# The Flat Earth Society

## Flat Earth Discussion Boards => Flat Earth Community => Topic started by: Pongo on February 28, 2015, 12:51:03 AM

Title: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on February 28, 2015, 12:51:03 AM
So, I was doing some flat-earth research for a colleague and I ran across some facts about CN Tower (http://en.wikipedia.org/wiki/CN_Tower) that, among other things, boasted about being able to see ~100 miles on a clear day (fact 12) (http://gocanada.about.com/od/canadiancities1/qt/15factscntower.htm).  This seemed fishy to me so I dusted off some triganometry and used arc cosine to find out how far you can see from the observation deck before the "round-earth" curves away.

Given Values
Height of the observation deck: 1,135 feet (http://en.wikipedia.org/wiki/CN_Tower)
Generous height of eye level: 6 feet (http://en.wikipedia.org/wiki/Template:Average_height_around_the_world)
Elevation of Toronto: 249 feet (http://en.wikipedia.org/wiki/Toronto)

h = Height of observation deck + Generous  height of eye level + Elevation of Toronto
R = Mean radius of the Earth (lol)

h= 1390 feet
R= 20,903,520 feet

Formula
How far you can see if you were on a sphere = R*ACOS(R/(R+h))

How far you can see if you were on a sphere = 20925524.9*ACOS(20925524.9/(20925524.9+1390))

How far you can see if you were on a sphere = 241,057.18 feet

How far you can see if you were on a sphere =  45.65 miles

45.65 miles isn't 100 miles... not by half.  The only possible way you can see landmarks 100 miles away, such as Niagara Falls as they boast, is if the earth is flat.  They even make note to say that it can only be seen on a clear day!  Or, exactly what you would expect to see on a flat-earth.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 12:29:51 AM
Wrong formula. Please do try harder.

For example, on a clear day standing outside, and individual can readily see over 93 million miles. Do bring your shades though.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Ghost of V on March 01, 2015, 12:32:20 AM
Wrong formula. Please do try harder.

For example, on a clear day standing outside, and individual can readily see over 93 million miles. Do bring your shades though.

I've seen you make this claim before, but you have yet to support it.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 12:39:39 AM
Wrong formula. Please do try harder.

For example, on a clear day standing outside, and individual can readily see over 93 million miles. Do bring your shades though.

I've seen you make this claim before, but you have yet to support it.
Just go outside tomorrow, If it's a clear day, you'll see that Pongo's formula is obviously the wrong one--in the light of day.

Hint: you must consider the relative height between the observer's location and the target's. He did not.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Ghost of V on March 01, 2015, 12:40:33 AM
Wrong formula. Please do try harder.

For example, on a clear day standing outside, and individual can readily see over 93 million miles. Do bring your shades though.

I've seen you make this claim before, but you have yet to support it.
Just go outside tomorrow, If it's a clear day, you'll see that Pongo's formula is obviously the wrong one--in the light of day.

Hint: you must consider the relative height between the observer's location and the target's. He did not.

This is easy for you to say, but difficult for you to demonstrate. Why can't you provide more conclusive evidence?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 12:43:56 AM
Wrong formula. Please do try harder.

For example, on a clear day standing outside, and individual can readily see over 93 million miles. Do bring your shades though.

I've seen you make this claim before, but you have yet to support it.
Just go outside tomorrow, If it's a clear day, you'll see that Pongo's formula is obviously the wrong one--in the light of day.

Hint: you must consider the relative height between the observer's location and the target's. He did not.

This is easy for you to say, but difficult for you to demonstrate. Why can't you provide more conclusive evidence?
Sure I can. I'll be happy to provide clear and convincing evidence, but only after the OP clearly supports his use of his formula. He who claims first must demonstrate first, right?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Ghost of V on March 01, 2015, 12:46:02 AM
Sure I can. I'll be happy to provide clear and convincing evidence, but only after the OP clearly supports his use of his formula. He who claims first must demonstrate first, right?

Fair enough.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 12:50:56 AM
Wrong formula. Please do try harder.

No. I'm sure you found some formula on Wikipedia or somewhere that shows you how to calculate the distance to the horizon so if you see another method, you think it's wrong. It's not wrong (other than the fact that the earth is flat, of course) and your post does nothing but show your ignorance of math in general. Please allow this Subway napkin (Eat Fresh) to crush your self-esteem and show everyone how little you know about math, round-earth theory, and basic research skills.

Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 01, 2015, 01:06:08 AM
How was the Italian B.M.T?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 01:14:02 AM
How was the Italian B.M.T?

It was awful. They were out of mustard so I had to sub in honey mustard. Big mistake.

On topic though, I hate it when I have to explain to round-earthers how their own damn model works. All the while they call me the dumb one! There's enough irony to write a catchy 90's song.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 01:16:52 AM
Wrong formula. Please do try harder.

No. I'm sure you found some formula on Wikipedia or somewhere that shows you how to calculate the distance to the horizon so if you see another method, you think it's wrong. It's not wrong (other than the fact that the earth is flat, of course) and your post does nothing but show your ignorance of math in general. Please allow this Subway napkin (Eat Fresh) to crush your self-esteem and show everyone how little you know about math, round-earth theory, and basic research skills.

So why would you use a formula for the distance to the astronomical horizon when calculation is to determine how far you can see? Like I said "wrong formula". Please do try harder.

Here's another hint. Consider why the wrong formula you used does not result in 93 million miles when looking to the sun. It's the wrong formula to calculate how far you can see.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 01:24:31 AM
Are you trying to be pedantic about the horizon and distance of the sun on the round-earth model? That's not what this thread is about.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 01:53:12 AM
Are you trying to be pedantic about the horizon and distance of the sun on the round-earth model? That's not what this thread is about.
This thread is about how far you can see from the observation deck of the CN Tower. You claim that your formula determines that one can never in RET see more than about 40 miles from the deck.

You fail to consider the height of the target (and the height of the foundation of the tower), assuming incorrectly that the target is on the horizon. The example of the sun at 93 million miles destroys your claim as the formula does not predict that.You fail miserably. Please do try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 01, 2015, 02:24:50 AM
Are you trying to be pedantic about the horizon and distance of the sun on the round-earth model? That's not what this thread is about.
This thread is about how far you can see from the observation deck of the CN Tower. You claim that your formula determines that one can never in RET see more than about 40 miles from the deck.

You fail to consider the height of the target (and the height of the foundation of the tower), assuming incorrectly that the target is on the horizon. The example of the sun at 93 million miles destroys your claim as the formula does not predict that.You fail miserably. Please do try harder.

1) In FET the sun is not 93 million miles away
2) Since we're being pedantic, its not 93 miles away in RET either
3) Given that the Sun is not on the Earth, your argument is irrelevant. Honestly, you're the RE'er in the room. If you're so certain you're obviously right, why do you have to resort to such idiotic arguments?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 02:41:48 AM
Are you trying to be pedantic about the horizon and distance of the sun on the round-earth model? That's not what this thread is about.
This thread is about how far you can see from the observation deck of the CN Tower. You claim that your formula determines that one can never in RET see more than about 40 miles from the deck.

