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Let there be $n$ circles that have radius $r$. They all touch each other at a tangent, so they could be arranged in a circle. If a circle with radius $R$ were to roll around the outside of this arrangement, by how much would it be rotated? If you have trouble visualising this, think of a one pound coin rolling around a circular arrangement of two-pound coins.

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    $\begingroup$ Are you asking the same thing as this? $\endgroup$ – Ng Chung Tak Feb 1 '17 at 19:41
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    $\begingroup$ @ Ng Chung Tak Yes, it looks the same question, with a higher degree of difficulty, with no more personal involvement. I propose to close this kind of question without work. $\endgroup$ – Jean Marie Feb 1 '17 at 22:32
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A circle rolling once around a circle of equal radius makes two rotations. I didn't expect this, but try it on a couple of coins with good milling. And it makes sense, since where a circle makes one rotation rolling along a straight line equal to its circumference, it will make two rotations rolling along the circumference of an equal circle. Working still with equal circles, one circle rolling around two tangent circles completes $2*\frac23*2 = \frac83$ rotations. This is because tangency makes one-third of each circle inaccessible to the rolling circle. A circle rolling around three tangent circles makes $3*\frac12*2= 3= \frac93$ rotations, since half of each circle is inaccessible. Four tangent circles yield $4*\frac{5}{12}*2 = \frac{10}{3}$ rotations. Five tangent circles make $5*\frac{33}{90}*2 = \frac{11}{3}$ rotations. For $n>1$, the numerator of the fraction increases by $1$ for each increase in the number of tangent circles. The number of rotations is $\frac{n+6}{3}$. The fraction of each circle accessible to, i.e. traversable by, the rolling circle = $\frac{240-\theta_n}{360}$, where $\theta_n$ is the angle, in degrees, of a regular n-gon. Checking with a 16-sided regular polygon whose vertices are centers of sixteen tangent circles, and $\theta_{16} = 157.5$ degrees, we get $\frac{22}{3}$ rotations, as the rule predicts. This is perhaps more of an investigation than a proof, and supposes the rolling circle equal to the tangent circles. Where the rolling circle's radius is unequal to those of the tangent circles, $\frac{n+6}{3}$ must be multiplied by the inverse of the ratio $\frac{R}{r}$. The ratio is inverted since a smaller circle makes more rotations on larger circles and fewer on smaller circles.

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When a circle of radius $R$ rolls without slipping around a circle of radius $r$, it rotates of an angle $2\pi(r/R+1)$.

In the case of $n$ circles of radius $r$ arranged around a central circle (see diagram below) the rolling (red) circle does not make a complete revolution around each blue circle, but only runs an arc corresponding to a central angle of $2(\alpha+\phi)$, where $\alpha=\pi/n$ and $\sin\phi=\cos\theta=r/(r+R)$.

In that path it then rotates of $2(\alpha+\phi)(r/R+1)$. Multiply that by the number of circles to get the rotation angle after a full revolution around all the circles: $\theta=2n(\alpha+\phi)(r/R+1)$, that is: $$ \theta=2\left(\pi+n\arcsin{r\over r+R}\right)\left({r\over R}+1\right). $$ enter image description here

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