[Bobby Shafto]

New video upload from JTolen Media 1:



Most of the footage is raw from left window seat of a LAX-MCO flight while passing over Arizona and New Mexico, view northward.

The intro is where the analysis is presented, along with a short evangelical statement regarding the flat earth movement. This will accumulate the usual accolades from the flat earth community, and I thought of posting it in the Flat Earth Medium forum. But I'd like to discuss it, so I'll post it here. I'm hoping JT will visit to answer some questions.

If he does show, I ask that the skeptics and naysayers of flat earth play nice. I'd really like to encourage YouTube video publishers to discuss their postings here. I know there's a view that The Flat Earth Society is a disinformation effort, but that's between the factions of flat earthers. I'd like to have a place other than YouTube comment section to  encourage discussion, debate, and critiques of these videos, including those defending the globe earth.

I really like this Infrared approach to imagery capture and wish I had the equipment to do the same. But I'm not willing to risk my modifying my cameras so I'm left to examining the products folks like JT are posting. 

[Tom Bishop]

Going through the comments, it was pointed out in YouTube that (paraphrasing) "We can see the curvature of the earth. The Earth is Round!!1!" and the following explanatory image was provided:



However, we can see from JTolan's IR observation, that there are mountains above that bright line of the "round horizon". Here is the important screenshot from the video:



The mountains are above that bright line of the horizon, telling us that it is not truly the earth's horizon we are looking at.

Secondly, here is the Earth Curve Calculator for those values:

https://dizzib.github.io/earth/curve-calc/?d0=475&h0=31000&unit=imperial

According to the Earth Curve Calculator, when looking over 475 miles at an altitude of 31,000 feet the amount hidden below the horizon is 44794.7290 feet below the "horizon".

Accounting for the height of the mountain:

44794.7290 feet - 14,000 feet = The top of the mountain should be 30794.729 feet, or 5.832 miles below the horizon.

JTolan further asserts that not only that, but the mountain is at the correct elevation it would need to be as if the earth were flat. I have not assessed this further assertion. Interesting.

He seems to be measuring the tops of the mountains in the foreground to determine that?

[Bobby Shafto]

I'm hoping JT will visit to answer some questions.

My first set of questions are about the graphic beginning at 1:55 mark:



Start with the insert:



The theoretical flat and globe earth horizons make sense, but I don't know how the -1.9° flat earth "dip" is derived, nor where the steeper -5.0° spherical earth "dip" angle comes from.

The answer may lie in the answer to the 2nd question, which is why does the shallow light ray have a sub-refractive curve upward away from the earth's surface near the "grazing angle" of the horizon, and what defines that "critical angle?"

[nickrulercreator]

Going through the comments, it was pointed out in YouTube that (paraphrasing) "We can see the curvature of the earth. The Earth is Round!!1!" and the following explanatory image was provided:



However, we can see from JTolan's IR observation, that there are mountains above that bright line of the "round horizon". Here is the important screenshot from the video:



The mountains are above that bright line of the horizon, telling us that it is not truly the earth's horizon we are looking at.

Secondly, here is the Earth Curve Calculator for those values:

https://dizzib.github.io/earth/curve-calc/?d0=475&h0=31000&unit=imperial

According to the Earth Curve Calculator, when looking over 475 miles at an altitude of 31,000 feet the amount hidden below the horizon is 44794.7290 feet below the "horizon".

Accounting for the height of the mountain:

44794.7290 feet - 14,000 feet = The top of the mountain should be 30794.729 feet, or 5.832 miles below the horizon.

JTolan further asserts that not only that, but the mountain is at the correct elevation it would need to be as if the earth were flat. I have not assessed this further assertion. Interesting.

He seems to be measuring the tops of the mountains in the foreground to determine that?

How can we confirm those are mountains? They could very well be artifacts, reflections, flares, etc.

[Bobby Shafto]

How can we confirm those are mountains? They could very well be artifacts, reflections, flares, etc.

