NavList:

Bill you wrote:

"Any body I view is on a great circle that passes through the body, my zenith
and nadir.  These great circles meet at my zenith and nadir.

When I observe one body (my only real experience to date) and if I could see
the great circle through the body, my zenith and nadir it would appear as a
straight line segment perpendicular to my horizon.  All refraction would be
linear and vertical."

 

Yes exactly. And all those arcs meet at the zenith. Back to the Sun's disk for a moment, since the sides of the Sun will be raised towards the zenith by refraction and since the lines are convering at the zenith, the width of the Sun necessarily is decreased, very slightly, by refraction. Likewise, two stars at the same altitude will be closer together as a result of refraction (though many books get this wrong).

 

And:
"Let's again assume the great circles are visible. If I look at two bodies
(say 45d apart) and point a camera at a point midway between them, the great
circle through that midpoint would appear to be a straight line.  The great
circles to the left and right through the bodies would appear as arcs from
the horizon to my zenith.  Refraction along those great circles would no
longer be linear as it is moving along the arc of the great circle to the
zenith, and would have x and y components."

 

Yes, when photographed (or mapped to any flat projection), those arcs would appear curved. But they're really still exactly vertical. And refraction is exactly vertical.

 

And:
"What prompted my question was a set of experiments using Spica and Arcturus
May 6 04:00 UT.  I calculated the angle by formula.  Then I determined the
Hc and azimuth of each star.  Using plane trig and a right triangle, I used
the difference in Hc as the opposite leg, and the difference in azimuth
(corrected by midpoint of Hc's using cosine) as the adjacent leg.  When I
calculated the hypotenuse, it was approx. 10% over the calculated distance.
This caused me to question if there was horizontal component to refraction
or if the opposite side was not indeed perpendicular, but rather slanted
toward the zenith.  Again, "towards the zenith" is the key."

 

The reason the numbers didn't work out here is simply that you can't solve spherical trig problems with plane trig equations. Over short distances (less than a few degrees), you can get good results (since a small section of a sphere is well-approximated by a plane) but not good enough for this sort of problem.

 

By the way, I don't think you were around when I posted a little thing on star-star sights last year. It's in the list archives "star-star sights" for April 6, 2004. It was designed as a preamble for some later stuff on lunars (see "Easy Lunars" April 28, 2004).