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用彎曲箭頭連接規則多邊形的頂點

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用彎曲箭頭連接規則多邊形的頂點

我有一個圖表，它只是一組均勻分佈在一個圓形周圍的節點。如何用圓形箭頭連接節點？

例如，假設我們有一個五邊形，每個頂點都有節點。顯然五邊形的頂點都在圓上。我想透過與圓「平行」的箭頭連接節點。 （所以總的來說它看起來像一個循環）

答案1

這是一種可能性：使用類型 的兩個節點regular polygon，一個內部節點，另一個外部節點，都放置在圓圈上。

\documentclass[a4paper,12pt]{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}

\newcommand{\polygonsides}{5}

\begin{document}
\begin{minipage}[t][0.3\textheight]{0.45\textwidth}
Clockwise [option \texttt{bend left}]:
\begin{flushleft}
\begin{tikzpicture}
% Internal cirlce with polygon
\draw[blue](0,0)circle(1.5cm);
\node[regular polygon, regular polygon sides=\polygonsides, minimum size=3cm, draw, name=x] at (0,0) {};
% Nodes on vertices
\foreach \corner in {1,2,...,\polygonsides}
\node[circle,ball color=blue] at (x.corner \corner){};

% External polygon
\node[regular polygon, regular polygon sides=\polygonsides, minimum size=4cm, draw=none, name=p] at (0,0) {};
% Invisible nodes on vertices
\foreach \corner in {1,2,...,\polygonsides}
\node at (p.corner \corner){};

% Connections
\foreach \source/\destination in {p.corner 1/p.corner 5,p.corner 5/p.corner 4,p.corner 4/p.corner 3,p.corner 3/p.corner 2,p.corner 2/p.corner 1}
\draw[-stealth,shorten <=0.2cm,shorten >=0.2cm,thick,blue](\source)to[bend left](\destination);
\end{tikzpicture}
\end{flushleft}
\end{minipage} 
\begin{minipage}[t][0.3\textheight]{0.55\textwidth}
Counterclockwise [option \texttt{bend right}]:
\begin{flushleft}
\begin{tikzpicture}
% Internal cirlce with polygon
\draw[red](0,0)circle(1.5cm);
\node[regular polygon, regular polygon sides=\polygonsides, minimum size=3cm, draw, name=x] at (0,0) {};
% Nodes on vertices
\foreach \corner in {1,2,...,\polygonsides}
\node[circle,ball color=red] at (x.corner \corner){};

% External polygon

\node[regular polygon, regular polygon sides=\polygonsides, minimum size=4cm, draw=none, name=p] at (0,0) {};

% Nodes on vertices
\foreach \corner in {1,2,...,\polygonsides}
\node at (p.corner \corner){};

% Connections
\foreach \source/\destination in {p.corner 1/p.corner 2,p.corner 2/p.corner 3,p.corner 3/p.corner 4,p.corner 4/p.corner 5,p.corner 5/p.corner 1}
\draw[-stealth,shorten <=0.2cm,shorten >=0.2cm,thick,red](\source)to[bend right](\destination);
\end{tikzpicture}
\end{flushleft}
\end{minipage} 
\end{document}

結果：

在此輸入影像描述

未繪製外部多邊形，因為它只是用作正確設定作為箭頭起點和終點的頂點的參考。當然，可以透過為該外部多邊形聲明更近的半徑來使箭頭更靠近多邊形：現在距離為1cm。以下有兩個範例：第一個箭頭是順時針方向，而第二個箭頭是逆時針方向。要實現這兩件事是必要的：

  • 使用bend left順時針和bend right逆時針選項；
  • 按順時針順序或逆時針順序連接該對source/destination，請記住節點是按逆時針順序編號的頂點：

在此輸入影像描述

答案2

緊湊版本（沒有自動進行週長計算，因為我不確定這是否是所要求的）

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric,decorations.markings}

\begin{document}
\begin{tikzpicture}[decoration={
    markings,
    mark=between positions 0 and 1 step 2.51376 cm with {\arrow{latex}}
    }
]
\draw[postaction={decorate}] (90:2cm) arc (-270:90:2cm);
\node[draw,red,regular polygon, regular polygon sides=5,minimum height=4cm] {};
\end{tikzpicture}
\end{document}

在此輸入影像描述

答案3

在下文中，我使用shapes.geometric定義名為 的形狀的庫regular polygon，以便無論多邊形有多少邊，您都可以獲得通用的解決方案。我還需要一個輔助巨集\pgfmathsetlenghtbetweenanchors來計算相同節點的兩個錨點之間的長度。

