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tikz中两点之间的同心圆段

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tikz中两点之间的同心圆段

我想在两个节点 A 和 B 之间画一条同心圆线段，线段的中心在 A 处。这些线段的高度应相同（如果可能）。到目前为止，我的代码如下：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}

%%% extract coordinate
\newdimen\XCoord
\newdimen\YCoord
\newcommand*{\ExtractCoordinate}[1]{\path (#1); \pgfgetlastxy{\XCoord}{\YCoord};}%

\begin{document}

\begin{tikzpicture}
    \def \h{30};
    \def \lw {2};
    \node[circle, fill] at (0,0) (A) {};
    \node[circle, fill] at (8,0) (B) {};

    % circle segments
    \foreach \x in {1,3,...,9}{
        \coordinate (x) at ($(A)!\x/10!(B)$);

        %%% find theta and radius
        \ExtractCoordinate{x};
        \def \ang {atan(\h/\XCoord)};
        \def \r {\XCoord};

        %%% draw arcs
        \draw[line width = \lw, red] (x) arc(0:\ang:\r);
        \draw[line width = \lw,red] (x) arc(0:-\ang:\r);
    };
\end{tikzpicture}

\end{document}

这只是一个特例。我怎样才能让它更通用，即对 A 和 B 的位置没有任何限制？

同心圆段

答案1

这是一个数学（仅限高中）解决方案：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}

%%% extract coordinate
\newdimen\XCoord
\newdimen\YCoord
\newcommand*{\ExtractCoordinate}[1]{\path (#1); \pgfgetlastxy{\XCoord}{\YCoord};}%


%#1=Number of waves
%#2 ans #3 point A and B
%#4 Angle of first wave
\def\NumSignalsFromToAngle#1#2#3#4{%
\def\NumberSignals{#1}
\ExtractCoordinate{#2}
\xdef\Xa{\XCoord}
\xdef\Ya{\YCoord}
\ExtractCoordinate{#3}
\xdef\Xb{\XCoord}
\xdef\Yb{\YCoord}
\pgfmathsetmacro\dist{10*sqrt((\Xb/10-\Xa/10)^2+(\Yb/10-\Ya/10)^2)}
\pgfmathsetmacro\step{\dist/\NumberSignals}
\pgfmathsetmacro\AngleFromAToB{\ifdim\Xb>\Xa atan((\Yb/10-\Ya/10)/(\Xb/10 -\Xa/10))\else \ifdim \Xb<\Xa 180+atan((\Yb/10-\Ya/10)/(\Xb/10 -\Xa/10))\else\ifdim\Ya>\Yb -90\else90\fi\fi\fi}
\foreach \i in {1,...,\NumberSignals}
{%
\coordinate(Point) at ($(#2)+({(\step pt)*(2*\i-1)/2*cos(\AngleFromAToB) },{(\step pt)*(2*\i-1)/2*sin(\AngleFromAToB)})$);
\pgfmathsetmacro\r{\step/2*(2*\i-1)}
%\l_1=\l_i => \angle_i=\angle_1/(2i-1)
\pgfmathsetmacro\angle{#4/(2*\i-1)}
\draw  (Point) arc (\AngleFromAToB:{\AngleFromAToB+\angle/2}:\r pt);
\draw  (Point) arc (\AngleFromAToB:{\AngleFromAToB-\angle/2}:\r pt);
}
}


\begin{document}

\begin{tikzpicture}
\draw[fill=green] (0,0) coordinate(Source) circle(0.2 cm);
\draw[fill=red] (6,2) coordinate(Target) circle(0.2 cm);
\NumSignalsFromToAngle{6}{Source}{Target}{90}

\draw[fill=green] (0,-3) coordinate(Source2) circle(0.15 cm);
\draw[fill=red] (4,-1) coordinate(Target2) circle(0.15cm);
\NumSignalsFromToAngle{5}{Source2}{Target2}{180}

\draw[fill=green] (10,-7) coordinate(Source3) circle(0.15 cm);
\draw[fill=red] (8,2) coordinate(Target3) circle(0.2 cm);
\NumSignalsFromToAngle{9}{Source3}{Target3}{60}

\draw[fill=green] (0,-12) coordinate(Source5) circle(1.31 mm);
\draw[fill=red] (1,-11) coordinate(Target5) circle(1.31mm);
\NumSignalsFromToAngle{3}{Source5}{Target5}{270}


