On Jan 6, 8:33 am, "warden [Andrew Leach - Maps API Guru]"
<[email protected]> wrote:
> On Jan 6, 3:48 pm, Markw65 <[email protected]> wrote:
> > Not really. 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.
>
> So where is the centre of gravity on that great circle? (29,60) may
> well be a reasonable answer for the spherical triangle drawn on the
> surface of the earth.
Not by any definition of "reasonable". The spherical "triangle" is
just a line (an arc, if you will). That point is nowhere near the arc.
> If you join the three points the incentre of the
> plane triangle is inside the earth.
Right - in 3D. But we're working on a 2 dimensional manifold. So the
CoG has to be *on* the manifold :-)
Not sure if there is a standard definition for CoG on a sphere, but
presumably its the point that minimizes the sum of the great-circle
distances to each of the vertices.
At any rate, its a point on the sphere, and it needs to be within the
the spherical triangle defined by the three vertices. Clearly 29,60
doesnt satisfy that :-)
>
> > I realize he said his points were all within 10 miles of each other -
>
> This is the crucial point.
No, its irrelevant, as I went on to show...
>
> > but you can still get the same problem with eg 89.99,0; 89.995,0;
> > 89.995,180)
>
> You're adding apples and oranges here, because one point is in a
> different hemisphere.
No. Those points are all within a few miles of the north pole.
> You need to normalise the points by shifting
> them around the great circle -- say by 40deg to {(49.99,0),(49.995,0),
> (50.005,0)}, do the calculation and arrive at (49.997,0) and then
> shift the result back to (89.997,0).
Which exactly proves my point!
The proposed equation is a reasonable approximation the the CoG over
most of the globe.
But near the poles its hopeless (indeed, if any part of the spherical
triangle defined by the three points comes close to either of the
poles, its hopeless).
> You can get the same result by
> saying the third point is (90.005,0) -- that is, more than 90deg from
> the Equator.
Right, and you could make the method fail miserably by adding 3600deg
to any of the coordinates - but then they're not normalized
coordinates.
In my example all the points had -90 < lat < 90 and -180 < lng <=180.
>
> Applying that method to the first example, the points are {(0,0),
> (87,0),(180,0)} and the result is (89,0). That's also a reasonable
> result,
Not just a reasonable result - the "correct" result (although only
because the three points lie on a great circle).
> but it doesn't reflect the same calculation as either of the
> first two goes with triangles.
Bingo!
So the proposed method of computing the center of gravity isnt
correct, is it?
Which as I said above, proves my point...
>
> Isn't spherical geometry fun?!
Yes.
Mark
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