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网格和圆圈上的平行线最佳实践？

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网格和圆圈上的平行线最佳实践？

我是 TeX/LaTeX 新手，但我正在快速学习。我即将尝试使用 LaTeX/TeX/TikZ 绘制下图。我已经（我认为）掌握了绘制圆、括号、直线、线段和节点等的丰富知识。

然而，上次我画的时候，我注意到我花了很多时间来微调参数，确保线条在正确的点与圆相交，确保线条平行等。

我需要绘制下图，因此我想首先询问是否有一种快速/简单的方法，或者在创建平行线（红色）时我应该记住什么，如下图所示：

在此处输入图片描述

我预见到的主要时间浪费如下：

  1. 确保紫线的长度完全正确，以免超出红线。

  2. 确保所有红线彼此平行。

  3. 确保我可以在线的相交处和/或圆上画出节点/点。

是否有推荐的特定方法可以确保上述几点不会浪费太多时间？我将不胜感激任何建议/示例。

谢谢！

PS 我也正在充分利用 TikZ。

编辑：

我附上了我的“技能”组合的一个最小示例。如您所见，我通常只需要精心调整坐标，确定事物相交的位置等。如果我想制作平行线，如上图红线所示，我也不知道如何不浪费时间。这是我的示例代码：

\documentclass[journal]{IEEEtran}
\usepackage{graphicx}
\usepackage{float}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{tikz}
\usetikzlibrary{matrix}
\usetikzlibrary{decorations.pathreplacing}
\usepackage{xspace} 
\usepackage{float}
\usepackage{capt-of}
\usepackage{cases}



\begin{document}

\newcommand{\var}{1.5}

\begin{tikzpicture}
%\begin{tikzpicture}

    \draw [help lines] (-4,-4) grid (4,4);

    % Draw the axes
    \draw [->,black, ] (-4,0) -- (4,0) ;
    \draw [->,black] (0,-4) -- (0,4) ;

    % Draw the circle
    \path [draw, ultra thick,  black] (0,0) circle (3);

    %Radial Lines    
    \draw[black](0:0)--(45:3);
    \draw[black](0:0)--(135:3);
    \draw[black](0:0)--(225:3);
    \draw[black](0:0)--(315:3);        

    %Wavefront normal
    \draw[blue](0:0)--(60:3);
    \draw[blue](0:0)--(240:3);

    \draw[blue](0,1) arc (90:60:1);
    \draw[blue] (0,0.95)arc(90:60:0.95);
    \node[] at (75:1.2)  {$\alpha$};

    \draw[blue] (0.7,1.2)arc(90:30:0.37);
    %\node[] at (50:2) {$\frac{pi}{4}-\alpha$};
    \draw[-latex](1.5,3)node[right]{$\frac{\pi}{4}-\alpha$}
        to[out=180,in=90] (50:1.5);

    %Points
    \path[fill, black](45:3) circle(0.1);
    \node[] at (45:3.3){1};
    \path[fill, black](135:3) circle(0.1);
    \node[] at (133:3.3){4};
    \path[fill,black](225:3) circle(0.1);
    \node[] at (225:3.3){3};
    \path[fill, black](315:3) circle(0.1);
    \node[] at (313:3.3){2};

    %Wavefronts
    \draw[blue, ultra thick](-.5,3.6347)--(3.5,1.3252);
    \draw[blue, ultra thick](-2.5,2.34)--(3.5,-1.1242);
    \draw[blue, ultra thick](-3.2,0.9509)--(2.8,-2.5132);
    \draw[blue, ultra thick](-3.2,-1.4985)--(0.2,-3.4615);

    %Right angle signs
    \draw[red](60:2.8)--(62:2.8)--(62:2.9);
    \draw[red](60:.68)--(68:.69)--(67:.79);




    % Done
\end{tikzpicture}


\end{document}

上面的代码现在给了我这个：

在此处输入图片描述

答案1

第一张 TikZ 图片展示了一种相当草率的绘制平行线的方法。第二个示例展示了一种更自动化的绘制割线的方法（可以\angle直接使用该值，但如果用户不知道矢量的方向，我想展示如何做到这一点。

第二个例子也表明，对于接近和calc， 并不是($(<p1>)!(<p3>)!(<p2>)$)很准确。当然，当 时甚至没有一条线。(<p1>)(<p2>)<p1> = <p2>

该intersections库可能能够更精确地找到矢量正交线和割线相交的点。

代码

\documentclass[tikz,convert=false]{standalone}
\usetikzlibrary{calc,through}
\tikzset{
  m*/.style args={#1:#2}{
    insert path={node [fill=green!50!black, outer sep=+0pt, shape=circle, inner sep=0pt, minimum size=+4pt,#1,label={#2}] {}}
  },
  m/.style={insert path={coordinate (#1)}},
  parLines/.style={draw=red,,shorten <=+-.5cm,shorten >=-.5cm},
  vertLines/.style={draw=purple,shorten >=.5\pgflinewidth},
  @splitLine/.code args={#1 -- #2}{\def\tikztotargetA{#1}\def\tikztotargetB{#2}},
  vert on/.style={
    to path={
      [@splitLine/.expanded={\tikztotarget}]
      -- ($(\tikztotargetA)!(\tikztostart)!(\tikztotargetB)$) node[right angle node,rotate=90*#1] {}\tikztonodes
    }
  },
  vert on/.default=0,
  right angle node/.style={
    at end,sloped,above,allow upside down,
    anchor=south east,
    shape=rectangle,
    inner sep=0pt,
    minimum size=3pt,
    append after command={
      \pgfextra\pgfinterruptpath\draw[right angle node path] (\tikzlastnode.south west) -- (\tikzlastnode.north west) -- (\tikzlastnode.north east);\endpgfinterruptpath\endpgfextra
    }
  },
  right angle node path/.style={draw,thin,black,-,shorten >=.4pt},
  secant/.style={
    to path={
      let \p{@dir}=(\tikztotarget), \n{@dir}={atan2(\x{@dir},\y{@dir})} in
      (node cs: name=\tikztostart, anchor=#1) [m*={name=m#1}:#1] -- (node cs: name=\tikztostart, anchor={2*(\n{@dir}-90)-#1}) [m=m#1'] \tikztonodes
    }
  }
}
\begin{document}
\begin{tikzpicture}[thick]
\clip (-2.5,-2.5) rectangle (2.5,2.5);
\draw[thin, ->] (0,0) [m*={black,minimum size=+3pt,name=O}:] -- node [sloped,above,inner sep=+1pt,font=\scriptsize] {$\vec v$} ++(150:1) coordinate (d);

\draw[thin, ->] (0,0) -- ([rotate=90]d)  [shorten >=-1.3cm];
\draw[thin, ->] (0,0) -- ([rotate=-90]d) [shorten >=-1.3cm];

\node [draw=blue, circle through={(2,0)}] (c) {};

\draw[parLines] (c.50)  [m*={name=m1}:1] -- ++([scale=2] d)    [m=m1'];
\draw[parLines] (c.140) [m*={name=m2}:2] -- ++([scale=-4] d)   [m=m2'];
\draw[parLines] (c.-30) [m*={name=m3}:3] -- ++([scale=4.5] d)  [m=m3'];
\draw[parLines] (c.210) [m*={name=m4}:4] -- ++([scale=-2.5] d) [m=m4'];

\draw[vertLines] (c.center) to[vert on] (m1 -- m1');
\draw[vertLines] (c.center) to[vert on] (m4' -- m4);
\end{tikzpicture}

\foreach \angle in {0,10,...,359}{% Careful, 36 pages!
\begin{tikzpicture}[thick]
\clip (-2.5,-2.5) rectangle (2.5,2.5);
\draw[thin, ->] (0,0) [m*={black,minimum size=+3pt,name=O}:] -- node [sloped,above,inner sep=+1pt,font=\scriptsize] {$\vec v$} ++(\angle:1) coordinate (d);

\draw[thin, ->] (0,0) -- ([rotate=90]d)  [shorten >=-1.3cm];
\draw[thin, ->] (0,0) -- ([rotate=-90]d) [shorten >=-1.3cm];

\node [draw=blue, circle through={(2,0)}] (c) {};

\draw[parLines] (c) to[secant=30]  (d);
\draw[parLines] (c) to[secant=140] (d);
\draw[parLines] (c) to[secant=-30] (d);
\draw[parLines] (c) to[secant=210] (d);


\draw[vertLines] (c.center) to[vert on] (m30 -- m30');
\draw[vertLines] (c.center) to[vert on] (m210' -- m210);
\end{tikzpicture}%
}
\end{document}

