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?

  • 2
    $\begingroup$ It depends on multiple factors such as altitude, humidity, temperature, air pollution (natural and industrial), etc. $\endgroup$ Commented Nov 27, 2020 at 3:23
  • 3
    $\begingroup$ Oh! and welcome to Astronomy Stack Exchange! :-) $\endgroup$ Commented Nov 27, 2020 at 3:23
  • 2
    $\begingroup$ 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 :-) $\endgroup$ Commented Nov 27, 2020 at 3:30
  • $\begingroup$ There must be, but I’m afraid (and sorry) that I don’t know it, hence I only comment instead of answer… $\endgroup$ Commented Nov 27, 2020 at 4:06
  • $\begingroup$ It mostly depends on the number of particulates in the air and so will vary from place to place and time to time. $\endgroup$
    – ProfRob
    Commented Nov 27, 2020 at 9:09

1 Answer 1


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.

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): $$ X(\theta) = \csc\left(90^\circ-\theta + \frac{244}{165+(90^\circ\!\!-\!\theta)^{1.1}} \right), $$ 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .