I can see curvature in this picture...

iWitness said that according the curvature data presented by NASA, it should be possible to observe curvature in this image:


I have a sneaking feeling that there is curvature in the image. If you draw a straight line on the horizon, and look closely, you can see the parts that stick out. But I may be wrong.

I can see curvature in the building as well (especially in the antenna).  To save Tom Bishop the bother, I'll tell you right off that a wide angle lens (but not a fish-eye) was used for that photo.

iWitness said that according the curvature data presented by NASA, it should be possible to observe curvature in this image:

I have a sneaking feeling that there is curvature in the image. If you draw a straight line on the horizon, and look closely, you can see the parts that stick out. But I may be wrong.

I have highlighted where you're making a mistake in red.

And I do observe curvature in the image.  In several areas of the picture, in fact.

It is very, very unlikely any curvature could be distinguished in that picture. 

For the sake of discussion, let's say that the horizon was 20 miles wide in that picture.  I've seen a chart on this site showing the drop in feet at varying distances (as listed in ENaG) and at 20 miles there would be 266 feet of drop.  Also consider, that the center of the horizon would be a high spot, so the sides would only drop half that distance from the peak in the center (ten miles in each direction from the center).  Based on the number of feet in a mile the drop from the curvature would be approximately 400 times less than the width.  Since the picture is 800 pixels wide, and the drop would only be half in each direction, it would mean that the horizon at the sides of the picture would be 1 pixel lower than the center. 

Considering the resolving power of the cameral lens, imperfections in focus, artifacts from jpeg processing, the fact that the building blocks a large part of the horizon, small differences in elevation of the terrain, foliage, buildings, etc., etc. it is virtually impossible that any curvature would be visibly detected in that picture.  Add to this the issue of any curvature introduced by lens distortion and any detected curvature is basically irrelevant in that picture

That 20 miles is just a random number for discussion.  If the horizon was 50 miles wide, the drop would be a larger portion of width, but it would still be something like 2-2.5 pixels of drop, still difficult to easily detect.

For a start, there is lens distortion in that image - look at the top of the tower on the left. It looks bent.
Secondly, I must have raised this with Iwitless (and others) until I am blue in the face, but there's no reason why you would see any sort of curvature to the side from such a low level. Look at it this way: if the horizon in the centre of the image is 30 miles away, the horizon at the edges of the picture is also 30 miles away, and so is the horizon beyond the frame of the picture, and so is the horizon behind the camera because the camera is aligned perpendicular to what we perceive as "down". If the horizon is the same distance away in every direction, why would you expect it to look different somewhere?
In order to see a curvature of the horizon you have to achieve a greater distance from it, where your eye can begin to perceive the slight change in perspective of the downward curvature of the horizon in the direction you're looking. This picture is nowhere near that altitude. And even then, if you were to align the camera exactly perpendicular to downwards, the horizon would still appear to be the same level all around. Take a 360 degree panned picture all the way round from the highest point you like, if you have the camera exactly level you'll still find the horizon matches up with itself. Even if you were high enough to perceive curvature.

For a start, there is lens distortion in that image - look at the top of the tower on the left. It looks bent.
Secondly, I must have raised this with Iwitless (and others) until I am blue in the face, but there's no reason why you would see any sort of curvature to the side from such a low level. Look at it this way: if the horizon in the centre of the image is 30 miles away, the horizon at the edges of the picture is also 30 miles away, and so is the horizon beyond the frame of the picture, and so is the horizon behind the camera because the camera is aligned perpendicular to what we perceive as "down". If the horizon is the same distance away in every direction, why would you expect it to look different somewhere?
In order to see a curvature of the horizon you have to achieve a greater distance from it, where your eye can begin to perceive the slight change in perspective of the downward curvature of the horizon in the direction you're looking. This picture is nowhere near that altitude. And even then, if you were to align the camera exactly perpendicular to downwards, the horizon would still appear to be the same level all around. Take a 360 degree panned picture all the way round from the highest point you like, if you have the camera exactly level you'll still find the horizon matches up with itself. Even if you were high enough to perceive curvature.

How come this image has lens distortion, but no picture of the Earth from a high altitude can have lens distortion?

Quote from: The Knowledge on April 09, 2012, 04:47:48 PM

For a start, there is lens distortion in that image - look at the top of the tower on the left. It looks bent.
Secondly, I must have raised this with Iwitless (and others) until I am blue in the face, but there's no reason why you would see any sort of curvature to the side from such a low level. Look at it this way: if the horizon in the centre of the image is 30 miles away, the horizon at the edges of the picture is also 30 miles away, and so is the horizon beyond the frame of the picture, and so is the horizon behind the camera because the camera is aligned perpendicular to what we perceive as "down". If the horizon is the same distance away in every direction, why would you expect it to look different somewhere?
In order to see a curvature of the horizon you have to achieve a greater distance from it, where your eye can begin to perceive the slight change in perspective of the downward curvature of the horizon in the direction you're looking. This picture is nowhere near that altitude. And even then, if you were to align the camera exactly perpendicular to downwards, the horizon would still appear to be the same level all around. Take a 360 degree panned picture all the way round from the highest point you like, if you have the camera exactly level you'll still find the horizon matches up with itself. Even if you were high enough to perceive curvature.

How come this image has lens distortion, but no picture of the Earth from a high altitude can have lens distortion?

Yet another straw man... Who said that no picture of the Earth from a high altitude can have lens distortion?

Also, objects on the horizon can make any alleged curvature almost impossible to make out. There are several objects on this horizon.

Quote from: The Knowledge on April 09, 2012, 04:47:48 PM

For a start, there is lens distortion in that image - look at the top of the tower on the left. It looks bent.
Secondly, I must have raised this with Iwitless (and others) until I am blue in the face, but there's no reason why you would see any sort of curvature to the side from such a low level. Look at it this way: if the horizon in the centre of the image is 30 miles away, the horizon at the edges of the picture is also 30 miles away, and so is the horizon beyond the frame of the picture, and so is the horizon behind the camera because the camera is aligned perpendicular to what we perceive as "down". If the horizon is the same distance away in every direction, why would you expect it to look different somewhere?
In order to see a curvature of the horizon you have to achieve a greater distance from it, where your eye can begin to perceive the slight change in perspective of the downward curvature of the horizon in the direction you're looking. This picture is nowhere near that altitude. And even then, if you were to align the camera exactly perpendicular to downwards, the horizon would still appear to be the same level all around. Take a 360 degree panned picture all the way round from the highest point you like, if you have the camera exactly level you'll still find the horizon matches up with itself. Even if you were high enough to perceive curvature.

How come this image has lens distortion, but no picture of the Earth from a high altitude can have lens distortion?

Nobody ever said that, least of all me. Proof that you either haven't read my thread on the subject or if you have, you didn't understand it.

Quote from: Pongo on April 10, 2012, 12:24:21 AM

Quote from: The Knowledge on April 09, 2012, 04:47:48 PM

For a start, there is lens distortion in that image - look at the top of the tower on the left. It looks bent.
Secondly, I must have raised this with Iwitless (and others) until I am blue in the face, but there's no reason why you would see any sort of curvature to the side from such a low level. Look at it this way: if the horizon in the centre of the image is 30 miles away, the horizon at the edges of the picture is also 30 miles away, and so is the horizon beyond the frame of the picture, and so is the horizon behind the camera because the camera is aligned perpendicular to what we perceive as "down". If the horizon is the same distance away in every direction, why would you expect it to look different somewhere?
In order to see a curvature of the horizon you have to achieve a greater distance from it, where your eye can begin to perceive the slight change in perspective of the downward curvature of the horizon in the direction you're looking. This picture is nowhere near that altitude. And even then, if you were to align the camera exactly perpendicular to downwards, the horizon would still appear to be the same level all around. Take a 360 degree panned picture all the way round from the highest point you like, if you have the camera exactly level you'll still find the horizon matches up with itself. Even if you were high enough to perceive curvature.

