Sky surface brightness vs magnitude limit visibility

I am wondering how we can calculate the magnitude limit of the celestial object, which could be visible at the given sky surface brightness conditions.

We already know that Venus and Jupiter are visible in the daylight. Occasionally also Mars too. I've read, that the magnitude visibility threshold is about -2 Mag for a full day. It changes, and as the Sun goes closer to the horizon it causes the light to diminish in the thick atmosphere. I found information, that the Arcturus star (-0,04Mag) was once visible even 24 minutes before sunset! If, for example, we consider the civil twilight. Is there a chance of seeing the objects with a magnitude of +3 or even +4? A good example can be the 12P/Pons-Brooks comet, which will pass the perihelion on late April 2024. Its maximum brightness is estimated to be about +4.1Mag, but its proximity to the Sun probably excludes it from good visual observation. I am aware, that these assumptions will be completely different for the naked-eye approaches than for observations made through binoculars or telescopes. I am rather interested in naked-eye observations, but I would be really happy to know the source, which would include some comparisons/tables regarding this.

Does anyone know the article/book or presentation, which would cover my question?

• This might be of interest: "The Visibility of Stars and Planets During Twilight" opg.optica.org/… Commented Apr 3, 2023 at 23:03
• – uhoh
Commented Apr 4, 2023 at 2:58
• @GregMiller this script is unfortunately not for free. If you could have the source for PDF download, that could be awesome! In fact thanks to your hint I found something similar: researchgate.net/publication/… and admittedly is a good source too.Thank you for your help Commented Apr 4, 2023 at 10:22
• @Geographos great find! Also potentially a bit helpful is Kastner, S. O. (1976) Calculation of the twilight visibility function of near-sun objects
– uhoh
Commented Apr 5, 2023 at 3:27
• @uhoh many thanks for this contribution. I wish I could bounty more questions, but I need more "points". I hope, all the answers will be useful to anyone who concerned :) Commented Apr 5, 2023 at 9:33

The human eye has an angular resolution of about 1 arcmin.

This means that the light coming from point like stars would be visible if it significantly exceeds the luminance of the sky over an area of 1 $$\mathrm{arcmin}$$ $$\times$$ 1 $$\mathrm{arcmin}$$.

Sky luminance is usually measured and reported in $$\mathrm{mag/arcsec}^2$$. It can be converted to as follows:

$$m$$ $$\mathrm{mag/arcmin}^2 = s$$ $$\mathrm{mag/arcsec}^2 - 2.5 \mathrm{log}_{10}(3600) = s - 8.89$$ [Further Read]

Therefore, the limiting magnitude of a sky with luminance $$s$$ $$\mathrm{mag/arcsec}^2$$ will be significantly more than $$s-8.89$$.

In practice, the subtraction factor is close to ~14. That is, in practice, the limiting magnitude of a sky with luminance $$s$$ $$\mathrm{mag/arcsec}^2$$ is about $$s-14$$.

The minimum sky luminance at the horizon, even during a total solar eclipse, is not known to exceed 14 $$\mathrm{mag/arcsec}^2$$.

So, the limiting magnitude will almost always remain negative as long as even a sliver of the Sun is visible. Post that, there will be a gradual increase towards the limiting magnitude of the night sky at that location.