绘制围道时，处理均匀地添加曲线上的箭头的最佳实践？

1 个回答

Sagittarius Rover 1小时前

这家伙很懒，什么也没写！

一个月前我在群里给出的方案是：

把全路径划分为多个子路径，预先保存中间顶点
“均匀”的添加箭头，只要保证子路径使用middle arrow=0.5的默认值即可
利用bengding,ext.arrows,ext.arrows-plus库提供的\arrow**{ext_Centered Stealth[bend]}}来准确的控制箭头的位置和正确的弯曲

\documentclass[tikz,border=3pt]{standalone}
\usepackage{lete-sans-math}
\usetikzlibrary{bending,arrows.meta,decorations.markings,ext.arrows,ext.arrows-plus}
\tikzset{
    middle arrow/.style={
        semithick,postaction={decorate},
        decoration={markings,mark=at position #1 with {\arrow**{ext_Centered Stealth[bend]}}}
    },middle arrow/.default={0.5}
}
\begin{document}
\begin{tikzpicture}[>=Stealth]
    \def\ang{17.5}\def\R{2.5}
    \pgfmathsetmacro{\r}{2*\R*sin(\ang/2)}
    \draw[thick,fill=gray!30]
        ({\R-\r},0) coordinate (A)
        arc (180:{90+\ang/2}:\r) coordinate (B)
        arc (\ang:{90-\ang}:\R) coordinate (C) 
        arc ({-\ang/2}:{-180+\ang/2}:\r) coordinate (D) 
        arc ({90+\ang}:{180-\ang}:\R) coordinate (E) 
        arc ({90-\ang/2}:0:\r) coordinate (F) 
        -- cycle;
    \path[middle arrow=0.4] (F) -- (A);
    \path[middle arrow] (A) arc (180:{90+\ang/2}:\r);
    \path[middle arrow] (B) arc (\ang:{90-\ang}:\R);
    \path[middle arrow=0.6] (C) arc ({-\ang/2}:{-180+\ang/2}:\r);
    \path[middle arrow] (D) arc ({90+\ang}:{180-\ang}:\R);
    \path[middle arrow] (E) arc ({90-\ang/2}:0:\r);
    \fill[radius=2pt] 
        (-\R,0) circle[] node[below] {$(-1,0)$}
        (+\R,0) circle[] node[below] {$(1,0)$}
        (0,+\R) circle[] node[above] {$(0,1)$}
        ;
    \draw[->] (-3,0) -- (3,0) node[right] {$\Re$};
    \draw[->] (0,-.5) node[above left] {$O$} -- (0,3.5) node[above] {$\Im$};
\end{tikzpicture}
\end{document}

不过这样其实似乎涉嫌重复绘制各段路径...

    \path[middle arrow] (A) arc (180:{90+\ang/2}:\r);
    \path[middle arrow] (B) arc (\ang:{90-\ang}:\R);
    \path[middle arrow=0.6] (C) arc ({-\ang/2}:{-180+\ang/2}:\r);
    \path[middle arrow] (D) arc ({90+\ang}:{180-\ang}:\R);
    \path[middle arrow] (E) arc ({90-\ang/2}:0:\r);