Curve on the horizon

[Dialogos]

I have a technical question to ask you all

In SpaceEngine, https://spaceengine.org/ ( what a great piece of software it is ), if you try to land on a planet ( say planet Earth ), and look at the horizon, you will see that the horizon is basically flat.

Let's say we turn off the clouds and atmosphere, to leave only the ground and water visible;
we will place our point of view on a ground practically at water level, and without hindrance all around (in the sense of placing it as if we were observing the horizon practically from the surface of the water );

if I then try to raise My observation point to 5km( 3.106856 mile ) above the ground, I will notice that a slight curvature on the sides of the horizon is already beginning to be seen; the higher I go, the more and more such curvature becomes visible, on both sides of My point of view ( as well as a slight bump in the center );

what I wonder, and I ask you all, is whether the SpaceEngine graphics engine produces some kind of image deformation, which could account for the curvature that one begins to see at the horizon already from 5km( 3.106856 mile ) height from the ground.

I thought that on the globe ( oblate ellipsoid ) the curvature at the horizon would begin to be seen from heights much greater than only 5km( 3.106856 mile );

if the SpaceEngine 3D engine does not produce any image deformation, on the sides of the screen, does it mean that the Earth's atmosphere is solely responsible for our not being able to see, in reality, the earth's curvature from such low heights ?

Or is there someone who claims that the Earth's curvature is visible, in reality, even from 5km( 3.106856 mile ) above the ground ?

I think it's just the Earth's atmosphere, the optical phenomena within it , the factor that should prevent you from seeing the curvature from such low heights, as instead you can see in SpaceEngine, by turning off the atmosphere, right ?

The field of view of one's point of view I don't think it has anything to do with it, in this, because if if I am observing a sphere, at the horizon ( so My point of view is on the sphere, at varying heights above the ground ), I will ALWAYS see, on the sides of My point of view, curvature, as well as a bump in the center of it; if the sphere is monstrously gigantic, it would require a greater height from the ground, to be able to begin to see these things, it is true; but here I refer to the dimensions of the planet Earth, and their supposedly mathematically correct representation in SpaceEngine, where, even at heights of 3-5km above the ground, without( switch off ) the atmosphere, the curvature on the sides can be seen well, distinctly.

The field of view in SpaceEngine can be modified ( and it behaves like a sort of linear zoom, compared to what you see; behaves correctly, this, I'm saying ) : if I set it as that of human sight ( and i think it is the default value ), there should therefore be no correspondence problems, between the hypothetical reality (the one where I get up from the ground at a height of 3-5km, 'turn off' the atmosphere, and I observe the horizon ), and the simulated reality inside SpaceEngine( where I do the same thing )


( All the above, is, you think ) correct ?

Thank you

 

[Ann K]

My rudimentary Italian fails me. What is the width of the field of view, and the angular span? Thanks.

 

[DavidB66]

I have a technical question to ask you all

If you don't already know it, I suggest you check out Walter Bislin's earth curve calculator/simulator here:

http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

By varying the parameters in the box below the visual display, such as observer height, field of view, and 'refraction', you can see how much visible curvature on the horizon the calculator predicts. I haven't checked the math myself, and it is probably beyond me, but I haven't seen anyone challenge it.

If you put in observer height of 5,000 m (5 km), and a wide field of view (say, 70 degrees), with either 'standard' refraction or zero refraction, you should see a slight curvature on the horizon. Maybe a bit less than in your attachment 1, but that could be due to a different field of view.

There is a common misconception (partly due to an old remark by Neil DeGrasse Tyson!) that curvature would not be visible until the observer is at least tens of miles above the ground. In fact it has been observed from aircraft much lower than this. There is a complication that the horizon is often obscured by clouds or haze, and in photographs there is the additional complication that wide angle lenses often introduced distortion.

There is a lot of discussion of these issues in various threads on this site.

 

[Mendel]

In SpaceEngine, https://spaceengine.org/ ( what a great piece of software it is ), if you try to land on a planet ( say planet Earth ), and look at the horizon, you will see that the horizon is basically flat.
[..]
if I then try to raise My observation point to 5km( 3.106856 mile ) above the ground, I will notice that a slight curvature on the sides of the horizon is already beginning to be seen; the higher I go, the more and more such curvature becomes visible, on both sides of My point of view ( as well as a slight bump in the center );

what I wonder, and I ask you all, is whether the SpaceEngine graphics engine produces some kind of image deformation, which could account for the curvature that one begins to see at the horizon already from 5km( 3.106856 mile ) height from the ground.

