Earth Curvature Simulation by Walter Bislins

We can measure the angle between any two lines of sight. This is often called the "visual angle", or the "angle subtended at the eye", or the "angular size" (when the two points are the extents of the object).

Importantly: this is entirely independent of the Field of View of the camera, or the eye.

A photo is flat. How we go from the real world to a flat surface is "projection" - which is literally like what it sounds like, projecting light onto a flat surface, like a movie projector projects an image onto a screen.

I'm beginning to realize some of the challenges in image interpretation. Your example was a great help. Here I kind of turned it around.

Suppose I'm watching a landscape where I see five tall buildings in my 60° horizontal FOV. They are in different distances but the angular spacing is the same 15°. I take a photo of the five buildings with some scenery on both sides. The focus is at the middle building.

In the nature I saw the buildings at the same angular distance 15°. So I might expect that they are equally spaced in the photo too. But they are not. If the distance of the outermost buildings 1 and 5 in the photo is 10 cm, the distance between the buildings 3 and 4 is 2.32 cm and between 4 and 5 it is 2.68 cm.

So just looking at the photo is hard to reason what the eye in reality saw. But can even the eye see "the reality"? In this respect, yes, because the human retina is not a flat disk.

In my previous post was an example "Tracy in ISS looking at the horizon". There I asked if the curve picture in Walter's calculator corresponds to what Tracy sees through the window. Now I think this was a little silly question, because a flat picture can never fully correspond what human eye sees.

The proper question could be: Are the options in calculator now correct, so that the curve picture corresponds Tracy's view the best possible way. There were other questions too. I would be thankful if you could comment them briefly. I guess it might help other uncertain readers too.
33.jpg
 

Mendel

Senior Member.
Now I think this was a little silly question, because a flat picture can never fully correspond what human eye sees.
No.

Stereoscopy aside, the flat picture does correspond to what the eye sees if your eye is at the correct distance to the picture, such that the frame edges maintain the 60⁰ viewing angle. If you move the picture closer or farther away, it will start to feel distorted. (This usually happens with wide angle and telephoto shots.)

In computer graphics it's much simpler to think of in terms of the "virtual image plane", and put that in front of the virtual camera.
2022-07-31_07-58-17.jpg
This is much easier to visualize, no image flipping, and fewer lines. Mathematically it's really the same.
If your astronaut painted her window to look like the outside view, she could not tell the difference as long as she didn't move her head.
 
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Stereoscopy aside, the flat picture does correspond to what the eye sees if your eye is at the correct distance to the picture, such that the frame edges maintain the 60⁰ viewing angle. If you move the picture closer or farther away, it will start to feel distorted. (This usually happens with wide angle and telephoto shots.)

If your astronaut painted her window to look like the outside view, she could not tell the difference as long as she didn't move her head.

Yes, you are right. I wasn't thinking clear.

So, to see my 10 x 15 cm picture without distortion you must look at correct distance. To get the 60° viewing angle to buildings 1 and 5 your eye's distance from the screen must be the same as the distance between buildings 1 and 5 on the screen.

I zoomed the picture to correspond to eye's distance 30 cm. I'm not sure if the distortion vanished or not. It wasn't large anyway and my picture might be quite inaccurate too.

In my other example astronaut Tracy sees the window at 40° viewing angle. I calculated that her eye's distance from the window must be about 1.374 times the width of the window. If this ratio is preserved there is no distortion in the pictures.

It means also that the square picture of the horizon in Bislin's calculator (see my previous post) must be watched from the right distance. E.g. if the square's side is 20 cm on the screen the eye's distance from the screen must be about 27.5 cm.
 

Mendel

Senior Member.
So, to see my 10 x 15 cm picture without distortion you must look at correct distance. To get the 60° viewing angle to buildings 1 and 5 your eye's distance from the screen must be the same as the distance between buildings 1 and 5 on the screen.
Yes, almost.
Check your trigonometry, it's equal sides for the triangle eye-building 1-building 5-eye. The orthogonal distance is sin(60⁰)×(building 1 to 5 distance), about 87%.
In my other example astronaut Tracy sees the window at 40° viewing angle. I calculated that her eye's distance from the window must be about 1.374 times the width of the window. If this ratio is preserved there is no distortion in the pictures.
Agreed.
 
