View across Utah Lake

edby

Member
This video Source: https://youtu.be/0V-tkCzfsT0
from ‘Taboo Conspiracy’ seems to provide apparently solid evidence of flat water. He claims to be filming across from Eagle Park in Saratoga Springs to a place on the other side of the lake where his wife is positioned with a flashlight. The flashlight is clearly visible, and goes off and on when he asks her by phone to turn it off and on.


He does not say where she is located but there is a clear view at 1:04 facing La Quinta Inn and its 3 gables, which places her at about Vineyard Beach on the East side of the lake. She also refers to some ‘steps’, and there are indeed steps down to the beach (about 30 of them, so about 20 feet on the assumption of a US 7” riser). He says the view is over 7 miles, which is consistent with Google Maps.


It is hard to fault this. As he points out, about 20 feet should be hidden. My only doubt is whether she is actually at the water’s edge, given that the light first appears when she is at the top of the steps.


Refraction would not explain why so little was hidden. Any ideas? He claims that anyone can repeat the experiment.

add: coordinates given from another version of the same video locations (says the Youtuber):
Lat: 40.31524628, Lng: -111.76501156
Lat: 40.33612475, Lng: -111.90493711
 
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deirdre

Senior Member.
Refraction would not explain why so little was hidden

what do you mean by this?

you need to count her height too. so if she is holding a heavy flashlight thing around waist level, we can estimate her height to be about 3 feet.

using his numbers (plus 3 feet for her):

2.PNG


so refraction can certainly cover 11 feet. it can cover 20 feet. it can cover more. :)

But... the video is odd. i only watched until 4:47.
First he can't find her even though she is supposedly shining the light at him and he is looking in the same place.

THEN, 3:35 he tells her to turn it off and i can still see a dancing light (although dimmer).
THEN at 4:47 when he tells her to turn it on, i see a reflection of the light above her. this repeats several times.
33.PNG


so.. while refraction could certainly account for seeing the light. i have to wonder if turning it off was too much of a pain so she was just aiming it downwards, then upwards :) which.. is what i would do if my husband said he couldn't see me.


But again I'm not watching 30 minutes of blurry, low volumn video. so if you would like to screengrab specific timestamps later int he video where all these weird anomalies aren't happening, that would be great.
 
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Mick West

Administrator
Staff member
Refraction would not explain why so little was hidden.
Sure it could, you can get a huge amount of refraction near the horizon when viewer and/or light are close to the water. Standard refraction does not apply there.
 

edby

Member
Right at the beginning he says the distance is 7.56 miles (which I checked), that camera height is 2.54 feet (which I can’t check). He claims the hidden amount is 20.98 feet, which agrees with the MB curve calculator. At 13:00 she claims the flashlight is next to the water.


According to MB, the refracted hidden is 16.99, which would not explain why the flashlight is close to water level yet still visible.


Someone called Jon McIntyre comments:


Even taking into account maximum refraction this observation is IMPOSSIBLE on the globe we're told that we live on. The standard formula for atmospheric refraction used by surveyors and electrical engineers in the field of radio frequency propagation is 14% of the curvature of the earth. What this means is that refraction can offset 14% of the effect of curvature. As an example, if a distant object is declined out of view by 100 ft. due to the curvature of the earth then refraction would raise it back up 14 ft.


Filming over the moisture dense layer above a lake may cause refraction amplitude to increase possible as high as 20% of the curvature of the earth or possibly even spike briefly up to 25%. In relation to the test being done here refraction could be lifting the spotlight back up 4 or maybe even as much as 5 feet. But the light should be lost behind 21 feet of curvature. Factoring in for maximum refraction the spotlight should still ABSOLUTELY be behind 15 to 16 feet of curvature and yet it is CLEARLY VISIBLE. This is not possible on the globe we have been told we live on with the curvature and light refraction amplitude that is accepted mainstream science. It's simply not possible.


Science has no explanation for this observation. Refraction simply does not bend light enough to explain being able to record a spotlight on the water's surface from a camera 7.5 miles away located 2.54 feet above the water's surface. On the globe this is impossible. That is a fact. If anyone seeks to argue this point please provide the literature which discusses an atmospheric refraction coefficient of .90 or 90% of the curve of the earth because this is what would be necessary for this observation to happen on the globe. And guess what? It does not exist. Look as far and wide as you can. You will not find it because it does not exist.

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Trailblazer

Moderator
Staff member
The amount of shimmering and heat haze in that video makes accurate measurement impossible. Clearly there is lots of refraction and distortion going on there. That's why curvature observations are much better done using larger, more distant objects, like islands or mountains. That way, the distortion close to the surface is far less important. Why try messing around looking for 20 hidden feet when you can go up a mountain and measure thousands of hidden feet?
 

edby

Member
The problem is (and here you only have to read the comments on that video) this kind of thing is extremely persuasive to many people.
 

Mick West

Administrator
Staff member
According to MB, the refracted hidden is 16.99, which would not explain why the flashlight is close to water level yet still visible.
Again, standard refraction does not apply in this situation. Standard refraction assumes a constant lapse rate from 18°C at the water. Even variants on standard refraction using different numbers are almost entirely irrelevant, as the refraction skimming the horizon is highly non-liner.

Have you looked at my refraction simulation?
https://www.metabunk.org/mirage/

Try changing the target distance to 7.5 miles,
Metabunk 2018-07-13 07-29-56.jpg
And change the image file to: toronto-cn-tower-geom.png

The flat earth view is:
Metabunk 2018-07-13 07-43-47.jpg

The magenta (pink) line is the bottom pixels of the image. Without refraction the view from 7.5 miles away at 2 feet looks like:

Metabunk 2018-07-13 07-45-45.jpg

With standard refraction
Metabunk 2018-07-13 07-54-20.jpg
it looks like:
Metabunk 2018-07-13 07-47-15.jpg

i.e everything raised up a bit

With a very very small temperature inversion (i.e. water colder than air),
Metabunk 2018-07-13 07-53-48.jpg
we get:
Metabunk 2018-07-13 07-49-41.jpg

Notice the curve on the red/green triangle.

A larger temperature inversion of about 1°C
Metabunk 2018-07-13 07-53-07.jpg
gives:
Metabunk 2018-07-13 07-51-31.jpg

Note the significant compression near the horizon. Because it's compressed, it actually looks like more is hidden.

A temperature inversion bends light downwards. It does not take much to bend it it around the curve of the Earth
 

Mick West

Administrator
Staff member
Here's a 100m (328ft) high view, that you can use in the refraction simulator:
https://www.metabunk.org/sk/Utah_Lake_100m.jpg


I added a diagonal red line near the shoreline, with a horizontal single pixel line at the bottom
Metabunk 2018-07-13 10-58-47.jpg

Use it like this:
Metabunk 2018-07-13 11-02-55.jpg
(click on RELOAD after setting up).

Then a moderate temperature gradient makes the shoreline visible
Metabunk 2018-07-13 11-04-09.jpg
Notice the compression of the 45° line. Also that you can see the bottom row of pixels - i.e. the waterline.
Metabunk 2018-07-13 11-04-48.jpg
Why is this? Look at the ray paths seen from the side:
Metabunk 2018-07-13 11-07-14.jpg
The vertical scale is exaggerated by a factor of 1000. The orange line represents the target image. The green is the earth (really the ocean, I should make it blue). The grey is the air. The whiter it is, the higher the refractive index. The dark band is the hotter air, which is less dense. So light curves away from it. The lower rays are all curved down until they hit the ocean surface, which makes a lot more of the surface visible, and means that nothing of the 7.5 mile distant beach is hidden, even from 2 feet - it's just compressed.

So if the water is colder than the air, then it's actually quite difficult to not have a line of sight to the distant shore -even if it's very compressed.
 
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edby

Member
It does not take much to bend it it around the curve of the Earth
But this is exactly the opposite of what the expert commenter is saying

Even taking into account maximum refraction this observation is IMPOSSIBLE on the globe we're told that we live on. The standard formula for atmospheric refraction used by surveyors and electrical engineers in the field of radio frequency propagation is 14% of the curvature of the earth. What this means is that refraction can offset 14% of the effect of curvature. As an example, if a distant object is declined out of view by 100 ft. due to the curvature of the earth then refraction would raise it back up 14 ft.


https://www.youtube.com/watch?v=0V-tkCzfsT0&lc=UginDkdn0XPeJHgCoAEC
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Mick West

Administrator
Staff member
But this is exactly the opposite of what the expert commenter is saying
Expert commenter? I'm an expert commenter too, so how are you going to determine who is correct?

Do you understand the ray diagram from above? The last image in my post before yours, and the text that follows it?
 

Mick West

Administrator
Staff member
And if you want an actual expert,
https://aty.sdsu.edu/explain/atmos_refr/horizon.html

Unfortunately, the refraction varies considerably from day to day, and from one place to another. It is particularly variable over water: because of the high heat capacity of water, the air is nearly always at a different temperature from that of the water, so there is a thermal boundary layer, in which the temperature gradient is far from uniform.

Worse yet, these temperature contrasts are particularly marked near shore, where the large diurnal temperature swings over the land can produce really large thermal effects over the water, if there is an offshore breeze. This is particularly bad news for anyone standing on the shore and wondering how far out to sea a ship or island might be visible.

It gets worse. While the dip of the horizon depends only on an average temperature gradient, and so can be found from just the temperatures at the sea surface and at the eye, the distance to the horizon depends on the reciprocal of the mean reciprocal of the temperature gradient. But the structure of thermal boundary layers guarantees that there will be large variations in the gradient, even in height intervals of a few meters. This means that on two different days with the same temperatures at the eye and the water surface (and, consequently, the same dip), the distance to the horizon can be very different.

In conditions that produce superior mirages, there are inversion layers in which the ray curvature exceeds that of the Earth. Then, in principle, you can see infinitely far — there really is no horizon.

Of course, we all know that visibility is limited by the clarity or haziness of the air. And the duct that (in principle) might allow you to see around the whole Earth doesn't really extend that far; it typically exists for some limited region, perhaps a few tens or a few hundreds of kilometers.

So the nice-looking formulae for calculating “the distance to the horizon” are really only rough approximations to the truth. You can consider them accurate to a few per cent, most of the time. But, occasionally, they will be wildly off, particularly if mirages are visible. Then it's common to see much farther than usual — a condition known as looming.
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[emphasis mine]
 

deirdre

Senior Member.
But this is exactly the opposite of what the expert commenter is saying

Even taking into account maximum refraction this observation is IMPOSSIBLE on the globe we're told that we live on. The standard formula for atmospheric refraction used by surveyors and electrical engineers in the field of radio frequency propagation is 14% of the curvature of the earth. What this means is that refraction can offset 14% of the effect of curvature. As an example, if a distant object is declined out of view by 100 ft. due to the curvature of the earth then refraction would raise it back up 14 ft.


https://www.youtube.com/watch?v=0V-tkCzfsT0&lc=UginDkdn0XPeJHgCoAEC
Content from External Source

here is the link your 'expert' quotes. Please pay attention to Metabunk's posting guidelines, if you see that he gave a source for his claim then give us the source.


Refraction is largely a function of atmospheric pressure and temperature gradients, which may cause the bending to be up or down by extremely variable amounts.
..........

The atmosphere refracts the horizontal line of sight downward, making the level rod reading smaller. The typical effect of refraction is equal to about 14% of the effect of earth curvature.

......

https://www.aboutcivil.org/curvature-and-refraction.html
Content from External Source
 

edby

Member
Expert commenter? I'm an expert commenter too, so how are you going to determine who is correct?

Do you understand the ray diagram from above? The last image in my post before yours, and the text that follows it?
Precisely, the question is how to determine who is correct. This is a classic FE argument. I did understand the ray diagram, and thanks for that. However, the FE reply would be that RE supporters always wheel out 'refraction' to avoid the problem that experiments made at a low level seem to confirm the FE hypothesis.

(I am simply repeating arguments I have heard before, which I find entirely unconvincing, but others do not).
 

deirdre

Senior Member.
However, the FE reply would be that RE supporters always wheel out 'refraction' to avoid the problem that experiments made at a low level seem to confirm the FE hypothesis.

so what? FEers always wheel out a made up, undocumented 'the reason 12,000 feet of that mountain is hidden, is because perspective makes the bottoms of things disappear'.

Your expert quotes
surveyors and electrical engineers in the field of radio frequency propagation
. But surveyors and engineers in the field of radio waves all know the Earth is round(ish). Things that make you go.. hmmmm.
 

DavidB66

Active Member
The FE claim is that no plausible amount of refraction can possibly account for a given amount of 'looming' at a given distance. I think the best answer to this would be with time-lapse video footage showing the allegedly implausible amount of refraction actually happening. For example this well-known video by Joshua Nowicki shows the shoreline of Chicago filmed across Lake Michigan in the course of a day, and it is evident that the tops of the buildings appear to move up and down by large amounts (probably 100 feet or more, but I haven't checked) over that period. Short of arguing that the footage is faked or manipulated in some way, this apparent movement can only be accounted for by atmospheric refraction, so any claim that this is impossible would be refuted by the facts. Source: https://www.youtube.com/watch?v=FTFEu-Tod7s


I don't know of any evidence directly applicable to the Utah lake case, but the principle is the same.
 

edby

Member
so what? FEers always wheel out a made up, undocumented 'the reason 12,000 feet of that mountain is hidden, is because perspective makes the bottoms of things disappear'.
They do indeed. The question is how to debunk these claims.

I think very long distance views of e.g. Toronto/Chicago, are a much stronger case for RE.
 

Mick West

Administrator
Staff member
Precisely, the question is how to determine who is correct.
It seems trivially simple here. One person claims that science says refraction cannot account for it because refraction can only do 14%. Yet their own reference contradicts them.

You seem to be really overthinking it. Why didn't you just look up the answer and point out to people that science says refraction varies a lot?
 

edby

Member
It seems trivially simple here. One person claims that science says refraction cannot account for it because refraction can only do 14%. Yet their own reference contradicts them.
Sorry, which was 'their own reference'? I checked what they said, and can't find a citation or reference.
 

Rory

Senior Member.
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Mick West

Administrator
Staff member
The 'expert commenter' Jon McIntyre has had a few little visits to Metabunk: most notably in a long thread where he demonstrated he didn't understand what refraction was:
I think he understood it's the bending of light, just not much beyond that.

I'd guess that most of us don't actually understand what refraction "is", as it's some kind of quantum wave/particle duality thing. Photons striving to fulfill Fermats principle for some reason known only to them. Snell's law, which only applies to abstract interfaces that don't exist. The unchanging speed of light slowing down then speeding up.

I was in a pub with a group of people including Leonard Tramiel recently, and the topic of "Crazy Rainbow Lady" came up, people laughed. As you know I don't like people to be mocked, so I mentioned that most people don't really understand why rainbows form. Leonard said "there's probably not more than two people at this table who understand rainbows". I thought he included me in that two. But then he said "I don't think you do", and proceeded to give a brief description of the complexity," which made me realize I didn't really. I just knew the basics of total internal reflection in raindrops, and light splitting like in a prism, and something about a critical angle.

We all have limits to our knowledge, and a significant challenge in explaining thing to people is in understand both your own limits and your friend's limits, and trying to expand the latter without overstepping the former.
 

deirdre

Senior Member.
I'd guess that most of us don't actually understand what refraction "is",
I think he, and most laymen, understand refraction bends light. we just don't understand 'by how much' is possible. Recently someone (maybe @Rory ?) linked a video debunker that had a link in the description to a calculator that you could 'slide' the refraction amount higher or lower and this showed the building going up and down. a little marker let you know as you slide the slider if that was considered "normal" or "moderate" or "severe" refraction. But now I cant find the link.

a chart would be helpful, for ex. "@30 miles standard refraction would move a building up x feet, moderate refraction would move that building up [range of] x feet, severe refraction could move that building up [range of]x feet".

I just bookmarked this post for ex https://www.metabunk.org/posts/209834/ that shows this pic, so I'm assuming .6 is not 'severe' refraction because the chances these guys caught a day that exhibited severe refraction is unlikely. yes?
upload_2017-8-23_17-14-31.png
 

Mick West

Administrator
Staff member
I think he, and most laymen, understand refraction bends light. we just don't understand 'by how much' is possible. Recently someone (maybe @Rory ?) linked a video debunker that had a link in the description to a calculator that you could 'slide' the refraction amount higher or lower and this showed the building going up and down. a little marker let you know as you slide the slider if that was considered "normal" or "moderate" or "severe" refraction. But now I cant find the link.
Maybe Bislings'
http://walter.bislins.ch/blog/index...h%3A+Wie+stark+ist+die+Kr%FCmmung+der+Erde%3F

The problem with this, and what I was trying to address with my mirage simulator (which I think I'll rename a "atmospheric refraction simulator") , is that you can't describe refraction with a single number. It's a lots of numbers, bigger numbers near the horizon, and smaller numbers higher up, with some variation in between, and the whole thing varies through the day.
 

deirdre

Senior Member.
yea that was it.

The problem with this, and what I was trying to address with my mirage simulator (which I think I'll rename a "atmospheric refraction simulator") , is that you can't describe refraction with a single number
oh. he is using the same formula for the base of the object as the top of the object?

I can't personally comprehend your simulator at all. I don't know what I'm looking at or what that graph on the left shows. :) Did you do a video tutorial?
 

Mick West

Administrator
Staff member
I can't personally comprehend your simulator at all. I don't know what I'm looking at or what that graph on the left shows. :)

That's what I was worried about :)

The graph on the left shows the air temperature at altitudes for the first 100 feet. The dots and green lines are just handles you can draw around to change the curve. I need to make it simpler. The problem is the whole thing is representing something that's a little hard to understand, and hence explain.

Did you do a video tutorial?
I will, but I need to improve it all a bit first.
 

deirdre

Senior Member.
The problem is the whole thing is representing something that's a little hard to understand, and hence explain.
hmm. maybe I read it on Khan Academy. I don't think its super hard to understand (unless I'm missing something), according to (I believe it was) Khan the light (they use a race car analogy) speeds up and slows down when it hits different humidities. and every time it slows down it turns one way and every time it speeds up it turns the other way because of puddles and wheel drag or something :) it's been a while since I watched that refraction vid.
 

Mick West

Administrator
Staff member
hmm. maybe I read it on Khan Academy. I don't think its super hard to understand (unless I'm missing something), according to (I believe it was) Khan the light (they use a race car analogy) speeds up and slows down when it hits different humidities. and every time it slows down it turns one way and every time it speeds up it turns the other way because of puddles and wheel drag or something :) it's been a while since I watched that refraction vid.
That's not a terrible analogy, and it's essentially how my simulator works. Here's the math:
Metabunk 2018-07-15 13-32-40.jpg

WAIT, ignore all that and just look at the diagram:
Metabunk 2018-07-15 13-33-27.jpg

What that's asking is if a photon of light is at point P, traveling in direction u-> then how much will it turn by the time it gets to P'?
Metabunk 2018-07-15 13-37-08.jpg
With the car analogy there's the left side going along the A line, and the right side going along the B line. B region is denser than A, so the B side is slowed down, so the car turns towards B

<complex>You can do the math fairly simply with and actual A and B line. The math in my diagram proves that you can also do it just with the gradient at a point, the math creates an A and B using the the gradient g one unit away along v (perpendicular to u), this is then solved for u' eliminating the perpendicular offset</complex>
 

Nth

Member
I'm personally inclined to agree with the refraction explanation. As discussed both here and over on the split-off thread RE: the Chicago skyline, what you can see over water varies massively (also, Jenna Fredo's Toronto skyline footage comes to mind). And not to disparage the video maker in any way, but the observed conditions are exceptionally blurry, making it a bit hard to tell exactly what is being seen. A smear of light above the horizon isn't exactly conclusive evidence.

Anyway, it could just be compressed up to the horizon by refraction, a la Mick's simulator. For those who are familiar, a guy going by the handle Mathias kp (on YouTube and Flickr) has done some great earth curvature observations on the Turning Torso Tower (https://en.wikipedia.org/wiki/Turning_Torso). He's built up a properly scaled 3-D model, and put this image together showing how the building's floors compress upwards near the horizon, particularly at low elevations relative to sea level.


One may also notice that he has calculated the refraction coefficients necessary for the observed obscured portion to match earth's stated dimensions from a purely geometric perspective. The highest value is .38 (Special!), much higher than what is typically taken as the surveyor's standard. Then again, given how clear it is that stuff is being compressed upwards, I would say that refraction is absolutely appropriate to invoke here, and arguably in the OP example.

The other notable thing is that the top parts of the tower are far less (if at all) distorted, probably because light from them is not coming through weird surface conditions, i.e. the same issue with the Bedford Level experiment back in the day. It'd be interesting to see this observation repeated in the A.R. Wallace style, incidentally something that Mathias has performed elsewhere.
 

Mick West

Administrator
Staff member
Anyway, it could just be compressed up to the horizon by refraction, a la Mick's simulator. For those who are familiar, a guy going by the handle Mathias kp (on YouTube and Flickr) has done some great earth curvature observations on the Turning Torso Tower (https://en.wikipedia.org/wiki/Turning_Torso).

Handy! Having a diagonal line of known slope is the best way of seeing the compression.
Here's a waterline to rooftop roughly parallel projection of that view:


You can plug it into the simulator.
https://www.metabunk.org/sk/Metabunk_2018-07-16_19-12-00.jpg

It's 623 feet, plus 12 feet for the road above water level, so 635 feet.

Settings:
Metabunk 2018-07-16 19-19-05.jpg

Metabunk 2018-07-16 19-19-24.jpg

Look at the compression of the lower floors.

You could probably match up the observations. To a degree.
 

Z.W. Wolf

Senior Member.
Going back to the video in question, it's important to understand what a duct is.

My emphasis.
https://aty.sdsu.edu/~aty/explain/atmos_refr/duct.html

A duct is an atmospheric structure that traps rays within a few minutes of arc of the astronomical horizon, so that they cannot escape from the atmosphere, but are periodically bent back down, so as to follow the curvature of the Earth. The bending is produced by a steep thermal inversion. The inversion can do this if its lapse rate is more negative than a critical value (near −0.11°/m, for typical conditions).
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Light rays can be refracted many times. It's an additive process. In this illustration the ray marked 0
is refracted twice.




With no knowledge of the phenomenon of atmospheric ducting, any talk of what refraction can or can't do is useless.



As a side note, an FE Believer is going to have a hard time with the concept of atmospheric ducting, because it is doctrine that there can be no difference between the astronomical horizon (so called eye level) and the true horizon.

 
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Mick West

Administrator
Staff member
Most of what we see probably isn't ducting - ducting creates a high distorted image with multiple inversion especially over long distances.

What we seem to be seeing is a simpler looming, with compression the closer you get to the horizon. The rays are simply bend without crossing:
 
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