题:

我能找到两条“线”之间的距离和该距离“线”的端点吗？ [ 1

我有什么:

目前，我可以使用这段代码成功获取两条线之间的距离(Adapted From Here 使用 Segment To Segment Section):

    public double lineNearLine(Line l1, Line l2)
    {
        Vector3D uS = new Vector3D { X = l1.Start.X, Y = l1.Start.Y, Z = l1.Start.Z };
        Vector3D uE = new Vector3D { X = l1.End.X, Y = l1.End.Y, Z = l1.End.Z };
        Vector3D vS = new Vector3D { X = l2.Start.X, Y = l2.Start.Y, Z = l2.Start.Z };
        Vector3D vE = new Vector3D { X = l2.End.X, Y = l2.End.Y, Z = l2.End.Z };
        Vector3D w1 = new Vector3D { X = l1.Start.X, Y = l1.Start.Y, Z = l1.Start.Z };
        Vector3D w2 = new Vector3D { X = l2.Start.X, Y = l2.Start.Y, Z = l2.Start.Z };
        Vector3D u = uE - uS;
        Vector3D v = vE - vS;
        Vector3D w = w1 - w2;
        double a = Vector3D.DotProduct(u, u);
        double b = Vector3D.DotProduct(u, v);
        double c = Vector3D.DotProduct(v, v);
        double d = Vector3D.DotProduct(u, w);
        double e = Vector3D.DotProduct(v, w);
        double D = a * c - b * b;
        double sc, sN, sD = D;
        double tc, tN, tD = D;
        if (D < 0.01)
        {
            sN = 0;
            sD = 1;
            tN = e;
            tD = c;
        }
        else
        {
            sN = (b * e - c * d);
            tN = (a * e - b * d);
            if (sN < 0)
            {
                sN = 0;
                tN = e;
                tD = c;
            }
            else if (sN > sD)
            {
                sN = sD;
                tN = e + b;
                tD = c;
            }
        }
        if (tN < 0)
        {
            tN = 0;
            if (-d < 0)
            {
                sN = 0;
            }
            else if (-d > a)
            {
                sN = sD;
            }
            else
            {
                sN = -d;
                sD = a;
            }
        }
        else if (tN > tD)
        {
            tN = tD;
            if ((-d + b) < 0)
            {
                sN = 0;
            }
            else if ((-d + b) > a)
            {
                sN = sD;
            }
            else
            {
                sN = (-d + b);
                sD = a;
            }
        }
        if (Math.Abs(sN) < 0.01)
        {
            sc = 0;
        }
        else
        {
            sc = sN / sD;
        }
        if (Math.Abs(tN) < 0.01)
        {
            tc = 0;
        }
        else
        {
            tc = tN / tD;
        }
        Vector3D dP = w + (sc * u) - (tc * v);
        double distance1 = Math.Sqrt(Vector3D.DotProduct(dP, dP));
        return distance1;
    }

我需要什么:

有什么方法可以从上面的代码中确定位移矢量“dP”的端点吗？如果没有，谁能建议一种更好的方法来找到最小距离和该距离的端点？

感谢您的阅读，提前感谢您的任何建议!

非常感谢@Isaac van Bakel 提供此解决方案背后的理论

这是我的完整代码:两条线之间的最短距离由以最短距离连接它们的线表示。

类:

1. Sharp3D.Math:我将此引用用于 Vector3D，但实际上任何 3D 矢量类都可以使用。最重要的是，如果逐个元素地进行减法，则甚至不需要向量。

2. Point3D:我的个人 Point3D 类(class)。想用多少就用多少。

   class Point3D
   {
       public double X { get; set; }
       public double Y { get; set; }
       public double Z { get; set; }
       public  Vector3D getVector()
       {
           return new Vector3D { X = this.X, Y = this.Y, Z = this.Z };
       }
   
