Michael Stachowsky

Ah, the rub is that I am obligated by department policy to hold both a midterm and an exam

@bendl: people seemed to fixate on the airship more than I had intended, but the question is all about the actual world and the magic system. The question is: when the people creating the magic through belief do not themselves believe in it, how does the magic system work? That is world based, not story based

@JMac: Yes. Once the magic is released it is released until the "Universal Universality" test I talked about in one of my comments below is failed. That's sort of the crux of the problem

@ITAlex Interesting concept that I hadn't thought of...I'll think on it now :-P

@ITAlex not necessarily. The belief has to be universal (with a "universal universality" [UU] test to be developed). So "airships fly" is a universal belief, but there is too much doubt (or evidence) to disprove a round Earth. There is, however, a risk that yes, if the flat earthers are convincing enough to pass the UU test, then sure, the earth becomes flat.

Oh I see, yes that makes sense. Definitely not going for "novel in the history of fiction" :-P

Can you clarify? Did I basically re-create something that is common?

Interesting comments, everyone. I am fairly sure that going from a 1atm N2/O2 atmosphere to a 0.2atm O2 atmosphere will require decompression. I like @ths 'ssuggestion, although that's for a different question (namely: how do you make a decompression chamber out of a shipwreck)

fair enough, although I expect that the astronauts required time to decompress and be brought down to that pressure beforehand, yes?

Agreed, but if one needs rescuing then they have to decompress, not so? The rescuers don't have this problem, the rescuees do.

I like the idea, but at 0.2 bar humans would experience decompression sickness unless they decompressed a lot, and to be honest I don't really know what would happen even then in such a low pressure. Since the point of the story is a rescue mission, I imagine that the characters won't have time to decompress. +1 for the idea, though

Howdy. Stack Exchange is a question/answer site, not a general posting site. Do you have a specific question to ask?

Indeed. No problem. I was glad to help!

What was it for?

Yeah, I thought that might the ticket. So you've shown that under certain conditions the time-optimal path is not the distance-optimal path. That's pretty interesting, actually

OK, but does that take more time (turning) than it does to go the shortest path?

The constant I suggested should make it so that they are both the same order of magnitude and neither should dominate, in theory anyway. I'm sure we can figure it out to make it a better scaling factor.

Interesting. I wonder if there is a practical use for angle minimization over distance

How do you mean? Did it find the longer path with the smaller angle?

so what do you end up with?

oh, good

Perhaps try pre-scaling your angles like this: Define $d_scale = \sqrt{(x_s-x_f)^2 + (y_s - y_f)^2}$, and then scale your angles by multiplying by $d_scale$

I think a potential issue is that the distance (Euclidean) might be a different unit than the angles. For instance, if your angles are at most 0 to pi, and your distance is 1e6, then the angles are practically irrelevant. However, if your distance is, say, 1e-3, then your angles dominate

You said you tried to minimize angle along with distance in your first post. Did you try that algorithm with the counter-example setup? What happens?

For instance, the minimal turning angle is zero if there is no actual travel goal to end up at

Ah, but without an actual Euclidean distance, it's unlikely that you'll end up at your goal, I think.

How are you setting it up? You'll have to consider either the absolute value of the angle or its square or something positive

It's a very interesting problem, though. I wonder what the 'shortest' path would look like if your only goal were to minimize turning angle?

I've edited the figure to demonstrate

I made the image imprecisely. If you move the square down so that it touches the rectangle, then I believe my turning angles are still correct. I know that the shortest path should be through the vertices, but I lack the paint skills to do it :-P