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I'm interested in knowing how low the sun could ever get in the sky and still be seen.

More technically: what's the minimum possible elevation of the sun at which it can still be seen above the visible horizon from a terrestrial vantage point? And where on Earth would you go to make such an observation?

Some clarification:

elevation is angular distance measured from the astronomical horizon (the great circle that is 90° from the local vertical).

For example, I imagine from a mountain top near the ocean, you might be able to view the sun at an elevation angle that is negative, below the astronomical horizon but above the visible horizon, and possibly by a decent amount. So then I thought Mt. Everest might be the spot, but locally I'm not sure how low the visible horizon is, because the surrounding area is at a high elevation too. So maybe I'm just looking for the highest angle possible between mountain peek and sea level, in other words, where is the lowest apparent visible horizon, but the sun may not necessarily ever rise or set in that direction.

I also looked around for a map of local visible horizon angles (measured relative to the astronomical horizon), but couldn't find one. I suppose such a thing could be constructed from a topographical map of the Earth, right? Maybe this would be the best way to "solve" this if nobody already knows.

I'm hoping the answer can ultimately be turned into a factoid of the following format (looking to fill in the unknowns in this):

"From the top of mount something-or-other, looking west, the visible horizon is x° below the astronomical horizon. From this vantage point, just before sunset, the edge of the sun could be observed to be f(x) sun diameters below eye-level."

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    $\begingroup$ probably from Mauna Kea. google.com/maps/@19.8210075,-155.47137,3a,75y,252.1h,82.63t/… $\endgroup$
    – James K
    Commented Apr 10, 2020 at 18:13
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    $\begingroup$ Beware atmospheric refraction: en.m.wikipedia.org/wiki/Atmospheric_refraction When near the horizon, the sun is usually refracted from somewhere other than it appears to be. $\endgroup$
    – Wayfaring Stranger
    Commented Apr 11, 2020 at 14:01
  • $\begingroup$ Barring refraction, as Way_Str pointed out, you want to get as high as possible with a direct LOS to the sunset via the lowest possible land/sea in between. Note that this will vary dramatically as the apparent latitude of sunset varies over a year. $\endgroup$
    – Carl Witthoft
    Commented Apr 13, 2020 at 17:44

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The lowest exposed point on land is the dead sea shores, at -413metres below sea level. so theres a good chance you could find a spot that lines up correctly to give you a view of a piece of the horizon well under sea level. I dont believe that elevating yourself is the key factor here though. At Around 35000 ft you can start noticing the curvature of the Earth, the horizon being 200 some odd miles away (Pythagorean theory -theorem?- can be used for this, several sources on google explain the steps to find your equation, and far better than I can. ) but the distance to the horizon grows rapidly. So you would certainly want to find the sweet spot of height /distance, too high and you wont know when the land has ascended back to sea level, as the terrain will become more vast and harder to judge detail or distance. Too close to the shore and your only seeing a few miles but the angle of sight wouldnt be the best possible for finding the lowest horizon. You would want to work out the math and find a spot that puts the dead sea shores as your visible horizon, bc if your standing there, watching the sun rise, the sun is rising over a horizon that isnt the lowest accessible location...

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    $\begingroup$ -1. I think you've misunderstood the question. The shore of the Dead Sea must obviously be lower than the surrounding land (otherwise the water wouldn't collect there) so this is exactly the opposite of what the question is asking, which is for the highest elevation. $\endgroup$
    – Chappo Hasn't Forgotten
    Commented Apr 14, 2020 at 0:22

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