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tikz，我如何旋转一个角度？

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tikz，我如何旋转一个角度？

（编辑两次）我已经向自己澄清了我的问题，请允许我重申我的帖子。

我希望将一个角度 Angle(ABC)（实际上是一个向量）绕另一个角度 \alpha 旋转。要旋转的角度要么以其度数 \beta 表示，要么以三个点表示。前一种情况确实很容易。对于后一种情况，我似乎需要包来tkz-euclide计算角度的度数。剩下的问题可能是：我可以不用吗tkz-euclide？

（请参见下文了解我仍然存在的一些小的美容问题）。

欢迎您提出任何进一步的建议。

%\documentclass[convert={density=300,size=1080x800,outext=.png}]{standalone}
\documentclass[12pt,a4paper]{scrartcl}  %%KOMA class
\setkomafont{sectioning}{\rmfamily\bfseries\boldmath}  %%
\usepackage{amsmath,tkz-euclide}
\usepackage{tikz}
\usetkzobj{all}
\usepackage{calculator}
\usepackage{pgfplots}
\usetikzlibrary{rulercompass}
\usetikzlibrary{intersections,quotes,angles}
\usetikzlibrary{calc}
\newcommand\round[2][2]{\pgfmathprintnumber[precision=#1]{#2}}

\begin{document}
    \pgfmathparse{30}
    \pgfmathsetmacro\anglealpha{\pgfmathresult}

    \pgfmathparse{170}
    \pgfmathsetmacro\anglea{\pgfmathresult}

    \pgfmathparse{150}
    \pgfmathsetmacro\angleb{\pgfmathresult} 

    \pgfmathparse{135}
    \pgfmathsetmacro\anglec{\pgfmathresult} 

    \begin{tikzpicture}[
    myangle/.style args= {#1/#2}{draw=black!55,<->,
        angle eccentricity=#1, angle radius=#2},
    myangle/.default = 1.05/5cm
    ]
    % grid
    \draw [color=black!5] (-3,0) grid (7,8);
    % axes
    \draw[name path=A]
    (0,0) coordinate[label=below left:{$(0,0)$}] (or) -- (0,8) node[below left] {$y$}
    (or) -- (7,0) coordinate[label=below left:$x$] (x)
    (or) -- (-3,0);
    % red line
    \path[ultra thick, draw=red]
    (0,2) coordinate (m) -- ++(\anglealpha:5) coordinate[label={[label distance=1mm]345:$ww$}] (ww);
            \draw
            let
            \p1=(0,2),\p2=(ww),  \n1={{veclen(\x2-\x1,\y2-\y1)}}
            in
            (\x1,\y1) -- (\n1,\y1) coordinate[label={[label distance=1mm]345:$w$}] (w) {}
            \pgfextra{\xdef\var{\n1}} ;

    % vectors C's
    \draw[name path=C0, ->] (w) to ["$\ell_0$", sloped, pos=0.8]  ++ (\anglea:7) coordinate (c0) node[left] {$C_0$};    
    \draw[name path=C1, ->] (w) to ["$\ell_1$", sloped,pos=0.85]  ++ (\angleb:6) coordinate (c1) node[left] {$C_1$};
    \draw[name path=C2, ->] (w) to ["$\ell_2$", sloped,pos=0.8]  ++ (\anglec:7) coordinate (c2) node[left] {$C_2$}; 

    % rotated vectors  
    \draw[red,dashed,->]    let \p1=($(w)-(c0)$),
                            \n1 = {veclen(\x1,\y1)} in
                            (ww) to ["$\ell_0$" ',text=red, pos=0.9] ++ (\anglea+\anglealpha:\n1) coordinate[label=left:$C_0^r$] (d0);

    \draw[red,dashed,->]    let \p1=($(w)-(c1)$),
                            \n1 = {veclen(\x1,\y1)} in
                            (ww) to ["$\ell_1$" ',text=red, pos=0.9] ++ (\angleb+\anglealpha:\n1) coordinate[label=left:$C_1^r$] (d1);

    \draw[red,dashed,->]    let \p1=($(w)-(c2)$),
                            \n1 = {veclen(\x1,\y1)} in
                            (ww) to ["$\ell_2$" ',text=red, pos=0.9] ++ (\anglec+\anglealpha:\n1) coordinate[label=left:$C_2^r$] (d2);

    % angles
    \draw pic[myangle, "$\alpha$"] {angle=w--m--ww} ;
    \path pic[myangle=1.5/1.5cm, "$\beta_0$" ] {angle=c0--w--m} ;
    \path pic[myangle=1.4/1.7cm, "$\beta_1$" ] {angle=c1--w--m} ;
    \path pic[myangle=1.3/1.9cm, "$\beta_2$"] {angle=c2--w--m} ;

