I have three belts that run over two wheels. 
Due to individual characteristics, the time for one turn, $x_i$, differs between the belts. Before starting the system, each belt is marked at the same spot (red dot). The system will stop once the position of the marked spots overlaps again (anywhere). The aim is to predict the runtime of the system only knowing the speed of the belts.
I don't really know how to solve this problem. My thoughts so far where to regard the system as a collection of three interval with length $x_i$ (representing the speed of the corresponding belt) times a constant $\alpha_i$ (representing the number of turns) and then estimate $\alpha_1$, $\alpha_2$ and $\alpha_3$ by the least common multiple of $x_1$, $x_2$ and $x_3$. But this attempt quickly turned out to be a fallacy because it ignores the "repeating character" of the system.
Now I feel quite at loss. I guess this is a (common) problem in physics/mechanics but I don't have any buzzword or anything. Any ideas?
Not the periods $x_i$, but the differences of the speeds $v_i$ have to be commensurable.
Let $0<v_1<v_2<v_3$ be the speeds of the belts, and $\ell$ their common length. The first time $t$ when the red dots on the first two belts coincide again is when the red dot on the second belt is lapping the red dot on the first belt. This means that $v_2 t=v_1 t+\ell$, or $$t={\ell\over v_2-v_1}=:t_2\ .$$ Henceforth the marks on these two belts coincide at all integer multiples of $t_2$. Similarly, the marks on the first and the third belt coincide at all integer multiples of $$t_3:={\ell\over v_3-v_1}\ .$$ We shall see a coincidence of all three marks (after $t=0$) iff the two numbers $t_2$, $t_3$ are commensurable, i.e. iff there are integers $q>p\geq1$ (in lowest terms) such that $$t_*:=pt_2=qt_3\ .$$ This is the case iff $${p\over q}={v_2-v_1\over v_3-v_1}\ .$$ If this condition is fulfilled all three marks coincide for the first time (after $t=0$) at time $t_*\>$.