I am trying to draw a spherical triangle $ABC$ and its polar counterpart $A'B'C'$ like in the picture
Is this possible using TiKz. There was a similar question in TikZ - Spherical triangles but without answer.
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Here is a possible way to do it. It has lot of calculations, but they are simple (I hope) and quite repetitive.
\documentclass {standalone}
\usepackage {tikz}
\usetikzlibrary{3d}
\usetikzlibrary{calc}
\usetikzlibrary{math}
% isometric axes
\pgfmathsetmacro\xx{1/sqrt(2)}
\pgfmathsetmacro\xy{1/sqrt(6)}
\pgfmathsetmacro\zy{sqrt(2/3)}
% some functions (cross products)
\tikzmath%
{%
function crossx(\mx,\my,\mz,\nx,\ny,\nz)
{% cross product, x coordinate, normailized
\pxx = \my\nz-\mz\ny;
\pyy = \mz\nx-\mx\nz;
\pzz = \mx\ny-\my\nx;
return {\pxx/sqrt(\pxx\pxx+\pyy\pyy+\pzz\pzz)};
};
function crossy(\mx,\my,\mz,\nx,\ny,\nz)
{% cross product, y coordinate, normailized
\pxx = \my\nz-\mz\ny;
\pyy = \mz\nx-\mx\nz;
\pzz = \mx\ny-\my\nx;
return {\pyy/sqrt(\pxx\pxx+\pyy\pyy+\pzz\pzz)};
};
function crossz(\mx,\my,\mz,\nx,\ny,\nz)
{% cross product, z coordinate, normailized
\pxx = \my\nz-\mz\ny;
\pyy = \mz\nx-\mx\nz;
\pzz = \mx\ny-\my\nx;
return {\pzz/sqrt(\pxx\pxx+\pyy\pyy+\pzz*\pzz)};
};
}
\newcommand{\greatcircle}[6] % pole x, y, z, color, two orientation factors (+1/-1)
{%
\coordinate (P) at (#1,#2,#3); % pole
\coordinate (N) at ($(0,0,0)!#61.25cm!(P)$); % these points are
\coordinate (S) at ($-1(N)$); % used to clip the
\coordinate (E) at ($(0,0,0)!-1.25cm!270:(P)$); % ellipses
\coordinate (W) at ($-1(E)$); % ...
\coordinate (NW) at ($(N)+(W)$);
\coordinate (NE) at ($(N)+(E)$);
\coordinate (SW) at ($(S)+(W)$);
\coordinate (SE) at ($(S)+(E)$);
\pgfmathsetmacro\ptheta{atan(#2/#1)} % pole, spherical coordinate theta
\pgfmathsetmacro\pphi {#5acos(#3)} % pole, spherical coordinate phi
\begin{scope}
\clip (W) -- (SW) -- (SE) -- (E) -- cycle;
\draw[rotate around z=\ptheta,rotate around y=\pphi,%
canvas is xy plane at z=0,#4] (0,0) circle (1);
\end{scope}
\begin{scope}
\clip (W) -- (NW) -- (NE) -- (E) -- cycle;
\draw[rotate around z=\ptheta,rotate around y=\pphi,%
canvas is xy plane at z=0,#4,densely dotted] (0,0) circle (1);
\end{scope}
}
\begin{document}
\begin{tikzpicture}[line cap=round,line join=round,scale=2,%
x={({-\xx cm,-\xy cm})},y={(\xx cm,-\xy cm)},z={(0 cm,\zy cm)}]
% points A, B, C in spherical coordinates
\def\atheta{55}
\def\aphi {30}
\def\btheta{10}
\def\bphi {85}
\def\ctheta{70}
\def\cphi {80}
% points A, B, C in cartesian coordinates
\pgfmathsetmacro\ax{cos(\atheta)sin(\aphi)}
\pgfmathsetmacro\ay{sin(\atheta)sin(\aphi)}
\pgfmathsetmacro\az{cos(\aphi)});
\pgfmathsetmacro\bx{cos(\btheta)sin(\bphi)}
\pgfmathsetmacro\by{sin(\btheta)sin(\bphi)}
\pgfmathsetmacro\bz{cos(\bphi)});
\pgfmathsetmacro\cx{cos(\ctheta)sin(\cphi)}
\pgfmathsetmacro\cy{sin(\ctheta)sin(\cphi)}
