this post was submitted on 14 Mar 2024
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I'm sure pirates knew the answer. Probably fighter pilots as well.

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[โ€“] Windex007@lemmy.world 11 points 8 months ago* (last edited 8 months ago) (1 children)

It depends on what you mean by "escape", and what you view as the alternative.

I suspect that the pursuer could never converge on the same instantaneous point, given sufficient initial distance (and orientation). At a certain distance, the prey could enter a stable orbit around the pursuer. I don't have a mathematical proof but I strongly suspect this to be the case,and I can envision the structure of a proof.

Could the prey infinitely extend the gap between themselves and the pursuer? No. I don't have the tooling to actually present such a proof, but of that one I am confident.

I think if you introduced concepts of obstacles and a "radius of escape" (where if the gap meets a threshold the predator is permanently foiled), then there are almost certainly scenarios where the prey could escape.

We actually see this scenario play out in nature all the time

[โ€“] Melatonin@lemmy.dbzer0.com 2 points 8 months ago (1 children)

WOW, really nice explanation.

[โ€“] Windex007@lemmy.world 3 points 8 months ago (1 children)

Thank you!

Although, I'm realizing that for completeness, there probably are mathematical constraints around the relationship between the required absolute values of turning speeds and movement speeds. They're kinda egde-casey for any practically imagined scenarios, but would come into play for a rigorous proof.

[โ€“] rufus@discuss.tchncs.de 1 points 8 months ago* (last edited 8 months ago) (1 children)

Concerning the proof, I'd consider that at any given point where both objects haven't converged yet, there has to be a next point that can be reached by the ship with the higher maneuverability but not by the faster ship. It's probably calculus from that point on and I'm not really good at that. If there's always such a possibility, the slower ship can always outmaneuver the other one. And seems to me like vectors in a polar coordinate system would be made for this.

Set vector1 equal to vector2 plus an arbitrary distance. See if there's a solution for phi2 < phi1.

[โ€“] Windex007@lemmy.world 2 points 8 months ago (1 children)

Yeah, I've got a similar thought.

There do exist scenarios where there would be a solution... For example if the base turning speed is 0, or the distance between them is already sufficiently small (and their relative orientations are aligned).

[โ€“] rufus@discuss.tchncs.de 1 points 8 months ago (1 children)

Sure. And I mean the "sufficiently small" distance is exactly the question. I mean it's not really an interesting question to ask if they're still 12 nautical miles apart... The initial distance isn't really of concern. It just has to work for any given initial state. And the next question is, are we talking about entering a ship or using cannons? Then it's either can the distance become 0 or can it get less than something.

[โ€“] Windex007@lemmy.world 2 points 8 months ago (1 children)

I agree they aren't practically interesting egde cases, because in order to hit them, they're no longer meaningful for realistically describing pirate ships. At very high turning speeds, it also ceases to matter that one is 3/4 than that the other either. But at that point, we're talking about pirate ships spinning like ballerinas across the seven seas.

[โ€“] rufus@discuss.tchncs.de 1 points 8 months ago* (last edited 8 months ago)

Feel free to extend that problem to fighter jets or ballerinas playing tag ๐Ÿ˜†

However, I'm pretty sure it's already solved. Doesn't seem difficult to prove and has had applications for centuries already. And I've played the Robots Game when I was 12 or something...