How come this image has lens distortion, but no picture of the Earth from a high altitude can have lens distortion?

Nobody ever said that, least of all me. Proof that you either haven't read my thread on the subject or if you have, you didn't understand it.

I started skimming after you called "lens distortion".  It was just so odd seeing RE'ers saying lens distortion,that I had to say something. I almost thought that you guys couldn't see lens distortion.

Look at the railing at the very bottom of the picture.

Are cameras made by the conspiracy (for distorting images) too?

Now there is some distortion in the image as noted by Tom, but it is still quite obvious that the earth is flat by viewing this picture.

If you are standing on top of a Ball being held down by Gravity then you are at the TOP of the ball. Meaning, everything is going to curve down away from you in all directions.



This photo was taken on top of the world trade center meaning the BASE is resting on TOP of the ball and everything should curve away from the base.

The horizon stretches at least 10 miles away if not 20 miles from the base as visible in the photo and you will note that the ground does not start to curve 8 inches after the first mile, or 32 inches after the 2nd mile or 6 feet ofter the 3rd mile. It definitely doesn't curve 66 feet after 10 miles.

IT IS FLAT!

In fact, the visible horizon or center of the horizon is NOT the highest point as mentioned by MrT. Since you are ON TOP of a Ball the ground should start curving away from where you are standing right now.

The only explanation that the horizon meets you at eye level is because you are standing on ONE BIG GIANT FLAT DISC.

Now there is some distortion in the image as noted by Tom, but it is still quite obvious that the earth is flat by viewing this picture.

If you are standing on top of a Ball being held down by Gravity then you are at the TOP of the ball. Meaning, everything is going to curve down away from you in all directions.

Iwitness, look at the line marked 72 inches in that drawing.  It's the same the length as the line going between 2 and 3, which is supposed to be a mile long.

If that picture was drawn to scale, it'd look completely flat.  The curvature is very slight.  Far slighter than the misleading drawings in EnAG would lead you to believe.  I feel it's important that you grasp this because I don't think you have yet.

Now there is some distortion in the image as noted by Tom, but it is still quite obvious that the earth is flat by viewing this picture.

If you are standing on top of a Ball being held down by Gravity then you are at the TOP of the ball. Meaning, everything is going to curve down away from you in all directions.



This photo was taken on top of the world trade center meaning the BASE is resting on TOP of the ball and everything should curve away from the base.

The horizon stretches at least 10 miles away if not 20 miles from the base as visible in the photo and you will note that the ground does not start to curve 8 inches after the first mile, or 32 inches after the 2nd mile or 6 feet ofter the 3rd mile. It definitely doesn't curve 66 feet after 10 miles.

IT IS FLAT!

In fact, the visible horizon or center of the horizon is NOT the highest point as mentioned by MrT. Since you are ON TOP of a Ball the ground should start curving away from where you are standing right now.

The only explanation that the horizon meets you at eye level is because you are standing on ONE BIG GIANT FLAT DISC.

My point was that the center of the horizon was the highest point of the horizon, not of the picture overall.  Also, my math shows that the drop would be something like one pixel difference.  It is not something that could be noticed in that picture.  The diagram you posted is way out of scale.  In the picture above you are talking about something like the point directly under the "T", to a few pixels from that spot.

The Empire State Building is only approximately 3 miles away in that picture, but it looks tiny, even though it's 1250 feet tall.  Do you really believe you would be able to detect the few feet of drop between the base of the tower the picture was taken from and the base of the Empire State Building?

Also, my math shows that the drop would be something like one pixel difference.  It is not something that could be noticed in that picture.

