我有 4 个点..我可以用这段代码画一个多边形
var p = new Polygon();
p.Points.Add(new Point(0, 0));
p.Points.Add(new Point(70, 0));
p.Points.Add(new Point(90, 100));
p.Points.Add(new Point(0, 80));
我如何使用 Silverlight 绘制适合此多边形的“椭圆”？
问题仍未得到解答!!!
最佳答案
一种方法是使用 QuadraticBezierSegment 或 BezierSegment。
例如，像这样:
<Path Stroke="Red" StrokeThickness="2" >
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="0,40">
<PathFigure.Segments>
<PathSegmentCollection>
<BezierSegment Point1="0,93"
Point2="90,117"
Point3="80,50"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="2" >
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="0,40">
<PathFigure.Segments>
<PathSegmentCollection>
<BezierSegment Point1="0,-13"
Point2="70,-17"
Point3="80,50"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Polygon Points="0,0 70,0 90,100 0,80"></Polygon>
这不是精确解，要精确解，您必须计算曲线的精确点并使用 4 QuadraticBezierSegment
编辑:
QuadraticBezierSegment 的例子
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="0,40">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="6,79"
Point2="45,90"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="45,90">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="80,91"
Point2="80,50"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="0,40">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="2,3"
Point2="35,0"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="35,0">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="72,8"
Point2="80,50"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Polygon Name="pol" Points="0,0 70,0 90,100 0,80" Stroke="Red" StrokeThickness="1"</Polygon>
它仍然是一个实验性的，不是计算点，但它非常准确。
编辑2: 您可以使用此图像和我的评论来计算曲线点:
曲线有起点、中点和终点。在此图像中，开始和结束点是 L,M,N,O ；中间是 W,X,Y,Z 。
例如我们如何计算点 L :
借助 y = k * x + b 行的方程，我们找到 AB,DC,AC,DB,AD 行的方程。我们如何通过 AC 和 DB 找到 R 。我们如何通过 AB 和 DC 找到 E 。之后我们找到线 ER 的方程以及 ER 和 AD 的交叉我们找到 L 。
我们如何计算点 W :
借助长度为 l = sqrt(sqr(x2 - x1) + sqr(y2 - y1)) 的等式，找到 AR 的长度。 AW = AR/(4*pi) 借助这个系数，以及直线方程和长度方程，求解平方方程后我们找到 W。
我们类似发现的其他点。
此算法不仅适用于具有平行线的多边形，但在这种情况下算法更容易。和长度系数相同。
借助此算法，我找到了您示例的 3 条曲线的点:
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="0,36">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="4.7,74.6"
Point2="39.9,88.9"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="39.9,88.9">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="83.43,92.7"
Point2="78.8,43.9"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="0,36">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="3.55,3.94"
Point2="31.8,0"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
而且它的曲线和椭圆线一模一样。下图:
您可以将此算法转化为“公式”，从而找到您的解决方案。
为此你需要 4 个函数:
1) 从 2 点的坐标中找到线系数
2) 找到 piont 的坐标如何从它的系数交叉 2 条线
3) 根据2个点的坐标求线段的长度
4) 从线系数和起点坐标中找到与此起点和此长度在一条直线上的点的坐标以及您在上一个函数中找到的长度除以 (4*pi)
编辑 3: 你可以优化这个解决方案，它有一些缺陷，如何平行线等。但它很快，如果你优化它可能会满足你的要求。
Xaml:
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure x:Name="pathleftdown" StartPoint="0,0">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment x:Name="bezleftdown" Point1="0,0"
Point2="0,0"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure x:Name="pathrigthdown" StartPoint="0,0">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment x:Name="bezrigthdown" Point1="0,0"
Point2="0,0"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure x:Name="pathleftup" StartPoint="0,0">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment x:Name="bezleftup" Point1="0,0"
Point2="0,0"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Path Stroke="Red" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure x:Name="pathrigthup" StartPoint="0,0">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment x:Name="bezrigthup" Point1="0,0"
Point2="0,0"
/>
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>
<Polygon Name="pol" Points="0,0 250,0 251,340 0,341" Stroke="Red" StrokeThickness="1"></Polygon>
<Button Content="Generate" Width ="80" Height="30" HorizontalAlignment="Right" VerticalAlignment="Top" Click="Button_Click"></Button>
和代码:
private class pointXY
{
public double x;
public double y;
}
private class lineKB
{
public double k;
public double b;
public bool flagXconst = false;
public double xConst = 0;
}
private lineKB GetLineFromPonts(pointXY A, pointXY B)
{
lineKB line = new lineKB();
if ((B.x - A.x) != 0)
{
line.k = (B.y - A.y) / (B.x - A.x);
line.b = A.y - A.x * line.k;
}
else
{
line.xConst = A.x;
line.flagXconst = true;
}
return line;
}
private pointXY GetPointFromLines(lineKB a, lineKB b)
{
pointXY point = new pointXY();
if (a.flagXconst)
{
point.x = a.xConst;
point.y = a.xConst * b.k + b.b;
}else
if (b.flagXconst)
{
point.x = b.xConst;
point.y = b.xConst * a.k + a.b;
}
else
{
point.x = (a.b - b.b) / (b.k - a.k);
point.y = a.k * point.x + a.b;
}
return point;
}
private double LengthOfLine(pointXY A, pointXY B)
{
return Math.Sqrt((B.x - A.x) * (B.x - A.x) + (B.y - A.y) * (B.y - A.y));
}
private pointXY GetMidlePoint(pointXY S, double l, lineKB line, bool leftright)
{
double b = -2 * S.x - 2 * line.k * (-line.b + S.y);
double a = (1 + line.k * line.k);
double c = (S.x * S.x - l * l + (-line.b + S.y) * (-line.b + S.y));
double d = b*b - 4 * a * c;
double x1 = (-b + Math.Sqrt(d)) / (2 * a);
double x2 = (-b - Math.Sqrt(d)) / (2 * a);
pointXY ret = new pointXY();
if (leftright)
if (x1 > S.x) ret.x = x1;
else ret.x = x2;
else
if (x1 < S.x) ret.x = x1;
else ret.x = x2;
ret.y = line.k * ret.x + line.b;
return ret;
}
private void Button_Click(object sender, RoutedEventArgs e)
{
pointXY A = new pointXY();
A.x = pol.Points[0].X;
A.y = pol.Points[0].Y;
pointXY B = new pointXY();
B.x = pol.Points[1].X;
B.y = pol.Points[1].Y;
pointXY C = new pointXY();
C.x = pol.Points[2].X;
C.y = pol.Points[2].Y;
pointXY D = new pointXY();
D.x = pol.Points[3].X;
D.y = pol.Points[3].Y;
lineKB AC = GetLineFromPonts(A, C);
lineKB BD = GetLineFromPonts(B, D);
pointXY R = GetPointFromLines(AC, BD);
lineKB AB = GetLineFromPonts(A, B);
lineKB BC = GetLineFromPonts(B, C);
lineKB CD = GetLineFromPonts(C, D);
lineKB DA = GetLineFromPonts(D, A);
pointXY E = GetPointFromLines(AB, CD);
lineKB ER = GetLineFromPonts(E, R);
pointXY L = GetPointFromLines(ER, DA);
pointXY N = GetPointFromLines(ER, BC);
pointXY F = GetPointFromLines(BC, DA);
lineKB FR = GetLineFromPonts(F, R);
pointXY M = GetPointFromLines(FR, AB);
pointXY O = GetPointFromLines(FR, CD);
pointXY W = GetMidlePoint(A, (LengthOfLine(A, R) / (4 * Math.PI)), AC, true);
pointXY X = GetMidlePoint(B, (LengthOfLine(B, R) / (4 * Math.PI)), BD, false);
pointXY Y = GetMidlePoint(C, (LengthOfLine(C, R) / (4 * Math.PI)), AC, false);
pointXY Z = GetMidlePoint(D, (LengthOfLine(D, R) / (4 * Math.PI)), BD, true);
pathleftup.StartPoint = new Point(L.x, L.y);
bezleftup.Point1 = new Point(W.x, W.y);
bezleftup.Point2 = new Point(M.x, M.y);
pathleftdown.StartPoint = new Point(L.x, L.y);
bezleftdown.Point1 = new Point(Z.x, Z.y);
bezleftdown.Point2 = new Point(O.x, O.y);
pathrigthdown.StartPoint = new Point(O.x, O.y);
bezrigthdown.Point1 = new Point(Y.x, Y.y);
bezrigthdown.Point2 = new Point(N.x, N.y);
pathrigthup.StartPoint = new Point(M.x, M.y);
bezrigthup.Point1 = new Point(X.x, X.y);
bezrigthup.Point2 = new Point(N.x, N.y);
}
关于c# - 4 点和椭圆，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/5451593/