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. If you join the three points the incentre of the
plane triangle is inside the earth.
> I realize he said his points were all within 10 miles of each other -
This is the crucial point.
> 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. 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). You can get the same result by
saying the third point is (90.005,0) -- that is, more than 90deg from
the Equator.
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, but it doesn't reflect the same calculation as either of the
first two goes with triangles.
Isn't spherical geometry fun?!
Andrew
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