It'd be around 0.06 pixels.  I did the math in another thread.

Quote from: MrT on April 12, 2012, 01:25:10 PM

Also, my math shows that the drop would be something like one pixel difference.  It is not something that could be noticed in that picture.

It'd be around 0.06 pixels.  I did the math in another thread.

I meant my math where I assumed a much larger than likely width of the visible horizon.  I was being generous. 

That's 4 miles base to base. Since you are on top of the WTC then your line of sight is farther away.

That's 4 miles base to base. Since you are on top of the WTC then your line of sight is farther away.

I believe it's about 3.5 miles to drive from one to the other, and it's not a straight line, so I estimated approximatly 3 miles straight line.  However, even assuming your "about 4 miles" is more accurate than my "approximately 3 miles" I still ask if you are claiming you would be able to discern the 6-10 feet of drop over that distance?  If the ground drops 6 feet in 600 feet it's not all that drastic (though you could probably tell you were looking down hill slightly).  If it drops 6 feet in 15,840 feet, and you are observing that drop from a height of 1300+ feet it would be impossible to detect the drop, particularly with buildings, slight elevation changes, etc., etc. 

You are correct, due to the height of the picture the line of site is actually further than the distance across the ground.  This would make discerning the drop even more difficult. 

There is not even an inch of curvature in that picture. The ground should be curving 8 inches after the first mile and curve more and more the further away you get. But it's a straight shot as far as you can see.

If the horizon behind the Empire state building is the "Bulge" of the curve then all the buildings would be titled with the bulge. Not Flat. Plus, if the ground curved away from you even slightly then the horizon would not meet you at eye level. That's the big kicker right there.

That's why the whole idea of gravity, spherical earth and orbits just don't work or make any sense. Maybe on drugs, but not in reality.

8 inches, over an entire mile.  63,360 inches in a mile, by the way.  It's clear that you still don't understand how slight the curvature is.  You have to be miles up before it becomes visible.  The only normal flights that went high enough were the concords.

There is not even an inch of curvature in that picture. The ground should be curving 8 inches after the first mile and curve more and more the further away you get. But it's a straight shot as far as you can see.

If the horizon behind the Empire state building is the "Bulge" of the curve then all the buildings would be titled with the bulge. Not Flat. Plus, if the ground curved away from you even slightly then the horizon would not meet you at eye level. That's the big kicker right there.

That's why the whole idea of gravity, spherical earth and orbits just don't work or make any sense. Maybe on drugs, but not in reality.

I have no idea how you are able to accurately descern whether there is "even an inch" of curvature in that picture.  That picture is absolutely not proof of lack of curvature at the scales involved.  I could not post a picture of a flower from 100 feet away and say there are obviously no pollen spores on the stamen, because there would be no way to discern something like that when the whole flower was only 8 pixels of the picture.

I notice you mention the horizon meeting you at eye level.  In that picture, assuming flat earth, the observer is 1300+ feet above the horizon, yet it seems to meet you at eye level.  Why would the 50-100 feet of drop due to earth curvature make the horizon suddenly not meet your gaze at eye level?

If the ground curves 8 inches in the first mile.... then that means the ground would get even LOWER the further away you get.

Pick a spot halfway between the world trade center and the empire state building... The ground should already by curving downwards.
But you will notice that the ground never starts curving and it is flat as far as you can see. The horizon even appears to rise ABOVE the empire state building into the distance.

That's because there is no curve and it's flat.

If the ground curves 8 inches in the first mile.... then that means the ground would get even LOWER the further away you get.

Pick a spot halfway between the world trade center and the empire state building... The ground should already by curving downwards.
But you will notice that the ground never starts curving and it is flat as far as you can see. The horizon even appears to rise ABOVE the empire state building into the distance.

That's because there is no curve and it's flat.

I am not sure what you base this assumption on.
Can you supply proof of this occurring?

