NavList:

   

Reply

Dear Robin

Yes we have worked out it is very much not a small circle. Indeed it
is (a subset of) the intersection of a quartic "cylinder" with the
sphere. The quartic can have one or two components and can get quite
triangular in shape.

You are correct that the stereographic projection is indeed 3rd order
(my colleague Gabor Megyesi, an algebraic geometer, just told me
this).

I am working on some animations in Mathematica. ... but I am a bit ill
at the moment so please be patient. Brain only firing on two
cylinders.  (My Covid 19 test kit has just arrived)

Best wishes

Bill

On Thu, 21 May 2020 at 17:17, Robin Stuart  wrote:
>
> It may be a bit late now but in order to determine whether a isoazimuth is a 
small circle I decided to construct a counter example. The plan was to obtain 
4 points lying on an isoazimuth. Choosing 3 of those points calculate the 
centre of the small circle on which they fall. Then choose a different 
combination of 3 out of the 4 points and calculate the centre on the circle 
that those fall on. If the centres are different then an isoazimuth is not a 
small circle.
>
> Circles great and small are conveniently handled analytically by 
stereographic projection onto the complex plane as described here. Circles on 
the sphere map to circles on the plane. Moreover since the Littrow projection 
is a simple analytic function of the stereographic projection this offers a 
consistent approach for the treatment of equal altitude circles and 
isoazimuths. In an earlier post by Bill Lionheart provided a link to a useful 
article which stated "If the complex z-plane is the Littrow map and the 
w-plane is the polar stereographic map, then z = w +1/w". After some time 
wrestling with this and trying to make it behave I realized there is a typo. 
It should read "z = w -1/w" and that makes all the difference.
>
> In any case here's the counter example generated by appealing to the 
properties of the Littrow projection
>
> Assume the GP of the body being observed is at 16°46.8'N 45°51.1'E. An observer at any of the points
> P1 = 10.306896°N 55.375875°E
> P2 = 0.248451°S 70.71412°E
> P3 = 25.433371°S 95.262638°E
> P4 = 48.38593°S 102.746085°E
>
> The points P1, P2 and P4 all lie on a small circle with centre or pole at
> c124 = 37.001705°S 28.211937°E
> this can be checked calculating their distances from the centre and seeing that they are all the same.
>
> The points P1, P2 and P3 all lie on a small circle with centre or pole at
> c123 = 43.700701°S 18.268654°E
>
> These centres are not the same hence isoazimuths are not in general small circles.
>
> Under stereographic projection onto the complex plane isoazimuths appear to 
satisfy 3rd order equations. It may still be possible to calculate the 
intersection of 2 isoazimuths or an isoazimuth with a small circle. This 
would be analogous to the algebraic solution of the double altitude problem 
discussed here.
>
>
> Robin Stuart
>
> 

   

Reply