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Mick West

Administrator
Staff member
If your astronaut painted her window to look like the outside view, she could not tell the difference as long as she didn't move her head.
That only makes the picture conform to what the eye sees at a specific position relative to the eye. It holds somewhat true for any type of projection with a focal point - like if the window was spherical, you could make the exact same observation.

This reminds me of the discussion regarding "distortion" in the Carter photo thread, which might be of interest to Pertti.
https://www.metabunk.org/threads/ex...arters-photo-not-fake-just-perspective.11724/

I say "somewhat" true, above, because it's ignoring focus and lighting issues. Consider this image of three books all the same LOS distance from the camera (from the above thread)


Layout:


Make it full screen, close one eye, and move the other eye towards the middle book until all three books look the same size (i.e. what I would see if my eye was where the camera originally was). For my monitor, this is about 5" (13cm). At that distance, the side books are at an acute angle, and I can't focus on anything. I'd need to blow up the image to several feet wide, and even then, I'd still have a difference in focus and view surface angle between the center and edge books.

But like I said in the other thread, it's somewhat semantic. The important thing is to understand what's actually going on.
 
Yes, almost.
Check your trigonometry, it's equal sides for the triangle eye-building 1-building 5-eye. The orthogonal distance is sin(60⁰)×(building 1 to 5 distance), about 87%.

Yes of course. It’s a "memory triangle" 90, 30, 60 degrees, 2:1: sqrt(3). So the distance is 5*sqrt(3) = 8.66 cm. How awkward of me, must be tired…

In Finnish the word "muistikolmio" means the two right triangles, the one above and the other 90, 45, 45 degrees, sqrt(2):1:1. The straight translation of the word is "memory triangle". Is there in English language some more proper word?
 

Mendel

Senior Member.
In Finnish the word "muistikolmio" means the two right triangles, the one above and the other 90, 45, 45 degrees, sqrt(2):1:1. The straight translation of the word is "memory triangle". Is there in English language some more proper word?
The 60⁰ triangle with equal sides is called 'equilateral triangle'.
Any triangle with two equal sides is called 'isosceles triangle'.
 

Rory

Senior Member.
In Finnish the word "muistikolmio" means the two right triangles, the one above and the other 90, 45, 45 degrees, sqrt(2):1:1. The straight translation of the word is "memory triangle". Is there in English language some more proper word?

In British English they're called "special right-angled triangles" - and more specifically "angle-based" (as opposed to "side-based").

In American English all right-angled triangles are called "right triangles".
 
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In British English they're called "special right-angled triangles" - and more specifically "angle-based" (as opposed to "side-based").
In American English right-angled triangles are called "right triangles".
To combine these two should I call them "special right triangles" ;) Or better yet "special rights triangles"?
 
That only makes the picture conform to what the eye sees at a specific position relative to the eye. It holds somewhat true for any type of projection with a focal point - like if the window was spherical, you could make the exact same observation.

This reminds me of the discussion regarding "distortion" in the Carter photo thread, which might be of interest to Pertti.
https://www.metabunk.org/threads/ex...arters-photo-not-fake-just-perspective.11724/

I say "somewhat" true, above, because it's ignoring focus and lighting issues. Consider this image of three books all the same LOS distance from the camera (from the above thread)

But like I said in the other thread, it's somewhat semantic. The important thing is to understand what's actually going on.

Suppose I'm in the middle of a circular temple. It's like that one with green background but has 24 pillars. I take a photo of the pillars with my wide-angle camera. At 150° horizontal FOV I just get 11 pillars in the photo.

