[Storm]

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?

[Storm]

This is hilarious.

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.

[Tumeni]

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?

[Storm]

This, here, is absolute NONSENSE.

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!!



I feel like I'm debating one of these:

[Storm]

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?!

[Tumeni]

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

[stack]

This, here, is absolute NONSENSE.

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.

[Tumeni]

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? 

[AATW]

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.

[Tumeni]

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;



How can the hill of 6 units intrude on the sightline from 8 units up to 9?

[Storm]

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.

[Storm]

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.



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.



Now, come with all your Ad Hominem attacks on my person, since you can't attack my arguments and let's all have it.

[Tumeni]

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;

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 ...






The observer can be 36 degrees beyond sunset or 36 degrees before sunrise and still see Venus.

[Tumeni]

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?

[Bikini Polaris]

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

[Tumeni]

Sundown -> Sunfar
Sunset -> Sunfar
Sunrise -> Sunnear

Explain, please

[Tumeni]

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.

[Bikini Polaris]

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"

[Tumeni]

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.

[GreatATuin]

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.