If the ground curves 8 inches in the first mile.... then that means the ground would get even LOWER the further away you get.

Pick a spot halfway between the world trade center and the empire state building... The ground should already by curving downwards.
But you will notice that the ground never starts curving and it is flat as far as you can see. The horizon even appears to rise ABOVE the empire state building into the distance.

That's because there is no curve and it's flat.

Sorry, no. 1) WTC is being rebuilt, so the suggested experiment is lame, and 2) you've again failed to do the math (or even the geometry) to support your outlandish claim that the expected curvature is detectable.

We've been over this before. The claims you continue to make on this point are fruitless.

iWitness, once again, you're not really understanding how slight the curvature is.

Over 20 miles, we'd have a 266 foot drop.  Wowsers that sounds like a lot!  Problem is, 20 miles is 105,600 feet.  That means the length of this drop is .25% of the length you're seeing.  I could draw a line 100 pixels long showing you what this curvature would look like, and it'd drop down one whole quarter of a pixel.  It'd look like a straight line.

Are you understanding this?  By the time you're far enough away to see these sorts of distances in their entirety, the curvature, as large as 266 feet sounds, is going to look insignificant.

Please, please think about this.  It's painful watching you make the same mistake over, and over, and over, and over, and over.

The formula for chords is so simple, it amazes me that they cant get it.  Every time I see a post like that they have a different number for how much it should "drop".

iWitness said that according the curvature data presented by NASA, it should be possible to observe curvature in this image:


I have a sneaking feeling that there is curvature in the image. If you draw a straight line on the horizon, and look closely, you can see the parts that stick out. But I may be wrong.

Hey guys.  According to Parmemides both the curvature and the horizon are illusions.  The extent of yourself.  Neither exists but for the observing witness...you,  Everything is you,  You are so much greater than the outline of your body.  How else can the flat, round or eliptical earth be conspiracies?   

Quote from: iWitness on April 12, 2012, 11:41:29 AM

Now there is some distortion in the image as noted by Tom, but it is still quite obvious that the earth is flat by viewing this picture.

If you are standing on top of a Ball being held down by Gravity then you are at the TOP of the ball. Meaning, everything is going to curve down away from you in all directions.



This photo was taken on top of the world trade center meaning the BASE is resting on TOP of the ball and everything should curve away from the base.

The horizon stretches at least 10 miles away if not 20 miles from the base as visible in the photo and you will note that the ground does not start to curve 8 inches after the first mile, or 32 inches after the 2nd mile or 6 feet ofter the 3rd mile. It definitely doesn't curve 66 feet after 10 miles.

IT IS FLAT!

In fact, the visible horizon or center of the horizon is NOT the highest point as mentioned by MrT. Since you are ON TOP of a Ball the ground should start curving away from where you are standing right now.

The only explanation that the horizon meets you at eye level is because you are standing on ONE BIG GIANT FLAT DISC.

My point was that the center of the horizon was the highest point of the horizon, not of the picture overall.  Also, my math shows that the drop would be something like one pixel difference.  It is not something that could be noticed in that picture.  The diagram you posted is way out of scale.  In the picture above you are talking about something like the point directly under the "T", to a few pixels from that spot.

The Empire State Building is only approximately 3 miles away in that picture, but it looks tiny, even though it's 1250 feet tall.  Do you really believe you would be able to detect the few feet of drop between the base of the tower the picture was taken from and the base of the Empire State Building?

Your sight is an illusion MrT.  The horizon doesn't begin at the perceived curvature beyond the Empire State Bldg, It begins at your observation of the whole photo.  The distance you create within the photo between any two points is equal to the distance between you and the photo itself.  Any horizon begins with you, and ends with you.  Time, distance, and measure are the ideas and devices you create in the attempt to separate your "self" from any other object to enable your abilty to create illusions and reality.  Always remember MrT, this is truly why the earth is flat, and all other ideas are really conspiracies.