In the photo the horizontal spacing of the pillars varies much. (See the sketch on the green area. The pillars are drawn as segments of line though their width should grow toward the edges)

I calculated the distance d to see the photo without distortion: d = a/(2*tan75°), where a is the distance of the outermost pillars in the photo.
a ––––– d
20 –––– 2.7
200 ––– 26.8
1000 –– 134.0

If the distance of the outermost pillars is 20 cm on the screen, I should look at the screen from the distance of 2.7 cm. This is quite impossible. At the distance 26.8 cm from the screen, the screen should be at least 2 m wide.

Maybe I should blow it up as a 10 m wide poster. Then I can watch it from 134 cm distance. Does the view now correspond what I saw in the tempel? I should make an empirical experiment to answer this.

The discussion in "Odd Looking Bidens/Carters Photo" thread is very interesting, thanks for the link Mick. It is yet mostly too technical to me (and sometimes little "semantic"). I must learn more to contribute. I understand it's complicated though.

For now, I am mostly interested in the question of how to get a picture of the Earth's horizon at different heights and different angles of view – as "true" as possible.

Bislin's calculator draws the picture of the horizon. This is just what I want. There is one pitfall though, as Mick said. To get the correct left-right drop angle (same as δ in formula (1)), the angle of view must be given in the form sqrt(2)*α, where α is the actual angle of view. Here are some correspondences:
10 –––– 14.14213562
20 –––– 28.28427125
30 –––– 42.42640687
40 –––– 56.56854249
50 –––– 70.71067812
60 –––– 84.85281374
64 –––– 90.50966799

If I want the angle δ at the viewing angle of 40 degrees, I need to put the value 56.56854249 in the calculator, and so on.

The maximum value of the angle of view slider in the calculator is 91.6 degrees. So I think the maximum size of the actual angle of view is about 64 degrees. I tried smaller angles too and they worked perfectly. 60 degrees is a safe choice though, and perhaps the best angle in demonstrating the curve to flat-Earthers.

Moreover I must first choose AspectRatio 1:1. Now I see a square in the drawing of the horizon. My interpretation is now that in this square is the "as true as possible" image of the horizon. I really hope I'm right.

In the picture below are some examples when the actual angle of view is 60 degrees. To get the most accurate view, you should look at the square from the distance of (a/2)*sqrt(3), where a is the length of square's side on the screen (about 86.6 % of a). This might be relevant in bigger heights, but hardly in demonstrating the curve to FE-people. ;)

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The heights in my previous picture might be hard to see. Here they are:
– Man on the beach: 2 m
– Like a curve on a running track: 166 m
– Aeroplane: 10 km
– Felix Baumgartner: 39 km
– ISS: 408 km

The angle of view can be selected by the slider. Then the maximum value is 91.6°. You can also type the value in the box. Then the maximum value is 160°. This correspond to the actual angle of view 113.1°

In the picture is an experiment with this maximum angle of view. There is ISS in the height of 408 km, and the actual angle of view is 113.1°. The left-right drop angle is correct i.e. same as δ in formula (1), but the horizon curve looks little odd. Looking closer from the distance "33 % of the length of square's side" doesn't help much. Maybe if I look even closer…?

So I think a good way to avoid distorted horizon curves is to keep the actual viewing angle at 60°.

ISS2.jpg
 
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84.8528? That's 60 * sqrt(2), it's simply the Pythagorean taking of the horizontal FOV, and converting it into the diagonal FOV, which is what the calculator expects. The result for the left/right drop angle is 0.157056

So that explains the discrepancy in the numbers. The pitfall here is that 84.8528 is NOT the diagonal FOV of a 60° wide square view. While the relationship between pixel distances is 1:1*sqrt(2), the angles don't scale linearly. To get the diagonal FOV from the horizontal FOV for a square aspect ratio, it's actually:

dfov = 2*atan(sqrt(2)*tan(hfov/2))

2*atan(sqrt(2)*tan(60/2 degrees)) in degrees = 78.46 degrees. (vs. 84.85). In Walter's calculator I think this only amounts to a quirk in inputting the FOV.

Now I think, that putting in the View∠ -box the value "horizontal FOV * sqrt(2)" resolves all my problems. I get the correct HorLftRgtDrop∠ and other values seem to be correct as well.

