The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Investigations => Topic started by: Storm on February 25, 2020, 06:52:27 PM
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(https://www.fortwortharchitecture.com/dfw-2005.jpg)
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?
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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?
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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?
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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.
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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.
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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
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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.
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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
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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?!!" ;D
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
(Smiley faces prove it's not a rant, by the way. ;))
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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.
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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.
(https://upload.wikimedia.org/wikipedia/commons/thumb/f/fa/Mount_Rainier_and_other_Cascades_mountains_poking_through_clouds.jpg/420px-Mount_Rainier_and_other_Cascades_mountains_poking_through_clouds.jpg)
(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."
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(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.
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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/ (https://en-us.topographic-map.com/maps/nct/Fort-Worth/)
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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?!!" ;D
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.
(https://i.ibb.co/h8KYVXz/Wheres-The-Curve.jpg)
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.
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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.
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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 ...
(https://i.imgur.com/njftNFU.jpg)
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(https://i.imgur.com/iHxcgX0.jpg)
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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 (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
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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
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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.
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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.
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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.
/quote]
Then derive the radius of the oblate spheroid or pear shape model with your simple school level geometry , because earth is not reckoned to be a perfect sphere . How does one use an imaginary value , which is not found in reality ,for R in an " exact equation for hidden height " calculation and expect an answer which is supposed to give a real value ?
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Then derive the radius of the oblate spheroid or pear shape model with your simple school level geometry , because earth is not reckoned to be a perfect sphere . How does one use an imaginary value , which is not found in reality ,for R in an " exact equation for hidden height " calculation and expect an answer which is supposed to give a real value ?
I don't need to, for others have done so already-
https://www.space.com/17638-how-big-is-earth.html
"The radius of Earth at the equator is 3,963 miles (6,378 kilometers) ... Earth is not quite a sphere. The planet's rotation causes it to bulge at the equator. Earth's polar radius is 3,950 miles (6,356 km) — a difference of 13 miles (22 km)."
For "real values", I refer back to my previous post, February 27, 2020, 08:16:20 PM.
I don't actually need a "hidden height", nor an exact determination of R, to show that the seas around the Forth are Not Flat.
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1. The higher the viewer is, THE MORE CURVATURE WOULD BE EVIDENT FROM LEFT TO RIGHT.
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
As far as I understand in the wiki they criticize Soundly's photos of Pontchartrain because, they say, if curvature was indeed *so* apparent, our planet would be quite small. I like that comment because it acknowledges the fact that you cannot just watch and see a round earth.
However, frontal curvature is somehow obvious when you start looking at it. In the picture you can notice three evidences:
- The abrupt stop of the horizon details (shouldn't they slowly disappear on a flat earth?)
- The fast downward decline of clouds.
- The cluster of buildings on the left popping out of nowhere, as if their base was hidden.
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1. The higher the viewer is, THE MORE CURVATURE WOULD BE EVIDENT FROM LEFT TO RIGHT.
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
As far as I understand in the wiki they criticize Soundly's photos of Pontchartrain because, they say, if curvature was indeed *so* apparent, our planet would be quite small. I like that comment because it acknowledges the fact that you cannot just watch and see a round earth.
However, frontal curvature is somehow obvious when you start looking at it. In the picture you can notice three evidences:
- The abrupt stop of the horizon details (shouldn't they slowly disappear on a flat earth?)
Why?
- The fast downward decline of clouds.
Clouds do not descend to the ground?
- The cluster of buildings on the left popping out of nowhere, as if their base was hidden.
There are objects in front of those buildings, including denser air at or near ground level.
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1. The higher the viewer is, THE MORE CURVATURE WOULD BE EVIDENT FROM LEFT TO RIGHT.
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
As far as I understand in the wiki they criticize Soundly's photos of Pontchartrain because, they say, if curvature was indeed *so* apparent, our planet would be quite small. I like that comment because it acknowledges the fact that you cannot just watch and see a round earth.
I agree, you cannot. But its curvature SHOULD be visible in certain scenarios.
It is not. .....Ever.
However, frontal curvature is somehow obvious when you start looking at it. In the picture you can notice three evidences:
- The abrupt stop of the horizon details (shouldn't they slowly disappear on a flat earth?)
Answer #1. This is due to terrain. We're talking about geography, here, not curvature of the Earth. This is evidenced by Totallackey's contribution of the topo map of the area in question. No Flat Earther ever tried to argue that there are no hills or mountains on the Earth. Texas is relatively flat, but there are rolling hills everywhere.
Using this link you can easily identify lower and higher elevations of terrain between where the photo was taken, roughly between Pecan Acres and Haslet, and the epicenter of Dallas. There is an area between Colleyville and Bedford that is roughly 650 feet, while Dallas is at roughly 428 feet. This is due to 'Terrain'. NOT Curvature of the Earth.
See for yourself.
https://en-us.topographic-map.com/maps/nct/Fort-Worth/
- The fast downward decline of clouds.
Answer #2. Though I don't see that at all in this photo, this is due to perspective, distance and the vanishing point/line. The same way that light poles appear to rise to the horizon/vanishing line while getting shorter and smaller. Clouds do the same thing as they disappear into the distance, but they shrink and move 'down' toward the horizon.
- The cluster of buildings on the left popping out of nowhere, as if their base was hidden.
(see Answer #1.)
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Clouds do not descend to the ground?
Physically, no, but they appear to do this in many instances;
(https://image.shutterstock.com/image-photo/full-frame-cloudy-sky-skyline-600w-1514054738.jpg)
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Answer #1. This is due to terrain
Remove terrain, and frontal curvature is still there.
Like this;
(https://i.imgur.com/njftNFU.jpg)
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Answer #1. This is due to terrain
Remove terrain, and frontal curvature is still there.
Like this;
(https://i.imgur.com/njftNFU.jpg)
That is a Flat Horizon vanishing line. It has been addressed many times.
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Vanishing line...
https://www.youtube.com/watch?v=VFhhCYYkILw
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I agree, you cannot. But its curvature SHOULD be visible in certain scenarios.
It is not. .....Ever.
What scenarios?
I mean, obviously you can see it from space but I guess you think those photos are fake?
This is due to terrain. We're talking about geography, here, not curvature of the Earth.
No terrain here. How is this explained in your model?
https://www.youtube.com/watch?v=MoK2BKj7QYk
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Answer #1. This is due to terrain
Remove terrain, and frontal curvature is still there.
Like this;
(https://i.imgur.com/njftNFU.jpg)
That is a Flat Horizon vanishing line.
... which has "frontal curvature"
Observation height - 210m
Lighthouse focal height 73m
Water beyond lighthouse clearly below sightline through top of lighthouse, so seas around the lighthouse cannot be flat.
If they were, the sightline from 210 through 73 would meet the water beyond the lighthouse. But it does not.
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That is a Flat Horizon vanishing line. It has been addressed many times.
Is this a "Flat Horizon vanishing line"?
Camera height 100m
Ship height 58m
Downward sightline, which should meet the water beyond the ship, IF the seas are flat. But it does not.
(https://i.imgur.com/fG5MK18.jpg?1)
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I agree, you cannot. But its curvature SHOULD be visible in certain scenarios.
It is not. .....Ever.
What scenarios?
I mean, obviously you can see it from space but I guess you think those photos are fake?
Photos [that should show] 600 feet of drop have been presented. Photos [that should show] half a mile of drop have been presented. When they are, it is claimed that they do not exist and they show nothing.
How can one argue with that kind of rebuttal when the evidence is plainly visible?
This is due to terrain. We're talking about geography, here, not curvature of the Earth.
No terrain here. How is this explained in your model?
I cannot explain a video/photo that somebody else has made. BUT I will say, in my personal opinion, that it is very likely doctored (photo-chopped) because the building does not 'ever' appear to lean away from the viewer.
Take a close look at the top of the building in each image. It is identical. Same height, same angle. Zero evidence of the building leaning away whatsoever. As it quite plainly should be were it sinking down around the other side of the "Frontal Curve" of the so-called round earth globe.
Don't ya think?
https://www.youtube.com/watch?v=MoK2BKj7QYk
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Clouds do not descend to the ground?
Physically, no, but they appear to do this in many instances;
(https://image.shutterstock.com/image-photo/full-frame-cloudy-sky-skyline-600w-1514054738.jpg)
Physically, clouds do descend to the ground.
Come on.
You know better.
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Answer #1. This is due to terrain. We're talking about geography, here, not curvature of the Earth. This is evidenced by Totallackey's contribution of the topo map of the area in question. No Flat Earther ever tried to argue that there are no hills or mountains on the Earth. Texas is relatively flat, but there are rolling hills everywhere.
I admit that I was hasty to conclude something without appreciating the terrain. But, by the same token, how can we say that the apparent horizontal flatness (I see that too!) is not hiding a curvature with a certain terrain configuration?
Answer #2. Though I don't see that at all in this photo, this is due to perspective, distance and the vanishing point/line. The same way that light poles appear to rise to the horizon/vanishing line while getting shorter and smaller. Clouds do the same thing as they disappear into the distance, but they shrink and move 'down' toward the horizon.
That wide white horizontal band on the background doesn't look to me as clouds vanishing due to perspective.
My points work better on the sea (as Tumeni pointed out), when I see a clear horizon on the sea I see the horizon abruptly changing into sky and clouds diving down faster than perspective. That of Texas is an interesting photo but maybe something more like frozen lakes give more clarity to the matter of flatness.
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Firstly, can you learn how to use the quote feature on here properly? It makes reading and replying to your messages really difficult. There's a Preview button so you can check how the post will look and you can edit your picture
Photos [that should show] 600 feet of drop have been presented. Photos [that should show] half a mile of drop have been presented. When they are, it is claimed that they do not exist and they show nothing.
I'm pretty sure no-one has claimed the photos don't exist. The initial image which kicked this thread off, the 600ft figure was stated before and as was explained previously that figure assumes a viewer height of 0. That's if you're looking from the ground and it also assumes a perfect sphere by the way, as in no terrain. And the photo is taken from up a hill. I already answered this point. From up a hill the drop is significantly less and you'd absolutely expect to see the distant buildings. And the point about curve left to right has been addressed too.
I cannot explain a video/photo that somebody else has made.
Why not? You're expecting us to do that with the photos you present.
BUT I will say, in my personal opinion, that it is very likely doctored (photo-chopped) because the building does not 'ever' appear to lean away from the viewer. Take a close look at the top of the building in each image. It is identical. Same height, same angle. Zero evidence of the building leaning away whatsoever. As it quite plainly should be were it sinking down around the other side of the "Frontal Curve" of the so-called round earth globe.
Once again you are failing to understand the scale of the earth. The earth is 24,900m in circumference. The furthest observer distance in that video is 30 miles. Simple maths will tell you that the angle the building is leaning away from you at that distance is less than half a degree. You would not be able to perceive that. FE people do often seem to struggle to understand the sheer size of the earth, when you understand how big it is it helps you understand what you "should" see in certain situations.
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I cannot explain a video/photo that somebody else has made.
Why not? You're expecting us to do that with the photos you present.
I present simple PHOTOGRAPHS. Not elaborate video constructions that have obviously been created using video editing software.
BUT I will say, in my personal opinion, that it is very likely doctored (photo-chopped) because the building does not 'ever' appear to lean away from the viewer. Take a close look at the top of the building in each image. It is identical. Same height, same angle. Zero evidence of the building leaning away whatsoever. As it quite plainly should be were it sinking down around the other side of the "Frontal Curve" of the so-called round earth globe.
