4点和椭圆
obe*_*iro 22 c# silverlight wpf drawing silverlight-4.0
我有4分..我可以画一个多边形使用这个代码
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"？
问题仍然没有发现!
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>
这不是确切的解决方案,因为确切地说,你必须计算曲线的精确点并使用4 QuadraticBezierSegment
编辑:
示例QuadraticBezierSegment
<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,而不是计算点,但它非常精确.
编辑2: 您可以使用此图像和我的评论计算曲线点:
曲线有一个起点,中间点和终点.在这个图像开始和结束piont是L,M,N,O; 和midle是W,X,Y,Z.
例如我们如何计算点数L:
在线方程的帮助下,y = k * x + b我们找到了线的方程AB,DC,AC,DB,AD.我们如何找到AC和.我们如何找到和.在那之后我们找到了线的方程以及我们如何发现的交叉.DBRABDCEERERADL
我们如何计算点数W:
蒙山方程长度帮助l = sqrt(sqr(x2 - x1) + sqr(y2 - y1))的发现长度AR.AW = AR/(4*pi)在求解方程后求解该系数,以及线方程和长度方程W.
其他我们发现的相似点.
该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.并且长度系数是相同的.
在这个算法的帮助下,我找到了你的例子的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>
它的曲线与椭圆线完全相同.图片如下:
您可以将此算法转换为"公式",以便找到您的解决方案.
你需要4个功能:
1)从2个pionts的坐标中找到线系数
2)找到piont的坐标如何从它的系数中交叉2条线
3)从2个点的坐标中找出段的长度
4)找到与此起点一致的piont的坐标,这个长度来自线系数和你在前一个函数中找到的起点和长度的坐标除以(4*pi)
编辑3: 您可以优化此解决方案,它有一些缺陷,如何平行线等.但它快速,如果您优化它可能满足您的要求.
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);
}
- 你把_quite_做了一些工作.谢谢你给我们最好的:) (3认同)
使用Jeff MI提供的信息创建了一个函数,该函数返回一个拟合在多边形中的椭圆:
Ellipse FitEllipse(Polygon poly)
{
double W0 = poly.Points[0].X;
double W1 = poly.Points[0].Y;
double X0 = poly.Points[1].X;
double X1 = poly.Points[1].Y;
double Y0 = poly.Points[2].X;
double Y1 = poly.Points[2].Y;
double Z0 = poly.Points[3].X;
double Z1 = poly.Points[3].Y;
double A = X0 * Y0 * Z1 - W0 * Y0 * Z1 - X0 * Y1 * Z0 + W0 * Y1 * Z0 - W0 * X1 * Z0 + W1 * X0 * Z0 + W0 * X1 * Y0 - W1 * X0 * Y0;
double B = W0 * Y0 * Z1 - W0 * X0 * Z1 - X0 * Y1 * Z0 + X1 * Y0 * Z0 - W1 * Y0 * Z0 + W1 * X0 * Z0 + W0 * X0 * Y1 - W0 * X1 * Y0;
double C = X0 * Y0 * Z1 - W0 * X0 * Z1 - W0 * Y1 * Z0 - X1 * Y0 * Z0 + W1 * Y0 * Z0 + W0 * X1 * Z0 + W0 * X0 * Y1 - W1 * X0 * Y0;
double D = X1 * Y0 * Z1 - W1 * Y0 * Z1 - W0 * X1 * Z1 + W1 * X0 * Z1 - X1 * Y1 * Z0 + W1 * Y1 * Z0 + W0 * X1 * Y1 - W1 * X0 * Y1;
double E = -X0 * Y1 * Z1 + W0 * Y1 * Z1 + X1 * Y0 * Z1 - W0 * X1 * Z1 - W1 * Y1 * Z0 + W1 * X1 * Z0 + W1 * X0 * Y1 - W1 * X1 * Y0;
double F = X0 * Y1 * Z1 - W0 * Y1 * Z1 + W1 * Y0 * Z1 - W1 * X0 * Z1 - X1 * Y1 * Z0 + W1 * X1 * Z0 + W0 * X1 * Y1 - W1 * X1 * Y0;
double G = X0 * Z1 - W0 * Z1 - X1 * Z0 + W1 * Z0 - X0 * Y1 + W0 * Y1 + X1 * Y0 - W1 * Y0;
double H = Y0 * Z1 - X0 * Z1 - Y1 * Z0 + X1 * Z0 + W0 * Y1 - W1 * Y0 - W0 * X1 + W1 * X0;
double I = Y0 * Z1 - W0 * Z1 - Y1 * Z0 + W1 * Z0 + X0 * Y1 - X1 * Y0 + W0 * X1 - W1 * X0;
double detT = A * E * I + B * F * G + C * D * H - A * F * H - B * D * I - C * E * G;
double J = (E * I - F * H) / detT;
double K = (C * H - B * I) / detT;
double L = (B * F - C * E) / detT;
double M = (F * G - D * I) / detT;
double N = (A * I - C * G) / detT;
double O = (C * D - A * F) / detT;
double P = (D * H - E * G) / detT;
double Q = (B * G - A * H) / detT;
double R = (A * E - B * D) / detT;
double a = J * J + M * M + P * P;
double b = J * K + M * N - P * Q;
double c = K * K + N * N - Q * Q;
double d = J * L + M * O - P * R;
double f = K * L + N * O - Q * R;
double g = L * L + O * O - R * R;
double Ex = (c * d - b * f) / (b * b - a * c);
double Ey = (a * f - b * d) / (b * b - a * c);
double Ea = Math.Sqrt(2.0 * (a * f * f + c * d * d + g * b * b - 2.0 * b * d * f - a * c * g) / ((b * b - a * c) * (Math.Sqrt((a - c) * (a - c) + 4.0 * b * b) - (a + c))));
double Eb = Math.Sqrt(2.0 * (a * f * f + c * d * d + g * b * b - 2.0 * b * d * f - a * c * g) / ((a * c - b * b) * (Math.Sqrt((a - c) * (a - c) + 4.0 * b * b) + (a + c))));
double phi = 0;
if (b == 0 && a < c) {
phi = 0;
} else if (b == 0 && a > c) {
phi = Math.PI / 2;
} else if (b != 0 && a < c) {
phi = (Math.PI / 2 - Math.Atan((a - c) / (2 * b))) / 2;
} else if (b != 0 && a > c) {
phi = (Math.PI / 2 - Math.Atan((a - c) / (2 * b))) / 2 + Math.PI / 2;
}
Ellipse el = new Ellipse();
el.Height = Ea * 2;
el.Width = Eb * 2;
el.RenderTransform = new RotateTransform(phi * 180 / Math.PI);
return el;
}
- 我一次只能跟踪4个无意义的变量名. (4认同)