NavList:

I've been trying to squeeze more information out of this -- but unsuccessfully. Perhaps you could advise.

First, γ CrA and τ Sgr lie pretty much on the same celestial meridian but they're about 60° from the vertical in the picture. Assuming the rectilinear projection and that the camera was held upright and the image wasn't rotated in post (Russell D. Sampson noted that the clouds are roughly horizontal), it places that meridian at about LHA = 60°. So we're looking southwest.

If we could determine the UT precisely enough, and thus the GHA of the meridian, we could get the approximate longitude from its LHA, too. Then we could search the Internet for pictures from that place on that day to see if the image comes up. I find it quite possible that the person behind the montage got the Milky Way photo from the Internet, too; if only we knew the search terms.

So, UT. The fastest-moving object I could find in the picture is Vesta. It makes some 15' of arc across the stars per day (3× more than Saturn). I attempted to estimate its RA+Dec from three nearby stars (HP88297, HP88125, HIP87782) using bilinear interpolation, to determine the UT when Vesta would be at the photographed location. However, at this almost subpixel scale, with blurry dots, the error bars in the resulting UT are fairly large (on the order of half a day). Of course, my resulting best guess, UT 6am on 13 June 2018, places the observer in the Eastern hemisphere (76°E) pretty much around the local noon. Not very good for taking pictures of the Milky Way.

It feels like there ought to be a way to figure this out, though. I think that if you knew where to look, you could probably spot an (artificial) satellite; Stellarium shows that there are quite a few, and that would settle the UT question very precisely. Or maybe there's another trick.

Matus