Once again you are failing to understand the scale of the earth. The earth is 24,900m in circumference. The furthest observer distance in that video is 30 miles. Simple maths will tell you that the angle the building is leaning away from you at that distance is less than half a degree. You would not be able to perceive that. FE people do often seem to struggle to understand the sheer size of the earth, when you understand how big it is it helps you understand what you "should" see in certain situations.
Oh, less than half a degree? Really?
I'd love to see these simple maths you boast of. Please, enlighten us.
Because the distant city in the original post pic is 45 miles away from the camera. Let's just see how many fractions of a degree those buildings are SUPPOSED to be leaning away. We'll be waiting patiently for your presentation.
Since you've never 'seen' the Earth from space with your own naked eyes, I'm curious how it is you've convinced yourself that YOU have grasped the "sheer size of it" so much better than the common laymen who starve below your grandeur.
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As far as I understand in the wiki they criticize Soundly's photos of Pontchartrain because, they say, if curvature was indeed *so* apparent, our planet would be quite small. I like that comment because it acknowledges the fact that you cannot just watch and see a round earth.
They say that because they don't understand the apparent foreshortening that occurs between visible features at great distances when viewed through high magnification.
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Vanishing line...
https://www.youtube.com/watch?v=VFhhCYYkILw
The ship is on the horizon, not past it. If it were past it, it would be obscured from the bottom up, and no amount of zooming in would restore it to full un-obscured height.
Oh, less than half a degree? Really?
I'd love to see these simple maths you boast of. Please, enlighten us.
Do you ever research anything? Earth's circumference, 24,901 miles divided by 360 (degrees) equals 69.169 miles. This means a building will be leaning away from you 1 degree at about 69 miles (*I'll round down instead). You should be able to figure it out from there.
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Oh, less than half a degree? Really?
I'd love to see these simple maths you boast of. Please, enlighten us.
Because the distant city in the original post pic is 45 miles away from the camera. Let's just see how many fractions of a degree those buildings are SUPPOSED to be leaning away. We'll be waiting patiently for your presentation.
Since you've never 'seen' the Earth from space with your own naked eyes, I'm curious how it is you've convinced yourself that YOU have grasped the "sheer size of it" so much better than the common laymen who starve below your grandeur.
Simple geometry, simple maths. Using approximate textbook figures, 360*69 = 24,840 circumference. 69 miles corresponds to 1 degree, since 360 make up a sphere or circle. Anything nearer than 69 miles leans by less than one degree away from you.
(https://i.imgur.com/XKgaFgk.jpg)
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Ok, AllAroundtheWorld or Model 29 or Tumeni or iamcpc or whatever your user name is TODAY,
excuse my ignorance and my lacking research, since you caught me,
BUT it seems to me that you keep using these maths and geometric formulas for a PERFECT SPHERE.
And, hmmm, I'm pretty sure that your Lord DeGrasse Tyson, and all the other cronies, have admitted repeatedly that the Earth is NOT a Perfect Sphere.
(https://upload.wikimedia.org/wikipedia/commons/thumb/4/4a/Geoid_undulation_10k_scale.jpg/220px-Geoid_undulation_10k_scale.jpg)
SO,......yeah,.......that'd make your simple maths, there, completely faulty.
Seeing that you have NO idea what portion of the Earth is more elongated, or more curved for that matter, it's sufficient to say that you CANNOT use that formula at all.
If the Earth is more oblate around the equator, (which doesn't match up with heat patterns there due to Earth's tilt and lack of alignment with the sun) then that means North to South might be 'flatter' in some areas, but not in others, and that it'd be more curved East to West in some areas, but not in others.
(https://external-preview.redd.it/SsopTaTieFXrLNZhVCgxHvCkW6mIbIDoXmi_VY7t_4w.jpg?auto=webp&s=216ead95cbdb6b193767653d8b50f41beade45da)
Again, Round Earthers can't provide any plausible proof of any of their rebuttal arguments. Just more abstract theory, garbed formulas, faulty maths and faulty logic.
Bottom line:
Round Earth wouldn't just lose in a court of law, it'd get laughed out of the court room before a trial could even ensue, due to a complete lack of supporting evidence.
Face it: You have NO case.
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Oh, less than half a degree? Really?
I'd love to see these simple maths you boast of. Please, enlighten us.
I see Tumeni already has, but I spent a while on this so rather than wasting that time I'll post this anyway:
(https://i.ibb.co/pjG2yB1/Geometry.jpg)
So, we are at point A and we are looking at a building at point B.
If we are living on a ball and the building is perpendicular to the ground then it will follow the line BD and will be leaning away from us at an angle.
I've drawn a line EF, parallel to the line AC.
So the angle the building is leaning away from us is EBD which I've called 'x'.
I've also drawn the lines from A and B to C which represents the centre of the circle/earth.
The angle at which those two lines meet at C is ACB which I've called 'y'.
But because AC and EF are parallel you should be able to see that x = y.
So now all we need to know is the circumference of the circle and the distance between A and B.
The circumference of earth = 24900 miles
The distance AB for the Turning Torso building is 30 miles.
So 30 / 24900 gives you the %age of the circle which AB covers.
All you need to do then is multiply that by 360 to get how many degrees y is, and therefore x.
(30 / 24900) x 360 = 0.4337 degrees.
To prove my logic is sound, what if the building was a quarter of the way around the world?
24900 / 4 = 6225.
So what would the angle be?
(6225/24900) x 360 = 90 degrees
Which is what you'd expect, a quarter of a circle.
Because the distant city in the original post pic is 45 miles away from the camera. Let's just see how many fractions of a degree those buildings are SUPPOSED to be leaning away. We'll be waiting patiently for your presentation.
I'll leave that as an exercise for the reader now I've given you the formula although now I've given you the value for 30 miles you should be about do 45 in your head
(hint, it'll be 50% more, so times the 30 mile value by 1.5).
I'm not having a go but come on, dude, this really was simple maths. I've probably over-complicated things, Tumeni's explanation was simpler and better. The fact that it had to be explained to you shows that you really don't understand stuff as well as you think you do. I note that you are now floundering around saying "Aha, but the experts say it's not a perfect sphere!". Well you're right.
For example, the WGS84 datum identifies the longest diameter of an ellipse (semi-major axis) as 6,378,137.0 m. Next, the semi-minor axis is 6,356,752.3 m.
https://earthhow.com/shape-of-the-earth/
Divide these values into one another and you get an amount of oblateness of 1.0034. Less than one third of a percent.
It's small enough to mean the above maths is close enough that the error is very small.
As I said elsewhere, a common mistake people make is to think that because a model is imperfect it is not useful.
The earth is not a perfect sphere but for the purposes of the maths above it is close enough that the margin of error is tiny.
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Ok, AllAroundtheWorld or Model 29 or Tumeni or iamcpc or whatever your user name is TODAY,
excuse my ignorance and my lacking research, since you caught me,
BUT it seems to me that you keep using these maths and geometric formulas for a PERFECT SPHERE.
And, hmmm, I'm pretty sure that your Lord DeGrasse Tyson, and all the other cronies, have admitted repeatedly that the Earth is NOT a Perfect Sphere.
(https://upload.wikimedia.org/wikipedia/commons/thumb/4/4a/Geoid_undulation_10k_scale.jpg/220px-Geoid_undulation_10k_scale.jpg)
What you posted is an exaggerated gravity view of earth, 10000 scale factor.
Referring to the image you posted, "The geoid is an equipotential surface, and as we see on this image of what a lumpy surface Earth would be if we only considered gravity. This image is exaggerated, but you can see that gravity isn't consistent across the entire topographic surface of the Earth.”
https://www.e-education.psu.edu/geog862/node/1799
This is the gravity image to scale:
(https://upload.wikimedia.org/wikipedia/commons/4/48/Geoid_undulation_to_scale.jpg)
Coincidentally, from the same wiki page you got the exaggerated gravity image. You really do need to research more.
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Simple geometry, simple maths. Using textbook figures, equatorial circumference is 24,901 miles, polar 24,860.
The distance of an arc of one degree is therefore between 69.17 miles and 69.05 (24901/360 and 24860/360)
The building exactly 69 miles away leans by one degree, when rounded to the nearest degree.
To a few decimal places, it's between 0.99754 and 0.99928 degrees.
Not enough difference for even the keenest-eyed human to make out a building "leaning away from them" ....
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I'm not having a go but come on, dude, this really was simple maths. I've probably over-complicated things, Tumeni's explanation was simpler and better. The fact that it had to be explained to you shows that you really don't understand stuff as well as you think you do. I note that you are now floundering around saying "Aha, but the experts say it's not a perfect sphere!". Well you're right.
Floundering? Please.
More evidence has been presented in support of Flat Earth than anything Round Earthers have EVER dreamed up.
--Polaris (North Star) never, ever, moving from its place in our sky.
--All the constellations staying exactly the same throughout all of recorded history despite the four 'alleged' and unfathomable movements of the Earth through space.
--The '50 mile long' Panama canal that is officially 'perfectly level' from end to end.
--The Bedford Level experiments.
--Countless photos and videos of horizons many miles long with zero curve.
--Crepuscular Rays indicating the close proximity and small size of the sun-disproving much of the myths about space and our solar system, etc.
--Mercury and Venus being visible in our night sky when they're 'supposedly' closer to the sun than Earth making it impossible for them to be seen when the Earth is turned 'AWAY' from the sun (i.e. night time).
The list goes on and on. No need to have any expertise in all the silly mathematical equations and formulas for proving empirical things wrong using metaphysical concepts.
And, just like Tumeni's response, here,......
Simple geometry, simple maths. Using textbook figures, equatorial circumference is 24,901 miles, polar 24,860.
The distance of an arc of one degree is therefore between 69.17 miles and 69.05 (24901/360 and 24860/360)
The building exactly 69 miles away leans by one degree, when rounded to the nearest degree.
To a few decimal places, it's between 0.99754 and 0.99928 degrees.
Not enough difference for even the keenest-eyed human to make out a building "leaning away from them" ....
......it is ALWAYS the same answer: "It's round, it's curved, but you will NEVER see it, nor find ANY irrefutable proof of it, no matter what."
Round Earth's arguments and rebuttals are textbook and utterly predictable.
Again, I present Nikola Tesla's quote referring to Einstein's Theory of Relativity due to its profound relevance.
"Einstein's Relativity Work is a magnificent mathematical garb which fascinates, dazzles and makes people blind to the underlying errors. The theory is like a beggar clothed in purple whom ignorant people take for a king...its exponents are brilliant men but they're metaphysicists rather than scientists."
-Nikola Tesla
Metaphysics - Abstract theory with no basis in reality
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I'm not having a go but come on, dude, this really was simple maths. I've probably over-complicated things, Tumeni's explanation was simpler and better. The fact that it had to be explained to you shows that you really don't understand stuff as well as you think you do. I note that you are now floundering around saying "Aha, but the experts say it's not a perfect sphere!". Well you're right.
Floundering? Please.
More evidence has been presented in support of Flat Earth than anything Round Earthers have EVER dreamed up.
--Polaris (North Star) never, ever, moving from its place in our sky.
--All the constellations staying exactly the same throughout all of recorded history despite the four 'alleged' and unfathomable movements of the Earth through space.
--The '50 mile long' Panama canal that is officially 'perfectly level' from end to end.
--The Bedford Level experiments.
--Countless photos and videos of horizons many miles long with zero curve.
--Crepuscular Rays indicating the close proximity and small size of the sun-disproving much of the myths about space and our solar system, etc.
--Mercury and Venus being visible in our night sky when they're 'supposedly' closer to the sun than Earth making it impossible for them to be seen when the Earth is turned 'AWAY' from the sun (i.e. night time).
