[stack]

This photograph of Dallas Fort Worth...



...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...



...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.



Tumeni refuses to create an image, very similar to this one...



..., 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.

[JSS]

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?

[JSS]

...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

[Storm]

None of what either of you keep saying makes the slightest dent in the validity of this photograph.



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...



...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...



...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.

[stack]

None of what either of you keep saying makes the slightest dent in the validity of this photograph.



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.

[JSS]

None of what either of you keep saying makes the slightest dent in the validity of this photograph.



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...



...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.




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.




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./



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.

[Storm]

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.



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,...



...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.

[ChrisTP]

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.

[JSS]

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."

[stack]

This image is the closest image in this entire thread to duplicating the dimensions in the photograph.



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':

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.

[Storm]

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.

[JSS]

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+&+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.

[Storm]

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.

[Storm]

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.

[JSS]

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.

[model 29]

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.

[ChrisTP]

Any leaning would also be in the direction away from the camera.

[Theglobe99]

Can anyone show me a flat earth version of the milky way?? Are a the other planets at aswell? I'm confused

[distance]

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.

[Tumeni]

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.