NavList:

Frank,

In further reply to your last post - (https://navlist.net/Southern-sky-CN-photo-puzzle-FrankReed-sep-2023-g54634) thanks for it - yes I am finding this digital picture "dynamic range"  very annoying, and here is not our only example of such horrific behavior where the picture stars respective brightnesses do not match at all the relative real world ones.

I am taking this opportunity to attempt getting a more refined Observer's Latitude if any possible.

Assuming that the "horizon" is flat and horizontal, and assuming that both Canopus, Sirius and Rigel are almost due South of the Observer, I am hereafter attempting to better determine the Observer's Latitude:

On my picture printed copy, Sirius and Canopus lie 61.5 mm apart.

This can can be used as a scaling factor given their difference in declination equal to 36.0°.

Hence 1mm on my printed copy equates to 0.59° .

Assuming that the vertical distances have the same scale on the picture, then :

Sirius ( δ = -16.7°) at 23.5 mm from the picture horizon, has an apparent height of 13.8° ,  which makes our Latitude equal to S 87.1° . Relative observation Weight equal to 2 because body [almost]  due South and above 10° (refraction ignored here).

Canopus ( δ = -52.7°) at 85.0 mm from the picture horizon, has an apparent height of  49.8°,  which makes our Latitude equal to S 87.1° . Relative observation Weight equal to 2 because body [almost]  due South and above 10° (refraction ignored here).

Rigel ( δ = -08.2°) at 8.0 mm from the picture horizon, has an apparent height of 4.7° ,  which makes our Latitude equal to S 86.5° . Relative observation Weight equal to 1 because body, although [almost] due South is low (refraction, which I am ignoring, cannot be ignored here).

Achernar ( δ = -57.1 °) at 97.0 mm from the picture horizon, has an apparent height of 56.8° ,  which makes our Latitude equal to S 89.7° . Relative observation Weight equal to 1 because body, although rather high (refraction can be ignored here) is significantly away from Azimuth 180°.

Hence our Latitude would be equal to 1/6 [ (-87.1° * 2) +  (-87.1° * 2) + (-86.5°) + (-89.7°) ] = - 87.4°

If we consider that this method yields a Latitude accurate to +/- 5°, then it would look that the Observer's Latitude is somewhere South of 82° S.

Coupled with the reasonable probability that this picture was actually taken from a US base station, a close look at the various existing Antartic stations, points and leads us - as advertised - towards the Admundsen-Scott South Pole Station. which, anyway lies in our "error circle".

Any comments welcome.

Kermit