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如何自动绘制这个圆弧（平面与球面的交点）？

tag-icon • tag-icon tikz-3dplot
如何自动绘制这个圆弧（平面与球面的交点）？

根据这里的答案我怎样才能自动纠正这张图片中的点 A 和 B？，我尝试用球体

\documentclass[tikz,border=1mm, 12 pt]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{backgrounds}
\begin{document}
\tdplotsetmaincoords{70}{50}
    \begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={R=25*sqrt(78)*(1/52);r=7*sqrt(3)*(1/3);
        alpha1(\th,\ph,\b)=\ph-asin(cot(\th)*tan(\b));%
        alpha2(\th,\ph,\b)=-180+\ph+asin(cot(\th)*tan(\b));%
        beta1(\th,\ph,\a)=90+atan(cot(\th)/sin(\a-\ph));%
        beta2(\th,\ph,\a)=270+atan(cot(\th)/sin(\a-\ph));%
    }]
    \path
    coordinate (O) at (0,0,0)
    coordinate (A) at (-3/2, {-13*sqrt(3)*(1/6)}, 0)
    coordinate (B) at (3/2, {-13*sqrt(3)*(1/6)}, 0)
    coordinate (C) at (4, {(1/3)*sqrt(3)}, 0)
    coordinate (S) at (0, 0, {(1/3)*sqrt(78)})
    coordinate (T) at  (0, 0, {-23*sqrt(78)*(1/156)});
    \begin{scope}[tdplot_screen_coords, on background layer]
    \draw[thick] (T) circle (R);
    \end{scope}

    \begin{scope}[canvas is xy plane at z={0}]
    \draw[dashed] (O) circle (r);
    \draw[thick] ({alpha1(\tdplotmaintheta,\tdplotmainphi,{atan(r/R)})}:r) arc({alpha1(\tdplotmaintheta,\tdplotmainphi,{atan(r/R)})}: {alpha2(\tdplotmaintheta,\tdplotmainphi,{atan(r/R)})}:r) ;
    \end{scope}
    \begin{scope}[on background layer]
        \foreach \v/\position in {T/above,O/below,A/below,B/below,C/below,S/right} {
        \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
    }
    \end{scope}
    \foreach \X in {A,B,C,O} \draw[dashed] (\X) -- (S); 
    \draw[dashed] (A) -- (B) -- (C) -- cycle;
     \end{tikzpicture}
\end{document}

在此处输入图片描述

结果不正确。如何使用此方法自动修复？

我用了另外一种方法。

\documentclass[tikz,border=1mm, 12 pt]{standalone}
\usepackage{fouriernc}
\usepackage{tikz-3dplot}
\usetikzlibrary{backgrounds}
\begin{document}
\tdplotsetmaincoords{70}{50}
    \begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={R=25*sqrt(78)*(1/52);r=7*sqrt(3)*(1/3);}]
    \path
    coordinate (O) at (0,0,0)
    coordinate (A) at (-3/2, {-13*sqrt(3)*(1/6)}, 0)
    coordinate (B) at (3/2, {-13*sqrt(3)*(1/6)}, 0)
    coordinate (C) at (4, {(1/3)*sqrt(3)}, 0)
    coordinate (S) at (0, 0, {(1/3)*sqrt(78)})
    coordinate (T) at  (0, 0, {-23*sqrt(78)*(1/156)});
    \begin{scope}[tdplot_screen_coords, on background layer]
    \draw[thick] (T) circle (R);
    \end{scope}
    \begin{scope}[canvas is xy plane at z=0]
    \draw[dashed] (\tdplotmainphi:r) arc(\tdplotmainphi:\tdplotmainphi+180:r);
    \draw[thick] (\tdplotmainphi:r)  arc(\tdplotmainphi:\tdplotmainphi-180:r)
    ;
    \end{scope}
            \foreach \v/\position in {T/above,O/below,A/below,B/below,C/right,S/right} {
        \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
    }
    \foreach \X in {A,B,C,O} \draw[dashed] (\X) -- (S); 
    \draw[dashed] (A) -- (B) -- (C) -- cycle;

    \end{tikzpicture}
\end{document}

在此处输入图片描述

答案1

链接的答案很好用，它对你的问题的应用也很好。让我从输出中重构你正在做的事情。你画一个球体，中心在T，它是不是原点和原点所在xy平面上的一个圆。圆的纬度是多少？它是不是 atan(r/R)，而纬度由 给出atan(T_z/r)，其中T_z是z的分量T。根据惯例，这里我们需要一个减号，这就是为什么 中有一个减号\pgfmathsetmacro{\myel}{-atan(23*sqrt(78)*(1/156)/r)}。这很可能就是Janis Lazovskis 的评论想告诉我们。如果你在这里问我，这个可以改变。（惯例是这样的，其他答案中不需要减号。）总的来说，在我看来这很好用。

