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. 

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:
$$
\varphi  = \tan \alpha  - \alpha  = \sqrt {\rho ^{\,2}  - 1}  - \arccos \left( {1/\rho } \right)
$$
