4

I am trying to draw a cylinder inside a sphere like this picture enter image description here

I tried:

\documentclass[12pt,border=3mm]{standalone}
\usepackage{fouriernc}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usepackage{tikz-3dplot}
\usetikzlibrary{calc,backgrounds}
\begin{document}
\tdplotsetmaincoords{70}{110}
\def\r{{2*sqrt(3)}}
\def\d{-60}
\begin{tikzpicture}
[scale=1,tdplot_main_coords]
\path
coordinate (O) at (0,0,0)
coordinate (I) at  (0,0,2)
coordinate (A') at  (0,\r,4)
coordinate (A) at  (0,\r,0);
\coordinate (B) at ($(O) + (\d:{2*sqrt(3)} and \r)$);
\coordinate (B') at ($(B)+(0,0,4)$);
\coordinate (O') at ($(O)+(0,0,4)$);
\draw[dashed] (A)--(A') (B) --(B') (O)--(O') (O)--(A) (I) --(A);
\foreach \v/\position in {I/left,O/below,O'/above,A/below,B/below,A'/left,B'/left} {
    \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
}
\begin{scope}[tdplot_screen_coords, on background layer]
\pgfmathsetmacro{\R}{4}%
%\pgfmathsetmacro{\r}{{2*sqrt(3)}}%
\fill[ball color=orange!70, opacity=1.0] (I) circle (\R);
\end{scope}
\tkzMarkRightAngle[size = 0.3](I,O,A);
\draw [thick] (B) arc (\d:90:\r);
\draw [thick, dashed] (A) arc (90:310:\r);
\draw [thick] (B') arc (\d:90:\r);
\draw [thick, dashed] (A') arc (90:310:\r);
\end{tikzpicture}
\end{document}

I got enter image description here

The cylinder look very bad. How can I repair it?

  • 1
    What is bad with your solution? For me its quite the same. What is missing? – Bobyandbob Nov 11 '17 at 17:03
  • Note that A and B are not the edges of the cylinder (x=\pm \r). – John Kormylo Nov 12 '17 at 15:39
  • @JohnKormylo I don't understand you Comment. Please help me. – minhthien_2016 Nov 13 '17 at 1:14
  • It turns out that A and B were NOT in the right place, albeit only about 2 degrees off. – John Kormylo Nov 13 '17 at 15:19
5

Points (A) (B) (A') and (B') are not the edges of the cylinder, but rather the points where the visible edge of the sphere (circle) and the visible edge of of the cylinder (ellipse) intersect, which depend on the viewing angles.
Points (C) and (D) represent the visible right and left edges.

math

\documentclass[12pt,border=3mm]{standalone}
\usepackage{fouriernc}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usepackage{tikz-3dplot}
\usetikzlibrary{calc,backgrounds}
\begin{document}
\def\myangle{70}%
\tdplotsetmaincoords{\myangle}{0}%

\begin{tikzpicture}[scale=1,tdplot_main_coords]
\pgfmathsetmacro{\r}{2*sqrt(3)}% sphere radius=4, cylendar height=4
\path% common coordinates
  (0,0,-2) coordinate (O)
  (0,0,0) coordinate (I)
  (0,0,2) coordinate (O')
  (O) ++(0:\r) coordinate (C)% right edge
  (O) ++(180:\r) coordinate (D);% left edge
\draw[dashed,thick] (C)-- ++(0,0,4) (D)-- ++(0,0,4);
\draw[dashed] (O)--(O');
\begin{scope}[tdplot_screen_coords, on background layer]
      \fill[ball color=orange!70, opacity=1.0] (I) circle (4);
\end{scope}

\pgfmathsetmacro{\mynumer}{2*cos(\myangle)}
\pgfmathsetmacro{\mydenom}{\r*sin(\myangle)}
\pgfmathsetmacro{\quadrant}{ifthenelse(abs(\mynumer)<abs(\mydenom), 0,
  ifthenelse(\mynumer>0, 1, 2))}% 0=side, 1=top, 2=bottom

\ifcase\quadrant
  \pgfmathsetmacro{\intercept}{asin(\mynumer/\mydenom)}
  \path
    (O) ++(-\intercept:\r) coordinate (A)
    (O) ++(-180+\intercept:\r) coordinate (B)
    (O') ++(\intercept:\r) coordinate (A')
    (O') ++(180-\intercept:\r) coordinate (B');
  \draw [thick] (B) arc (-180+\intercept:-\intercept:\r);
  \draw [thick, dashed] (A) arc (-\intercept:180+\intercept:\r);
  \draw [thick] (B') arc (-180-\intercept:\intercept:\r);
  \draw [thick, dashed] (A') arc (\intercept:180-\intercept:\r);

  \foreach \v/\position in {I/left,O/below,O'/above,A/below,B/below,A'/left,B'/left} {
    \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
  }
\or% top
  \path
    (C) coordinate (A)
    (D) coordinate (B)
    (A) ++(0,0,4) coordinate (A')
    (B) ++((0,0,4) coordinate (B');
  \draw [thick] (O') circle (\r);
  \draw [thick, dashed] (O) circle (\r);

  \foreach \v/\position in {I/left,O/below,O'/above,A/below,B/below,A'/left,B'/left} {
    \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
  }
\or% bottom
  \path
    (C) coordinate (A)
    (D) coordinate (B)
    (A) ++(0,0,4) coordinate (A')
    (B) ++((0,0,4) coordinate (B');
  \draw [thick,dashed] (O') circle (\r);
  \draw [thick] (O) circle (\r);

  \foreach \v/\position in {I/left,O/below,O'/above,A/below,B/below,A'/left,B'/left} {
    \draw[draw =black, fill=black] (\v) circle (1.2pt) node [\position=0.2mm] {$\v$};
  }
\fi

\draw[dashed] (O)--(A) (I) --(A);
\tkzMarkRightAngle[size = 0.3](I,O,A);
\end{tikzpicture}
\end{document}

demo

  • Is there a code true for every cylinder? – minhthien_2016 Nov 15 '17 at 15:44
  • Yes and no. You need one solution for small \phi and another for large \phi. Then you need to calculate \theta using pgfmath (I just used a calculator). – John Kormylo Nov 15 '17 at 15:48
  • I shall try to write a general code. When I see difficult, I need your help. Can you help me? – minhthien_2016 Nov 15 '17 at 16:12
  • The tricky bit is the test for h \cot\phi/r_x. pgfmath has a ifthenelse command, but it basically only affects \pgfmathresult. LaTeX has \ifdim\len<1pt\relax ... \else ... \fi construction but you will need to append pt to the number. – John Kormylo Nov 15 '17 at 16:27
  • New thought: use \pgfmathparse or \pgfmathsetmacro to assign 0,1,2 and possibly 3 then use \ifcases\pgfmathresult\relax ... \else ... \else ... \fi to implement the images. BTW, \relax means "Stop looking for more digits. This number is done." – John Kormylo Nov 16 '17 at 13:55

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