tikz - 圆形路径上的切线矩形

Question

我想将矩形放置在圆形路径上，使它们与圆相切。

如果我使用极坐标和连通线，最终会得到如下结果：

我希望矩形部分保持真正的矩形。如果我将矩形命令与旋转选项一起使用，则很难将矩形沿圆形路径放置。

是否可以将矩形命令与极坐标一起使用？

编辑：我的糟糕尝试如下：

\documentclass[a4paper,10pt]{scrreprt}
\usepackage{tikz}
\begin{tikzpicture}[scale=1]
\foreach \a in {0, 45, 90, 135, 180}{
\path[draw] (\a:1) -- (\a:2) -- (\a + 10:2) -- (\a + 10:1) -- cycle;
}
\end{document}

Answer 1

\documentclass[tikz, border=2mm]{standalone}

\begin{document}

\begin{tikzpicture}

\draw circle (3cm);

\draw (30:3cm) node[draw, anchor=south, rotate=-60, minimum width=1cm] {A};

\draw (60:3cm) node[draw, anchor=south, rotate=-30, minimum width=1cm] {B};

\draw (215:3cm) node[draw, anchor=south, rotate=125, minimum width=1cm] {C};
\end{tikzpicture}

\end{document}

Answer 2

我认为这样做可以：

\documentclass[tikz,border=5]{standalone} 
\usetikzlibrary{math}
\begin{document} 
\begin{tikzpicture} 
\tikzmath{%
  \r = 1;
  \a = 10; % angle of chord forming base of rectangle
  \s = 15; % angle between rectangle centers 
  \R = \r * cos(\s / 2);
  \n = int(360 / \s);
  \x = \r / 2 * sin(\a);
}

\draw [help lines] circle [radius=\r];
\foreach \i in {1, ..., \n}
  \draw [help lines] 
    (0,0) -- (\i*\s-\a/2:\r+1/2) (0,0) -- (\i*\s+\a/2:\r+1/2);

\foreach \i in {1, ..., \n}
  \draw [shift=(\i*\s:\R),rotate=\i*\s+270] 
    (-\x, 0) rectangle (\x, 1/8);
\end{tikzpicture}
\end{document}

Answer 3

您可以使用\foreach循环在圆圈周围绘制矩形。

\documentclass{article}

\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[
   rectangle node/.style={
      minimum width = \w,
      minimum height = \h,
      fill = gray,
      inner sep = 0pt,
      anchor = center,
   },
]
   % Variables
   \def\N{36}% Number of rectangles - 1
   \def\h{0.2cm}% height of rectangles
   \def\w{0.4cm}% width of rectangles
   \def\R{3}% Radius
   % draw rectangles
   \foreach \n in {0,...,\N} {
      \node [rectangle node, rotate=\n*(360/(\N+1))+90] at ({\n*360/(\N+1)}:\R) {};
   }
\end{tikzpicture}
\end{document}

我定义了一些变量来设置矩形的大小（\w和\h），矩形的数量（\N+1）和圆的半径（\R）。