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Tikz 坐标计算

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Tikz 坐标计算

考虑以下 tikz 图片。我需要放置线段 re-D，以便其尺寸为线段 ut-C 尺寸的 8/9。现在我将其放置在 ut 和 ut' 之间 8/9 的随机距离处。我应该使用什么函数（相交？正弦，余弦？）以及如何为 tikz 写下它？

\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}

谢谢你，

如何设置线段 re-D 使得它的长度等于 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}

@Fritz 解决方案

答案1

您可以使用该库来计算该路径与另一条路径在高度上intersection的交点（第一张图片中的红色路径）。(C)-(c)8/9C

\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);

使用交叉库

或者你可以自己使用语法let和截距定理，你可能在学校学过（第二张图片）。

\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);

使用 let 语法

\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}

答案2

这是用于intersection of计算交点 D 的另一种解决方案，而不是使用交点库。

在此处输入图片描述

代码如下：

\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 (leftD) at ($ (C)!1.0/9!(ut) $);
  \coordinate (rightD) at (leftD-|c);
  \coordinate (D) at (intersection of leftD--rightD and C--c);
  \draw[red](D)node[above]{D}--(D|-ut)node[below]{re};

  \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}

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