\pgfmathsetmacro\cz{cos(\cphi)});
% polar points P, Q, R in cartesian coordinates
\pgfmathsetmacro\px{crossx(\ax,\ay,\az,\bx,\by,\bz)}
\pgfmathsetmacro\py{crossy(\ax,\ay,\az,\bx,\by,\bz)}
\pgfmathsetmacro\pz{crossz(\ax,\ay,\az,\bx,\by,\bz)}
\pgfmathsetmacro\qx{crossx(\cx,\cy,\cz,\ax,\ay,\az)}
\pgfmathsetmacro\qy{crossy(\cx,\cy,\cz,\ax,\ay,\az)}
\pgfmathsetmacro\qz{crossz(\cx,\cy,\cz,\ax,\ay,\az)}
\pgfmathsetmacro\rx{crossx(\bx,\by,\bz,\cx,\cy,\cz)}
\pgfmathsetmacro\ry{crossy(\bx,\by,\bz,\cx,\cy,\cz)}
\pgfmathsetmacro\rz{crossz(\bx,\by,\bz,\cx,\cy,\cz)}
% triangles
\greatcircle{\ax}{\ay}{\az}{red} { 1}{1}
\greatcircle{\bx}{\by}{\bz}{red} { 1}{1}
\greatcircle{\cx}{\cy}{\cz}{red} { 1}{1}
\greatcircle{\px}{\py}{\pz}{blue}{-1}{1}
\greatcircle{\qx}{\qy}{\qz}{blue}{ 1}{1}
\greatcircle{\rx}{\ry}{\rz}{blue}{-1}{1}
% sphere and axes
\draw (0,0,0) circle (1 cm);
\draw[gray,dashed] (0,0,0) -- (1,0,0);
\draw[gray,dashed] (0,0,0) -- (0,1,0);
\draw[gray,dashed] (0,0,0) -- (0,0,1);
\draw[gray,-latex] (1,0,0) -- (1.5,0,0) node (X) [left] {$x$};
\draw[gray,-latex] (0,1,0) -- (0,1.5,0) node (Y) [right] {$y$};
\draw[gray,-latex] (0,0,1) -- (0,0,1.5) node (Z) [above] {$z$};
% points
\fill[blue] (\ax,\ay,\az) circle (0.5pt) node [above] {$A$};
\fill[blue] (\bx,\by,\bz) circle (0.5pt) node [below] {$B$};
\fill[blue] (\cx,\cy,\cz) circle (0.5pt) node [below left] {$C$};
\fill[red] (\px,\py,\pz) circle (0.5pt) node [below right] {$C'$};
\fill[red] (\qx,\qy,\qz) circle (0.5pt) node [left] {$B'$};
\fill[red] (\rx,\ry,\rz) circle (0.5pt) node [above left] {$A'$};
\fill[gray] (0,0,0) circle (0.5pt) node [left] {$O$};
\end{tikzpicture}
\end{document}
A little explanation. First, for simplicity I am working with a unit sphere and the data points are provided in spherical coordinates. Then the key here are the poles. Given two points on the sphere, we are looking for the great circle that passes through them, and for the point that would be the north pole if said circumference were the equator. Let A(ax,ay,az) and B(bx,by,bz) be the points, then we can compute the cross product (sorry, there is no MathJax):
| i j k |
v = A x B = | ax ay az |
| bx by bz |
The above vector v will give us the direction to find the north pole, and if we normalize it (dividing for its module) the new vector u will point exactly at the pole (C' in my drawing). But that is not all, because this vector u in spherical coordinates will provide the angles for a new canvas in tikz 3d that contains the equator (the great circle which passes through A and B). Due to the perspective we see that circle as an ellipse and its semi-major axis is perpendicular to the projection of u, so then again, the vector can be used to clip the ellipses and separate the visible and non-visible parts.
Juan Castaño
- 28,426
tikz-3dplot-circleofsphereyou can answer both questions. – Apr 01 '21 at 14:18