Question

我有一个由两个PointD定义的线段（一个自定义类型用于处理双精度而不是浮点数，因为C＃中的数学在双精度数中起作用）：p1（.541,41.929），p2（363.736,88.48） - 假设这条线会倾斜。

在不可预测的时间间隔内，我需要设置一个垂直于线段的偏移点，可变距离 - 只需说，0.125,0.25等。

似乎所有的例子都假设你总是需要这个垂直点基于中点。

我已经解决了p3 :(

private PointD subPt(PointD a, PointD b, double p)
    {
        double x = (1 - p) * a.X + p * b.X;
        double y = (1 - p) * a.Y + p * b.Y;

        return new PointD(x, y);
    }

0.25的返回点是（91.339,53.566）。我已经在cad中验证了这一点，足以满足我的要求。

我无法弄清楚我的生活，就是如何在距离线条的位置得到一个垂直点5（只是为了使它成圆形）。

我期待的结果是p4（90.705,58.526）

Answer 1

我已经提出了一个有效的C＃实现：

/// <summary>
/// Create a perpendicular offset point at a position located along a line segment.
/// </summary>
/// <param name="a">Input. PointD(x,y) of p1.</param>
/// <param name="b">Input. PointD(x,y) of p2.</param>
/// <param name="position">Distance between p1(0.0) and p2 (1.0) in a percentage.</param>
/// <param name="offset">Distance from position at 90degrees to p1 and p2- non-percetange based.</param>
/// <param name="c">Output of the calculated point along p1 and p2. might not be necessary for the ultimate output.</param>
/// <param name="d">Output of the calculated offset point.</param>
private void PerpOffset(PointD a, PointD b, double position, double offset, out PointD c, out PointD d)
{
    //p3 is located at the x or y delta * position + p1x or p1y original.
    PointD p3 = new PointD(((b.X - a.X) * position) + a.X, ((b.Y - a.Y) * position) + a.Y);

    //returns an angle in radians between p1 and p2 + 1.5708 (90degress).
    double angleRadians = Math.Atan2(a.Y - b.Y, a.X - b.X) + 1.5708;

    //locates p4 at the given angle and distance from p3.
    PointD p4 = new PointD(p3.X + Math.Cos(angleRadians) * offset, p3.Y + Math.Sin(angleRadians) * offset);

    //send out the calculated points
    c = p3;
    d = p4;
}

此外，这是我的PointD课程。基本上只是更容易存储和使用x，y值作为双精度，并包含一个属性来返回PointF用于图形输出目的。

public class PointD
{
    public double X { get; set; }
    public double Y { get; set; }

    public float Xf { get { return (float)X; } }
    public float Yf { get { return (float)Y; } }

    public PointF PointF { get { return new PointF(Xf, Yf); } }

    public PointD()
    { }

    public PointD(double cX, double cY)
    {
        X = cX;
        Y = cY;
    }

    public override string ToString()
    {
        return string.Format("[{0}, {1}]", X, Y);
    }
}

C＃偏移垂直于给定线段的点

1 个答案: