TikZ style that shifts a path perpendicular

Question

Is it possible to define a TikZ style that shifts a path perpendicularly? I.e., the resulting path should be obtained from the given path by shifting every point, say, 0.1cm to the right, orthogonal to the direction of the path at this point.

This should work for arbitrary paths (not just straight lines). Also, I do not just want to draw a line along the shifted path, but also do clipping or filling, etc. So using the "raise=..." key together with a decoration as described in "perpendicular shift" for paths won't work I think.

Answer 1

If I understood your question correctly, this is natively possible with TikZ's double option which allows you to draw a double line. Here is a slightly modified example from the Graphic Parameters: Double Lines and Bordered Lines section of the 3.1.5b manual.

\documentclass[tikz,border=5mm]{standalone}

\begin{document}
\begin{tikzpicture}
\draw[very thick,double] (0,0) arc (180:90:1cm);
\draw[very thick,double distance=5pt,double=green!50] (1,0) arc (180:90:1cm);
\draw[thin,double distance=.1cm] (2,0) arc (180:90:1cm);
\end{tikzpicture}
 
\end{document}

Answer 2

Some examples from page 151 of the pgf book maybe point in the right direction

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}

\begin{tikzpicture}
\draw [help lines] (0,0) grid (3,2);
\coordinate (a) at (1,0);
\coordinate (b) at (3,1);
\draw (a) -- (b);
\coordinate (c) at ($ (a)!.25!(b) $);
\coordinate (d) at ($ (c)!1cm!90:(b) $);
\draw [<->] (c) -- (d) node [sloped,midway,above] {1cm};
\end{tikzpicture}
\begin{tikzpicture}
\draw [help lines] (0,0) grid (3,2);
\coordinate (a) at (0,1);
\coordinate (b) at (3,2);
\coordinate (c) at (2.5,0);
\draw (a) -- (b) -- (c) -- cycle;
\draw[red] (a) -- ($(b)!(a)!(c)$);
\draw[orange] (b) -- ($(a)!(b)!(c)$);
\draw[blue] (c) -- ($(a)!(c)!(b)$);
\end{tikzpicture}
\end{document}