   }
   

3. 线路:我的个人线路类(class)。想用多少就用多少。

   class Line
   {
       public string Name { get; set; }
       public Point3D Start { get; set; } = new Point3D();
       public Point3D End { get; set; } = new Point3D();
       public double Length
       {
           get
           {
               return Math.Sqrt(Math.Pow((End.X - Start.X), 2) + Math.Pow((End.Y - Start.Y), 2));
           }
       }
   }
   

函数:

1. ClampPointToLine :我编写的夹紧函数，用于将点夹紧到一条线上。

   public Point3D ClampPointToLine(Point3D pointToClamp, Line lineToClampTo)
   {
       Point3D clampedPoint = new Point3D();
       double minX, minY, minZ, maxX, maxY, maxZ;
       if(lineToClampTo.Start.X <= lineToClampTo.End.X)
       {
           minX = lineToClampTo.Start.X;
           maxX = lineToClampTo.End.X;
       }
       else
       {
           minX = lineToClampTo.End.X;
           maxX = lineToClampTo.Start.X;
       }
       if (lineToClampTo.Start.Y <= lineToClampTo.End.Y)
       {
           minY = lineToClampTo.Start.Y;
           maxY = lineToClampTo.End.Y;
       }
       else
       {
           minY = lineToClampTo.End.Y;
           maxY = lineToClampTo.Start.Y;
       }
       if (lineToClampTo.Start.Z <= lineToClampTo.End.Z)
       {
           minZ = lineToClampTo.Start.Z;
           maxZ = lineToClampTo.End.Z;
       }
       else
       {
           minZ = lineToClampTo.End.Z;
           maxZ = lineToClampTo.Start.Z;
       }
       clampedPoint.X = (pointToClamp.X < minX) ? minX : (pointToClamp.X > maxX) ? maxX : pointToClamp.X;
       clampedPoint.Y = (pointToClamp.Y < minY) ? minY : (pointToClamp.Y > maxY) ? maxY : pointToClamp.Y;
       clampedPoint.Z = (pointToClamp.Z < minZ) ? minZ : (pointToClamp.Z > maxZ) ? maxZ : pointToClamp.Z;
       return clampedPoint;
   }
   

2. distanceBetweenLines :返回表示两条线之间最短距离的线的函数。如果无法解决，则返回 null。

   public Line distBetweenLines(Line l1, Line l2)
   {
       Vector3D p1, p2, p3, p4, d1, d2;
       p1 = l1.Start.getVector();
       p2 = l1.End.getVector();
       p3 = l2.Start.getVector();
       p4 = l2.End.getVector();
       d1 = p2 - p1;
       d2 = p4 - p3;
       double eq1nCoeff = (d1.X * d2.X) + (d1.Y * d2.Y) + (d1.Z * d2.Z);
       double eq1mCoeff = (-(Math.Pow(d1.X, 2)) - (Math.Pow(d1.Y, 2)) - (Math.Pow(d1.Z, 2)));
       double eq1Const = ((d1.X * p3.X) - (d1.X * p1.X) + (d1.Y * p3.Y) - (d1.Y * p1.Y) + (d1.Z * p3.Z) - (d1.Z * p1.Z));
       double eq2nCoeff = ((Math.Pow(d2.X, 2)) + (Math.Pow(d2.Y, 2)) + (Math.Pow(d2.Z, 2)));
       double eq2mCoeff = -(d1.X * d2.X) - (d1.Y * d2.Y) - (d1.Z * d2.Z);
       double eq2Const = ((d2.X * p3.X) - (d2.X * p1.X) + (d2.Y * p3.Y) - (d2.Y * p2.Y) + (d2.Z * p3.Z) - (d2.Z * p1.Z));
       double[,] M = new double[,] { { eq1nCoeff, eq1mCoeff, -eq1Const }, { eq2nCoeff, eq2mCoeff, -eq2Const } };
       int rowCount = M.GetUpperBound(0) + 1;
       // pivoting
       for (int col = 0; col + 1 < rowCount; col++) if (M[col, col] == 0)
           // check for zero coefficients
           {
               // find non-zero coefficient
               int swapRow = col + 1;
               for (; swapRow < rowCount; swapRow++) if (M[swapRow, col] != 0) break;
   
               if (M[swapRow, col] != 0) // found a non-zero coefficient?
               {
                   // yes, then swap it with the above
                   double[] tmp = new double[rowCount + 1];
                   for (int i = 0; i < rowCount + 1; i++)
                   { tmp[i] = M[swapRow, i]; M[swapRow, i] = M[col, i]; M[col, i] = tmp[i]; }
               }
               else return null; // no, then the matrix has no unique solution
           }
   
