[Tex/LaTex] Tikz coordinate calculation

coordinatespositioningtikz-pgf

Consider the followinting tikz picture. I need to place the segment re-D so that its measure will be 8/9ths the measure of the segment ut-C. Now I just placed it at a random distance of 8/9 between ut and ut'.
What function should I use (intersect? sin, cos?) and how do I write it down for tikz?

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


\begin{tikzpicture}[scale=2]


\coordinate (o) at (0,0);
\coordinate [label=below:ut] (ut) at (2,0);
\coordinate [label=below:ut'] (ut') at (4,0);


\coordinate [label=above:C] (C) at (2, 2.4);
\coordinate [label=above:c] (c) at (4, 1.2);
\coordinate (O) at ($ (c)!2!(C) $);

\draw (ut) -- (ut');
\draw (ut) -- (C);
\draw (C) -- (c);
\draw (ut') -- (c);

\coordinate [label=below:re] (re) at ($ (ut)!8.0/9!(ut') $);

\coordinate [label=above:D] (D) at ($ (C)!8.0/9!(c) $);

\draw (re) --(D);

\node [fill=black, inner sep=1pt] (c') at ($ (ut)!1.0/6!(C) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!2.0/6!(C) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!3.0/6!(C) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!4.0/6!(C) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!5.0/6!(C) $) {};

\node [fill=black, inner sep=1pt] (c') at ($ (ut')!1.0/3!(c) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut')!2.0/3!(c) $) {};
\draw (ut') -- (c);



\end{tikzpicture}
\end{document}

Thank you,

How to set the segment re-D so that its lenght is equal to 8/9 ut-C?

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,decorations,decorations.text}

\begin{document}

\begin{tikzpicture}[scale=5]



\coordinate (o) at (0,0);
\coordinate [] (ut) at (2,0);
\node  [xshift=-0.8ex, yshift=1.5ex] at (ut) {u};
\node  [xshift=0.65ex, yshift=1.6ex] at (ut) {t};
\coordinate (fa') at (4,0);
\node  [xshift=1.2ex, yshift=1.5ex] at (fa') {fa};


\coordinate [label=above:C] (C) at (2, 2.4);
\coordinate [label=above:c] (c) at (4, 1.2);
\coordinate (O) at ($ (c)!2!(C) $);

\draw (ut) -- (fa');
\draw (ut) -- (C);
\draw (C) -- (c);
\draw (fa') -- (c);

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (fa') ++({-(8/9*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate  (re)
         ++(0, {8/9*\y1}) coordinate  (D);

\draw (re) -- (D);
\node  [xshift=-0.7ex, yshift=1.5ex] at (re) {r};
\node  [xshift=0.7ex, yshift=1.5ex] at (re) {e};

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (fa') ++({-(64/81*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate  (mi)
         ++(0, {64/81*\y1}) coordinate  (E);
\draw (mi) --(E);
\node  [xshift=-1.2ex, yshift=1.5ex] at (mi) {m};
\node  [xshift=0.45ex, yshift=1.75ex] at (mi) {i};

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (fa') ++({-(3/4*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate  (fa)
         ++(0, {3/4*\y1}) coordinate  (F);
\draw (fa) -- (F);
\node  [xshift=-0.7ex, yshift=1.9ex] at (fa) {f};
\node  [xshift=0.7ex, yshift=1.5ex] at (fa) {a};

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (fa') ++({-(2/3*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate  (sol)
         ++(0, {2/3*\y1}) coordinate [label=above:G] (G);
\draw  (sol) -- (G);
\node  [xshift=-0.5ex, yshift=1.5ex] at (sol) {s};
\node  [xshift=1.2ex, yshift=1.9ex] at (sol) {ol};

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (fa') ++({-(16/27*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate  (la)
         ++(0, {16/27*\y1}) coordinate (A);
\draw  (la) -- (A);

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (fa') ++({-(128/243*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate  (si)
         ++(0, {128/243*\y1}) coordinate (B);
\draw  (si) -- (B);

\node [fill=black, inner sep=1pt] (c') at ($ (ut)!1.0/6!(C) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!2.0/6!(C) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!3.0/6!(C) $) {};
\node [xshift=-0.5em, yshift=-2em] (c') at ($ (ut)!3.0/6!(C) $) {6};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!4.0/6!(C) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (ut)!5.0/6!(C) $) {};

\node [fill=black, inner sep=1pt] (c') at ($ (fa')!1.0/3!(c) $) {};
\node [xshift=0.5em, yshift=-2em] (c') at ($ (fa')!1!(c) $) {3};
\node [fill=black, inner sep=1pt] (c') at ($ (fa')!2.0/3!(c) $) {};
\draw (fa') -- (c);

\node [fill=black, inner sep=1pt] (c') at ($ (sol)!1.0/4!(G) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (sol)!2.0/4!(G) $) {};
\node [fill=black, inner sep=1pt] (c') at ($ (sol)!3.0/4!(G) $) {};
\node [xshift=-0.5em, yshift=-2em] (c') at ($ (sol)!3.0/4!(G) $) {4};

\draw[postaction={decorate,decoration={text along path,text align=center,text={Diapason},raise=5pt}}](ut)to[bend right=800] (fa'){};
\draw[postaction={decorate,decoration={text along path,text align=center,text={Diapente},raise=5pt}}](ut)to[bend right=800] (sol){};
\draw[postaction={decorate,decoration={text along path,text align=center,text={Diatessaron},raise=5pt}}](sol)to[bend right=800] (fa'){};



\end{tikzpicture}
\end{document}

The @Fritz solution

Best Answer

You can either use the intersection library to calculate the intersection of the path (C)-(c) with another path at height 8/9C (the red path in the first picture).

