怎样改善身材？

Question

我想要绘制彼此相切且与大圆相切的小圆，如图所示。

但是，我的图形并没有像我希望的那样出现，小圆圈实际上与大圆圈相切。

\documentclass[12pt]{article}

\usepackage{tikz}   

\begin{document}

\begin{center}

\begin{tikzpicture}[scale=1.2]
\draw[ultra thick, red!95] (0,0) circle [radius=2.9];
\draw[ultra thick, blue!90] (3.4,2) circle [radius=1];
\draw[ultra thick, yellow!90] (2,3.4) circle [radius=1];
\draw[ultra thick, green!90] (0,4) circle [radius=1];
\draw[ultra thick, cyan] (-2,3.4) circle [radius=1];
\draw[ultra thick, black] (-3.4,2) circle [radius=1];
\draw[ultra thick, green!90] (-4,0) circle [radius=1];
\draw[ultra thick, blue!90] (-3.4,-2) circle [radius=1];
\draw[ultra thick, cyan] (-2,-3.4) circle [radius=1];
\draw[ultra thick, blue!90] (0,-4) circle [radius=1];
\draw[ultra thick, yellow!95] (2,-3.4) circle [radius=1];
\draw[ultra thick, black] (3.4,-2) circle [radius=1];
\draw[ultra thick, green!90] (4,0) circle [radius=1];
\node at (2.2,2.2) {$\mathbf{S_{1}}$};
\node at (0.8,3)   {$\mathbf{S_{2}}$};
\node at (-0.8,3)  {$\mathbf{S_{3}}$};
\node at (-2.2,2.2){$\mathbf{S_{4}}$};
\node at (-3,0.8)  {$\mathbf{S_{5}}$};
\node at (-3,-0.8) {$\mathbf{S_{6}}$};
\node at (-2.2,-2.2){$\mathbf{S_{7}}$};
\node at (-0.8,-3){$\mathbf{S_{8}}$};
\node at (0.8,-3)   {$\mathbf{S_{9}}$};
\node at (2.2,-2.2) {$\mathbf{S_{10}}$};
\node at (3,-0.8) {$\mathbf{S_{11}}$};
\node at (3,0.8) {$\mathbf{S_{12}}$};
\end{tikzpicture}

\end{center}

\end{document}

Answer 1

这里有一个限制。您不能拥有任意大小的半径，以及任意数量的圆，并且让它们都像您描述的那样相切。

下面的代码确保它们相切，并根据给定的大圆半径计算小圆所需的半径。

\documentclass[12pt]{article}

\usepackage{tikz}   
\usetikzlibrary{calc}

\begin{document}

\begin{center}

\begin{tikzpicture}[scale=1.2]

\def\numcircles{12}
\def\bigradius{2.9}
% define the colors beforehand so they're in an array and easy to use
\def\circlecolors{blue!90, yellow!90, green!90, cyan, black, green!90, blue!90, cyan, blue!90, yellow!95, black, green!90}

% I'm doing trigonometry here....
\pgfmathsetmacro\s{sin(360/(2*\numcircles))}
\pgfmathsetmacro\smallradius{\bigradius * \s / (1-\s)}%

\draw[ultra thick, red!95] (0,0) circle [radius=\bigradius];

% Do the circles & labels in a loop, instead of one-by-one.
\foreach\circcolor [count=\i,
        evaluate={\i as \angle using {(3-\i)*360/\numcircles}}] in \circlecolors {
    % this "evaluate" statement gives us the angle. The "(3-\i)" 
    % is to shift the starting position of the labels to match the ones in
    % your picture. The negative is to ensure it goes counter-clockwise.

    % first draw the circle ....
    \draw[ultra thick,color=\circcolor] (\angle:{\smallradius+\bigradius}) circle [radius={\smallradius}];

    % and then add the label at an offset equal to half the angle
    \node  at ({\angle+180/\numcircles}:{\bigradius+0.2}){$S_{\i}$};
}
\end{tikzpicture}

\end{center}
\end{document}

得到下图的结果：

线条重叠，但可以通过减少\smallradius线条宽度的数量来消除这种情况。

至于三角函数，只需画两个与大圆相切的小圆，并将它们的中心用线段连接起来。设“r”为大圆半径，“a”为小圆中心之间的角度，“x”为小圆半径。应该立即得到sin(a / 2) = x / (x+r)，我刚刚解出了 x。

