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    Re: STARS below ?? declination are never visible.
    From: Frank Reed
    Date: 2017 Jan 26, 09:55 -0800

    Herman Dekker, you wrote:
    "Just a simple question, on internet I did not find the answer. I live on 52° North at sealevel. I can never see Stars (above the horizon)  that have a S declination more then ?? °"

    I suspect the answer you were looking for was the simple geometric one: 38° S. It's just 90 minus your latitude. I find that it's most useful to have a good understanding of the meridian altitude of the celestial equator, which is the same thing numerically; its 38° high. Then you can easily picture altitudes above and below there as easy offsets in declination. For example, if the celestial equator crosses the meridian with an altitude of 38° (as it does from 52°N), and you learn of a comet with a declination of 30°S, you can easily see that the comet's maximum altitude is 8° above the southern horizon. Likewise it's easy to work out the Sun's maximum and minimum altitudes during the year: you take the meridian altitude of the celestial equator and add/subtract 23.5°. Easy and quick.

    Getting into more detail: Refraction gives you an extra half a degree, but extinction takes that back (in a complicated way). Gary LaPook already linked the extinction tables from "ICQ" which I highly recommend reviewing (see his post). For quick reference, here's the average extinction table (in magnitudes for various elevations above sea level (h in km), somewhat abbreviated (note that z is zenith distance):

         z     h = 0    h = 0.5    h = 1     h = 2     h = 3
        10      0.29      0.24      0.21      0.16      0.13
        40      0.37      0.31      0.27      0.21      0.17
        60      0.56      0.48      0.41      0.32      0.26
        70      0.82      0.70      0.60      0.47      0.39
        75      1.08      0.92      0.79      0.62      0.51
        76      1.15      0.98      0.84      0.66      0.54
        77      1.24      1.05      0.91      0.71      0.58
        78      1.34      1.13      0.98      0.76      0.63
        79      1.45      1.23      1.06      0.83      0.68
        80      1.59      1.34      1.16      0.91      0.74
        81      1.75      1.48      1.28      1.00      0.82
        82      1.94      1.65      1.42      1.11      0.91
        83      2.19      1.86      1.60      1.25      1.03
        84      2.50      2.12      1.83      1.43      1.17
        85      2.91      2.46      2.13      1.66      1.36
        86      3.45      2.93      2.53      1.97      1.62
        87      4.23      3.59      3.10      2.42      1.99
        88      5.41      4.59      3.96      3.09      2.54
        89      7.38      6.26      5.40      4.22      3.46
        90     11.24      9.53      8.23      6.42      5.28

    You can see that the rule Geoffrey Kolbe suggested works reasonably well for observers 2 km above sea level or in unusually dry air (maybe his desert experience at work!). Right at sea level, under average atmospheric conditions, extinction is considerably more severe. Reading old logbooks, navigators occasionally noted in the 19th century that they could see the North Star from about latitude 3° N. With a polar distance back then of about a degree, this implies that they were observing it when its altitude was about 4°. Now the corrected apparent magnitude of Polaris is 1.97, and if we check the sea level extinction table above, at a zenith distance of 86° we find 3.45, which implies that the "extincted" magnitude of the star would be just 5.43. Average observers can detect stars as faint as 6.0 roughly, so that's about right. The North Star would be visible, barely, when on the high side of its circum-polar "orbit" around the north celestial pole from a latitude of 3° N. Using that standard, and trusting the extinction table, we can make a list of minimum altitudes for stars of various magnitudes at sea level:

    mag.  min alt.
    4        14°
    3          7°
    2          4°
    1          2.7°
    0          1.8°
    -1         1.1°
    -5         0°

    This table would apply for naked eye visibility by an average observer from a dark sight at sea level under average atmospheric conditions. So if there's some supernova that appears in southern skies at declination 38.5° S with an apparent magnitude of, let's say, -10, then you'll be able to see it as it peaks above the horizon at meridian transit from 52° N. Similarly, if you want to see a star of magnitude 2, then you will need it to be 4° above the southern horizon. Since the geometric declination there is 38° S, you will only be able to see second magnitude stars with declinations above 34° S. That's with all the conditions above. Change any of them, and you'll have different outcomes. For example, if you use binoculars, you'll immediately take your limiting magnitude down by 3 or 4. On the other hand, if you're observing from a sight with typical "suburban" light pollution, you will lose 2 or 3 magnitudes.

    So what's the easy, modern way to test all this out (in daylight!)? Although it has had its flaws and bugs over the years, the open source project Stellarium has become a stable high-quality desktop planetarium tool, and it's free for download from stellarium.org. Just set it to your observing location and then have a look: what declination of stars are visible above the southern horizon? The product includes both refraction and extinction. You can directly see which stars will be visible on the meridian above your southern horizon. You can also "turn off" the atmosphere to see the simple geometric solution. One catch: the standard settings in Stellarium appear to use an extinction coefficient for high altitude. Go into the settings, and you can change the extinction from 0.13 to 0.25. That's appropriate for sea level. 

    By the way, the very rapid change in extinction near the horizon also helps explain the Venus UFO phenomenon. If you have a clear eastern horizon, and Venus is near maximum brightness, while it's rising through the first couple of degrees of altitude (where no stars are visible), it transitions from a moderately bright red object to a brilliant white object in just 10 or 15 minutes. During those minutes it barely changes its altitude and azimuth creating the illusion of something fast and large coming straight at the observer. Apparently this has startled more than a few pilots at high altitude in pre-dawn hours.

    Frank Reed
    ReedNavigation.com
    Conanicut Island, New England
       
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