I've wondered: if you stood at a high peak such as Pikes Peak in Colorado or Mauna Kea in Hawaii each at 14,000 feet, how far could you see to the horizon (maximum line of sight distance) allowing for atmospheric transparency?

The formula for calculating maximum line of sight distance d is

d = SQRT (2Rh + h EXP 2) where R is radius of Earth and h is height above sea level

an approximation when h << R is d = SQRT (13h) using metric values

For a 14,000 foot peak that would give about 200 miles maximum line of sight distance. Would the atmosphere allow you to see that far?

Of course the Space Shuttle 160 miles up can see clearly down to the ground on a clear day as it's looking straight down through the thinnest part of the atmosphere

When I lived SW of Carrizzo, New Mexico in 1954, the visabiliity to the top of a 12,003 foot mountan was about 60 miles. I think, less from the top of the mountain to the 4000 foot plain. I got to use a quality telescope for missile tracking and could just barely see cars moving on route 54 about 7 miles away. Most of the difficulty was the distortion produced by the hot air rising from the desert. Few locations have clearer air. Neil

University of Hawaii Institute for Astronomy: Mauna Kea Weather Center Webcams

City of Colorado Springs: Pikes Peak Panoramic Webcam

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I can see 13 billion light years just by looking up on a clear night. Beat that...

From the snowline on Mt. Hood, I could see the Three Sisters about 100 miles away.

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"What you think you thought you saw you did not see." Agent J, MiB - Manhatten Bureau

When you watch the moon rise or set you are seeing through the refraction of light in our atmosphere something like 385,000 km. At that low altitude the atmospheric distortion forces a lower magnification lens be used.
What is it you are asking?
If it is as I understand. You want to know of Earths atmosphere's clarity.
Given heat distortions and pollution levels vary a great deal. Your location is important.
On those high alt mountain tops of the Andes where the air is clear and dry and as clean as it can be.. They have built some mighty fine instruments for looking...
You have the numbers and can do the math... how far across the surface of this curved globe can you see... If and when you get a clear day... or, ask an eagle.

By metric values, we mean h in meters gives you d in kilometersThe formula for those units is d = SQRT (1.5h), h in feet, d in miles. But, regardless of which formula I use, I get about 145 miles, not 200.

Curiously enough, the question of atmospheric transparency was heavily investigated by the U.S. Navy back in the days of (only) optical spotting for battleships. Remember those big white shelters on the tops of the cage masts on the battleships at Pearl Harbor? Those were the 'spotting tops', where the gunnery control people could locate and identify targets. For practical purposes, in clear weather, they could 'spot' out to about 40 miles. Of course, most of the time conditions were less than perfect, oftentimes a lot less.

You might check the distances between stations in France's visual telegraph system, from the Napoleonic era. As I recall, it was like 8 to 15 miles between the sites.

Along the coast in California the visibility occasionally exceeds the theoretical limit. We lived in Santa Rosa, and I have seen from there, at only 40 feet above sea level, mountains 50 plus miles away.

As a pilot who's flown at altitudes from sea level to 49,999 feet (weren't technically allowed to reach 50,000 feet, as we weren't wearing pressure suits...), I can attest to having seen geographical features nearly 200 miles away while at 14,000 feet.

However, those features were themselves mountains, the peaks of which were poking up beyond the dense and dust-laden thicker part of the atmosphere.

When it comes to seeing some on the ground (sea-level, or close), I think the best I've seen is around 90 miles. That was on an extremely clear day after a cold front blew through the area, dragging with it an air mass from frostier regions which don't suffer much from dust.

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I am Mugs, of the Alien clan of Usa, Nordamerica, a Terran, of Sol.

Perception isn't reality. It's merely an abstraction thereof, and quite often not a very good one at that.

I am human. Fully human.

The 208 mile figure LOS distance was for Mt. Everest. Wow! Meaning from atop Mt. Everest you could see atop another 29,000 foot peak about 400 miles away.

A more precise formula is d = SQRT (2Rh + h EXP 2) where R is Earth's radius in km and h is your height in km above sea level. It's a more precise formula to use for satellites in orbit where h is a greater fraction of R.

Which 208 figure?

Close to the surface of the earth the refraction by the atmosphere is important. The King tracking rate for telescopes is for correcting for the apparent movement of objects as they move from near the horizon. You can effectively see around the surface of the earth.

Need help deriving a formula. Need to calculate the amount of visual cutoff of a building viewed from a distance who's bottom appears below the horizon due to the Earth's curvature yet we can still see the top.

The amount of cutoff as a function of increasing distance is not linear as the Earth is round. As we move the building further and further out, the more the cutoff due to Earth curvature increases (not a linear variation).

A person 1.7 meters off the ground or water surface can see straight line distance about 4.7 km to the horizon.

d = SQRT {2*6391 km*0.0017 + 0.0017 EXP 2} Let's assume a 1000 foot high skyscraper is 100 km away and you're eyeballing it 1.7 meters off the water surface.

How do we calculate what percentage of that building's height, as seen from 100 km away and 1.7 meters up from sea level (h), appears to our eyes above the horizon and how much of the building is hidden below the horizon due to the Earth's curvature?

Is there a formula?