How can I draw an arc from point A -> B on a 3D sphere in TikZ?

Question

I'm basing myself on http://www.texample.net/tikz/examples/spherical-and-cartesian-grids/ and now I need to add a arc from point P to Q. What I was able to get so far:

\begin{tikzpicture}[scale=1,every node/.style={minimum size=1cm}]
%% some definitions

\def\R{4} % sphere radius

\def\angEl{25} % elevation angle
\def\angAz{-100} % azimuth angle
\def\angPhiOne{-110} % longitude of point P
\def\angPhiTwo{-45} % longitude of point Q
\def\angBeta{30} % latitude of point P and Q

%% working planes

\pgfmathsetmacro\H{\R*cos(\angEl)} % distance to north pole
\LongitudePlane[xzplane]{\angEl}{\angAz}
\LongitudePlane[pzplane]{\angEl}{\angPhiOne}
\LongitudePlane[qzplane]{\angEl}{\angPhiTwo}
    \LongitudePlane[nzplane]{\angEl}{-86}
\LatitudePlane[equator]{\angEl}{0}
\fill[ball color=white!10] (0,0) circle (\R); % 3D lighting effect
\coordinate (O) at (0,0);
\coordinate[mark coordinate] (N) at (0,\H);
\coordinate[mark coordinate] (S) at (0,-\H);

\DrawLongitudeCircle[\R]{\angPhiOne} % pzplane
\DrawLongitudeCircle[\R]{\angPhiTwo} % qzplane
\DrawLatitudeCircle[\R]{\angBeta}
\DrawLatitudeCircle[\R]{0} % equator
%labelling north and south
\node[above=8pt] at (N) {$\mathbf{N}$};
\node[below=8pt] at (S) {$\mathbf{S}$};
    \draw[-,dashed, thick] (N) -- (S);

%setup coordinates P and Q
\path[pzplane] (0:\R) coordinate (P);
\draw[->] (O) -- node[above=4pt] {$\overrightarrow{P}$} (P);
\path[qzplane] (\angBeta:\R) coordinate (Q);
\draw[->] (O) -- node[above=2pt] {$\overrightarrow{Q}$} (Q);
\path[nzplane] (153:\R) coordinate (N);
\draw[->,color=red] (O) -- node[right=2pt] {$\overrightarrow{N}$} (N);
\draw (P) arc (-110:-45:\R) (Q);    
\end{tikzpicture}

Is there a possibility to make an arc from P to Q with the center O?

Upon request, the full compilable sourcecode: http://pastebin.com/m809Jwp7

Answer 1

To give a correct answer, we need to define cross product and vector product (this work is done with metapost in cahier gutemberg 48 but it's in french)

I don't have enough time to define all these macros but it's possible to find a way to draw the arc. First we know that the arc PQ (blue) is in the plane OPQ and is a part of a circle of center O and radius OP. So I search a Coordinate system xyz with x=OP and y=OA'. A is a point of the equator of longitude = -20. Why ? because I want OP and OA radius of the equator and OP perpendicular at OA. Then I need to find A' of longitude-20 and latitude >30 but I need to calculate the value.

Update How to determine the latitude of A' ?

In the next pictures, H is the projection of Q on the plane (OPA). It's possible to calculate PH with two sides (OP=1 and OH=0.866) I find 1.001. Then the lines PH and OA have an intersection at the point I. Now i calculate OI=1.238 and PI=1.591. J is a point of the line OA' and I is the projection of J on the plane (OPA). P, Q, J are aligned and IJ= 0.795. IJ/OI=0.641=tan(32.7). The latitude of A' is 32.7. Now I can draw the circle of radius 1 that passes through P and A' with center O.

Now I need to draw the circle of center O and radius 1. The circle passes through P and A' but also by Q. I draw the diameter POP' and QOQ'.

Todo : Calculus to determine correctly the latitude of A', cross product to determine N'. A macro to place a point with longitude and latitude.

In my code, I redefined personal macro with names that I understand correctly.

\documentclass[11pt]{scrartcl}
\usepackage{tikz}
\usetikzlibrary{calc}
\tikzset{%
    add/.style args={#1 and #2}{
        to path={%
 ($(\tikztostart)!-#1!(\tikztotarget)$)--($(\tikztotarget)!-#2!(\tikztostart)$)%
  \tikztonodes},add/.default={.2 and .2}}
}  


\tikzset{%
  mark coordinate/.style={inner sep=0pt,outer sep=0pt,minimum size=2pt,
    fill=black,circle}%
}

\newcommand\pgfmathsinandcos[3]{%
  \pgfmathsetmacro#1{sin(#3)}%
  \pgfmathsetmacro#2{cos(#3)}%
}
\newcommand\LongitudePlane[2][current plane]{%
  \pgfmathsinandcos\sinEl\cosEl{\Elevation} % elevation
  \pgfmathsinandcos\sint\cost{#2} % azimuth
  \tikzset{#1/.estyle={cm={\cost,\sint*\sinEl,0,\cosEl,(0,0)}}}
}
\newcommand\LatitudePlane[2][current plane]{%
  \pgfmathsinandcos\sinEl\cosEl{\Elevation} % elevation
  \pgfmathsinandcos\sint\cost{#2} % latitude
  \pgfmathsetmacro\ydelta{\cosEl*\sint}
  \tikzset{#1/.estyle={cm={\cost,0,0,\cost*\sinEl,(0,\ydelta)}}} %
}
\newcommand\DrawLongitudeCircle[1]{
  \LongitudePlane{#1}
  \tikzset{current plane/.prefix style={scale=\R}}
  \pgfmathsetmacro\angVis{atan(sin(#1)*cos(\Elevation)/sin(\Elevation))} %
  \draw[current plane,thin,black]  (\angVis:1)     arc (\angVis:\angVis+180:1);
  \draw[current plane,thin,dashed] (\angVis-180:1) arc (\angVis-180:\angVis:1);
}%

\newcommand\DrawLatitudeCircle[1]{
  \LatitudePlane{#1}
  \tikzset{current plane/.prefix style={scale=\R}}
  \pgfmathsetmacro\sinVis{sin(#1)/cos(#1)*sin(\Elevation)/cos(\Elevation)}
  \pgfmathsetmacro\angVis{asin(min(1,max(\sinVis,-1)))}
  \draw[current plane,thin,black] (\angVis:1) arc (\angVis:-\angVis-180:1);
  \draw[current plane,thin,dashed] (180-\angVis:1) arc (180-\angVis:\angVis:1);
}%

\newcommand\DrawPointOnSphere[3]{%
\pgfmathsinandcos\sinLoM\cosLoM{#1}  
\pgfmathsinandcos\sinLaM\cosLaM{#2}
} 


\begin{document}
  \null\vfill
\begin{center}
  \begin{tikzpicture}
  \def\R{4} % sphere radius
  \def\Elevation{25} % elevation angle
  \def\angleLongitudeP{-110} % longitude of point P
  \def\angleLongitudeQ{-45} % longitude of point Q
  \def\angleLatitudeQ{30} % latitude  Q    ; 0 latitude of P 
  \def\angleLongitudeA{-20} % longitude of point A

  \pgfmathsetmacro\H{\R*cos(\Elevation)} % distance to north pole
  \LongitudePlane[PLongitudePlane]{\angleLongitudeP}
  \LongitudePlane[QLongitudePlane]{\angleLongitudeQ}
  \LongitudePlane[ALongitudePlane]{\angleLongitudeA}   

  \fill[ball color=white!10] (0,0) circle (\R); % 3D lighting effect
  \coordinate (O) at (0,0);
  \coordinate[] (N) at (0,\H);
  \coordinate[] (S) at (0,-\H);

  \DrawLongitudeCircle{\angleLongitudeP} % PLongitudePlane
  \DrawLongitudeCircle{\angleLongitudeQ} % QLongitudePlane
  \DrawLongitudeCircle{\angleLongitudeA} 
  \DrawLatitudeCircle{\angleLatitudeQ}
  \DrawLatitudeCircle{0} % equator
  \DrawLongitudeCircle{0}
  %setup coordinates P and Q
  \path[ALongitudePlane] (0:\R) coordinate (A);
  \path[ALongitudePlane] (32.5:\R) coordinate (A'); 
   \path[ALongitudePlane] (122.5:\R) coordinate (N');  
  \path[PLongitudePlane] (0:\R) coordinate (P);
  \draw[dashed,add= 1 and 0] (O) to  (P); 
  \path[QLongitudePlane] (\angleLatitudeQ:\R) coordinate (Q);
  \draw[dashed,add= 1 and 0] (O) to  (Q) ;
  \path[QLongitudePlane] (0:\R) coordinate (B);
  \draw [dashed] (O) --  (B) ;
  \draw [dashed] (O) --  (N) ;  

\foreach \v in {A,O,N,S,P,Q,A',B,N'} {\coordinate[mark coordinate] (v) at (\v);
\node [above] at (\v) {\v};} 
 \begin{scope}[ x={(P)}, y={(A')}, z={(N')}]     
          \draw[dashed,fill opacity=.3] circle (1);  
          \draw[blue] ( 0:1) arc (0:68:1) ;
          \draw[] ( 68:1) arc (68:115:1) ;
          \draw[] (-55:1) arc (-55:0:1);
          \draw[red,->](0,0,0)--(0,0,1); 
          \draw[red,->](0,0,0)--(0,1,0); 
          \draw[red,->](0,0,0)--(1,0,0);      
 \end{scope} 
\end{tikzpicture}   
\end{center}
\vfill 


\end{document}  

Answer 2

The trick is to rotate the coordinate system. This bit of code shows the great circle passing through your particular values for P and Q, as well as a blue arc from P to Q.

\begin{scope}[rotate around={30:(0,0)}]
    \DrawLatitudeCircle[\R]{11}
    \draw[current plane,blue,thick] (240:1) arc (240:310:1);
\end{scope}

(I dropped a couple of your drawing commands to make the result a bit clearer. Obviously all the parameters are specific to this problem and, less obviously, were chosen visually.)