You fail to consider the height of the target (and the height of the foundation of the tower), assuming incorrectly that the target is on the horizon. The example of the sun at 93 million miles destroys your claim as the formula does not predict that.You fail miserably. Please do try harder.

1) In FET the sun is not 93 million miles away
2) Since we're being pedantic, its not 93 miles away in RET either
3) Given that the Sun is not on the Earth, your argument is irrelevant. Honestly, you're the RE'er in the room. If you're so certain you're obviously right, why do you have to resort to such idiotic arguments?
1) That's irrelevant.
2) Really? See:
The Sun is at an average distance of about 93,000,000 miles (150 million kilometers) away from Earth.
3) Why is the argument irrelevant? It points directly to Pongo's sophomoric error.

So why doesn't Pongo's RET formula predict that an observer on the deck of the CN Tower can see 93 million miles? Well, to answer my own question: because he's using the wrong formula, not considering the height of the target or of the Tower's foundation.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 02:45:28 AM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 02:49:36 AM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 02:55:59 AM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.

Again, because of the pedantry, what do you think h represents in the diagram?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 02:59:10 AM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.

Again, because of the pedantry, what do you think h represents in the diagram?
As you said, h is "Height of the observation deck = 1,118 feet", though you mis-typed it as 1128 in your calculation. You really do need to try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 03:06:07 AM
Actually, I mistyped it in the description, but good catch. I'll fix that. Reguardless, why do you suppose that the height of the observation deck does not include the tower's foundation  ???
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 03:12:41 AM
Actually, I mistyped it in the description, but good catch. I'll fix that. Reguardless, why do you suppose that the height of the observation deck does not include the tower's foundation  ???
Because of the international standard as chronicled here:
[For buildings] Height is measured from the sidewalk of the main entrance to the structural top of the building including penthouse and tower.

Oh, and while you're correcting your OP, please add the height of the observer, say 5 feet.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 03:36:27 AM
So you want the underground parts added?  Your arguments are insane. I understand that you can't see that because, by definition, you cannot see your own insanity. It's always reassuring when the round-earthers come up with crap like this as their primary arguments. Makes me believe that my dream of seeing flat-earth theory widely accepted in my lifetime is possible.

Also, I added 6 feet for the observer in the original calculation. I didn't mention it because it effects the outcome so slightly that only someone who is insane would bring it up.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 03:39:28 AM
So you want the underground parts added?  Your arguments are insane. I understand that you can't see that because, by definition, you cannot see your own insanity. It's always reassuring when the round-earthers come up with crap like this as their primary arguments. Makes me believe that my dream of seeing flat-earth theory widely accepted in my lifetime is possible.

Also, I added 6 feet for the observer in the original calculation. I didn't mention it because it effects the outcome so slightly that only someone who is insane would bring it up.
So here's the correct diagram. Now apply your math correctly and you'll see where your went wrong. Happy learning.

(http://i.imgur.com/X6ARxIk.png)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 03:48:56 AM
Why aren't you accounting for the foundation of the tower? I was told that was super necessary...
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 03:50:50 AM
Why aren't you accounting for the foundation of the tower? I was told that was super necessary...
I did. It's in the altitude of the base of the tower.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 05:42:27 PM
Of course, lol. Still, none of this makes the arccosine formula incorrect.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 01, 2015, 06:09:36 PM
Are you trying to be pedantic about the horizon and distance of the sun on the round-earth model? That's not what this thread is about.
This thread is about how far you can see from the observation deck of the CN Tower. You claim that your formula determines that one can never in RET see more than about 40 miles from the deck.

You fail to consider the height of the target (and the height of the foundation of the tower), assuming incorrectly that the target is on the horizon. The example of the sun at 93 million miles destroys your claim as the formula does not predict that.You fail miserably. Please do try harder.

1) In FET the sun is not 93 million miles away
2) Since we're being pedantic, its not 93 miles away in RET either
3) Given that the Sun is not on the Earth, your argument is irrelevant. Honestly, you're the RE'er in the room. If you're so certain you're obviously right, why do you have to resort to such idiotic arguments?
1) That's irrelevant.
2) Really? See:
The Sun is at an average distance of about 93,000,000 miles (150 million kilometers) away from Earth.
3) Why is the argument irrelevant? It points directly to Pongo's sophomoric error.

So why doesn't Pongo's RET formula predict that an observer on the deck of the CN Tower can see 93 million miles? Well, to answer my own question: because he's using the wrong formula, not considering the height of the target or of the Tower's foundation.

1) This entire argument is irrelevant
2) Nope. It's more like 92,956,000 miles away. Lrn2pedantic
3) No. It doesn't. Pongo's formula only applies to things on the Earth's surface. Obviously. It's about the curvature of the Earth hiding things, not some fundamental law of RET.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 07:22:07 PM
Of course, lol. Still, none of this makes the arccosine formula incorrect.
Again, you use that formula incorrectly. It does not predict how far one can see on an RET. Please do try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 07:28:40 PM
...
1) This entire argument is irrelevant
2) Nope. It's more like 92,956,000 miles away. Lrn2pedantic
3) No. It doesn't. Pongo's formula only applies to things on the Earth's surface. Obviously. It's about the curvature of the Earth hiding things, not some fundamental law of RET.
1) You're right. Your argument that FET distances matter in Pongo's proof about RET is indeed irrelevant. Thanks for agreeing.
2) Are you incorrectly claiming that no visible part of the sun is ever 93,000,000 miles from earth? Please do try harder.
3) So since the observation deck is not on the earth's surface, you obvious agree that he's used the wrong formula, right? Why would Pongo have 'h' be a height above the earth's surface, if you were right? Please do try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 09:11:27 PM
Oh, and just to clarify. The "Niagara Falls" that can be seen from the CN Tower's Observation Deck is just the taller buildings of the City of Niagara Falls. See: http://www.tripadvisor.com/LocationPhotoDirectLink-g155019-d155483-i32613999-CN_Tower-Toronto_Ontario.html (http://www.tripadvisor.com/LocationPhotoDirectLink-g155019-d155483-i32613999-CN_Tower-Toronto_Ontario.html)

This photo [of from the] CN Tower is courtesy of TripAdvisor

So T is actually much longer. Shall I assume that FEers have had enough debunking of their sophomoric proof of a FE in this thread? Please do try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 01, 2015, 09:32:24 PM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.

Again, because of the pedantry, what do you think h represents in the diagram?
As you said, h is "Height of the observation deck = 1,118 feet", though you mis-typed it as 1128 in your calculation. You really do need to try harder.
It doesn't matter. In order to see 100 miles on a flat earth, you need to be at a height of 6660 feet. In other words if you can see 100 miles from the top of the CN Tower, the earth is flat, case closed.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 09:36:28 PM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.

Again, because of the pedantry, what do you think h represents in the diagram?
As you said, h is "Height of the observation deck = 1,118 feet", though you mis-typed it as 1128 in your calculation. You really do need to try harder.
It doesn't matter. In order to see 100 miles on a flat earth, you need to be at a height of 6660 feet. In other words if you can see 100 miles from the top of the CN Tower, the earth is flat, case closed.
Wrong. Please read the thread and try harder. There's nothing in RET to say that you can't see 100 miles from a tower of 1000 feet or so. You must consider where you are and where you're looking to predict how far you can see.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 09:43:42 PM
You don't even plug numbers into your equation! For all you know, it could yield the same results, but we don't know because all you're capable of saying is, "wrong formula," and, "do try harder."  All you can really say is that the observer can see further because the sun is x million miles away (something you've yet to demonstrate). And that's not even the furthest thing you can see in the round-earth model when you take stars into account. When your primary arguments are "nuh uh," and, "the world is round because you can see the sun," you had best rethink your stance.

