绘制穿过3个点的椭圆

下面是如何使用 DrawEllipse 方法从旋转、短轴和两个顶点。

首先,我们计算 Size 边界的 Rectangle :

鉴于 Points A and B 坐在长度较短的一侧 smallSize 我们用一个小毕达哥拉斯得到了长边:

int longSide = (int)(Math.Sqrt((A.Y - B.Y) * (A.Y - B.Y) + (B.X - A.X) * (B.X - A.X)));

因此:

Size size = new System.Drawing.Size(longSide, smallSize);

接下来我们需要旋转角度:

float angle = -(float)(Math.Atan2(A.Y - B.Y, B.X - A.X) * 180f / Math.PI);

这将使事情更容易得到中心 Point C :

Point C = new Point((A.X + B.X)/ 2, (A.Y + B.Y)/ 2);

我们最不想要的是一个例程,它绘制给定形状的椭圆 大小 ,旋转 C 从一个角度来看:

void DrawEllipse(Graphics G, Pen pen, Point center, Size size, float angle)
{
    int h2 = size.Height / 2;
    int w2 = size.Width / 2;
    Rectangle rect = new Rectangle( new Point(center.X - w2, center.Y - h2), size );

    G.TranslateTransform(center.X, center.Y);
    G.RotateTransform(angle);
    G.TranslateTransform(-center.X, -center.Y);
    G.DrawEllipse(pen, rect);
    G.ResetTransform();
}

[![在此处输入图像描述][1]][1]

这里有一个小试验台,将所有这些结合在一起:

Point A = new Point(200, 200); // *
Point B = new Point(500, 250);
int smallSize = 50;


void doTheDraw(PictureBox pb)
{
    Bitmap bmp = new Bitmap(pb.Width, pb.Height);

    float angle = -(float)(Math.Atan2(A.Y - B.Y, B.X - A.X) * 180f / Math.PI);
    int longSide = (int)(Math.Sqrt((A.Y - B.Y) * (A.Y - B.Y) + (B.X - A.X) * (B.X - A.X)));
    Point C = new Point((A.X + B.X) / 2, (A.Y + B.Y) / 2);
    Size size = new System.Drawing.Size((int)longSide, smallSize);

    using (Pen pen = new Pen(Color.Orange, 3f))
    using (Graphics g = Graphics.FromImage(bmp))
    {
        // a nice background grid (optional):
        DrawGrid(g, 0, 0, 100, 50, 10,
            Color.LightSlateGray, Color.DarkGray, Color.Gainsboro);

        // show the points we use (optional):
        g.FillEllipse(Brushes.Red, A.X - 4, A.Y - 4, 8, 8);
        g.FillRectangle(Brushes.Red, B.X - 3, B.Y - 3, 7, 7);
        g.FillEllipse(Brushes.Red, C.X - 5, C.Y - 5, 11, 11);

        // show the connection line (optional):
        g.DrawLine(Pens.Orange, A, B);

        // here comes the ellipse:
        DrawEllipse(g, pen, C, size, angle);
    }
    pb.Image = bmp;
}

网格是一个很好的助手:

void DrawGrid(Graphics G, int ox, int oy, 
              int major, int medium, int minor, Color c1, Color c2, Color c3)
{
    using (Pen pen1 = new Pen(c1, 1f))
    using (Pen pen2 = new Pen(c2, 1f))
    using (Pen pen3 = new Pen(c3, 1f))
    {
        pen2.DashStyle = DashStyle.Dash;
        pen3.DashStyle = DashStyle.Dot;
        for (int x = ox; x < G.VisibleClipBounds.Width; x += major)
            G.DrawLine(pen1, x, 0, x, G.VisibleClipBounds.Height);
        for (int y = oy; y < G.VisibleClipBounds.Height; y += major)
            G.DrawLine(pen1, 0, y, G.VisibleClipBounds.Width, y);

        for (int x = ox; x < G.VisibleClipBounds.Width; x += medium)
            G.DrawLine(pen2, x, 0, x, G.VisibleClipBounds.Height);
        for (int y = oy; y < G.VisibleClipBounds.Height; y += medium)
            G.DrawLine(pen2, 0, y, G.VisibleClipBounds.Width, y); 

        for (int x = ox; x < G.VisibleClipBounds.Width; x += minor)
            G.DrawLine(pen3, x, 0, x, G.VisibleClipBounds.Height);
        for (int y = oy; y < G.VisibleClipBounds.Height; y += minor)
            G.DrawLine(pen3, 0, y, G.VisibleClipBounds.Width, y);
    }
}

注意,我做了 A, B, smallSide 类级别的变量,以便我可以在测试期间修改它们,(我做到了 * )..

如你所见,我添加了一个 TrackBar 使 smallside 动态为了更有趣,我添加了这个 MouseClick 事件:

private void pictureBox1_MouseClick(object sender, MouseEventArgs e)
{
    if (e.Button.HasFlag(MouseButtons.Left)) A = e.Location;
    else B = e.Location;
    doTheDraw(pictureBox1);
}

我希望我的一些逻辑可以帮助你实现你想要实现的目标。你们提出的下一个问题是尝试添加更多信息,不要太笼统。

计算相似度变换(平移和旋转),使前两点 (-a, 0) 和 (a, 0) . [这可以通过矩阵变换或复数来实现。]

对第三个点应用相同的变换。现在,简化椭圆的方程是

x² / a² + y² / b² = 1.

您只需确定参数 b 通过插入第三个点的简化坐标,

b = y / √(1 - x²/a²).

现在你有了

• 中心,
• 轴的方向,
• 长轴的长度,
• 短轴的长度。