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渐变填充旋转的矩形

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渐变填充旋转的矩形

我有一个带有各种线条的图形，需要在一侧有渐变，如下图所示：

在此处输入图片描述

渐变的方向应该垂直于红色粗线，从线本身的深绿色开始，然后变得完全透明（现在是黑色）。

我如何修复渐变的位置（因为它的起点似乎在矩形之外）并使黑色完全透明（使用transparent似乎无法正常工作）？我知道我应该使用画布变换，但我无法正确执行此操作，因为我无法再使用我保存的坐标。

以下是用于生成图像的 MWE：

\documentclass{article}

\usepackage{tikz,pgf}
\usetikzlibrary{shapes,positioning,decorations.pathreplacing,intersections,calc}
\usepackage{color}
\definecolor{darkgreen}{rgb}{0,0.4,0}

%Length and angle for a segment in tikz
\makeatletter      
\newcommand{\getAngle}[2]{%
    \pgfmathanglebetweenpoints{\pgfpointanchor{#1}{center}}
                              {\pgfpointanchor{#2}{center}}
    \global\let\angle\pgfmathresult % we need a global macro 
}
\makeatother

\begin{document}
\begin{tikzpicture}
    \draw[->] (0,0) coordinate(origin)
    ++(-.2,0) -- ++(6.5,0) node[right]{$x_1$};
    \draw[->] (origin) ++(0,-.2) -- ++(0,5.5) node[above]{$x_2$};

    \draw[blue,shorten >= -.1cm, shorten <=-.1cm] (4,0) coordinate(v1_1) -- (4,5) coordinate(v1_2); % eq 1
    \path ($(v1_1)!-.1cm!(v1_2)$) coordinate(v1_1) ($(v1_2)!-.1cm!(v1_1)$) coordinate(v1_2);

    \draw[red,shorten >= -.1cm, shorten <=-.1cm] (0,5) coordinate(v2_1) -- (4,5) coordinate(v2_2);% eq 2
    \path ($(v2_1)!-.1cm!(v2_2)$) coordinate(v2_1) ($(v2_2)!-.1cm!(v2_1)$) coordinate(v2_2);

    \path[shorten >= 1.1cm, shorten <=-.1cm] (6,0) coordinate(v3_1) -- (2,6) coordinate(v3_2); % eq 3
    \path ($(v3_1)!-.1cm!(v3_2)$) coordinate(v3_1) ($(v3_2)!1.1cm!(v3_1)$) coordinate(v3_2);
    \getAngle{v3_1}{v3_2}
    \begin{scope}[rotate=\angle]
        \path (v3_2) -- ++(90:.7) coordinate(v3_3) (v3_1) -- ++(90:.7) coordinate(v3_4);
        % Here black should become transparent
        \shade[left color=black,right color=darkgreen,shading angle=\angle] (v3_3) rectangle (v3_1);
    \end{scope}
    \draw[red,ultra thick] (v3_1) -- (v3_2);
\end{tikzpicture}
\end{document}

答案1

我怀疑你正在寻找这个：

在此处输入图片描述

tikz我稍微重写了你的 mwe。我使用具有矩形形状和库的节点而不是矩形fadings。我还删除了所有未使用的坐标。

\documentclass[tikz, margin=3pt]{standalone}
\usetikzlibrary{calc,
                fadings,
                scopes}

\usepackage{xcolor}
\definecolor{darkgreen}{rgb}{0,0.4,0}

%Length and angle for a segment in tikz
\newcommand{\getAngle}[2]{%
    \pgfmathanglebetweenpoints{\pgfpointanchor{#1}{center}}
                              {\pgfpointanchor{#2}{center}}
    \global\let\angle\pgfmathresult % we need a global macro
}

