Question

我有两个线段，在开始/结束点用3D点表示。

行：

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

坐标X，Y和Z的3D点数仅为3倍。

3DPoint：

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

问题：

我可以找到两条“线”与该距离“线”的端点之间的距离。 [ Here is an Image to Better Illustrate What I am trying to Achieve 1

我有什么：

目前，我可以使用此代码成功获取两行之间的距离（Adapted From Here使用段到段部分）：

    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;
    }

我需要的是什么：

有没有办法从上面的代码中确定位移矢量'dP'的端点？如果没有，有人能建议一种更好的方法来找到最小距离和该距离的终点吗？

感谢您阅读，并提前感谢任何建议！

非常感谢@Isaac van Bakel对此解决方案背后的理论

这是我的代码完成：两条线之间的最短距离由在最短距离处连接它们的线表示。

类：

Sharp3D.Math：我将此参考用于Vector3D，但实际上任何3D矢量类都可以使用。除此之外，如果按元素执行减法元素，则甚至不需要向量。

Point3D：我的个人Point3D课程。随意随意使用。

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 };
    }

}

专栏：我的个人专线课程。随意随意使用。

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));
        }
    }
}

功能

ClampPointToLine：我写的夹紧功能是将一个点夹到一条线上。

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;
}

distanceBetweenLines：返回表示两条线之间最短距离的线的函数。如果无法解析，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 };
}

实施

Line shortestDistanceLine = distBetweenLines(l1, l2);

结果：

到目前为止，这在我的测试中是准确的。如果传递两条相同的行，则返回null。我感谢任何反馈！

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。

如果我们有一条带有已知点p1和p2的线l1，以及一条带有已知点p3和p4的线l2：

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

d2也一样。

让我们在线l1，l2上取点，其中v实际上垂直于方向。我们将分别称为i1和i2这两个点。

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。

现在您可以将其替换回您的点积等式：

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

这已经将我们的方程数减少到2（两个点积方程），有两个未知数（m和n），所以你现在可以解决它们了！

一旦你有m和n，你可以通过回到i1和i2的原始计算找到坐标。

如果你只想在p1-p2和p3-p4之间的段上的点的最短距离，你可以在这些坐标范围之间夹住i1和i2，因为最短距离总是尽可能接近垂线

找到两个3D线段之间的最短距离

1 个答案: