Question about high resolution photos of the sea horizon

Greetings,

if you take a photo of the sea horizon under these conditions:

• >= 4k resolution
• sharp
• no fog or clouds overlapping
• no panorama mode or wide-angle / bulls-eye lense

you can draw at your PC a tangent or tendon, revealing the curvature of the horizon.

So my question would be: How do you explain this with the FET?

If you don't have pictures form your last holiday you can just take any image from google following the same requirements. Then you can use freeware like Gimp to draw the line. Or if the image is already well aligned, you can also just zoom in with your smartphone and then move the highest point of the horizon towards the upper edge of your screen and then move the image only horizontal. The horizon will be lower at the sides of the image.

Kind regards
Souleon

Greetings,

if you take a high resolution photo of the sea horizon, you can draw at your PC a straight line in parallel to the horizon,

Do you mean tangent?

Quote from: Souleon on May 01, 2019, 12:50:59 PM

Greetings,

if you take a high resolution photo of the sea horizon, you can draw at your PC a straight line in parallel to the horizon,

Do you mean tangent?

Yes tangent or tendon. Thanks, I edited it.

Greetings,

if you take a sharp high resolution photo (4k) of the sea horizon, you can draw at your PC a tangent or tendon, revealing the curvature of the horizon. If you don't have pictures form your last holiday you can just take any sharp >4k sea images from google, without fog or clouds overlapping. Then you can use freeware like Gimp to draw the line. Or if the image is already well aligned, you can also just zoom in with your smartphone and then move the highest point of the horizon towards the upper edge of your screen and then move the image only horizontal. The horizon will be lower at the sides of the image.
Souleon

You claim, "The horizon will be lower at the sides of the image" but show no examples, why?

So my question would be: How do you explain this with the FET?

Have you actually done this yourself?
If not I suggest that you do it first by taking your own photos from no more than 2 m above sea level and prove your case before asking, "How do you explain this with the FET?".

I can show you examples or make a video, if you want. But it will be much more convincing if you do it yourself, won't it?

Peace.

Forgot that I'm never supposed to post in this forum.

I can show you examples or make a video, if you want. But it will be much more convincing if you do it yourself, won't it?

Peace.

Photos I've taken look quite flat but you made the claim so over to you.

Yes they look flat, because earth is quite huge.

Ok please do the following (takes <1 min):

1. Go to e.g. https://images.wallpaperscraft.com/image/sea_horizon_ship_117671_3840x2160.jpg
2. Zoom in to maximum.
3. Move your section so that the highest point of the horizon (in the example a bit left to the ship) is at the upper edge of your browser window or screen.
4. Move your section only horizontal to the left and afterwards to the right.

What do you see?

Yes they look flat, because earth is quite huge.

Ok please do the following (takes <1 min):

1. Go to e.g. https://images.wallpaperscraft.com/image/sea_horizon_ship_117671_3840x2160.jpg
2. Zoom in to maximum.
3. Move your section so that the highest point of the horizon (in the example a bit left to the ship) is at the upper edge of your browser window or screen.
4. Move your section only horizontal to the left and afterwards to the right.

What do you see?

I see a Wallpaperscraft image of unknown origin taken from an unknown height above the ocean - that's what I see!

Remember that I asked:

Have you actually done this yourself?
If not I suggest that you do it first by taking your own photos from no more than 2 m above sea level and prove your case before asking, "How do you explain this with the FET?".

Remember that I asked:

Quote from: rabinoz on May 01, 2019, 02:07:53 PM

Have you actually done this yourself?
If not I suggest that you do it first by taking your own photos from no more than 2 m above sea level and prove your case before asking, "How do you explain this with the FET?".

Yes, of course. That was the first thing, I tested and it worked. Do you want to post a picture you made yourself? But it should meet the requirements, that I edited into the first post of this thread. I can do the drawing afterwards, if you trust me – or, if you don't, you can do the drawing by yourself.

Quote from: rabinoz on May 02, 2019, 01:06:33 AM

Remember that I asked:

Quote from: rabinoz on May 01, 2019, 02:07:53 PM

Have you actually done this yourself?
If not I suggest that you do it first by taking your own photos from no more than 2 m above sea level and prove your case before asking, "How do you explain this with the FET?".

Yes, of course. That was the first thing, I tested and it worked.

But that picture that you posted was taken from some unknown height - certainly not 2 m. It's your thread so it's up to you to present your evidence.

If I post a picture made by me, you can just say again that it is fake, which would not bring us any step further. Therefore, I ask you again: Do you want to post a picture made by you or someone you trust?

If I post a picture made by me, you can just say again that it is fake, which would not bring us any step further. Therefore, I ask you again: Do you want to post a picture made by you or someone you trust?

I'm still waiting for your effort and I'll compare it to photos that I have personally taken.

OP you have posted an assertion that clearly can't just be taken as axiomatic, and you've not provided evidence for it.

I'm moving this to debate.

OP, please back up your claim.

Yeah, I have to agree. Post your picture as an example to show this. Anyone who calls it a fake then has to post their own pictures.

Of course, at this point you could have everyone posting fake pictures.

I'd give up though, FEers aren't interested in evidence, if they were they wouldn't be FEers.

you can draw at your PC a tangent or tendon, revealing the curvature of the horizon.

What tangent are you drawing?
The curvature across the horizon in such cases is insignificant and what you are far more likely to see is distortion from the camera not being perfect.

1. Go to e.g. https://images.wallpaperscraft.com/image/sea_horizon_ship_117671_3840x2160.jpg
What do you see?

I see a ship which is roughly 87 px long. This ship is a fairly small one. I will be generous and say it is 15 m long.
This means each pixel is roughly 17 cm.
This means the entire 3840 px wide image is roughly 740 m accross.
That means to go from the centre to the edge (to calculate the expected drop) you have a distance of roughly 370 m.
Then using the simple formula to calculate drop:
h=d²/(2 R)
You get a drop of roughly 1 cm. This means from the boat, following the great circle for those roughly 370 m you would get ~1 cm of drop.
However that is the great circle. The horizon obscured that.
This means the drop expected in the photo is less.
1.27 cm is less than a pixel.
As such, the expected curvature of the horizon given a round Earth of radius 6371 km is less than a pixel (less than 0.1 pixels in fact) and thus should not be observed.

As such, I see an image with significant distortion, likely barbell distortion, but as I'm not a camera expert and am not sure on exactly what the different types of distortion refer to, feel free to correct me on just what type of distortion it is.

If you wish to disagree and claim it is actually the curvature of Earth, then being lazy and assuming we are seeing the great circle, then these ~2 pixels of drop equate to 0.34 m of drop and thus a radius of 20 km.
So are you claiming Earth is a ball with a radius of 20 km?

Also, going by your reasoning, this photo:
https://images.pexels.com/photos/7321/sea-water-ocean-horizon.jpg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260
clearly Earth must be concave with us living on the inside of a ball.

Dear JackBlack, thanks for your feed back. About your calculation with the ship: You assumed that the ship has the same distance as the horizon (from the camera's point of view). However, in fact the ship is still far away from the horizon (look again closely).

Also, going by your reasoning, this photo:
https://images.pexels.com/photos/7321/sea-water-ocean-horizon.jpg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260
clearly Earth must be concave with us living on the inside of a ball.

I have to admit, that taking any image from google does not proof anything, as I thought in the beginning. So, I learned something new, thanks!

As I understand it now, the reason is a possible barrel-distortions of some lenses:
https://www.mathworks.com/matlabcentral/fileexchange/37980-barrel-and-pincushion-lens-distortion-correction

You can see for the barrel-distortion that a horizontal line above the center is bent convexly and below the center concavely. The latter happened in your example. The more "wide-angle" the camera lens is, the stronger is this distortion.
The problem with Google pictures is that we don't know, if they were sectioned afterwards. Therefore, we cannot know, if the horizon was above or below the center of the picture during the shot.

So, I chose a self-taken (with the Samsung S7) and unsectioned picture, where the horizon is below the picture center in order to rule out the lense-distortion being the reason for the convex curvature. Then I did the drawing again (and blurred my wife ).
Here is the result: https://i.ibb.co/9shgyTC/curvature-horizon.jpg
red = horizontal line, touching the horizon.
yellow = tendon between the intersections of the horizon with the image sides.
Both lines reveal the curvature and lens distortion can be ruled out.

Thanks for reading

Dear JackBlack, thanks for your feed back. About your calculation with the ship: You assumed that the ship has the same distance as the horizon (from the camera's point of view). However, in fact the ship is still far away from the horizon (look again closely).

Quote from: JackBlack on May 02, 2019, 05:49:25 AM

Also, going by your reasoning, this photo:
https://images.pexels.com/photos/7321/sea-water-ocean-horizon.jpg?auto=compress&cs=tinysrgb&dpr=3&h=750&w=1260
clearly Earth must be concave with us living on the inside of a ball.

I have to admit, that taking any image from google does not proof anything, as I thought in the beginning. So, I learned something new, thanks!

As I understand it now, the reason is a possible barrel-distortions of some lenses:
https://www.mathworks.com/matlabcentral/fileexchange/37980-barrel-and-pincushion-lens-distortion-correction

You can see for the barrel-distortion that a horizontal line above the center is bent convexly and below the center concavely. The latter happened in your example. The more "wide-angle" the camera lens is, the stronger is this distortion.
The problem with Google pictures is that we don't know, if they were sectioned afterwards. Therefore, we cannot know, if the horizon was above or below the center of the picture during the shot.

So, I chose a self-taken (with the Samsung S7) and unsectioned picture, where the horizon is below the picture center in order to rule out the lense-distortion being the reason for the convex curvature. Then I did the drawing again (and blurred my wife ).
Here is the result: https://i.ibb.co/9shgyTC/curvature-horizon.jpg
red = horizontal line, touching the horizon.
yellow = tendon between the intersections of the horizon with the image sides.
Both lines reveal the curvature and lens distortion can be ruled out.

Thanks for reading

Cool picture. I'm seeing a problem on the right side, almost like the camera isn't entirely level. As well, I'd make the yellow line straight. It is not, a couple of stair-stepped pixels.

Overall, it's almost impossible to see a curve at such a low altitude no matter how clear the day. I would say not very convincing.

Why does the camera need to be leveled perfectly to show the curvature?

Dear JackBlack, thanks for your feed back. About your calculation with the ship: You assumed that the ship has the same distance as the horizon (from the camera's point of view). However, in fact the ship is still far away from the horizon (look again closely).

Yes, I did assume that, as it is quite close to the horizon.
Yes, the extra distance to the horizon would make the horizon slightly longer, but not by enough to make the curvature of the horizon observed in the photo to be reasonable.
Assuming the horizon is a great circle would require the horizon to be roughly 27 km away.

I have to admit, that taking any image from google does not proof anything, as I thought in the beginning. So, I learned something new, thanks!

Glad to have been able to help.

As I understand it now, the reason is a possible barrel-distortions of some lenses:
https://www.mathworks.com/matlabcentral/fileexchange/37980-barrel-and-pincushion-lens-distortion-correction

You can see for the barrel-distortion that a horizontal line above the center is bent convexly and below the center concavely. The latter happened in your example. The more "wide-angle" the camera lens is, the stronger is this distortion.
The problem with Google pictures is that we don't know, if they were sectioned afterwards. Therefore, we cannot know, if the horizon was above or below the center of the picture during the shot.

Yes, however notice that is only one type of distortion. Another type is pincushion, which works in the opposite way (and there are other types as well), and there is the potential of software inside the camera to try and self-correct.

Really the only way to do it is to have a reference line, exactly where the horizon would be. Without that you want the horizon as close as possible to the centre or as close to level as possible.


If you want a good reference image for testing, print a square grid and then take pictures of it from various distances and from various angles and see what they look like. (or have one on your computer screen and take a picture of that).

Really the only way to do it is to have a reference line, exactly where the horizon would be. Without that you want the horizon as close as possible to the centre or as close to level as possible.


If you want a good reference image for testing, print a square grid and then take pictures of it from various distances and from various angles and see what they look like. (or have one on your computer screen and take a picture of that).

I did the following just now:

• took a picture of our kitchen tiles with the same camera.
• used the "multiply" layer filling method in photoshop to use the gaps/joints between the tiles as reference grid.
• rotated the pictures (horizon and tiles) for leveling correction by less than 0.2°.
• Placed red lines at the horizon and through the tile joints next to the horizon.

Here is the result:
https://ibb.co/nkXt6Fg

I do not try to convince you about this picture. I am trying to point you towards a method how you can easily check the truth for yourself (not quantitatively but at least qualitatively). As I understood the main argument of FET is that earth "looks" flat. You can understand this method as a tool for more precise looking compared to the naked eyes.

Best regards

I did the following just now:

• took a picture of our kitchen tiles with the same camera.
• used the "multiply" layer filling method in photoshop to use the gaps/joints between the tiles as reference grid.
• rotated the pictures (horizon and tiles) for leveling correction by less than 0.2°.
• Placed red lines at the horizon and through the tile joints next to the horizon.

Did you notice how I said from various distances at various angles?
A single picture wont cut it.
Was it all set up the same, with the same zoom and focal length used by your picture of the horizon?

Using a similar method to before, your image shows a roughly 3 px drop across 2560 px.
To make this match the great circle of Earth (to show curvature, again the actual drop expected will be less) you would need a horizon length in the image of roughly 60 km to produce a drop of ~70 m.
Assuming you have a FOV of roughly 120 degrees, this means the actual horizon length around you would need to be roughly 180 km and thus the horizon would need to be 28.5 km away. This requires you to be standing roughly 64 m above sea level. From the picture you don't appear to be anywhere near that high. So again, the actual curvature of Earth is highly unlikely to be what you are seeing here.

You are looking for such a minor difference you won't be able to see it without any insane resolution and a very good camera.

You are far better off trying to measure if the horizon is at eye level.

Dear JackBlack,

I understood now that you are not a FETler defending FET as I thought in the beginning lol. 
I cannot follow your calculation, but while searching for the formulas in the internet I found the amazing blog made by Walter Bislin. Did you know this?

Here is a link to his earth curvature simulator, followed by the formulas and grid overlay to ISS pictures etc.: http://walter.bislins.ch/bloge/index.asp?page=Flat-Earth%3A+Finding+the+Curvature+of+the+Earth
By entering the height of my picture at the beach (around 7 meters from a bay watch tower) a curvature is indeed not visible so strong as in my photo. I still dont really understand why I got this strong distortion at the beach and not at the kitchen Both were taken with 1x zoom and auto focus, while at the beach the focus point was most likely near infinity and in kitchen also not so far away from that. At least it looks sharp with at infinity and gets already unsharp by moving towards makro by around 10%.

Anyways at the page from Walter, you have to click also on the buttons "Curve", "Causew"... to see some animations and comparisons to photos.
Or here: Comparison between FE model and globe model and photos:
http://walter.bislins.ch/bloge/index.asp?page=Comparison+of+Globe+and+Flat-Earth+Model+Predictions+with+Reality

I recommend to check out also the other sub pages, it's really interesting.

Thanks for the lesson
Peace

Walter Bilsin's site is a great resource. We use his calculator a lot:

http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

It's great for estimating curve and flat expectations with multiple objects. Built in refraction slider, etc.

Walter Bilsin's site is a great resource. We use his calculator a lot:

http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

It's great for estimating curve and flat expectations with multiple objects. Built in refraction slider, etc.

Have you seen this that gives the number of pixels of "curvature" to expect in a photo? FEI Horizon Calculator - he labels "curvature" as "Sagitta".

From 2 m above the horizon is so close to straight that it's quite negligible.

"Sagitta" is the formal name for the distance from a chord to an arc at its mid-point.

I understood now that you are not a FETler defending FET as I thought in the beginning lol. 
I cannot follow your calculation, but while searching for the formulas in the internet I found the amazing blog made by Walter Bislin. Did you know this?

No, I'm not a FEer.

The calculation is an approximation based upon Pythagoras for a great circle.
This image might help

Basically the horizon is a point on Earth (marked by a black h) which you can extend a line tangent to Earth towards the observer (marked by a black o).
This line has a length of d. The observer is at a height of h above the Earth, and Earth has a radius of R.
This gives a right angle triangle and thus by Pythagoras we know:
R²+d²=(R+h)²
R²+d²=R²+2Rh+h²
d²=2Rh+h²
d²/(2R)=h+h²/(2R)

If we now note that h²/(2R) is basically nothing as R is much larger than h (unless you are very high, in space), this can be approximated as:
d²/(2R)=h

And this is where the FE saying of 8 inches per mile squared comes from.
I was being lazy and using excel solver to figure it out.
But to do it properly we really need a scaling factor, say k.
So to change the formula, now d and h are measured in px and k is in say m/px, we get:
(kd)²/(2R)=kh
kd²/(2R)=h
k=2Rh/(d²)

That lets you figure out the scale factor and then find out what the dip and length would need to be.
Does that make sense?

The problem is that this focuses on the great circle which will be hidden by the horizon.

As for the site, while I haven't spent much time on it, I have seen it used here and had a cursory look at it before.
It does a much better job that what I have been doing as I have been focusing on the great circle, not the horizon.

I still dont really understand why I got this strong distortion at the beach and not at the kitchen Both were taken with 1x zoom and auto focus, while at the beach the focus point was most likely near infinity and in kitchen also not so far away from that. At least it looks sharp with at infinity and gets already unsharp by moving towards makro by around 10%.

There are a few possibilities.

Perhaps the simplest is that the tiles aren't actually that straight and you just can't easily tell by eye. Most people wouldn't notice a slight curve in the tiles, and depending upon how they were laid, there might be some curve.

Another is auto-correction by the software of the camera.
With some cameras, rather than trying to have the lens be perfect they choose to distort the image in software to try and make it look better. In some cases this will be applying a "simple" transformation based upon the known properties of the lens. Then there is the lazy way based upon a more complex algorithm which tries to fit something in the photo and then applies a distortion based upon that. It could be that it uses the latter algorithm and fitted the grid better than the horizon.

Another issue is distance. The actual distance to the object can affect what kind of distortion you have based upon the changing focal point.

The final issue I can think of (which I overlooked before) is that the line of the tiles doesn't quite match the line of the horizon.
The tiles are a straight line, the horizon is a circle. These get distorted differently.

Thanks for the lesson
Peace

Your welcome.
Glad I could help.

Yes they look flat, because earth is quite huge.

Ok please do the following (takes <1 min):

1. Go to e.g. https://images.wallpaperscraft.com/image/sea_horizon_ship_117671_3840x2160.jpg
2. Zoom in to maximum.
3. Move your section so that the highest point of the horizon (in the example a bit left to the ship) is at the upper edge of your browser window or screen.
4. Move your section only horizontal to the left and afterwards to the right.

What do you see?

Paste that image into Photoshop, Photopaint, PaintShop or GIMP, keep the vertical size and shrink it horizontally to 2% or something like that.

Quote from: Souleon on May 02, 2019, 12:14:36 AM

Yes they look flat, because earth is quite huge.

Ok please do the following (takes <1 min):

1. Go to e.g. https://images.wallpaperscraft.com/image/sea_horizon_ship_117671_3840x2160.jpg
2. Zoom in to maximum.
3. Move your section so that the highest point of the horizon (in the example a bit left to the ship) is at the upper edge of your browser window or screen.
4. Move your section only horizontal to the left and afterwards to the right.

What do you see?

Paste that image into Photoshop, Photopaint, PaintShop or GIMP, keep the vertical size and shrink it horizontally to 2% or something like that.

That would be an elegant, but yet just another way to see distortion of the lense with unkown magnitude and not the real curvature. I am still searching for an easy round earth prove that everyone can do without high cost and that cannot be debunked. The horizontal curvature near see level isn't one, unfortunately. Maybe it will be future, with even better cameras.

Quote from: Macarios on May 05, 2019, 10:01:49 PM

Quote from: Souleon on May 02, 2019, 12:14:36 AM

Yes they look flat, because earth is quite huge.

Ok please do the following (takes <1 min):

1. Go to e.g. https://images.wallpaperscraft.com/image/sea_horizon_ship_117671_3840x2160.jpg
2. Zoom in to maximum.
3. Move your section so that the highest point of the horizon (in the example a bit left to the ship) is at the upper edge of your browser window or screen.
4. Move your section only horizontal to the left and afterwards to the right.

What do you see?

Paste that image into Photoshop, Photopaint, PaintShop or GIMP, keep the vertical size and shrink it horizontally to 2% or something like that.

That would be an elegant, but yet just another way to see distortion of the lense with unkown magnitude and not the real curvature. I am still searching for an easy round earth prove that everyone can do without high cost and that cannot be debunked. The horizontal curvature near see level isn't one, unfortunately. Maybe it will be future, with even better cameras.

The horizontal curvature is so small because the horizon is exactly the same distance from the observer all 360° around.
So when looking at the horizon we are looking at the edge of a circle from a very shallow angle and that slight curve is what this on-line calculator shows"
Flat Earth Insanity: FEI Horizon Calculator. What he calls "Sagitta, horizon centered" and "Sagitta, camera level" is what we would commonly loosely call "curvature".
There is more detail on the horizon in:

Proving the Earth is not Flat - Part 1 - The Horizon by VoysovReason

But when looking at the curvature in the direction away from you, foreshortening automatically magnifies the curvature as in Soundly's Lake Pontchartrain video's, for example:

Flat Earth Debunked - Lake Pontchartrain Causeway - Three Lakeway Center - 15th Floor by Soundly. He shows this sort of curvature:

From: Can you show an example of some perpendicular object viewed from far away that is seemingly tilted due to the curvature of Earth? - Doug Houser's answer.

But if you are "still searching for an easy round earth proof that everyone can do without high cost and that cannot be debunked" it might be hard to find anything that convinces a dedicated flat-earther.

They will invariably come up with unprovable hypotheses "explaining" anything you can come up with.