Pictured in Figure 1, there are two equally sized, eccentrically mounted circles with radius r=25, turning around centerpoints with distance a=100 with eccentricity e=10. The numerical values are given for clarity only. The line at the top connects both circles tangentially. The circles spin around their centers with a phase difference of 180°, meaning the lines connecting the circles' centers to the centers of rotation are always parallel, but point in opposite directions. There are (objectively poorly drawn) angles α and β, with α being the angle between the horizontal and tangential line. I'm now trying to find an equation for α as a function of β.
While I expected this to be sinusoidal, I let a CAD program do some computations and according to Figure 2, it seems just about not to be.
I'd appreciate any help!
Consider a rotated diagram that keeps the tangent line horizontal. Also the angle $\alpha$ can describe the following triangle
In the picture above the known distances $a$ and $e$ are fixed, as well as the angle $\beta$ at any instant. The result of the kinematics is the distance $b$ between the centers and the angle $\alpha$.
Describe the (x,y) coordinates of the pivot points relative to the center (between the two circles) as
$$\begin{aligned}b/2+e\cos\beta & =a/2\cos\alpha & & \text{x-axis}\\ e\sin\beta & =a/2\sin\alpha & & \text{y-axis} \end{aligned} $$
and solve for
$$\begin{aligned}b= & \sqrt{a^{2}-4e^{2}\sin^{2}\beta}-2e\cos\beta\\ \tan\alpha & =\frac{2e\sin\beta}{b+2e\cos\beta} \end{aligned}$$
Note that for $e \ll a$ you have approximately
$$ \tan( \alpha) \approx \frac{2 e}{a} \sin \beta$$ or
$$ \tan( \alpha) \approx \frac{2 e}{a} \sin \beta + \frac{4 e^3}{a^3} \sin^3 \beta$$
Based on the above if you plotted the angle $\beta$ versus $\tan \alpha$ you would see a much better match for the sine curve based on the first approximation above.