1) How we know standard atmospheric refraction is a real phenomena and 2) Why you used a value of 7/6*R for it [in the Metabunk curve calculator]

Atmospheric refraction is measurable when observing stars. Stars visually appear to rotate around the Earth once ever sidereal day - and they do this at a constant speed. However as they approach the horizon they seem to slow down a little, basically staying visible longer than they should. When they reach the horizon their actual calculated position is a bit over 0.5° below the horizon - a bit more than the diameter of the sun. This amount matches what we have empirically found about the refractive index of air at various temperatures and pressures, and what we have empirically found about the change in temperature and pressure at various altitudes in the atmosphere, and what we have empirically determined to be the radius of the Earth. The refraction of terrestrial objects differs from stars in that only part of the atmosphere, whereas astronomical observation passes through the entire atmosphere. The value of 7/6*R is just a rough approximation of the effect of refraction on the apparent radius (R) of the Earth. It's very approximate as the atmosphere is quite variable. Basically though you do the same calculations, but the larger radius gives you the reduced curvature. I initially got the 7/6 value from: http://aty.sdsu.edu/explain/atmos_refr/horizon.html The main ways in which observations will differ from the standard refraction are that the temperature near the ground often varies greatly from a few meters above. Also the lapse rate (how quickly the air cools) may not be linear, especially at low altitudes. This can result in vertical stretching and/or compression of the image. If you want more details, including empirical observations, I recommend this set of tests by the USGS: https://www.ngs.noaa.gov/PUBS_LIB/ResultsOfLevelingRefractionTests_by_NGS_TR_NOS92_NGS22.pdf

Here's a great video showing the effects of refraction on stars, the sun, and the moon as they get closer to the horizon. Source: https://vimeo.com/188149183

There's always some refraction yes, because of the pressure gradient of the atmosphere. The actual amount and nature of this can vary - like if there's a temperature inversion. https://en.wikipedia.org/wiki/Horizon#Effect_of_atmospheric_refraction

I made a video analyzing the relative slowdown of stars as they get closer to the horizon: Source: https://www.youtube.com/watch?v=m-xXhrTG3Sk Summed up in this image: What I did was track one star and put a red dot (and green horizontal line) where it was every one second. I then took the first two positions and used them to extrapolate expected positions at a constant speed. The results show that the star is already slowing down a bit, and the slowdown rapid increases as you get closer to the horizon.

Looking for more derivations of the 7/6 (1.1666) value. https://en.wikipedia.org/wiki/Atmospheric_refraction For a standard atmosphere at mean sea level, P= 1013.25, T=288.15°K (15°C), dT/dH = 6.5/1000. Hence, k = 503*1013.25/288.15^2*(0.0343 + 6.5/1000) = 0.25, so Reff = 1.333*R, as opposed to 1.166R Is the seconds equation supposed to use the "smaller" k? Did I make a mistake? Reference 11 is attached, in which Andy Young says: 1/6 = 0.16666, giving Reff of 1/(1-1/6)*R = 1.2*R. The 7/6 is just a ballpark, but it would be good to to verify the derivation of this number (and clarify which "k" is used. It's probably in Dr. Young's document.

Different people use different numbers. I wonder if the calculator should show a range. Maybe even generate a diagram of horizon lines. http://www.aboutcivil.org/curvature-and-refraction.html So they say "14% of the effect of earth curvature.", is this the same as using a radius of 1.14*R? Not really clear what they mean there.

I think they got that wrong for a start. Even if the temperature is constant with height the density will drop due to pressure drop. A typical lapse rate, which is limited to the dry adiabatic rate, would decrease refraction somewhat.

But if you've got a warm region lower down the expansion of the air due to heating reduces the density, cause the bend upwards. Hence some types of mirages.

But the air is unstable, and convection and mixing generally occurs. A hot road is slightly different in that inversion can be stable in a very thin layer, the thermal boundary layer, I think.

It might be nice if the calculator had a range of values indicating extreme refraction but I have used a method of changing the value of Radius in my own calculator. My problem has been in determining if that value is achievable in real life. Your calculator is very valuable for evaluating flat earth claims

The problem with a range is that the more extreme examples of refraction are not at all linear. i.e. you can't simply say there's 1/6 or more visible. What often happens is that the view is over cool water which mostly bends the light paths near the horizon down over the curve and creates a highly compressed image of behind-the-horizon objects. So you can't really reduce that to a simple number.

The effect of refraction looking at the tops of mountains (even small ones!) should be more predictable than objects closer to sea level, as the lapse rate will tend to average out over a greater height difference between object and observer. If the top of a mountain pokes through the top of the planetary boundary layer (~1-2km.} there is often an inversion there, which might possibly increase the range at which at which the tops of mountains can be seen.

The equation cited is from Hirt, et al, 2010, and appears to have been empirically derived from a long series of simultaneous reciprocal observations and back-calculating the refraction component from those observations. They then built an empirical relationship between the temperature gradient and the refraction correction. I haven't read their paper, but the technique of calculating refraction corrections from simultaneous reciprocal observation appears to have the earth's radius as one of its assumptions, so FE doing a deep dive into it will find a somewhat circular reasoning. One reason you are getting much higher k value than is normally used is you plugged in a temperature inversion when you used a positive value for dT/dh. In "standard conditions" the temperature drops as you get higher, so dT/dh is negative. Using dT/dh = -0.0065 gets to within about 3% of the 1/6 of curvature drop rule of thumb for standard refraction. If you use a very strong negative gradient, the value of k goes negative, which is what you see with a road "mirroring" the sky on a hot sunny day. A derivation from basic principles of a similar refraction correction equation (with an added humidity gradient correction) is found in Torge's "Geodesy". In the 2nd edition (1991) the derivation is in section 4.3.1, and the entire section is in Google Books' preview of the book. In the 3rd edition it is section 5.1, 5.1.1, and 5.1.2, but in Google Books' preview two pages of these sections are omitted, so you will have to look for those pages in a library. The two derivations differ slightly in presentation, and the third edition relies on a new set of atmospheric index of refraction equations adopted by the International Association of Geodesy in 1999, but the end result is the same, as several of the coefficients have been rounded and the newer index of refraction model was in general agreement with the previously used equations. Torge's derivation ends with equation 4.40 (2nd ed) or 5.19a (3rd ed), which doesn't quite give it in a form where k is calculated directly. Baselga 2014 takes that and plugs it into one of the preceding equations in Torge's derivation to present The Baselga paper is a good background document that describes the different approximations and techniques surveyors have used for atmospheric refraction correction. "Practical Formulas for the Refraction Coefficient" 2014 by Baselga et. al. This paper also has an example of how significantly refraction corrections can vary with different atmospheric conditions, even at a single site over the course of 24 hours.

I just signed up to make a suggestion on your curve calculation page. The diagram below does not show the level line that would measure the drop in your chart. I hope you will consider adding the drop line as this is a major hurdle in explaining drop vs bulge to flat earthers.

Here's an interesting home demonstration someone did, where they created a layer of cool air which brought into view something that was hidden by a horizon: