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创建双凹透镜

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创建双凹透镜

我已经尝试了好几个小时了。。有人能帮我吗？我无法用颜色填充镜头的“形状”。

\documentclass[border=12pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{intersections}

\begin{document}

\begin{tikzpicture}[scale=1]

\begin{scope}[>=latex]
\draw [->] (-3,0) -- (3,0);
\draw [->] (0,-1.25) -- (0,1.5);
\end{scope}
\definecolor{lensBlue}{RGB}{217,232,250}



\path [name path=arc1](-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\path [name path=arc2](1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];
\path [name path=rect](-0.9,-0.9) rectangle (0.9,0.9);


\path [name intersections={of = arc1 and rect}];
\coordinate (A)  at (intersection-1);
\coordinate (B)  at (intersection-2);

\path [name intersections={of = arc2 and rect}];
\coordinate (C)  at (intersection-1);
\coordinate (D)  at (intersection-2);

\draw (A) -- (D);
\draw (B) -- (C) ;

\draw (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];


\end{tikzpicture}

\end{document}

在此处输入图片描述

答案1

以下是两种可能的选择：

在此处输入图片描述

在第一种情况下，我习惯fill=white用白色涂掉弧形部分中的粉红色填充。但是，如果您想要填充，您可以像第二种情况一样进行操作。轴的绘制被移到末尾，以便它位于顶部。您也可以backgrounds使用库来实现相同的效果on background layer。

代码：

\documentclass[border=3pt]{standalone}

\usepackage{tikz} %pgf-tikz pakcage
\usetikzlibrary{intersections}

\definecolor{lensBlue}{RGB}{217,232,250}

\begin{document}
\begin{tikzpicture}

\path [name path=arc1, draw=none](-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\path [name path=arc2, draw=none](1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];
\path [name path=rect, draw=none](-0.9,-0.9) rectangle (0.9,0.9);

\path [name intersections={of = arc1 and rect}];
\coordinate (A)  at (intersection-1);
\coordinate (B)  at (intersection-2);

\path [name intersections={of = arc2 and rect}];
\coordinate (C)  at (intersection-1);
\coordinate (D)  at (intersection-2);

\draw [brown, ultra thick] (A) -- (D);
\draw [brown, ultra thick] (B) -- (C) ;

\fill [red!50] (A) -- (D) -- (C) -- (B) -- cycle;

\draw [blue, thick, fill=white] (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick, fill=white] (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];

% axis
\begin{scope}[>=latex]
    \draw [->] (-3,0) -- (3,0);
    \draw [->] (0,-1.25) -- (0,1.5);
\end{scope}
\end{tikzpicture}
\begin{tikzpicture}

\path [name path=arc1, draw=none](-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\path [name path=arc2, draw=none](1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];
\path [name path=rect, draw=none](-0.9,-0.9) rectangle (0.9,0.9);

\path [name intersections={of = arc1 and rect}];
\coordinate (A)  at (intersection-1);
\coordinate (B)  at (intersection-2);

\path [name intersections={of = arc2 and rect}];
\coordinate (C)  at (intersection-1);
\coordinate (D)  at (intersection-2);

\draw [brown, ultra thick] (A) -- (D);
\draw [brown, ultra thick] (B) -- (C) ;

\fill [red!50] (A) -- (D) -- (C) -- (B) -- cycle;

\draw [blue, thick, fill=yellow!50] (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick, fill=yellow!50] (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];

% axis
\begin{scope}[>=latex]
    \draw [->] (-3,0) -- (3,0);
    \draw [->] (0,-1.25) -- (0,1.5);
\end{scope}
\end{tikzpicture}

\end{document}

答案2

两种方法也

A) 无图书馆交叉口

我们绘制并填充一个矩形，然后添加两个圆弧。

\documentclass[border=3pt]{standalone}
\usepackage{tikz} 

\begin{document}
\begin{tikzpicture}

\filldraw [red!50,draw=brown, ultra thick] (-1,-0.9) rectangle (1,0.9);

\draw [blue, thick, fill=white] (-1.9,-1.5) 
      arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick, fill=white] (1.9,1.5) 
      arc[start angle=90, end angle=270,radius=1.5];

\begin{scope}[>=latex]
    \draw [->] (-3,0) -- (3,0);
    \draw [->] (0,-1.25) -- (0,1.5);
\end{scope}
\end{tikzpicture}

\end{document}

B）与图书馆交叉口相连，但外部没有填充

我们直接填充中心部分。我们需要这样做来确定由圆弧中心 I 和圆弧的一点（此处为 D）定义的角度。

\documentclass[border=3pt]{standalone}
\usepackage{tikz} 
\usetikzlibrary{intersections}

\begin{document}
\begin{tikzpicture}
 \coordinate (I)  at (1.9,0);     
\path [name path=arc1, draw=none](-1.9,-1.5) 
     arc[start angle=-90, end angle=90,radius=1.5];
\path [name path=arc2, draw=none](1.9,1.5)   
     arc[start angle=90, end angle=270,radius=1.5];
\path [name path=rect, draw=none](-0.9,-0.9) rectangle (0.9,0.9);

\path [name intersections={of = arc1 and rect}];
\coordinate (A)  at (intersection-1);
\coordinate (B)  at (intersection-2);

\path [name intersections={of = arc2 and rect}];
\coordinate (C)  at (intersection-1);
\coordinate (D)  at (intersection-2);

\pgfmathanglebetweenpoints{\pgfpointanchor{I}{center}}{%
                           \pgfpointanchor{D}{center}}   
      \let\tmpan\pgfmathresult 

 \fill[red!50] (A)--(D)  
        arc[start angle=\tmpan, end angle=360-\tmpan,radius=1.5] -- (B)
        arc[start angle=\tmpan-180, end angle=180-\tmpan,radius=1.5] ; 

\draw [brown, ultra thick] (A) -- (D);
\draw [brown, ultra thick] (B) -- (C) ;
\draw [blue, thick] (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick] (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];

\begin{scope}[>=latex]
    \draw [->] (-3,0) -- (3,0);
    \draw [->] (0,-1.25) -- (0,1.5);
\end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案3

用户界面：

\const{Major}{3}% semi major
\const{Minor}{2}% semi minor
\const{Xo}{3.25}% distance from origin to ellipse center
\const{ThetaD}{50}

在此处输入图片描述

\documentclass[pstricks,border=12pt,nomessages]{standalone}
\usepackage{pstricks-add,fp}

\def\const#1#2{%
    \expandafter\FPeval\csname#1\endcsname{round(#2:17)}%only 17 digits after . is allowed
    \pstVerb{/#1 \csname#1\endcsname\space def}%
}

% user defined data
\const{Major}{3}% semi major
\const{Minor}{1.5}% semi minor
\const{Xo}{3.25}% distance from origin to ellipse center
\const{ThetaD}{60}

% internal used data
\const{AlphaD}{180-ThetaD}
\const{Alpha}{pi-ThetaD*pi/180}% supplementary angle relation
\const{DeltaAx}{Major*cos(Alpha)}
\const{DeltaAy}{Minor*sin(Alpha)}
\const{DeltaBx}{-Major}
\const{DeltaBy}{0}


\psset
{
    unit=1.5cm,
    linecolor=red,
}

\def\quadrant#1{%!
        \psscalebox{#1}{
        \pnode(!Xo 0){O}
        \pnode[!DeltaAx DeltaAy](O){A}
        \pnode[!DeltaBx DeltaBy](O){B}
        \pnode(0,0){Origin}
        \pscustom*[linecolor=yellow,origin={O}]
        {
            \psline(Origin)(Origin|A)(A)
            \psellipticarc(O)(!Major Minor){(A)}{(B)}
        }
        \psellipticarc[linestyle=dashed,linecolor=gray,origin={O},dimen=middle](O)(!Major Minor){90}{(A)}
        \psline(Origin|A)(A)
        \psellipticarc[origin={O},dimen=middle](O)(!Major Minor){(A)}{(B)}        
       }}

\begin{document}

\begin{pspicture}[showgrid=false](-\Xo,-\Minor)(\Xo,\Minor)
    \psforeach{\i}{1 1,1 -1,-1 -1,-1 1}{\quadrant{\i}}
    \psaxes[labels=none,ticks=none,linecolor=gray]{->}(0,0)(-\Xo,-\Minor)(\Xo,\Minor)
\end{pspicture} 

\end{document}

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