Example: Rooty helix

Published 2008-12-07 | Author: Felix Lindemann

The idea of the rooty-helix is very simple. One starts e.g. with the length of 1, adds a right angle with the length of 1 and the hypotenuse equals sqrt{2}. If one continues with sqrt{2} repeating the procedure (adding a right angle with the length of 1) the hypotenuse is sqrt{3}. And so on and so on. At some point due to the iterations some triangles are overpainted. Because of this, one repaints the overpainted triangles

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Rooty helix

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% Rooty helix
% Author: Felix Lindemann



\xdefinecolor{darkgreen}{RGB}{175, 193, 36}

    \foreach \y in {86,38,15}{
        \coordinate (a) at (0,0);
        \coordinate (b) at (0:1);
        \foreach \x in {1,...,\y}{%
            \coordinate (c) at ($ (b)!1cm!270:(a) $);
                \draw[fill=\couleur!\thecntShader] (a)--(b)--(c)--cycle;
            \node[fill=white,draw,circle,inner sep=1pt] at (c)
            \coordinate (b) at (c);
    \node[fill=white,draw,circle,inner sep=1pt] at (0:1) {$\sqrt{1}$};



  • #1 Thiago de Melo, December 11, 2008 at 7:37 p.m.

    Hello. How can I use \cos(x) and/or \sin(x)? I would like to load a point at (x,\cos(x)), for specifics values of x, like x=\pi/3.

    Any idea? Thanks.

  • #2 Kjell Magne Fauske, December 11, 2008 at 7:55 p.m.

    You could try something like this:

        \foreach \x in {0,0.2,...,{pi/3}}
            \draw (\x,{cos(\x r)}) circle (2pt);
  • #3 paul, January 30, 2009 at 11:32 a.m.

    It isn't a rooty helix. There is no anything like \x^(0.5). I wrote the similar example of non-rooty helix:

    \xdefinecolor{darkgreen}{RGB}{175, 193, 36}
        \coordinate (B) at (0,0);
        \foreach \x in {100,99,...,1}{
            \coordinate (C) at (B);
            \coordinate (B) at ({sin(0.2*\x r)*0.2*\x},{cos(0.2*\x r)*0.2*\x});
            \filldraw[fill=darkgreen!\x, draw=gray!\x] (B) -- (C) -- (0,0) -- cycle;
  • #4 Felix Lindemann, October 27, 2009 at 6:35 p.m.

    @paul ... of course it is a rooty Helix... take a segment of any length (lets assume the length to be \sqrt{x} ) if you attach a lenght of 1 in a right angle in one of the segements ends and you connect the two open ends the segment betweet these ends will have the length of l= \sqrt{(\sqrt{x})^2+1} = \sqrt{x+1} q.e.d.

    best regards felix

  • #5 Håvard Berland, October 12, 2010 at 10:25 p.m.

    Felix, your image looks so good, so I copied it to Latex' wikipedia page, at http://en.wikipedia.org/wiki/LaTeX

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