This question revolves around an amusement park teacup ride, where riders spin themselves in a teacup that is on top of a large spinning disk. I'm interested particularly in the line of site of riders and when they see one another. See the animation below or mess around with it on desmos.
For this question there are only two 2D cups $A$ and $B$, each spinning at an independent constant rate. $A$ spins at $\omega_A$, $B$ at $\omega_B$, and the larger disc $\omega \ [\text{rad/s}]$. $\theta_A$ and $\theta_B$ correspond to the field of view angle, which for this problem $\theta_A = \theta_B$. The rider is located halfway between the center and edge of their "cup".
Questions: If the two cups spin on the duration $[0,t]$,
1) How many times will the riders see each other ($A$ sees $B$ or $B$ sees $A$) and how many times will they "meet eyes" ($A$ sees $B$ and $B$ sees $A$)? Can this be written in terms of $t$, $\theta$, and $\omega$?Assume the riders are facing opposite directions at $t=0$.
2) Is it guaranteed, if $\omega_A \neq \omega_B \neq 0$, that there will be a time $t$ when both riders are looking at each other?
$A$ "sees" $B$ means rider $B$ is located within or on the edge of the triangular field of view of rider $A$.
Context: I came across this when working on a VR simulation of a teacup ride. I am aware I could simply check for plane intersections or use raycasting to count the passes, but I'm curious to see if there is a closed form solution for general $\theta$ and $\omega$.
Firstly, let's transform into a rotating frame of reference with the larger circle, rotating at rate $\omega$. In this frame both the smaller circles rotate around a fixed point at rates I will call $\omega_A$ and $\omega_B$, which may be different notation to what you asked in your question.
The rotation time for the smaller circles is $T_A=\omega_A/(2 \pi)$ and $T_B=\omega_B/(2 \pi)$.
1) The number of times disk A looks at the centre of disk B is $T/T_A$ rounded either up or down, depending on the starting orientation of the disk A. To get the precise answer, calculate the amount of time it takes to rotate until the centre of disk B comes into view, this is the time at which the count goes up from 0 to 1, the count increases by 1 every $T_A$. The problem of 'how many times does disk A look at person B' is much much harder, because if disk B is rotating much faster than disk A then the person will pass in and out of the vision range on each pass. If this is what you want then the problem is tractable, but hard.
2) No. For example if the disks are rotating at the same speed and A starts looking at B but B starts looking directly away from A, (and the view angles are less than 180°) then they will be in sync and never look at each other. More generally if $T_A/T_B$ is rational, the view angle sufficiently small, and the starting orientations are correct then they will never see each other at the same time.