This, sun-distance argument is so asinine that you not only embarrass yourself, but all of round-earth theory as well.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 09:50:54 PM
You don't even plug numbers into your equation! For all you know, it could yield the same results, but we don't know because all you're capable of saying is, "wrong formula," and, "do try harder."  All you can really say is that the observer can see further because the sun is x million miles away (something you've yet to demonstrate). And that's not even the furthest thing you can see in the round-earth model when you take stars into account. When your primary arguments are "nuh uh," and, "the world is round because you can see the sun," you had best rethink your stance.

This, sun-distance argument is so asinine that you not only embarrass yourself, but all of round-earth theory as well.
I made only the claim that you're using the wrong formula, not that I had calculated an answer using the right formula. You, not I, made the claim in the OP that you've proven an FE. You have the responsibility to deal with the critique and to try harder to save your proof--or you could whine that I need to do your work for you like you are in your last post.

You build a straw man. I never said 93 million miles was as far as you can see on a RET earth. Please pay attention and try harder.

I have demonstrated that the observer on a RET earth can indeed see 93 million miles. Get over it. Your use of the formula for estimating the distance to the astronomical horizon over a still ocean to determine how far you can see on RET is obviously wrong. You're not even looking over an ocean.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 01, 2015, 09:58:00 PM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.

Again, because of the pedantry, what do you think h represents in the diagram?
As you said, h is "Height of the observation deck = 1,118 feet", though you mis-typed it as 1128 in your calculation. You really do need to try harder.
It doesn't matter. In order to see 100 miles on a flat earth, you need to be at a height of 6660 feet. In other words if you can see 100 miles from the top of the CN Tower, the earth is flat, case closed.
Wrong. Please read the thread and try harder. There's nothing in RET to say that you can't see 100 miles from a tower of 1000 feet or so. You must consider where you are and where you're looking to predict how far you can see.
Erm, you can use any calculator on the internet to tell you how high you must be and how far the horizon will be. Like the one below.
http://www.ringbell.co.uk/info/hdist.htm

Bellowing wrong in every thread doesn't make us wrong. It just means you can't work out these things for yourself.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 10:00:52 PM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.

Again, because of the pedantry, what do you think h represents in the diagram?
As you said, h is "Height of the observation deck = 1,118 feet", though you mis-typed it as 1128 in your calculation. You really do need to try harder.
It doesn't matter. In order to see 100 miles on a flat earth, you need to be at a height of 6660 feet. In other words if you can see 100 miles from the top of the CN Tower, the earth is flat, case closed.
Wrong. Please read the thread and try harder. There's nothing in RET to say that you can't see 100 miles from a tower of 1000 feet or so. You must consider where you are and where you're looking to predict how far you can see.
Erm, you can use any calculator on the internet to tell you how high you must be and how far the horizon will be. Like the one below.
http://www.ringbell.co.uk/info/hdist.htm

Bellowing wrong in every thread doesn't make us wrong. It just means you can't work out these things for yourself.
Again, calculating the distance to the astronomical horizon over a still ocean is not the way to determine how far one can see from the CN Tower Observation Deck. Please pay attention and try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 10:05:35 PM
Oh, and just to clarify. The "Niagara Falls" that can be seen from the CN Tower's Observation Deck is just the taller buildings of the City of Niagara Falls. See: http://www.tripadvisor.com/LocationPhotoDirectLink-g155019-d155483-i32613999-CN_Tower-Toronto_Ontario.html (http://www.tripadvisor.com/LocationPhotoDirectLink-g155019-d155483-i32613999-CN_Tower-Toronto_Ontario.html)

This photo [of from the] CN Tower is courtesy of TripAdvisor

So T is actually much longer. Shall I assume that FEers have had enough debunking of their sophomoric proof of a FE in this thread? Please do try harder.

Now your argument is, "because I can see Niagra Falls, then d must be further than 40 miles because the earth is round and I could only see it if d were further than 40 miles."  Shall I pick apart your logic there or can you see the flaws for yourself?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 01, 2015, 10:12:14 PM
Because you're being ultra-pedantic, can you describe what you mean by "foundation?"
The foundation, the base, of the CN Tower is above sea level thus R is not just 20,925,524.9 feet, as you incorrectly stated. The height of that side of the triangle is more than R+h. You really do need to try harder.

Again, because of the pedantry, what do you think h represents in the diagram?
As you said, h is "Height of the observation deck = 1,118 feet", though you mis-typed it as 1128 in your calculation. You really do need to try harder.
It doesn't matter. In order to see 100 miles on a flat earth, you need to be at a height of 6660 feet. In other words if you can see 100 miles from the top of the CN Tower, the earth is flat, case closed.
Wrong. Please read the thread and try harder. There's nothing in RET to say that you can't see 100 miles from a tower of 1000 feet or so. You must consider where you are and where you're looking to predict how far you can see.
Erm, you can use any calculator on the internet to tell you how high you must be and how far the horizon will be. Like the one below.
http://www.ringbell.co.uk/info/hdist.htm

Bellowing wrong in every thread doesn't make us wrong. It just means you can't work out these things for yourself.
Again, calculating the distance to the astronomical horizon over a still ocean is not the way to determine how far one can see from the CN Tower Observation Deck. Please pay attention and try harder.
At this point you are offering nothing. You are just whinging. We used round earth maths and the facts don't match the theory. Its all over, FET comes out on top again.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 01, 2015, 10:13:17 PM
If you solve for h in my formula and allow d to be 100, as round-earthers claim, you arrive at the same 6660 foot height required that Thork found in his online calculator. So I guess I'm wrong, the formula is wrong, the internet is wrong, trigonometry is wrong, and only Gulliver is right; all alone laughing at us lunatics.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 10:27:20 PM
If you solve for h in my formula and allow d to be 100, as round-earthers claim, you arrive at the same 6660 foot height required that Thork found in his online calculator. So I guess I'm wrong, the formula is wrong, the internet is wrong, trigonometry is wrong, and only Gulliver is right; all alone laughing at us lunatics.
Again, you are wrong. The formula does not calculate how far you can see. It's quite obvious.

For the confused... Consider two towers each 1000 feet high on the near shore of a still ocean separated by 50 miles. Does the astronomical horizon formula that Pongo and Thork used consider that the height of both towers which makes it possible to see the top of the other from each? No. both Pongo and Thork fail, again.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 01, 2015, 10:31:35 PM
If you solve for h in my formula and allow d to be 100, as round-earthers claim, you arrive at the same 6660 foot height required that Thork found in his online calculator. So I guess I'm wrong, the formula is wrong, the internet is wrong, trigonometry is wrong, and only Gulliver is right; all alone laughing at us lunatics.
Again, you are wrong. The formula does not calculate how far you can see. It's quite obvious.

For the confused... Consider two towers each 1000 feet high on the near shore of a still ocean separated by 50 miles. Does the astronomical horizon formula that Pongo and Thork used consider that the height of both towers which makes it possible to see the top of the other from each? No. both Pongo and Thork fail, again.
Are you saying there are two CN Towers? ::)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 01, 2015, 10:36:18 PM
If you solve for h in my formula and allow d to be 100, as round-earthers claim, you arrive at the same 6660 foot height required that Thork found in his online calculator. So I guess I'm wrong, the formula is wrong, the internet is wrong, trigonometry is wrong, and only Gulliver is right; all alone laughing at us lunatics.
Again, you are wrong. The formula does not calculate how far you can see. It's quite obvious.

For the confused... Consider two towers each 1000 feet high on the near shore of a still ocean separated by 50 miles. Does the astronomical horizon formula that Pongo and Thork used consider that the height of both towers which makes it possible to see the top of the other from each? No. both Pongo and Thork fail, again.
Are you saying there are two CN Towers? ::)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 01, 2015, 10:43:38 PM
If you solve for h in my formula and allow d to be 100, as round-earthers claim, you arrive at the same 6660 foot height required that Thork found in his online calculator. So I guess I'm wrong, the formula is wrong, the internet is wrong, trigonometry is wrong, and only Gulliver is right; all alone laughing at us lunatics.
Again, you are wrong. The formula does not calculate how far you can see. It's quite obvious.

For the confused... Consider two towers each 1000 feet high on the near shore of a still ocean separated by 50 miles. Does the astronomical horizon formula that Pongo and Thork used consider that the height of both towers which makes it possible to see the top of the other from each? No. both Pongo and Thork fail, again.
Are you saying there are two CN Towers? ::)
Why would I consider it? If you are on the CN Tower, you cannot see for 100 miles on a round earth because there isn't another hypothetical tower over 4000ft high within 100 miles to look at.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 02, 2015, 01:01:29 AM
... there isn't another hypothetical tower over 4000ft high within 100 miles to look at.
To the contrary, by definition, there are hypothetical towers of every height everywhere.

Quote from: http://www.merriam-webster.com/dictionary/hypothetical
: involving or based on a suggested idea or theory : involving or based on a hypothesis

: not real : imagined as an example
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 02, 2015, 05:24:35 AM
...
1) This entire argument is irrelevant
2) Nope. It's more like 92,956,000 miles away. Lrn2pedantic
3) No. It doesn't. Pongo's formula only applies to things on the Earth's surface. Obviously. It's about the curvature of the Earth hiding things, not some fundamental law of RET.
1) You're right. Your argument that FET distances matter in Pongo's proof about RET is indeed irrelevant. Thanks for agreeing.
2) Are you incorrectly claiming that no visible part of the sun is ever 93,000,000 miles from earth? Please do try harder.
3) So since the observation deck is not on the earth's surface, you obvious agree that he's used the wrong formula, right? Why would Pongo have 'h' be a height above the earth's surface, if you were right? Please do try harder.

The CN Tower levitates above the Earth? I had no idea. Damn, I need to visit Seattle some time. That must be a sight to behold.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 02, 2015, 06:22:45 AM
...
1) This entire argument is irrelevant
2) Nope. It's more like 92,956,000 miles away. Lrn2pedantic
3) No. It doesn't. Pongo's formula only applies to things on the Earth's surface. Obviously. It's about the curvature of the Earth hiding things, not some fundamental law of RET.
1) You're right. Your argument that FET distances matter in Pongo's proof about RET is indeed irrelevant. Thanks for agreeing.
2) Are you incorrectly claiming that no visible part of the sun is ever 93,000,000 miles from earth? Please do try harder.
3) So since the observation deck is not on the earth's surface, you obvious agree that he's used the wrong formula, right? Why would Pongo have 'h' be a height above the earth's surface, if you were right? Please do try harder.

The CN Tower levitates above the Earth? I had no idea. Damn, I need to visit Seattle some time. That must be a sight to behold.
You seen to be really confused. No, the CN Tower does not levitate above the earth. Why would you visit Seattle to see a Toronto landmark?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 02, 2015, 11:36:35 AM
Despite your appeal to create a scenario that doesn't exist, it isn't going to prove anything. This is what the Zetetic method is all about. If there are any huge Towers within 100 miles of the CN Tower, we'd be delighted to discuss them with you. Being as there aren't, it cannot be possible to see 100 miles on a round earth from the CN Tower, so Pongo's assertion they are making flat earth claims stands.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 02, 2015, 01:31:30 PM
Despite your appeal to create a scenario that doesn't exist, it isn't going to prove anything. This is what the Zetetic method is all about. If there are any huge Towers within 100 miles of the CN Tower, we'd be delighted to discuss them with you. Being as there aren't, it cannot be possible to see 100 miles on a round earth from the CN Tower, so Pongo's assertion they are making flat earth claims stands.
The Skylon Tower in Niagara Falls, Ontario is 520 feet tall and boasts quite the view as well.  This is not to mention that the city of Niagara Falls is on an escarpment, so the observation deck of the Skylon Tower is about 775 feet above the bottom of the falls.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 02, 2015, 03:08:08 PM
...
1) This entire argument is irrelevant
2) Nope. It's more like 92,956,000 miles away. Lrn2pedantic
3) No. It doesn't. Pongo's formula only applies to things on the Earth's surface. Obviously. It's about the curvature of the Earth hiding things, not some fundamental law of RET.
1) You're right. Your argument that FET distances matter in Pongo's proof about RET is indeed irrelevant. Thanks for agreeing.
2) Are you incorrectly claiming that no visible part of the sun is ever 93,000,000 miles from earth? Please do try harder.
3) So since the observation deck is not on the earth's surface, you obvious agree that he's used the wrong formula, right? Why would Pongo have 'h' be a height above the earth's surface, if you were right? Please do try harder.

The CN Tower levitates above the Earth? I had no idea. Damn, I need to visit Seattle some time. That must be a sight to behold.
You seen to be really confused. No, the CN Tower does not levitate above the earth. Why would you visit Seattle to see a Toronto landmark?

Well, if it's in space I figured that it would be easiest to see from the space needle. I assume they don't offer space flights up there to the public.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 02, 2015, 04:15:07 PM
...
1) This entire argument is irrelevant
2) Nope. It's more like 92,956,000 miles away. Lrn2pedantic
3) No. It doesn't. Pongo's formula only applies to things on the Earth's surface. Obviously. It's about the curvature of the Earth hiding things, not some fundamental law of RET.
1) You're right. Your argument that FET distances matter in Pongo's proof about RET is indeed irrelevant. Thanks for agreeing.
2) Are you incorrectly claiming that no visible part of the sun is ever 93,000,000 miles from earth? Please do try harder.
3) So since the observation deck is not on the earth's surface, you obvious agree that he's used the wrong formula, right? Why would Pongo have 'h' be a height above the earth's surface, if you were right? Please do try harder.

The CN Tower levitates above the Earth? I had no idea. Damn, I need to visit Seattle some time. That must be a sight to behold.
You seen to be really confused. No, the CN Tower does not levitate above the earth. Why would you visit Seattle to see a Toronto landmark?

Well, if it's in space I figured that it would be easiest to see from the space needle. I assume they don't offer space flights up there to the public.
You remain confused. Why do you infer that it's in space? How did you figure that it would be easiest to see from the Space Needle? You really do need to try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 02, 2015, 04:24:30 PM
Despite your appeal to create a scenario that doesn't exist, it isn't going to prove anything. This is what the Zetetic method is all about. If there are any huge Towers within 100 miles of the CN Tower, we'd be delighted to discuss them with you. Being as there aren't, it cannot be possible to see 100 miles on a round earth from the CN Tower, so Pongo's assertion they are making flat earth claims stands.
The Skylon Tower in Niagara Falls, Ontario is 520 feet tall and boasts quite the view as well.  This is not to mention that the city of Niagara Falls is on an escarpment, so the observation deck of the Skylon Tower is about 775 feet above the bottom of the falls.
::)

And can you see the Skylon Tower from the CN Tower? Or are you just trying to make this thread go 20 pages to bury the facts as usual?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 02, 2015, 04:56:26 PM
Despite your appeal to create a scenario that doesn't exist, it isn't going to prove anything. This is what the Zetetic method is all about. If there are any huge Towers within 100 miles of the CN Tower, we'd be delighted to discuss them with you. Being as there aren't, it cannot be possible to see 100 miles on a round earth from the CN Tower, so Pongo's assertion they are making flat earth claims stands.
The Skylon Tower in Niagara Falls, Ontario is 520 feet tall and boasts quite the view as well.  This is not to mention that the city of Niagara Falls is on an escarpment, so the observation deck of the Skylon Tower is about 775 feet above the bottom of the falls.
::)

And can you see the Skylon Tower from the CN Tower? Or are you just trying to make this thread go 20 pages to bury the facts as usual?
Why don't you try Google before asking inane questions? See http://commons.wikimedia.org/wiki/File:CN_Tower_as_seen_from_skylon_tower,_Niagara_Falls.jpg (http://commons.wikimedia.org/wiki/File:CN_Tower_as_seen_from_skylon_tower,_Niagara_Falls.jpg)

(http://i.imgur.com/3bXu928.jpg)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 02, 2015, 05:11:01 PM
Despite your appeal to create a scenario that doesn't exist, it isn't going to prove anything. This is what the Zetetic method is all about. If there are any huge Towers within 100 miles of the CN Tower, we'd be delighted to discuss them with you. Being as there aren't, it cannot be possible to see 100 miles on a round earth from the CN Tower, so Pongo's assertion they are making flat earth claims stands.
The Skylon Tower in Niagara Falls, Ontario is 520 feet tall and boasts quite the view as well.  This is not to mention that the city of Niagara Falls is on an escarpment, so the observation deck of the Skylon Tower is about 775 feet above the bottom of the falls.
::)

And can you see the Skylon Tower from the CN Tower? Or are you just trying to make this thread go 20 pages to bury the facts as usual?
Why don't you try Google before asking inane questions? See http://commons.wikimedia.org/wiki/File:CN_Tower_as_seen_from_skylon_tower,_Niagara_Falls.jpg (http://commons.wikimedia.org/wiki/File:CN_Tower_as_seen_from_skylon_tower,_Niagara_Falls.jpg)

(http://i.imgur.com/3bXu928.jpg)
And how far are the two apart? The answer is 30 miles. It proves nothing. Another huge red herring from Markjo as I pointed out at the beginning. We need a huge tower 100 miles away that can be seen. They claim you can see 100 miles. What can you see 100 miles away? ::)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 02, 2015, 05:33:22 PM
And how far are the two apart? The answer is 30 miles. It proves nothing. Another huge red herring from Markjo as I pointed out at the beginning. We need a huge tower 100 miles away that can be seen. They claim you can see 100 miles. What can you see 100 miles away? ::)
No, it's over 40 miles. Why did you ask the question if you thought that answer wold be a "huge red herring"?

(http://i.imgur.com/DIJkJ2q.png)

Oh, and you can see the sun 93 million miles away.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Thork on March 02, 2015, 05:36:49 PM
Not 100 miles, is it.

Also the sun isn't 93 million miles away, it is 800 miles up as documented in our FAQ. And that is another red herring as you are trying to see things on the ground to prove earth round, not in space. You aren't able to construct, follow or reason an argument. That's two threads now where I've lost the patience to keep spelling out the same point in several ways. I'm sorry but I don't think I will be responding to your queries any more. It adds nothing to the site.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 02, 2015, 05:46:09 PM
Not 100 miles, is it.

Also the sun isn't 93 million miles away, it is 800 miles up as documented in our FAQ. And that is another red herring as you are trying to see things on the ground to prove earth round, not in space. You aren't able to construct, follow or reason an argument. That's two threads now where I've lost the patience to keep spelling out the same point in several ways. I'm sorry but I don't think I will be responding to your queries any more. It adds nothing to the site.
93 million miles is more than 100 miles. Remember Pongo OP deals with RET, not FET, so switching horses in mid-stream underscores your failure. Why would you consider the tops of buildings in New York as "on the ground"? Moving the goalposts underscores your failure even more.

Oh, and the FAQ does not document that the sun is 800 miles up. Please do try harder. (It only puts forth an outlandish claim.)

Do feel free to run away each and every time you fail. You just draw even more attention to your failure, and I like that.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 02, 2015, 10:28:04 PM
Not 100 miles, is it.

Also the sun isn't 93 million miles away, it is 800 miles up as documented in our FAQ. And that is another red herring as you are trying to see things on the ground to prove earth round, not in space. You aren't able to construct, follow or reason an argument. That's two threads now where I've lost the patience to keep spelling out the same point in several ways. I'm sorry but I don't think I will be responding to your queries any more. It adds nothing to the site.
93 million miles is more than 100 miles. Remember Pongo OP deals with RET, not FET, so switching horses in mid-stream underscores your failure. Why would you consider the tops of buildings in New York as "on the ground"? Moving the goalposts underscores your failure even more.

Oh, and the FAQ does not document that the sun is 800 miles up. Please do try harder. (It only puts forth an outlandish claim.)

Do feel free to run away each and every time you fail. You just draw even more attention to your failure, and I like that.

It doesn't matter that it's (several hundred thousand miles less than) 93 million miles in RET, because it's in space. Pongo's math shows how far away something can be without being hidden by the supposed curvature of the Earth. The sun would not be hidden by the curvature of the Earth because it is in the sky. Unless you are in a cave, the sky is not generally hidden by the Earth.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 02, 2015, 10:51:03 PM
It doesn't matter that it's (several hundred thousand miles less than) 93 million miles in RET, because it's in space. Pongo's math shows how far away something can be without being hidden by the supposed curvature of the Earth. The sun would not be hidden by the curvature of the Earth because it is in the sky. Unless you are in a cave, the sky is not generally hidden by the Earth.
Why would you expect the tops of buildings, to be hidden according to Pongo's incorrect use of the formula to determine the distance to the astronomical horizon? It's like saying you can't see Earl Hindman behind a fence, even if he's taller than the fence. You must consider the target's height to determine whether you can see it. Pongo does not. He fails.

