5

I asked myself: Is there a simple way to transform the solution "small circles of a sphere" from @John Kormylo (code below) into cones, placed in the sphere by the two angles theta and phi?

Hint: I saw some cone-solutions here; but I think they have not been complicated placed by the two spheric angles theta and phi.

Hint: I want to draw this picture (with 8 cones):

enter image description here

Probably one cone would be sufficient, so that I can do the rest if necessary.

MWE from @JohnKormylo:

enter image description here

\documentclass[margin=5mm, tikz]{standalone}
\usepackage{mathtools}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{arrows,calc,backgrounds}
\begin{document}

\pgfmathsetmacro{\R}{3} %  
\pgfmathsetmacro{\a}{1.5} %  

\pgfmathsetmacro{\r}{sqrt(\R*\R-\a*\a} %  
%\pgfmathsetmacro{\Alpha}{atan(\r/\a)}   
\pgfmathsetmacro{\Alpha}{acos(\a/\R)} %  

\pgfkeys{/tikz/savevalue/.code 2 args={\global\edef#1{#2}}}

\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[
tdplot_main_coords,
>=latex, font=\footnotesize,
]

\coordinate[label=$Z$] (Z) at (0,0,0); 


\pgfmathsetmacro{\Teta}{90} %  measured to the z-axis
\pgfmathsetmacro{\Phi}{50} %   measured to the x-axis

\tdplotsetrotatedcoords{50}{90}{0}
\begin{scope}[tdplot_rotated_coords]
  \coordinate[label=$A$] (A) at (0,0,\R); 
  \coordinate[label=$M$] (M) at (0,0,\a); 
  \draw[red, thick] (M) circle[radius=\r];
\end{scope}

\draw[thick] (Z) -- (A);
\draw[red, thick] (Z) -- (M);

% Point P of direction vector p
\pgfmathsetmacro{\xP}{\R*sin(\Teta-\Alpha)*cos(\Phi)} % 
\pgfmathsetmacro{\yP}{\R*sin(\Teta-\Alpha)*sin(\Phi)} % 
\pgfmathsetmacro{\zP}{\R*cos(\Teta-\Alpha)} % 
\coordinate[label=$P$] (P) at (\xP,\yP,\zP); 
\draw[thick] (Z) -- (P);
\draw[->] (M) -- (P);

\path let              
\p0 = (M), % Center
\p1 = (P),
\n1 = {veclen(\y1-\y0,\x1-\x0)},    \n2={atan2(\y1-\y0,\x1-\x0)}
in    [savevalue={\Radius}{\n1}, savevalue={\angle}{\n2}];
\pgfmathsetmacro{\RadiusP}{\Radius/28.4528} % wipe of 'pt' 

% Point Q of direction vector q
\pgfmathsetmacro{\xQ}{\R*sin(\Teta)*cos(\Phi-\Alpha)} % 
\pgfmathsetmacro{\yQ}{\R*sin(\Teta)*sin(\Phi-\Alpha)} % 
\pgfmathsetmacro{\zQ}{\R*cos(\Teta)} % 
\coordinate[label=$Q$] (Q) at (\xQ,\yQ,\zQ); 
\draw[thick] (Z) -- (Q);
\draw[->] (M) -- (Q);

\path let              
\p0 = (M), % Center
\p1 = (Q),
\n1 = {veclen(\y1-\y0,\x1-\x0)},    \n2={atan2(\y1-\y0,\x1-\x0)}
in    [savevalue={\Radius}{\n1}, savevalue={\angle}{\n2}];
\pgfmathsetmacro{\RadiusQ}{\Radius/28.4528} % wipe of 'pt' 

%OLD      
% 3D Small Circle
%\foreach \t in {0,...,360}{
%\pgfmathsetmacro{\rp}{cos(\t)*\r/\RadiusP} %  
%\pgfmathsetmacro{\rq}{sin(\t)*\r/\RadiusQ} %  
%\coordinate[label=$$] (X) at ($(M)+\rp*(P)-\rp*(M)+\rq*(Q)-\rq*(M)$); 
%\draw[red] (X) circle (1pt); 
%}

% Sphere
\begin{scope}[tdplot_screen_coords, on background layer]
\fill[ball color= gray!20, opacity = 0.3] (Z) circle (\R); 
\end{scope}


\begin{scope}[-latex, shift={(Z)}, xshift=0*2.1*\R cm, yshift=0*0.1*\R cm]
\foreach \P/\s/\Pos in {(5,0,0)/x/right, (0,5,0)/y/below, (0,0,5)/z/right} 
\draw[] (0,0,0) -- \P node (\s) [\Pos, pos=0.9,inner sep=2pt]{$\s$};

\node[above=1cm, align=left, font=\normalsize] at (z){Equation of a 3D-circle: \\
$\vec{x}  = \vec{m} + r \cos(t) \cdot \vec{p} + r \sin(t) \cdot \vec{q}
~~\text{(with $t = 0\dots 2\pi$)}$
};
\end{scope}

