Jesse3959
Member
I'm not sure if I posted this already but here's a measurement I made of the water surface horizon through a theodolite from about 56 feet above sea level: Source: https://youtu.be/IqAZuqSSmfw
Measuring the Curvature of the Horizon with a Level
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DavidB66
Senior Member
Flat Earth debunker Bobby Shafto, who does some great work, has successfully replicated Rory's 'horizonomatic' method from a height of nearly 500 meters and obtained a nice clear result. See his short YouTube video here:
Source: https://www.youtube.com/watch?v=KWUAuAvzxyc
Source: https://www.youtube.com/watch?v=KWUAuAvzxyc
henrikmorsing
Member
To take lens distortion in to account, make sure the horizon is exactly in the middel of the picture. Otherwise lens distortion is a possibility. Especially when using a wide angle lens. Also be aware of Adobe Lightroom has build in lens- and chromatic aberration correction for many cameras and this is actually changing the image slightly.
henrikmorsing
Member
Squeezing the picture, the horizon is clearly curved. Photo from sailing across the Atlantic Sea. Notice the cloud front going behind the horizon. I was standing on a sail boat approximately 3 meters above sea level. Squeezing the picture, the horizon is clearly curved.
Rory
Closed Account
Is the horizon in the exact centre of the image?
Better to have a straight line in there to compare to, like earlier posts in this thread.
That's pretty much a requirement, really. Even if you get it in the center, that does not guarantee that there is no distortion.
Pertti Niukkanen
Member
About three years ago I came across Rory's Youtube video where he demonstrated this compressing/stretching method. It must have been the same video as mentioned in Mod Edit above: "Here's a great explanation of this technique by @Rory".
Sadly "this video is no longer available because the account associated with this video has been closed". Have you really deleted all your FE debunking videos? What a loss!
I was very fond of this simple method and presented it in the Finnish FE debunking group. Now we know it in Finland as "Rory's method".
It was great fun to find the curve also in flat-earthers photos. Here is the cover photo of Eric Dubay's Facebook page. https://www.facebook.com/ericdubaz. The resolution is quite low and hardly allows great stretching. Yet I tried 10-, 20- and 40-times vertical stretch. The last picture is heavily blurred. To me at least "It's not flat folks".
Of course, this is not proof of anything. As Rory said, "It better have a straight line to compare it to." And Mick: "It's pretty much a requirement, really. Even if you get it in the middle, it doesn't guarantee there won't be distortion."
I guess we can still do "horizonspotting" with just the camera. Whenever you see a sharp horizon (even at low altitude), take a picture. Keep the horizon in the center (or perhaps a little below it to avoid the claims of barrel-distortion). Then use the Rory's method.
One photo hardly proves anything, but maybe dozens of them do. Perhaps this could be a statistical evidence or at least a hint to your flat-earther friend too.
Sadly "this video is no longer available because the account associated with this video has been closed". Have you really deleted all your FE debunking videos? What a loss!
I was very fond of this simple method and presented it in the Finnish FE debunking group. Now we know it in Finland as "Rory's method".
It was great fun to find the curve also in flat-earthers photos. Here is the cover photo of Eric Dubay's Facebook page. https://www.facebook.com/ericdubaz. The resolution is quite low and hardly allows great stretching. Yet I tried 10-, 20- and 40-times vertical stretch. The last picture is heavily blurred. To me at least "It's not flat folks".
Of course, this is not proof of anything. As Rory said, "It better have a straight line to compare it to." And Mick: "It's pretty much a requirement, really. Even if you get it in the middle, it doesn't guarantee there won't be distortion."
I guess we can still do "horizonspotting" with just the camera. Whenever you see a sharp horizon (even at low altitude), take a picture. Keep the horizon in the center (or perhaps a little below it to avoid the claims of barrel-distortion). Then use the Rory's method.
One photo hardly proves anything, but maybe dozens of them do. Perhaps this could be a statistical evidence or at least a hint to your flat-earther friend too.
Rory
Closed Account
I know! It is a little sad - especially this one and a few others.
Unfortunately I had another YouTube account (where I posted Tetris videos) and I received a copyright strike. Just as I was thinking of deleting it I received another copyright strike and they terminated that account and three others that were linked to it, including my main one with all the flat earth videos (as well as some popular guitar tutorials).
I did protest that they were supposed to send three copyright strikes and tried all other avenues but, alas, didn't get anywhere. It was basically like talking to a robot - but one programmed to impersonate a brickwall.
That's very cool though!
And I think I still most of the files I used to make my videos, available here:
Rory's flat earth file collection
Includes lots of lots of horizonomatic and eye level photos and videos from Ibiza.
Pertti Niukkanen
Member
Nearly two years ago Mr Sensible conducted a high altitude balloon experiment MAGE II. The goal was among other things to show the curvature of the earth at an altitude of 39000 m (same height as Felix Baumgartner stratosphere jump). The altitude was almost reached.
In his video "MAGE II - What shape is the Earth?" Mr Sensible explains how the curve can be detected despite the expected barrel distortion of the camera. There are two taut strings in front of the camera. They form the reference lines to compare the horizon curve with. See from time 32:24 on
Source: https://www.youtube.com/watch?v=dVOLj1je0lk&t=1944s
Below are screenshots of a situation where the horizon is clearly below the center of the image. In this case, barrel distortion only reduces the curvature of the horizon, but the curvature is still visible. But it's quite certain that even this is not enough for flat-earthers.
There may be doubts about the straightness of these strings (how tight they really are, etc.) Maybe a metal grid (perpendicular thin threads very precisely made) firmly attached in front of the camera would do better. So the barrel distortion (or some other distortion) could be instantly seen.
Walter Bislin has done an extensive analysis on the MAGE video: "M.A.G.E. - Mission Above Globe Earth, Image Analysis", http://walter.bislins.ch/bloge/index.asp?page=M.A.G.E.+-+Mission+Above+Globe+Earth,+Image+Analysis. That's where his calculator's photo-analytical abilities come into their own.
In his video "MAGE II - What shape is the Earth?" Mr Sensible explains how the curve can be detected despite the expected barrel distortion of the camera. There are two taut strings in front of the camera. They form the reference lines to compare the horizon curve with. See from time 32:24 on
Source: https://www.youtube.com/watch?v=dVOLj1je0lk&t=1944s
Below are screenshots of a situation where the horizon is clearly below the center of the image. In this case, barrel distortion only reduces the curvature of the horizon, but the curvature is still visible. But it's quite certain that even this is not enough for flat-earthers.
There may be doubts about the straightness of these strings (how tight they really are, etc.) Maybe a metal grid (perpendicular thin threads very precisely made) firmly attached in front of the camera would do better. So the barrel distortion (or some other distortion) could be instantly seen.
Walter Bislin has done an extensive analysis on the MAGE video: "M.A.G.E. - Mission Above Globe Earth, Image Analysis", http://walter.bislins.ch/bloge/index.asp?page=M.A.G.E.+-+Mission+Above+Globe+Earth,+Image+Analysis. That's where his calculator's photo-analytical abilities come into their own.
Rory
Closed Account
I'm sure those images do show the curve and probably the string thing works really well. Just one thing:
It's not always true that a line below mid-frame will curve convexly - some cameras actually have pincushion distortion rather than barrel distortion, so it's always good to know which:
https://en.wikipedia.org/wiki/Distortion_(optics)#Radial_distortion
One camera I was using (on a phone) actually had the third type, mustache distortion.
It's not always true that a line below mid-frame will curve convexly - some cameras actually have pincushion distortion rather than barrel distortion, so it's always good to know which:
https://en.wikipedia.org/wiki/Distortion_(optics)#Radial_distortion
One camera I was using (on a phone) actually had the third type, mustache distortion.
Pertti Niukkanen
Member
Mick West: "Of course, the horizon is curved on a disk earth too."
Rory: "Do we have the maths for this somewhere?"
I just made a kind of "Flat Earth Curve Calculator" with the Finnish version of Excel. In the English version, you have to make a couple of substitutions in the formulas:
PII() -> PI()
SQUARE ROOT -> SQRT
Note that in the Finnish version of Excel, the decimal separator is a comma.
The input data is:
R = the radius of the disk, put in the cell A2
h = the height of the observer above the center of the disk, put in the cell B2
α (deg) = the angle of view (horizontal FOV), put in the cell C2
As output data you get δ (deg), φ (deg) AC and s%. Let me explain.
δ (deg) is the most important result. It's called "Horizon Curve Angle" or "Left-Right Drop Angle". The derivation of angle δ is below. In Excel you must type the formula in the cell
D2:
=(ATAN(A2/B2)-ACOS(B2/(NELIÖJUURI((1-(SIN(C2*PII()/360))^2)*(A2^2+B2^2)))))*180/PII()
φ (deg) is "Horizon Dip Angle" (angle between eye level and the horizon). It's simply = 90° – β1. Type in cell
E2:
=90-ATAN(A2/B2)*180/PII()
AC is the distance of the horizon seen from point A. Type in cell
F2:
=NELIÖJUURI(A2^2+B2^2)
s% is the tricky one.
I don't know programming or computer graphics. Still, it would be fun to see a flat Earth horizon curve on the screen (like in Bislin's calculator). At a certain viewing angle, you see the horizon curve as an angular segment with a chord and a sagitta (bulge). Now the value s% tells how many percent of the chord the sagitta is on the screen.
These s% numbers are usually very small. For example, if Felix Baumgartner were 39 km above the FE North Pole, the number would be 0.026 (see the first image). So the sagitta is only 0.026 percent of the chord he sees at 60 degrees field of view. It's just a straight line.
If Felix could rise 4748 km height, he could see the curve at 90 degrees field of view like this:
The ratio s% = 5 gives three points. So I can fit an arc of a circle through these points. (There is only one such circle).
I'm not sure about my "graphical method". It hardly has any use. Fortunately, you can ignore it. Other values are not related to it. Also my formula might be wrong. Anyway, here it is. So type in the cell
G2:
=100/(2*TAN(C2*PII()/360)/TAN(D2*PII()/180))
----------------
So here is my Flat Earth Curve Calculator. I would be happy if someone tried it to see if it works.
Here, the observer is always in the center of the disc. So the situation is the same in every direction. Looking from some other place makes it harder. The Horizon Dip Angle will definitely vary depending on the direction. How the angle δ varies, I have no idea.
Rory: "Do we have the maths for this somewhere?"
I just made a kind of "Flat Earth Curve Calculator" with the Finnish version of Excel. In the English version, you have to make a couple of substitutions in the formulas:
PII() -> PI()
SQUARE ROOT -> SQRT
Note that in the Finnish version of Excel, the decimal separator is a comma.
The input data is:
R = the radius of the disk, put in the cell A2
h = the height of the observer above the center of the disk, put in the cell B2
α (deg) = the angle of view (horizontal FOV), put in the cell C2
As output data you get δ (deg), φ (deg) AC and s%. Let me explain.
δ (deg) is the most important result. It's called "Horizon Curve Angle" or "Left-Right Drop Angle". The derivation of angle δ is below. In Excel you must type the formula in the cell
D2:
=(ATAN(A2/B2)-ACOS(B2/(NELIÖJUURI((1-(SIN(C2*PII()/360))^2)*(A2^2+B2^2)))))*180/PII()
φ (deg) is "Horizon Dip Angle" (angle between eye level and the horizon). It's simply = 90° – β1. Type in cell
E2:
=90-ATAN(A2/B2)*180/PII()
AC is the distance of the horizon seen from point A. Type in cell
F2:
=NELIÖJUURI(A2^2+B2^2)
s% is the tricky one.
I don't know programming or computer graphics. Still, it would be fun to see a flat Earth horizon curve on the screen (like in Bislin's calculator). At a certain viewing angle, you see the horizon curve as an angular segment with a chord and a sagitta (bulge). Now the value s% tells how many percent of the chord the sagitta is on the screen.
These s% numbers are usually very small. For example, if Felix Baumgartner were 39 km above the FE North Pole, the number would be 0.026 (see the first image). So the sagitta is only 0.026 percent of the chord he sees at 60 degrees field of view. It's just a straight line.
If Felix could rise 4748 km height, he could see the curve at 90 degrees field of view like this:
The ratio s% = 5 gives three points. So I can fit an arc of a circle through these points. (There is only one such circle).
I'm not sure about my "graphical method". It hardly has any use. Fortunately, you can ignore it. Other values are not related to it. Also my formula might be wrong. Anyway, here it is. So type in the cell
G2:
=100/(2*TAN(C2*PII()/360)/TAN(D2*PII()/180))
----------------
So here is my Flat Earth Curve Calculator. I would be happy if someone tried it to see if it works.
Here, the observer is always in the center of the disc. So the situation is the same in every direction. Looking from some other place makes it harder. The Horizon Dip Angle will definitely vary depending on the direction. How the angle δ varies, I have no idea.
Pertti Niukkanen
Member
The idea of AE projection (azimuthal equidistant projection) is that the distances measured from a certain center point are preserved. The "semi-official" FE map is an AE map North Pole as the center point. So on the FE map, the distances measured from the North Pole are the same as on the globe map. In general, all distances measured along meridians are correct on the FE map.
Hardly anything else is correct. The areas of the Northern and Southern Hemispheres are of course the same in reality, but on the FE map the area of the southern "semi puck" is three times the area of the northern one. The FE map magnifies all areas and all distances (except those measured along meridians). Here are some examples of these "stretch factors":
– Earth's surface area: 2.46
– Northern Hemisphere area: 1.23
– Southern Hemisphere area: 3.70
– Length of the Equator: 1.57
– Distance in east-west direction at latitude 60°: 1.04
– Distance in east-west direction at latitude -60°: 5.19
– Distance from Perth to Sydney: 2.52
In Finland the FE map does not have very large distortions. In Helsinki (latitude 60), the east-west distance is only approx. 4% too long, further north even less. Distortions increase in the southern hemisphere. The most dramatic is the situation at the South Pole, where one point stretches into a circle about 125,700 km long.
Below left are FE and GE to scale. The diameter of the FE disk is 40000 km, which is also the circumference of the Earth.
Below right is a magnified screen shot from Bislin's calculator. There are flat Earth and Globe side by side seen from a very long distance (70,000 km). http://walter.bislins.ch/bloge/inde...53-9-9-9-1~0.0343-10-10.00373383-1~86.204-9-4.
The ratio in these comparisons is about the same. So we can conclude that the radius of the flat Earth is 20000 km in Bislin's calculator too.
Another useful calculator made by Bislin is "Creating Flight Plans for Flat Earth", http://walter.bislins.ch/bloge/index.asp?page=Creating+Flight+Plans+for+Flat+Earth. It is easy to compare FE and GE distances by providing the coordinates of the locations. You get the coordinates e.g. from Google Maps by right-clicking on a place. Then click on the coordinates in the context menu and they will be copied to the clipboard in decimal format.
Most flat-earthers admit the existence of research stations located on the coast of Antarctica. Such are Finland's Aboa (-73.048,-13.420) and USA's McMurdo (-77.842,166.688). The direct route between these stations happens to go very precisely through the South Pole.
The image on the left shows the distance between these stations on the globe map (green segment). It is about 3240 km. That's roughly the distance between Detroit, Michigan and Sacramento, California.
On the right is the same journey on FE map. The shortest route between the stations goes through the North Pole. The length of the trip will be approx. 36,790 km.
Yes, the journey from Aboa to McMurdo could be made via the North Pole on the globe also. The length of this trip would also be 36,790 km. This is because the journeys are made along meridians, so both maps give the same values.
As a joint project between Aboa and McMurdo, a research flight from one station to another could be carried out. Leading figures of the FE community would also be included, agreeing to all their demands (research equipment, food, etc.) A stopover could be made at the South Pole in Amundsen Scott station. There could the researchers spend a night watching the midnight sun. (The station does have accommodation for tourists.)
Could this journey show to the flat-earthers which map was followed on the trip? Was the distance 3200 km or 37000 km? Perhaps something could also be deduced from the landscapes below.
If the budget of the research stations is not enough for this public education project, one could ask some of the world's many billionaires as a sponsor. They might be interested in the publicity value of the trip, especially if their own press people were going along. "Flat Earth supporters at the South Pole admiring Midnight Sun" sounds a very attractive tabloid headline. 
Hardly anything else is correct. The areas of the Northern and Southern Hemispheres are of course the same in reality, but on the FE map the area of the southern "semi puck" is three times the area of the northern one. The FE map magnifies all areas and all distances (except those measured along meridians). Here are some examples of these "stretch factors":
– Earth's surface area: 2.46
– Northern Hemisphere area: 1.23
– Southern Hemisphere area: 3.70
– Length of the Equator: 1.57
– Distance in east-west direction at latitude 60°: 1.04
– Distance in east-west direction at latitude -60°: 5.19
– Distance from Perth to Sydney: 2.52
In Finland the FE map does not have very large distortions. In Helsinki (latitude 60), the east-west distance is only approx. 4% too long, further north even less. Distortions increase in the southern hemisphere. The most dramatic is the situation at the South Pole, where one point stretches into a circle about 125,700 km long.
Below left are FE and GE to scale. The diameter of the FE disk is 40000 km, which is also the circumference of the Earth.
Below right is a magnified screen shot from Bislin's calculator. There are flat Earth and Globe side by side seen from a very long distance (70,000 km). http://walter.bislins.ch/bloge/inde...53-9-9-9-1~0.0343-10-10.00373383-1~86.204-9-4.
The ratio in these comparisons is about the same. So we can conclude that the radius of the flat Earth is 20000 km in Bislin's calculator too.
Another useful calculator made by Bislin is "Creating Flight Plans for Flat Earth", http://walter.bislins.ch/bloge/index.asp?page=Creating+Flight+Plans+for+Flat+Earth. It is easy to compare FE and GE distances by providing the coordinates of the locations. You get the coordinates e.g. from Google Maps by right-clicking on a place. Then click on the coordinates in the context menu and they will be copied to the clipboard in decimal format.
Most flat-earthers admit the existence of research stations located on the coast of Antarctica. Such are Finland's Aboa (-73.048,-13.420) and USA's McMurdo (-77.842,166.688). The direct route between these stations happens to go very precisely through the South Pole.
The image on the left shows the distance between these stations on the globe map (green segment). It is about 3240 km. That's roughly the distance between Detroit, Michigan and Sacramento, California.
On the right is the same journey on FE map. The shortest route between the stations goes through the North Pole. The length of the trip will be approx. 36,790 km.
Yes, the journey from Aboa to McMurdo could be made via the North Pole on the globe also. The length of this trip would also be 36,790 km. This is because the journeys are made along meridians, so both maps give the same values.
As a joint project between Aboa and McMurdo, a research flight from one station to another could be carried out. Leading figures of the FE community would also be included, agreeing to all their demands (research equipment, food, etc.) A stopover could be made at the South Pole in Amundsen Scott station. There could the researchers spend a night watching the midnight sun. (The station does have accommodation for tourists.)
Could this journey show to the flat-earthers which map was followed on the trip? Was the distance 3200 km or 37000 km? Perhaps something could also be deduced from the landscapes below.
If the budget of the research stations is not enough for this public education project, one could ask some of the world's many billionaires as a sponsor. They might be interested in the publicity value of the trip, especially if their own press people were going along. "Flat Earth supporters at the South Pole admiring Midnight Sun" sounds a very attractive tabloid headline.
captancourgette
Active Member
Nice Diagram, I wonder how the flat earthers explain flying from perth of sydney takes half the time of flying from london to new york, even though on their map London -> new york is a shorter distance
Pertti Niukkanen
Member
You are right. Those posts are quite off-topic in this thread. Moreover I have more like this in my mind. So I made a new thread with a broader title "Comparing flat Earth and spherical Earth from a geometric point of view", https://www.metabunk.org/threads/co...l-earth-from-a-geometric-point-of-view.12591/
I copied these two posts into the new thread, but didn't delete them from here. The moderator may delete them if it is the right thing to do in this case. I hope all the geometrically oriented FE debunkers find this new thread.
henrikmorsing
Member
Yes, I should have the horizon in center, however I thought it was close enough to illustrate the curvature + the cloud front going over and behind.
Not sure I agree there will be distortion of a line going exactly through the center of the lens, however it can be difficult to hit center exactly without guidelines in the camera viewfinder. About distortion in the image center, that's not the case in any of the 3 types of distortions I know from photography: Barrel (Away from center), Pincushion (into center), Mustache (Away from center). If the lens is "normal" circular type and the camera sensor is centered, the distortion will be evenly distributed in both horizontal and vertical direction from the center. But just a tiny bit off center will have some distortion depending of the lens, camera and software used. Some cameras, especially mobile ones, has lens correction build in the software. So you can't trust the output image represent real world 100%. That's important if you use a photo to measure angels and distances. They will be different in the edges vs the center. However, the effect in most cases is minor (except fish eye, super wide angle, anamorphic etc.). When you import RAW images in Adobe Lightroom, some lens correction and aromatic abbreviation correction is added automatically. Adobe has build in correction profiles for many camera and lens brands.
Check the distortion of the specifik camera used for the picture: I tool a photo to check the specifik distortion of the exact same camera, settings and workflow used for the photo (Sony RX10m3, JPG, 24 mm, Import to Lightroom + added control lines in Photoshop). The image shows at slight "barrel" distortion. Tiny, but visible.
Conclusion: Distortion of this specifik picture is of the type "barrel". The horizontal lines UNDER the image center is distorted downwards, away from the center. So even when there is a slight distortion downwards in the center, the image shows a horizon curved upwards. The image shows a real curve.
Fun fact: Special "bellow" lenses used for product- and macro photography has lens that can be moved independently of the sensor. They can change the angle of the focal plane and change the distortion. This is why they are most used for studio product photography and macro.
Distortion types:
https://en.wikipedia.org/wiki/Distortion_(optics)
Adobe lens correction profiles:
https://helpx.adobe.com/uk/camera-raw/kb/supported-lenses.html
Adobe Lightroom setting:
Camera + Lens distortion check:
