当 x 和 y 尺度不同时，如何使用“倾斜”？

Question

在下面的例子中，我想将文本对齐到切线的右侧，由于比例不同，倾斜度是错误的。我尝试使用“变换形状”，但文本变得难以阅读（太小），我当然可以添加比例（根据 x 和 y 而不同），但这会变成一个复杂的代码。

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

\begin{document}

\begin{tikzpicture}[xscale=0.15,yscale=0.4]

\draw [thin,xstep=5,ystep=1,blue,dashed] (0,0) grid (50,12);
\draw[-latex,thick] (0,0) -- (51,0) node[below]{$t$ en \si{\second}};

\draw[-latex, thick] (0,0) -- (0,13) node[above right]{$v$ en \si{\mm\per\second}};

\draw[domain=0:50] plot (\x,{8*(1-exp(-\x/10))});
\draw[dashed,red,thick] (0,8)-- (50,8)node[above left]{$s_\infty=G\cdot U_0=\SI{8}{\mm/\second}$};
\draw[dashed,red,thick] (0,0) coordinate(O)-- node[sloped,above, pos=0.8]{Tangente en (0,0)}(10,8);
\draw[thick, red] (10,0) -- (10, {8*(1-exp(-1))}) coordinate(T0) -- (T0-|O)node[above]{$\num{0.63}\cdot G\cdot U_O$};
\node[below] at (10,0){10};
\node[left] at (0,10){10};
\end{tikzpicture}

\end{document}

Answer 1

一种方法是使用decoration：

\draw[dashed,red,thick,
    postaction={decorate},
    decoration={
        markings,
        mark=at position 1.0 with {
            \node [blue, transform shape, anchor=base, shift={(10pt,2pt)}] 
                {Tangente en (0,0)};
        },
    },
] (0,0) coordinate(O) -- (10,8);

代码：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary {calc}
\usepackage{siunitx}
\usetikzlibrary{decorations.markings}

\begin{document}
\begin{tikzpicture}[xscale=0.15,yscale=0.4]

\draw [thin,xstep=5,ystep=1,gray,dashed] (0,0) grid (50,12);
\draw[-latex,thick] (0,0) -- (51,0) node[below]{$t$ en \si{\second}};

\draw[-latex, thick] (0,0) -- (0,13) node[above right]{$v$ en \si{\mm\per\second}};

\draw[domain=0:50] plot (\x,{8*(1-exp(-\x/10))});
\draw[dashed,red,thick] (0,8)-- (50,8)node[above left]{$s_\infty=G\cdot U_0=\SI{8}{\mm/\second}$};
\draw[dashed,red,thick,
    postaction={decorate},
    decoration={
        markings,
        mark=at position 1.0 with {
            \node [blue, transform shape, anchor=base, shift={(10pt,2pt)}] 
                {Tangente en (0,0)};
        },
    },
] (0,0) coordinate(O) -- (10,8);
\draw[thick, red] (10,0) -- (10, {8*(1-exp(-1))}) coordinate(T0) -- (T0-|O)node[above, anchor=east]{$\num{0.63}\cdot G\cdot U_O$};
\node[below] at (10,0){10};
\node[left] at (0,10){10};
\end{tikzpicture}%
\end{document}

Answer 2

另一种方法是修复\pgftransformlineattime。

人们也应该修正\pgftransformarcaxesattime并\pgftransformcurveattime遵循同样的想法。

\documentclass{article}
\usepackage{tikz}

\makeatletter

\def\pgftransformlineattime#1#2#3{%
  \pgfgettransformentries\aa\ab\ba\bb\notimportantx\notimportanty%%% ⬅️⬅️⬅️ new
  \pgf@process{#2}%
  \pgf@xb=\pgf@x% xb/yb = start point
  \pgf@yb=\pgf@y%
  \pgf@process{#3}%
  \pgf@xc=\pgf@x% xc/yc = end point
  \pgf@yc=\pgf@y%
  \pgftransformshift{\pgfpointlineattime{#1}{\pgfqpoint{\pgf@xb}{\pgf@yb}}{\pgfqpoint{\pgf@xc}{\pgf@yc}}}%
  \ifpgfresetnontranslationattime%
    \pgftransformresetnontranslations%
  \fi%
  \ifpgfslopedattime%
    \advance\pgf@xc by-\pgf@xb%
    \advance\pgf@yc by-\pgf@yb%
    % Now that we get the tangent vector without transformation
    % It suffices to apply the non-shift part of the transformation
    {%%% ⬇️⬇️⬇️ new
        \pgfsettransformentries\aa\ab\ba\bb{0pt}{0pt}
        \pgf@pos@transform{\pgf@xc}{\pgf@yc}
        \global\pgf@xc\pgf@xc
        \global\pgf@yc\pgf@yc
    }%%% ⬆️⬆️⬆️ new
    % OK, now xc and yc are correctly tangent to the line
    % Continue the usual routine
    \ifpgfallowupsidedownattime%
    \else%
      \ifdim\pgf@xc<0pt%
        \pgf@xc=-\pgf@xc%
        \pgf@yc=-\pgf@yc%
      \fi%
    \fi%
    \pgf@x=\pgf@xc%
    \pgf@y=\pgf@yc%
    \pgfpointnormalised{}% x/y = normalised vector
    \pgf@ya=-\pgf@y%
    \pgftransformcm%
    {\pgf@sys@tonumber{\pgf@x}}{\pgf@sys@tonumber{\pgf@y}}%
    {\pgf@sys@tonumber{\pgf@ya}}{\pgf@sys@tonumber{\pgf@x}}{\pgfpointorigin}%
  \fi%
}
\makeatother

\begin{document}
\begin{tikzpicture}[xscale=2, yscale=3, xshift=40, yshift=50]
    \draw (0, 0) -- node[sloped, above] {sloped correctly?} (2, 2);
\end{tikzpicture}%
\end{document}