TikZ: Understanding the usage of calc library

Your approach does not work because you try to use { and } where you should use ($ and $). You can definitely do that without auxiliary coordinates and actually even without calc.

\documentclass{article}
\usepackage{tikz,mathtools,amssymb}
\usetikzlibrary{shapes,arrows,positioning,calc}

\begin{document}

\tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style  = {coordinate}, 
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}

\begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
     \node [sum] (sum2) {};
    \node [block, right = 1cm of sum2](ractuator){$\frac{2}{s+2}$};
    \node [block, right = 1cm of ractuator,] (vdynamics) {$\frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
    \node [block, right = 1cm of vdynamics,] (integrator) {$\frac{1}{s}$};
    \node [output, right = 1.5cm of integrator] (output) {};
    \node [block] (yaw) at 
    ($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
    %
    \draw [->] (ractuator) -- (vdynamics);
    \draw [->] (vdynamics) -- (integrator);
    \draw [->] (integrator) -- node[name=heading]{$\Psi(s)$} (output)
    coordinate[midway] (aux);
    \draw (aux) |- (yaw) -| (sum2);
\end{tikzpicture}

\bigskip

\begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
     \node [sum] (sum2) {};
    \node [block, right = 1cm of sum2](ractuator){$\frac{2}{s+2}$};
    \node [block, right = 1cm of ractuator,] (vdynamics) {$\frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
    \node [block, right = 1cm of vdynamics,] (integrator) {$\frac{1}{s}$};
    \node [output, right = 1.5cm of integrator] (output) {};
    \node [block] (yaw) at 
    ([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
    %
    \draw [->] (ractuator) -- (vdynamics);
    \draw [->] (vdynamics) -- (integrator);
    \draw [->] (integrator) -- node[name=heading]{$\Psi(s)$} (output)
    coordinate[midway] (aux);
    \draw (aux) |- (yaw) -| (sum2);
\end{tikzpicture}

\end{document}

I don't know how complex expression can be understood by calc but instead of trying to understand how to write such expression, I think it's easier to use an auxiliar coordinate and solve the problem:

\documentclass{article}
\usepackage{tikz,mathtools,amssymb}
\usetikzlibrary{shapes,arrows,positioning,calc}

\begin{document}

\tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style  = {coordinate}, 
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}

\begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
     \node [sum] (sum2) {};
    \node [block, right = 1cm of sum2](ractuator){$\frac{2}{s+2}$};
    \node [block, right = 1cm of ractuator,] (vdynamics) {$\frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
    \node [block, right = 1cm of vdynamics,] (integrator) {$\frac{1}{s}$};
    \node [output, right = 1.5cm of integrator] (output) {};
    \coordinate (aux) at ($(integrator.east)!.5!(output)$);
    \node [block] (yaw) at ([yshift=-2cm]$(aux)!0.5!(sum2)$) {C};
    \draw (aux) |- (yaw);
    \draw (yaw)-|(sum2);
    %
    \draw [->] (ractuator) -- (vdynamics);
    \draw [->] (vdynamics) -- (integrator);
    \draw [->] (integrator) -- node[name=heading]{$\Psi(s)$} (output);
    \end{tikzpicture}

\end{document}

Here, to simplify the code, I'll replace (integrator) with (A), (output) with (B) and (sum2) with (C). There is two things not right with ($ (A) + 0.5*{ (B)-(A) }!0.5!(C) $).

• First, I don't think you can use {}, with the calc package, for the coordinate part. For me, it only works for with scalar. So ($ {2+2}*(A) $) will compute, but not ($ 2*{(A)+(B)} $) (or am I wrong?)

• The second thing is that this formula doesn't seem to correspond to the point you want. I kind of get that you want to start from (A), "move" to the middle of [AB] and continue like that, but you mix relative (B-A) and absolute positioning (C). One right formula would have been ($ { (A) + 0.5*{(B)-(A)} }!0.5!(C) $). But because tikz can't do the computation, you'll have to give the expanded formula: ($ .25*(A) + .25*(B) + .5*(C)$).

One other way to do it is ($ (A) !.5! (B) !.5! (C) $). Here, we take the middle of (A) and (B), and then the middle of the result and (C).

I hope this will answer your interrogations!

You can test the three solutions here (the last one with temporary coordinate):

\documentclass[tikz,margin=10pt]{standalone}
\usetikzlibrary{calc}

\begin{document}

\begin{tikzpicture}[line width=1]

  \draw[black!10] (0,0) grid (4,4);
  \node (A) at (1,1) {A};
  \node (B) at (3,1) {B};
  \node (C) at (2,3) {C};

  %% 1
  \draw[red] ($ (A) !.5! (B) !.5! (C) $) circle (.05);
  %% 2
  \draw[orange] ($ .25*(A) + .25*(B) + .5*(C) $) circle (.1);
  %% 3
  \coordinate (foo) at ($ (A) !.5! (B) $);
  \draw[yellow] ($ (foo) !.5! (C) $) circle (.15);

\end{tikzpicture}
\end{document}

which gives