As to compute how much a circle rotates when 'rolled' along a curve in $\mathbb{R}^2$, the most intuitive way to me to find the number of rotations is:

$S/C+W/(2\pi)$

- $S$ is the arc-length of the curve
- $C$ is the circumference of the circle
- $W$ is the total curvature of the curve

However, this seems to agree with:

$T/C$

- $T$ is the arc-length of path of the center of the circle

Can anyone intuitively explain why the latter works as well?

Also I'm wondering whether $T/C$ still works if the circle is replaced by some closed curve (whereby $C$ is the arc-length of the closed curve and $T$ is the arc-length of path of the mass-center of the closed curve). **Edit**: After writing down the integrals, I think a sensible generalization might (rather than the mass-center) have more to do with the center of the osculating circle of the close curve at its current intersection with the curve it's being rolled on.

$\ $

**Edit:** In other words:

Let $\ c:[a,b]\to\mathbb{R}^2\ $ be some smooth curve along which a circle with radius $\text{abs}(r)$ is being rolled.

Let $\ \text{center}:[a,b]\to\mathbb{R}^2\ $ be the center of the circle given by:
$$\text{center}(t)=c(t)+r\frac{\{c_2'(t),-c_1'(t)\}}{||c'(t)||_2}$$
That is the sign of $r$ determines on which side of the curve the circle is being rolled.

Expressed with integrals, the formulas for the total rotation are

$S/C+W/(2\pi)={\large\int_a^b}\dfrac{||c'(t)||_2}{2r\pi}dt+ {\huge\int_{\large a}^{\large b}}\dfrac{\det{\left( \begin{array}{cc} c_1'(t) & c_2'(t) \\ c_1''(t) & c_2''(t) \\ \end{array} \right)}}{||c'(t)||_2^2\cdot (2\pi)}dt$

$T/C = {\large\int_a^b}\dfrac{||\text{center}'(t)||_2}{2\,\text{abs}(r)\pi}\cdot\text{sign}\left(\dfrac{1}{r}+\dfrac{\det{\left( \begin{array}{cc} c_1'(t) & c_2'(t) \\ c_1''(t) & c_2''(t) \\ \end{array} \right)}}{||c'(t)||_2^3}\right)dt$

both of which are influenced by the signs of $r$ and the curvature determinant.