A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Gary LaPook
Date: 2017 Jan 25, 02:39 -0800
A body at declination 38 south will have a zenith distance of 90 degrees when it crosses your meridian (90 - 52 = 38). This means its altitude would be zero. just on the horizon for a minute as it crosses the meridian and then not be visible. But, since there is an atmosphere (you can breathe after all), extenction dims a star's light near the horizon so that even if it were Sirius, magnitude -1.5, you couldn't see it when it was less than 1 degree above the horizon. But since Sirius is only 16 degrees south this doesn't apply to you at 52 north. For a 2.5 magnitude star then, you won't see it unless it is higher than 3 degees making the zenith distance 87 degrees when it crosses your meridian. 87-52 = 35 south declination is the farthest south declination that you might be able to see a 2.5 maginitude star right on your meridian and even lower for dimmer stars. For a 5th magnitude star it would have to be 15 degrees above the horizon, ZD 75 degrees. (75-52 = 23 south declination.).