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如何使用交点为我的简单实例绘制适当的圆弧

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如何使用交点为我的简单实例绘制适当的圆弧

我想通过圆弧连接点(s)和(t)线。你能帮我实现吗？！我试过这种方法，\draw (s) arc (t);但错了！简单地说，我需要正确的圆弧，但不能延伸到与定义的路径重合的第一和第二条线，称为血腥圆弧。

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\usepackage{rotating}
\usetikzlibrary{calc,intersections}
\tikzset{
HH/.style={thick}}

\def\scalefactor{2}

\begin{document}

\begin{tikzpicture}[scale=\scalefactor]
\draw[help lines] (0,0) grid (7,5);
\path[red,thick,name path= bloody arc]
  ([shift={(0:3)}]2,1) arc (0:90:3);
\draw[HH,name path=first line] (2,1)--(5,2);
\draw[HH,name path=second line] (2,1)--(6,5);

\path[red,name intersections={of=first line and bloody arc, by=s}];
\path[red,name intersections={of=second line and bloody arc, by=t}];
\draw[HH] (s)--(t);
% \draw (s) arc (t);
\end{tikzpicture}
\end{document}

答案1

使用 tkz-euclide

这个问题的可接受答案在 TikZ 中绘制由三点定义的圆弧的最简单方法是什么？演示如何使用中心和两个点绘制圆弧，这正是您要实现的。但是它使用的tkz-euclide包目前仅以法语记录，尽管据说将来会发布英语手册。

\tkzDrawArc这是修改后的代码，使用from绘制圆弧tkz-euclide。另外，请注意，圆弧与原始线条略有重叠，因为交点位于线条的中心。

第一弧的特写

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usetikzlibrary{calc,intersections}
\tikzset{HH/.style={thick}}

\def\scalefactor{2}

\begin{document}

\begin{tikzpicture}[scale=\scalefactor]
\draw[help lines] (0,0) grid (7,5);
\path[draw, red,thick,name path= bloody arc]
  ([shift={(0:3)}]2,1) arc (0:90:3);
\draw[HH,name path=first line] (2,1)--(5,2);
\draw[HH,name path=second line] (2,1)--(6,5);

\path[red,name intersections={of=first line and bloody arc, by=s}];
\path[red,name intersections={of=second line and bloody arc, by=t}];
\draw[HH] (s)--(t);
\path (2,1) coordinate (center); % Where the two lines intersect
\tkzDrawArc[HH,color=black](center,s)(t) % Draw arc counterclockwise from (s) to (t).
\end{tikzpicture}
\end{document}

使用剪辑

或者，如果您不想使用tkz-euclide，您可以使用clip路径将弧线限制在两条线之间的区域。遗憾的是，您无法在剪切路径上指定任何其他选项，因此您必须两次提供线条的坐标：一次用于剪切，一次用于实际绘制两条线。

第二弧的特写

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}
\tikzset{HH/.style={thick}}

\def\scalefactor{2}

\begin{document}

\begin{tikzpicture}[scale=\scalefactor]
    \draw[help lines] (0,0) grid (7,5);
    \begin{scope}
        \clip (6,5) -- (2,1) -- (5,2); % Only for clipping
        \path[red,draw,thick,name path= bloody arc]
            ([shift={(0:3)}]2,1) arc (0:90:3);
    \end{scope}
    \draw[HH] (6,5) -- (2,1)--(5,2); % Actually drawing the lines
\end{tikzpicture}
\end{document}

答案2

您还可以使用该calc库：

\documentclass[tikz, border=5]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}
\tikzset{HH/.style={thick}}
\def\scalefactor{2}
\begin{document}
\begin{tikzpicture}[scale=\scalefactor]
\draw[help lines] (1,0) grid (7,5);
\draw[red,thick, name path=bloody arc]  ([shift={(0:3)}]2,1) arc (0:90:3);
\draw[HH, name path=first line] (2,1)--(5,2);
\draw[HH, name path=second line] (2,1)--(6,5);
\path[red,name intersections={of=first line and bloody arc, by=s}];
\path[red,name intersections={of=second line and bloody arc, by=t}];
\coordinate (u) at (2,1);
\draw [green] let \p1=($(s)-(u)$),\p2=($(t)-(u)$),\n1={atan2(\y1,\x1)},
\n2={atan2(\x2,\y2)} in (s) arc (\n1:\n2:3);
\end{tikzpicture}
\end{document}

