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

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

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

<p>Observed elevation would be $\alpha_{observed}=\alpha_{technical}+\Delta\alpha_{reflect}$</p>

<p><em>Example</em></p>

<ul>
<li>a point technically (without atmosphere) 34.9' (minutes of arc) below horizon, <em>appears</em> on the horizon.</li>
<li>a point technically on the horizon, <em>appears</em> 28.1' above the horizon.</li>
</ul>

CC BY-SA 4.0
2018-06-12T00:11:47.633Z
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.
PaulS
CC BY-SA 4.0
2018-06-17T13:06:52.387Z
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?
Khaled
CC BY-SA 4.0
2018-06-17T22:04:03.190Z
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
PaulS
CC BY-SA 4.0
2018-06-11T23:46:35.730Z
2018-06-11T23:46:35.730Z
PaulS
<p>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.</p>

CC BY-SA 4.0
2018-06-06T07:48:56.950Z
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
Khaled
CC BY-SA 4.0
2018-06-06T07:51:39.803Z
http://www.hko.gov.hk/m/article_e.htm?title=ele_00493
Khaled
CC BY-SA 4.0
2018-06-06T16:41:13.593Z
The observer at a height H gains more dip angle at the horizon than an observer at sea level. See please http://www.siranah.de/html/sail040o.htm
Khaled
<positional-astronomy><atmospheric-effects>
Formula for sun's atmospheric refraction below the horizon
Khaled
# Formula for sun's atmospheric refraction below the horizon

(Asked by: username_0 on 2018-06-04)

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.

**Tags:** positional-astronomy, atmospheric-effects
> **Comments:**
> **username_0**: 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  
> **username_0**: http://www.hko.gov.hk/m/article_e.htm?title=ele_00493  
> **username_0**: The observer at a height H gains more dip angle at the horizon than an observer at sea level. See please http://www.siranah.de/html/sail040o.htm  

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## 1 answers

### Answer by username_1 on 2018-06-11. Score: 2

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.

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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?  
> **username_1**: 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  

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