The list goes on and on. No need to have any expertise in all the silly mathematical equations and formulas for proving empirical things wrong using metaphysical concepts.
- It has, it does, and it will move relevant to us.
- They have, they do and they will move relevant to us. They are very, very far away.
- ‘Level’ to to the center of the earth, correct.
- The most famous and well documented of which, Hampden v Wallace, FE lost and Hampden kind of went mad.
- Earth is large, massive, in fact. Lots of images from on high show curvature. Google it.
- A near small sun runs you into a slew of issues; seasons, eclipses, equinoxes, sunsets/sunrises to name a few. Once you work those out, let us know.
- Huh?
And, just like Tumeni's response, here,......
Simple geometry, simple maths. Using textbook figures, equatorial circumference is 24,901 miles, polar 24,860.
The distance of an arc of one degree is therefore between 69.17 miles and 69.05 (24901/360 and 24860/360)
The building exactly 69 miles away leans by one degree, when rounded to the nearest degree.
To a few decimal places, it's between 0.99754 and 0.99928 degrees.
Not enough difference for even the keenest-eyed human to make out a building "leaning away from them" ....
......it is ALWAYS the same answer: "It's round, it's curved, but you will NEVER see it, nor find ANY irrefutable proof of it, no matter what."
Round Earth's arguments and rebuttals are textbook and utterly predictable.
Again, I present Nikola Tesla's quote referring to Einstein's Theory of Relativity due to its profound relevance.
"Einstein's Relativity Work is a magnificent mathematical garb which fascinates, dazzles and makes people blind to the underlying errors. The theory is like a beggar clothed in purple whom ignorant people take for a king...its exponents are brilliant men but they're metaphysicists rather than scientists."
-Nikola Tesla
Metaphysics - Abstract theory with no basis in reality
Lot’s of folks back in the day didn’t buy into relativity, some still don’t. So far, it has passed a century of experimentation.
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Mercury and Venus being visible in our night sky when they're 'supposedly' closer to the sun than Earth making it impossible for them to be seen when the Earth is turned 'AWAY' from the sun (i.e. night time).
Not a proof of Flat Earth, nor a disproof of Globe.
The observer can be close to the terminator line, and Mercury and/or Venus can be at up to 90 degrees from the Earth, relative to the Sun, not between Earth and Sun, allowing the observer an adequate sightline to see either.
Go model it in 3D to see what I mean. Or look on my YouTube for the explanatory videos. Channel name "Tumeni Hits"
If you want to talk photos, look at ANY photo of a ship, taken from a position onshore, with the observer higher than the highest point of the ship. If at any time you can see clear sky behind and beyond the top of the ship, you have the proof that the seas are Not Flat.
Like this;
(https://i.imgur.com/EP5H9Ty.jpg)
(https://i.imgur.com/fG5MK18.jpg?1)
In both cases, the observer is above the highest point of the ship. If the seas were flat, the sightline from observer SHOULD meet the water. But it does not. Conclusion; the seas CANNOT be flat. A flat sea is geometrically inconsistent with these observations.
Tesla quotes don't prove anything, especially since science has moved on since Tesla's day
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Floundering? Please.
Replying with over exaggerated models and resorting to the "you're an alt" accusation. Yep, that's floundering.
More evidence has been presented in support of Flat Earth than anything Round Earthers have EVER dreamed up.
--Polaris (North Star) never, ever, moving from its place in our sky.
--All the constellations staying exactly the same throughout all of recorded history despite the four 'alleged' and unfathomable movements of the Earth through space.
--The '50 mile long' Panama canal that is officially 'perfectly level' from end to end.
--The Bedford Level experiments.
--Countless photos and videos of horizons many miles long with zero curve.
--Crepuscular Rays indicating the close proximity and small size of the sun-disproving much of the myths about space and our solar system, etc.
--Mercury and Venus being visible in our night sky when they're 'supposedly' closer to the sun than Earth making it impossible for them to be seen when the Earth is turned 'AWAY' from the sun (i.e. night time).
The list goes on and on.
Polaris is really far away. Learn how scale works.
The constellation do change over time. Who said they didn't?
The Panama canal uses a series of locks to raise ships to the higher elevations of the lakes along its course, but each body of water is 'level' from one end to the other. That's how it works on a globe.
Bedford Level.... classic example of refraction courtesy of cooler air just above the water.
With each point of the horizon an equal distance from the camera, and not being very far from the surface, left to right curvature will not be very apparent.
Crepuscular rays also indicate a very small sun right in the tops of the trees from what I've photographed in the past. Or, maybe you should learn how perspective works.
Mercury can be visible about an hour before sunrise and an hour after sunset, and Venus visible much longer before sunrise and after sunset. Try looking at a scale diagram of the orbits of the inner planets and draw some lines from the orbit of Venus and see just where it can intersect Earth.
Sure, the list goes on and on with all sorts of stuff flat Earthers don't understand.
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Floundering? Please.
Yes. You're flailing.
You started this thread by saying "look at this photo, how can you see the distant buildings when there should be a 600ft drop and why can't you see any left to right curve?"
I patiently explained that the 600 foot figure was one given an observer height of 0. But the observer height isn't zero, the photo is taken from up a hill. So yes, you should be able to see the taller buildings in the distance.
And you shouldn't expect to see any curve left to right over 30 miles. I have explained that with a diagram. The earth is very big. See reply 13.
Then I presented the Turning Torso video to which your weak response was that it was probably doctored and that the buildings should be leaning away from us on a globe earth. It is honestly astonishing to me that you couldn't work out that the amount of lean at that distance should be less than half a degree and thus not something one should be able to discern.
You haven't addressed any of these explanations, you just move to a different topic.
Your entire argument in this thread is "the earth must be flat because if it was round then ..."
The problem is the "..." is consistently you not understanding stuff. Not being about to use a curve calculator and account for viewer height, not understanding that the scale of the earth means you wouldn't expect to see a curve in a 30 mile horizon left to right and the angle of lean away from you would be very small.
And now you've ignored all the explanations presented so far and just produced another list of "the earth must be flat because if it was round then ...". Again, the "..."s are either you not understanding things or just being wrong about things. Other people have tried to explain some of those things to you so I won't elaborate but if you need more detail on any of them I can try and explain more.
Your problem is you don't understand the globe earth heliocentric model well enough to debunk it.
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I think we can prove or disprove flat or round earth by measuring the distance between two objects,silos poles or cranes.The tops should be farther apart than the bottom.
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I think we can prove or disprove flat or round earth by measuring the distance between two objects,silos poles or cranes.The tops should be farther apart than the bottom.
In theory, yes.
In practice, it's not as easy as it sounds.
The Earth is very big, so you'll need your two objects to be a) very tall and b) very far apart. You'll obviously also need them to be perfectly vertical.
The towers of the Humber Bridge (https://en.wikipedia.org/wiki/Humber_Bridge), in its day the longest in the world, are 155m tall and 1410m apart. Still, the difference is "only" 36mm between the top and the bottom. Good luck measuring on your own that with the required accuracy.
There are easier, more practical ways to show the Earth's curvature. However, be aware that no matter how you show it, someone can and will come up with ad hoc arguments to claim it's wrong.
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I would be willing to buy a long range laser measuring device like a bosch GLR825 accuracy to1/25 of in.and some plumb bobs if anyone wants to see what happens.
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I would be willing to buy a long range laser measuring device like a bosch GLR825 accuracy to1/25 of in.and some plumb bobs if anyone wants to see what happens.
I'm afraid it wouldn't be good enough. At such short distances, the expected difference would probably be within measurement errors margins, making the experiment inconclusive.
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Sure, the list goes on and on with all sorts of stuff flat Earthers don't understand.
Hmmm....wow. You're right.
So, let's recap what Flat Earthers don't understand.
Here's your proof of round earth.
https://www.youtube.com/watch?v=MoK2BKj7QYk
And, let's see, you said "The furthest observer distance in that video is 30 miles.", right?
Ok, so that means the far right building, being 190 meters/623.6 feet tall, will disappear approximately 2/3 of its height, or 415 feet, from view at that distance (30 miles). With no obstructions, no geography in the way, viewer at roughly sea level, blah, blah, blah.
BUT,....on the same world, there's this........
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
With the buildings in the city (Dallas, Tx.) to the far left in the image being '45' miles from the observer. Those buildings having no greater height than 915 ft., the tallest.
Now:
-the photo was taken at a height of roughly 800-850 ft elevation;
-the elevation of the city in question being roughly 430 ft;
-and a rise in geography between the two at roughly 650 ft in the area of Colleyville. (see Topo map)
At 45 miles, there 'should' be a drop/curvature of 1,350 feet, on a PERFECT SPHERE.
So, tell me, and all the other poor Flat Earthers who just can't seem to grasp the complicated math and divine science, just HOW it's possible to see the ENTIRETY of those buildings.
The only portion of those buildings that is out of view is the bottom approximate 200 feet, which is explained by the ridge of geography in the Colleyville area on the Topo Map.
Here's the Topo Map to prove all those elevations.
https://en-us.topographic-map.com/maps/nct/Fort-Worth/
Do you really expect everyone to believe that the ~400 ft. difference in elevation between Dallas and the observer accounts for the plain visibility of that entire downtown city skyline when your own maths and CAD diagrams, etc. confirm that there 'should' be a THIRTEEN HUNDRED AND FIFTY foot drop in curvature at that distance?
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Your problem is you don't understand the globe earth heliocentric model well enough to debunk it.
While we're offering each other constructive criticism,
You don't understand it well enough to prove it.
I'm impressed you used the correct word, there, in reference to Round Earth.
Definition of debunk
transitive verb
: to expose the sham (see SHAM entry 1 sense 2) or falseness of
debunk a legend
https://www.merriam-webster.com/dictionary/debunk
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Re; Texas photo.
Let us presume the base level from which all the heights are measured is a flat plane.
Let us examine the geometry of the hills and buildings, how that would fit with that flat plane, and whether that geometry accords with what we see in the photo. Shall we?
-
-the photo was taken at a height of roughly 800-850 ft elevation
I'll take your word for that. But OK, it was certainly taken up a hill. Let's be nice and say 800 feet.
At 45 miles, there 'should' be a drop/curvature of 1,350 feet, on a PERFECT SPHERE.
So, tell me, and all the other poor Flat Earthers who just can't seem to grasp the complicated math and divine science, just HOW it's possible to see the ENTIRETY of those buildings.
Firstly, I don't think you are seeing the entirety of them. But you are certainly seeing most of them.
Because - and I have explained this before - you need to take viewer height into account
Look! When you use a curve calculator it asks for the viewer height:
With a height of 800ft - I've gone for the lower end of your estimate - you get these results:
(https://i.ibb.co/Dtg5kBk/Curve-Calculator.jpg)
From 800ft less than 72 feet is hidden so yes, if there's a building over 900 feet in height then you'd expect to see most of it.
The reason viewer height makes a difference should be obvious but I drew this if it helps:
(https://i.ibb.co/JKwb7xF/viewerheight.jpg)
The red line is you with a viewer height very close to the ground. You can only see the very top of the distant building.
The green line is you looking from up a hill. From that height you can see further over the curve and would see most of the building.
This is also why the horizon (which is simply you seeing the edge of the earth's curve, on a flat earth you wouldn't get a sharp horizon line) gets further away as your viewer height increases. Here's the curve calculator I used
https://dizzib.github.io/earth/curve-calc/
Do you really expect everyone to believe that the ~400 ft. difference in elevation between Dallas and the observer accounts for the plain visibility of that entire downtown city skyline when your own maths and CAD diagrams, etc. confirm that there 'should' be a THIRTEEN HUNDRED AND FIFTY foot drop in curvature at that distance?
No, I don't expect anyone to believe that.
And it is YOUR maths which predicts 1,350feet. Maths you have done incorrectly as I have explained - not for the first time in this thread.
I do understand it well enough to prove it if you are willing to actually listen. I mean, I can't "prove" it in the way I could prove a mathematical theorem, but I certainly understand the globe earth model to explain where you are going wrong. If you're going to keep making the same mistakes despite me doing so then I honestly don't know what I can do about that.
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Re; Texas photo.
Let us presume the base level from which all the heights are measured is a flat plane.
Let us examine the geometry of the hills and buildings, how that would fit with that flat plane, and whether that geometry accords with what we see in the photo. Shall we?
(https://i.imgur.com/Bv9PLBx.jpg)
Shall we calculate the angles, and see where the downward sightline should lead?
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Here's what I came up which jives with what AATW & Tumeni came up with. In short, it seems that Storm is not considering the Observer's height.
(https://i.imgur.com/BhDZwjC.jpg)
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The photographer says we can just see the ball of Reunion Tower above the intervening land.
The intervening land is lower than the photographer, yielding a downward sightline over the peak of that land. A downward sightline should reach the area of Reunion Tower somewhere below 650 feet, then. But it does not.
Why would that be?
The top of the tower is only just visible above the intervening land, and the top of the tower, at 991 feet, is above the photographer, who is presumed at 800 - 850.
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And it is YOUR maths which predicts 1,350feet. Maths you have done incorrectly as I have explained - not for the first time in this thread.
Let's check.
Distance squared, multiplied by 8 (inches of curve per mile), divided by 12 (to render feet).
45 miles (45x45)
8 inches per mile (multiplied by 8 )
12 inches in a foot (divided by 12)
What do you come up with?
Maybe it's your calculator. Or maybe you don't understand Round Earth MYTHOLOGY well enough to properly PROVE it.
(Hint: It's really hard to prove something that has no inherent TRUTH.)
Who's floundering?
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This is hilarious.
(https://i.imgur.com/Bv9PLBx.jpg)
Doesn't account for the 1,350 foot drop in curvature of the target building AT ALL.
I guess you hoped everyone would forget that part.
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This is hilarious. Didn't read my previous post AT ALL
Re; Texas photo.
Let us presume the base level from which all the heights are measured is a flat plane.
Let us examine the geometry of the hills and buildings, how that would fit with that flat plane, and whether that geometry accords with what we see in the photo. Shall we?
Why should my diagram "account for" curvature when it is explicitly based on a presumption of a flat plane?
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This, here, is absolute NONSENSE.
(https://i.imgur.com/BhDZwjC.jpg)
This is exactly the kind of computer generated fantasy abstract cartoon foolery that'd get laughed out of a court of law.
Without the magical math theorems and computer-aided Disney nonsense, there doesn't seem to be any other way to refute the Plane TRUTH of the Flat Earth we ALL inhabit.
BUSTED!!
(https://ohmy.disney.com/wp-content/uploads/2014/08/Disney-Characters-Grumpier-Than-Grumpy.png)
I feel like I'm debating one of these:
(https://cdn.clipart.email/0cf2bab6ff99fee7970da124922a10b2_clip-art-graphic-of-a-friendly-desktop-computer-cartoon-character-_450-450.jpeg)
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Why should my diagram "account for" curvature when it is explicitly based on a presumption of a flat plane?
Right. I mean Why address the gigantic hole in your argument--the missing curve?
We don't need you to explain WHY we can see the buildings perfectly on a Flat Plane. We know WHY we can see them--because it's a Flat Plane!
Your task is to answer WHY they're not hidden from view. Remember?!
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Why should my diagram "account for" curvature when it is explicitly based on a presumption of a flat plane?
Right. I mean Why address the gigantic hole in your argument--the missing curve?
We don't need you to explain WHY we can see the buildings perfectly on a Flat Plane. We know WHY we can see them--because it's a Flat Plane!
Your task is to answer WHY they're not hidden from view. Remember?!
This is hilarious.
You don't see that the observation in the photo is inconsistent with the sightlines which would apply, IF the plane below the various land features were presumed to be flat?
If it were flat, the photographer would be looking up from 850 to 991, and the 650 hill in the middle would not intrude on this sightline
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This, here, is absolute NONSENSE.
(https://i.imgur.com/BhDZwjC.jpg)
This is exactly the kind of computer generated fantasy abstract cartoon foolery that'd get laughed out of a court of law.
Without the magical math theorems and computer-aided Disney nonsense, there doesn't seem to be any other way to refute the Plane TRUTH of the Flat Earth we ALL inhabit.
BUSTED!!
Actually, it's your data. Is your data incorrect? And it seems you don't have a rebuttal of any sort. Just stomping your feet and exclaiming the world is flat - Well done. I'm sure that would play nicely in a court of law.
Why don't you think about the data you provided and come up with a cogent explanation and actually defend your position rather than just whining.
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Your task is to answer WHY they're not hidden from view. Remember?!
.. but they are (hidden from view)
The photographer explicitly states that we can only see the ball of the Reunion Tower, which stands 991 feet above (I presume) MSL.
The photographer is presumed to be at 800 - 850
Something is obscuring some 950 feet of the Reunion Tower, and the ground below it. How can it be the 650 feet of intervening ground? How did that grow so much it can intrude on the upward sightline from 800 to 991?
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Who's floundering?
It's still you, dude.
You are just completely ignoring the error you made, despite me explaining it.
I don't know if the images I made are showing up - lackey had a problem seeing them.
I've attached the relevant image. It demonstrates why viewer height makes a difference.
There is nothing wrong with your maths, as such. It's just that it assumes a viewer height of 0
(Represented by the red line in the image).
But the viewer height isn't 0. You said yourself, he's up a hill - you gave the figure 800-800ft.
From that height you'd be able to see most of the distant buildings, you'd only expect 70-odd feet to be hidden
That's why this curve calculator:
https://dizzib.github.io/earth/curve-calc/?d0=30&h0=10&unit=imperial
Asks you about the viewer height. The answer to that changes the amount of stuff hidden by the curve of the earth at a certain distance.
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Your task is to answer WHY they're not hidden from view. Remember?!
.. but they are (hidden from view)
The photographer explicitly states that we can only see the ball of the Reunion Tower, which stands 991 feet above (I presume) MSL.
The photographer is presumed to be at 800 - 850
Something is obscuring some 950 feet of the Reunion Tower, and the ground below it. How can it be the 650 feet of intervening ground? How did that grow so much it can intrude on the upward sightline from 800 to 991?
Again, let us presume a flat plane, as follows;
(https://i.imgur.com/tK9UKu0.jpg)
How can the hill of 6 units intrude on the sightline from 8 units up to 9?
-
Your task is to answer WHY they're not hidden from view. Remember?!
.. but they are (hidden from view)
The photographer explicitly states that we can only see the ball of the Reunion Tower, which stands 991 feet above (I presume) MSL.
The photographer is presumed to be at 800 - 850
Something is obscuring some 950 feet of the Reunion Tower, and the ground below it. How can it be the 650 feet of intervening ground? How did that grow so much it can intrude on the upward sightline from 800 to 991?
You are way off. The Reunion Tower is only 561 feet tall.
https://en.wikipedia.org/wiki/Reunion_Tower
Now, look at it this way:
Observer = ~800-850 ft
Colleyville terrain obstruction = ~600-650 ft
Dallas = ~450 ft
If you consider the elevation of Dallas, ~450 feet, as simply 0 feet, then consider the point of the observer to be ~400 feet above that and the elevation in between, the Colleyville area, to be ~200 feet above that zero point of elevation, then the photo looks "EXACTLY" as it should look.
Isn't that interesting?
You are purposely ignoring the roughly 100 feet of trees and buidlings above the elevation of the Colleyville 600-650 foot elevation. And the ball of the Reunion Tower is roughly 100 feet, top to bottom. So, it's actually exactly as it should be.
And for those who keep saying the observer should be looking UP, that is just a bunch more nonsense. You are looking at a city that is FORTY FIVE MILES AWAY! The tallest building is 915 feet, not nine THOUSAND feet.
The horizon is at eye-level. That is where the buildings are.
Now, since you are STILL deflecting from the obvious, here, I'd like to issue you a challenge.
Fire up your CAD diagram machine and create a diagram of your ROUND EARTH depiction of that photo, from the observer to the target city--using ACCURATE dimensions, showing ACCURATE curvature/drop for that distance (1350 feet), with the observer on the left and the target on the right - just like you've been showing with a FLAT ground line and make this one with a ROUND ground line. Wow us with your results.
Show us all how it's possible to see that city if the Earth is a ball.
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Mercury can be visible about an hour before sunrise and an hour after sunset, and Venus visible much longer before sunrise and after sunset. Try looking at a scale diagram of the orbits of the inner planets and draw some lines from the orbit of Venus and see just where it can intersect Earth.
Sure, the list goes on and on with all sorts of stuff flat Earthers don't understand.
No.
I'm sorry.
That just doesn't cut it.
(https://lh3.googleusercontent.com/proxy/dwUNc9sBDsgHHEjPrhtrdfrUhPxZXf1tC0yRHy6Ul9Z77yKi91wii8mzRNgfBa5uIFJiJghqUPD6G56QsElQmCjinpSGyCcfI5hJ43Kd8DU7KdSHul8y3sXHCfUUGC6M_ZdKbllnptxzH57jU1DvqK48MSt-1tE-qkJYEBKwu3MYviMixZ0sK9F_cOQT820oYLZPKPh5)
Venus moves at 78,341 mph on its orbit around the sun, while Earth moves 67,000 mph. Venus' orbit is only 67 million miles around the sun, while Earth's orbit is 92.96 million.
https://earthsky.org/astronomy-essentials/visible-planets-tonight-mars-jupiter-venus-saturn-mercury
Yet, in the southern U.S., Venus is visible at 45 degrees -- WAY up in the night sky, after sundown for at least two months STRAIGHT. Tonight, Venus is visible for as long FOUR HOURS after sunset. That means I can see it til ONE O'CLOCK in the morning.
How is this possible when the Earth is turned AWAY FROM THE SUN COMPLETELY at night time??
There's not a CAD-diagram spell you can CAST that'll explain THAT one away.
(https://i.pinimg.com/originals/a6/ab/c6/a6abc6e048962dcc8e837a4219a1a5ab.jpg)
Now, come with all your Ad Hominem attacks on my person, since you can't attack my arguments and let's all have it.
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That just doesn't cut it.
Venus' orbit is only 67 million miles around the sun, while Earth's orbit is 92.96 million.
https://earthsky.org/astronomy-essentials/visible-planets-tonight-mars-jupiter-venus-saturn-mercury
From that page- "Mercury’s greatest elongation in the morning sky happens on the same date as Venus’ greatest elongation in the evening sky: March 24, 2020. "
Greatest elongation. Do you know what that means? That means that if you draw a line between Earth and Sun, and another between Venus and Sun, these lines are perpendicular. Venus is at 90 degrees to the Earth, with the Sun at the angular point. From a discussion I previously had on this subject, at another time;
(https://i.imgur.com/safUBEA.jpg)
(https://i.imgur.com/9zWKCPu.jpg)
Yet, in the southern U.S., Venus is visible at 45 degrees -- WAY up in the night sky, after sundown for at least two months STRAIGHT. Tonight, Venus is visible for as long FOUR HOURS after sunset. That means I can see it til ONE O'CLOCK in the morning.
How is this possible when the Earth is turned AWAY FROM THE SUN COMPLETELY at night time??
