使用桥梁和淡入淡出跨越贝塞尔曲线

Question 1

这座桥的建造并不太难：

绘制一条曲线A，并构造另一条曲线的路径，B同时保存它。
计算和inter的交点。AB
在其周围画一个圆圈inter。
绘制时使用反向夹子，B以便周围留有间隙inter。
计算圆和的交点B。
利用这些交点画出圆弧，同时确保它是一座桥。

代码：

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{intersections,angles,calc}
\begin{document}
\begin{tikzpicture}
 \newcommand{\bridgeRadius}{.5ex}
 \begin{scope}
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \path[name path=B,save path=\pathB] (0,0) to[bend left] (2,2);
  \path[name intersections={of=A and B,by=inter}];
  \path[name path=C] (inter) circle[radius=\bridgeRadius];
  \path[name intersections={of=C and B,name=i}];
  \begin{scope}
   \clip (inter) circle[radius=\bridgeRadius] (current bounding box.south west) |-
    (current bounding box.north east) |- cycle;
   \draw[use path=\pathB,-latex]; 
  \end{scope}
  \path let \p1=($(i-1)-(i-2)$),\n1={scalar(int(sign(\x1)))} in
   \ifnum\n1=-1
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-2--inter--i-1}
   \else 
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-1--inter--i-2}
   \fi; 
 \end{scope} 
 \begin{scope}[xshift=3cm]
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \path[name path=B,save path=\pathB] (2,2) to[bend left] (0,0);
  \path[name intersections={of=A and B,by=inter}];
  \path[name path=C] (inter) circle[radius=\bridgeRadius];
  \path[name intersections={of=C and B,name=i}];
  \begin{scope}
   \clip (inter) circle[radius=\bridgeRadius] (current bounding box.south west) |-
    (current bounding box.north east) |- cycle;
   \draw[use path=\pathB,-latex]; 
  \end{scope}
  \path let \p1=($(i-1)-(i-2)$),\n1={scalar(int(sign(\x1)))} in
   \ifnum\n1=-1
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-2--inter--i-1}
   \else 
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-1--inter--i-2}
   \fi; 
 \end{scope} 
\end{tikzpicture}
\end{document}

这是一种做同样事情的风格。（我并不是说这种用法特别直观，但你可以根据自己的需要进行调整。）它会进行一些健全性检查，例如验证路径是否真的相交。

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{intersections,angles,calc}
\makeatletter
\tikzset{bridge/.code args={#1 over #2}{%
 \path[name path=tmp@bridge@path@B] #1;
 \tikzset{name intersections={of=#2 and tmp@bridge@path@B,
    by=tmp@i@0,total=\tmp@i@total}}
 \ifnum\tmp@i@total=0
  \typeout{These paths do not intersect. No bridge, sorry.}%
 \else
  \path[name path=tmp@bridge@path@C] (tmp@i@0) 
    circle[radius=\pgfkeysvalueof{/tikz/bridge radius}]; 
  \tikzset{name intersections={of=tmp@bridge@path@B and tmp@bridge@path@C,
    by={tmp@i@1,tmp@i@2},total=\tmp@i@total}}
  \ifnum\tmp@i@total=2
   \begin{scope}
    \clip (tmp@i@0) circle[radius=\pgfkeysvalueof{/tikz/bridge radius}]
     (current bounding box.south west) |-
     (current bounding box.north east) |- cycle;
    \draw[bridge-style] #1; 
   \end{scope}  
   \path let \p1=($(tmp@i@1)-(tmp@i@2)$),\n1={scalar(int(sign(\x1)))} in
    \ifnum\n1=-1
     pic [draw,line cap=round, angle radius=\pgfkeysvalueof{/tikz/bridge radius},
        bridge-arc] {angle=tmp@i@2--tmp@i@0--tmp@i@1}
    \else 
     pic [draw,line cap=round, angle radius=\pgfkeysvalueof{/tikz/bridge radius},
        bridge-arc] {angle=tmp@i@1--tmp@i@0--tmp@i@2}
    \fi; 
  \else
   \typeout{Given the bridge radius \pgfkeysvalueof{/tikz/bridge radius}, 
    the path is not suited to construct a nice bridge. No bridge, sorry.}%
  \fi   
 \fi 
},bridge radius/.initial=.5ex,
    bridge style/.code={\tikzset{bridge-style/.style={#1}}},
    bridge arc/.code={\tikzset{bridge-arc/.style={#1}}},
    bridge style={},bridge arc={}}
\makeatother
\begin{document}
\begin{tikzpicture}
 \begin{scope}
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \tikzset{bridge={(0,0) to[bend left] (2,2)} over A}
 \end{scope} 
 \begin{scope}[xshift=3cm]
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \tikzset{bridge style={-latex},bridge={(2,2) to[bend left] (0,0)} over A}
 \end{scope} 
\end{tikzpicture}
\end{document}