\end{tikzpicture}
\end{document} 
9

This is not a full answer in the sense that the following does not work "out of the box" for arbitrary view angles. The bases of the cones, i.e. the circles, are drawn in appropriate planes, where the respective rotation angles are determined with the macro \RotationAnglesForPlaneWithNormal, which is explained here. The tricky part is to figure out where the boundaries of the cone attach to the circles. This requires either to determine intersections and to distinguish several cases, or an analytic computation. The good news is that one can infer the angles from the slopes that the coordinate axes have in the local planes (which are determined in the scope with \pgftransformreset), the bad news is that a fully automatic solution requires considerable effort (distinguish more cases) so I just made some choices by hand. Also the ordering 0,1,2,3 works out here "by accident", if you change the view angles too much this is no longer the appropriate ordering. However, for this configuration it works. The strange list \LstNormals just contains the vertices of a tetrahedron, and I got them from Wikipedia.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot} 
\usetikzlibrary{intersections}
\newcommand{\RotationAnglesForPlaneWithNormal}[5]{%\typeout{N=(#1,#2,#3)}
    \pgfmathtruncatemacro{\itest}{ifthenelse(abs(#3)==1,0,1)}
    \ifnum\itest=0
            \xdef#4{0}   
            \xdef#5{0}
    \else
    \foreach \XS in {1,-1}
    {\foreach \YS in {1,-1}
        {\pgfmathsetmacro{\mybeta}{\XS*acos(#3)} 
            \pgfmathsetmacro{\myalpha}{\YS*acos(#1/sin(\mybeta))} 
            \pgfmathsetmacro{\ntest}{abs(cos(\myalpha)*sin(\mybeta)-#1)%
                +abs(sin(\myalpha)*sin(\mybeta)-#2)+abs(cos(\mybeta)-#3)}
            \ifdim\ntest pt<0.1pt
            \xdef#4{\myalpha}   
            \xdef#5{\mybeta}
            \fi
    }}
    \fi
} 
\begin{document}
\tdplotsetmaincoords{110}{60}
\begin{tikzpicture}[tdplot_main_coords]
\xdef\LstNormals{{{sqrt(8/9), 0, -1/3},%
{-sqrt(2/9), sqrt(2/3), -1/3},%
{-sqrt(2/9), -sqrt(2/3), -1/3},%
{0, 0, 1}}}
\pgfmathsetmacro{\R}{3} %  
\pgfmathsetmacro{\a}{1.5} %  
\pgfmathsetmacro{\r}{sqrt(\R*\R-\a*\a} %  
 \path (0,0,0) coordinate (O);
 \foreach \myind in {0,1,2,3}
 {\pgfmathsetmacro{\myNx}{\LstNormals[\myind][0]}
  \pgfmathsetmacro{\myNy}{\LstNormals[\myind][1]}
  \pgfmathsetmacro{\myNz}{\LstNormals[\myind][2]}
  \RotationAnglesForPlaneWithNormal{\myNx}{\myNy}{\myNz}{\tmpalpha}{\tmpbeta} 
  \typeout{\myNx,\tmpalpha,\tmpbeta}
  \tdplotsetrotatedcoords{\tmpalpha}{\tmpbeta}{0}
  \begin{scope}[tdplot_rotated_coords,canvas is xy plane at z=\r,local bounding
  box=loc]
   \path[name path=circle] (0,0) circle[radius=\a];
   \path[overlay,name path=test] (0,0) -- (O);
   \path (1,0) coordinate (Xloc) (0,1) coordinate (Yloc) (0,0) coordinate (Oloc);
   \begin{scope}
   \pgftransformreset
   \path let \p1=($(Xloc)-(Oloc)$),\p2=($(Yloc)-(Oloc)$),
   \n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)}
   in (Xloc) -- (Oloc) -- (Yloc) (Oloc) node{\myind}
   \pgfextra{\xdef\myxi{\n1}\xdef\myeta{\n2}};
   \end{scope}
   \path[name intersections={of=circle and test,total=\iNum}]
   \pgfextra{\xdef\iNum{\iNum}};
   \ifnum\iNum>0
    \ifnum\myind=1
    \draw[fill=blue] (-\myxi+90:\a) -- (O) -- (-\myeta-90:\a);
    \else
    \draw[fill=blue] (-\myxi+90:\a) -- (O) -- (-\myxi-90:\a);
    \fi
    \draw[fill=blue!30](0,0) circle[radius=\a];
   \else
    \draw[fill=blue](0,0) circle[radius=\a];
   \fi
  \end{scope}
 }
 \path[ball color=gray,opacity=0.2,tdplot_screen_coords] (O) circle[radius=\R];
\end{tikzpicture}
\end{document}