The odd looking horizon curve in my previous picture is not so odd after all. To see the correct curve you must select "Show Left-Right Drop". Then the correct curve is above the red line segment between the two little triangles. My mistake was to take the whole side of the square as the base (or chord).

"The angles don't scale linearly" as you said. That is true, but somehow the calculator seems to take care of that. Maybe it has something to do with my mistake mentioned above?

Giving the angles of horizontal FOV and Left-Right Drop I can calculate the ratio in which the chord (red line segment) and the sagitta should be seen on the screen. When I measured these lengths on the screen, the ratio was quite the same as the calculated value. (In a great contrast with using the side of the square as chord.)

To get the most accurate view you should watch at the curve from the right distance. If the horizontal FOV is 60 degrees the correct distance is (a/2)*sqrt(3), where a is the length of the red line segment on the screen (about 86.6 % of a).

Maybe all my troubles with the calculator are now over? :)korjaus.jpg
 
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Above, only 1:1 aspect ratio is used. You can also use other ratios, but they require their own coefficient:

Ratio 1.jpg

The "off ratio" is first measured roughly on the screen. Then I increased the accuracy so that I got the most correct value for the Left-Right Drop angle. That seems to work perfectly. Here are the coefficients k = c/a with greater precision:

AspectRatio –– k
off ––––––––– 1.2018502
1:1 ––––––––– 1.4142136
3:2 ––––––––– 1.2018504
2:3 ––––––––– 1.8027756
4:3 ––––––––– 1.2500000
3:4 ––––––––– 1.6666667
16:9 –––––––– 1.1473475
9:16 –––––––– 2.0397289

When AspectRatio is off, I get the most space to show the horizon. That might be a benefit in demonstrating The Curve to flat-earthers.

Below is Felix Baumgartner jumping from 39 km height. The horizontal FOV is 60°, so I type in the View∠-box the value 60*1.2018502 = 72.111. Do you think "that stuff is flat"? ;)
Ratio 2.jpg
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@Mendel. Thanks for the tip. Here is the Baumgartner setup:
http://walter.bislins.ch/bloge/inde...66937-9-9-9-6~0.0343-23.1822605-1~24.9-9-31-1
 
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FatPhil

Senior Member.
Below is Felix Baumgartner jumping from 39 km height. The horizontal FOV is 60°, so I type in the View∠-box the value 60*1.2018502 = 72.111. Do you think "that stuff is flat"? ;)

The earth doesn't come with nice neat grid markings, or a horizontal line to compare the horizon against. Given that this is a question of perception, deliberately making it easy to see invalidates the question. How would a typical FE-er, when presented with a VR experience of being in Baumgartner's position perceive what he is seeing? Is it possible that they might say "that stuff is flat"? That you or I are sure we would be able to perceive the curve says very little about how they'd process the situation. What happens if you put them inside a Zorb (or other spherical cavity), and get them to draw what they see. Does the earth take up exactly half of the sphere, or do they stop short?
 

Mendel

Senior Member.
Below is Felix Baumgartner jumping from 39 km height. The horizontal FOV is 60°, so I type in the View∠-box the value 60*1.2018502 = 72.111. Do you think "that stuff is flat"?
Bislin provided an option to save your setup to a URL ("Get App URL"); it'd be great if you posted that along with any screenshots.
 

Mendel

Senior Member.
What happens if you put them inside a Zorb
51TbjfnWjZL._AC_SY450_.jpgbubble-soccer-mieten.jpg
I can't imagine what other projection you thought I was implying.
I didn't see you imply anything, except maybe that they are meant to paint the inside of the sphere, and you also put them down in a place with little vegetation (unobstructed horizon)

if you float the Zorb in the ocean, the horizon will come up pretty much exactly half way, which is to be expected in both cosmologies
 
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FatPhil

Senior Member.
I didn't see you imply anything, except maybe that they are meant to paint the inside of the sphere, and you also put them down in a place with little vegetation (unobstructed horizon)

They're painting what they see through the VR simulation of Baumgartner's view, obviously, that's what the whole paragraph was about.
 

