\tikzdeclarecoordinatesystem 问题

Question

我使用 \tikzdeclarecoordinatesystem 创建了一个新的坐标系。当 scale=1 时编译为 true，否则编译为 false。为什么？你能帮助我吗？非常感谢。这是我的代码。

\documentclass[tikz, border=2mm]{standalone}
\usetikzlibrary{calc, intersections}
\makeatletter
\tikzset{/tikz/giaodiemthangtron cs/.cd,
    thang line initial coordinate/.store in=\tikz@giaodiem@onA,
    thang line final coordinate/.store in=\tikz@giaodiem@onB,
    tron line initial coordinate/.store in=\tikz@giaodiem@andC,
    tron line final coordinate/.store in=\tikz@giaodiem@andD,
    thang/.style args={#1--#2}{
        thang line initial coordinate=#1,
        thang line final coordinate=#2,
    },
    tron/.style args={#1--#2}{
        tron line initial coordinate=#1,
        tron line final coordinate=#2,
    },
    point/.store in=\point,
}
%
\tikzdeclarecoordinatesystem{giaodiemthangtron}{
    \tikzset{/tikz/giaodiemthangtron cs/.cd,#1}
    \pgfextra
        {
     \pgfextract@process\pgf@giaodiem@onA{\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@onA)\relax}%
        \pgfextract@process\pgf@giaodiem@onB{\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@onB)\relax}%
        %
        \pgfextract@process\pgf@giaodiem@andC{\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@andC)\relax}%
        \pgfextract@process\pgf@giaodiem@andD{\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@andD)\relax}%
        %rCD
        \pgfpointdiff{\pgf@giaodiem@andC}{\pgf@giaodiem@andD}%
        \pgfmathveclen@{\pgfmath@tonumber\pgf@x}{\pgfmath@tonumber\pgf@y}%
        \edef\pgf@bankinhCD{\pgfmathresult pt}
        % get intersections
        \pgfintersectionofpaths
        {%AB
        \pgfpathmoveto{\pgf@giaodiem@onA}
        \pgfpathlineto{\pgf@giaodiem@onB}
        %
        %\pgfgetpath\temppath
        %\pgfsetpath\temppath
        }
        {%circle
        \pgfpathcircle{\pgf@giaodiem@andC}{\pgf@bankinhCD}
        }
        \pgfextract@process\pgf@giao{
            \pgfpointintersectionsolution{\point}
            }
            \pgfcoordinate{giaodiem}{\pgf@giao}%
            }
        \tikz@scan@one@point\pgfutil@firstofone(giaodiem)%
    \relax
    }
\makeatother
\begin{document}
\begin{tikzpicture}[font=\small, scale=1]
\path(0,0)coordinate(A)(5,1)coordinate(B) (2,1)coordinate(C) (1,2)coordinate(D)
(giaodiemthangtron cs: thang=A--B, tron=C--D, point=1) coordinate(M)
(giaodiemthangtron cs: thang=A--B, tron=C--D, point=2) coordinate(N);
\draw[blue] let \p1 =($(C)-(D)$), \n1 = {veclen(\x1,\y1)} in (C) circle (\n1)
    (A)--(B);
\foreach \p/\pos in {A/-90, B/-90,C/45,D/135,M/-135, N/-45}
    \fill(\p)circle(1.pt) node[shift={(\pos:8pt)}]{$\p$};
\end{tikzpicture}
\end{document}

Answer 1

由 设置的（非恒等）变换矩阵scale=2被应用两次，一次由 施加\pgfintersectionofpaths，一次由 施加\pgfcoordinate。

\pgf@giao在下面的例子中，通过将逆变换矩阵（一次）应用于、\pgfintersectionofpaths和之间来解决问题\pgfcoordinate。

此外，最后两行\tikzdeclarecoordinatesystem{giaodiemthangtron}{...}被注释掉，因为

• \pgf@process{...}已经在全球范围内设置\pgf@x，\pgf@y并且
• 有点奇怪的是，一个坐标节点（此处giaodiem）会根据每个坐标（此处）自动（重新）定义一次又一次(giaodiemthangtron cs: <key-vals>)。

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

\makeatletter
\tikzset{/tikz/giaodiemthangtron cs/.cd,
    thang line initial coordinate/.store in=\tikz@giaodiem@onA,
    thang line final coordinate/.store in=\tikz@giaodiem@onB,
    tron line initial coordinate/.store in=\tikz@giaodiem@andC,
    tron line final coordinate/.store in=\tikz@giaodiem@andD,
    thang/.style args={#1--#2}{
        thang line initial coordinate=#1,
        thang line final coordinate=#2,
    },
    tron/.style args={#1--#2}{
        tron line initial coordinate=#1,
        tron line final coordinate=#2,
    },
    point/.store in=\point,
}

