☰

绘制庞加莱圆盘的等距线

tag-icon • tag-icon tikz-pgf
绘制庞加莱圆盘的等距线

尝试绘制庞加莱圆盘的“等距离”线，就像这篇文章中那样：https://math.stackexchange.com/a/1343730/390217

我尝试使用双曲函数，但不是那样，结果不太正确： 错误的庞加莱圆盘

这是代码，您可以通过操作值和函数来获得一些漂亮的图片。

\documentclass[border=5pt]{standalone}
\usepackage{tkz-euclide}
\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{fp}
\usepackage{comment}

\begin{document}
\begin{tikzpicture}[scale=3]
  \tkzDefPoint(0,0){O}
  \tkzDefPoint(1,0){A}
  \tkzDrawCircle(O,A)
  \draw (A) -- (-1,0);
  \draw (0,1) -- (0,-1);
\foreach \nt in {0.1,0.2,...,1}{
\pgfmathsetmacro\nx{cosh(\nt)}
\pgfmathsetmacro\ny{sinh(\nt)}
\pgfmathsetmacro\mx{(-\nx)}
\pgfmathsetmacro\my{(-\ny)}
  \tkzClipCircle(O,A)
  \tkzDefPoint(\nx,\ny){z1}
  \tkzDefPoint(\nx,\my){z2}
  \tkzDefPoint(\mx,\my){z3}
  \tkzDefPoint(\mx,\ny){z4}
  \tkzDefPoint(\ny,\nx){z5}
  \tkzDefPoint(\my,\nx){z6}
  \tkzDefPoint(\my,\mx){z7}
  \tkzDefPoint(\ny,\mx){z8}
  \tkzDrawCircle[orthogonal through=z1 and z2](O,A)
  \tkzDrawCircle[orthogonal through=z3 and z4](O,A)
  \tkzDrawCircle[orthogonal through=z1 and z4](O,A)
  \tkzDrawCircle[orthogonal through=z2 and z3](O,A)
  \tkzDrawCircle[orthogonal through=z5 and z6](O,A)
  \tkzDrawCircle[orthogonal through=z7 and z8](O,A)
  \tkzDrawCircle[orthogonal through=z5 and z8](O,A)
  \tkzDrawCircle[orthogonal through=z6 and z7](O,A)
}
\end{tikzpicture}
\end{document}

有人知道我是否可以通过调整函数或值来获得正确的结果吗？我不知道如何通过投射抛物面的类时线来获得它，就像上面提到的帖子中解释的那样……

感谢您的帮助/建议

答案1

问题解决了 ！

我联系了上述图片的作者，他慷慨地将使用的代码发送给了我。这是计算机生成的非常长的坐标点列表，所以我使用 tikz-euclide 和 symetries 对其进行了调整，以尽可能减少它。结果是完美的。以下是感兴趣的人可以查看的代码：

\documentclass[border=5pt]{standalone}
\usepackage{tkz-euclide}
\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{fp}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{10pt}%

\begin{document}

\begin{tikzpicture}
\pgfsetlinewidth{0.4pt}
\useasboundingbox (0in,0in) rectangle (6in,6in);
%circle
\def\rad{3in}
\node (O) at (3in,3in) []{};
\draw (O) circle (\rad);
\node (A) at (6in,3in) []{};
\pgfsetlinewidth{0.8pt}
%arcs
\draw (3in,0)--(3in,6in); %vertical
\draw (0,3in)--(6in,3in); %horizontal
\foreach \x/\y in {
2.27947/0.210234,
1.67325/0.431525,
1.22212/0.705443,
0.906042/0.973135,
0.687819/1.20952,
0.53563/1.40973,
0.427272/1.57697,
0.34825/1.71661,
0.289239/1.83382,
0.244185/1.93299,
0.209092/2.01764,
0.181261/2.09053,
0.158834/2.15381,
0.140503/2.20918,
0.125327/2.25798}{
  \tkzClipCircle(O,A)
\pgfmathsetmacro\ny{6-\y}
  \node (z1) at (\x in,\y in) []{};
  \node (z2) at (\x in,\ny in) []{};
  \node (z3) at (\y in,\x in) []{};
  \node (z4) at (\ny in,\x in) []{};
\pgfmathsetmacro\nx{6-\x} 
  \node (z5) at (\y in,\nx in) []{};
  \node (z6) at (\ny in,\nx in) []{};
  \node (z7) at (\nx in,\y in) []{};
  \node (z8) at (\nx in,\ny in) []{};
  \tkzDrawCircle[orthogonal through=z1 and z2](O,A)
  \tkzDrawCircle[orthogonal through=z3 and z4](O,A)
  \tkzDrawCircle[orthogonal through=z5 and z6](O,A)
  \tkzDrawCircle[orthogonal through=z7 and z8](O,A)
}
\end{tikzpicture}
\end{document}

结果 ： 庞加莱圆盘

相关内容