由多个像素的缩短的线 [英] Shorten a line by a number of pixels

问题描述

我画使用.NET GDI +业务对象的自定义图。在其他方面，该图由所连接的对象的几行。

在一个特定的情况下，我需要缩短一个线的像素的具体数量，让我们说10个象素，即发现在该位于10个像素的行的结束点之前的行中的点

想象与半径的研究的= 10个像素，并且与开始点的线（X1，Y1）和结束点（X2，Y2）的圆。圆的中心位于线的终点，如以下图示。

我如何计算出标有红色圆圈的点，即圆和直线的交点？这将使我行的新终点，10个像素缩短了。

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解决方案

感谢您的答案，从中我能够放在一起以下步骤。我把它命名为LengthenLine，因为我觉得它更自然地传递像素负数，如果我想行缩短了。

具体而言，我试图把一个函数，可以带圆角画一条线，可以发现，<一个href="http://stackoverflow.com/questions/1805582/net-gdi-drawing-lines-with-rounded-corners">here.

 公共无效LengthenLine（的PointF的startPoint，裁判的PointF端点浮动pixelCount）
{
  如果（startPoint.Equals（终点））
    返回; //不是一条线

  双DX = endPoint.X  -  startPoint.X;
  双DY = endPoint.Y  -  startPoint.Y;
  如果（DX == 0）
  {
    // 垂线：
    如果（endPoint.Y＆LT; startPoint.Y）
      endPoint.Y  -  = pixelCount;
    其他
      endPoint.Y + = pixelCount;
  }
  否则，如果（DY == 0）
  {
    //横线：
    如果（endPoint.X＆LT; startPoint.X）
      endPoint.X  -  = pixelCount;
    其他
      endPoint.X + = pixelCount;
  }
  其他
  {
    //非水平非垂直线：
    双长度=的Math.sqrt（DX * DX + DY * DY）;
    双刻度=（长+ pixelCount）/长;
    DX * =规模;
    DY * =规模;
    endPoint.X = startPoint.X + Convert.ToSingle（DX）;
    endPoint.Y = startPoint.Y + Convert.ToSingle（DY）;
  }
}
 

解决方案

找到方向向量，即让位置向量是（用花车）B =（X2，Y2）和A =（X1，Y1），然后AB = B - A。由它的长度（的Math.sqrt（X * X + Y * Y））除以归一化的载体。然后乘以方向矢量AB的原始长度减去圆的半径，并添加回线起始位置：

 双DX = X2  -  X1;
双DY = Y2  -  Y1;
双长度=的Math.sqrt（DX * DX + DY * DY）;
如果（长度大于0）
{
    DX / =长度;
    DY / =长度;
}
DX * =长度 - 半径;
DY * =长度 - 半径;
INT X3 =（INT）（X1 + DX）;
INT Y3 =（INT）（Y1 + DY）;
 

编辑：修正了code，aaand固定在最初的解释（以为你想线路从圆的中心，其周边出门：P）

I'm drawing a custom diagram of business objects using .NET GDI+. Among other things, the diagram consists of several lines that are connecting the objects.

In a particular scenario, I need to shorten a line by a specific number of pixels, let's say 10 pixels, i.e. find the point on the line that lies 10 pixels before the end point of the line.

Imagine a circle with radius r = 10 pixels, and a line with start point (x1, y1) and end point (x2, y2). The circle is centered at the end point of the line, as in the following illustration.

How do I calculate the point marked with a red circle, i.e. the intersection between circle and line? This would give me the new end point of the line, shortening it by 10 pixels.

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Solution

Thank you for your answers from which I was able to put together the following procedure. I named it LengthenLine, since I find it more natural to pass a negative number of pixels if I want the line shortened.

Specifically, I was trying to put together a function that could draw a line with rounded corners, which can be found here.

public void LengthenLine(PointF startPoint, ref PointF endPoint, float pixelCount)
{
  if (startPoint.Equals(endPoint))
    return; // not a line

  double dx = endPoint.X - startPoint.X;
  double dy = endPoint.Y - startPoint.Y;
  if (dx == 0)
  {
    // vertical line:
    if (endPoint.Y < startPoint.Y)
      endPoint.Y -= pixelCount;
    else
      endPoint.Y += pixelCount;
  }
  else if (dy == 0)
  {
    // horizontal line:
    if (endPoint.X < startPoint.X)
      endPoint.X -= pixelCount;
    else
      endPoint.X += pixelCount;
  }
  else
  {
    // non-horizontal, non-vertical line:
    double length = Math.Sqrt(dx * dx + dy * dy);
    double scale = (length + pixelCount) / length;
    dx *= scale;
    dy *= scale;
    endPoint.X = startPoint.X + Convert.ToSingle(dx);
    endPoint.Y = startPoint.Y + Convert.ToSingle(dy);
  }
}

解决方案

Find the direction vector, i.e. let the position vectors be (using floats) B = (x2, y2) and A = (x1, y1), then AB = B - A. Normalize that vector by dividing by its length ( Math.Sqrt(x*x + y*y) ). Then multiply the direction vector AB by the original length minus the circle's radius, and add back to the lines starting position:

double dx = x2 - x1;
double dy = y2 - y1;
double length = Math.Sqrt(dx * dx + dy * dy);
if (length > 0)
{
    dx /= length;
    dy /= length;
}
dx *= length - radius;
dy *= length - radius;
int x3 = (int)(x1 + dx);
int y3 = (int)(y1 + dy);

Edit: Fixed the code, aaand fixed the initial explanation (thought you wanted the line to go out from the circle's center to its perimeter :P)

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