    \path pic[myangle=1.5/1.5cm, "$\beta_0$" ] {angle=d0--ww--m} ;
    \path pic[myangle=1.4/1.7cm, "$\beta_1$" ] {angle=d1--ww--m} ;
    \path pic[myangle=1.3/1.9cm, "$\beta_2$" ] {angle=d2--ww--m} ;  
    % circles w, ww
    \draw[fill=red!5]   (ww) circle [radius=0.15] ;
    \draw[fill=red!5]   (w)  circle [radius=0.15];
    \end{tikzpicture}


    \begin{tikzpicture}[
    myangle/.style args= {#1/#2}{draw=black!55,<->, angle eccentricity=#1, angle radius=#2},
    myangle/.default = 1.05/5cm]
    % grid
    \draw [color=black!5] (-3,0) grid (7,8);
    % axes
    \draw[name path=A] (0,0) coordinate[label=below left:{$(0,0)$}] (or) -- (0,8) node[below left] {$y$}
    (or) -- (7,0) coordinate[label=below left:$x$] (x)
    (or) -- (-3,0);
    % red line
    \path[ultra thick, draw=red]
    (0,2) coordinate[label=below left:{$(0,2)$}] (m) -- ++(\anglealpha:5) coordinate[label={[label distance=1mm]345:$ww$}] (ww);
    \draw pic[myangle, "$\alpha$", <->] {angle=w--m--ww} ;
    \draw
        let
        \p1=(0,2),\p2=(ww),  \n1={{veclen(\x2-\x1,\y2-\y1)}}
        in
        (\x1,\y1) -- (\n1,\y1) coordinate[label={[label distance=1mm]345:$w$}] (w) {}
        \pgfextra{\xdef\var{\n1}} ;

    \draw[name path=C3, ->] (w) to ["$\ell_3$", sloped, pos=0.8] (0.25,7) coordinate (c3) node[left] {$C_3$};

        \tkzFindAngle(c3,w,m)
        \tkzGetAngle{angleOne};
        \FPround\angleOne\angleOne{0}

        \tkzMarkAngle[size=.5](c3,w,m)
        \tkzLabelAngle[pos=.82](c3,w,m){\tiny $\angleOne^\circ$}        

    % angle 0 from (ww) is line to the right, angle 90 is vertical, angle 180 is line on the left, 180+\alpha is red line, so 180+\alpha - \beta is the rotation of w--C3
    \pgfmathparse{(180+\anglealpha)-\angleOne}
    \pgfmathsetmacro\angletot{\pgfmathresult}

    \draw[red,dashed,->]    
            let \p1=(w),\p2=(c3),
            \n1 = {veclen(\x2-\x1, \y1-\y2)} in
        (ww) to ["$\ell_3$",text=red, pos=0.8] ++(\angletot:\n1) coordinate[label=left:$C_3^r$] (d0);


        \tkzFindAngle(m,ww,d0)
        \tkzGetAngle{angleTwo}
        \FPround\angleTwo\angleTwo{0}   
        \pgfmathparse{360-\angleTwo}   % return decimal part .0 . How do I delete it?
        \pgfmathsetmacro\angleTwo{\pgfmathresult}

        \tkzMarkAngle[size=.5](d0,ww,m)
        \tkzLabelAngle[pos=-.82](d0,ww,m){\tiny $\angleTwo^\circ$}              
        % circles w, ww
        \draw[fill=red!5]   (ww) circle [radius=0.15] ;
        \draw[fill=red!5]   (w)  circle [radius=0.15];

        \end{tikzpicture}
\end{document}

在此处输入图片描述

我仍然有几个外观问题：

  • 我如何获取 46而不是46.0来自\pgfmathparse{360-\angleTwo}？
  • 确实，为什么\tkzFindAngle(m,ww,d0)返回314，和\tkzFindAngle(m,ww,d0)分别-314（而不是46）？
  • 我怎样在靠近弧顶的地方写标签 \betas ？

答案1

这是你想要的吗？

在此处输入图片描述

完整代码如下：

\documentclass[convert={density=300,size=1080x800,outext=.png}]{standalone}
%\documentclass[12pt,a4paper]{scrartcl}  %%KOMA class
%\setkomafont{sectioning}{\rmfamily\bfseries\boldmath}  %%
\usepackage{tikz}
\usepackage{calculator}
\usepackage{pgfplots}
\usetikzlibrary{rulercompass}
\usetikzlibrary{intersections,quotes,angles}
\usetikzlibrary{calc}
\newcommand\round[2][2]{\pgfmathprintnumber[precision=#1]{#2}}
\begin{document}

\begin{tikzpicture}
\draw [color=black!5] (0,0) grid (9,5); 
\draw (9,0) coordinate (a) node[right, below] {x}
-- (0,0) coordinate (b) node[left] {(0,0)} 
-- (0,5) coordinate (c) node[left] {y};
\path  (0,0) coordinate (ad)   -- (30:5cm) coordinate (dd);
\draw (ad) -- (dd) coordinate[pos=0.455](c1);
\coordinate (c2) at ($(c1)!2*2.365 cm!(dd)$); 
\draw[->,  ultra thick, red]
let
\p1=(ad),\p2=(c2),  \n1={{veclen(\x2-\x1,\y2-\y1)}}
in
(\x1,\y1) -- (\x2,\y2)
\pgfextra{\xdef\var{\n1}} ;

