"Viewsheds" of any location can be calculated and matched with photographs using "GeoImageViewer", an application I wrote a couple of years ago. Any feature in the image can be interactively identified in a mapview and vice versa, including the boundary of the viewshed. As has been mentioned in the comments, it is essential to include atmospheric refraction in the calculation, at least for distances above ~100km.

[1] https://hdersch.github.io/Viewing.html

It is interesting to also consider dry land which is below sea level. "For an average height person standing on a beach, their horizon is around 4.5km away." For an average height person standing at sea level, near a large depression such as the Dead Sea, for example, you may be able to see land at a much greater distance.

I doubt if this factor would impact your findings for the longest sight lines on Earth, but perhaps for some local findings.

Also, applying the same concept to other planets has some surprising results. The tallest mountain in the solar system, Olympus Mons, has very gentle slopes. So gentle that from the peak all you can see is Olympus Mons itself.

Related, there’s a Reddit post tracking the furthest photos of the Sydney skyline:

https://www.reddit.com/r/sydney/comments/1lzke5t/the_actual_...

Google Earth map: https://earth.google.com/web/@-32.84964163,149.89452785,590....

I'm sure it's nearly an academic distinction, but:

> Basically, for any given region, we find its highest point and assume that there is a perfectly placed sibling peak of the same height that is mutually visible.

Shouldn't you always add 335km to the horizon distance to account for the possibility of Everest (i.e. a taller sibling peak) being on the other side of the horizon?

I've been wondering about using OpenStreetMap data to achieve something similar, like getting the longest uninterrupted (by buildings) distance between two viewpoints (the Nikon Coolpix P900/P1000/P1100 cameras or a telescope are pretty good for this due to their extreme focal lengths), or maybe something like "most ice cream shop dense area" https://wiki.openstreetmap.org/wiki/Key:amenity

Putting the typical airplane cruise altitude (10km) into the distance to horizon calculator https://www.ringbell.co.uk/info/hdist.htm results in a "reverse" distance of 357km to the horizon. So that's how far typical airplanes can be seen when standing at sea level!

I was recently looking into this for roguelike-like visibility calculations. The term I came across was "isovist" not "viewshed", but they seem identical? The former appears to be used a lot in architecture.

Calculating a 2d isovist is (relatively) simple, for 3d it seems much more complex. Is this simplified by the fact that it's dealing with a height map?

A kid once asked me “How far can humans see?”. I told him you can see the Andromeda Galaxy 2.5 million light years away, but when he said he meant on Earth I didn’t know the answer.

Oh that project is right up my alley, as I did the same! Back in 2021, for fun, I wrote an algorithm to find the longest sightline on earth. And I did find a previously undiscovered sightline.

My code was relatively unoptimized, it ran for 95 days on an 8-core 16-thread AMD Ryzen 5750G to crunch all the data. I have never published my code or results. I should really do it...

I downloaded a digital earth model of about 30 gigabytes from http://www.viewfinderpanoramas.org/Coverage%20map%20viewfind... which had a resolution of 3 seconds, so I had the elevation of each 90x90 meters "tile" of land. I wrote a simple algo in C that: finds potential viewpoints (mountain peaks or plateaus above a minimum ), calculates the longest sightline 360° around you, taking into account an atmospheric refraction coefficient (I used 0.13 which seems to be a default used by some panorama tools like https://www.udeuschle.de). It's really basic trigonometry: look one tile ahead of you, and the next one, etc, until you find the one with the highest vertical angle of view. Stop one sightline exploration when reaching a certain maximum distance (750 km in my implementation) when it becomes mathematically impossible to find a visible tile beyond this distance. And I found the longest sightline:

In Kyrgyzstan from Pik Dankova (41.059167,77.683333) which is at 5977.5 meters (in my DEM data), you can see 538.1 km into China if you look toward bearing 169.7° (roughly south) as you see some distant minor peak at 36.295364,78.75593 which is 6444.0 meters high. However this sightline is already known. But it's only theoretical. No one ever encountered good enough atmospheric clear-sky conditions to observe it from Pik Dankova. Using the panorama tool maintained by this cool German guy, you can verify it: https://www.udeuschle.de/panoramas/panqueryfull.aspx?mode=ne...

However one notable finding my tool gave me was it discovered the second longest sightline that is competely unknown up until this day AFAIK:

In Colombia, from Pico Cristóbal Colón (10.838333,-73.687500) which is at 5668.5 meters (in my DEM data), you can see 502.6 km over the plains of the Caribbean region all the way to the Colombian Andes if you look toward bearing 206.2° (roughly south-south-west) as you see some distant peak at 6.777286,-75.692304 which is 3347.0 meters high. Here you can verify it here: https://www.udeuschle.de/panoramas/panqueryfull.aspx?mode=ne...

I am not sure if this Comlombian sighline has ever been observed.

As I said I should really publish my code and results. And at some point I would like to optimize it and re-run it using a higher-precision DEM with 30x30 meters tiles instead of 90x90 meters. I think there are a few 30-meter DEMs available but I need to find the highest-quality one.

Nitpick:

"resolution of 3 arc-seconds (~100m²)"

This resolution is equivalent to tiles of 8500 m² not 100 m². I think the author confused tile edge length (92 meters exactly) with tile area.

> I find these TVS heatmaps eerily beautiful.

The first one somehow looked like something violent going on. Limbs? Bodies?

Since you asked about alternate packing solutions:

https://news.ycombinator.com/item?id=43305920

I don't know enough about the subject to have any useful suggestions, but I'd be interested in your take on how a project like yours would work differently if you were to choose a different solution.

As proposed in the article, calculating the area weights all points in the viewshed equally. I wonder if it makes sense to give more weight to far-away points? Or maybe less weight? Not sure! But would be interesting to explore and see if it gives anything.

Does the source dataset includes with enough precision human made buildings?

Will it allow to determine the farthest point theorically visible from the top of the Eiffel Tower?

If you consider atmospheric refraction it might be more complicated...

Nice work! Do the tiles point to actual heatmap?

You have a very interesting project. What happens if you put your maps and elevation data on a 'sphere', to model the world as 3D? You can then leverage a GPU to render a view from one vertex to then determine what is furthest away yet visible from that vertex. The vertices can then be given some type of heatmap colour and an interactive tool made.

A full GIS tool isn't needed for this, with Blender and three.js there are all of the tools needed.

Ofc my home country's name was misspelled :/