NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
George's example, revisited
From: Alexandre Eremenko
Date: 2006 Jun 17, 05:36 -0400
From: Alexandre Eremenko
Date: 2006 Jun 17, 05:36 -0400
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.