They do align with the expected bearing. I believe he's captured what he's claiming. And it is remarkable. My questions lie with how he's interpreting what he's captured.

I have low hopes that he's going to visit though, based on his response on YouTube. I'm not going to engage on that platform and it seems he has little or no interest of defending his argument here.

But I do think flat earth skeptics do have to deal with his having "seen" over the visual horizon. Trying to refute that he's managed to sight 13-14K' Colorado Rockies peaks from 30K' over southern Arizona isn't going to work as a counter-argument. Because he did, just as he captured more of San Jacinto from Malibu than the earth curve calculators predict.

[Bobby Shafto]

The mountains are above that bright line of the horizon, telling us that it is not truly the earth's horizon we are looking at.

I agree with you Tom. (Shocking!)

At that zoomed-in view, the space above the mountain (Humphrey's Peak near Flagstaff) is dark with those wispy flecks of white above. But zoomed out, that space brightens and blends with the more distant white. With increased focal length and narrower field of view we can see that the remaining white isn't the horizon but rather the tops of surface clouds lying beyond the peak.

I also noticed that it depends on my monitor, but those white flecks are not clouds. They are, in fact, snow caps on a mountain range in the distance. With some contrast adjustment I can make out the faint land rise in the space below the elevation of those flecks.

He's definitely captured distant terra firma. What I challenge is his triangulation bearings and deduced distances. But I'll save that for now. For the time being, I agree with Tom.

Secondly, here is the Earth Curve Calculator for those values:

https://dizzib.github.io/earth/curve-calc/?d0=475&h0=31000&unit=imperial

According to the Earth Curve Calculator, when looking over 475 miles at an altitude of 31,000 feet the amount hidden below the horizon is 44794.7290 feet below the "horizon".

Accounting for the height of the mountain:

44794.7290 feet - 14,000 feet = The top of the mountain should be 30794.729 feet, or 5.832 miles below the horizon.

JTolan further asserts that not only that, but the mountain is at the correct elevation it would need to be as if the earth were flat. I have not assessed this further assertion. Interesting.

He seems to be measuring the tops of the mountains in the foreground to determine that?

I'm glad Tom noticed this because this was going to be my next question for JT had he chosen to revisit and discuss this video. I don't understand JT's vertical gauging here. Had he indexed it to angular units and figured elevation with distance, that would make sense to me, but this method is strange to me and, unless I'm misinterpreting, I'd say it's wrong. (And that's before contesting his distance parameters.)

[JTolan_Media1]

Hello gentlemen, 

I dropped by to see the discussion, isn't this stuff exciting?


So, how did I create that scale?  (I thought its obvious but maybe not)

I took the difference in height between the two peaks of Humphrey's Mtn, and created the scale based on that difference.  (I got the elevations from Google Maps with Terrain data turned on.  It's fairly accurate, but do consider that there is a slight downward looking angle of about 0.9 deg to Humphrey's Peak, per flat surface assumption) 

Shockingly, the snow covered mountain peaks in the distance seem to come up to the correct elevation for a shallow angle of observation.   

The actual angle to the distant Colorado Mountains (flat earth surface assumption) is equal to  atan( (31000 - 14000)/(5280*500mi) ) = 0.37 deg     

I assume the mountains are about 14000 ft high (judging by the cluster of Mountain peaks seen and checking with Google Maps terrain data), and the observation distance is about 480 mi.   Humphrey's Peak is at about 200 miles away, and the size of the snow capped Colorado Mountains, relative to Humphrey's peak seem to be about the right size per the laws of perspective. (i.e if twice the distance away, it should be half as small)

By chance, or can we say providence, this turned out to be an excellent experiment, because two tall mountain peaks, that rise high above the more dense part off the atmosphere with its associated distortion, have lined up at that particular instant and we can judge their heights relative to each other.   Shocking isn't it!   I'm still reeling from this realization.

-JT

[Bobby Shafto]

Hello gentlemen, 

I dropped by to see the discussion, isn't this stuff exciting?