\begin{document}
    \begin{tikzpicture}[
shorten <>/.style = {shorten <=#1, shorten >=#1}
                        ]
% axis
\draw[->] (-.2,0) -- ++ (6.5,0) node[right] {$x_1$};
\draw[->] (0,-.2) -- ++ (0,5.5) node[above] {$x_2$};
%
\draw[blue,shorten <>=-.1cm]    (4,0) -- (4,5); % eq 1, coordinate (v2_1) isn't used ...
\draw[red, shorten <>=-.1cm]    (0,5) -- (4,5); % eq 2, coordinate (v2_2) isn't used ...
% shape coordinates
\coordinate (v3_1) at (6,0);
\coordinate (v3_2) at (2,6);
% coordinates position correction
\path   ($(v3_1)!-.1cm!(v3_2)$) coordinate(v3_1)
        ($(v3_2)!1.1cm!(v3_1)$) coordinate(v3_2);
\getAngle{v3_1}{v3_2}
\scoped[transform canvas={rotate around={\angle:(v3_2)}}]
\path   let \p1 = ($(v3_1)-(v3_2)$),
            \n1 = {veclen(\y1,\x1)} in
        node [minimum width=\n1, minimum height=7mm,
              path fading=north,
              top color=black, bottom color=darkgreen,
              anchor=south west,
             ]  at (v3_2) {};
\draw[red,ultra thick] (v3_1) -- (v3_2);
    \end{tikzpicture}
\end{document}

注意：由于使用的transform canvas节点形状溢出了图像边界框。

答案2

这可能不是最佳选择，因为我不是 LaTeX 专家，但下面是我设法实现我想要的结果的方法（目标是对所有其他线条执行相同的操作，但对于垂直线和水平线来说更容易）：

\usetikzlibrary{fadings} % Put in preamble
\begin{tikzpicture}
    \draw (0,0) coordinate(origin);

    \path[shorten >= -.1cm, shorten <=-.1cm] (4,0) coordinate(v1_1) -- (4,5) coordinate(v1_2); % eq 1
    \path[shorten >= -.1cm, shorten <=-.1cm] (0,5) coordinate(v2_1) -- (4,5) coordinate(v2_2);% eq 2
    \path[shorten >= .5cm, shorten <=-.1cm] (6,0) coordinate(v3_1) -- (2,6) coordinate(v3_2); % eq 3

    \path (origin) ++(-.1,0) coordinate(v4_1) (v1_1) ++(.1,0) coordinate(v4_2);
    \fill[opacity=.5,gray,path fading=north] (v4_1) ++(0,.7) rectangle (v4_2);
    \path (origin) ++(0,-.1) coordinate(v5_1) (v2_1) ++(0,.1) coordinate(v5_2);
    \fill[opacity=.5,gray,path fading=east] (v5_1) ++(.7,0) rectangle (v5_2);

    \path ($(v1_1)!-.1cm!(v1_2)$) coordinate(v1_1) ($(v1_2)!-.1cm!(v1_1)$) coordinate(v1_2);
    \fill[opacity=.5,blue,path fading=west] (v1_2) ++(-.7,0) rectangle (v1_1);
    \draw[blue] (v1_1) -- (v1_2);

    \path ($(v2_1)!-.1cm!(v2_2)$) coordinate(v2_1) ($(v2_2)!-.1cm!(v2_1)$) coordinate(v2_2);
    \fill[opacity=.5,red,path fading=south] (v2_2) ++(0,-.7) rectangle (v2_1);
    \draw[red] (v2_1) -- (v2_2);

    \path ($(v3_1)!-.1cm!(v3_2)$) coordinate(v3_1) ($(v3_2)!.5cm!(v3_1)$) coordinate(v3_2);
    \getAngle{v3_1}{v3_2}
    %Rotated rectangle with gradient
    \begin{scope}[transform canvas={rotate around={\angle:(v3_1)}}]
        %Rotate the canvas to simplify gradient drawing
        % Inverse transform the global coordinate defined outside the rotate scope
        \path ([rotate around={-\angle:(v3_1)}] v3_2) coordinate(v3_2_in);
        \path (v3_2_in) -- ++(90:.7) coordinate(v3_3);
        % Fill a rectangle (with one side over the desired line) with the desired color, then fade the opposite side
        \fill[opacity=.5,darkgreen,path fading=north] (v3_3) rectangle (v3_1);
    \end{scope}
    \draw[darkgreen,name path=v3] (v3_1) -- (v3_2);

    \draw[->,name path=ax1] ++(-.2,0) -- ++(6.5,0) node[right]{$x_1$};
    \draw[->] (origin) ++(0,-.2) -- ++(0,5.5) node[above]{$x_2$};

    \path[name intersections={of=v3 and ax1,by=ax1_6}];
    \path (origin |- v2_2) node[left]{5} (origin -| v1_2) node[below]{4} (ax1_6) node[below]{6};
\end{tikzpicture}

在此处输入图片描述

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