Mendel

Senior Member.
They're painting what they see through the VR simulation of Baumgartner's view, obviously, that's what the whole paragraph was about.
ok, so you want to put their eyes at the center of a glass sphere at 39 km height, paint the visible horizon on the inside of the sphere (as seen by them), and look for horizon drop?

I don't really see what doing this virtually would achieve

and there are easier ways to demonstrate horizon drop
 

FatPhil

Senior Member.
ok, so you want to put their eyes at the center of a glass sphere at 39 km height, paint the visible horizon on the inside of the sphere (as seen by them), and look for horizon drop?

I don't really see what doing this virtually would achieve

and there are easier ways to demonstrate horizon drop

They already had the VR goggles on. Keeping them on is easier than ballooning them to the stratosphere.
 
In the previous picture the segment of horizon really corresponds 60° horizontal FOV. To avoid those helping lines I didn't use "Show Left-Right Drop". I kept it in mind though and cut off the excess to get the correct segment AB.

So here is my artistic vision of what Baumgartner might have seen in a 60° horizontal FOV. To get the most "realistic" view your eye and the points A and B on the screen must form an equilateral triangle (triangle with equal sides) perpendicular to the screen.

FELIX CGI2.jpg
 
I just noticed one horrible sign error in my old post. I can't edit or delete it, so I write the whole post anew. I really hope my formula (2) is now correct.

==================================

What is the "left-right drop angle" δ on a flat Earth?

If R is the radius of the disk, and h the observer's height above the center of the disk, you can calculate the "left-right drop angle" δ of the horizon in a given field of view α.

One question may arise: What is the horizon in the FE model? It must be the edge of the disk. What else could it be?

I made the calculation in the case α = 60°. You can choose the radius R (though in my sketch the disk is the "standard" flat Earth model where the radius is 20000 km).

In the picture below is the derivation of formula (2). You can put the formula in Excel:
Select the cell C2 and type the formula

=(ATAN(A2/B2)-ACOS((2*B2)/NELIÖJUURI(3*(A2^2+B2^2))))*180/PII()

In English version replace PII() -> PI() and NELIÖJUURI -> SQRT

Some examples.
Keeping the height constant h = 10, we get
R –––––– δ (deg)
20 ––––– 4.52588
50 ––––– 1.77854
100 –––– 0.88709
500 –––– 0.17728
1000 ––– 0.08864
20000 –– 0.00443

Keeping the radius constant R = 20000, we get
h –––––– δ (deg)
10 ––––– 0.0044
20 ––––– 0.0089
39 ––––– 0.0173
100 –––– 0.0443
1000 ––– 0.443
2000 ––– 0.887
5000 ––– 2.227

The latter figures correspond to the flat Earth model. For example Felix Baumgartner would see on his jump The Edge curving by δ = 0.0173 degrees = 1.04 arc minutes.

Here the angle δ is calculated only above "the North Pole". So δ is the same in all directions. Looking in some other location must give different values in different directions. That's what I guess but I'm not sure. The drop of the horizon is certainly different.

5 oikea.jpg
 
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This is my latest "Flat Earth Curve Calculator". It's just a generalization of the previous one. Now you can select the viewing angle α too.

Below is the derivation of formula (3). You can put the formula in Excel:
Select the cell D2 and type the formula

=(ATAN(A2/B2)-ACOS(B2/(NELIÖJUURI((1-(SIN(C2*PII()/360))^2)*(A2^2+B2^2)))))*180/PII()

In English version replace PII() -> PI() and NELIÖJUURI -> SQRT

As an example, Felix again above the North Pole. Now we can calculate how much "more curve" he sees as his field of view expands from 60 degrees to 90 degrees. The Left-Right Drop Angle δ increases to about 2.7 times.

I still can't calculate angle δ anywhere else but on the North Pole. Well, it can be good that there are challenges. Maybe they can slow down my old age dementia. ;) 5 laajennus.jpg
 
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