Atmospheric hazing on an infinite plane

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  • #1
Ryan_m_b
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A question came up with some friends recently that googling hasn't turned up a satisfying answer to. Simply put: If the world was flat and infinite (ignoring all the stupidity of this) how far could you see from sea level before the atmosphere itself is preventing any light reaching you? Obviously if Earth was flat then someone standing on the west coast of Portugal isn't going to see someone on the eastern seaboard with the naked eye, but assuming they had a telescope would they be able to see through the thousands of kilometres of atmosphere? Is there a simple way of figuring out an answer to this question? A lot of people have asked this question online but there's a lot of guesstimates and unexplained answers which aren't helpful.
 

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  • #2
Mister T
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A question came up with some friends recently that googling hasn't turned up a satisfying answer to. Simply put: If the world was flat and infinite (ignoring all the stupidity of this) how far could you see from sea level before the atmosphere itself is preventing any light reaching you?
Do you know the answer to this question: On planet Earth, how far can someone see before the atmosphere itself prevents light from reaching you?

The effect of Earth's curvature can be removed by considering situations such as standing on top of a tall mountain or structure, and looking at some far-away structure of comparable height.

I suppose that some light from a distant object may reach the observer, but is it enough light for the observer to be able to see the object? And it seems to be heavily dependent on the brightness of the object, whether it's a source of light, or simply a reflector of light.
 
  • #4
jbriggs444
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Assuming a typical temperature and density profile, what viewing angle would you need to use to see the surface of a planar earth at a particular distance?
 
  • #5
russ_watters
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Assuming a typical temperature and density profile, what viewing angle would you need to use to see the surface of a planar earth at a particular distance?
I'm not sure what you are getting at with that. What does viewing angle have to do with this?

The practical test case I would think would be seeing a tall building or mountain from another tall building or mountain.

E.G. New York, Philly and Boston lie roughly on a line, all very near sea level; 80 miles between Philly and NYC and 180 between NYC and Boston.

It doesn't look to me like there is any land above 200' in between Philly and NYC, but maybe 800' between Boston and NYC.

A quick google tells me a thumb rule is the square root of the eye height in feet times 1.23 gives the horizon distance in miles.

The Freedom Tower observation deck in NYC is 1,370' in elevation, for a horizon distance of 50 miles.

The Comcast Tech Center in Philly is about 1000 ft tall (not certain of the highest floor), for a horizon distance of 39 miles.

Thus on a clear day, you should be able to see the top of the Philly skyline from the Freedom Tower even on a round Earth.

The Hancock Tower in Boston is about 790' (occupied to the top), for a horizon distance of 35 miles. But due to the height of the Comcast Center, you should be able to see it over the obstructions from Philly on a clear day, if the Earth were flat. Certainly from the Freedom Tower.

None of this of course will matter to flat Earthers. Being able to see this effect is totally mundane and ancient: Noting that distant ships are "hull down" on the horizon is standard mariner practice and as far as I know has been for millenia, so you really shouldn't need to scale it up. And it is of course standard practice for navigators to calculate how far away they should be able to see lighthouses.

If you google for it, you can indeed find "hull down" photos of skylines....though at least one seems to be from a flat earther who doesn't seem to notice what he's actually proven.
 
  • #6
jbriggs444
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'm not sure what you are getting at with that. What does viewing angle have to do with this?
Due to atmospheric refraction, you will need to look up slightly. Your line of sight will follow an arc, not a straight line. The path will be high in the middle and low at the ends.

Ballpark figure is 35 minutes of arc for sunrise/sunset on a curved earth. On a planar earth, I'm not sure how to calculate it.
 
  • #7
Merlin3189
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Maybe look at calculations for graded index fibres?
 
  • #9
tech99
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Maybe look at calculations for graded index fibres?
The problem is similar to that of planning a microwave link. There is always ray bending going on due to the atmospheric refraction gradient. Also, absorption, which for microwaves we simply express in dB per km. Not sure why optics has to have an entirely different set of terminology.
Information on ray bending is given in Astro Navigation tables. From memory we assume about half the Sun's diameter, or 15 minutes of arc.
 
