NavList:

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.).

See:

http://fer3.com/arc/m2.aspx/Star-rise-set-visibility-LaPook-oct-2014-g28867

http://fer3.com/arc/m2.aspx/Refraction-FrankReed-aug-2005-w25210

http://fer3.com/arc/m2.aspx/Latitude-Polaris-visibility-Rudzinski-oct-2014-g28939

http://www.icq.eps.harvard.edu/ICQExtinct.html

gl