C＃绘制弧连得3分 [英] C# Drawing Arc with 3 Points

问题描述

我需要使用GraphicsPath的并具有初始，中间和最终指向划出一道弧线。电弧必须通过他们。

我试过.DrawCurve和.DrawBezier但结果是不完全的弧线。

我能做些什么

解决方案：

在一对夫妇的代码编写的时间我设法画什么，我想用这个算法（给予3点A，b，C和A GraphicsPath的路径）：

 双D = 2 *（AX  -  CX）*（CY  -  BY）+ 2 *（BX  -  CX）*（AY  -  CY）; 
双M1 =（Math.Pow（a.X，2） -  Math.Pow（c.X，2）+ Math.Pow（a.Y，2） -  Math.Pow（c.Y，2））; 
双M2 =（Math.Pow（c.X，2） -  Math.Pow（b.X，2）+ Math.Pow（c.Y，2） -  Math.Pow（b.Y，2））; 
双NX = M1 *（c.Y  -  b.Y）+ 2 *（c.Y  -  a.Y）; 
双NY = M1 *（b.X  -  c.X）+ 2 *（a.X  -  c.X）; 
双CX = NX /天; 
双CY = NY /天; 
双DX = CX  -  a.X; 
双DY = CY  -  a.Y; 
双距离=的Math.sqrt（DX * DX + DY * DY）; 
矢量VA =新的向量（a.X  -  CX，a.Y  -  CY）; 
向量V =新的向量（b.X  -  CX，b.Y  -  CY）; 
矢量Vc =新的向量（c.X  -  CX，c.Y  -  CY）; 
矢量x轴=新的向量（1，0）; 
浮动由startAngle =（浮点）Vector.AngleBetween（x轴，VA）; 
浮动sweepAngle =（浮点）（Vector.AngleBetween（VA，VB）+ Vector.AngleBetween（VB，VC））; 
 path.AddArc（
（浮点）（CX  - 距离），（浮点）（CY  - 距离），
（浮点）（距离* 2），（浮点）（距离* 2 ），
 startAngle开始，sweepAngle）; 
  

解决方案

我会使用 DrawArc（ ）由ANC_Michael建议。要查找到3点经过圆弧要计算由点形成的三角形的外接圆。

一旦你的外接圆计算边框与使用最小/最大尺寸DrawArc 使用圆（中心+/-半径）。现在计算你开始和使外接圆上的原点（由-circumcenter翻译）为中心的翻译点停止角度，并采取与X轴的归一化的开始和结束向量的点积：

 双startAngle开始= Math.Acos（VectorToLeftPoint.Dot（X轴））; 
双stopAngle = Math.Acos（VectorToRightPoint.Dot（X轴））; 
  

注意 DrawArc 顺时针期望的角度从X轴，所以你应该添加 Math.PI 如果计算出的向量是x轴的上方。这应该是足够的信息来调用 DrawArc（）

编辑：这种方法会找到一个圆形弧不一定是'最适合'弧根据您的预计终结点行为。

I need to draw an arc using GraphicsPath and having initial, median and final points. The arc has to pass on them.

I tried .DrawCurve and .DrawBezier but the result isn't exactly an arc.

What can I do?

SOLUTION:

After a couple of hours of code writing I managed to draw what I wanted with this algorithm (give 3 Point a,b,c and a GraphicsPath path):

double d = 2 * (a.X - c.X) * (c.Y - b.Y) + 2 * (b.X - c.X) * (a.Y - c.Y);
double m1 = (Math.Pow(a.X, 2) - Math.Pow(c.X, 2) + Math.Pow(a.Y, 2) - Math.Pow(c.Y, 2));
double m2 = (Math.Pow(c.X, 2) - Math.Pow(b.X, 2) + Math.Pow(c.Y, 2) - Math.Pow(b.Y, 2));
double nx = m1 * (c.Y - b.Y) + m2 * (c.Y - a.Y);
double ny = m1 * (b.X - c.X) + m2 * (a.X - c.X);
double cx = nx / d;
double cy = ny / d;
double dx = cx - a.X;
double dy = cy - a.Y;
double distance = Math.Sqrt(dx * dx + dy * dy);
Vector va = new Vector(a.X - cx, a.Y - cy);
Vector vb = new Vector(b.X - cx, b.Y - cy);
Vector vc = new Vector(c.X - cx, c.Y - cy);
Vector xaxis = new Vector(1, 0);
float startAngle = (float)Vector.AngleBetween(xaxis, va);
float sweepAngle = (float)(Vector.AngleBetween(va, vb) + Vector.AngleBetween(vb, vc));
path.AddArc(
    (float)(cx - distance), (float)(cy - distance),
    (float)(distance * 2), (float)(distance * 2), 
    startAngle, sweepAngle);

解决方案

I would use DrawArc() as suggested by ANC_Michael. To find an arc that passes through 3 points you want to calculate the circumcircle of the triangle formed by the points.

Once you have the circumcircle calculate a bounding box for the circle to use with DrawArc using the min/max dimensions (center +/- radius). Now calculate your start and stop angles by translating the points so that the circumcircle is centered on the origin (translate by -circumcenter) and take the dot-product of the normalized start and end vectors with the X-axis:

double startAngle = Math.Acos(VectorToLeftPoint.Dot(XAxis));
double stopAngle = Math.Acos(VectorToRightPoint.Dot(XAxis));

Note that DrawArc expects angles clockwise from the X-axis so you should add Math.PI if the calculated vector is above the x-axis. That should be enough information to call DrawArc().

Edit: This method will find a circular arc and not necessarily the 'best fit' arc depending on your expected endpoint behavior.

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