Alternatively what eye properties can be modified to give the feeling of living on a flat, but infinite plane rather than being on a really tiny planet?

model of two buildings on a tiny planet

Edit (16 hours after posting): Thank you for all the answers so far- lots of great explanations, suggestions and inspiration! More context: I'm making a looping 2D animation, where camera is traveling through imaginary city and I want viewers to think that what they are looking at is just a 5 point perspective (fisheye lens). My approach is to first model that city in 3D on a sphere to avoid duplication and make camera looping easier - unfortunately this means that I have double the roundness - 1 from ultra wide lens, and the other from the fact that it's a small planet. That's where I decided to ask here what else could be done from a world building point of view to partially combat the latter.

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    $\begingroup$ Please define what the context is for "appear flat". The formula for the distance to the horizon, for example, is at en.wikipedia.org/wiki/Horizon#Distance_to_the_horizon and from it, we can deduce that to have the distance to the horizon be infinite (which is a property of being on a flat plane) would require a planet of infinite radius. $\endgroup$ Feb 11 at 0:30
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    $\begingroup$ What @GrumpyYoungMan said. Humans used to think the world was flat, and it is not. What a truly flat world would look like is unintuitive. $\endgroup$
    – jdunlop
    Feb 11 at 0:56
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    $\begingroup$ Add fog. That's it. $\endgroup$
    – PcMan
    Feb 11 at 4:59
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    $\begingroup$ We need more context here. Are you assuming a barren planet, or are there obstacles blocking line of sight (trees, hills, ...)? Are we trying to trick an animal's intuition or are we considering that an intelligent species might measure the planet's shape? $\endgroup$
    – Flater
    Feb 11 at 12:58
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    $\begingroup$ I suspect this is not the correct SE given that your actual question is about CGI. That said do you need a spherical "world"? Or would toroidal work? You can "fake" a toroidal world with portals. $\endgroup$
    – Matthew
    Feb 11 at 18:37

With the right optical effects it could be quite small. Some predictions about the planet Venus (from I think the 1960s) suggested that the high concentration of CO2 plus the predicted variation in atmospheric density with depth would be such that, if you looked horizontally along the surface of Venus, then the light would be continuously refracted along a circular path of approximately the same diameter as the surface of the planet.

Assume such an exotic atmosphere exists with refraction radius r on a planet of radius R.

If r>R (or r -ve) then the planet looks 'spherical' like the earth does from sea-level. For example, the horizon appears to be a tiny bit below horizontal and ships appear to drop below the horizon as they move away. (If r >>R or -ve, that accentuates the effect to make the horizon appear closer/lower and the planet appear smaller).

If r<R then the planetary surface would appear to be bowl-like. So, for example the horizon would appear a small distance above horizontal and you could see distant cities from 'above'.

If r=R then the world would appear flat. The horizon would be exactly horizontal and ships moving away would remain in perpendicular view until they faded into the obscurity resulting from atmospheric absorption. The world wouldn't look like an 'infinite' plane though, as if the atmosphere was clear enough to see 'forever' then you could observe the back of your own head in the distance.

enter image description here

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    $\begingroup$ whoa! That's amazing $\endgroup$ Feb 11 at 17:41
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    $\begingroup$ This is very cool, although for the OP's particular use case, I suspect this would complicate rendering software, as that usually relies on light rays moving in straight lines. $\endgroup$ Feb 11 at 18:21
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    $\begingroup$ For example, the horizon appears to be a tiny bit below horizontal and ships appear to drop below the horizon as they move away. For this reason, perhaps among others, a lot of people already knew the world was round, before 1492. (And even then, Columbus didn't circumnavigate it obviously, so he still wouldn't have proven it directly through his voyages.) $\endgroup$ Feb 11 at 22:28
  • $\begingroup$ en.wikipedia.org/wiki/Prisoners_of_Power inhabitants believe they live inside a sphere. $\endgroup$ Feb 12 at 3:49
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    $\begingroup$ @candied_orange: I interpreted it as meaning concave (i.e. "curving downwards"). $\endgroup$
    – Kevin
    Feb 12 at 17:47

Assuming nothing is more than 1.8m above the surface; Radius of 606km

This assumes:

  • Your human-sized observers have human eyes.
  • Every human eye stays at or below 1.8m altitude at all times.
  • No-one ever compares shadows in different places on the planet. Or compares the stars.
  • No-one ever builds anything taller than 1.8m high.

So - the human eye at 20/20 vision is accurate to about 1/60th of a degree at 20ft:

enter image description here

The "20ft" (6.1m) is where the eye muscles that focus your eye for near vision relax - it's just "anything greater than 6.1m has a 1/60 degree accuracy".

So we're trying to find the radius of the circle made by the intersection of two points anywhere along these two lines (which become tangents), where the angle between them is 0.016 degrees:

enter image description here

Having not done geometry for 2 decades, I messed around with some diagrams for several minutes before giving up and plugging it into FreeCAD's sketching tool. Which promptly failed to solve it in all sorts of crazy ways relating to floating point accuracy that I find very interesting but nobody else will.

Experimentation and spoon-feeding the solver eventually gave me the answer that given a radius of 606km, and a requirement that no eye ever gets 1.8m above sea level, the world will always appear flat within the human eyes vertical resolution.

Even if they invent lasers to precisely mark where the horizon "should" be, they won't be able to detect it.

If they only have people 1.8m high and walk around and look at each other, they still won't be able to tell, as fellow humans will be less than 1/60th of a degree when they disappear over the horizon.

How to calculate something more realistic from this - ie. Lets add buildings!

