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I have searched in Google and found interesting articles for atmospheric refraction influence on sun's elevation above the horizon (from 0° till 90°). eg. Bennett G.G. etc. I would like to know if there is a formula that approximates the refraction when the sun's center is below the horizon (from 0° to ca. -3°) and taking in account for temperature and air pressure.

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  • $\begingroup$ Arguably, once the refracted value of elevation is below 0 degrees, the object is not visible, so computing refraction could be pointless. However, if you assume an perfectly transparent Earth that itself doesn't refract light, you might be able to answer this question. $\endgroup$
    – user21
    Commented Jun 5, 2018 at 17:31
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    $\begingroup$ Conventionally, to calculate the times of sunset or sunrise, it is calculated for an elevation -0.833 degree below the horizon, 16' as mean value for sun's semi-diameter plus 34' as mean refraction value for the sun at this position under mean weather temperature and air pressure. That's all assuming the observer's location at a height of 0 meter above the horizon, so your claiming is false, sometimes one still see the sun till -1 degree below the horizon $\endgroup$
    – Khaled
    Commented Jun 6, 2018 at 7:48
  • $\begingroup$ hko.gov.hk/m/article_e.htm?title=ele_00493 $\endgroup$
    – Khaled
    Commented Jun 6, 2018 at 7:51
  • $\begingroup$ OK, but if you're elevated, won't that change the value of refraction as well? I think the 34 minutes at the horizon applies to the observer's horizon, even if the observer is elevated. $\endgroup$
    – user21
    Commented Jun 6, 2018 at 15:56
  • $\begingroup$ The observer at a height H gains more dip angle at the horizon than an observer at sea level. See please siranah.de/html/sail040o.htm $\endgroup$
    – Khaled
    Commented Jun 6, 2018 at 16:41

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Without taking temperature and pressure into account, from my old code I found this approximate formula: (angles in degrees)

$\Delta \alpha_{reflect} = tan[(90°-\alpha_{technical}) \times 0.9679] \times 0.023626$

If this calculated reflection angle exceeds $\frac{34.9}{60}$ degrees, use that.

Observed elevation would be $\alpha_{observed}=\alpha_{technical}+\Delta\alpha_{reflect}$

Example

  • a point technically (without atmosphere) 34.9' (minutes of arc) below horizon, appears on the horizon.
  • a point technically on the horizon, appears 28.1' above the horizon.
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    $\begingroup$ I have a more precise (still empirical) set of formulas/steps that takes temperature and pressure into account, it is still as code - needs some reformatting. $\endgroup$
    – PaulS
    Commented Jun 12, 2018 at 0:11
  • $\begingroup$ Thanks Paul, I plotted this as a graph, from elevation 0 to 3 degrees below the horizon, it changes dramastically. I would know the range of use this formula to be a good approximation of the refraction, I would also know, in which programming language have you your code? $\endgroup$
    – Khaled
    Commented Jun 17, 2018 at 13:06
  • $\begingroup$ It is valid from (technical angle) -0.581666 degrees to +90.0 degrees (which will indicate an observed angle of +0.0 degrees to +90.0 degrees. Code was in c $\endgroup$
    – PaulS
    Commented Jun 17, 2018 at 22:04

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