First, all the illustrations on the page you quoted show it low in the sky, and nowhere does it state it to be at 45 degrees elevation above the horizon.
Don't you get a clue from the specific time that it can be seen, just after sunset? A whole hemisphere of the Earth may be on the opposite side of the Sun, but A - you're not at the farthest point from the Sun at all times, and B - you are not obliged to look away from the Sun
Plug the distance and elongation figures into a simple pair of diagrams, and ...
(https://i.imgur.com/CtmSOw4.jpg)
(https://i.imgur.com/mxG2ny6.jpg)
The observer can be 36 degrees beyond sunset or 36 degrees before sunrise and still see Venus.
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You are way off. The Reunion Tower is only 561 feet tall.
561 ft above the ground it stands on. So you need to add that ground level to the height of the tower for the tower's height above MSL. Mean Sea Level (or a similar reference level). Then consider the height of the observer above that level, and the height of the intervening hills, also measured from the same level. If all these are set out, on a flat plane, it can be seen that the hills inbetween cannot intrude on the observer's sightline to the Reunion Tower, such that only the ball of the tower can be seen. In fact, the whole tower should be visible if all this is set out on a flat plane... (EDIT correction; MOST OF the tower. The top 361 feet or so)
Now, look at it this way:
Observer = ~800-850 ft
Colleyville terrain obstruction = ~600-650 ft
Dallas = ~450 ft
If you consider the elevation of Dallas, ~450 feet, as simply 0 feet, then consider the point of the observer to be ~400 feet above that and the elevation in between, the Colleyville area, to be ~200 feet above that zero point of elevation, then the photo looks "EXACTLY" as it should look.
No, it does not, for the photographer states that most of the 561 feet of the Reunion Tower is HIDDEN BY THE HILL, and it is
561 ft ABOVE your zero reference level for Dallas.... so the photographer would be looking up from 450 to 561, with a 200 hill between - a hill which cannot intrude on his sightline
You are purposely ignoring the roughly 100 feet of trees and buidlings above the elevation of the Colleyville 600-650 foot elevation.
Just using the figures you quoted to start with. Now you want to change them? Makes no difference, anyway
And for those who keep saying the observer should be looking UP, that is just a bunch more nonsense. You are looking at a city that is FORTY FIVE MILES AWAY! The tallest building is 915 feet
So if the observer is at 800, and he's looking at something 915, he's looking up, isn't he? Makes no difference how far away, if it's all on a flat plane, and the city hasn't shrunk into the ground .....
He would be looking level if he looked at something of 800, and down if he looked at something of 700. No?
The horizon is at eye-level. That is where the buildings are.
Eye level is the observer's height. At least one of the buildings is 561 feet high, so the whole of it cannot be AT eye level ...
I'd like to issue you a challenge.
Fire up your CAD diagram machine and create a diagram of your ROUND EARTH depiction of that photo, from the observer to the target city--using ACCURATE dimensions, showing ACCURATE curvature/drop for that distance (1350 feet), with the observer on the left and the target on the right - just like you've been showing with a FLAT ground line and make this one with a ROUND ground line. Wow us with your results.
Don't have CAD, but ... why should I? I've shown that it's totally inconsistent with a flat plane. You tried to redefine the reference level for the plane, such that Dallas = zero level, but even that does not fit with the photo.
What IS the reference level for all the heights you have quoted for the land? Are they consistently defined from one level? If so, what is that reference level?
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Yet, in the southern U.S., Venus is visible at 45 degrees -- WAY up in the night sky, after sundown for at least two months STRAIGHT. Tonight, Venus is visible for as long FOUR HOURS after sunset. That means I can see it til ONE O'CLOCK in the morning.
Sundown -> Sunfar
Sunset -> Sunfar
Sunrise -> Sunnear
-
Sundown -> Sunfar
Sunset -> Sunfar
Sunrise -> Sunnear
Explain, please
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And for those who keep saying the observer should be looking UP, that is just a bunch more nonsense. You are looking at a city that is FORTY FIVE MILES AWAY! The tallest building is 915 feet, not nine THOUSAND feet.
What difference does the distance make? The city does not grow smaller with distance. The buildings have fixed heights, as does the land.
The observer is at 800 to 850
The land at the city is 430, with the Reunion Tower a further 561 above this. That makes a total of 991.
Is 991 greater than 850? If so, the observer has an upward sightline to the top of the Reunion Tower. Eye level must be at 850.
When last I checked, 991 was greater than 850.
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Sundown -> Sunfar
Sunset -> Sunfar
Sunrise -> Sunnear
Explain, please
FEs all agree that the Sun hovers above us, so as they correctly prefer the term "atmolayer" to "atmosphere", they should also avoid sundown, sunset, sunrise... all terms suggesting a vertical movement of the Sun. E.g., it's difficult for me to parse a FE reasoning like "the sun rise in the atmosphere, over the line of a flat horizon", it would be much better to read " the sun nears in the atmolayer, over a vanishing flat horizon"
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And for those who keep saying the observer should be looking UP, that is just a bunch more nonsense. You are looking at a city that is FORTY FIVE MILES AWAY! The tallest building is 915 feet, not nine THOUSAND feet.
What difference does the distance make? The city does not grow smaller with distance. The buildings have fixed heights, as does the land.
To further make the point, imagine you construct two towers of 100m tall, one at the ocean's edge, the other some distance out to sea. If you look from the top of one to the top of the other, the line of sight connecting the two is, assuming a perfectly flat sea, parallel to the sea. Each top is 100m above sea level
No matter what distance you put between the two towers, they do not shrink in size. They remain 100m tall. So your line of sight, regardless of distance between the two, is always parallel to the presumed flat surface of the sea.
(https://i.imgur.com/yTv0C7E.jpg)
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Yet, in the southern U.S., Venus is visible at 45 degrees -- WAY up in the night sky, after sundown for at least two months STRAIGHT. Tonight, Venus is visible for as long FOUR HOURS after sunset. That means I can see it til ONE O'CLOCK in the morning.
No, you can't. The sun sets much earlier than 9pm. I tried on https://www.heavens-above.com/skychart2.aspx with Austin, TX. With that data I get a sunset at about 7:45pm with Venus at about 46º and disappearing over the horizon around 11:20pm. Note that you could actually see Venus even higher in the sky earlier in the day if you know where to look: it's bright enough to be visible in broad daylight.
How is this possible when the Earth is turned AWAY FROM THE SUN COMPLETELY at night time??
cc8e837a4219a1a5ab.jpg[/img]
This is actually perfectly consistent: Venus is currently at its maximum elongation of about 47º with the Sun. Being able to see it with the Sun well below the horizon is therefore totally expected, exactly how long depends on your latitude. Seen from Earth, Venus more or less appears to follow the Sun on the ecliptic, which means it will rise and set a few hours later than the Sun.
Now, come with all your Ad Hominem attacks on my person, since you can't attack my arguments and let's all have it.
Your arguments are invalid.
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If the observer is at 800-850 feet, with an intervening hill between him and the target buildings of 650 feet, and we know the distance between them, then simple school-level trig will tell us the downward angle from observer to intervening hill, and by extension from that, the height H at which that sightline should meet the buildings - IF the underlying plane and reference level for all the heights is perfectly flat.
Like this;
(https://i.imgur.com/7po0465.jpg)
Storm, I invite you do the calcs for us ...
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Yet, in the southern U.S., Venus is visible at 45 degrees -- WAY up in the night sky, after sundown for at least two months STRAIGHT. Tonight, Venus is visible for as long FOUR HOURS after sunset. That means I can see it til ONE O'CLOCK in the morning.
Can you tell us what latitude you are at in the mid-northern latitudes, and what time sunset is for that particular latitude in March on a particular day that you claim to see Venus at 1am?
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I know this is an old topic but I just thought I would throw it out there the first time I read this post I looked up how to distance of curve of earth in a mile. I thin it was only 8 inch fall making 30 miles distans only 240 inches not 600 ft. Or 240 ft
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I know this is an old topic but I just thought I would throw it out there the first time I read this post I looked up how to distance of curve of earth in a mile. I thin it was only 8 inch fall making 30 miles distans only 240 inches not 600 ft. Or 240 ft
The correct answer is for 30 miles the drop will be 600.18 feet.
The formula for estimating the curvature of the Earth you are trying to remember is "Eight Inches Per Mile Squared" that is also used in the 1800's FE bible by Samuel Rowbotham so is used almost exclusively instead of the correct math by most of the Flat Earth community.
The problem is this is wrong. That 8 inches per mile squared function plots a parabola, not a circle. It's obvious when you note that the FE formula doesn't include PI, which is essential for calculating anything involving circles.
It gives close results for some distances but with very large values it becomes completely inaccurate. It's used a lot in back of the envelope engineering calculations because it's good enough for a few miles and works well as a rough estimate for more, bu eventually it stops becoming useful.
See the image below for how that formula works. It fits pretty well for a while, but then starts to go wrong and gets worse very fast.
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Yeah I apologise 2 all. My math was wrong. Still in using the calculated drop for 30 miles at the distance of the photo that drop is imperceptible being it falls equally in both directions. Perception of what ur eyes are seeing is up to ur brain. That should explain it. Better.
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I'd like to issue you a challenge.
Fire up your CAD diagram machine and create a diagram of your ROUND EARTH depiction of that photo, from the observer to the target city--using ACCURATE dimensions, showing ACCURATE curvature/drop for that distance (1350 feet), with the observer on the left and the target on the right - just like you've been showing with a FLAT ground line and make this one with a ROUND ground line. Wow us with your results.
Don't have CAD, but ... why should I? ...
Exactly. Why should you make it THAT obvious that you're completely wrong.
The fact is ... you can't. Plain and simple.
If you were to post the diagram that you've been challenged to create, everyone who saw it would see immediately that the very photograph this thread is all about 100% proves the Flat Earth.
The formula for estimating the curvature of the Earth you are trying to remember is "Eight Inches Per Mile Squared" that is also used in the 1800's FE bible by Samuel Rowbotham so is used almost exclusively instead of the correct math by most of the Flat Earth community.
The problem is this is wrong. That 8 inches per mile squared function plots a parabola, not a circle. It's obvious when you note that the FE formula doesn't include PI, which is essential for calculating anything involving circles.
It gives close results for some distances but with very large values it becomes completely inaccurate. It's used a lot in back of the envelope engineering calculations because it's good enough for a few miles and works well as a rough estimate for more, bu eventually it stops becoming useful.
See the image below for how that formula works. It fits pretty well for a while, but then starts to go wrong and gets worse very fast.
None of this makes any difference at all. It's all completely irrelevant because NEITHER formula works one single bit on the Earth that we all live on.
Not once has either curvature formula ever been proven to reflect the reality of the physical plane that we all inhabit.
Furthermore:
Though there were a handful of responses, not one person has tried to explain why Venus can be seen every single night in a row, for going on THREE MONTHS now, at 45 degrees in the sky after dark.
Not ONE diagram or explanation has even come close to explaining this tom-foolery.
Now, somebody will try to claim that Earth and Venus are completely in-sync with each other despite the fact that I've already posted their 'far-from-close' orbit speeds and orbit circumference distances.
So, ....... what's the explanation?
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I know this is an old topic but I just thought I would throw it out there the first time I read this post I looked up how to distance of curve of earth in a mile. I thin it was only 8 inch fall making 30 miles distans only 240 inches not 600 ft. Or 240 ft
The correct answer is for 30 miles the drop will be 600.18 feet.
The formula for estimating the curvature of the Earth you are trying to remember is "Eight Inches Per Mile Squared" that is also used in the 1800's FE bible by Samuel Rowbotham so is used almost exclusively instead of the correct math by most of the Flat Earth community.
