Measuring the Curvature of the Horizon with a Level

The camera is tilted.

My tripod has a level bubble, it was not my intention to measure the curvature of the horizon that day, but the hidden of the lighthouse, so I didn't checked the bubble. I'll make sure to show you guys that the camera is levelled before I get back there.
 
My tripod has a level bubble, it was not my intention to measure the curvature of the horizon that day, but the hidden of the lighthouse, so I didn't checked the bubble. I'll make sure to show you guys that the camera is levelled before I get back there.

You were on the beach, so no curve would be visible. From that altitude above sea level the horizon is both very close, and essentially flat. And it's never tilting from one side of an image to the other - it only curves up in the middle. Hence your photos are not level.
 
What are you basing the horizontal (yellow) line on? It looks more to me as if the camera is just a bit tilted.
Weirdly, zooming in and overlaying a grid appears to show that the lighthouse is pretty much bang on vertical. And rotating the image so that the horizon is horizontal in the frame puts the lighthouse at a noticeable slant.
 
Weirdly, zooming in and overlaying a grid appears to show that the lighthouse is pretty much bang on vertical. And rotating the image so that the horizon is horizontal in the frame puts the lighthouse at a noticeable slant.
No it doesn't.

(Visual interpretation without pictures can be refuted without pictures)



There's a slight lean illusion there due to the lighting. The left side is illuminated, so less visible.
 
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Double weird. When I first looked it really appeared slanted; but I rechecked on my Slant Detector 5000 (TM) and it's not there now.

Must be a bit o' that ol' Mandela Effect. ;)
 
Had another go at photographing the curve of the horizon today, with this setup from about 300 feet above sea level:

unaltered.jpg

The results, to quote Professor Denzil Dexter, were disappointing:

above centre.jpg
1. Above centre of frame

slightly above centre.jpg
2. Very slightly above centre of frame

below centre.jpg
3. Below centre of frame

Disappointing, but some interesting distortions.
 
As has been mentioned in this thread and elsewhere, it is often said that from the normal cruising height of a commercial airliner - between 30,000 and 36,000 feet - the horizon is usually too obscure for its curvature to be definitely visible. I was therefore interested to see a 'flat earth' video which appears to get round this problem by using an infrared filter (as I think was also suggested somewhere in this thread). The video is here:
Source: https://www.youtube.com/watch?v=1s6n7j-vC2s&t=566s
In case the link is not valid, the uploader is JTolan Media1, and the title is "Infrared Horizon from 33,000 feet, Astounding Flat Earth Phenomena!"
What is really astounding is that despite the claims of the uploader, and the chorus of adulation from his flat-earth subscribers, the infra-red footage actually shows pretty clear curvature throughout, except when the camera is zoomed in to a narrow field of view of the horizon. It is even very noticeable in the opening 'title' shot of the video. Of course, the curvature is small, but it looks consistent with what should be expected from a height of 11,000 meters in Walter Bislin's simulator. (It would be interesting for someone with the technical ability to overlay the two for comparison.)
It is a puzzling psychological phenomenon when large numbers of people apparently literally cannot see evidence that conflicts with their beliefs. I can only surmise that they have grossly exaggerated ideas of the amount of curvature that (according to 'globers') they ought to see, so that when they do not see that exaggerated amount, they fail to observe the smaller amount of curvature that is actually there in front of their eyes.
 
I suggested that it would be interesting to overlay a shot of the horizon in JTolan's video with the expected curvature from Walter Bislin's simulator. I don't have the tech ability to do this, but out of interest I did the following.
First I took a screencap from the video. Using the basic Windows picture viewer tools I inserted a green dot on the horizon at the left, then another one at the centre. (I used green because it seemed to give the best contrast.) These two dots determine a straight line. If the horizon were flat, the straight line would continue along the horizon. But in fact it rises above the horizon, because the horizon is curved. I inserted two green dots on the right, on the same straight line, which clearly rise above it.
I then did the same with Walter Bislin's simulator, using a height of 10,000 meters and a field of view of 70 degrees. This gave a good match with the shot from the video.
The two images are here, if this works (just noticed one of them shows some bookmarks. Fortunately nothing very incriminating, so I will leave them) :

Screenshot (595)_LI.jpg Screenshot (596)_LI.jpg
 
I expressed the hope that someone would overlay the horizon in the video onto the predictions of Walter Bislin's curvature simulator for the same height, as I didn't have the ability to do that myself. My hope has been more than satisfied, because Walter Bislin himself has done it! In his page here the simulation for a viewing height of 33,000 feet (10,058 meters) and a field of view of 75 degrees is superimposed on the same shot from J Tolan's video: http://walter.bislins.ch/bloge/index.asp?page=Curvature+App:+Simulation+of+Globe-Earth+and+Flat-Earth&state=-49-1Click--into--image--left--to--right--to--change--opacity!-210058.4-128.149668-10.44-10.05-51-9-9-8~0.0343-10-1262.00752-1~50.3796-634-1Infrared_Horizon_from_33000_ft.jpg-10.77257785-179-1Credit:--<a--href="https://www.youtube.com/watch?v=1s6n7j~vC2s">JTolan--Media1</a>-1#App

The result is a near-perfect match.

[compare]
download (3).png download (2).png
[/compare].
 
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I came across this French video, which at first sight is one of the best general flat-earth-debunks I have seen. There are a lot of good things in it, but especially relevant to this thread are the last 5 minutes or so, where the narrator discusses the curve of the horizon, and uses the lateral compression method to reveal curvature in several photos. A lot of the video is self-explanatory from the graphics, without any knowledge of French. For those with a little French, it may help to slow the video down to .75 x normal speed. The auto-generated French subtitles may be also of some use, but are not to be trusted: at one point they show the narrator talking about zombies, when he is really talking about les ombres (shadows)!


Source: https://www.youtube.com/watch?v=5ZdWRC57ems
 
There are a lot of good things in it, but especially relevant to this thread are the last 5 minutes or so, where the narrator discusses the curve of the horizon, and uses the lateral compression method to reveal curvature in several photos.

Those could have been really nice. But without the straight edge, how can we be sure the curve isn't being caused by barrel distortion?
 
I can often "see" the horizon while at 35k feet on a commercial flight....depends on the weather and clarity. Sometimes it's blurred or obscured by haze.

I still like the simple method TWCobra mentioned in THIS POST.
......a partially-filled water bottle, held sideways, so it's water is "level".
It does not match with the horizon...meaning, the earth curves down, relative to your leveled position.
It's a simple test that anyone flying can do.
In fact they hand-out the leveling tool to anyone.
(water bottles) :p

Bottle 3.jpeg
 
I can often "see" the horizon while at 35k feet on a commercial flight....depends on the weather and clarity. Sometimes it's blurred or obscured by haze.

I still like the simple method TWCobra mentioned in THIS POST.
......a partially-filled water bottle, held sideways, so it's water is "level".
It does not match with the horizon...meaning, the earth curves down, relative to your leveled position.
It's a simple test that anyone flying can do.
You could also do it with one of the plastic cups. Just ask for no ice.

However that’s the dip, not a visible curve. Of course it’s a consequence of the curve, but a. It OT for this thread
 
Wolfie6020 is a well-known flat earth debunker on YouTube and Qantas pilot. Here's a still from a video he posted flying at 46,000 feet:

Screenshot (2).png
Source: www.youtube.com/watch?v=9DDwx18JT9Y

It seems almost academic to apply the compression technique to something that shows the curve of the earth so clearly. But since there are some straight lines in there I figured, why not?

wolfie window shot.jpg

I'd say that's pretty clear that the curve isn't being caused by the camera or the window.
 
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Two and a half years and still no convincing shot of the curve of the horizon? :(

I'm in a great spot to do it but I've never owned a good enough camera.
 
Two and a half years and still no convincing shot of the curve of the horizon? :(

I'm in a great spot to do it but I've never owned a good enough camera.

Isn't Critical Think's video a convincing shot of the curve? I don't see how you could look down opposite ends of a tube and see the horizon at both ends if it wasn't curved
 
Indeed. Lots of convincing proofs of the curve of the horizon - but what I mean is the ol' straight edge 'n' compression technique, which is kind of lovely. :)
 
I posted a video on youtube briefly summarising how to take shots of the curvature of the horizon, and hopefully it will inspire some people to do this. A few responses, however, have shown that some don't think it's possible, given that the general understanding is that the horizon is 'flat'.

As Mick and others have explained, that's true - and yet, there is also a curve. The simplest way I can think of to explain it is to imagine a tiny being hovering above a coin. The coin is a 'flat disk', yet the edge of the coin is curved, and can be photographed.

4e coin.jpg
Compressing images from Google Earth provides theoretical support for this, as does Walter Bislin's round earth/flat earth simulator:

upload_2019-1-19_10-42-21.png

This shows a flat earth horizon on the left and globe earth horizon on the right, for a viewer at an elevation of 656 feet with a field of view of 60°.

Here's the vertically stretched flat earth horizon:

bislins flat.jpg
And here's the one for the globe:

bislins globe.jpg

Due to the resolution of the screenshot, it's not as smooth a curve as we would expect from a good quality shot of the actual horizon, but it's clearly there.
 
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Took some photos today with two straight edges framed around the horizon. Elevation was about 575 feet above sea level. Here's a small selection:

IMG_1968.jpg
IMG_1989.jpg

The curve matches both the simulated curve from Walter Bislin's website and the expected pixel difference between the centre of the curve and the edge (~4).
 
This is the original of the first of those images, unaltered, if anybody wants to check it:

IMG_1968.JPG
 
I'd be interested to know what people make of this video:


Source: https://www.youtube.com/watch?v=CJhLuQ9rvyQ&t=310s

This shows a guy maybe a few hundred feet above sea level looking out to the horizon. He has a thin wire or string stretched between two screws for his straight line, and his pan across appears to reveal the curve of the horizon.

To me, though, it seems that there's way more curve than would be possible from that elevation, like so:

upload_2019-3-27_18-55-24.png

The dark straight line is his string/wire, with the horizon above it. At the left and right of his pan the two meet.

It looks like the horizon, but it can't be, right?
 
This shows a guy maybe a few hundred feet above sea level looking out to the horizon. He has a thin wire or string stretched between two screws for his straight line, and his pan across appears to reveal the curve of the horizon.
horizon drop is the same in any compass direction, panning can not amplify it in any way if the camera does not change height.
even taut wires or strings sag, which a "billiard queue" lengthwise shot along the wire would probably have revealed. With knowledge of the tension, the sag can be predicted.
What happens when you apply your compression technique, can you see the wire curving?
 
williamlynn said: I am looking for a formula to calculate the deflection(sag) in steel wire that is supported between two points with a known tension. I want to be able to calculate the deflection(sag) at any point along the wire.

Well it depends upon a lot of stuff, like the properties and temperature of the wire, and the span/sag ratio. The wire will take the shape of the catenary (hyperbolic functions, kind of tough to calculate manually), however, if the sag is relatively small in comparison to the span (say the sag is less than about 10 percent of the span), then the catenary curve is very closely approximated by a parbolic curve using the following equation: T = w l^2 / ( 8 d ) , where w is the weight of the wire per unit length, l is the horizontal span between supports, d is the sag, and T is the horizontal tension in the cable. So, given T, l, and w, you can easily calculate d (the sag at the low point of the curve). If it is less than 10% or so of l, you plot the parabolic curve, and can then get sags at various points using the parabolic properties of that curve. Temperature and live loadis will affect sag/tension values, but if you're just looking at one tension at a given temperature under the wire dead load weight only, you need not go further.

Reference https://www.physicsforums.com/threads/wire-sag-formula.220848/
Content from External Source
 
Yeah, you can calculate sag, if you know all the factors. But you don't, other than it's light wire under high tension, and a short span, so the result is going to be "not a lot".

Looking along it shows it looks straight.
Metabunk 2019-03-27 16-37-26.jpg
 
Can you the measure "sag" by using a small laser between the wire attached ends ?

In fact, it may be possible to correct the sag somewhat, by a few supports to the wire, using the laser to verify the corrections....... if needed.
 
What happens when you apply your compression technique, can you see the wire curving?

Doing that, it appears straight.

I've messaged the video maker - a very reasonable man - and he said "though I am only 70 metres above sea level [the 8 foot string is] being observed from only 2ft from the string so the angle of horizon I am looking at is about 120 degrees. That is a lot of horizon and my string is taught and straight."

By my math, 2 feet away from an 8 foot length would give him a 127° FOV, and using Walter Bislin's site, a horizon that would look like this:

curve.jpg
 
@Mick West Is there a way to use your curve calculator for this footage? Given that it's a panoramic shot, what would the 'image width' be set at?
 
Thank you, @Mick West and @Rory . I had been looking for a lengthwise shot in the video, but didn't notice this one. You both show the wire to be reasonably straight, and Rory shows there is a horizon curve to seen, via the simulator.

My understanding is that for the panoramic shot to display the horizon curvature, the photograph would have to have been taken in one piece, and this still then be zoomed and panned for the video.

On the other hand, if you take a panoramic shot of a taut wire by turning the camera, the wire ends will be more distant and thus closer to the horizon when you turn to look at them directly. If that thought is correct, then Rory should be able to do a turn-the-camera panorama of his horizonamatic from close up and observe the same effect.
 
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