[Storm]

This photograph has been undisputed since its presentation in a Fort Worth chat forum in June of '09.

It is a photo of Dallas and Fort Worth in the same shot...30 miles apart....on the same FLAT plane.

There is plenty discussion about the details of the photo in the threads noted below, as well as other photos to substantiate the target cities in the photo.

http://www.fortwortharchitecture.com/forum/index.php?showtopic=626&hl=\Dallas+&+Fort+Worth+Skylines\

http://www.fortwortharchitecture.com/forum/index.php?showtopic=3770




Anybody care to dispute, or disprove, the photo?

[Tumeni]

Anybody care to dispute, or disprove, the photo?

What do you think it proves, since you invite a disproof?

The photographer states he is photographing from his "favourite hilltop", but there's no indication of his camera height. I searched both pages for "elevation" and "height", but there's no mention of. He actually seems to want to keep it a secret.

How can we meaningfully discuss, without this data?

[AATW]

I’d be interested to know what Storm thinks one should see on a globe earth that is not seen in that photo.
The horizon appears flat unless you are at serious altitude so what would one expect to see in this photo if the earth is flat that you wouldn’t see if the earth is a globe?

[Storm]

What do you think it proves, since you invite a disproof?

See thread title.

How can we meaningfully discuss, without this data?

All the data needed is in the photo, itself. You don't need the elevation, nor height, to see clearly what is evident, there.

[Storm]

I’d be interested to know what Storm thinks one should see on a globe earth that is not seen in that photo.
The horizon appears flat unless you are at serious altitude so what would one expect to see in this photo if the earth is flat that you wouldn’t see if the earth is a globe?

You don't have to be at "serious altitude" to see that the horizon is flat for well over 30 miles in that photo. At that distance, NASA's data for the circumference of Earth, and the equation to find drop (8 inches multiplied by distance[miles] squared, then divided by 12 to get feet) would put that cluster of buildings in the distance (seen at left in photo) 600 feet below the horizon.

AND, at that distance, you would ABSOLUTELY 'see' some curvature IF the Earth were the shape and dimensions we are told it is.

[ChrisTP]

looks to me like the closest town is in front of the horizon and the furthest is behind it, maybe I'm missing something but that suggests curvature? It's coming up 2am here and I'm getting tired so if the numbers here are wrong do say, but I was particularly generous in favour of proving flat earth with the numbers by adding 20 miles from the photographer to the first town and only assuming he's at ground level instead of a hill, I looked up the biggest building in dallas and rounded it down for the targets size. You can still see the target even at that distance, seemingly. Using this website;

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

[AATW]

You don't have to be at "serious altitude" to see that the horizon is flat for well over 30 miles in that photo. At that distance, NASA's data for the circumference of Earth, and the equation to find drop (8 inches multiplied by distance[miles] squared, then divided by 12 to get feet) would put that cluster of buildings in the distance (seen at left in photo) 600 feet below the horizon.

Yes. 600 feet...at a viewer height of 0.
But he says in the thread you linked to that he was up a hill.
I don't know how high the hill was but if you put in a viewer height of 200 feet then the hidden height is just over 107 feet. Perfectly feasible you'd see taller buildings then. And he might well have been higher.

This is the thing - you need to know what the viewer height was before you can properly analyse the photo. Given 200 feet doesn't seem that high for a hill I'd say it's perfectly plausible you'd see the distant buildings but without knowing the exact location and height it's hard to tell.

[Tumeni]

You don't have to be at "serious altitude" to see that the horizon is flat for well over 30 miles in that photo. At that distance, NASA's data for the circumference of Earth, and the equation to find drop (8 inches multiplied by distance[miles] squared, then divided by 12 to get feet) would put that cluster of buildings in the distance (seen at left in photo) 600 feet below the horizon.

....but surely the figure depends absolutely on the height of the observer above the zero feet reference level? And the author explicitly says he was on his "favourite hilltop"



You've provided a single photograph which, without data, is inconclusive. Off the top of my head, I can cite five separate and distinct YoUTube channels which show conclusively that the lands and seas around Sydney, Aus; Firth of Forth, Scotland; and the English Channel, are Not Flat. And three of these are video authors who are clearly flat-earthers. The other two include myself, and my own videos, but last time I posted links to them, I was bounced to AR for "selling myself" or something like .... 

Since this sub-forum is "Flat Earth Investigations", I can outline for you the method which you can follow to determine this (investigate it) for yourself, and I can outline how each individual has shown this to be so.

Are you interested?  Y/N

[Storm]

Every one of you is completely glossing over (IGNORING) the simple, and abundantly obvious, FACT that there is no curvature anywhere in this photo, or any of the photos that have been presented, from the left to the right of the photo.


You want to use the "height-of-the-viewer" argument, but completely ignore the facts that:
1. The higher the viewer is, THE MORE CURVATURE WOULD BE EVIDENT FROM LEFT TO RIGHT.
2. We're talking about a 3 DIMENSIONAL PLANE, whether it's spherical or flat.
3. You have to consider both: The curvature from the viewer to the target AND the curvature from left to right on the horizon.

You agree that 600 feet of drop is accurate for a 30 mile distance, but you ignore the fact that those two cities are 30 miles apart FROM LEFT TO RIGHT IN THE PHOTO.

Once again,..........a resounding..........."WHERE.......IS.........THE CURVE?!!"








(Smiley faces prove it's not a rant, by the way. )

[ChrisTP]

if the planet was a cylinder and you were looking from one end to the other you might see a left to right curve visibly... Otherwise, disc or ball, given the sheer size of the earth you won't be seeing a left to right curve from a hilltop.

[Storm]

if the planet was a cylinder and you were looking from one end to the other you might see a left to right curve visibly... Otherwise, disc or ball, given the sheer size of the earth you won't be seeing a left to right curve from a hilltop.

[/b]

Nor from a mountaintop.


(34 mi from left mountain to right mountain--so roughly 50-60 mi left to right in photo)

I guess it's just one of those things you have to accept like a little child when your parents tell you something that makes no sense at all like,

"You're never, ever, gonna see any curve--no matter how high you are or what the circumstances may be, ...........but it's a ball. Trust me."

[Pete Svarrior]

(Smiley faces prove it's not a rant, by the way. )

They don't, and throwing in snarky comments in the middle of your post is unlikely to make it better. Your tone continues to be inappropriate for this forum, and if you can't get that in check, your time here will be brief.

[totallackey]

Anybody care to dispute, or disprove, the photo?

What do you think it proves, since you invite a disproof?

The photographer states he is photographing from his "favourite hilltop", but there's no indication of his camera height. I searched both pages for "elevation" and "height", but there's no mention of. He actually seems to want to keep it a secret.

How can we meaningfully discuss, without this data?

A possible height for the hill in question is 800+ ft.

ETA: This photo was obviously taken from the northwest, looking southeast toward Fort Worth, with Dallas to the east.

A topographical map query shows that elevations of the area to the northwest of Fort Worth are indeed at or around 800-900 ft.
https://en-us.topographic-map.com/maps/nct/Fort-Worth/

[AATW]

You agree that 600 feet of drop is accurate for a 30 mile distance

From a viewer height of 0, which this photo was not taken from.

but you ignore the fact that those two cities are 30 miles apart FROM LEFT TO RIGHT IN THE PHOTO.
Once again,..........a resounding..........."WHERE.......IS.........THE CURVE?!!"

The diameter of earth is 7,917 miles. So I went into Paint.NET and drew a circle 791x791 (should have been 792 really, but let's not split hairs.
Here's the top of that circle with two lines 40 pixels apart.



That represents 400 miles - 10 pixels to a mile. Where's the curve?
tl;dr - the earth is really big. At normal scales and altitudes you can't see the curve left to right. You need to be very high up for that.

[Tumeni]

Every one of you is completely glossing over (IGNORING) the simple, and abundantly obvious, FACT that there is no curvature anywhere in this photo, or any of the photos that have been presented, from the left to the right of the photo

 ... yet you have failed to show that there should be.

