Draw a sphere whose latitude is the base circle of a cone

Question

I want to draw a sphere whose latitude is the circle of the base of a cone.

This is my code

\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}

You will need to do some math if you want to draw this manually in 2D. Here the sphere and the cone base share the same diameter, hence the sphere intersects the cone. You have to define the diameter of the cone base and the diameter of the shpere, in the first place. Take a look at this question and its answers, which may give you hints on how to start. — SebGlav, Jul 24 '21 at 09:38

SebGlav · Accepted Answer · 2021-07-24T12:25:31.230

Here's a starting point with the basic computing:

\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}

EDIT

Following a doubt of OP in the comments, when the radius of the cone base is too small, the ellipse is too fat and doesn't reflect the exact reality of a 3D view. It's then possible to change the width of the ellipse to correct this issue:

% 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);

Here's what you have with r=3 and the factor is 0.1.

+1, Very good! I was doing essentially the same, but you were faster than me!!! :-P — Juan Castaño, Jul 24 '21 at 10:44
@SebGlav. I tried your code with \def\r{3}, the result seems not nice. — minhthien_2016, Jul 24 '21 at 12:16
Of course, because, with a too small radius for the cone base, the ellipse is too fat. Try and change the 0.2 factor by 0.1 in both commands like (0:-180:\r cm and 0.2*\r cm). — SebGlav, Jul 24 '21 at 12:21

minhthien_2016 · Answer 2 · 2021-08-07T05:40:39.720

You can use this code. I use 3dtools to draw it.

\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}

If you use R=5;r=2.5, you get the picture like this.

\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}

With r =2, we get

Draw a sphere whose latitude is the base circle of a cone

2 Answers2

EDIT