The problem is this is wrong. That 8 inches per mile squared function plots a parabola, not a circle. It's obvious when you note that the FE formula doesn't include PI, which is essential for calculating anything involving circles.
It gives close results for some distances but with very large values it becomes completely inaccurate. It's used a lot in back of the envelope engineering calculations because it's good enough for a few miles and works well as a rough estimate for more, bu eventually it stops becoming useful.
See the image below for how that formula works. It fits pretty well for a while, but then starts to go wrong and gets worse very fast.
(https://i.imgur.com/6h1tkZo.jpg)
Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?
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Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?
What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.
Can you show me where I claimed he used it to measure over thousands of miles? I never said that.
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None of this makes any difference at all. It's all completely irrelevant because NEITHER formula works one single bit on the Earth that we all live on.
Not once has either curvature formula ever been proven to reflect the reality of the physical plane that we all inhabit.
The "curvature formula" is just basic geometry about the curvature of a sphere. Nothing special about it. I'm not sure how you can claim it's never been proven. The thousands of 'sinking ship' style observations match the expected curve. Bridges long enough have to have their supports at slight angles to account for it. Millions of pictures from space match the "curvature formula" by showing a sphere. All the thousands of satellites, several space stations and hundreds of people, and a few monkeys and dogs that were launched into orbit depend on the Earth being a sphere.
It's basic geometry.
Could you please show me your examples of where the "curvature formula" does not work?
Thanks.
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Though there were a handful of responses, not one person has tried to explain why Venus can be seen every single night in a row, for going on THREE MONTHS now, at 45 degrees in the sky after dark.
Not ONE diagram or explanation has even come close to explaining this tom-foolery.
Now, somebody will try to claim that Earth and Venus are completely in-sync with each other despite the fact that I've already posted their 'far-from-close' orbit speeds and orbit circumference distances.
So, ....... what's the explanation?
You can see Venus for about 9 months out of the year before becoming too hard to see due to appearing too close to the sun from our perspective, so seeing it for 3 months in a row is hardly surprising.
I'm not sure what you mean by 45 degrees. Venus will be at 0 degrees when it rises on the horizon, go up and then back down to 0 when it sets. Not knowing your location I can't give any specifics on how high that might currently be, and it will change nightly.
You asked for an explanation. This is a pretty good animation of the phases and why you see them.
https://earthsky.org/astronomy-essentials/tracking-venus-in-late-2019-and-early-20
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Could you please show me your examples of where the "curvature formula" does not work?
Thanks.
Oh, if you insist...
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
...it does a smash-up job of working here.
The city to the left of the image is 45 miles from the observer. Should be 1,350 feet below the horizon based on Rowbotham's curvature formula that he took directly from the scientific-stated (claimed) circumference of the Earth.
I'd ... say it's not workin there too well.
I'm not sure what you mean by 45 degrees. ... and it will change nightly.
Well, it means when somebody in the southern U.S. is perpendicular to the flat ground, the very bright star Venus is at a 45 degree angle from their person. If they look straight up at a 90 degree angle, the star is at 45 degrees.
And, no, it doesn't change each night. Not in any perceptible way. It's been at the same angle (roughly) for months now; every single night after dark. Brightest star in the sky; hard to miss.
Many of the explanations would only make sense if it were way down right on top of the horizon line at sundown. Anybody observing it will confirm that is not the case.
And it remains visible in the night sky for 3-4 hours after sundown.
The only way that is possible is for Venus to be somewhere on the opposite side of the Earth from the sun; it would need to have an orbit outside the Earth's orbit; meaning a further orbit from the sun than the Earth.
And, regardless of that particular detail, the FACT that it has been visible so high in the night sky, at roughly the SAME ANGLE, for months now, demands that it be in some sort of sync with the Earth. Which has already been proven to be impossible per their widely differing orbit speeds and orbit circumferences.
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Could you please show me your examples of where the "curvature formula" does not work?
Thanks.
Oh, if you insist...
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
...it does a smash-up job of working here.
The city to the left of the image is 45 miles from the observer. Should be 1,350 feet below the horizon based on Rowbotham's curvature formula that he took directly from the scientific-stated (claimed) circumference of the Earth.
I'd ... say it's not workin there too well.
We went through this already. Unless the data you provided was incorrect, here's the deal in reality. If you don't like the answer, then formulate with facts why that is the case, don't just stomp your feet.
This, here, is absolute NONSENSE.
(https://i.imgur.com/BhDZwjC.jpg)
This is exactly the kind of computer generated fantasy abstract cartoon foolery that'd get laughed out of a court of law.
Without the magical math theorems and computer-aided Disney nonsense, there doesn't seem to be any other way to refute the Plane TRUTH of the Flat Earth we ALL inhabit.
BUSTED!!
Actually, it's your data. Is your data incorrect? And it seems you don't have a rebuttal of any sort. Just stomping your feet and exclaiming the world is flat - Well done. I'm sure that would play nicely in a court of law.
Why don't you think about the data you provided and come up with a cogent explanation and actually defend your position rather than just whining.
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The city to the left of the image is 45 miles from the observer. Should be 1,350 feet below the horizon based on Rowbotham's curvature formula that he took directly from the scientific-stated (claimed) circumference of the Earth.
I'd ... say it's not workin there too well.
This has been debunked multiple times, I'm not going to duplicate it. Please re-read all the times it's been explained.
I'm not sure what you mean by 45 degrees. ... and it will change nightly.
Well, it means when somebody in the southern U.S. is perpendicular to the flat ground, the very bright star Venus is at a 45 degree angle from their person. If they look straight up at a 90 degree angle, the star is at 45 degrees.
And, no, it doesn't change each night. Not in any perceptible way. It's been at the same angle (roughly) for months now; every single night after dark. Brightest star in the sky; hard to miss.
Many of the explanations would only make sense if it were way down right on top of the horizon line at sundown. Anybody observing it will confirm that is not the case.
And it remains visible in the night sky for 3-4 hours after sundown.
The only way that is possible is for Venus to be somewhere on the opposite side of the Earth from the sun; it would need to have an orbit outside the Earth's orbit; meaning a further orbit from the sun than the Earth.
And, regardless of that particular detail, the FACT that it has been visible so high in the night sky, at roughly the SAME ANGLE, for months now, demands that it be in some sort of sync with the Earth. Which has already been proven to be impossible per their widely differing orbit speeds and orbit circumferences.
Do you have any proof that it reaches exactly 45% every night and doesn't change? Have you been taking a series photos all these months? All I see is you claiming a fact with zero evidence, just you stating it is. What date did you first observe Venus at 45 degrees? From what latitude? What is it at now?
You are making claims based on a misunderstanding of how angles and orbits work, I'm providing some more information to help below.
I also know you're wrong from my own observations, I do amateur astrophotography and have never once pointed my telescope where science predicted a planet to be and have it not be there. I've taken pictures of Venus, even got shots of it crossing the sun.
Please read the link I posted, and read the new one below. Here is another that explains it. It has a diagram of why we see Venus, and from what angles. It also explains why it will vary in height from 19% to 46%, but that also changes based on your position on the Earth.
https://in-the-sky.org/article.php?term=venus
Some quotes and images from that page for reference:
(https://in-the-sky.org/imagedump/book/innerPlanets_venusOvertake2.png)
The journey that Venus makes from the evening sky to the morning sky is much quicker than the reverse journey. This is because Venus's orbit is so close to our own, so the points of greatest elongation are much closer to inferior conjunction than to superior conjunction.
(https://in-the-sky.org/imagedump/innerplanets/inclination_ecliptic.png)
The inclination of the ecliptic to the horizon changes over the course of the year, affecting how high planets close to the Sun appear in the sky.
(https://in-the-sky.org/imagedump/innerplanets/greatest_elongations_soa_venus_east.png)
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Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?
What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.
Can you show me where I claimed he used it to measure over thousands of miles? I never said that.
Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine for measuring the distances in the book.
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Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?
What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.
Can you show me where I claimed he used it to measure over thousands of miles? I never said that.
Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine to measure the distances in the book.
Again, you are making up things I said and then arguing against them. Classic straw man. You again completely ignored my question so I will repeat.
Where did I state that Rowbotham was measuring thousands of miles. Quote it. <-- I'm asking you a question right here.
Otherwise, please admit that I never said that and stop arguing like I did.
I'm waiting for that quote.
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Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?
What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.
Can you show me where I claimed he used it to measure over thousands of miles? I never said that.
Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine to measure the distances in the book.
Again, you are making up things I said and then arguing against them. Classic straw man. You again completely ignored my question so I will repeat.
Where did I state that Rowbotham was measuring thousands of miles. Quote it. <-- I'm asking you a question right here.
Otherwise, please admit that I never said that and stop arguing like I did.
I'm waiting for that quote.
That's what I asked you. You said the equation is wrong because it becomes inaccurate over thousands of miles. Thousands of miles were not tested, however. Therefore the equation is correct.
Are you assuming that math did not exist in the 1800's and that this was not known?
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Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?
What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.
Can you show me where I claimed he used it to measure over thousands of miles? I never said that.
Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine to measure the distances in the book.
Again, you are making up things I said and then arguing against them. Classic straw man. You again completely ignored my question so I will repeat.
Where did I state that Rowbotham was measuring thousands of miles. Quote it. <-- I'm asking you a question right here.
Otherwise, please admit that I never said that and stop arguing like I did.
I'm waiting for that quote.
That's what I asked you. You said the equation is wrong because it becomes inaccurate over thousands of miles. Thousands of miles were not tested, however. Therefore the equation is correct.
Are you assuming that math did not exist in the 1800's and that this was not known?
I'm really sorry you are having so much trouble understanding this. I'm not sure how simpler I can make this, but will try, again.
I never mentioned Rowbotham's tests. I mentioned his use of an approximation. A parabola is not a circle. These are all indisputable facts.
You were the one that brought up his experiments. Nobody else mentioned it before that. Only you. It was NEVER mentioned before your comment. Can I make this any clearer?
Please, take a little time to read and understand this before replying. If you need something clarified, ask.
But if you insist on demanding references to things, you must quote where I said such things. Quote me, or please quit claiming I said it. Understand that you can only argue about things I said, not thing you imagined I said.
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I didn't say that you said anything. I asked: "Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?"
The answer is "No" and gives us everything we need to know about the applicability of the math used in this work.
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I didn't say that you said anything. I asked: "Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?"
The answer is "No" and gives us everything we need to know about the applicability of the math used in this work.
That is the very definition of a straw man argument. You ask a question that has nothing at all to do with the discussion, answer it, and claim victory.
Can you not see what you did here?
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This photograph of Dallas Fort Worth...
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
...has not been "gone thru", it has not been dealt with and it most certainly has not been "DEBUNKED."
Not one single person in this entire thread has explained away why the majority of the city of Dallas can be seen CLEARLY from 45 miles away, from just north of Fort Worth, which is 30 miles away from Dallas; when it should be 1,350 feet below the horizon.
Nobody has argued that math.
These are facts.
There is no more proof NEEDED, and there is no satisfactory proof that can be presented online to convince those who refuse to accept what their eyes plainly see.
The computer generated images from Stack of this...
(https://i.imgur.com/BhDZwjC.jpg)
...are absolutely worthless, and have zero basis in reality. Especially the last one.
JSS demands photos of all my observations over time, as well as further evidence, while JSS provides nothing concrete and expects this 'computer generated' image to serve as undebatable evidence of Venus' dynamics in relation to Earth.
(https://in-the-sky.org/imagedump/book/innerPlanets_venusOvertake2.png)
Tumeni refuses to create an image, very similar to this one...