\tikzdeclarecoordinatesystem{giaodiemthangtron}{
  \tikzset{/tikz/giaodiemthangtron cs/.cd,#1}
  % extract A and B
  \pgfextract@process\pgf@giaodiem@onA
    {\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@onA)\relax}
  \pgfextract@process\pgf@giaodiem@onB
    {\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@onB)\relax}
  % extract C and D
  \pgfextract@process\pgf@giaodiem@andC
    {\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@andC)\relax}
  \pgfextract@process\pgf@giaodiem@andD
    {\tikz@scan@one@point\pgfutil@firstofone(\tikz@giaodiem@andD)\relax}
  % get radius rCD
  \pgfpointdiff{\pgf@giaodiem@andC}{\pgf@giaodiem@andD}
  \pgfmathveclen@{\pgfmath@tonumber\pgf@x}{\pgfmath@tonumber\pgf@y}
  \edef\pgf@bankinhCD{\pgfmathresult pt}
  % calculate intersections
  \pgfintersectionofpaths
    {%AB
      \pgfpathmoveto{\pgf@giaodiem@onA}
      \pgfpathlineto{\pgf@giaodiem@onB}
    }
    {%circle
      \pgfpathcircle{\pgf@giaodiem@andC}{\pgf@bankinhCD}
    }
  \pgfextract@process\pgf@giao
    {\pgfpointintersectionsolution{\point}}
  % cancel the applied transformation
  {%
    \pgftransforminvert
    \pgf@process{\pgfpointtransformed{\pgf@giao}}
  }
  % get intersections
  % \pgfcoordinate{giaodiem}{}
  % \tikz@scan@one@point\pgfutil@firstofone(giaodiem)\relax
}
\makeatother


\begin{document}
\begin{tikzpicture}[font=\small, scale=1]
  \draw[help lines] (0,0) grid (5,3);
  \path
    (0,0) coordinate (A)
    (5,1) coordinate (B)
    (2,1) coordinate (C)
    (1,2) coordinate (D)
    (giaodiemthangtron cs: thang=A--B, tron=C--D, point=1) coordinate (M)
    (giaodiemthangtron cs: thang=A--B, tron=C--D, point=2) coordinate (N);
  \draw[blue] 
    let \p1 =($(C)-(D)$), 
        \n1 = {veclen(\x1,\y1)}
    in
       (C) circle (\n1)
       (A)--(B);
  \foreach \p/\pos in {A/-90, B/-90,C/45,D/135,M/-135, N/-45}
    \fill (\p) circle (1.pt) node[shift={(\pos:8pt)}] {$\p$};
\end{tikzpicture}

\begin{tikzpicture}[font=\small, scale=2]
  \draw[help lines] (0,0) grid (5,3);
  \path
    (0,0) coordinate (A)
    (5,1) coordinate (B)
    (2,1) coordinate (C)
    (1,2) coordinate (D)
    (giaodiemthangtron cs: thang=A--B, tron=C--D, point=1) coordinate (M)
    (giaodiemthangtron cs: thang=A--B, tron=C--D, point=2) coordinate (N);
  \draw[blue] 
    let \p1 =($(C)-(D)$), 
        \n1 = {veclen(\x1,\y1)}
    in
       (C) circle (\n1)
       (A)--(B);
  \foreach \p/\pos in {A/-90, B/-90,C/45,D/135, M/-135, N/-45}
    \fill (\p) circle (1.pt) node[shift={(\pos:8pt)}] {$\p$};
\end{tikzpicture}

\begin{tikzpicture}[font=\small, xslant=.5]
  \draw[help lines] (0,0) grid (5,3);
  \path
    (0,0) coordinate (A)
    (5,1) coordinate (B)
    (2,1) coordinate (C)
    (1,2) coordinate (D)
    (giaodiemthangtron cs: thang=A--B, tron=C--D, point=1) coordinate (M)
    (giaodiemthangtron cs: thang=A--B, tron=C--D, point=2) coordinate (N);
  \draw[blue] 
    let \p1 =($(C)-(D)$), 
        \n1 = {veclen(\x1,\y1)}
    in
       (C) circle (\n1)
       (A)--(B);
  \foreach \p/\pos in {A/-90, B/-90,C/45,D/135, M/-135, N/-45}
    \fill (\p) circle (1.pt) node[shift={(\pos:8pt)}] {$\p$};
\end{tikzpicture}
\end{document}