Nothing new, but a diffent way to look at the question:
First consider a cirlce with an inside hexagon (in general an n-figure).
Then the circle will roll the distance of the circumference PULS an angle wihich is due
to tilting at the edges. Thus, the additional angle (the $+1$ you look for) amounts in total to
$$6 \cdot \frac{2\pi}{6} = 2\pi, (n \cdot \frac{2\pi}{n} = 2\pi)$$
explaining the extra turn - no matter how many edges you take for your approximation.
Second consider an infinitisimal approach: For $x \in [0, R)$,
$f(x) = \sqrt{R^2-x^2}$ is a circle with radius $1$. Later, we also need
$$f'(x) = \frac{-x}{\sqrt{R^2-x^2}}$$
A tangent vector $\vec{T}$ at $x$ is
$$\vec{T}=\begin{pmatrix}1\\f'(x)\end{pmatrix}$$ and $\vec{T'}$ at $x+\Delta x$ is
$$\vec{T}=\begin{pmatrix}1\\f'(x+\Delta x)\end{pmatrix} = \begin{pmatrix}1\\f'(x)+f''(x)\cdot \Delta x + \frac{f''(x)}{2}\Delta x^2\end{pmatrix}$$.
Now, the cosine of the angle $\Delta \alpha$ between the to vectors is
$$\cos(\Delta \alpha)=\frac{\vec{T}\cdot \vec{T'}}{\left|\vec{T}\right|\cdot \left|\vec{T'}\right|}$$
To first order this is
$$h(\Delta x) = \frac{1+\frac{f'f''\Delta x}{1+f'^2}+\frac{0.5f'f'''\Delta x^2}{1+f'^2}}{\left(1+\frac{f''^2\Delta x^2}{1+f'^2}+\frac{2f'f''\Delta x}{1+f'^2}+\frac{f'f'''\Delta x^2}{1+f'^2}\right)^{1/2}}$$
Use the following abbreviations for a Talor-series (which equals $\cos(x) = 1 - x^2/2$):
$$N=\frac{1}{1+f'^2}, a = Nf'f'', b = 0.5Nf'f''', c = Nf''^2, d = c + 2b$$
Then
$$h(\Delta x) = \frac{1 + a\Delta x + b \Delta x^2}{\left(1 + 2a \Delta x + d \Delta x^2\right)^{1/2}}$$
$$h'(\Delta x) = \frac{(a^2-c)\Delta x + 3ab \Delta x^2 + bd \Delta x^3}{\left(1 + 2a \Delta x + d \Delta x^2\right)^{3/2}}$$ and
$$h''(\Delta x = 0) = \frac{(a^2-c)\Delta x + 3ab \Delta x^2 + bd \Delta x^3}{\left(1 + 2a \Delta x + d \Delta x^2\right)^3/2}$$
Finally we obtain
$$1 - 0.5\frac{f''^2}{\left(1+f'^2\right)^2} \Delta x^2 = 1 - \frac{\Delta \alpha^2}{2}$$
or
$$\int_{x_0}^{x_1} \frac{f''}{1+f'^2}dx = \arctan(f'(x_1)) - \arctan(f'(x_0))$$.
This again gives for a full circle $8\cdot\frac{pi}{4} = 2 \pi$. Furthmore we observe, that in the end, only the difference between the angles at $x_0$ and $x_1$ play a role.