\fill[black] (\var,0) circle (2.5pt)
pic["$\alpha$", draw=black!55, thick, <<->>, angle eccentricity=1.1, angle radius=\var]{angle=a--b--dd} ;

\draw[->,   black] (0,0)coordinate(o) -- (\var ,0) coordinate (w);
\draw[->,   black] (w) -- (0 ,3.5)coordinate(ww) [fill=black] node[left] {$C$};

\draw  [fill=red!5] (w) circle [radius=0.25]  node[xshift=-0.35cm, yshift=-0.45cm] {$w$}
pic["$\beta$", draw=black,  <-, angle eccentricity=1.1, angle radius=3cm]{angle=w--o--c2};

\draw  [fill=red!5] (c2) circle [radius=0.25]  node[xshift=-0.35cm, yshift=-0.45cm] {$ww$};
\end{tikzpicture}

\end{document}

答案2

如果我理解正确的话，那么您正在寻找类似这样的东西：

![在此处输入图片描述

即，您喜欢在w--b--ww具有两个不同半径（外半径\alpha和内半径\beta）的之间绘制角度。为此，我认为没有必要对轨迹半径进行任何特殊的计算。此外，一些定义的坐标是重复的或未使用的（在下面的 MWE 中，我删除了所有这些坐标）：

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{angles, quotes}

\begin{document}
    \begin{tikzpicture}[
myangle/.style args= {#1/#2}{draw=black!55,->,
                             angle eccentricity=#1, angle radius=#2},
myangle/.default = 1.05/5cm
                        ]
% grid
\draw [color=black!5] (0,0) grid (7,4);
% axes
\draw (0,0) coordinate[label=below left:{$(0,0)$}] (b) 
            -- (0,4) node[below left] {$y$}
      (b)   -- (7,0) node[below left] {$y$};
% red line
\path[ultra thick, draw=red]   
    (b) -- (30:5) coordinate[label={[label distance=1mm]345:$ww$}] (ww)
           ( 0:5) coordinate[label={[label distance=1mm]345:$w$ }] (w);
% arrow C
\draw[->] (w) -- (ww -| b) node[left] {$C$};
% angles
\draw pic[myangle, "$\alpha$"] {angle=w--b--ww} ;
\path pic[myangle=1.1/2.5cm, "$\beta$" ] {angle=w--b--ww} ;
% circles w, ww
\draw[fill=red!5]   (ww) circle [radius=0.25] ;
\draw[fill=red!5]   (w)  circle [radius=0.25];
    \end{tikzpicture}
\end{document}

附录： 从您对问题的编辑中，我可以看出，在第一次尝试中我误解了您（但代码非常简单 :-) ）。 现在代码不再那么简单了……因为需要计算由y-axis 和 line的交点决定的矢量长度C，对吗？：

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{angles, calc, intersections, quotes}

\begin{document}
    \begin{tikzpicture}[
myangle/.style args= {#1/#2}{draw=black!55,<->,
                             angle eccentricity=#1, angle radius=#2},
myangle/.default = 1.05/5cm
                        ]
% grid
\draw [color=black!5] (-20,0) grid (7,4);
% axes
\draw[name path=A]
    (0,0) coordinate[label=below left:{$(0,0)$}] (b)
          -- (0,4) node[below left] {$y$}
    (b)   -- (7,0) coordinate[label=below left:$x$] (x);
% red line
\path[ultra thick, draw=red]
    (b) -- (30:5) coordinate[label={[label distance=1mm]345:$ww$}] (ww)
           ( 0:5) coordinate[label={[label distance=1mm]345:$w$ }] (w);
% arrow C
\path[name path=C] (w) -- ++ (150:6);
\draw [name intersections={of=A and C, by={c}},->] 
    (w)  to ["$\ell$" ',sloped,pos=0.4]  (c) node[left] {$C$};
% arrow ?
\draw[red,dashed,->]    let \p1=($(w)-(c)$),
                            \n1 = {veclen(\x1,\y1)} in
                        (ww) to ["$\ell$" ',text=black] ++ (180:\n1) coordinate[label=left:?] (d);
% angles
\draw pic[myangle, "$\alpha$"] {angle=w--b--ww} ;
\path pic[myangle=1.3/1.5cm, "$\beta$" ] {angle=c--w--b} ;
\path pic[myangle=1.3/1.5cm, "$\beta$" ] {angle=d--ww--b} ;
% circles w, ww
\draw[fill=red!5]   (ww) circle [radius=0.25] ;
\draw[fill=red!5]   (w)  circle [radius=0.25];
    \end{tikzpicture}
\end{document}

![![在此处输入图片描述

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