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.
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).
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.
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.
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.
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).
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.
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.