编辑: (抱歉,我不知道我应该编辑我的帖子并且不添加其他答案。)
编辑2:
感谢大家的回复。
如果有人感兴趣的话,这是代码:
\documentclass{scrreprt}
\usepackage{amssymb,amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.pathreplacing,angles,quotes}
\begin{document}
\begin{tikzpicture}[>=latex]
\coordinate (o);
\draw [->] (-.5,0) -- (7,0) coordinate (a) node [right] {$\alpha$};
\draw [->] (0,-.5) -- (0,6) coordinate (jb) node [above] {j$\beta$};
\begin{scope}[rotate=30]
\draw [draw=orange,->] (-.5,0) -- (6,0) coordinate (d) node [right] {$d$};
\draw [draw=orange,->] (0,-.5) -- (0,6) coordinate (jq) node [above] {j$q$};;
\end{scope}
\path (2.4,5.5) coordinate (z) node [above left] {$\underline{z}$};
\draw [draw=black,->,line width=0.6mm] (o) -- (z);
\draw [decorate,decoration={brace,amplitude=3mm}] (0,0) -- node[sloped,above=3mm] {$|\underline{z}|$} (2.4,5.5);
\draw [draw=blue,dotted]
($(o)!(z)!(jb)$) coordinate (zb) node [left] {$z_\beta$} -- (z) --
($(o)!(z)!(a)$) coordinate (za) node [below] {$z_\alpha$};
\draw [draw=orange,dashed]
($(o)!(z)!(jq)$) coordinate (zq) node [left] {$z_q$} -- (z) --
($(o)!(z)!(d)$) coordinate (zd) node [below] {$z_d$};
\draw [draw=blue,dotted]
(za) -- ($(z)!(za)!(zd)$) coordinate (x) -- (z);
\draw [draw=blue,dotted]
(za) -- ($(o)!(za)!(zd)$) coordinate (x) -- (zd);
\draw [decorate,decoration={brace,mirror,amplitude=1.2mm}] (x) -- node[sloped,above=-6mm,font=\footnotesize] {$z_\alpha \sin\gamma$} (za);
\draw [decorate,decoration={brace,mirror,amplitude=3mm}] (za) -- node[sloped,above=-9mm,font=\small] {$z_\beta \sin\gamma$} ($(z)!(za)!(zd)$);
\draw [decorate,decoration={brace,amplitude=3mm}] (z) -- node[sloped,above=2mm,font=\small] {$z_\beta \cos\gamma$} ($(z)!(za)!(zd)$);
\draw [decorate,decoration={brace,amplitude=3mm}] (o) -- node[sloped,above right=1mm of d,font=\small] {$z_\alpha \cos\gamma$} ($(za)!(zd)!(x)$);
\draw [draw=black,->] (0:1.2) arc (0:26:1.4) node [left,pos=0.38,font=\small] {$\gamma$};
\draw [draw=black,->] (2.4,4.5) arc (-180:0:0.29cm) node [left,pos=1.3,font=\small] {$\gamma$};
\end{tikzpicture}
\end{document}
答案1
仅通过右括号即可找到解决方案,我让你修改其他的。
备注,您的示例代码不起作用,它缺少库(我删除了德语)
你会注意到,我在节点中添加了文本的旋转角度(90 + 30)并修改了定位
对于偏移量,我删除了括号路径中的加注。
\documentclass{scrreprt}
%\usepackage[ngerman]{babel}
\usepackage{amssymb,amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.pathreplacing,angles,quotes}
\begin{document}
\begin{tikzpicture}[>=latex]
\coordinate (o);
\draw [->] (-.5,0) -- (6,0) coordinate (a) node [right] {$\alpha$};
\draw [->] (0,-.5) -- (0,6) coordinate (jb) node [above] {j$\beta$};
\begin{scope}[rotate=30]
\draw [draw=orange,->] (-.5,0) -- (6,0) coordinate (d) node [right] {$d$};
\draw [draw=orange,->] (0,-.5) -- (0,6) coordinate (jq) node [above] {j$q$};;
\end{scope}
\path (2.4,5) coordinate (z) node [above left] {$\underline{z}$};
\draw [draw=black,->,line width=0.8mm] (o) -- (z);
\draw [draw=blue,dotted]
($(o)!(z)!(jb)$) coordinate (zb) node [left] {$z_\beta$} -- (z) --
($(o)!(z)!(a)$) coordinate (za) node [below] {$z_\alpha$};
\draw [draw=orange,dashed]
($(o)!(z)!(jq)$) coordinate (zq) node [left] {$z_q$} -- (z) --
($(o)!(z)!(d)$) coordinate (zd) node [below] {$z_d$};
\draw [draw=blue,dotted]
(za) -- ($(z)!(za)!(zd)$) coordinate (x) -- (z);
\draw [draw=blue,dotted]
(za) -- ($(o)!(za)!(zd)$) coordinate (x) -- (zd);
\draw [decorate,decoration={brace,mirror,raise=3pt}] (x) -- node[left,font=\small] {$z_\alpha \sin\gamma_N$} (za);
\draw [decorate,decoration={brace,mirror,raise=3pt}] (za) -- node[right,font=\small] {$z_\beta \sin\gamma_N$} ($(o)!(za)!(zd)$);
\draw [decorate,decoration={brace}] (z) -- node[,rotate={30+90},below=0.5em,font=\small] {$z_\beta \cos\gamma_N$} ($(o)!(z)!(d)$);
\draw [decorate,decoration={brace,raise=3pt}] (o) -- node[left,font=\small] {$z_\alpha \cos\gamma_N$} ($(za)!(zd)!(x)$);
\draw [draw=black, ->] (0:1.4) arc (0:30:1.4) node [left,pos=0.3,font=\small] {$\gamma_N$};
\draw [draw=black, ->] (z) arc (0:-30:1.4) node [left,pos=0.3] {$\gamma_\text{N}$};
\end{tikzpicture}
\end{document}
答案2
当使用该库时,事情会变得容易得多,calc
如以下(不完整的)示例所示:
\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[>=latex]
\coordinate (O);
\draw [->] (-.5,0) -- (5,0) coordinate (a) node [below] {$\alpha$};
\draw [->] (0,-.5) -- (0,5) coordinate (jb) node [left] {$j_\beta$};
\begin{scope}[rotate=30]
\draw [draw=orange, ->] (-.5,0) -- (5,0) coordinate (d) node [below] {$d$};
\draw [draw=orange, ->] (0,-.5) -- (0,5) coordinate (jq) node [left] {$j_q$};;
\end{scope}
\path (2,4) coordinate (Z) node [above right] {$Z$};
\draw [draw=blue, dotted]
($(O)!(Z)!(jb)$) coordinate (Zb) node [left] {$Z_\beta$} --
(Z) --
($(O)!(Z)!(a)$) coordinate (Za) node [below] {$Z_\alpha$};
\draw [draw=orange, dashed]
($(O)!(Z)!(jq)$) coordinate (Zq) node [left] {$Z_q$} --
(Z) --
($(O)!(Z)!(d)$) coordinate (Zd) node [below] {$Z_d$};
\draw [draw=blue, dotted]
(Za) -- ($(Z)!(Za)!(Zd)$) coordinate (x) -- (Z);
\end{tikzpicture}
\end{document}