On Jan 7, 12:35 am, marcelo <[email protected]> wrote:
> On Jan 6, 10:49 pm, Markw65 <[email protected]> wrote:
>
> > > No.
> > > "A spherical triangle is a figure formed on the surface of a sphere by
> > > three great circular arcs intersecting pairwise in three vertices"
> > > From:http://mathworld.wolfram.com/SphericalTriangle.html
>
> > Thats why I put the "" around "triangle" - because its degenerate.
>
> It's not "degenerate". It's a different shape altogether, called a
> "lune".
If you're going to pick nits, no, its not a lune, because a lune is
defined by *two* great circles, and there's only one in my example
At least, I assumed that was obvious. Apparently assumed one went
around the equator. But the area defined by three points on a sphere
has to be the *smallest* possible area - otherwise you never know
whether you're looking at the inside or the outside, amongst other
things!.
I'll admit that I should have put the third point at 1,180 to avoid
that problem. Again, it doesnt change anything as far as the original
problem goes.
>
> > But to be honest, Im not sure your definition is particularly
> > authoritative - simply because it *does* exclude degenerate triangles.
>
> Search for other definitions until you find a site you trust and show
> us a link.
> All the definitions I can find refer to three great circles, not more
> not less.
I'll admit that I couldnt find one. But its doesnt affect the argument
at all. And since you appear to agree that 0,0; 87,0; 0,180-epsilon
for any epsilon!=0 is a triangle, that pretty much shows that 0,0;
87,0; 0,180 is a degenerate triangle by the definition of degenerate!
But its not worth arguing about. Feel free to let me know you
disagree. I have nothing more to say on this, since its totally
irrelevant to my point.
> > Doesnt really affect anything either way. Just move the last point to
> > 179.99999999999, and everyone's happy.
>
> Yes, it does.
> If you move the last point away from the anti-meridian, as you suggest
> here, then you have a triangle instead of a lune.
whatever :-)
>
> In any case, this discussion started when you challenged Mike's method
> for finding the center of gravity, saying
> ---------------
> "Suppose the three points are (lat,lng) 0,0; 87,0; 0,180.
> Your "center of gravity" would be 29,60 - which isnt even on the same
> great circle. "
> ---------------
> But what "same great circle" are you referring to?
> "same great circle" as what?
The same great circle that all three points lie on! What else?
I was trying to create an example where it was totally obvious that
there was a problem - without realizing that people would get so
caught up in the terminology. I deliberately avoided the pole (because
I knew people would get hung up on that), but hadnt anticipated the
other nits people would pick :-)
Try this one:
0,0; 87,0; .0001,179.9999
Now we have a very thin spherical triangle, which is approximately a
line from 0,0 to 0, 180 going over the north pole. But the "center" as
given by mikes formula is at 29,60. How can that *possibly* be the
correct answer?
Or, if you're being confused by the curvature of the earth and saying
"but the real center of gravity lies inside the earth so its hard to
know what the correct answer is", then try my second example (again,
adjusting the points slightly).
89.99,0; 89.995,0; 89.995,179
Now we have an approximately flat triangle. And the "center" by mike's
method is at 89.9933333,59.666666. Which is several miles outside of
the triangle
> I don't see any problem with Mike's method.
Still?
Mark
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