Creating an involute curve

Question

I need assistance in drawing this diagram in Geogebra or Tikz. Its for my thesis "Modelling of an involute spur gear via finite element method".

Welcome to TeX SX! Mathematically, what are the curves? The lower one looks like a circular arc. — Bernard, Sep 03 '16 at 09:14
see this answer (http://tex.stackexchange.com/a/58791/5355) if you want to draw à gear with involute
the answer below with only one involute: — rpapa, Sep 03 '16 at 09:36

score 2 · Answer 1 · answered Sep 03 '16 at 15:27

There are some differences with yours, but the concept is the same. The tangents and the lower arc are calculated properly, however, the top arcs are just cosmetic. It's really hard to build arcs there since the various tangents have different lengths.

Maybe someone else can help in that department.

Output

enter image description here

Code

\documentclass[margin=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc, intersections}

\tikzset{
    dot/.style 2 args={fill, circle, inner sep=1pt, label={#1:#2}}
}

\newcommand\Radius{6}

\begin{document}
\begin{tikzpicture}

\begin{scope}[rotate=90,yscale=-1,xscale=1]
\draw[name path=quartarc, thick] (\Radius,0) arc (0:90:\Radius) coordinate[pos=0] (C0); 

\foreach \angle [
        count=\n, 
        evaluate=\n as \xn using int(5-\n)
    ] in 
    {20,45,60,80,90}{%
    \path[name path={tan\n}] (0,0) -- (\angle:{\Radius+1});
    \path[name intersections={of={tan\n} and quartarc,by={P\n}}];
    \pgfmathsetmacro\pointA{cos(\angle)*\Radius}
    \pgfmathsetmacro\pointB{sin(\angle)*\Radius}
    \pgfmathsetmacro\ArcLen{(2*pi*\Radius/360)*\angle}
    \pgfmathsetmacro\pointX{\pointA+sin(\angle)*\ArcLen}
    \pgfmathsetmacro\pointY{\pointB-cos(\angle)*\ArcLen}    
    \coordinate (C\n) at (\pointX,\pointY);
}
\end{scope}

\foreach \j [remember=\j as \lastj (initially 0)] in {0,...,5}{
    \draw[thick] (C\lastj) edge[bend left=7.5] (C\j);
}
%
\node[dot={below}{$N_0$}] at (C0) {}; 
\foreach \x in {1,...,5}{
    \draw[thick] (P\x) -- (C\x) coordinate[pos=0,dot={}{}] coordinate[pos=1,dot={}{}] (a\x);
    \ifnum\x<4\relax
    \node[dot={180-(15*\x)}{$M_\x$}] at (C\x) {};
    \node[dot={255-(15*\x)}{$N_\x$}] at (P\x) {};
    \fi
}

\draw (40:\Radius) --++ (225:3cm) node[fill=white] {$E$};
\draw ($(a3)!.5!(a4)+(110:1mm)$) --++ (110:2cm) node[fill=white] {$I$};

\end{tikzpicture}
\end{document}

score 1 · Answer 2 · answered Sep 03 '16 at 09:34

here is a solution

\documentclass[border=5mm,tikz]{article}
\usepackage{mwe}
\usepackage{tikz}
\usetikzlibrary{positioning,calc}

\begin{document}


 \def\Rb{3}
 \def\AngleT{1}

\def\xxt{\Rb*(cos(\t r)+\t*sin(\t r))}
\def\yyt{\Rb*(sin(\t r) - \t*cos(\t r))}

\begin{tikzpicture}

\draw (0,0) circle (\Rb); 
\draw[domain=-0:\AngleT,smooth,variable=\t,thick,red,thick]
plot ({atan2(\xxt,\yyt)}:{\Rb*sqrt(1+\t^2)}  )coordinate(ff);
\draw (0,0) -- ({90-\AngleT*180/3.14159} :\Rb)--(ff);
\end{tikzpicture}

\end{document}

Creating an involute curve

2 Answers2

Output

Code