(https://i.imgur.com/Bv9PLBx.jpg)
..., of the photograph in question, but on a curved plane -- which would prove the entire position of the OP and lay this entire thread to rest.
So, all that is going on here is people talking in circles.
The photograph that started this entire thread is YET to be explained, from a round earth perspective, simply because it cannot be explained away.
Once again, in a court of law, the ONLY piece of evidence that would hold up to legal scrutiny is the photograph that this thread is based upon.
Nothing else that has been presented to the contrary holds one single drop of water.
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This photograph of Dallas Fort Worth...
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
...has not been "gone thru", it has not been dealt with and it most certainly has not been "DEBUNKED."
Not one single person in this entire thread has explained away why the majority of the city of Dallas can be seen CLEARLY from 45 miles away, just north of Fort Worth, which is 30 miles away from Dallas; when it should be 1,350 feet below the horizon.
Nobody has argued that math.
These are facts.
There is no more proof NEEDED, and there is no satisfactory proof that can be presented online to convince those who refuse to accept what their eyes plainly see.
The computer generated images from Stack of this...
(https://i.imgur.com/BhDZwjC.jpg)
...are absolutely worthless, and have zero basis in reality. Especially the last one.
JSS demands photos of all my observations over time, as well as further evidence, while JSS provides nothing concrete and expects this 'computer generated' image to serve as undebatable evidence of Venus' dynamics in relation to Earth.
(https://in-the-sky.org/imagedump/book/innerPlanets_venusOvertake2.png)
Tumeni refuses to create an image, very similar to this one...
(https://i.imgur.com/Bv9PLBx.jpg)
..., of the photograph in question, but on a curved plane -- which would prove the entire position of the OP and lay this entire thread to rest.
So, all that is going on here is people talking in circles.
The photograph that started this entire thread is YET to be explained from a round earth perspective simply because it cannot be explained away.
Once again, in a court of law, the ONLY piece of evidence that would hold up to legal scrutiny is the photograph that this thread is based upon.
Nothing else that has been presented to the contrary holds one single drop of water.
You not excepting the explanations doesn't mean that explanations have not been provided or are incorrect. Again, just stomping your feet and saying "it can't be explained" with no concrete rebuttal as to why the explanations are not acceptable to you is the same as me saying, "The earth is a globe because I said so." In a court of law, the explanations provided would win out because they are evidence backed by science. You have simply offered an opinion - Opinions don't hold a lot of weight in a court of law.
Unless you have some evidence to provide with the properly calculated explanations, e.g., accounting for height of the observer versus height of object, I'm afraid you don't really have a case.
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JSS demands photos of all my observations over time, as well as further evidence, while JSS provides nothing concrete and expects this 'computer generated' image to serve as undebatable evidence of Venus' dynamics in relation to Earth.
All right, your question was, and I quote...
Though there were a handful of responses, not one person has tried to explain why Venus can be seen every single night in a row, for going on THREE MONTHS now, at 45 degrees in the sky after dark.
Not ONE diagram or explanation has even come close to explaining this tom-foolery.
So you asked me to explain the position of Venus in the sky. I'm sorry that all the diagrams confused you, but I gave you quite a detailed explanation of why we see Venus and how it moves. I even linked a video showing where Venus will appear and why. It clearly explained why we can see Venus for months at a time, and showed high high it will rise every night. I'm not sure why you are upset that I provided the information you asked for, and if you didn't understand it that's not anything I can help with. Learning is something you have to do on your own, I can't force you to understand anything, just show you the information.
I asked for your observations because you made a claim, and did not provide any evidence behind it. How can you expect to prove your case if you refuse to share anything backing it up? How for example, did you measure the 45 degrees? And did you record what time each of these sightings were?
I'm seriously not sure how to provide you with what you want. You asked for me to explain how Venus worked, yet got angry when I provided diagrams showing just that.
What is it exactly that you would accept as proof?
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(https://www.fortwortharchitecture.com/dfw-2005.jpg)
...it does a smash-up job of working here.
The city to the left of the image is 45 miles from the observer. Should be 1,350 feet below the horizon based on Rowbotham's curvature formula that he took directly from the scientific-stated (claimed) circumference of the Earth.
There are a number of mistakes you are making here. The biggest one is your formula is only valid for a viewpoint directly on the surface. The photographer was 500 feet higher than Dallas which makes the drop 200 feet instead, far less than the 921 foot tall building shown, so we should be seeing the top 700 feet of the tallest building.
So the picture does indeed, show what is expected, the tops of very tall buildings as seen from a high hill. Storm did a more detailed analysis, but I could get all the data required easily enough from the source of the photo that lists it's location.
If you're curious, the location is somewhere near that hill to the left in this Google Maps link.
https://www.google.com/maps/@32.7787272,-97.5498708,3a,54.6y,96.83h,88.35t/data=!3m6!1e1!3m4!1sYl9iHCzA0ozWUXL3pI6kGA!2e0!7i13312!8i6656
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None of what either of you keep saying makes the slightest dent in the validity of this photograph.
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
If you don't want to accept the Truth, that's a personal issue, but even if the observer were at TWO THOUSAND feet elevation, this image would NOT look like this on a Round Earth. You still wouldn't see this much of the buildings due to their leaning away at forty five miles distance; and they also wouldn't appear straight up and down like that either.
All this image proves...
(https://in-the-sky.org/imagedump/book/innerPlanets_venusOvertake2.png)
...is that Venus is very visible during the day time when THAT side of the Earth is toward the sun and Venus.
Look closely at the line that bisects the black Earth dot. That line demarcates the day/light side of Earth from the night/dark side. With that particular trajectory, shown by the dotted line, it is clear to see that Venus would only be visible just above the horizon at sundown. NOT at roughly 45 degrees, way up in the night sky, for hours after sundown.
And it doesn't matter if you want rock-solid proof of the 45 degree angle; anybody in the southern U.S. can simply look up into the sky for themselves tonight and see CLEARLY what I am claiming.
Absolutely ZERO proof needed here. They can go prove it to themselves.
So go ahead and create an image of the photograph like this one which has a flat plane surface...
(https://i.imgur.com/Bv9PLBx.jpg)
...and make the ground curve, accurate to the proper dimensions in that area of Texas, and we can all put this one to rest.
Because it's Game, set, match and Checkmate when that is presented.
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None of what either of you keep saying makes the slightest dent in the validity of this photograph.
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
If you don't want to accept the Truth, that's a personal issue, but even if the observer were at TWO THOUSAND feet elevation, this image would NOT look like this on a Round Earth. You still wouldn't see this much of the buildings due to their leaning away at forty five miles distance; and they also wouldn't appear straight up and down like that either.
Do you believe that elevation has no bearing on how far you can see? How much would the buildings be leaning 'away'? You're simply saying things without evidence. Again, just your opinion and if you don't think observer height makes a difference then even your opinion is half-baked. Back up your claims with data like everyone else has.
...and make the ground curve, accurate to the proper dimensions in that area of Texas, and we can all put this one to rest.
Because it's Game, set, match and Checkmate when that is presented.
We already have. You were checkmated pages ago.
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None of what either of you keep saying makes the slightest dent in the validity of this photograph.
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
If you don't want to accept the Truth, that's a personal issue, but even if the observer were at TWO THOUSAND feet elevation, this image would NOT look like this on a Round Earth. You still wouldn't see this much of the buildings due to their leaning away at forty five miles distance; and they also wouldn't appear straight up and down like that either.
It would look exactly like that, and they would look straight up and down too. The angle they would be leaning you be far too small to see at this distance with the naked eye.
This is basic geometry, but the scale we are working at is something the human mind has trouble imagining.
So go ahead and create an image of the photograph like this one which has a flat plane surface...
(https://i.imgur.com/Bv9PLBx.jpg)
...and make the ground curve, accurate to the proper dimensions in that area of Texas, and we can all put this one to rest.
You complained the last time I showed you diagrams and math, but I will try one more time.
I made some images to show what the curve looks like at that scale, as requested.
The first is the Earth and a rectangle that is 45 miles long, and 900 feet high. This is to scale, and the ground curves, but very very little in this picture because the Earth is extremely large. It has a circumference of 30,000 miles so 4 miles is just a tiny tiny part of it. If you look REALLY close you can see the curve of the earth just barely cuts into the rectangle in the center.
(https://i.imgur.com/NW7bCQ2.jpg)
Here is a closeup of the right side of the rectangle. You can see the curve just start to cut into the bottom, but if it wasn't for the rectangle above it you would be hard pressed to tell this from a straight line.
(https://i.imgur.com/eaTFV9y.jpg)
Just for scale, here is the Earth zoomed out. I made the rectangle MUCH taller just so it would show up, otherwise it would be an invisible sliver. You can see just how small a part of the Earth it covers. This is how much 45 miles is on the surface. This is why it's so hard to see it's curvature directly, it's just so large. It's why the lean you expect to see is invisible, at that scale there is hardly any separation between the angled of two upright structures./
(https://i.imgur.com/YEtMwKP.jpg)
I hope this helped someone understand. The Earth is big. Really big. So big it's hard to grasp it. If the Earth were small enough to show a visible curve from the surface, then it would be an extremely small world after all.
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That doesn't even come close to what you were asked to present.
I find it interesting that there are EIGHTEEN computer-generated images used in this entire thread, some more complex than others, yet nobody will present the simple image that they've been challenged to present.
This image is the closest image in this entire thread to duplicating the dimensions in the photograph.
(https://lh3.googleusercontent.com/proxy/sJPGM_vY_ECiNE_qXPzJM39rfodnGfEXbd2i_VCmqzr_D3hY83YZbbUJQfDIxIYgf5ak4YjiWCKCr-YKDyVSkeKXjU222Zg3hqMUs5Bpr0eM)
This image puts the observer at 'Object A' and the city of Dallas at 'Object B.'
Since Dallas' elevation is the lowest, being ~430 feet, it is the zero elevation line in this image, with the ~200 ft. middle ridge of Colleyville (Tangent Point) at ~200 feet, and the observer at ~400 feet. This image leaves out the high ~200 foot ridge of geography, that would be present at the Tangent Point; this addition would only allow the top THIRD (33%) of the tallest building in the image (915 ft.) to be visible.
As you can plainly see,...
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
...that is not the case.
You can even use this Curvature Calculator to prove this.
https://dizzib.github.io/earth/curve-calc/?d0=45&h0=400&unit=imperial
You claim I don't have a case?
You're right - 'cause this case is closed.
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Storm, this one you're using;
https://dizzib.github.io/earth/curve-calc/?d0=45&h0=400&unit=imperial
the graphic does not change no matter what numbers you put in, or at least it doesn't for me. I think it's just a static diagram to show what the numbers are for.
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That doesn't even come close to what you were asked to present.
Take this as a life-lesson, you can't demand strangers spoon-feed you everything, especially if you ignore all the explanations and work they already did. It's not my problem if you can only understand one specific diagram.
You made zero comments on what, if anything was wrong with my diagrams, you just dismissed them. That doesn't make me want to spend even more time helping explain.
Since Dallas' elevation is the lowest, being ~430 feet, it is the zero elevation line in this image, with the ~200 ft. middle ridge of Colleyville (Tangent Point) at ~200 feet, and the observer at ~400 feet. This image leaves out the high ~200 foot ridge of geography, that would be present at the Tangent Point; this addition would only allow the top THIRD (.33%) of the tallest building in the image (915 ft.) to be visible.
That is assuming your numbers are correct, but we are all guessing about the exact location of the photographer and nobody has provided a detailed topographic graph of the land between that spot and the city.
Dallas is at about 450ft. The observer, as best we can guess is about 850 feet.