The sun, and the tops of buildings, are high enough to be seen. Why doesn't Pongo's formula predict both appearances? Because it's the wrong formula. He fails. Deal with it.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 03, 2015, 03:39:28 AM
It doesn't matter that it's (several hundred thousand miles less than) 93 million miles in RET, because it's in space. Pongo's math shows how far away something can be without being hidden by the supposed curvature of the Earth. The sun would not be hidden by the curvature of the Earth because it is in the sky. Unless you are in a cave, the sky is not generally hidden by the Earth.
Why would you expect the tops of buildings, to be hidden according to Pongo's incorrect use of the formula to determine the distance to the astronomical horizon? It's like saying you can't see Earl Hindman behind a fence, even if he's taller than the fence. You must consider the target's height to determine whether you can see it. Pongo does not. He fails.

The sun, and the tops of buildings, are high enough to be seen. Why doesn't Pongo's formula predict both appearances? Because it's the wrong formula. He fails. Deal with it.

It doesn't account for whether or not the object is glowing, either. Both of these things are irrelevant.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 03, 2015, 04:16:44 AM
It doesn't matter that it's (several hundred thousand miles less than) 93 million miles in RET, because it's in space. Pongo's math shows how far away something can be without being hidden by the supposed curvature of the Earth. The sun would not be hidden by the curvature of the Earth because it is in the sky. Unless you are in a cave, the sky is not generally hidden by the Earth.
Why would you expect the tops of buildings, to be hidden according to Pongo's incorrect use of the formula to determine the distance to the astronomical horizon? It's like saying you can't see Earl Hindman behind a fence, even if he's taller than the fence. You must consider the target's height to determine whether you can see it. Pongo does not. He fails.

The sun, and the tops of buildings, are high enough to be seen. Why doesn't Pongo's formula predict both appearances? Because it's the wrong formula. He fails. Deal with it.

It doesn't account for whether or not the object is glowing, either. Both of these things are irrelevant.
Why do you think that the formula does not consider whether the object (target) is glowing? I think Pongo's use is equally inappropriate for either glowing or not glowing objects. It's simply the wrong formula to use to determine whether you can see an object on a clear day.

You probably should review EnaG. Even Rowbotham in Experiment 1 predicts that the height of the target flag affects whether it could be seen. A 16-foot high object could be seen, but a 10-foot high object could not, according to R's interpretation of RET. (Oh, the irony of having to point to R. to demonstrate the confusion of modern FEers.)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 03, 2015, 04:47:54 AM
It doesn't matter that it's (several hundred thousand miles less than) 93 million miles in RET, because it's in space. Pongo's math shows how far away something can be without being hidden by the supposed curvature of the Earth. The sun would not be hidden by the curvature of the Earth because it is in the sky. Unless you are in a cave, the sky is not generally hidden by the Earth.
Why would you expect the tops of buildings, to be hidden according to Pongo's incorrect use of the formula to determine the distance to the astronomical horizon? It's like saying you can't see Earl Hindman behind a fence, even if he's taller than the fence. You must consider the target's height to determine whether you can see it. Pongo does not. He fails.

The sun, and the tops of buildings, are high enough to be seen. Why doesn't Pongo's formula predict both appearances? Because it's the wrong formula. He fails. Deal with it.

It doesn't account for whether or not the object is glowing, either. Both of these things are irrelevant.
Why do you think that the formula does not consider whether the object (target) is glowing? I think Pongo's use is equally inappropriate for either glowing or not glowing objects. It's simply the wrong formula to use to determine whether you can see an object on a clear day.

You probably should review EnaG. Even Rowbotham in Experiment 1 predicts that the height of the target flag affects whether it could be seen. A 16-foot high object could be seen, but a 10-foot high object could not, according to R's interpretation of RET. (Oh, the irony of having to point to R. to demonstrate the confusion of modern FEers.)

Okay, but why is any of this relevant to whether or not you can see 100 miles from the top of the CN Tower?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 03, 2015, 05:13:47 AM
It doesn't matter that it's (several hundred thousand miles less than) 93 million miles in RET, because it's in space. Pongo's math shows how far away something can be without being hidden by the supposed curvature of the Earth. The sun would not be hidden by the curvature of the Earth because it is in the sky. Unless you are in a cave, the sky is not generally hidden by the Earth.
Why would you expect the tops of buildings, to be hidden according to Pongo's incorrect use of the formula to determine the distance to the astronomical horizon? It's like saying you can't see Earl Hindman behind a fence, even if he's taller than the fence. You must consider the target's height to determine whether you can see it. Pongo does not. He fails.

The sun, and the tops of buildings, are high enough to be seen. Why doesn't Pongo's formula predict both appearances? Because it's the wrong formula. He fails. Deal with it.

It doesn't account for whether or not the object is glowing, either. Both of these things are irrelevant.
Why do you think that the formula does not consider whether the object (target) is glowing? I think Pongo's use is equally inappropriate for either glowing or not glowing objects. It's simply the wrong formula to use to determine whether you can see an object on a clear day.

You probably should review EnaG. Even Rowbotham in Experiment 1 predicts that the height of the target flag affects whether it could be seen. A 16-foot high object could be seen, but a 10-foot high object could not, according to R's interpretation of RET. (Oh, the irony of having to point to R. to demonstrate the confusion of modern FEers.)

Okay, but why is any of this relevant to whether or not you can see 100 miles from the top of the CN Tower?
R. approach is the correct one for the challenge. He seems to have used the correct formula to determine how high off the water the target would have be be visible from 6 (vice 100) miles away and inches (vice 1,118 feet + the height of the base of the CN Tower above Lake Ontario) off the water. Note that R. correctly measures height (of both observer's eyes and the top of the distant object) from the water, unlike FEers in this thread in their fail. (That mistake alone is enough to invalidate Pongo's lame proof.)

If R. can get it right, and took the time to write down the results, modern FEers should be able to do even better. So how high off of Lake Ontario would an object 100 miles away across the Lake have to be to be visible on a clear day from the Observation Deck of the CN Tower? Get the right formula. Do the calculation. Show that you can see an object shorter than that and then we can further consider your proof that the Earth is not round. So far though, FEers have failed miserably.

I've already drawn the correct diagram for you even. Now please do better.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 03, 2015, 03:28:43 PM
All you can do is repeat "wrong formula" ad nauseum.  If it's the wrong formula then prove it.  I posted a perfectly good formula and I will not address each person that isn't happy with the results.  If you want to make the claim that I used an incorrect formula then the onus is on you to back that claim up.  Perhaps if you just tried a bit harder...
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 03, 2015, 08:10:17 PM
All you can do is repeat "wrong formula" ad nauseum.  If it's the wrong formula then prove it.  I posted a perfectly good formula and I will not address each person that isn't happy with the results.  If you want to make the claim that I used an incorrect formula then the onus is on you to back that claim up.  Perhaps if you just tried a bit harder...
I have done more than just repeat "wrong formula". For example, I pointed out your error in measuring the height without the altitude of the base of the CM Tower. I've provided a diagram to assist you in understanding your error. I've provided a reference from EnaG showing that R. comes up with a different way to determine what you can see, not using the same formula as you claim is correct.

It's your claim that you've proven that the earth is not round. Show us one reference that states that your formula, especially without considering the height about water, is correct.