I thought that on the globe ( oblate ellipsoid ) the curvature at the horizon would begin to be seen from heights much greater than only 5km( 3.106856 mile );

You can compare:

Source: https://youtu.be/JkrIm0ZUyJY


With a real camera, we'd have to consider thr possibility of barrel distortion, but your sea level screenshot has a perfectly straight horizon, so that's fine.

The simulator might be cutting off (clipping) at certain distance, or be not as precise as needed.

And it does come down to the field of view, if that is wider than a typical camera, the curvature will look more pronounced. conversely, the more you zoom in on a curve, the straighter it looks.


Source: https://imgur.com/gallery/8dvnDn6

 

[FatPhil]

The 173+ degree "Dim.Apparente" looks very bogus, but like Ann I may be misinterpreting the Italian. That's what you'd expect if you were hovering at 5km altitude on an 170km diameter planet:
? 2*atan(170/2/5)*180/Pi
173.267

 

[Mendel]

The 173+ degree "Dim.Apparente" looks very bogus,

looks to me like the angular size of the object, it varies with the distance to the observer as expected, and is useful to know when comparing the apparent size of the sun with the apparent size of the moon etc.

 

[FatPhil]

looks to me like the angular size of the object, it varies with the distance to the observer as expected, and is useful to know when comparing the apparent size of the sun with the apparent size of the moon etc.

That was obviously what the calculation that you snipped was calculating.

Edit: Except it wasn't - I was misinterpreting the other intputs to my equations - the correct calculation with working is down below.

 

[Mendel]

looks to me like the angular size of the object,

but I think it's not correct

from the OP:

from https://www.metabunk.org/curve/ :

if we don't care about the atmosphere, horizon dip of 2.29⁰ on both sides should make the angular size of the planet 180⁰-2×2.29⁰=2×87.71⁰=175.42⁰. The simulation's 173⁰20' is too small?

 

[Mendel]

That was obviously what the calculation that you snipped was calculating.

it wasn't

That's what you'd expect if you were hovering at 5km altitude on an 170km diameter planet:
? 2*atan(170/2/5)*180/Pi
173.267

r=170/2=85
85/5 is the aspect you get when hovering 5 km above a 170 km diameter circle

a 6000 km diameter planet has a non-refracted angular size of 173.33⁰ at 5090m altitude, according to curve calculator (horizon dip 3.335⁰)

 

[FatPhil]

from https://www.metabunk.org/curve/ :
View attachment 58190
if we don't care about the atmosphere, horizon dip of 2.29⁰ on both sides should make the angular size of the planet 180⁰-2×2.29⁰=2×87.71⁰=175.42⁰. The simulation's 173⁰20' is too small?

Confirmed:
? acos(6371/(6371+5.09))
0.0399600
? acos(6371/(6371+5.09))*180/Pi
2.28954
? asin(6371/(6371+5.09))*180/Pi*2
175.421

Of course, when you're talking about the incredibly wide, what also becomes important is the perspective mapping used, in particular with a wide field of view. For display on a flat screen, I'd hope a planar mapping is used, and that would do nothing to help the small curve be made more visible. However, if a cylindrical perspective (very unlikely) or spherical perspective (more likely) is used, the edges will be shrunk inwards, and the curve will be more apparent.

 

[Mendel]

However, if a cylindrical perspective (very unlikely) or spherical perspective (more likely) is used,

spherical perspective produces distortion, that is not in evidence in the screenshots (straight lines are straight)

 

[FatPhil]

spherical perspective produces distortion, that is not in evidence in the screenshots (straight lines are straight)

Woh, from the thumnails, it wasn't clear that there were any straight lines. I only looked at the 5_01km image, as that was the one that was repeatedly mentioned in the text, and that has no straight lines. Indeed, the antialiasing persuades me that the lines in the 1_50m image are absolutely straight, and cannot imagine what they would represent in anything apart from planar perspective. So that's indeed most likely the projection in use, as hoped.