Because trig of really small numbers can be simplified to pass-through functions, you can scale the radius and the feature height linearly (within reason). So if you're largest ship is 18m tall, to not notice something fishy when it sails off, the planet needs to have a radius of 6060km. Same if you have a building 18m high and start walking away from it.

If the largest structure is 360m above the spheres surface, then the planet needs to be 12120km in radius for a human not notice it shrinking over the horizon.

  • $\begingroup$ If the observers you are trying to fool are creatures like eagles, who have better than human distance vision and can fly, this must be adjusted accordingly, which makes the necessary radius much larger. $\endgroup$
    – Ryan C
    Feb 11 at 14:44
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    $\begingroup$ This relies on the eye's resolution, but all this really shows is that you can't pick apart two 1.8m tall objects on top of one another at some distance. But the problem here is that the object disappears entirely when it goes over the horizon. It's very, very easy for the human eye to detect simple presence/absence - picking out stars in the sky is a trivial task, despite the fact that they can occupy less than one millionth of a degree of visual angle. The fact that the person is only 1/60th of a degree when they go over the horizon does not mean you can't see them disappear. $\endgroup$ Feb 11 at 18:39
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    $\begingroup$ This is the direct answer to the question that I interpreted. The others at the moment take different approaches. Anyway, you might also want to mention the actual radius of Earth for comparison. 6,371km average, according to en.wikipedia.org/wiki/Earth_radius $\endgroup$
    – AaronD
    Feb 11 at 20:34
  • $\begingroup$ @NuclearHoagie I saw a Flat Earth debunking video that took advantage of that; by smoothly moving a camera up from human eye level down to just above the surface of the water, you could see ships disappearing over the horizon. $\endgroup$
    – nick012000
    Feb 12 at 4:51
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    $\begingroup$ @nick012000 Right, the idea behind this answer is that at some point, the object is so distant that your eye does not have sufficient resolution to perceive the object getting any smaller as it goes over the horizon - at this distance, a 1m object and a 2m object are indistinguishable. The problem is, this does not mean you can't tell the difference between a 1m object and the object not being visible at all. We can perceive objects trillions of km away, an object doesn't become invisible when it's 1/60th of a degree of visual angle. $\endgroup$ Feb 12 at 14:03

Limit the range of vision!

A.K.A. Just add Fog!

A planet appears flat if the local area is flat, and the horizon does not obviously subtend less than 180 degrees.

Example:For a normal human, the distance of the horizon on Earth is about 5km, and the subtended angle is 0.1 degrees less than 180 = 179.9 degrees, and the Earth appears completely flat.

But if your maximum vision range never exceeds about 2km, you could stare all day and not notice the horizon's curvature. Not the easily visible effects, such as a ship "sinking" below the horizon as it sail away. Before the from is obvious, the object is out of sight range!

As long as the maximum visible range is less than the distance at which curvature is detectable, that curvature becomes invisible to the human eye.

I suspect you could take this to the extreme where your planet is only about 1km in radius (curvature visible at 50m), but if the fog limits your vision to 30m, you will be oblivious!!


An infinitely distant horizon is not visible as such, because of atmospheric scattering. The farthest humans can see, even on dry days in high deserts (which have the clearest skies there are), the practical limit for seeing from one mountain to the next is at most about 150 miles, at least according to historical US weather service reports from Arizona and New Mexico. Near the coasts, or on humid days, maximum clear air visibility is a lot less. On Mars it would be significantly farther, since the atmosphere there is much thinner (though that has lots of other consequences). You could use that distance and the horizon formula to calculate the radius necessary to cause the horizon to be at least that far away, but that only works if you can limit how high any observer can climb or fly.

Consider the example of a ship with a tall mast. Anyone who watches one sail away over the horizon sees it sink down as it moves farther away, disappearing from bottom to top while still of appreciable apparent size. This is proof the oceans of actual earth are curved, not flat, because they are hiding only the bottom of all distant ships, exactly as a hill does. On a truly flat world, the ship would keep getting smaller until it all got too tiny to see at all, but your last glimpse would still see the ship all the way from bottom to top. On a world curved the other way, as on the inside of a cylinder or sphere, you could look over distant mountains to see valleys and oceans on the other side, if the air were clear enough or the curvature sufficiently steep.

  • $\begingroup$ "the practical limit for seeing from one mountain to the next is at most about 130 miles" I've personally seen as far as 250km (but these were exceptional weather conditions), so this limit might not be a hard one. $\endgroup$
    – Polygnome
    Feb 11 at 13:59
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    $\begingroup$ @Polygnome 250 km is 155 miles, so I'd call that consistent with the uncertainty of "about 130". And yes, it's a probability distribution, where larger distances are increasingly unlikely. $\endgroup$
    – Ryan C
    Feb 11 at 14:40

If the world has topography, pretty small.

For a casual observer, one of the simplest methods of proving the earth's roundness is to watch a ship disappear over the horizon, and to note that the bottom of the ship disappears before the top. The reason this example is always conducted with a ship and not some other vehicle is because the ocean is very big and very flat. Watching someone walk over the horizon in hilly terrain just means that they've crested a hill, and is not itself evidence of the earth's curvature. If there are no large flat areas on the planet, it won't be possible to notice the curvature by watching someone travel over the horizon. You could probably figure it out if you're able to catalogue mountain heights and calculate that the angles of how distant mountains get obscured don't quite work out, but it won't be nearly as obvious as on a perfectly spherical planet. I'm not sure exactly how small you could go before the curvature becomes obvious (it'll depend on height and closeness of the mountains), but since mountains ruin the horizon trick on earth, you can certainly go smaller than the size of our current planet.


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