BDanielMayfield wrote: ... And also how bright can auroras get?
This is a harder question, but I believe I have an answer you can relate to.
Like other natural phenomenon, auroral brightness does have a standard, and it has a logarithmic scale. It is the International Brightness Coefficient (IBC) standard. (You might find S&T article interesting: An Aurora Watcher's Guide
where brightness and other characteristics are discussed)The levels range from 0 - 4, on a logarithmic scale:
0 ≡ subvisual, detected only with instruments,
1 ≡ comparable to the Milky Way; whitish with no discernible color,
2 ≡ comparable to moonlit cirrus clouds; color is barely identifiable (usually yellow-green),
3 ≡ like brightly lit cirrus or moonlit cumulus clouds; colors are evident,
4 ≡ much brighter than three; may cast discernible shadows; good colors, many of them fleeting
The unit of auroral brightness is the Rayleigh
intensity unit. This allows a quantitative assessment of sky brightness, and referencing this paper: Auroral contribution to sky brightness for optical astronomy on the Antarctic Plateau
, a calculation of the brightness for an IBC 4 aurora using a V-Band (Green) filter, you get ≈+14 Magntudes/arcsec2
sky brightness. So, what does this mean?
The equivalent brightness of the full moon ≈ 3.5 Magntudes/arcsec2
, so the full moon brightness per unit area ≈10.5 magnitudes brighter than the brightest aurora! Asked another way, how big would a uniformly bright, IBC 4 aurora need to be such that its integrated brightness equals a full moon (mag = -12.5)? If you were directly under a uniformly bright, circular IBC 4 aurora with a diameter about 65°, you would experience a full-moon lighting condition!
By the way, this same aurora would not be visible with the naked-eye during the daytime. As a thought experiment, consider increasing the moons diameter by 130x without changing the integrated brightness - it would not be visible with the bare eye.
Thanks alter-ego. And nice skill too; locating a known star and measuring the elapsed time from its rotation angle.
Thank you! I love solving problems by using old tools in new ways