\documentclass[tikz,border=1mm, 12 pt]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{backgrounds}
\begin{document}
\tdplotsetmaincoords{70}{50}
    \begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={R=25*sqrt(78)*(1/52);r=7*sqrt(3)*(1/3);
        alpha1(\th,\ph,\b)=\ph-asin(cot(\th)*tan(\b));%
        alpha2(\th,\ph,\b)=-180+\ph+asin(cot(\th)*tan(\b));%
        beta1(\th,\ph,\a)=90+atan(cot(\th)/sin(\a-\ph));%
        beta2(\th,\ph,\a)=270+atan(cot(\th)/sin(\a-\ph));%
    }]
    \path
    coordinate (O) at (0,0,0)
    coordinate (A) at (-3/2, {-13*sqrt(3)*(1/6)}, 0)
    coordinate (B) at (3/2, {-13*sqrt(3)*(1/6)}, 0)
    coordinate (C) at (4, {(1/3)*sqrt(3)}, 0)
    coordinate (S) at (0, 0, {(1/3)*sqrt(78)})
    coordinate (T) at  (0, 0, {-23*sqrt(78)*(1/156)});
    \begin{scope}[tdplot_screen_coords, on background layer]
    \draw[thick] (T) circle (R);
    \end{scope}

    \begin{scope}[canvas is xy plane at z={0}]
    \draw[dashed] (O) circle (r);
    \pgfmathsetmacro{\myel}{-atan(23*sqrt(78)*(1/156)/r)}
    \typeout{\myel}
    \draw[thick] ({alpha1(\tdplotmaintheta,\tdplotmainphi,{\myel})}:r) 
    arc({alpha1(\tdplotmaintheta,\tdplotmainphi,{\myel})}: 
    {alpha2(\tdplotmaintheta,\tdplotmainphi,{\myel})}:r) ;
    \end{scope}
    \begin{scope}[on background layer]
        \foreach \v/\position in {T/above,O/below,A/below,B/below,C/below,S/right} {
        \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
    }
    \end{scope}
    \foreach \X in {A,B,C,O} \draw[dashed] (\X) -- (S); 
    \draw[dashed] (A) -- (B) -- (C) -- cycle;
     \end{tikzpicture}
\end{document}

在此处输入图片描述

一个可能更优雅的解决方案是重新定义alpha1和，alpha2以便处理减号，并T使用代码提取的 z 分量。

\documentclass[tikz,border=1mm, 12 pt]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{backgrounds}
\makeatletter
% retrieves the 3D coordinates
\def\RawCoord(#1){\csname tikz@dcl@coord@#1\endcsname}%
\def\scalprod#1=#2.#3;{%
\edef\coordA{\RawCoord#2}%
\edef\coordB{\RawCoord#3}%
\pgfmathsetmacro\pgfutil@tmpa{scalarproduct({\coordA},{\coordB})}
\edef#1{\pgfutil@tmpa}}%
\makeatother 
\newcommand{\spaux}[6]{(#1)*(#4)+(#2)*(#5)+(#3)*(#6)}  
\pgfmathdeclarefunction{scalarproduct}{2}{% scalar product of two 3-vectors
  \begingroup%
  \pgfmathparse{\spaux#1#2}%
  \pgfmathsmuggle\pgfmathresult\endgroup}  
\begin{document}
\tdplotsetmaincoords{70}{50}
    \begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={R=25*sqrt(78)*(1/52);r=7*sqrt(3)*(1/3);
        alpha1(\th,\ph,\b)=\ph+asin(cot(\th)*tan(\b));%
        alpha2(\th,\ph,\b)=-180+\ph-asin(cot(\th)*tan(\b));%
        beta1(\th,\ph,\a)=90+atan(cot(\th)/sin(\a-\ph));%
        beta2(\th,\ph,\a)=270+atan(cot(\th)/sin(\a-\ph));%
    }]
    \path (0,0,0) coordinate (O)
        (-3/2, {-13*sqrt(3)*(1/6)}, 0) coordinate (A) 
        (3/2, {-13*sqrt(3)*(1/6)}, 0) coordinate (B) 
        (4, {(1/3)*sqrt(3)}, 0) coordinate (C) 
        (0, 0, {(1/3)*sqrt(78)}) coordinate (S) 
        (0, 0, {-23*sqrt(78)*(1/156)}) coordinate (T)
        (0,0,1) coordinate(Z);
    \begin{scope}[tdplot_screen_coords, on background layer]
    \draw[thick] (T) circle (R);
    \end{scope}

    \begin{scope}[canvas is xy plane at z={0}]
    \draw[dashed] (O) circle (r);
    \scalprod\myz=(T).(Z); % z component of T
    \pgfmathsetmacro{\myel}{atan(-1*\myz/r)}
    \draw[thick] ({alpha1(\tdplotmaintheta,\tdplotmainphi,{\myel})}:r) 
    arc({alpha1(\tdplotmaintheta,\tdplotmainphi,{\myel})}: 
    {alpha2(\tdplotmaintheta,\tdplotmainphi,{\myel})}:r) ;
    \end{scope}
    \begin{scope}[on background layer]
        \foreach \v/\position in {T/above,O/below,A/below,B/below,C/below,S/right} {
        \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
    }
    \end{scope}
    \foreach \X in {A,B,C,O} \draw[dashed] (\X) -- (S); 
    \draw[dashed] (A) -- (B) -- (C) -- cycle;
     \end{tikzpicture}
\end{document}

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