為了繪製弧線，我使用給定的語法這裡我需要如上所述的兩個角之間的角度這裡。

在此輸入影像描述

\documentclass{standalone}

\usepackage{tikz}
\usetikzlibrary{shapes.geometric}

\begin{document}

\makeatletter
\def\pgfmathsetlenghtbetweenanchors#1#2#3#4{%
  % #1: length
  % #2: node
  % #3: first anchor
  % #4: second anchor
  \pgfpointdiff{%
    \pgfpointanchor{#2}{#3}}{%
    \pgfpointanchor{#2}{#4}}%
  \pgfmathparse{veclen(\pgf@x,\pgf@y)}%
  #1=\pgfmathresult pt}
\makeatother

\newlength\nagonradius

\begin{tikzpicture}
  \def\nsides{5}%
  \node[regular polygon,
        regular polygon sides = \nsides,
        draw,
        rotate=25,
        minimum size = 3cm] (\nsides-agon) {};

  \pgfmathsetlenghtbetweenanchors\nagonradius{\nsides-agon}{center}{%
    corner 2}% 
  \foreach \n [remember = \n as \m (initially \nsides)] in
    {1,...,\nsides} {%
      \pgfmathanglebetweenpoints{%
        \pgfpointanchor{\nsides-agon}{center}}{%
        \pgfpointanchor{\nsides-agon}{corner \m}}%
      \let\anglem\pgfmathresult
      \pgfmathanglebetweenpoints{%
        \pgfpointanchor{\nsides-agon}{center}}{%
        \pgfpointanchor{\nsides-agon}{corner \n}}%
      \let\anglen\pgfmathresult
      \ifdim\anglen pt < \anglem pt
        \pgfmathparse{\anglen + 360}%
        \let\anglen\pgfmathresult
      \fi
      \draw[->,red] ([shift={(\anglem:\nagonradius)}]\nsides-agon.center) arc
        (\anglem:\anglen:\nagonradius);} 

\end{tikzpicture}

\begin{tikzpicture}
  \def\nsides{7}%
  \node[regular polygon,
        regular polygon sides = \nsides,
        draw,
        rotate=25,
        minimum size = 3cm] (\nsides-agon) {};

  \pgfmathsetlenghtbetweenanchors\nagonradius{\nsides-agon}{center}{%
    corner 2}% 
  \foreach \n [remember = \n as \m (initially \nsides)] in
    {1,...,\nsides} {%
      \pgfmathanglebetweenpoints{%
        \pgfpointanchor{\nsides-agon}{center}}{%
        \pgfpointanchor{\nsides-agon}{corner \m}}%
      \let\anglem\pgfmathresult
      \pgfmathanglebetweenpoints{%
        \pgfpointanchor{\nsides-agon}{center}}{%
        \pgfpointanchor{\nsides-agon}{corner \n}}%
      \let\anglen\pgfmathresult
      \ifdim\anglen pt < \anglem pt
        \pgfmathparse{\anglen + 360}%
        \let\anglen\pgfmathresult
      \fi
      \draw[->,red] ([shift={(\anglem:\nagonradius)}]\nsides-agon.center) arc
        (\anglem:\anglen:\nagonradius);} 

\end{tikzpicture}


\end{document}

答案4

這是一個使用以下解決方案arc：

在此輸入影像描述

和代碼：

\documentclass{standalone}
\usepackage{tikz}
\pgfmathsetmacro{\radius}{2}
\begin{document}
\foreach \nbn in {3,5,7,9}{
  \begin{tikzpicture}
    \pgfmathsetmacro{\angle}{360/\nbn}
    % center
    \node {\nbn{} nodes};
    % draw nodes
    \foreach \i in {1,...,\nbn}{
      \node at (\angle*\i:\radius) {$N_\i$};
    }
    % draw arrows (clockwise)
    \foreach \i in {1,...,\nbn}{
      \draw[-latex,very thick,red!50!black]
      ({\angle*(\i+.2)}:\radius-.2)
      arc (\angle*(\i+.2):\angle*(\i+1-.2):\radius-.2);
    }

    % draw arrows (anticlockwise)
    \foreach \i in {1,...,\nbn}{
      \draw[-latex,very thick,green!50!black]
      ({\angle*(\i-.2)}:\radius+.2)
      arc (\angle*(\i-.2):\angle*(\i-1+.2):\radius+.2);
    }
  \end{tikzpicture}
}
\end{document}

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