Answer 2

类似这样的情况？也许你需要进行一些计算来将线宽从半径中排除。然后它就变成了一个不难解决的数学问题。

下面pic可以画一个大圆，以及围绕大圆任意一个半径和线宽相同的小圆。

选项

• r：大圆的半径
• num：小圆圈的数量
• small width：小圆圈的线宽
• big width：大圆的线宽
• pos：大圆边界到节点的距离
• color 0：大圆圈的颜色
• color i(i >= 1): 第 i 个小圆圈的颜色

第一个小圆位于 上0 deg。如果有 n 个小圆，则第二个小圆位于 上，360/n deg依此类推。

第一个节点位于第一个和第二个小圆圈之间，picname-1依此类推。

\documentclass[tikz, border=1cm]{standalone}
\usetikzlibrary{fpu}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand {\myfor} { m +m } {
  \int_step_variable:nNn {#1} \l_for_tl {\def\forvar{\tl_use:N \l_for_tl}#2}
}
\ExplSyntaxOff

% predefine
\myfor{50}{
  \pgfkeys{/multicircles/color \forvar/.initial=black}
}
\makeatletter
\newlength\mc@r
\newlength\mc@R
\newlength\mc@rR
\newlength\mc@node

\tikzset{
  mc/.pic={
    \pgfkeys{/pgf/fpu, /pgf/fpu/output format=fixed}
    \pgfmathsetlength{\mc@R}{\mc@radius-\mc@big@width/2}
    \pgfmathsetmacro{\mc@ang}{360/\mc@num}
    \pgfmathsetmacro{\mc@cos}{90-\mc@ang/2}
    \pgfmathsetlength{\mc@r}{cos(\mc@cos)/(1-cos(\mc@cos))*\mc@radius-\mc@small@width/2}
    \pgfmathsetlength{\mc@rR}{\mc@radius/(1-cos(\mc@cos))}
    \pgfmathsetlength{\mc@node}{\mc@pos*\mc@rR*cos(\mc@ang/2)+(1-\mc@pos)*\mc@R}
    \pgfkeys{/pgf/fpu=false}
    \draw [\mc@color@main, line width=\mc@big@width] (0, 0) circle (\mc@R);
    \foreach \i in {1,...,\mc@num} {
      \def\temp@color{\pgfkeysvalueof{/multicircles/color \i}}
      \draw [\temp@color, line width=\mc@small@width] ({(\i-1)*\mc@ang}:\mc@rR) circle (\mc@r);
      \coordinate (-\i) at ({(\i-0.5)*\mc@ang}:\mc@node);
    }
  },
  /multicircles/.search also=/tikz,
  /multicircles/.cd,
  color 0/.store in=\mc@color@main,
  color 0=black,
  num/.store in=\mc@num, num=6,
  r/.store in=\mc@radius, r=1cm,
  big width/.store in=\mc@big@width,
  big width=1.6pt,
  small width/.store in=\mc@small@width,
  small width=1.6pt,
  pos/.store in=\mc@pos, pos=.3,
}
\makeatother
\newcommand{\multicircles}[1][]{
  \pic [pic type=mc, /multicircles/.cd,#1];
}
\newcommand{\multicirclesset}[1]{\pgfqkeys{/multicircles}{#1}}
\multicirclesset{
  colors/.code args={#1!#2!#3}{
    \myfor{#2}{
      \pgfmathparse{\forvar/#2*100}
      \edef\temp{
        \noexpand\multicirclesset{color \forvar=#1!\pgfmathresult!#3}
      }
      \temp
    }
  }
}

\begin{document}
\begin{tikzpicture}
  \multicircles[
    r=2cm, num=6, color 0=red!60,
    big width=3pt, small width=3pt,
    colors=red!6!blue,
    at={(0, 0)},
    name=a,
  ]
  \foreach \i in {1,...,6} {
    \node [font=\small] at (a-\i) {$S_{\i}$};
  }
  \multicircles[
    r=3cm, num=12, color 0=red!60,
    big width=2pt, small width=4pt,
    colors=yellow!12!cyan,
    at={(12, 0)},
    name=b,
  ]
  \foreach \i in {1,...,12} {
    \node [font=\tiny] at (b-\i) {$S_{\i}$};
  }
  \multicircles[
    r=4cm, num=19, color 0=red!60,
    big width=8pt, small width=5pt,
    colors=green!19!teal,
    at={(24, 0)},
    name=c,
    pos=-.5,
  ]
  \foreach \i in {1,...,19} {
    \node at (c-\i) {$S_{\i}$};
  }
\end{tikzpicture}
\end{document}