Next message: Alexandre E Eremenko: "Re: George's example, corrected"
George's example of a "shifted circle" was the following.
Begin with the circle centered at
(Lat, Long)=(0,0)
of radius 60 degrees.
Shift every point 1 degree N.
The resulting figure is not a circle,
and the best fitting circle deviates by more than 14 miles from it.
All this is correct.
Indeed, choose 4 points on the original circle as follows:
A=(0,60), B=(0,-60), C=(60,0) and D=(-60,0).
They are shifted to the new points
a=(1,60), b=(1,-60), c=(61,0) and d=(-59,0).
Now consider the circle best fitting these four points.
It is clear from symmetry that the center of this best
fitting circle should have longitude 0.
The equidistant point from c and d is x=(1,0).
However, the distance from this point x to a (or to b)
is about 50d46' (of a great circle). Simple trig is used to
compute this.
So the deviation of our figure from any circle is about 14
miles, as George stated.
This shows:
a) I was wrong when I wrote that the "size of the circle
is irrelevant, but only relevant is its distance from
the poles". Size of the circle is also relevant.
b) It is a bad idea to use position circles of large radius
for running fixes. In any case, it is wrong to think that
a shift of a position circle will be close to a circle.
c) The only case when the use of position circle
(as opposite to a position LINE on a Mercator chart) is
justified is when the circle is small.
Thus all statements of George on the subject (except that one cannot draw
a circle through 3 points) were correct.
Alex.
