# Labels perpendicular to a c

I am new to all this. I intend to make scales some of which are curved. I would like to have the tick marks and the corresponding numbers erected perpendicularly to a given 2D curve, e.g. a circle - think of the face of a clock. How can I do that?

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Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – Christian Hupfer Aug 16 '14 at 3:42

I am not doing anything proper today!

\documentclass[dvipsnames,x11names]{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\filldraw [fill=Goldenrod!30] (2,0)arc[radius=2cm,start angle=0,end angle=180] -- (-2,0) -- cycle;
\fill[IndianRed3] (0.1,0)arc[radius=0.1cm,start angle=0,end angle=180] -- (-0.1,0) -- cycle;
\foreach \angle/\label in {10/1,20/2,30/3,40/4,50/5,60/6,70/7,80/8,90/9,100/10,110/11,120/12,130/13,140/14,150/15,160/16,170/17}{
\draw (\angle:1.8cm) -- (\angle:2cm);
\path (\angle:1.7cm)  -- node[rotate=\angle-90,pos=0.02,font=\tiny]{\textsf{\label}}(\angle:2cm);
}
\foreach \angle in {1,2,...,179}{
\draw[thin] (\angle:1.9cm) -- (\angle:2cm);
}
\draw[IndianRed3, thick] (0,0) -- (68:1.6cm);
\end{tikzpicture}
\end{document}


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Depending of the path you can use the sloped option to put the node perpendicular to path. For example

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[allow upside down]
\draw (0,2) arc (90:-270:2) node foreach \p in {1,...,12} [pos=\p/12,below,sloped] {\p};
\draw (0,2) arc (90:-270:2) node foreach \p in {1,...,12} [pos=\p/12,sloped] {\tikz\draw (0,0)--(0,1ex);};
\end{tikzpicture}
\end{document}


produces

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