0

I'm trying to calculate the optimal speed (minimizing time) to intercept another object that is moving in the same circular path as me. Here's the catch, there are some constraints:

  • There is a mininum value for the time ($t_{min}$).
  • My speed has a minimum ($\omega_{1min}$) and a maximum ($\omega_{1max}$) value.

My initial position is $\alpha_1$ and the target's initial position $\alpha_2$.

$$\Delta_0=\alpha_2-\alpha_1$$

I don't care how many "turns" each object does before the encounter. So the final difference is $2{\pi}N$. Where $N$ is any integer number including 0 and negative values. So here is my main equation: $$\Delta=\Delta_0-(\omega_1-\omega_2)t$$ solving for $\omega_1$ and replacing $\Delta$: $$\omega_1=\omega_2+\frac{\Delta_0-2{\pi}N}t$$ Of course my first approach is to solve using $t=t_{min}$ and $N=0$. But what happens if the resulting angular speed is below the lower limit ($\omega_1<\omega_{1min}$) or above the upper limit ($\omega_1>\omega_{1max}$)? Should I increase $N$? Decrease $N$? Use $\omega_{1min}$ or $\omega_{1max}$ and solve for $t$?

There are so many possibilities that I don't know which one I should I apply under each condition. I expect that the solution is some sort of iteration of steps, which is fine for me because this is going to be converted into programming code.

Notes: I don't know the relation between $\omega_2$ and $\omega_{1min/max}$, but I do know all of them are positive:

$$0<\omega_2$$

$$0<\omega_{1min}<\omega_{1max}$$

I also realize that in some cases there may be multiple solutions, in such cases I would choose the solution where $N$ is closer to zero.

TShiong
  • 1,257
  • I'd look at this the other way around: What's the range of possible relative distances $\Delta$ I can attain at time $t_{\mathrm{min}}$? Is there a multiple of $2\pi$ in that set? If not, how much more time will it take until going at either $\omega_{1\mathrm{min}}$ or $\omega_{2\mathrm{max}}$ reaches a point $\Delta=2\pi N$? – aschepler Feb 03 '22 at 23:09
  • Thanks @aschepler! I ended up using your approach. Finding the right $N$ when I had some multiples of $2\pi$ in the $\Delta$ range was not super straight forward, but eventually I solved it. – kiel814 Feb 06 '22 at 03:30

0 Answers0