画一个球体，其纬度是圆锥的基圆

Question

我想画一个球体，它的纬度是圆锥底面的圆。

这是我的代码

\begin{tikzpicture}[scale=1, font=\footnotesize, line join=round, line cap=round, >=stealth]
\def\a{1}

\def\b{0.4}

\def\h{3}

\pgfmathsetmacro\g{asin(-\b/\h)}

\pgfmathsetmacro\xo{\a *cos(\g)}

\pgfmathsetmacro\yo{\b *sin(\g)}

\draw[dashed] (\xo,\yo) arc (\g:180-\g:{\a} and {\b});

\draw (\xo,\yo) arc (\g:-180-\g:{\a} and {\b}) -- (-90:\h) --cycle;

\begin{scope}

\draw[dashed] (\xo,\yo) arc (\g:-180-\g:{\a});

\draw (\xo,\yo) arc (\g:180-\g:{\a});

\end{scope}

\end{tikzpicture}

Answer 1

以下是基本计算的起点：

\begin{document}

    \begin{tikzpicture}
        \def\R{5}   % R is the radius of the sphere
        \def\r{4.8} % r is the radius of the cone base
        \pgfmathsetmacro\h{sqrt(\R*\R-\r*\r)} 
        \pgfmathsetmacro\a{asin(\r/\R)}
        \pgfmathsetmacro\H{\r*tan(\a)}
        
        %\draw[red,thick] (0,0) -- (-90+\a:\R);
        
        \path   (-\r,-\h) coordinate (S) --++ (\r,0) coordinate (K) --++ (\r,0) coordinate (T)
                (0,0) coordinate (O) --++ (0,-\h-\H) coordinate (U);
        
        % Wireframe
        \draw   (T) arc (-90+\a:270-\a:\R);
        \draw   (S) -- (U) -- (T)
                (T) arc (0:-180:\r cm and 0.2*\r cm);
        \draw [dashed]  (S) -- (T)
                        (K) -- (U)
                        (T) arc (0:180:\r cm and 0.2*\r cm)
                        (S) arc (270-\a:270+\a:\R);
    \end{tikzpicture}

\end{document}

编辑

回应 OP 在评论中的疑问，当圆锥底部的半径太小时，椭圆会太粗，无法反映 3D 视图的准确现实。然后可以更改椭圆的宽度来纠正此问题：

% Wireframe
        \draw   (T) arc (-90+\a:270-\a:\R);
        \draw   (S) -- (U) -- (T)
                (T) arc (0:-180:\r cm and 0.1*\r cm);
        \draw [dashed]  (S) -- (T)
                        (K) -- (U)
                        (T) arc (0:180:\r cm and 0.1*\r cm)
                        (S) arc (270-\a:270+\a:\R);

这是您所拥有的r=3，因子是 0.1。

Answer 2

你可以使用这个代码。我用的是3d工具来绘制它。

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
\begin{tikzpicture}[3d/install view=%
    {phi=110,psi=0,theta=70},line join = round, line cap = round,c/.style={circle,fill,inner sep=1pt},
    declare function={R=4;r=3.5;h= sqrt(R*R- r*r);a=asin(r/R);H=r*tan(a); d = h + H;}] 
    \path (0,0,0) coordinate (O) 
    (0,0,-h) coordinate (H)
    (0,0,-d) coordinate (T);
    \path (H)    pic[3d/cone/inner/.style={save named path=cone,draw=none}]{3d/cone={r=r,h=-H}};
    \path[save named path=sph,3d/screen coords] (O) circle[radius=R];
    \tikzset{3d/draw ordered paths={cone,sph}} 
\path foreach \p/\g in {O/90,H/0,T/-90}
    {(\p)node[c]{}+(\g:2.5mm) node{$\p$}};
    \draw[3d/hidden] (T) --(O);
\end{tikzpicture}
\end{document}    

如果您使用R=5;r=2.5，您将得到像这样的图片。

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
    \begin{tikzpicture}[3d/install view=%
        {phi=110,psi=0,theta=70},line join = round, line cap = round,c/.style={circle,fill,inner sep=1pt},
        declare function={R=4;r=3.5;h= sqrt(R*R- r*r);a=asin(r/R);H=r*tan(a); d = h + H;}] 
        \path (0,0,0) coordinate (O) 
        (0,0,-h) coordinate (H)
        (0,0,-d) coordinate (T);
        \path (H)    
        pic[3d/cone/inner/.style={save named path=icone,draw=none},
        3d/cone/outer/.append style={save named path=ocone},
        ]{3d/cone={r=r,h=-H}};
        \path[save named path=sph,3d/screen coords] (O) circle[radius=R];
        \tikzset{3d/draw ordered paths={icone,sph,ocone}} 
        \path foreach \p/\g in {O/90,H/0,T/-90}
        {(\p)node[c]{}+(\g:2.5mm) node{$\p$}};
        \draw[3d/hidden] (T) --(O);
    \end{tikzpicture}
\end{document}

有了r =2，我们得到