Comparing flat Earth and spherical Earth from a geometric point of view has an article called "The Reason We Cannot See Earth's Curvature When Standing on a Beach", There is an example where the observer height is 2 m and the field of view is 65°.


Putting these values into Walter Bislin's calculator we must notice that in the View∠-box is the diagonal FOV. But 65° is the horizontal FOV. So you must type into the View∠-box the value 65*1.2018502 = 78.1203.

Without refraction, we get the values as follows:


These calculations in the -article are correct:
– Distance to the horizon = 5048 m (HorDistView v)
– Length of the visible horizon = 5425 m (HorLftRgtWidth)

The visible bulge (sagitta) of the horizon (HorLftRgtDrop) is 0.742757 m. But says, "If the length of the horizon is 5048 m, then the bulge in the middle is 0.577 m." This is clearly not correct.

The bulge calculated by is the sagitta of the circular segment of the Earth's great circle when the chord is 5425 m. You can see this with the Metabunk calculator when the distance is 5.425 km:
"Sagitta (or 'Bulge') is the amount of rise of the earth's curve from a straight line connecting two points on the surface. 'Sagitta' r-sqrt(4*r*r - d*d)/2, = 0.58 meters",

This is the sagitta we have been talking about when trying to show the curvature of the Earth. It does not depend on the observer. The horizon bulge (HorLftRgtDrop) is the sagitta of the horizon ring, seen at a certain height above the center of the ring. These are two different things.

In this example the "Earth curve sagitta" is 58 cm and the "Horizon curve sagitta" is 74 cm. They are almost the same. So the sentence in the meme is right: "The bulge is only about 0.01% the length of the visible horizon". However, the reasoning is wrong.

At the observer height 5 m we get the chord 8577 m and the horizon curve sagitta 1.86 m. Then the bulge is about 0.02% the length of the visible horizon. Yet we might say that the conclusion is true: "We can’t see Earth’s curvature from the Earth surface itself, not because “there’s no curvature”, but because the curvature is too small for us to perceive."

Ultimately, the conclusion is based on a misconception that confuses the "Sagitta of the Horizon curve" with the "Sagitta of the Earth curve". So should correct the meme and the article, especially since Erik Yde Lauritsen pointed out the error in the comment section already four years ago. (Although I don't know where he gets the bulge 62.6 cm from, instead of 74.3 cm.)
When looking at the horizon, you're looking at a circular ring in which you are at the center (more precisely, slightly above the center). You can compare this to a hula hoop. When you raise the hoop to eye level, you won't see any curvature. When you lower the hoop to the level of your chest, you will see it curved – and most importantly, you will see the same curvature in every direction. But if you look at a hula hoop standing upright against a wall, you'll see the hoop continue to curve as you move your head.

So when you look at the horizon, you are not looking at the curved great circle of the Earth, but at the horizon circle, in the middle of which you are standing. If you look down (in the depth direction), you can see the real curved surface of the Earth, but it is difficult to calculate this curvature directly.

On the left in the picture the red man is watching at the curving horizon. He is standing on a spherical cap. It's easier to see the curving of the horizon if we cut that cap off. The "Sagitta of the Horizon curve" is marked with the letter k. On the right in the picture is demonstration how difficult it is to see the "Sagitta of the Earth curve" s directly.Image1.jpg
Suppose the red man above looking at the horizon would only see a small part of the horizon and nothing else. He may remember the old exercise in school geometry: "Given an arc of a circle, complete the circle". But if he doesn't know where he is, he doesn't know in which plane the arc is either.

Below is another example. There is the home curve of the running track. A small part of it is lit with LEDs. In complete darkness, the gray man can only see this part of the arc, so he can complete the circle in many ways in his mind. He may imagine seeing a large circle in front of him.

Perhaps the natural way for a human is to see the circle as standing perpendicular to the gaze. In this example the gray man's interpretation was wrong but also human. So maybe the misconception on site is a human error too. ;)

@Rory. Some of these images are probably from your video that is no longer available. So thank you!
Truthly McTruthface demonstrates horizon curve for a planet with a radius of 1.31 miles and a surface area 1/3 the size of Washington DC:

Truthly's video is aptly titled "The Flat Earth Video That NO ONE Can Debunk!", because there is nothing to debunk. Its purpose is somewhat obscure. Is it a clever parody, is it mocking the FE people, or is it a real invitation for them to study and "beat them at their own game"?