  • #10
Ryan_m_b
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Atmospheric extinction is the issue:

https://www.skyandtelescope.com/astronomy-resources/transparency-and-atmospheric-extinction/

It's wavelength and of course weather dependent, but it should be possible to make some assumptions and calculate an answer.
This is really useful, thanks :) It seems like multiples of airmass equivalent to 10km straight up reduce magnitude in a linear fashion. So even if you could in theory see for thousands of kilometers any light source is going to be hundreds of times dimmer.

Here is a log linear model for atmospheric extinction brief report. http://faculty2.ucmerced.edu/snewsam/papers/Graves_WACV11_UsingVisibility.pdf
Follow the cited papers if this does not float your boat, See Section5: Prediction model
Awesome :) thanks
 
  • #11
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This is an easy-model I used to simulate hazing conditions in image processing ...


About Using Subtractive Filters Model (Beer-Lambert)

I will explain it with an example and with the maths that supports it

Suposse you are sitted at the table and you have a Rubik Cube (RC) on the table at 1 mts from your point of view (at the table level), looking directly to the RC

The “light” coming from the RC is 100 (at its maximum level, because there is no mist at all)

You have 10 films with thickness of 1 cms each one, all of the same material, that has the property of obsorving 25% of the incident light, then you insert the first film in front of the RC in the line of your vision, the light that you’ll perceive coming from the RC is not 100, is …

100 x (1 – 25%) = 75

When you insert the second film is not 75, is …

75 x (1 – 25%) = 56

When you insert the third film is not 56, is …

56 x (1 – 25%) = 42

Resuming this into formulas :

I = IO (1-K)n

Where I is the actual LIGHT INTENSITY after the original level (IO) has passed through n filters, where each filters absorbs K% of the incident light

IO = 100
K = 25%
N = 3

An obviously the result is again 42

Now trying to re-express the formula, call

B = (1-K)

Then the formula could be written

I = IO Bn

Now if the films have the same thickness, therefore the same K, “n” is a measure of the distance, so call it “d” instead of “n”

I = IO Bd

Now and at last we can express the factor “B” in other form :

B = e

Solving the equation :

α = - ln (B) “α” is the natural logarithm of B

So we have the final expression (The Beer Lambert-Law, it’s no an empirical law)

I = IO e-αd

This means that the light coming from the object follows an EXPONENTIAL curve, function of the distance

α is the absortion coefficient per meter, it depends on the material and is function of the wavelength

When you analyze mainly an absorptive media you can use the equivalent α, so you can analize the whole Light Intensity independently of the wavelength using the equivalent α

Beer-Lambert Law is strictly true for absorptive media
Media should be at least non highly dispersive (scattering) media
Media must be homogeneous in the interaction volume
Media should not contain fluorescence or phosphorescence phenomena
Media should not be a highly concentrated solution


Visibility Series, RGB Math

Scope

Excellent solution to face fog, mist or haze conditions on an image using RGB Colors, applying Beer- Lambert law, Koschmeider Equation​

Variables and Parameters

D : Visual Range from the observer where contrast decay to 2% of his real value [mts] (Weathercast)
d : Distance from the observation point to the object in the mist
b : Attenuation Factor, sets the contrast decay velocity
XV : RGB-Channel Value for the target color, when X = D [0;255] (the color of the mist horizon)
XO : RGB-Channel Value for the object color seeing by the observer at a distance d
X : RGB-Channel Value for the object local color seeing by the observer at a null distance
G : Gamma correction factor in image processing​

Equations

b = 3.912 / D
xO = (x – xV) e-bd + xV
XO = 255 (xO)1/G
x = (X / 255)G Linear RGB [0,255]
X = 255 (x)1/G Standard Not Linear RGB [0,255]

Observations

This formula should be applied to each RGB channel (R,G,B) to find the color seen by the observer
Typos like x means Linear-RGB, X means Not-Linear-RGB​
 

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