You might consider using your approach to get R. answer in Experiment 1. I bet you can't reproduce his results.

Why is high school math so hard for you? Please do try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 03, 2015, 09:53:26 PM
It doesn't matter that it's (several hundred thousand miles less than) 93 million miles in RET, because it's in space. Pongo's math shows how far away something can be without being hidden by the supposed curvature of the Earth. The sun would not be hidden by the curvature of the Earth because it is in the sky. Unless you are in a cave, the sky is not generally hidden by the Earth.
Why would you expect the tops of buildings, to be hidden according to Pongo's incorrect use of the formula to determine the distance to the astronomical horizon? It's like saying you can't see Earl Hindman behind a fence, even if he's taller than the fence. You must consider the target's height to determine whether you can see it. Pongo does not. He fails.

The sun, and the tops of buildings, are high enough to be seen. Why doesn't Pongo's formula predict both appearances? Because it's the wrong formula. He fails. Deal with it.

It doesn't account for whether or not the object is glowing, either. Both of these things are irrelevant.
Why do you think that the formula does not consider whether the object (target) is glowing? I think Pongo's use is equally inappropriate for either glowing or not glowing objects. It's simply the wrong formula to use to determine whether you can see an object on a clear day.

You probably should review EnaG. Even Rowbotham in Experiment 1 predicts that the height of the target flag affects whether it could be seen. A 16-foot high object could be seen, but a 10-foot high object could not, according to R's interpretation of RET. (Oh, the irony of having to point to R. to demonstrate the confusion of modern FEers.)

Okay, but why is any of this relevant to whether or not you can see 100 miles from the top of the CN Tower?
R. approach is the correct one for the challenge. He seems to have used the correct formula to determine how high off the water the target would have be be visible from 6 (vice 100) miles away and inches (vice 1,118 feet + the height of the base of the CN Tower above Lake Ontario) off the water. Note that R. correctly measures height (of both observer's eyes and the top of the distant object) from the water, unlike FEers in this thread in their fail. (That mistake alone is enough to invalidate Pongo's lame proof.)

If R. can get it right, and took the time to write down the results, modern FEers should be able to do even better. So how high off of Lake Ontario would an object 100 miles away across the Lake have to be to be visible on a clear day from the Observation Deck of the CN Tower? Get the right formula. Do the calculation. Show that you can see an object shorter than that and then we can further consider your proof that the Earth is not round. So far though, FEers have failed miserably.

I've already drawn the correct diagram for you even. Now please do better.

Nobody is talking about how high something would have to be to be been except for you. The rest of us are talking about how far away something of a presumably minimal height could be seen from the top of the CN tower. This entire argument is a red herring.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 03, 2015, 10:43:38 PM
Nobody is talking about how high something would have to be to be been except for you. The rest of us are talking about how far away something of a presumably minimal height could be seen from the top of the CN tower. This entire argument is a red herring.
Wonderful! Now just tell me how you decided that the target object must be of minimal height. Surely you're not claiming that
Quote from: http://en.wikipedia.org/wiki/CN_Tower
On a clear day, it is possible to see up to 100 to 120 km (62 to 75 mi) away, to the city of Rochester across Lake Ontario in the United States, the mist rising from Niagara Falls, or the shores of Lake Simcoe.
is talking about an object of minimal height, are you? Where in that sentence do you see a word in any way similar to "minimal"?

Why did R. use a 10-foot (not minimal height) flag in Experiment 1?

I'm talking about how high would an object have to be to be seen 100 (75?) miles away on a clear day from the observation deck of the CN Tower.. Why would we be discussing anything different? Don't infer "minimal" for your convenience in another's sentence please. Please do try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 03, 2015, 11:11:49 PM
It's your claim that you've proven that the earth is not round.

OMG, it was my claim that you can't see 100 miles from the CN tower and I posted a freaking proof with it. You claim it's incorrect then YOU have to demonstrate it. If I stand outside a museum and shout evolution is wrong because of x, it's not the job of the museumists to come outside and prove me wrong. If I make the claim that they are wrong then I would have demonstrate why; just as you do.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 03, 2015, 11:22:45 PM
It's your claim that you've proven that the earth is not round.

OMG, it was my claim that you can't see 100 miles from the CN tower and I posted a freaking proof with it. You claim it's incorrect then YOU have to demonstrate it. If I stand outside a museum and shout evolution is wrong because of x, it's not the job of the museumists to come outside and prove me wrong. If I make the claim that they are wrong then I would have demonstrate why; just as you do.
No you have not posted a proof. You used the wrong formula. There is NOTHING in RET that supports your outlandish claim. Shouting that you've proven something doesn't shift your burden of proof. You still need to defend your proof. Here: why did you use that formula? Can you find a single RET source that states that the formula determines how far you can see on a clear day? Even R. disagrees with you. Even Jon Davis tried this and failed. You must explain the reason you don't include the height off the water of either the source or the target. By the way, you make an inherent error by omitting both. You fail.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 04, 2015, 01:56:36 AM
Nobody is talking about how high something would have to be to be been except for you. The rest of us are talking about how far away something of a presumably minimal height could be seen from the top of the CN tower. This entire argument is a red herring.
Wonderful! Now just tell me how you decided that the target object must be of minimal height. Surely you're not claiming that
Quote from: http://en.wikipedia.org/wiki/CN_Tower
On a clear day, it is possible to see up to 100 to 120 km (62 to 75 mi) away, to the city of Rochester across Lake Ontario in the United States, the mist rising from Niagara Falls, or the shores of Lake Simcoe.
is talking about an object of minimal height, are you? Where in that sentence do you see a word in any way similar to "minimal"?

Why did R. use a 10-foot (not minimal height) flag in Experiment 1?

I'm talking about how high would an object have to be to be seen 100 (75?) miles away on a clear day from the observation deck of the CN Tower.. Why would we be discussing anything different? Don't infer "minimal" for your convenience in another's sentence please. Please do try harder.

There isn't a single person on this board, apart from you, who cares whether or not a skyscraper can be seen from 100 miles away when viewed from the CN tower. We're talking about whether or not standing on the top of the CN tower moves the horizon to more than 100 miles away, which it does not.

Honestly, why are you being so obtuse about this? You can just take the easy route and say that the CN tower is lying about how far you can see. There's no sense in trying to disprove basic trigonometry by yelling louder than your opponent.

Also, John Davis is has an H in his name.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 04, 2015, 02:14:49 AM
There isn't a single person on this board, apart from you, who cares whether or not a skyscraper can be seen from 100 miles away when viewed from the CN tower. We're talking about whether or not standing on the top of the CN tower moves the horizon to more than 100 miles away, which it does not.
Who claimed that the horizon is 100 miles away as seen from the CN Tower? ???
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 04, 2015, 02:48:58 AM
There isn't a single person on this board, apart from you, who cares whether or not a skyscraper can be seen from 100 miles away when viewed from the CN tower. We're talking about whether or not standing on the top of the CN tower moves the horizon to more than 100 miles away, which it does not.
Who claimed that the horizon is 100 miles away as seen from the CN Tower? ???

I'm sorry, but I'm not going to argue about arguing about an argument with you, Markjo. This is getting too meta.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 04, 2015, 03:30:08 AM
We're talking about whether or not standing on the top of the CN tower moves the horizon to more than 100 miles away, which it does not.

