16

I am trying to calculate the center of mass of a few coordinates and the logical way of adding up and then dividing results in an error.

\documentclass[a4paper,14pt]{scrartcl}
\usepackage{tikz}
\usepackage{pgfplots,pgfplotstable}
\usetikzlibrary{calc}
\usepackage{calc}
\pgfplotsset{compat=newest}

\usepackage[active,tightpage]{preview}
\setlength\PreviewBorder{2pt}


\begin{document}
\begin{preview}

\begin{tikzpicture}
\coordinate (m1) at (0,0);
\filldraw [blue] (m1) circle (5);
\draw [blue] (m1)  circle (75);
\draw [blue] (m1)  circle (200);

\coordinate (f1) at ($(m1) + (0,10)$);
\filldraw [red] (f1) circle (3);

\coordinate (m2) at (45:150);
\filldraw[blue] (m2) circle (5);
\draw[blue] (m2) circle (15);

\coordinate (m3) at ($(m2) +(10:30)$);
\filldraw[blue] (m3) circle (5);

\coordinate (m4) at ($(m1) +(200:65)$);
\filldraw[blue] (m4) circle (5);

\coordinate (com) at ($ ((m1)+(m2)+(m3)+(m4)+(f1))/5 $);



\end{tikzpicture}
\end{preview}
\end{document}

I couldn't find any examples like this online. Adding up the five coordinates work fine. Thanks for the help!

17

1st option: You may use a multiplication instead of a division (the sum of the weights must be one):

\coordinate (com) at ($.2*(m1)+.2*(m2)+.2*(m3)+.2*(m4)+.2*(f1)$);

or

\pgfmathsetmacro\weight{1/5}
\coordinate (com) at ($\weight*(m1)+\weight*(m2)+\weight*(m3)+\weight*(m4)+\weight*(f1) $);

2nd option: You may use barycentric cs (the barycentric coordinate system) with weights:

\coordinate (com) at (barycentric cs:m1=.2,m2=.2,m3=.2,m4=.2,f1=.2);

or

\coordinate (com) at (barycentric cs:m1=1,m2=1,m3=1,m4=1,f1=1);

A complete example:

enter image description here

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\coordinate (m1) at (0,0);
\filldraw [blue] (m1) circle (5);
\draw [blue] (m1)  circle (75);
\draw [blue] (m1)  circle (200);

\coordinate (f1) at ($(m1) + (0,10)$);
\filldraw [red] (f1) circle (3);

\coordinate (m2) at (45:150);
\filldraw[blue] (m2) circle (5);
\draw[blue] (m2) circle (15);

\coordinate (m3) at ($(m2) +(10:30)$);
\filldraw[blue] (m3) circle (5);

\coordinate (m4) at ($(m1) +(200:65)$);
\filldraw[blue] (m4) circle (5);

\coordinate (com) at ($.2*(m1)+.2*(m2)+.2*(m3)+.2*(m4)+.2*(f1) $);
\filldraw[green] (com) circle (5);

\coordinate (com1) at (barycentric cs:m1=.2,m2=.2,m3=.2,m4=.2,f1=.2);
\filldraw[black] (com1) circle (3);
\end{tikzpicture}
\end{document}
11

You cannot perform arithmetic (division) on node names. You have to deal with their x and y part separately, which would require the use of the let ... in construct.

It is much easier to use the barycentric coordinate system, as follows:

\coordinate (com) at (barycentric cs:m1=0.5,m2=0.5,m3=0.5,m4=0.5,f1=0.5);
\filldraw[yellow] (com) circle (5);

and get:

Result

For completness, this is a solution which uses the aforementioned let ... in syntax:

\path let
   \p1 = (m1),
   \p2 = (m2),
   \p3 = (m3),
   \p4 = (m4),
   \p5 = (f1),
   \n1 = {(\x1+\x2+\x3+\x4+\x5)/5},
   \n2 = {(\y1+\y2+\y3+\y4+\y5)/5}
 in coordinate (com) at (\n1, \n2);

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