At my work we have $2$ machines that are fitted end to end so that an array tape goes in machine $1$ from one end and it gets read $8 \cdot 144 \, \mathrm{mm}$ in $15$ minutes. When comes out of the machine $1$, the tape then can be fed to the machine $2$ which reads the same length$(8 \cdot 144 \, \mathrm{mm})$ in $7$ minutes. Before starting the process we will know how many array sections we are gonna read, thus what is the total length of the tape (e.g., $100 \text{ arrays} \cdot 144 \, \mathrm{mm} = 14400 \, \mathrm{mm}$).
So the question is, is there a way we can derive a mathametical formula to know after how many array being read by the machine $1$ if we start reading machine $2$ we will never have a sceraio when machine $2$ run out of array and try to pull the tape out of machine $1$? Lets say a safe gap of minimum $5$ arrays always should be there between these $2$ machines.
Update: So, thanks to Tobi's awesome answer I was able to use app inventor and develop a very minimalistic app, that can tell me the number of tape to wait for before i can start reading it with machine 2.
Though I'm not good at maths but always love to share things in return, so, I'm more than happy to add the image of the code that I have written, so that whoever is looking for a solution like this can get a fair idea of not only how to work it out but also how to implement it as end to end.
Once more: Thanks Tobi.
Let the length of the tape be $D_T$
Let the speed of the first tape(slower one) be $S_1$
Let the speed of the second tape(faster one) be $S_2$
Let the length of the tape for which your request is satisfied be $D_x$
Let the total time for which tape2 will run be $t$
Distance = speed * time
$$D_T = S_2 t \tag{1} $$
$$ D_x+ S_1 \cdot t\gt S_2 \cdot t + 5 \tag{2} $$
$$t= D_T / S_2 \tag{3} $$
Collect like-terms in (2)
$$D_x \gt S_2 \cdot t - S_1 \cdot t + 5, $$
$$D_x \gt t(S_2 - S_1) + 5\tag{4}$$
From (1) $$ t = D_T /S_2 \tag{5} $$
Sub (5) into (4)
$$ D_x \gt {D_T} / {S_2} *(S_2 - S_1) + 5 \tag{6} $$
Plug the values from your question into (6)
$$D_x \gt D_T \cdot \frac{7} {8} \cdot (\frac{8}{7} - \frac{8}{15}) + 5, $$
$$ D_x \gt (\frac{8}{15})D_T + 5 \tag{7} $$
That's your final answer.