We don't know the observer/camera height. We don't know what type of lens and/or camera was used. We don't know what level of telephoto or zoom was required to frame the shot.

You can't simply assert "there should be left to right curvature visible" without some basis.

[Tumeni]

Perhaps should be in a new thread, but since at least one other poster above has introduced other photos, here's a screen grab from a YouTuber's video, taken looking over the Firth of Forth in Scotland, looking out from the Hopetoun Monument to the Isle of May and beyond.

The relevant data -

Observation height 210m
Lighthouse focal height - 73m, some 28km from the observer
Ship height - unknown, distance unknown, but it is seen in the video to pass between the Isle and the mast.
Inch Cape Met Mast (to the right of the ship) - 169.5m, some 50km or so from the observer

Without further input from me, what does this tell you, if you presume the seas to be flat, and all elevations measured from that flat sea level?

EDIT - oops, need links to the pictures ...



- -

[iCare]

This photograph has been undisputed since its presentation in a Fort Worth chat forum in June of '09.

Why would it be disputed (from a FE/RE point of view)?
It's a panoramic shot in an architecture forum.
If anything would be discussed it would likely be details of buildings, where it was taken from (actually a point of the linked discussion), which type camera was used ...

It is a photo of Dallas and Fort Worth in the same shot...30 miles apart....on the same FLAT plane.

Given that the difference in elevation is over 200 ft. (Dallas approx. 430 ft, Fort Worth approx. 650 ft, elevation of camera unknown), how is this a flat plane?
Obviously in a relatively small area, local features of terrain are more significant then "global" (FE or RE) characteristics.
As pointed out by others, 30 miles isn't a large distance compared to the size of the world.

Anybody care to dispute, or disprove, the photo?

Let me point you to the TFES FAQ:
https://wiki.tfes.org/Flat_Earth_-_Frequently_Asked_Questions#There_are_many_pictures_on_the_Internet_and_in_other_media_depicting_the_Earth_as_being_round._Why_do_these_not_disprove_the_Flat_Earth_Theory.3F
It points out that - for several reasons - TFES "do not lend much credibility to photographic evidence."
Also, I know some professional photographers and it's really amazing what different effects/impressions they can create by means of perspective, choice of lens, settings (aperture, shutter speed, ...), filters, ...

This pictures may very well be authentic ... but by itself, it doesn't prove or dispute anything.
Based on this photograph I wouldn't decide either way.

iC

[somerled]

It's interestingly comical to see that  the advanced calculator uses the most archaic perfect sphere model , R=6371km in it's "exact equations for hidden heights " section .

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

[Tumeni]

It's interestingly comical to see that  the advanced calculator uses the most archaic perfect sphere model , R=6371km in it's "exact equations for hidden heights " section .

Why is that "comical"?

The circumference was, according to the textbooks, determined by two different methods, and different teams and individuals, in the 1600s and 1700s.

Radius can be derived from circumference with simple school-level geometry.

[AATW]

It's interestingly comical to see that  the advanced calculator uses the most archaic perfect sphere model , R=6371km in it's "exact equations for hidden heights " section .

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

One thing which a lot of people seem to struggle with is the idea that a model can be imperfect and still useful.

For many engineering projects the earth can be assumed to be flat, for example - that is NOT evidence that it is flat, simply that the error in not accounting for curvature is too small to be significant so for the sake of simplicity it can be disregarded.

Anyone who has done any physics or mechanics should understand this. When you learn about Newtonian mechanics and how bodies move you generally disregard air resistance for example. For certain things you might need to consider that just like for certain engineering projects the curvature of earth has to be considered. It depends what you’re doing.

In brief - for most practical purposes and for the sake of simplicity an earth curve calculator can use a model of a spherical earth and give perfectly usable results. The oblateness of earth is pretty subtle, if you look at photos of earth from more distant satellites you can’t actually tell it’s not a perfect circle, you have to count the pixels horizontally and vertically to notice.

People are rightly criticised on here for bowling in and attacking FE theories without first understanding them, it would help discussion if some FE people took more time to understand the heliocentric globe earth model.