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It's certainly not always the case. In planar Euclidean geometry, for any three points, if the triangle formed by those three points has all its angles less th
by jameshart 4y ago
It's certainly not always the case.
In planar Euclidean geometry, for any three points, if the triangle formed by those three points has all its angles less than 120 degrees, then the sum of the distances from the three points to certain points inside the triangle will be smaller than the sum of the shortest two sides. Only if one of the angles exceeds 120 degrees will one of the three points always have the shortest total travel distance.
The relevant concept is the 'Fermat point', which is the point that minimizes total distance from the vertices of a triangle.
In spherical geometry, it gets more complicated, but the same logic applies to spherical triangles that aren't too large. You can definitely find sets of three points on a sphere, such that points inside the triangle are the one which minimizes travel distance.
- Kwpolska 4y agoDoes this still hold if you count the flown distance twice, because the meeting participants would probably like to get back home after it ends?
- winnit 4y agoOf course, the lengths are just doubled. If someone travels twice as far to get to someone’s home they also have to travel twice as far back. If the two met in the middle instead there would be four journeys but each would be half as long (2+2=1+1+1+1).
- wiml 4y agoThe carbon cost of a flight isn't a linear function of its distance, though — takeoff is much more expensive than cruise, and on long flights there's the cost of hauling the fuel itself, rocket-equation style. (Presumably the data underlying this website accounts for all that.)
- seabass-labrax 4y agoIndeed; furthermore, some air traffic control regions have adopted a direct routing strategy, replacing the conventional routing strategy of published airways. This means that a flight from, say, Prague to Bucharest should be ever so slightly more efficient than the equivalent distance from London to Venice because of the direct routing.
- jameshart 4y agoAll of which complexity just means that air travel carbon costs are a slightly more complex noneuclidean metric than just ‘spherical geometry’. But you should still be able to define a metric over the graph of airports which is based on costs. And what Fermat points tell us is that even in well behaved metric spaces like the plane there can be cases where minimizing distance means going to a node other than one of the starting point. So in more complex spaces it seems likely that that remains the case. Which was all the op was asking.
- abdullahkhalids 4y agoI would expect that aircraft emissions have a fixed cost + a distance dependent variable cost. Coupled with the fact that we are optimizing over a discrete set of locations, it is very possible that the optimal location is at one of the corners of the triangle.
- jameshart 4y ago… sure, but also possible that it won’t always be.
- akira2501 4y agoThey have variable costs that aren't dependent on the distance. Some planes take off in 10 minutes, some planes have to sit for 45 before they can take off. This all also assumes that you will always be able to fly directly to your destination on the selected day. If you end up diverting or getting delayed, or if certain routes make that _more_ likely, then this wouldn't be a win at all.