NavList:

Bob,

I'm no expert, but here's what I found:

I'm not sure what the exact time and location of the sight was, so I'll use 53°33.0'N, 113°28.1'W (Edmonton, AB, Canada) and 2015, Mar. 26 0300UT for this example. The moon's GHA was 147°42.1' with a declination of N18°07.7'. This yields an LHA of 34°14.0'. We can calculate the geocentric altitude of the center of the moon with the N.A. equation:

asin(sin(Dec.)·sin(Lat.)+cos(Dec.)·cos(LHA)·cos(Lat.)) = 45°48.9' = "Hc"

We can calculate the amount of refraction with:

-0.0167°/tan(H+7.32/(H+4.32)) = -1.0'

The horizontal parallax of the moon was 56.9'. We can use that figure to calculate the parallax in altitude:

0°56.9'·cos(H) = 39.7'

Since we are going from "Hc" to "Ha", we will add the refraction and subtract the parallax in altitude:

45°48.9' + 1.0' - 39.7' = 45°10.2' = "Ha"

Entering the N.A. with that altitude, we get a main correction of 50.4' and an additional correction of 3.3' (for UL, HP=57.0'). Total correction is 53.7'. Subtracting 15' from that, we get 38.7'. Adding that to 45°10.2' yields 45°48.9' = "Hc"

So, from what I can tell, your logic appears sound. :) Hope that helps!

Cheers,

Sean C.