sketching semicirles and horizontal lines

I want to sketch this parallel lines and semicircles between them sth like the attached figure... would you plz help me?

-
Questions about how to draw specific graphics that just post an image of the desired result are really not reasonable questions to ask on the site. Please post a minimal compilable document showing that you've tried to produce the image and then people will be happy to help you with any specific problems you may have. See minimal working example (MWE) for what needs to go into such a document. – Paul Gessler Jul 27 '14 at 22:40
actually I can oly draw parallel lines but I cant draw semicircles between them...I dont want to draw this graphic with other programs like Microsoft office or paint etc because I am typing my thesis with latex and it is better to have my graphics drawn by it too. I would so grateful if you can guide me. – hamed Jul 27 '14 at 22:57
Are the circular arcs supposed to have common centres? If so wouldn't the down arrows be straight lines? – Thruston Jul 27 '14 at 23:23
You can do this sort of diagram in Metapost, or Asymptote, or Tikz easily enough, but you just need to pick one and read the manuals! – Thruston Jul 27 '14 at 23:24
yes the circular arcs have the common center... If i learn how to draw one of this arcs i will be able to draw the rest...But unfortunately I am amateur! :( – hamed Jul 27 '14 at 23:50

Another (possibly better) complete solution using intersections and calc libraries with the aid of a \foreach construct can be as follows:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,arrows,positioning,intersections}
\begin{document}
\def\dis{1cm}
\begin{tikzpicture}[line width=0.7pt,>=latex',node distance=\dis]
\node (o) at (0*\dis,8*\dis) {$\ast$};
\path [draw,name path=bigarc1, bend right=10,->,shorten >=10pt](o.center)to([xshift=2.5*\dis,yshift=-7*\dis]o);
\path [draw,name path=bigarc2, bend left=10,->,shorten >=10pt](o.center)to([xshift=-2.5*\dis,yshift=-7*\dis]o);
\foreach \y/\c in {7/1,6/2,5/3,4/4,3/5,2/6}{
\path [draw,name path=Line-\c] (-3*\dis,\y*\dis)--(3*\dis,\y*\dis) node[right=0.2*\dis]{$V_{\c}$};
\path [name intersections={of = bigarc1 and Line-\c}];
\coordinate (p) at (intersection-1);
\path [name intersections={of = bigarc2 and Line-\c}];
\coordinate (q) at (intersection-1);
\draw [bend left,shorten >=-5pt,shorten <=-5pt](p)to(q);
}
\end{tikzpicture}
\end{document}


The output of such a code is as expected:

I completed another solution based on the idea of Sigur as an alternative.

The complete code (following Sigur's idea) is as follows:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,intersections,arrows,positioning}
\begin{document}
\begin{tikzpicture}[>=latex']
\node  (p-0) at (0,0) {$\ast$};
\node  (q-0) at (0,0) {};
\foreach \raft/\rbef in {1/0,2/1,3/2,4/3,5/4,6/5}{
\path [draw, name path global/.expanded =L-\raft](-6,-\raft)--(6,-\raft) node [right] {$V_{\raft}$};
\path [draw, name path global/.expanded =A-\raft,thick] (-120:1.1*\raft) arc (-120:-60:1.1*\raft);
\path [name intersections/.expanded ={of ={L-\raft} and {A-\raft}}];
\coordinate (p-\raft) at (intersection-1);
\coordinate (q-\raft) at (intersection-2);
}
\draw [->,shorten >=-10pt](p-0.center)--(p-1)--(p-2)--(p-3)--(p-4)--(p-5)--(p-6);
\draw [->,shorten >=-10pt](q-0.center)--(q-1)--(q-2)--(q-3)--(q-4)--(q-5)--(q-6);
\end{tikzpicture}
\end{document}


The output of that code is the following figure:

But it is slightly different from the required figure in the question.

-
thank so much for your favor dear Mr @AboAmmar. It was so useful and I am so glad that I could get my answer as best as possible. I am so interested to learn more about drawing such graphics with latex when I see how beautiful this graphics are made. I learned so many points with your help. – hamed Jul 28 '14 at 2:44
You can mark my answer accepted if see it more helpful. – AboAmmar Jul 28 '14 at 2:58
But these arcs do not have a common center... – Thruston Jul 28 '14 at 17:39
@Thruston-- For equidistant concentric circles, the intersection points with equidistant parallel lines should be aligned to a straight line. This is not the case in the requested picture of the question. If you observe the two long down curves you see what I mean. The two long down-paths may be curves or segments of a straight line, but surely not a straight line. – AboAmmar Jul 28 '14 at 22:52
We are not racing towards the most accurate, everybody is stating their craziest ideas :) so don't worry too much about correctness. Actually the crazier, the better it gets. Sometimes the crazy answer opens up a whole angle and people build up on it. – percusse Jul 29 '14 at 1:23

Partial solution (using tikz)

\documentclass{report}
\usepackage{amsthm,amsmath,amssymb}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\node at (0,0) {$\ast$};
\foreach \r in {1,2,...,6}{
\draw (-6,-\r)--(6,-\r)node[right]{$v_{\r}$};
%\draw (-135:\r) circle (1pt);
\draw[thick] (-135:1.2*\r) arc (-135:-45:1.2*\r);
}
\end{tikzpicture}

\end{document}


The syntax for the arc needs the initial point P to determine the ray OP and then makes use of the second argument which is the initial angle and the final angle and the radius.

Another image to explain the angles (in polar coordinates (angle:radius)):

-
Could you possibly explain the syntax of (-135:1.2*\r) arc (-135:-45:1.2*\r)? Not the \r bit but the specification of the arc. I keep reading the manual on this and I realise that drawing lines is pretty much the most straightforward thing there is but, for whatever reason, I find it almost impossible to understand. (I end up just guessing by trial-and-error which is not very efficient.) I more-or-less understand Metapost's syntax for this, if that helps. – cfr Jul 28 '14 at 0:09
wooow... its exactly whatever I want... I was able to draw the parallel lines but i could not draw the intersecting arc...thanks so so so much for your help. – hamed Jul 28 '14 at 0:15
updated to show the angles and polar coordinates. Note that I used 20% bigger radius otherwise the lines would be tangent. – Sigur Jul 28 '14 at 0:22
Thanks. Why do you need only one angle specified for the first point but two fro the second point, though? Why isn't (-45:1.2*\r) enough for the second? – cfr Jul 28 '14 at 2:45