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Now you live in a neighborhood where all the houses, including yours, are constantly wildly moving in all directions. Who's your closest neighbor then?

Dylan16807 8y ago
All directions? No idea. Each house has a ~circular track and has a different frequency? Then unless you're asking about right this second it's the two houses on the adjacent tracks, or whichever one of those has the closest track to mine. They live on that ring. It's really easy to figure out neighbors with concentric rings.
- jobigoud 8y ago
  > Then unless you're asking about right this second it's the two houses on the adjacent tracks Yeah, when someone is asking what thing is closest to them they usually mean closest right now, or closest at any random point in time. Not the thing that just so happen to be closest for a very limited amount of time. Thought experiment: imagine there was a twin Earth on the same orbit but the opposite side of the Sun. According to your metric this is the closest neighbor of Earth. What's strange is that you still use the average in your metric. You use Venus average distance to the Sun to get its orbital radius and declare it the closest ring track. Which leads me to a second thought experiment: imagine there was a planet with a very eccentric orbit making it pass very close to Earth at a single point in time, closer to any other body, but its average distance to the Sun is larger than Venus. What is the closest then? Third thought experiment: imagine we are no longer talking about planet orbits but about a merry-go-round where all rings are moving at a fixed angular speed. You are on a horse on ring 3 and there are other people scattered around. You are asked who is closest to you. Do you still use the ring distance as a metric or the regular euclidean distance to whichever person is closest?
  - Dylan16807 8y ago
    When someone says "closet neighbor", they don't mean the guy from three doors over that's walking down the sidewalk in front of you. > twin Earth Yeah, I'd probably say that's a closer neighbor than Venus or Mars. If Venus would stay on the same side of the sun as Earth I would consider it over anti-Earth, but because it goes almost as far away those brief sweeps don't do enough to overcome an orbital match. The delta-v to get to anti-Earth is almost zero. You might be able to convince me to treat an anti-Earth differently, but that doesn't change my answer on Venus vs. Mercury. > you still use the average in your metric No, I use the orbits being concentric near-circles. > second thought experiment If it's always between the orbits of Venus and Mars then it easily wins. If it's criss-crossing then the answer gets more complicated. I freely admit this logic doesn't apply when the orbits aren't concentric near-circles. I'm not trying to solve every case ever. > fixed angular speed If all the angular speeds are fixed, then you'd use the constant distance. This logic is a fallback for when you don't have a constant distance. Something that takes patterns into account, making it work better than an average.
- rejschaap 8y ago
  It gets interesting when the tracks are elliptical, but I assume that is what you mean with "~circular" and "ring".
int_19h 8y ago
I think you missed the point of my analogy. Houses aren't planets, they are orbits; humans inside them are the planets. Orbits don't wildly move in all directions.

Dylan16807 8y ago

All directions? No idea. Each house has a ~circular track and has a different frequency? Then unless you're asking about right this second it's the two houses on the adjacent tracks, or whichever one of those has the closest track to mine. They live on that ring. It's really easy to figure out neighbors with concentric rings.

jobigoud 8y ago
> Then unless you're asking about right this second it's the two houses on the adjacent tracks Yeah, when someone is asking what thing is closest to them they usually mean closest right now, or closest at any random point in time. Not the thing that just so happen to be closest for a very limited amount of time. Thought experiment: imagine there was a twin Earth on the same orbit but the opposite side of the Sun. According to your metric this is the closest neighbor of Earth. What's strange is that you still use the average in your metric. You use Venus average distance to the Sun to get its orbital radius and declare it the closest ring track. Which leads me to a second thought experiment: imagine there was a planet with a very eccentric orbit making it pass very close to Earth at a single point in time, closer to any other body, but its average distance to the Sun is larger than Venus. What is the closest then? Third thought experiment: imagine we are no longer talking about planet orbits but about a merry-go-round where all rings are moving at a fixed angular speed. You are on a horse on ring 3 and there are other people scattered around. You are asked who is closest to you. Do you still use the ring distance as a metric or the regular euclidean distance to whichever person is closest?
- Dylan16807 8y ago
  When someone says "closet neighbor", they don't mean the guy from three doors over that's walking down the sidewalk in front of you. > twin Earth Yeah, I'd probably say that's a closer neighbor than Venus or Mars. If Venus would stay on the same side of the sun as Earth I would consider it over anti-Earth, but because it goes almost as far away those brief sweeps don't do enough to overcome an orbital match. The delta-v to get to anti-Earth is almost zero. You might be able to convince me to treat an anti-Earth differently, but that doesn't change my answer on Venus vs. Mercury. > you still use the average in your metric No, I use the orbits being concentric near-circles. > second thought experiment If it's always between the orbits of Venus and Mars then it easily wins. If it's criss-crossing then the answer gets more complicated. I freely admit this logic doesn't apply when the orbits aren't concentric near-circles. I'm not trying to solve every case ever. > fixed angular speed If all the angular speeds are fixed, then you'd use the constant distance. This logic is a fallback for when you don't have a constant distance. Something that takes patterns into account, making it work better than an average.
rejschaap 8y ago
It gets interesting when the tracks are elliptical, but I assume that is what you mean with "~circular" and "ring".

int_19h 8y ago

I think you missed the point of my analogy. Houses aren't planets, they are orbits; humans inside them are the planets. Orbits don't wildly move in all directions.