Answer

这座桥的建造并不太难：

绘制一条曲线A，并构造另一条曲线的路径，B同时保存它。
计算和inter的交点。AB
在其周围画一个圆圈inter。
绘制时使用反向夹子，B以便周围留有间隙inter。
计算圆和的交点B。
利用这些交点画出圆弧，同时确保它是一座桥。

代码：

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{intersections,angles,calc}
\begin{document}
\begin{tikzpicture}
 \newcommand{\bridgeRadius}{.5ex}
 \begin{scope}
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \path[name path=B,save path=\pathB] (0,0) to[bend left] (2,2);
  \path[name intersections={of=A and B,by=inter}];
  \path[name path=C] (inter) circle[radius=\bridgeRadius];
  \path[name intersections={of=C and B,name=i}];
  \begin{scope}
   \clip (inter) circle[radius=\bridgeRadius] (current bounding box.south west) |-
    (current bounding box.north east) |- cycle;
   \draw[use path=\pathB,-latex]; 
  \end{scope}
  \path let \p1=($(i-1)-(i-2)$),\n1={scalar(int(sign(\x1)))} in
   \ifnum\n1=-1
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-2--inter--i-1}
   \else 
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-1--inter--i-2}
   \fi; 
 \end{scope} 
 \begin{scope}[xshift=3cm]
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \path[name path=B,save path=\pathB] (2,2) to[bend left] (0,0);
  \path[name intersections={of=A and B,by=inter}];
  \path[name path=C] (inter) circle[radius=\bridgeRadius];
  \path[name intersections={of=C and B,name=i}];
  \begin{scope}
   \clip (inter) circle[radius=\bridgeRadius] (current bounding box.south west) |-
    (current bounding box.north east) |- cycle;
   \draw[use path=\pathB,-latex]; 
  \end{scope}
  \path let \p1=($(i-1)-(i-2)$),\n1={scalar(int(sign(\x1)))} in
   \ifnum\n1=-1
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-2--inter--i-1}
   \else 
    pic [draw,line cap=round, angle radius=\bridgeRadius] {angle=i-1--inter--i-2}
   \fi; 
 \end{scope} 
\end{tikzpicture}
\end{document}