       // elimination
       for (int sourceRow = 0; sourceRow + 1 < rowCount; sourceRow++)
       {
           for (int destRow = sourceRow + 1; destRow < rowCount; destRow++)
           {
               double df = M[sourceRow, sourceRow];
               double sf = M[destRow, sourceRow];
               for (int i = 0; i < rowCount + 1; i++)
                   M[destRow, i] = M[destRow, i] * df - M[sourceRow, i] * sf;
           }
       }
   
       // back-insertion
       for (int row = rowCount - 1; row >= 0; row--)
       {
           double f = M[row, row];
           if (f == 0) return null;
   
           for (int i = 0; i < rowCount + 1; i++) M[row, i] /= f;
           for (int destRow = 0; destRow < row; destRow++)
           { M[destRow, rowCount] -= M[destRow, row] * M[row, rowCount]; M[destRow, row] = 0; }
       }
       double n = M[0, 2];
       double m = M[1, 2];
       Point3D i1 = new Point3D { X = p1.X + (m * d1.X), Y = p1.Y + (m * d1.Y), Z = p1.Z + (m * d1.Z) };
       Point3D i2 = new Point3D { X = p3.X + (n * d2.X), Y = p3.Y + (n * d2.Y), Z = p3.Z + (n * d2.Z) };
       Point3D i1Clamped = ClampPointToLine(i1, l1);
       Point3D i2Clamped = ClampPointToLine(i2, l2);
       return new Line { Start = i1Clamped, End = i2Clamped };
   }
   

实现:

Line shortestDistanceLine = distBetweenLines(l1, l2);

结果:

到目前为止，这在我的测试中是准确的。如果传递了两条相同的行，则返回 null。感谢任何反馈!

Answer 1

两条斜线(不相交的线)之间的最短距离是与两条斜线垂直的线的距离。

如果我们有一条直线 l1，已知点 p1 和 p2，直线 l2 已知点 p3 和 p4:

The direction vector of l1 is p2-p1, or d1.
The direction vector of l2 is p4-p3, or d2.

因此我们知道我们正在寻找的向量 v 垂直于这两个方向向量:

d1.v = 0 & d2.v = 0

或者，如果您愿意:

d1x*vx + d1y*vy + d1z*vz = 0

对于 d2 也是如此。

让我们取直线 l1、l2 上的点，其中 v 实际上垂直于方向。我们将这两个点分别称为 i1 和 i2。

Since i1 lies on l1, we can say that i1 = p1 + m*d1, where m is some number.
Similarly, i2 = p3 + n*d2, where n is another number.

由于 v 是 i1 和 i2 之间的向量(根据定义)，我们得到 v = i2 - i1。

这给出了 v 的 x,y,z 向量的替换:

vx = i2x - i1x = (p3x + n*d2x) - (p1x + m*d1x)

等等。

您现在可以将其代回您的点积方程:

d1x * ( (p3x + n*d2x) - (p1x + m*d1x) ) + ... = 0

这将我们的方程数量减少到 2 个(两个点积方程)和两个未知数(m 和 n)，所以您现在可以求解它们了!

一旦有了 m 和 n，就可以通过返回 i1 和 i2 的原始计算来找到坐标。

如果你只想要 p1-p2 和 p3-p4 之间线段上点的最短距离，你可以将 i1 和 i2 夹在这些坐标范围之间，因为最短距离总是尽可能接近垂直线.