Involute of a circle - what is the separation distance? It seems like a simple enough question. For the involute of a circle, what is the separation distance between successive turns?
Is this derivation correct?
Parametric formula for the y-coordinate:
$ y = r(Sin(\theta) - \theta Cos(\theta)) $
Differentiating:
$ \frac{dy}{d\theta} = r \theta Sin(\theta) $
Which has roots at $ \theta = \pi n, n\in \mathbb{Z} $
Taking every other $n$, since those are successive turns:
$ y = r(Sin(\theta) - \theta Cos(\theta)) $
simplifies to
$ y = -r \pi n $
where $n$ is even and $n \ge 0$.
Therefore the spacing between successive turns, $D$ is:
$ D = 2 \pi r $
Is that even close to right? Is it that simple? I guess it makes intuitive sense based on the circumference of the circle. And some plots I've made bear it out. But I'd like to know for sure. 
 A: We can represent the parametric equations of the circle involute, by factoring out the 
radius $r$ of the generating circle, which is just a scale factor.
$$
\left\{ \begin{gathered}
  x = X/r = \cos \theta  + \theta \sin \theta  \hfill \\
  y = Y/r = \sin \theta  - \theta \cos \theta  \hfill \\ 
\end{gathered}  \right.\quad \left\{ \begin{gathered}
  \frac{{dx}}
{{d\theta }} = \theta \cos \theta  \hfill \\
  \frac{{dy}}
{{d\theta }} = \theta \sin \theta  \hfill \\ 
\end{gathered}  \right.\quad \left| {\;0 \leqslant \theta  \in \;\mathbb{R}} \right.
$$
Let's indicate by $P = \left( {x ,\;y } \right)$ the point on the curve and by $T = \left( {\cos \theta ,\;\sin \theta } \right)$
the point on the circle from where the theter evolves.
Then 
$$
\left| {TP} \right| = \sqrt {\left( {x - \cos \theta } \right)^{\,2}  + \left( {y - \sin \theta } \right)^{\,2} }  = \theta 
$$
$$
\left| {OP} \right| = \sqrt {x^{\,2}  + y^{\,2} }  = \sqrt {1 + \theta ^{\,2} } 
$$
and
$$
\mathop {TP}\limits^ \to  \; \bot \;\mathop {OT}\limits^ \to  \quad \quad \vec v = \left( {dx/d\theta ,\;dy/d\theta } \right)\; \bot \;\mathop {TP}\limits^ \to  
$$
which means that $T$ is also the center of curvature.  
Now, by putting $dy/d\theta=0$ you find the $y$ of the points where the tangent to the line is horizontal, 
so $D/r$ gives the spacing between consecutive horizontal tangents to the line.
If you plug $\theta=\pi n$ into the parametric equation for $x$ you get $x=1,-1,1,..$,
that means that the points with horizontal tangent lie on the verticals $x=1$ (for $y<=0$) and $x=-1$ (for $0<y$),
which is in accordance with the results found above.
However , over $x=0$ (as well as over any radial line from $O$) the intersection points
 are not equally spaced, although the distance tends towards $2 \pi $.
To find them, we make the subtitution
$$
\left\{ \begin{gathered}
  \frac{1}
{{\sqrt {1 + \theta ^{\,2} } }} = \cos \alpha  \hfill \\
  \frac{\theta }
{{\sqrt {1 + \theta ^{\,2} } }} = \sin \alpha  \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \tan \alpha  = \theta 
$$
so that in the polar coordinates $\rho$ and $\varphi$ we have:
$$
\left\{ \begin{gathered}
  \tan \alpha  = \theta  \hfill \\
  \rho  = \sqrt {1 + \theta ^{\,2} }  = \frac{1}
{{\cos \alpha }} \hfill \\
  \tan \varphi  = \tan \left( {\theta  - \alpha } \right)\quad  \Rightarrow \quad \varphi  = \tan \alpha  - \alpha  \hfill \\ 
\end{gathered}  \right.
$$
and we can express $\varphi$ in terms of $\rho$ as:
$$
\varphi  = \tan \alpha  - \alpha  = \sqrt {\rho ^{\,2}  - 1}  - \arccos \left( {1/\rho } \right)
$$