Truthly's YouTube career three years ago is quite short. In two months, he released five videos. The one mentioned here is the first one. It has 18758 views. The last one has only 248 views.

Maybe the call to study science was too much for FE people, or maybe Truthly himself got bored with the subject. Anyway, I think he is not a genuine flat-earther, but a mere troll is he neither. I especially liked his example using miniature model of the Earth. He could be an inspiring science teacher with his demonstration skills.

Truthly .jpg
Last edited:
A miniature model of the Earth can show many geometric features and optical phenomena of the real Earth. (Unfortunately, the same cannot be said for its ability to describe gravitational phenomena.)

Let's say that the Earth model is made on a scale of 1:10,000. In this model, the radius is 637.1 m and a 180 cm tall person is 0.18 mm.

A ball with a radius of 637 m is impossible to build. So is a hemisphere or a dome, though there are buildings taller than it. But to demonstrate what I have in mind, we need only a spherical cap of the ball.

A circular area on the Earth with diameter of 1100 km has the sagitta of 23.78 km (for example the circular area on US map below). In the model this cap's diameter is 110 m, the radius 55 m and the sagitta 2.378 m.

Let's say we are building a round Science Center (building with a circular ground plan). The roof of this building is the aforementioned spherical cap with a diameter of 110 m (see image below). In the drawing, the height of the building is about 25 m. This number is unimportant. The important thing is that the shape and size of the roof cap are geometrically exactly right.

In the center of the building will be an elevator that takes the visitor to the roof. There, the visitor can make observations about the distance and curvature of the horizon, hidden heights, etc. Refraction is not a problem as it is in the real world. Perhaps the scale equivalents are also quite easy to remember: 1 km – 10 cm and 10 m – 1 mm.

These science centers are being built all over the world all the time. Maybe some architect will be interested in this roof idea. But what exciting or educational experiences could such a roof offer? I have some ideas, but better to leave them for later.

Where did I get the roof diameter value of 110 m? It's just the diameter of Globen's "dome". Globen (officially Avicii Arena) is a sports arena in Stockholm. It is the largest hemispherical building on Earth.

As a miniature model of the Earth, Globen is too small (radius = 55 m) for simulations of hidden heights, etc. For these purposes, the roof of the Science Center is much better (radius = 637 m), although it is only a small spherical cap (and not a hemispherical building like Globen).



Senior Member.
One nice demonstration I did was to take a long piece of flexible plastic (about 5 metres long) and use it to recreate a photograph of mountaintops. When the plastic was flat the scale model mountain peaks didn't match the photo, but when it was curved by the appropriate amount they did.

Unfortunately the video I made is lost but I think there may be some photos somewhere.
Some more thought of the Science Center model. Here are examples of the scale:

Man on the beach
– Earth: Eye height 2 m, Horizon distance 5.05 km
– Model: Eye height 0.2 mm, Horizon distance 50.5 cm

On top of the skyscraper of the future
– Earth: Eye height 1 km, Horizon distance 112.9 km
– Model: Eye height 10 cm, Horizon distance 11.3 m

– Earth: Eye height 10 km, Horizon distance 357 km
– Model: Eye height 1 m, Horizon distance 35.7 m

Felix Baumgartner's jump
– Earth: Eye height 39 km, Horizon distance 706 km
– Model: Eye height 3.9 m, Horizon distance 70.6 m

– Earth: Eye height 408 km, Horizon distance 2316 km
– Model: Eye height 40.8 m, Horizon distance 231.6 m

The "Man on the beach" example has to be left out because the eye height of 0.2 mm is certainly too small to adjust. Also, Felix Baumgartner's jump and ISS have to be excluded because they have horizon distances greater than the roof radius. (To get a horizon distance of 55 m, the height of the eye should be about 2.38 m. This is the same as the sagitta of the roof!).

Let's view the skyscraper example.

Let's say you are looking at the sea from an eye height of 1000 m. A very large ship is heading towards the horizon. The height of the ship is 100 m. (This height is about as much exaggerated as the height of the skyscraper.) You see the ship on top of horizon at the distance of 112.9 km. Then it starts to disappear from the bottom up. At a distance of 148.6 km, it is completely out of sight. If the ship stops right there, you can see it in its entirety again if you rise to 1733 m (by helicopter?)