That leaves the observer 400 feet above Dallas. Plugging that and 45 miles into the calculator says there should be a drop of 280 feet.
So 280 feet of the 561 foot tower is about 50% of the tower.
Even the photographer is saying you can only see the top of the building. As far as I can see, the photo matches both our numbers. Hardly the slam-dunk case you are claiming.
From the photographers post:
"Dallas zoomed in. You can see Cityplace on the far left and the ball of Reunion Tower peeking above the hill on the right. For the first time that I have taken this shot, it is clear enough to identify the buildings."
(https://i.imgur.com/1lewgsQ.png)
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This image is the closest image in this entire thread to duplicating the dimensions in the photograph.
(https://lh3.googleusercontent.com/proxy/sJPGM_vY_ECiNE_qXPzJM39rfodnGfEXbd2i_VCmqzr_D3hY83YZbbUJQfDIxIYgf5ak4YjiWCKCr-YKDyVSkeKXjU222Zg3hqMUs5Bpr0eM)
This image puts the observer at 'Object A' and the city of Dallas at 'Object B.'
Since Dallas' elevation is the lowest, being ~430 feet, it is the zero elevation line in this image, with the ~200 ft. middle ridge of Colleyville (Tangent Point) at ~200 feet, and the observer at ~400 feet. This image leaves out the high ~200 foot ridge of geography, that would be present at the Tangent Point; this addition would only allow the top THIRD (.33%) of the tallest building in the image (915 ft.) to be visible.
You said here that the observer was between 800-850':
(https://www.fortwortharchitecture.com/dfw-2005.jpg)
With the buildings in the city (Dallas, Tx.) to the far left in the image being '45' miles from the observer. Those buildings having no greater height than 915 ft., the tallest.
Now:
-the photo was taken at a height of roughly 800-850 ft elevation;
-the elevation of the city in question being roughly 430 ft;
-and a rise in geography between the two at roughly 650 ft in the area of Colleyville. (see Topo map)
Which is it? 400' or 800-850'?
And remember, the tallest building, BofA Ctr, is 921' tall. And located in Dallas sitting at approximately 482' MSL meaning that the building top is 1400' above MSL.
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Well, interestingly, the image I loaded (two different ways - and checked the preview) has been removed and when I attempt to pull it back up using the exact URL, Google displays a blank screen with a big: 'Error 403 (Forbidden)!!1' right in the middle of the screen.
Never seen THAT in my life.
Stack, I think JSS understands what I'm doing with the elevation equations. I'm "simplifying" by making the lowest of the 3 elevations a 'zero' point of elevation so that the only relevant information can easily be dealt with and more easily comprehended by all reading.
The sea level elevation makes absolutely NO DIFFERENCE to the equation and serves only to complicate, and max out the elevation numbers, to the fullest so that the entire concept is more difficult to understand.
I continue to simplify, you continue to complicate and obfuscate.
Who's shocked?
Oldest trick in the book when hiding the Truth is the ultimate objective.
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Well, interestingly, the image I loaded (two different ways - and checked the preview) has been removed and when I attempt to pull it back up using the exact URL, Google displays a blank screen with a big: 'Error 403 (Forbidden)!!1' right in the middle of the screen.
Never seen THAT in my life.
I don't know why your computer keeps having troubles with images but it doesn't matter, it's just a closeup of the image you posted. The picture works fine for me and everyone else too I'm sure, it's just an imgur link. You can also find it in the original artist post.
http://www.fortwortharchitecture.com/forum/index.php?showtopic=626&hl=\Dallas+&amp;+Fort+Worth+Skylines\
Remember at the start you were saying the city should be 1350 feet below the horizon and invisible, but that was not taking ~450ft of elevation into effect. The correct answer is that Dallas should be about 250 feet below the horizon, and that's just what the picture shows.
All our current guesswork says you should see between 50% and 33% of the tops of the buildings and that's what I'm seeing in the picture. We certainly are not seeing the street level, considering we only see the top of one of the more recognizable buildings.
Again, the picture is showing me exactly what we should expect to see, the top half of Dallas skyscrapers.
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Storm, this one you're using;
https://dizzib.github.io/earth/curve-calc/?d0=45&h0=400&unit=imperial
the graphic does not change no matter what numbers you put in, or at least it doesn't for me. I think it's just a static diagram to show what the numbers are for.
Thanks, Chris.
I did notice that, but the calculator does show that almost a third of the building (280.4869 ft.) is below line of sight even without the raised area of land in between.
When you use 400 for the 'Eye height' and 45 for the 'Target Distance', then add ~200 feet to the horizon line (Stack, that's 650 Colleyville elevation [without the extra height of buildings/structures, trees, etc.] minus 450 the elevation of Dallas.), that changes the angle of sight from the observer, at h0 on the diagram, and raises it quite significantly. Making only about the top 33% of the tallest building even plausibly visible.
It's simple and I think Stack understands it.
Make a curved diagram refuting it, Stack. Be sure to stay true to the elevations and FORGET about the ridiculous sea level inflated numbers used to confuse everyone.
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I don't know why your computer keeps having troubles with images but it doesn't matter, it's just a closeup of the image you posted. The picture works fine for me and everyone else too I'm sure, it's just an imgur link. You can also find it in the original artist post.
Remember at the start you were saying the city should be 1350 feet below the horizon and invisible, but that was not taking ~450ft of elevation into effect. The correct answer is that Dallas should be about 250 feet below the horizon, and that's just what the picture shows.
All our current guesswork says you should see between 50% and 33% of the tops of the buildings and that's what I'm seeing in the picture. We certainly are not seeing the street level, considering we only see the top of one of the more recognizable buildings.
Again, the picture is showing me exactly what we should expect to see, the top half of Dallas skyscrapers.
Nope. Wrong again.
The image I posted in reply #106 was a diagram with Object A, B and a Tangent point in the center on a curved DIAGRAM representing Earth.
It's missing right where there's a hole in my post with three lines of missing content.
What I've been saying is that the drop/curvature at 45 miles is 1,350 feet. That is correct.
What you are (not surprisingly) ignoring, and striving oh-so-desperately to get all to forget, is that there is a ridge of elevated geography in Colleyville that is ~650 feet high. With THAT present, my friend, on a CURVED EARTH, you would NOT be able to see what you are seeing in that photograph.
And, I might add, among the few curious onlookers who have followed this thread, that has been made abundantly clear.
But, I'll keep dancin with ya. Round and round we go, where this dead horse of a thread ends, nobody knows.
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What I've been saying is that the drop/curvature at 45 miles is 1,350 feet. That is correct.
Nope. Wrong again. That is the drop if you are viewing from the same elevation. Since the photographer was 450 feet higher, you have to enter 450 into that on-line calculator for the viewing height. When you do that, it says it's only a 241 foot drop. Since the tallest building is 900ft tall, it's not surprising we can see it, with the bottom section covered by the hill. Exactly what we see.
Remember, that calculator requires you to put the height of the observer in for accurate numbers. If you are high up, you can see further.
What you are (not surprisingly) ignoring, and striving oh-so-desperately to get all to forget, is that there is a ridge of elevated geography in Colleyville that is ~650 feet high. With THAT present, my friend, on a CURVED EARTH, you would NOT be able to see what you are seeing in that photograph.
I'm not forgetting anything. In fact, let me show you a topographical chart going from the area of the photographer directly to the base of the tallest building in that image, the Bank of America Plaza.
So looking this image, on a Flat Earth you could see the streets of the city, but in the picture the tallest building's base is behind the hill. It's below the hill because of the Earth's curve. If the earth was flat you would see the city streets. Even your own diagram shows that on a flat earth you would have a sight line to the street, and you can clearly see hills blocking the bottom of the city. Because of the drop.
(https://i.imgur.com/Bxl3tr5.png)
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You still wouldn't see this much of the buildings due to their leaning away at forty five miles distance; and they also wouldn't appear straight up and down like that either.
You really should learn how the globe works. At 70 miles the buildings would be leaning away about 1 degree. It would not be very noticeable.
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Any leaning would also be in the direction away from the camera.
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Can anyone show me a flat earth version of the milky way?? Are a the other planets at aswell? I'm confused
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(https://www.fortwortharchitecture.com/dfw-2005.jpg)
Anybody care to dispute, or disprove, the photo?
I am not a scientist, but I would like to offer this to the people who argue that the Earth looks flat to us:
The Earth is HUGE. Compared to the Earth, we are very insignificant. The curveature of the Earth is most likely to subtle for us to observe. So that is why to us, it would appear as though the Earth is 'flat'. We are not looking at the whole picture. We see our point of view only.
Also, if you look at the photograph with the city to the far right, you will notice that you cannot see at street view to the buildings out in the distance that are towards the center of the photo. There are no land features blocking the view in this photograph. You will even see how the mountains are slightly dipped below the horizion.
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So go ahead and create an image of the photograph like this one which has a flat plane surface...
IMG - see quoted post above
...and make the ground curve, accurate to the proper dimensions in that area of Texas, and we can all put this one to rest.
Well, it's my image, but if you look at the banlist, you'll see that I got banned at around the time you asked for this.
Back in a little while with what you asked for.
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I lack the space and the tools to do it, but my method would yield something like this;
(https://i.imgur.com/NeomF4v.jpg)
Take a steel tape measure or similar semi-rigid material, of length 6.371 metres. This corresponds to the mean radius of Earth of 6,371km, according to the textbooks.
Fix it at one end, such that you can rotate it around this end, and with a pen or pencil attached to the other, draw an arc on paper. This is a scaled-down Earth surface arc, such that 6,371km is scaled down to 6.371 metres. 1km = 1mm at this scale.
Measure out 7.24cm of this arc. This is equivalent to the 45 miles or so from the observer to the Reunion Tower.
72.4km = 72.4mm at scale = 7.24 cm
Draw the heights of observer and Reunion Tower at each end of this measured arc, and the intermediate hill at the appropriate distance from each, between them, and join them with a sightline (the blue line in my example).
Herein lies a problem at this scale, for each height (Ho = 259m, Hi = 198m, and Hr = 302m), is less than 1km, and hence at this scale will be less than 1mm. Not practical to draw them by hand.
So whilst I have the method, I lack the space and equipment to achieve this by hand. Would need a 63.71 metre tape measure to draw the arc
You can see how it works, though...
Perhaps JSS could use the digital drawing tool he used above to do this...?
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OK. If the creator of this string actually read all of the string from the source, the poster of the picture tells you almost exactly where he took the picture from--Near Tarrent County/Parker County lines and White settlement Road. This makes the distance from where the photo was taken and Dallas to be approximately 43.3 miles. Another thing the initiator of this string failed to mention is that the altitude of the photo's location and Dallas are strikingly different. This is not taken at sea level, kids.
I took 10 random elevations around the county line/White Settlement road intersection (ranging from 787 -925 feet) and took the lowest of the values (just to put myself at the greatest disadvantage possible) of 787 feet. Just to remind you, the highest elevation was 925 feet and I BET the photographer was a lot closer to that elevation since he said it was his "favorite hill", but lets go with the lower. Now the elevation of Dallas off of Google is 430 feet, but as you will see, it does not matter. I used an altitude website to estimate altitude for 10 random areas around downtown Dallas area and the values I got ranged from 400-479 with the average being 438 feet, so pretty dang close. Using zero refraction and an Earth curve calculator, the hidden distance from 787 feet high was a mere 52.2 feet--above sea level. Using standard refraction, this dropped to a measly 3 feet.
So can you see Dallas from that vantage point? On a clear day, YES, every day and twice on Sunday. So no, this is not proof of a flat earth. Next…