这是一种做同样事情的风格。（我并不是说这种用法特别直观，但你可以根据自己的需要进行调整。）它会进行一些健全性检查，例如验证路径是否真的相交。

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{intersections,angles,calc}
\makeatletter
\tikzset{bridge/.code args={#1 over #2}{%
 \path[name path=tmp@bridge@path@B] #1;
 \tikzset{name intersections={of=#2 and tmp@bridge@path@B,
    by=tmp@i@0,total=\tmp@i@total}}
 \ifnum\tmp@i@total=0
  \typeout{These paths do not intersect. No bridge, sorry.}%
 \else
  \path[name path=tmp@bridge@path@C] (tmp@i@0) 
    circle[radius=\pgfkeysvalueof{/tikz/bridge radius}]; 
  \tikzset{name intersections={of=tmp@bridge@path@B and tmp@bridge@path@C,
    by={tmp@i@1,tmp@i@2},total=\tmp@i@total}}
  \ifnum\tmp@i@total=2
   \begin{scope}
    \clip (tmp@i@0) circle[radius=\pgfkeysvalueof{/tikz/bridge radius}]
     (current bounding box.south west) |-
     (current bounding box.north east) |- cycle;
    \draw[bridge-style] #1; 
   \end{scope}  
   \path let \p1=($(tmp@i@1)-(tmp@i@2)$),\n1={scalar(int(sign(\x1)))} in
    \ifnum\n1=-1
     pic [draw,line cap=round, angle radius=\pgfkeysvalueof{/tikz/bridge radius},
        bridge-arc] {angle=tmp@i@2--tmp@i@0--tmp@i@1}
    \else 
     pic [draw,line cap=round, angle radius=\pgfkeysvalueof{/tikz/bridge radius},
        bridge-arc] {angle=tmp@i@1--tmp@i@0--tmp@i@2}
    \fi; 
  \else
   \typeout{Given the bridge radius \pgfkeysvalueof{/tikz/bridge radius}, 
    the path is not suited to construct a nice bridge. No bridge, sorry.}%
  \fi   
 \fi 
},bridge radius/.initial=.5ex,
    bridge style/.code={\tikzset{bridge-style/.style={#1}}},
    bridge arc/.code={\tikzset{bridge-arc/.style={#1}}},
    bridge style={},bridge arc={}}
\makeatother
\begin{document}
\begin{tikzpicture}
 \begin{scope}
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \tikzset{bridge={(0,0) to[bend left] (2,2)} over A}
 \end{scope} 
 \begin{scope}[xshift=3cm]
  \draw[name path=A] (1,0) to[bend right] (0,2);
  \tikzset{bridge style={-latex},bridge={(2,2) to[bend left] (0,0)} over A}
 \end{scope} 
\end{tikzpicture}
\end{document}

Question 2

我不知道路径衰减，但这里有一个桥梁概念证明，它是相对自动化的。它使用最新版本（截至 2021 年 2 月）的spath3该包现在位于 TeXLive 服务器上。

代码如下：

\documentclass{article}
%\url{https://tex.stackexchange.com/q/581540/86}

\usepackage{tikz}
\usetikzlibrary{spath3,intersections}

\begin{document}
\begin{tikzpicture}
\path[spath/save=over] (0,0) to[out=60,in=150] (3,3);
\path[spath/save=under] (0,3) to[out=30,in=120] (3,0);
\path[spath/save=arc] (0,0) arc[radius=1cm, start angle=180, delta angle=-180];
\tikzset{
  spath/split at intersections with={over}{under},
  spath/insert gaps after components={over}{8pt}{1},
  spath/join components with={over}{arc},
  spath/split at intersections with={under}{over},
  spath/insert gaps after components={under}{4pt}{1},
}
\draw[spath/restore=over];
\draw[spath/restore=under];
\end{tikzpicture}
\end{document}

结果如下：

这里发生的情况如下：

定义了三条路径：两条主路径和一条弧线（即桥梁）
从过路径的角度找到交点，并在该点处分割过路径。
路径在该分割点的两侧缩短（沿其路径）。
将弧形路径拼接到上方路径中以填补这个间隙（并将路径连接在一起以形成单一的连续路径）。
现在，下路径在与此（新的）上路径相交的地方被分割，并且在此分割点处插入了一个较小的间隙。（如果我知道路径淡入淡出，这大概可以用来淡化此点处的路径，而不是分割并插入间隙。）
一旦完成所有这些，路径就会被呈现。

虽然此图中只有一个交点，但相同的代码可以处理多个交点。需要额外做的是确保所有弧都是“桥梁”而不是“隧道”。

Answer