# Mathematics of a roller chain animation

Given two circular sprockets \(C_1(M_1, r_1)\) and \(C_2(M_2, r_2)\) with their midpoint and radius, we want to find a way to put a chain around them and animate the links like this:

Lets start with some trivial observations. The distance \(d\) between the two midpoints is \(d=|M_2-M_1|\). \(s\) is the difference between the two radii \(r_1\) and \(r_2\), which will be zero if the circles are of the same size and negative if circle \(C_1\) has a greater radius than \(C_2\). \(\theta\) is the rotation angle between the two circles and can be calculated with \(\theta=\tan^{-1}\left(\frac{M_2^y - M_1^y}{M_2^x - M_1^x}\right)\). The points $T_1, T_2$ and the opposite points $T_1', T_2'$ can be calculated as follows:

$$T_1:= C_1 + Rot(\frac{1}{2}\pi + \phi + \theta) r_1$$

$$T_2:= C_2 + Rot(\frac{1}{2}\pi + \phi + \theta) r_2$$

$$T_1':= C_1 + Rot(\frac{3}{2}\pi - \phi + \theta) r_1$$

$$T_2':= C_2 + Rot(\frac{3}{2}\pi - \phi + \theta) r_2$$

## Calculate the chain length

**Case 1)** \(r_2 > r_1\): The triangle \((M_1, M_2, S)\) has a right angle at point \(S\) and the line \((M_1, S)\) is parallel to line \((T_1, T_2)\) and also has the same length \(t\), which is \(t=|T_2 - T_1|\) or easier \(t=\sqrt{d^2 - s^2}\) using Pythagorean theorem. \(\phi\) can be calculated using the law of sines as \(\phi=\sin^{-1}\left(\frac{s}{d}\right)\) or with the inverse tangent again, \(\phi=\tan^{-1}\left(\frac{s}{t}\right)\).

Having these parameters, the length of the chain can be calculated with the two straight lines plus the two arcs around the circles. The length of the arcs is

\[l_1 = 2\pi\cdot r_1 \cdot \frac{\pi - 2\phi}{2\pi} = r_1(\pi - 2\phi)\]

\[l_2 = 2\pi\cdot r_2 \cdot \frac{\pi + 2\phi}{2\pi} = r_2(\pi + 2\phi)\]

And the total length of the chain is given by

\[l = l_1 + l_2 + 2t\]

**Case 2)** \(r_1 > r_2\): This case is analogue to case 2 with flipped indices.

**Case 3)** \(r_1 = r_2\): As stated before, if the radii have the same length, \(s\) will become zero. \(t\) equals \(d\), which can be calculated with \(t=\sqrt{d^2 - s^2}\) from case 1. \(\phi\) is zero, which is given by the tangent from case 1 again. Thus the length for case 3 is already handled with case 1.

Conclusion: Since \(s\) loses the negative sign when being squared, we only need case 1 to calculate the length.

## Calculate the position of the links

Lets say we have \(n\) links. It follows that the length of every link is \(\overline{l} = \frac{l}{n}\). If we open the chain and put it straight on the table, we can segment the chain as illustrated in this sketch:

That means we start at point \(T_2\) go down on the arc of length \(l_2\), continue for length \(t\), go up the arc on the left with length \(l_1\) and back to point \(T_2\). If we now get an arbitrary point $c'$, we can calculate \(c' \mod l\) to keep it circular on the chain. The initial distribution of all links on the line is \(k\cdot \overline{l}\, \forall k<n\).

We have four segments, which will be handled separately now. For each case, we calculate a parameter \(p\in[0,1]\), which represents the position in percent on the given segment. Additionally, we have the current position \(c:= k\cdot\overline{l} + z\, \forall k<n, z\in\mathbb{R}\) to calculate the point \(N_k\), the position of the kth link. \(z\) is a free parameter to scroll.

**Case 1)** \(c < l_2\):

\(p:= \frac{c}{l_2}\)

\(N_k:= M_2 + Rot(\frac{1}{2}\pi + \phi + \theta - p (\pi + 2 \phi)) r_2\)

**Case 2)** \(l_2\leq c < l_2 + t\):

\(p:= \frac{c - l_2}{t}\)

\(N_k:= T_2' + Rot(\pi - \phi + \theta) \cdot t \cdot p\)

**Case 3)** \(l_2 + t\leq c < l_2 + t + l_1\):

\(p:= \frac{c - l_2 - t}{l_1}\)

\(N_k:= M_1 + Rot(\frac{3}{2}\pi - \phi + \theta - p (\pi - 2 \phi)) r_1\)

**Case 4)** \(l_2 + t + l_1\leq c < l\):

\(p:= \frac{c - l_2 - t - l_1}{t}\)

\(N_k:= T_1 + Rot(\phi + \theta) \cdot t \cdot p\)

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