enter image description here

Alternative: One can let TikZ numerically find the contour, see here.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot} 
\usetikzlibrary{intersections}
\newcommand{\RotationAnglesForPlaneWithNormal}[5]{%\typeout{N=(#1,#2,#3)}
    \pgfmathtruncatemacro{\itest}{ifthenelse(abs(#3)==1,0,1)}
    \ifnum\itest=0
            \xdef#4{0}   
            \xdef#5{0}
    \else
    \foreach \XS in {1,-1}
    {\foreach \YS in {1,-1}
        {\pgfmathsetmacro{\mybeta}{\XS*acos(#3)} 
            \pgfmathsetmacro{\myalpha}{\YS*acos(#1/sin(\mybeta))} 
            \pgfmathsetmacro{\ntest}{abs(cos(\myalpha)*sin(\mybeta)-#1)%
                +abs(sin(\myalpha)*sin(\mybeta)-#2)+abs(cos(\mybeta)-#3)}
            \ifdim\ntest pt<0.1pt
            \xdef#4{\myalpha}   
            \xdef#5{\mybeta}
            \fi
    }}
    \fi
} 
\begin{document}
\tdplotsetmaincoords{110}{60}
\begin{tikzpicture}[tdplot_main_coords]
\xdef\LstNormals{{{sqrt(8/9), 0, -1/3},%
{-sqrt(2/9), sqrt(2/3), -1/3},%
{-sqrt(2/9), -sqrt(2/3), -1/3},%
{0, 0, 1}}}
\pgfmathsetmacro{\R}{3} %  
\pgfmathsetmacro{\a}{1.5} %  
\pgfmathsetmacro{\r}{sqrt(\R*\R-\a*\a} %  
 \path (0,0,0) coordinate (O);
 \foreach \myind in {0,1,2,3}
 {\pgfmathsetmacro{\myNx}{\LstNormals[\myind][0]}
  \pgfmathsetmacro{\myNy}{\LstNormals[\myind][1]}
  \pgfmathsetmacro{\myNz}{\LstNormals[\myind][2]}
  \RotationAnglesForPlaneWithNormal{\myNx}{\myNy}{\myNz}{\tmpalpha}{\tmpbeta} 
  %\typeout{\myNx,\tmpalpha,\tmpbeta}
  \tdplotsetrotatedcoords{\tmpalpha}{\tmpbeta}{0}
  \begin{scope}[tdplot_rotated_coords,canvas is xy plane at z=\r,local bounding
  box=loc]
   \path[name path=circle] (0,0) circle[radius=\a];
   \path[overlay,name path=test] (0,0) -- (O);
   \path (1,0) coordinate (Xloc) (0,1) coordinate (Yloc) (0,0) coordinate (Oloc);
   \path[name intersections={of=circle and test,total=\iNum}]
   \pgfextra{\xdef\iNum{\iNum}};
   \ifnum\iNum>0
    \begin{scope}
     \pgftransformreset 
     \path let \p1=($(Oloc)-(O)$),\n1={mod(720+atan2(\y1,\x1),360)} in
      \pgfextra{\xdef\oldmax{\n1}\xdef\oldmin{\n1}};
    \end{scope}
    \typeout{\myind,\oldmax}
    \foreach \XX in {0,1,...,359}       
    {\path ($(\XX:\r)-(O)$) coordinate (aux1) ($(\XX:\r)-(Oloc)$) coordinate
    (aux2);
    \pgftransformreset
    \path let \p1=(aux1),%\p2=(aux2),
    \n1={atan2(\y1,\x1)} in
    \pgfextra{\pgfmathtruncatemacro{\itest}{ifthenelse(sin(\n1-\oldmin)<0,0,1)}
     \ifnum\itest=0
     \xdef\oldmin{\n1}
     \xdef\oldanA{\XX}
     \fi
     \pgfmathtruncatemacro{\itest}{ifthenelse(sin(\oldmax-\n1)<0,0,1)}
     \ifnum\itest=0
     \xdef\oldmax{\n1}
     \xdef\oldanB{\XX}
     \fi};
     }
    \draw[fill=blue] (\oldanA:\a) -- (O) -- (\oldanB:\a);
    \draw[fill=blue!30](0,0) circle[radius=\a];
   \else
    %\message{\myind: no intersections}
    \draw[fill=blue](0,0) circle[radius=\a];    
   \fi
  \end{scope}
 }
 \path[ball color=gray,opacity=0.4,tdplot_screen_coords] (O) circle[radius=\R];
\end{tikzpicture}
\end{document}

enter image description here

  • @ user121799 Mhh, I have something like \foreach \w in {5,10, ..., 360} \draw [red] (M) -- + (\w: \ r) - (Z); tested (where M is the center of the circle, r the circle radius and Z the origin of the coordinates). This is more of an approximation, but could work. – cis Aug 8 at 19:08
  • @cis Well, this still requires you to apply 3d ordering somehow. And the "somehow" is the tricky part. – user121799 Aug 8 at 19:13
  • @ user121799 Yes, that's right... – cis Aug 8 at 19:14
  • @cis It is actually not particularly hard to get candiate angles. – user121799 Aug 8 at 19:51
  • @cis Your proposal can also be made work by just numerically finding the right attachment points. An example of this kind can be found here. – user121799 Aug 8 at 23:41

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