我在这里找到了这个 PDCA 循环(计划、执行、检查、行动)
http://www.texample.net/tikz/examples/pdca-cycle/
% PDCA cycle
% Author: tikzanfaenger, Helmut, and Bartman
\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{decorations.text}
\definecolor{mygray}{RGB}{208,208,208}
\definecolor{mymagenta}{RGB}{226,0,116}
\newcommand*{\mytextstyle}{\sffamily\Large\bfseries\color{black!85}}
\newcommand{\arcarrow}[3]{%
% inner radius, middle radius, outer radius, start angle,
% end angle, tip protusion angle, options, text
\pgfmathsetmacro{\rin}{1.7}
\pgfmathsetmacro{\rmid}{2.2}
\pgfmathsetmacro{\rout}{2.7}
\pgfmathsetmacro{\astart}{#1}
\pgfmathsetmacro{\aend}{#2}
\pgfmathsetmacro{\atip}{5}
\fill[mygray, very thick] (\astart+\atip:\rin)
arc (\astart+\atip:\aend:\rin)
-- (\aend-\atip:\rmid)
-- (\aend:\rout) arc (\aend:\astart+\atip:\rout)
-- (\astart:\rmid) -- cycle;
\path[
decoration = {
text along path,
text = {|\mytextstyle|#3},
text align = {align = center},
raise = -1.0ex
},
decorate
](\astart+\atip:\rmid) arc (\astart+\atip:\aend+\atip:\rmid);
}
\begin{document}
\begin{tikzpicture}
\fill[even odd rule,mymagenta] circle (1.5);
\node at (0,0) [
font = \mytextstyle,
color = white,
align = center
]{
PDCA\\
Cycle
};
\arcarrow{ 85}{ 3}{ PLAN }
\arcarrow{270}{357}{ DO }
\arcarrow{182}{269}{ CHECK }
\arcarrow{176}{ 96}{ ACT }
\end{tikzpicture}
\end{document}
也就是说,我想删除产生拼图效果和首选方向的弧形箭头对象的起点和终点。
答案1
写下答案可能比将信息分散在太多评论中更加清晰。
\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{decorations.text}
\definecolor{mygray}{RGB}{208,208,208}
\definecolor{mymagenta}{RGB}{226,0,116}
\newcommand*{\mytextstyle}{\sffamily\Large\bfseries\color{black!85}}
\begin{document}
\begin{tikzpicture}
\fill[mymagenta] circle (1.35);
\node at (0,0) [
font = \mytextstyle,
color = white,
align = center
]{
PDCA\\
Cycle
};
\pgfmathsetmacro{\mydist}{2}
\pgfmathsetmacro{\Radius}{2}
\foreach \X [count=\Y] in {PLAN,DO,CHECK,ACT}
{\pgfmathtruncatemacro{\itest}{sign(sin(180-\mydist-90*\Y))}
\ifnum\itest>0
\draw[mygray,line width=1cm,postaction={decoration = {
text along path,
text = {|\mytextstyle|\X},
text align = {align = center},
raise = -1.0ex
},
decorate}] (180-\mydist-90*\Y:\Radius) arc(180-\mydist-90*\Y:90+\mydist-90*\Y:\Radius);
\else
\draw[mygray,line width=1cm,postaction={decoration = {
text along path,
text = {|\mytextstyle|\X},
text align = {align = center},
raise = -1.0ex
},
decorate}] (90+\mydist-90*\Y:\Radius) arc(90+\mydist-90*\Y:180-\mydist-90*\Y:\Radius);
\fi
}
\end{tikzpicture}
\end{document}
答案2
这轮图我写的包,可以使用。
初始设置为value=\WCvarA
、slices style=\WCvarB
和data=\WCvarC
。对于此图,我们设置value=1
为使每个切片具有相同的大小。我们还将键设置data
为空。
弧线中的文本由键指定。此文本的方向取决于键中使用的arc data
弧线中间的角度。此文本的位置由键决定。\WCmidangle
arc data dir
arc data pos
切片之间的间隙是通过密钥获得的gap
。
\documentclass[border=6pt]{standalone}
\usepackage{wheelchart}
\usetikzlibrary{decorations.text}
\definecolor{mygray}{RGB}{208,208,208}
\definecolor{mymagenta}{RGB}{226,0,116}
\begin{document}
\begin{tikzpicture}
\sffamily\Large\bfseries\color{black!85}
\fill[mymagenta] (0,0) circle[radius=1.5];
\wheelchart[
arc data=\WCvarA,
arc data dir={\WCmidangle<180?1:-1},
arc data pos=0.5,
data=,
gap,
middle=PDCA\\Cycle,
radius={1.7}{2.7},
slices style=mygray,
value=1
]{PLAN,DO,CHECK,ACT}
\end{tikzpicture}
\end{document}