输出

在此处输入图片描述

在此处输入图片描述

答案2

使用 tkz-euclide 解决方案，我们得到平行线

  \tkzDefLine[parallel=through B](a,A)     \tkzGetPoint{b}

新直线经过 B 点并且与 Aa 点平行，我们得到该直线上的点 b 点。

 \documentclass{article}
 \usepackage{tkz-euclide}
 \usetkzobj{all}

 \begin{document}
 \begin{tikzpicture}
    % init the picture
    \tkzInit[xmin=-5,xmax=5,ymin=-5,ymax=5]
    \tkzGrid \tkzClip
    % definition of points  
    \tkzDefPoint(45:3){A}  
    % it was possible to use random points something like
    % \tkzGetRandPointOn[circle = center O radius 3 cm]{A}  
    \tkzDefPoint(0,0){O}
    \tkzDefPoint(135:3){B}  
    \tkzDefPoint(225:3){C}  
    \tkzDefPoint(315:3){D}  
     \tkzDefPoint(60:3){M} 
     \tkzDefPoint(240:3){N} 
    \tkzDefPoint(-1,4){a} 
    % definition of parallel lines
    \tkzDefLine[parallel=through B](a,A)     \tkzGetPoint{b}    
    \tkzDefLine[parallel=through C](a,A)     \tkzGetPoint{c}    
    \tkzDefLine[parallel=through D](a,A)     \tkzGetPoint{d}
    % orthogonal projection of O on the first line  
    \tkzDefPointBy[projection=onto A--a](O)  \tkzGetPoint{H}  
    % we can avoid the next line and use tkzInterLL 
    \tkzDefPointBy[projection=onto C--c](O)  \tkzGetPoint{K}     
    \tkzInterLL(K,H)(b,B)                    \tkzGetPoint{L}  
    % now we can draw         
    \tkzDrawCircle[R,thick](O,3 cm)
    \tkzDrawLines[add=1 and 1,color=blue](a,A B,b C,c)
    \tkzDrawLine[add=2 and 1,color=blue](D,d)   
     \tkzDrawSegments(A,C B,D)
    \tkzDrawLine[add=2 and 1,color=red](O,H)
     % mark right angles    
    \tkzMarkRightAngles[fill=lightgray](a,H,O C,K,O B,L,O)
     % draw points
    \tkzDrawPoints(A,B,C,D,O)
    % mark angles
     \tkzMarkAngle[mark= ,arc=ll,size=1 cm,mkcolor=blue](H,O,B)
     \tkzLabelAngle[pos=1.25](H,O,B){$\alpha$}
     \tkzMarkAngle[mark= ,arc=l,size=2.25 cm,mkcolor=blue](A,O,H)
     \tkzLabelAngle[pos=2.5,below=10pt](A,O,H){$\frac{\pi}{4}-\alpha$}
    % label points
     \tkzLabelPoint[right](A){$1$} \tkzLabelPoint[right](D){$4$}
     \tkzLabelPoint[above](B){$2$}  \tkzLabelPoint[below](C){$3$}
 \end{tikzpicture}
 \end{document}

在此处输入图片描述

答案3

使用 PSTricks 可以算是最佳实践。下面的代码应该足够清晰，所以剩下的部分故意留给你做家庭作业。

在此处输入图片描述

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}

\begin{document}
\begin{pspicture}[showgrid](-3,-3)(3,3)
    \pnodes{O}(0,0)(2,0)
    \pstCircleOA{O0}{O1}
    \pnodes{P}(-3,3)(3,1)
    \pcline[nodesepA=1.5,nodesepB=.5](P0)(P1)
    \pstInterLC[PointSymbolB=*]{P0}{P1}{O0}{O1}{I0}{I2}
    \pstProjection{P0}{P1}{O0}
    \psline(O0)(O0')
    \pstRightAngle[RightAngleSize=0.15]{P0}{O0'}{O0}
    \pstTranslation[DistCoef=0.5]{O0'}{O0}{P0,P1}
    \pcline[nodesepA=1.5,nodesepB=.5](P0')(P1')
    \pstInterLC[PointSymbolA=*]{P0'}{P1'}{O0}{O1}{I0'}{I1'}
\end{pspicture}
\end{document}

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