\path[draw,red, name path=target] ($ (ut)!8.0/9!(C) $)
    node [label=left:8/9] {} -- ++(1,0);
\path[name intersections={of=target and C-c}]
    (intersection-1) coordinate[label=above:D](D)
    (intersection-1 |- ut) coordinate[label=below:re](re);

Using the intersection library

Or you could do the calculations yourself using the let syntax and the intercept theorem, which you have probably learned in school (second picture).

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (ut') ++({-(8/9*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate [label=below:re] (re)
         ++(0, {8/9*\y1}) coordinate [label=above:D] (D);

Using the let syntax

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc, intersections}
\begin{document}

\begin{tikzpicture}[scale=2]
\coordinate (o) at (0,0);
\coordinate [label=below:ut] (ut) at (2,0);
\coordinate [label=below:ut'] (ut') at (4,0);

\coordinate [label=above:C] (C) at (2, 2.4);
\coordinate [label=above:c] (c) at (4, 1.2);
\coordinate (O) at ($ (c)!2!(C) $);

% Help path. Style draw,red is only for illustration.
\path[draw,red, name path=target] ($ (ut)!8.0/9!(C) $)
    node [fill=black, inner sep=1pt, label=left:8/9] {}
    -- ++(1,0);

\draw (ut) -- (ut');
\draw (ut) -- (C);
\draw[name path=C-c] (C) -- (c);
\draw (ut') -- (c);

\path[name intersections={of=target and C-c}]
    (intersection-1) coordinate[label=above:D](D)
    (intersection-1 |- ut) coordinate[label=below:re](re);

\draw (re) --(D);
\draw (ut') -- (c);
\end{tikzpicture}

\begin{tikzpicture}[scale=2]
\coordinate (o) at (0,0);
\coordinate [label=below:ut] (ut) at (2,0);
\coordinate [label=below:ut'] (ut') at (4,0);

\coordinate [label=above:C] (C) at (2, 2.4);
\coordinate [label=above:c] (c) at (4, 1.2);
\coordinate (O) at ($ (c)!2!(C) $);

\draw (ut) -- (ut');
\draw (ut) -- (C);
\draw (C) -- (c);
\draw (ut') -- (c);

% Use basic school geometry to calculate intersection.
\node [fill=black, inner sep=1pt, label=left:8/9] at ($ (ut)!8.0/9!(C) $) {};

\path let \p1=($(C)-(ut)$), \p2=($(c)-(ut)$) in
    (ut') ++({-(8/9*\y1 / (\y1-\y2) - 1)  * \x2},0)
        coordinate [label=below:re] (re)
         ++(0, {8/9*\y1}) coordinate [label=above:D] (D);

\draw (re) --(D);
\draw (ut') -- (c);

\end{tikzpicture}

\end{document}

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[Tex/LaTex] How to define the default vertical distance between nodes

I don't really get the question so I hope this is what you wanted. If you include a full document (such that we copy paste and see the problem on our systems) things are much more easier.

Here, you can change the default setting within a scope but your block style had a node distance which was resetting every time it is issued. I've made it 2mm such that we can see the difference easier.

\documentclass[tikz]{standalone}
\usetikzlibrary{arrows,shapes.geometric,positioning}

\begin{document}
\begin{tikzpicture}[decision/.style={diamond, draw, text width=4.5em, text badly centered, node distance=3.5cm, inner sep=0pt},
                    block/.style   ={rectangle, draw, text width=6em, text centered, rounded corners, minimum height=4em, minimum height=2em},
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    \node [block] (1st) {1st};
    \node [block, right= of 1st]  (2nd1) {2nd1};
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    \node [block, below= of 2nd1] (2nd3) {2nd3};
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    \node [block, above= of 3rd1] (3rd2) {3rd2};
    \node [block, above= of 3rd2] (3rd3) {3rd3};
\end{scope}
    \node [block, below= of 3rd1] (3rd4) {3rd4};
    \node [block, below= of 3rd4] (3rd5) {3rd5};

    \path [line] (1st) -- (2nd1);
    \path [line] (2nd1) -- (2nd2);
    \path [line] (2nd1) -- (2nd3);
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\end{tikzpicture}
\end{document}

enter image description here

[Tex/LaTex] Using current path position in coordinate calculation

I found a solution in this TeX-SX answer:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\makeatletter
\newcommand\currentcoordinate{\the\tikz@lastxsaved,\the\tikz@lastysaved}
\makeatother
\begin{document}
\begin{tikzpicture}
  \node (aNodeWithALongName) at (1,1) {a};
  \node (bNodeWithALongName) at (1,-1) {b};
  \node (c) at (0,0) {c};

  \draw[->] (aNodeWithALongName) -- (\currentcoordinate -| c);
  \draw[->] (bNodeWithALongName) -- (\currentcoordinate -| c);

  \draw[->] ($(aNodeWithALongName)!.5!(bNodeWithALongName)$) -- (\currentcoordinate -| c);
  \draw[->] (bNodeWithALongName) -- ++(0, 1cm) -- (\currentcoordinate -| c);
\end{tikzpicture}
\end{document}

Best Answer

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[Tex/LaTex] How to define the default vertical distance between nodes

[Tex/LaTex] Using current path position in coordinate calculation

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