Question

我正在使用C＃中的一个简单工具。我有三个点使两条线在P点相遇。所以PP1和PP2。我想在会合点处对线进行倒角，使得距线d1从线PP1修剪，距离d2从线PP2修剪，然后连接修剪线。我有问题，因为我无法得到确切的结果。任何想法都是我的代码中的问题。谢谢 Result1 Result2

private void Chamfer(Graphics g,PointF P,PointF P1,PointF P2,double d1,double d2)
    {

        //Vector 1 Length
        double PP1 = Math.Sqrt((Math.Pow((P.X - P1.X), 2) + Math.Pow((P.Y - P1.Y), 2)));
        //Vector 2 Length
        double PP2 = Math.Sqrt((Math.Pow((P.X-P2.X), 2) + Math.Pow((P.Y - P2.Y), 2)));
        //Slopes & Angles
        double m1 = (P.Y-P1.Y) / (P.X - P1.X);
        double angle1 = Math.Atan(m1) * (180 / Math.PI);
        double m2 = (P.Y - P2.Y) / (P.X - P2.X);
        double angle2 = Math.Atan(m2) * (180 / Math.PI);
        //Coordinates of points of Chamfer

        if(P1.X>P.X && P1.Y > P.Y)
        {
            int Pd1X = Convert.ToInt32(P.X + d1 * Math.Cos(angle1));
            int Pd1Y = Convert.ToInt32(P.Y + d1 * Math.Sin(angle1));
            Chp1 = new Point(Pd1X, Pd1Y);
        }

        else if (P1.X > P.X && P1.Y < P.Y)
        {
            int Pd1X = Convert.ToInt32(P.X + d1 * Math.Cos(angle1));
            int Pd1Y = Convert.ToInt32(P.Y - d1 * Math.Sin(angle1));
            Chp1 = new Point(Pd1X, Pd1Y);
        }

        else if (P1.X < P.X && P1.Y < P.Y)
        {
            int Pd1X = Convert.ToInt32(P.X - d1 * Math.Cos(angle1));
            int Pd1Y = Convert.ToInt32(P.Y - d1 * Math.Sin(angle1));
            Chp1 = new Point(Pd1X, Pd1Y);
        }
        else if (P1.X < P.X && P1.Y > P.Y)
        {
            int Pd1X = Convert.ToInt32(P.X - d1 * Math.Cos(angle1));
            int Pd1Y = Convert.ToInt32(P.Y + d1 * Math.Sin(angle1));
            Chp1 = new Point(Pd1X, Pd1Y);
        }

        if (P2.X > P.X && P2.Y > P.Y)
        {
            int Pd2X = Convert.ToInt32(P.X + d2 * Math.Cos(angle2));
            int Pd2Y = Convert.ToInt32(P.Y + d2 * Math.Sin(angle2));
            Chp2 = new Point(Pd2X, Pd2Y);
        }
        else if (P2.X > P.X && P2.Y < P.Y)
        {
            int Pd2X = Convert.ToInt32(P.X + d2 * Math.Cos(angle2));
            int Pd2Y = Convert.ToInt32(P.Y - d2 * Math.Sin(angle2));
            Chp2 = new Point(Pd2X, Pd2Y);
        }
        else if (P2.X < P.X && P2.Y < P.Y)
        {
            int Pd2X = Convert.ToInt32(P.X - d2 * Math.Cos(angle2));
            int Pd2Y = Convert.ToInt32(P.Y - d2 * Math.Sin(angle2));
            Chp2 = new Point(Pd2X, Pd2Y);
        }
        else if (P2.X < P.X && P2.Y > P.Y)
        {
            int Pd2X = Convert.ToInt32(P.X - d2 * Math.Cos(angle1));
            int Pd2Y = Convert.ToInt32(P.Y + d2 * Math.Sin(angle1));
            Chp2 = new Point(Pd2X, Pd2Y);
        }

        Pen penPre = new Pen(Color.Green);
        penPre.DashStyle = System.Drawing.Drawing2D.DashStyle.Dot;
        g.Clear(this.BackColor);
        g.DrawLine(Pens.Black, P1, Chp1);
        g.DrawLine(penPre, P1, P);
        g.DrawLine(penPre, P2, P);
        g.DrawString("P1", this.Font, Brushes.Red, new Point(Convert.ToInt32(P1.X + 2), Convert.ToInt32(P1.Y - 2)));
        g.DrawLine(Pens.Black, P2, Chp2);
        g.DrawString("P2", this.Font, Brushes.Red, new Point(Convert.ToInt32(P2.X + 2), Convert.ToInt32(P2.Y - 2)));
        g.DrawString("P", this.Font, Brushes.Red, new Point(Convert.ToInt32( P.X + 3),Convert.ToInt32( P.Y - 2)));

        g.DrawLine(Pens.Black, Chp1, Chp2);

    }

我在特定坐标处获得的内容，我的代码正常工作，您可以从附带的照片中看到。如果我没有这些黄金坐标，我无法判断。 Result from golden coordinates

Answer 1

 //Vector 1 Length
    double PP1 = Math.Sqrt((Math.Pow((P.X - P1.X), 2) + Math.Pow((P.Y - P1.Y), 2)));

    //unit direction vector
    upx1 = (P1.X - P.X) / PP1
    upy1 = (P1.Y - P.Y) / PP1

    // chpoint coordinates
    chx1 = P.X + upx1 * d;
    chy1 = P.Y + upy1 * d;

   // now the same for PP2

使用C＃给出两条线的倒角

1 个答案: