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Tikz-以特定角度放置矩形

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Tikz-以特定角度放置矩形

我试图将一个矩形放在一个具有一定半径的假想圆的顶部，该圆的中心位于另一个矩形的中间。到目前为止，我尝试过以下方法

\documentclass{article}

\usepackage{kerkis}
\usepackage{tikz}

\begin{document}
 \begin{tikzpicture}
  % Incident Beam
  \draw[->, thick] (0,0) -- (2,0);
  % Target : Boron + Au --- The center of the imaginary circle is in the center of these rectangles
  \draw[gray!30,fill=gray!30] (6,1) rectangle (6.5,-1);
  \draw[gray!70,fill=gray!70] (6.5,1) rectangle (8.5,-1);
  % Scattering Chamber - The imaginary circle
  %\draw[thick,red,->,dashed] ([shift=(-120:5cm)]7.25,0) arc (0:100:-1cm);
  %\draw[dashed, gray!50] (7.25,0) circle (7cm);
  % Telescopes --- I want to place this rectangle on 170 degrees on top of the imaginary circle
  \draw[black!70, fill=black!70, rotate=30] (2,-2) rectangle (2.5, -2.5); 
 \end{tikzpicture}
\end{document}

我的输出是

在此处输入图片描述

如何将这个黑色矩形以一定角度放置在虚线圆的上方？

答案1

如果想象圆的中心在(7.25,0)，并且圆的半径为 7cm，则可以在 处插入一个旋转的矩形节点(7.25,0)+(170:7)：

\path(7.25,0)+(170:7)
  node[
    fill=black!70,
    minimum width=.5cm,minimum height=1cm,% rectangle
    rotate=80 % 170°-90°=80°
  ]{};

在此处输入图片描述

代码：

\documentclass[margin=5mm]{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
  % Incident Beam
  \draw[->, thick] (0,0) -- (2,0);
  % Target : Boron + Au --- The center of the imaginary circle is in the center of these rectangles
  \draw[gray!30,fill=gray!30] (6,1) rectangle (6.5,-1);
  \draw[gray!70,fill=gray!70] (6.5,1) rectangle (8.5,-1);
  % Scattering Chamber - The imaginary circle
  %\draw[thick,red,->,dashed] ([shift=(-120:5cm)]7.25,0) arc (0:100:-1cm);
  \draw[dashed, red] (7.25,0)coordinate(M) circle [radius=7cm];
  % Telescopes --- I want to place this rectangle on 170 degrees on top of the imaginary circle
  \path(7.25,0)+(170:7)
    node[fill=black!70,
    minimum width=.5cm,minimum height=1cm,% rectangle
    rotate=80 % 170°-90°=80°
    ]{};
\end{tikzpicture}
\end{document}

答案2

您还可以使用节点，

\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}
\node[draw,dotted,minimum size=3cm,circle] (a) {};
\node[fill,minimum size=2mm,rotate=55,inner sep=0] at (a.140) {};
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案3

另一个想法：对填充的正方形使用倾斜节点。这样，您不必指定旋转该正方形的角度，它将自动计算为与圆的半径对齐：

\begin{tikzpicture}
  \tikzset{
    my square/.style={
       fill=orange!70, 
       minimum size=2mm, 
       pos=1, 
       sloped
    }
  }

  \draw [dotted] (0,0) circle(1.5cm);
  \foreach \angle in {0,30,...,360}
    \foreach \distance in {0.9,1.2,1.5}
      \path (0,0) -- (\angle:\distance) 
         node[my square] {};
\end{tikzpicture}

结果

更新

回答评论中的附加问题，可以绘制一个矩形，给出节点的最小宽度和高度。但是，将宽度保留为具有默认值的参数可能很有用：

\begin{tikzpicture}
  \tikzset{
    my rectangle/.style={
       fill=orange!70, 
       minimum width=#1,   % The width is a parameter
       minimum height=2mm, 
       pos=1, 
       sloped
    },
    my rectangle/.default = 4mm,   % Width when not specified
  }

  \draw [dotted] (0,0) circle(1.5cm);

  % Some rectangles with default width
  \foreach \angle in {0,30,...,330}
     \path (0,0) -- (\angle:1.5) node[my rectangle] {};

  % Other rectangles with given width (and different color)
  \foreach \angle in {0,90,...,330}
     \path (0,0) -- (\angle:1.5) node[my rectangle=7mm, red] {};
\end{tikzpicture}

结果2

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