<p>Here's a partial answer:</p>
<p>As mentioned in the comments, the transmission of the atmosphere depends quite a lot on local factors. But given a transmission <span class="math-container">$T_0(\lambda)$</span> at zenith, the transmission at an angle <span class="math-container">$\theta$</span> from zenith can be written
<span class="math-container">$$
T(\lambda,\theta) = T_0(\lambda)\times X(\theta),
$$</span>
where <span class="math-container">$X$</span> is the <a href="https://en.wikipedia.org/wiki/Air_mass_(astronomy)" rel="nofollow noreferrer">air mass</a>.</p>
<p>The air mass can be approximated as <span class="math-container">$X=\sec\theta$</span> up to <span class="math-container">$\theta\simeq70^\circ$</span>, but as you approach the horizon, the calculation becomes increasingly model-dependent. A relatively simple expression is given by <a href="https://ui.adsabs.harvard.edu/abs/2002DIO....12....3P" rel="nofollow noreferrer">Pickering (2002)</a>:
<span class="math-container">$$
X(\theta) = \csc\left(90^\circ-\theta + \frac{244}{165+(90^\circ\!\!-\!\theta)^{1.1}} \right),
$$</span>
where <span class="math-container">$\theta$</span> is measured in degrees, and <span class="math-container">$\csc x\equiv 1/\!\sin x$</span>. This model doesn't take into account refraction (which requires ray-tracing), so may be off by half a degree or so when the Sun is at the horizon.</p>
<p>If <span class="math-container">$S_0(\lambda)$</span> is the un-transmitted Solar spectrum, the observed spectrum is then
<span class="math-container">$$
S_\mathrm{obs}(\lambda,\theta) = S_0(\lambda)\times T_0(\lambda)\times X(\theta).
$$</span></p>
<p>Calculating the color that the human eye perceives, given a spectrum, is not trivial, but I recently outlined an approch in <a href="https://astronomy.stackexchange.com/a/40014/5264">this answer</a>.</p>

CC BY-SA 4.0
2020-11-27T14:40:52.727Z
2020-11-27T20:16:56.223Z
2020-11-27T20:16:56.223Z
pela
<p>I would like to know at what angle in the sky the sun changes from a white yellow to orange and red. I know that after sunset it's considered twilight until 18° beneath the horizon. So how many degrees above the horizon generally before the sun changes color?</p>

CC BY-SA 4.0
2020-11-27T03:23:17.593Z
It depends on multiple factors such as altitude, humidity, temperature, air pollution (natural and industrial), etc.
Pierre Paquette
CC BY-SA 4.0
2020-11-27T03:23:49.210Z
Oh! and welcome to Astronomy Stack Exchange! :-)
Pierre Paquette
CC BY-SA 4.0
2020-11-27T03:30:58.973Z
Ok, so is there a way mathematically (maybe a resource you could point me to) to tell how the light will scatter based on the degree of the sun in the sky along with those other factors? Maybe it doesn't have to be exact but could just be based on averages of real world examples? And thank you :-)
Peyton Hanel
CC BY-SA 4.0
2020-11-27T04:06:34.487Z
There must be, but I’m afraid (and sorry) that I don’t know it, hence I only comment instead of answer…
Pierre Paquette
CC BY-SA 4.0
2020-11-27T09:09:33.810Z
It mostly depends on the number of particulates in the air and so will vary from place to place and time to time.
ProfRob
<the-sun><sunset><twilight>
what degree in the sky does the sun change color?
Peyton Hanel
# what degree in the sky does the sun change color?

(Asked by: username_0 on 2020-11-27)

I would like to know at what angle in the sky the sun changes from a white yellow to orange and red. I know that after sunset it's considered twilight until 18° beneath the horizon. So how many degrees above the horizon generally before the sun changes color?

**Tags:** the-sun, sunset, twilight
> **Comments:**
> **username_1**: It depends on multiple factors such as altitude, humidity, temperature, air pollution (natural and industrial), etc.  
> **username_1**: Oh! and welcome to Astronomy Stack Exchange! :-)  
> **username_0**: Ok, so is there a way mathematically (maybe a resource you could point me to) to tell how the light will scatter based on the degree of the sun in the sky along with those other factors? Maybe it doesn't have to be exact but could just be based on averages of real world examples? And thank you :-)  
> **username_1**: There must be, but I’m afraid (and sorry) that I don’t know it, hence I only comment instead of answer…  
> **username_2**: It mostly depends on the number of particulates in the air and so will vary from place to place and time to time.  

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## 1 answers

### Answer by username_3 on 2020-11-27. Score: 4

Here's a partial answer:

As mentioned in the comments, the transmission of the atmosphere depends quite a lot on local factors. But given a transmission $T\_0(\lambda)$ at zenith, the transmission at an angle $\theta$ from zenith can be written
$$
T(\lambda,\theta) = T\_0(\lambda)\times X(\theta),
$$
where $X$ is the [air mass](https://en.wikipedia.org/wiki/Air_mass_(astronomy)).

The air mass can be approximated as $X=\sec\theta$ up to $\theta\simeq70^\circ$, but as you approach the horizon, the calculation becomes increasingly model-dependent. A relatively simple expression is given by [Pickering (2002)](https://ui.adsabs.harvard.edu/abs/2002DIO....12....3P):
$$
X(\theta) = \csc\left(90^\circ-\theta + \frac{244}{165+(90^\circ\!\!-\!\theta)^{1.1}} \right),
$$
where $\theta$ is measured in degrees, and $\csc x\equiv 1/\!\sin x$. This model doesn't take into account refraction (which requires ray-tracing), so may be off by half a degree or so when the Sun is at the horizon.

If $S\_0(\lambda)$ is the un-transmitted Solar spectrum, the observed spectrum is then
$$
S\_\mathrm{obs}(\lambda,\theta) = S\_0(\lambda)\times T\_0(\lambda)\times X(\theta).
$$

Calculating the color that the human eye perceives, given a spectrum, is not trivial, but I recently outlined an approch in [this answer](https://astronomy.stackexchange.com/a/40014/5264).


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