So, how did I create that scale?  (I thought its obvious but maybe not)

Thanks, but the issue isn't with the how. The how WAS obvious. It's why you would consider such an approach valid that's the question.

I took the difference in height between the two peaks of Humphrey's Mtn, and created the scale based on that difference.  (I got the elevations from Google Maps with Terrain data turned on.  It's fairly accurate, but do consider that there is a slight downward looking angle of about 0.9 deg to Humphrey's Peak, per flat surface assumption) 

I understand that, and with that you built a scale of vertical distance, but it's scaled for a distance of 200 miles away. Your scale would show where the summit of a 13,900' mountain that was at around 200 miles, not one that's 480 miles away. What I can't figure out is how you determine from that scale to say a 14,000' peak 480 miles away is aligned to the right height. You need to work out that angular dimensions to scale the linear measurements at 200 miles with those at 480 miles away.

Shockingly, the snow covered mountain peaks in the distance seem to come up to the correct elevation for a shallow angle of observation.   

What's the rationale for reaching this conclusion? It's not found from your reasoning above.

The actual angle to the distant Colorado Mountains (flat earth surface assumption) is equal to  atan( (31000 - 14000)/(5280*500mi) ) = 0.37 deg     

This I can agree with. I call this "dip" angle and, using 480 miles vice 500, I got 0.38°. So we're in the same ballpark there.

I assume the mountains are about 14000 ft high (judging by the cluster of Mountain peaks seen and checking with Google Maps terrain data), and the observation distance is about 480 mi.   Humphrey's Peak is at about 200 miles away, and the size of the snow capped Colorado Mountains, relative to Humphrey's peak seem to be about the right size per the laws of perspective. (i.e if twice the distance away, it should be half as small)

I follow. 480 miles is 2.4x 200 miles, so a 480-mile distant object of the same height as a 200-mile distant object should appear 0.42x as small. But has your scale shown that?'
 

By chance, or can we say providence, this turned out to be an excellent experiment, because two tall mountain peaks, that rise high above the more dense part off the atmosphere with its associated distortion, have lined up at that particular instant and we can judge their heights relative to each other.   Shocking isn't it!   I'm still reeling from this realization.

-JT

I hope I've conveyed to you why I don't follow your amazement.

For illustration, here's a rough diagram, not to scale but just showing the geometrical relationships. The linear values (mileage, heights) that I derived from your video vary a little bit from what you came up with, but just for the sake of this point, I'll use your numbers.



The angle circled in red we basically agree upon.
The angle circled in blue we're apart by only a tenth of a degree (you said -0.9°).

Without knowing where "eye level" is to measure declination angle from level, and not having a ground reference to work up from, we could take your approach and use the delta between the two Humphrey's peaks and, with a known distance, work out the angular difference and build a scale based on that, just like you did with the analysis of San Jacinto imagery using Le Meridien Delfina hotel. And then using that angular measure, see if that marries with what the geometry predicts for a 480-mile distant elevation.

I don't think it does. I agree that the distant peak shouldn't appear above the nearer peak based on spherical earth calculations with no refraction adjustments or even with standard atmospheric refraction. For whatever reason, the mere fact that we can see that distant peak means the earth appears to be less spherical than one with a radius of 3959 miles.

However, that doesn't mean "flat?" Comparing angular ratios, that distant peak is still lower than would be expected if flat earth geometry is assumed. Neither the curved earth calculation nor the flat earth calculation predict what you captured in your video. It's somewhere in between a flat earth and a globe earth of the size claimed. The challenge is to figure out why that is. Why would a flat earth appear convex? Why would a globe earth appear less convex?

The point of this particular question I had with your video was that I didn't agree with your declaration that the distant mountain aligned at the expected height for a flat earth. I didn't see how you could draw that conclusion from the way you appeared to be calculating. Hoping you can see what I'm saying and either confirm or tell me what I'm getting wrong and why you believe your method was correct.

------

Even if we can't come to a meeting of the minds on this point, I'm hoping you'll stick around to address a few other inputs I have.

Oh, and while I'm at it, on another topic we got to talking about your Salton Sea cell tower sighting mentioned in your Malibu/San Jacinto video. Though there was some doubt from certain quarters, we think we id'ed the cell tower (T-Mobile tower in Salton City), but we couldn't figure out where your vantage had been to produce that picture. If you could resolve that for us, it would be much appreciated. And, of course, if we got the wrong cell tower, let us know that too. Thanks.

[Tom Bishop]

Here is my attempt at explaining the scale in the image:

Imagine that we had twenty identical clones of Michael Jordan, who have a height of 6 feet, 6 inches. The twenty clones are are lined up in a row, spaced far apart from each other—it doesn't matter how far—stretching into the distance. If we align ourselves with them and also align our eye to look right along the top of their shiny bald heads we will see the top of their bald heads perfectly lined up and stretching into the distance.



Correct?

From a First Person View, as the Michael Jordan basketball players recede from the observer they will get smaller, and also ascend in height due to perspective, since all things get smaller and ascend to the eye as they recede. What is important is that the top of the heads of the Michael Jordan clones continue to align with our eye when they are lined up in front of us in First Person View.

Next, lets say that we replace the first Michael Jordan in the line in front of us with Muggsy Bogues, who stands at a shorter 5 feet, 3 inches. Muggsy Bogues is significantly shorter than the Michael Jordan clones, and makes an odd imperfection along the top of our eyesight of the Michael Jordan heads.

However, if Muggsy Bogues were holding a ruler in his hand vertically, we could know how much to draw on a picture to get him to the artificial height of 6 feet, 6 inches, where the top of a man's head should align at that height.



MJ 3 - 20 are standing behind MJ 2, and are hidden, shrinking and ascending into the distance with the top of their heads matching MJ 2 and the artificial height of MB 1.

This is what JTolan did. He turned a Muggsy Bogues into a Michael Jordan. The mountain in the foreground provides the "ruler" where the top of the larger mountain behind it should align. The plane is at a slightly shallow angle from parallel for the distances involved, which might modify things a bit, as JTolan is talking about, but this appears to be his reasoning for why the scale should align with the mountain.

[Bobby Shafto]

Mugsy is shorter than MJ. But if each had a 12" ruler their rulers are the same length.

If you see Mugsy holding a ruler close by and MJ holding a ruler in the distance, Mugsy's ruler will appear bigger. If you want to measure something at MJ's distance using the ruler Mugsy is holding, you have to adjust the scale to account for that.

The scale in the video is based off vertical distance at the range of Humphreys Peak. To use that at the range of the Rockies, you must account for that distance in translating that scale.

The scale should be in angular units rather than linear distance. 14000' subtends a smaller angle at 380 miles than it does at 200 miles.

[stack]

Just poking around on the google terrain map:



Probably means nothing, just scratching my head with this one.

Also, it would be great to know which flight and on what day he shot this enroute from LAX to MCO. I looked up about a dozen historical LAX to MCO flights, Delta, American, JetBlue and United and all the flight paths great circle north of PHX about 50 miles south of Mt Humphreys, not south of PHX as shown in the video. (Flightradar24.com) But again, not evidence of anything, just curiosity.

[Humble B]

However, we can see from JTolan's IR observation, that there are mountains above that bright line of the "round horizon". Here is the important screenshot from the video:



The mountains are above that bright line of the horizon, telling us that it is not truly the earth's horizon.....

I'm sorry to disappoint you, but those little white spots above the horizon are not Colarado mountains 475 miles away, but a double refected miror image of the same mountaintops as visible in the foreground. An airplane has double glassed windows. The outer side of the inner window is refleting light back towards the outer window, and the inner side of the outer window is reflecting that back towards the observer. Normally by daylight such a reflection is not visible because the reflected light is too weak to see, unless, as in this case, the background is pitch black.

So no proof here the earth is flat... For such a proof we need much more and much better observation, done by clear daylight and clear enough to identify those mountains without any doubt. Just two vague spots of lights on a double glassed window by night is not enough evidence to throw away 25 ages of research and knowledge about the shape of the Earth.

[Tom Bishop]

Mugsy is shorter than MJ. But if each had a 12" ruler their rulers are the same length.

If you see Mugsy holding a ruler close by and MJ holding a ruler in the distance, Mugsy's ruler will appear bigger. If you want to measure something at MJ's distance using the ruler Mugsy is holding, you have to adjust the scale to account for that.

The scale in the video is based off vertical distance at the range of Humphreys Peak. To use that at the range of the Rockies, you must account for that distance in translating that scale.

I disagree with this, but consider this slightly modified, and perhaps clearer, analogy: Our eyes are at 6'6" in height. Michael Jordan is an unknown distance away, but his height is known: 6'6". The top of Michael Jordan's head is lined up with our eye level.

We have an object in the foreground, in front of Michael Jordan, which is also an unknown distance away, that just happens to reach a height of 6 feet, with a ruler painted upon it that shows us what 6 inches looks like. We can use that ruler to draw an imaginary line 6 inches above the height of that foreground object. Should not that line line up with the top of Michael Jordan's head? Our eye level is still at 6'6", after all.

Does it not follow as well that it does not matter how far those bodies are away from us?

Granted, the plane is at a small shallow angle from parallel for the distances involved, as mentioned by JTolan; but the point is that there is certainly no reason why it should not be possible to use a ruler in the foreground to tell us where things should be in the background, provided that we know the heights of the bodies involved.

[JTolan_Media1]

yes, Tom got it

Our eyes are at 6'6" in height. Michael Jordan is an unknown distance away, but is height is known: 6'6". The top of Michael Jordan's head is lined up with our eye level.

We have an object in the foreground, in front of Michael Jordan, which is also an unknown distance away, that just happen to reach a height of 6 feet, with a ruler painted upon it that shows us what 6 inches looks like. We can use that ruler to draw an imaginary line 6 inches above the height of that foreground object. Should not that line line up with the top of Michael Jordan's head? Our eye level is still at 6'6", after all.

Bobby, you are correct with your diagram as well, if there were no atmospheric bending and the propagation medium homogeneous. 

Regarding your statement here:

I understand that, and with that you built a scale of vertical distance, but it's scaled for a distance of 200 miles away. Your scale would show where the summit of a 13,900' mountain that was at around 200 miles, not one that's 480 miles away. What I can't figure out is how you determine from that scale to say a 14,000' peak 480 miles away is aligned to the right height. You need to work out that angular dimensions to scale the linear measurements at 200 miles with those at 480 miles away.

Note I did not say anything about size, just that the mountains appear to be at the "correct elevation for a shallow observation angle."    The mountains do appear smaller and shrunken, but just happen to be at the correct elevation for a shallow observation angle, that's all.   This tells us the propagation path is almost parallel to the ground from the distant mountain peak.

Remember, propagation in a stratified medium is a bit different then in a homogeneous medium.

Without getting into too many details on electromagnetic wavefronts and propagation through stratified or inhomogeneous mediums, just take a look at this graphic for a simple visual explanation.  Also remember Huygens principle which states that every point on a wavefront is a source of wavelets.

-JT

[Bobby Shafto]

... consider this slightly modified, and perhaps clearer, analogy: Our eyes are at 6'6" in height. Michael Jordan is an unknown distance away, but is height is known: 6'6". The top of Michael Jordan's head is lined up with our eye level.

We have an object in the foreground, in front of Michael Jordan, which is also an unknown distance away, that just happen to reach a height of 6 feet, with a ruler painted upon it that shows us what 6 inches looks like. We can use that ruler to draw an imaginary line 6 inches above the height of that foreground object. Should not that line line up with the top of Michael Jordan's head? Our eye level is still at 6'6", after all.

That works, if you can perform the survey at the level of the target elevation. In that case, the alignment is doing the work of perspective translating for you.

Does it not follow as well that it does not matter how far those bodies are away from us?

Yes, as long as you're observing level with the target elevation. Is that analogous to the situation JT is surveying of nearer Humphreys Peak and the farther supposed Rockies?

Granted, the plane is at a small shallow angle from parallel at the distances involved, as mentioned by JTolan; but my point is that there is certainly no reason why it should not be possible to use a ruler in the foreground to tell us where things should be in the background, provided that we know the heights of the bodies involved.

Sure, as long as your alignment is level. This approach becomes more flawed the greater the angle.

In this scenario, the viewpoint was not 14,000' as your analogy would assume. It was 31,000'.  The expectation that adding 1600' to 10,400' summits should would make them align with 14,000' Rockies in the distance, when viewed from an elevation of 31,000' makes no sense. To derive a prediction of "correct elevation", you have to take angular displacement into account. You do that, inherently, when you perform a level alignment. But if you can't do that, you have to do the math. You can't just take a linear ruler and expect it to work if you aren't so aligned.

[Bobby Shafto]

Regarding your statement here:

I understand that, and with that you built a scale of vertical distance, but it's scaled for a distance of 200 miles away. Your scale would show where the summit of a 13,900' mountain that was at around 200 miles, not one that's 480 miles away. What I can't figure out is how you determine from that scale to say a 14,000' peak 480 miles away is aligned to the right height. You need to work out that angular dimensions to scale the linear measurements at 200 miles with those at 480 miles away.

Note I did not say anything about size, just that the mountains appear to be at the "correct elevation for a shallow observation angle."    The mountains do appear smaller and shrunken, but just happen to be at the correct elevation for a shallow observation angle, that's all. 

I still don't know how you reach that conclusion. If Tom got it right, then were you flying at 14,000'? Because that's what would be required for Tom to be right. And if so, then yes, your method would work.

But given that you had a (flat earth) 1° declined angle toward Humphreys Peak and a (flat earth) 0.4° declined angle toward the distant peaks, how do you know the distant peaks were at the correct, expected elevation?  They appear too low to me; NOT at the correct elevation.

Using the heights and and distances we've been discussing, here's how the two nearer indexing peaks should be expected to appear if one did as Tom described and viewed them from eye level of the background peak:



Go ahead and to a pixel measurement version of your scaling method and you'll find it works. Just like Tom said. And I agree. Using your method, the background peak would appear at the correct elevation.

But change observer elevation from 14,000' to 31,000'



Now try your approach. It won't work. If those distant peaks are, in fact, 14000' Rockies, why would they appear low as if being viewed from 14,000'? Why would you expect them to be at an alignment where you could use a 14000' eye-level gauging method like Tom's?

And just for good measure, here's a view from 150'



Obviously, converting Humphreys Peak in this scenario to a 14K elevation, you can't expect it to align with a distant 14k peak.

Personally, what I think you've done, JT, is find confirmation of a biased presumption that they would line up. So when you constructively added the delta elevation to the gauge peaks of Humphreys and found it to be coincident with those cloud-like snowcaps, you didn't examine it any further.

I agree that what you see and recorded were the Rockies. I disagree that they appear at the correct elevation for flat earth. You're method of affirming that claim is in error.

[JTolan_Media1]

Guys,

Here's a graphic to better explain the perspective. (see below)


Question:    In CASE 2,   can I tell the absolute height of the lower mountain at 300 mi, based on a vertical scale derived from it's two peaks AT THAT DISTANCE, and knowing my observational height?   

Answer:   Yes


Note:  I tried to keep the mountains proportional, i.e., the one at 400 miles is half the one at 200 mile, and the one at 1000 mile half the one at 500 mile.


So even though the same size mountains appear smaller proportional to their distance from the observer, we can still tell their absolute height relative to each other if the observation angle is shallow, albeit with some slight error. 

Bobby, do you now comprehend why this observation is so astonishing?


-JT

[edit:  your previous analysis with the red and yellow towers seems correct, except you forgot about the atmospheric refraction I mentioned in my previous post]

[Bobby Shafto]

Guys,

Here's a graphic to better explain the perspective. (see below)

A key element of your graphic:


But I think this particular criticism is exhausted. I made my point. You've made your defense.

And if the objective of the scale was to verify identity of those snowcaps as the Rockies, then it doesn't matter because for other reasons I agree that's what they are. What I'm disagreeing with is that the elevation is what would be expected if on a flat earth when viewed from FL310, which you believe. Though it IS surprising to me that they are above the horizon, in challenge to a spherical earth of radius r=3959 miles, they are too low for a flat earth.

If there's some propagation reason for them to be appearing lower, fine. I'd like to get into that as well. But so far we've only been dealing with straight geometry and I assess the 'right elevation' conclusion to be faulty.

I had a list of other issues (mostly refining your assessments of distances and lines of bearing), but let's jump to the properties of light propagation over the earth. I'd really like to understand how that works in your model.

[Bobby Shafto]

[edit:  your previous analysis with the red and yellow towers seems correct, except you forgot about the atmospheric refraction I mentioned in my previous post]

I promise I haven't forgotten about atmospheric refraction. I'm hoping to get to that next.

But what I gather is that your expected elevation of those Rocky peaks included a predicted adjustment for refraction, then, that would produce the equivalence of a 14,000' eye-level view and permit this -- I'll call it the Bishop method -- approach:



Because that doesn't work from a vantage point of 31,000'. But if you anticipated atmospheric refraction would turn the scenario on the left into the scenario on the right:



...okay.  I don't know how you could have arrived at such an expectation unless it is after-the-fact 'expectation', but at least it's explained.

Moving on to how atmospheric refraction.

[Bobby Shafto]

Since we're (maybe) moving on from Tom's issue, let's get back to my original question posed since it deals with refraction and that's where we seem to be heading now.

My first set of questions are about the graphic beginning at 1:55 mark:



Start with the insert:



The theoretical flat and globe earth horizons make sense, but I don't know how the -1.9° flat earth "dip" is derived, nor where the steeper -5.0° spherical earth "dip" angle comes from.

The answer may lie in the answer to the 2nd question, which is why does the shallow light ray have a sub-refractive curve upward away from the earth's surface near the "grazing angle" of the horizon, and what defines that "critical angle?"

And here's that upward curving light ray path again:



If that's due to refraction of a non-homogeneous, stratified air mass, then we're looking at a situation of denser air resting over rare air:



I'd consider that to be an anomalous situation, with the resulting sub-refraction being an anomalous propagation of light, quite deviated from "standard."  With enough of a refractive index, you could have enough bending to depress the apparent elevation of a distant object to the extent of what we apparently see in those mountain images (and modeled with the red/yellow bars). And differences in the angle of incidence (Θ^1) between the upper and lower ray traces will have a distorting effect as well. (Whether it's stretching or squashing depends on which ray has the greater Θ^1.)

If I'm capturing this all correctly, then what you have for a flat earth is an upward refraction phenomenon that gives the earth a more 'spherical' appearance, as if distant objects are declined in elevation and/or your observation altitude is lower than actual. Compare that with a spherical earth model in which downward refraction of light accounts for the earth having a "flatter" appearance than a no-atmosphere earth of radius=r calculates, allowing distant objects to appear raised in elevation and/or observation altitude to seem higher than actual.

For some reason, a predominant theme in flat earth evangelism is to reject the effects of atmospheric refraction to explain why the curve earth calculators don't match with predictions of what should be observed on a spherical earth. Yet, what I'm seeing here is the invoking of refraction (in the opposite direction) to explain why a flat earth no-refraction geometry doesn't match with observation. If refraction's a reality, then it applies in analyzing and assessing both flat and spherical models.

But I digress and am soap boxing.  I'd like to understand what the basis is for believing light increasingly refracts away from the earth at shallower angles (greater angles of incidence).