In the model, the height of your eye is 10 cm and the height of the ship is 1 cm. The ship is on the top of the horizon at a distance of 11.3 m. It has completely disappeared at a distance of 14.9 m. If you stop it there, you can see it wholly again at eye height of 17.3 cm.

These are the numbers that the globe model gives. But the goal is to verify them with your own eyes. So there must be a very precise mechanism that can be used to adjust the viewing height and the movement of the "ship". In order to see the small ship clearly, you must have also some kind of telescope connected at the viewing height adjuster. (You can also use larger ships).

Maybe there's a computer that tells the ship's distance. You can stop the ship's movement by pressing a button, for example when you see it on the top of horizon. Then the computer tells you the distance and gives also the reference number calculated by globe model, so you can compare them.

A sceptical flat-earther can be given the opportunity to go on the roof and measure the distances with his own measuring tape. There is also a Nikon P1000 available, with which he can try to zoom the half-lost ship back.

In this example the horizon is at a distance of 11.3 m and the ship vanishes at a distance of 14.9 m. Yet the horizon looks completely flat. Is this proof that the roof is shaped like a paper towel roll? No, and neither is the Earth.

Last edited:
Concorde's maximum cruising altitude is about 18000 m. Putting this value into Bislin's curve calculator, you get the following values with a horizontal FOV = 60° (

Now the DipHeight b = 17949.3 m. This is the drop (the amount the curve of the earth drops away from horizontal level) at the horizon distance d. It is also the same as the sagitta of the Earth curve at the chord length 2d. (HorDistZ p = b+h. So I think this value is a bit redundant.)

At the distance of the horizon, the drop (and the sagitta) seems to be very closely the same as the height of the observer. This is quite interesting.

I made a calculation to compare the length/sagitta ratio. Here I marked, as before, the sagitta with the letter s. The simple formula below in the red frame gives the ratio h/s as a function of the horizon distance d (HorDistX d).


The table lists some h/s (=h/b) ratios. For example, Concorde's height is about 0.3% larger than the sagitta.

d –––––––––––– s ––––––––––– h –––––––––– h/s
1 km –––––––– 0.08 m –––––– 0.08 m –––––– 1.0000000
5 km –––––––– 1.96 m –––––– 1.96 m –––––– 1.0000003
20 km –––––– 31.39 m ––––– 31.39 m –––––– 1.0000049
50 km ––––– 196.20 m –––– 196.21 m –––––– 1.0000308
200 km ––––– 3.140 km –––– 3.142 km ––––– 1.0004931
350 km ––––– 9.621 km –––– 9.636 km ––––– 1.0015124
478 km ––––– 17.96 km –––– 18.01 km ––––– 1.0028265 (Conc.)
700 km ––––– 38.57 km –––– 38.81 km ––––– 1.0060912 (Baum.)
1500 km ––– 179.10 km ––– 184.28 km ––––– 1.0289247
2177 km ––– 383.49 km ––– 408.05 km ––––– 1.0640478 (ISS)
3000 km ––– 750.53 km ––– 850.76 km ––––– 1.1335358
4000 km ––– 1412.2 km ––– 1814.4 km ––––– 1.2847881

During his 39 km jump, Felix Baumgartner saw the horizon about 700 km away. The drop at this distance is also about 39 km. If he had said, "The drop at the distance to the horizon is the same as my altitude", he would have made only a 0.6% error.

If an astronaut on ISS said the same thing, he/she would make a 6% error. But there the distance to the horizon is about 2200 km. At small distances sagitta s (= drop b) and the observer's height h are practically the same.


Senior Member.
One nice demonstration I did was to take a long piece of flexible plastic (about 5 metres long) and use it to recreate a photograph of mountaintops. When the plastic was flat the scale model mountain peaks didn't match the photo, but when it was curved by the appropriate amount they did.

Unfortunately the video I made is lost but I think there may be some photos somewhere.
you lost your youtube account? jeez. there is another demo in that thread, you had linked
does it work?


Senior Member.
you lost your youtube account? jeez.

Yep, been gone over 2.5 years. A linked account I had got a couple of copyright strikes and everything connected to it was terminated. Supposed to be three strikes before they do that but all attempts to communicate with them hit brick walls.

there is another demo in that thread, you had linked
does it work?

Yep, that's the one. Has Bobby Shafto's version and some details of mine, including scale and spreadsheet with the calculations. Thanks for finding. :)

Latest posts