\draw[fill=green] (5,-12) coordinate(Source6) circle(1.5 mm);
\draw[fill=red] (5,-9) coordinate(Target6) circle(1.5mm);
\NumSignalsFromToAngle{3}{Source6}{Target6}{45}

\draw[fill=green] (7,-8) coordinate(Source7) circle(1.5 mm);
\draw[fill=red] (7,-11) coordinate(Target7) circle(1.5 mm);
\NumSignalsFromToAngle{4}{Source7}{Target7}{120}

\end{tikzpicture}

\end{document}

输出：

在此处输入图片描述

答案2

这里有一个提议，就是绘制完整的圆圈，但将它们夹在从 A 到 B 的盒子上（高度为 1 厘米，可以调整）。

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\makeatletter % from https://tex.stackexchange.com/a/412901/121799
\newcommand{\Distance}[3]{% % from https://tex.stackexchange.com/q/56353/121799
\tikz@scan@one@point\pgfutil@firstofone($#1-#2$)\relax  
\pgfmathsetmacro{#3}{veclen(\the\pgf@x,\the\pgf@y)/28.45274}
}
\makeatother 
\begin{document}

\begin{tikzpicture}
    \node[circle, fill] at (0,0) (A) {};
    \node[circle, fill] at (8,2) (B) {};
    \path (A) -- (B) \foreach \X [count=\n]in {0.1,0.3,0.5,0.7,0.9} {coordinate[pos=\X]
    (X\n)};
    \begin{scope}
    \clip ($ (A)!0.5cm!90:(B) $) -- ($ (B)!-0.5cm!90:(A) $)
    --($ (B)!0.5cm!90:(A) $) -- ($ (A)!-0.5cm!90:(B) $) -- cycle;
    \foreach \n in {1,...,5}
    {\Distance{(A)}{(X\n)}{\rad}
    \draw[red,thick] (A) circle (\rad);
    }
    \end{scope}
\end{tikzpicture}

\end{document}

在此处输入图片描述

答案3

类似于 marmot 的解决方案：在区域内绘制波浪clipped。但波浪是经过expanding waves装饰绘制的。

\documentclass[tikz,border=2mm]{standalone} 
\usetikzlibrary{decorations.pathreplacing}

\begin{document}
\begin{tikzpicture}[
    wave/.style={
        decorate,
        decoration={
            expanding waves,
            angle=10,
            segment length=2mm}}]

    \node [circle, fill] at (0,0) (A) {};
    \node [circle, fill] at (8,0) (B) {};

     \begin{scope}
     \clip ([yshift=-2mm]A.south east) rectangle ([yshift=2mm]B.north east);
    \draw[wave] (A) -- (B);
    \end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

在此代码中angle定义了线条如何围绕中心生长以及segment length线条之间的距离。

或者，您可以将路径应用clipping到您的解决方案中：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}

%%% extract coordinate
\newdimen\XCoord
\newdimen\YCoord
\newcommand*{\ExtractCoordinate}[1]{\path (#1); \pgfgetlastxy{\XCoord}{\YCoord};}%

\begin{document}

\begin{tikzpicture}
    \def \h{30};
    \def \lw {2};
    \node[circle, fill] at (0,0) (A) {};
    \node[circle, fill] at (8,0) (B) {};

    %define clipping area
    \begin{scope}
    \clip ([yshift=-2mm]A.south east) rectangle ([yshift=2mm]B.north west);
    % circle segments
    \foreach \x in {1,3,...,9}{
        \coordinate (x) at ($(A)!\x/10!(B)$);

        %%% find theta and radius
        \ExtractCoordinate{x};
        \def \ang {atan(\h/\XCoord)};
        \def \r {\XCoord};

        %%% draw arcs
        \draw[line width = \lw, red] (x) arc(0:\ang:\r);
        \draw[line width = \lw,red] (x) arc(0:-\ang:\r);
    };
    \end{scope}
\end{tikzpicture}

\end{document}

在此处输入图片描述

答案4

不知道这是否是你真正想要的，但我会选择

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,mindmap,snakes}
\begin{document}
\begin{tikzpicture}
\node[circle, fill] at (0,0) (A) {};
\node[circle, fill] at (8,0) (B) {};
\path (A) edge [snake=expanding waves,segment length=5mm,segment angle=10,draw] (B);
\end{tikzpicture}
\end{document}

导致

在此处输入图片描述

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