在此处输入图片描述

但是这里有一个circular arc around“路径”，我认为它能正确地完成任务，但为了提高速度使用了一些内部命令。

\documentclass[tikz, border=5]{standalone}
\usetikzlibrary{calc,intersections}
\makeatletter
\tikzset{%
  circular arc around/.style={
    to path={%
      \pgfextra{%
        \tikz@scan@one@point\pgf@process#1%
        \pgf@xa=-\pgf@x\pgf@ya=-\pgf@y
        \pgf@xb=-\pgf@x\pgf@yb=-\pgf@y
        \tikz@scan@one@point\pgf@process(\tikztostart)%
        \advance\pgf@xa by\pgf@x%
        \advance\pgf@ya by\pgf@y%
         \tikz@scan@one@point\pgf@process(\tikztotarget)%
       \advance\pgf@xb by\pgf@x%
       \advance\pgf@yb by\pgf@y%
       \pgfmathveclen{\pgf@xa}{\pgf@ya}\let\tikz@radius=\pgfmathresult%
       \pgfmathatantwo{\the\pgf@ya}{\the\pgf@xa}%
       \pgfmathMod@{\pgfmathresult}{360}%
       \let\tikz@angle@a=\pgfmathresult%
       \pgfmathatantwo{\the\pgf@yb}{\the\pgf@xb}%
       \pgfmathMod@{\pgfmathresult}{360}%
       \let\tikz@angle@b=\pgfmathresult%    
       \edef\tikz@to@arc@path{ arc(\tikz@angle@a:\tikz@angle@b:\tikz@radius pt) }\show\tikz@to@arc@path
      } \tikz@to@arc@path
    }  
  }
}
\begin{document}
\begin{tikzpicture}[scale=1/2]

\draw [help lines] (0,0) grid (8,8);
\coordinate (o) at (4,4);
\tikzset{shift=(o)}


\foreach \i in {0.5,1,...,3.5}
  \draw [very thick] (o) circle [radius=\i];

\foreach \i [evaluate={\a=rnd*45+\i;\b=\a+rnd*20+30;}] in {0,90,180,270}{
\coordinate  (p) at (\a:4);
\coordinate  (q) at (\b:4);

\fill [fill opacity=0.25] (o) -- (p) to [circular arc around=(o)] (q) -- cycle;

\foreach \c [count=\r from 1] in {red, yellow, pink, green, orange, purple, blue}{
 \coordinate (s) at ($(o)!\r*0.5cm!(p)$);
 \coordinate (t) at ($(o)!\r*0.5cm!(q)$);
 \draw [very thick,densely dotted, \c] (s) to [circular arc around=(o)] (t); 
}
}
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案3

如果您已经知道之间的交点bloody arc，first/second line您可以重复bloody arc，但clip使用适当的矩形。

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\usepackage{rotating}
\usetikzlibrary{calc,intersections}
\tikzset{
HH/.style={thick}}

\def\scalefactor{2}

\begin{document}

\begin{tikzpicture}[scale=\scalefactor]
\draw[help lines] (0,0) grid (7,5);
\draw[red,thick,name path= bloody arc]
  ([shift={(0:3)}]2,1) arc (0:90:3);
\draw[HH,name path=first line] (2,1)--(5,2);
\draw[HH,name path=second line] (2,1)--(6,5);

\path[red,name intersections={of=first line and bloody arc, by=s}];
\path[red,name intersections={of=second line and bloody arc, by=t}];
\draw[HH] (s)--(t);
\begin{scope}
\clip (s) rectangle (t);
\draw[green,thick,name path= bloody arc]
  ([shift={(0:3)}]2,1) arc (0:90:3);
\end{scope}
% \draw (s) arc (t);
\end{tikzpicture}
\end{document}

在此处输入图片描述

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