I'm sorry, but I'm not going to argue about arguing about an argument with you, Markjo. This is getting too meta.
<Whiplash> Ts, you made an outlandish claim and ran away from it in your very next post. How low can you go?

No one has made the claim that "standing on the top of the CN tower moves the horizon to more than 100 miles away", except you in that post, though that's all Pongo's errant proof disputed.

Pongo and you built a straw man: the horizon can't be as far away as the objects you can see. And now want credit for knocking it over. It's easy to win a straw man argument for FEers, isn't it?

Again, wrong formula. I renew my challenge: show any RET claim that Pongo's use, of the formula to estimate the distance to the horizon, shows how far you can see from a tower on the RE. Attacking that RET is wrong about something RET doesn't even say is simply juvenile.

Oh, and I've already produced attributed photographs showing that you can indeed see objects more that 41 miles away from the CN Tower, so, no, I will not claim that fact to be false. Integrity is important.

You really should consider pp's advice to Thork on another issue: Don't demonstrate that you can't solve high-school level physics problems, or at least not so often.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 04, 2015, 03:38:07 AM
There isn't a single person on this board, apart from you, who cares whether or not a skyscraper can be seen from 100 miles away when viewed from the CN tower. We're talking about whether or not standing on the top of the CN tower moves the horizon to more than 100 miles away, which it does not.
Who claimed that the horizon is 100 miles away as seen from the CN Tower? ???

I'm sorry, but I'm not going to argue about arguing about an argument with you, Markjo. This is getting too meta.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 04, 2015, 02:25:29 PM
Oh, and I've already produced attributed photographs showing that you can indeed see objects more that 41 miles away from the CN Tower, so, no, I will not claim that fact to be false. Integrity is important.

Right... you can see objects more than 41 miles away from CN Tower because the earth is flat.  What do you think we're discussing in this thread?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 04, 2015, 05:19:00 PM
Oh, and I've already produced attributed photographs showing that you can indeed see objects more that 41 miles away from the CN Tower, so, no, I will not claim that fact to be false. Integrity is important.

Right... you can see objects more than 41 miles away from CN Tower because the earth is flat.  What do you think we're discussing in this thread?
We are discussing "Proving a Flat-Earth Using Round-Earth Maths". Now if you'd be so kind as to address the repeated challenge: What Round-Earth Maths require that you not be able to see more than 41 miles away from CN Tower, you might make progress. Please do try harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tom Bishop on March 05, 2015, 01:32:48 AM
Calculating the drop

The Earth is a sphere with a radius of 3963 miles at sea level. Converting that to feet we get: 3963 * 5280 ft to a mile = 20924640 ft above the center of the earth. Add 1,118 feet for the observation deck and height of the observer (Pongo's figure) and we are now 20925758 ft above the center of the earth.

The distance being looked across is 41.15 miles. Converting that to feet we get 100 * 5280 ft to a mile = 217272 ft

(http://i62.tinypic.com/a1lmw9.png)

Using the theorem of Pythagoras we can use:

a^2 = 20925758^2 + 217272^2 = 437934554996548

When we square root that figure we get a = 20926885.9364 ft

Thus your position is 20926885.9364 - 20924640 (radius of the earth) = 2245.9364 feet above the surface of the earth.

Hence, after 41.15 miles, the earth would drop 2245.9364 feet.

The tallest building in the world is the Burj Khalifa at 2,717 ft and the second tallest building in the world is the Shanghai Tower at 2,073 ft. I submit that the Burj Khalifa is not in Niagara Falls. Therefore the earth is flat.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 05, 2015, 01:34:21 AM
Tom, the distance to the horizon (a point tangent to the surface of the round earth from your eye level) is not the same as the "drop" that you are calculating.  At an elevation of 1118 feet, the horizon is approximately 41 miles (http://www.ringbell.co.uk/info/hdist.htm) away.  Please note that the horizon is not necessarily the limiting factor to how far away you can see things (ex. masts of partially sunken ships, tall buildings, etc.).
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 05, 2015, 01:39:19 AM
Calculating the Drop

The Earth is a sphere with a radius of 3963 miles at sea level. Converting that to feet we get: 3963 * 5280 ft to a mile = 20924640 ft above the center of the earth. Add 1,118 feet for the observation deck and height of the observer (Pongo's figure) and we are now 20925758 ft above the center of the earth.

The distance being looked across is 41.15 miles. Converting that to feet we get 100 * 5280 ft to a mile = 217272 ft

...

Using the theorem of Pythagoras we can use:

a^2 = 20925758^2 + 217272^2 = 437934554996548

when we square root that figure we get a = 20926885.9364 ft

thus your position is 20926885.9364 - 20924640 (radius of the earth) = 2245.9364 feet above the surface of the earth.

Hence, after 41.15 miles, the earth would drop 2245.9364 feet.

The tallest building in the world is the Burj Khalifa at 2,717 ft and the second tallest building in the world is the Shanghai Tower at 2,073 ft. I submit that the Burj Khalifa is not in Niagara Falls. Therefore the earth is flat.
Again, wrong formula. Again, FEers ignore that Lake Ontario, the CN Tower, and the target object are not on the surface of the hypothetical sphere. They have altitude more than their height. Do you really think that Lake Ontario is at sea level? Please do try harder. Why do you believe that you can't see beyond the drop? Do you think the photons get too tired to travel over Lake Ontario? Please think.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 05, 2015, 01:48:15 PM
Again, wrong formula. Again, FEers ignore that Lake Ontario, the CN Tower, and the target object are not on the surface of the hypothetical sphere. They have altitude more than their height. Do you really think that Lake Ontario is at sea level? Please do try harder. Why do you believe that you can't see beyond the drop? Do you think the photons get too tired to travel over Lake Ontario? Please think.

I know you can't do your own maths Gulliver, so I did them for you!  I adjusted the post to reflect your concerns about sea level.  Here's a synopsis.

Given Values
Height of the observation deck: 1,135 feet
Generous height of eye level: 6 feet
Elevation of Toronto: 249 feet

Mean Radius of the Earth at sea level (lol) = 20,903,520 feet

h = Height of observation deck + Generous height of eye level + Elevation of Toronto
R = Mean radius of the Earth (lol)

h= 1390 feet
R= 20,903,520 feet

Formula
How far you can see if you were on a sphere = R*ACOS(R/(R+h))

How far you can see if you were on a sphere = 20925524.9*ACOS(20925524.9/(20925524.9+1390))

How far you can see if you were on a sphere = 241,057.18 feet

How far you can see if you were on a sphere =  45.65 miles

Looks like you got yourself a few more miles!  Still, you have over 54 more miles to account for, but as I know you'll not do any of your own math and just tell us to try harder, I guess we'll just have to leave it as it.

Flat-Earth victory!
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 05, 2015, 02:05:11 PM
That's how far the horizon is, not how far you can see.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 05, 2015, 06:10:23 PM
That's how far the horizon is, not how far you can see.

I agree. Since the earth is flat, the distance to the horizon is irrelevant.

However, let's continue on with the delusion that globularism is a possible scenario here. In order to see something from the top of the CN tower, the distance to the horizon from the top of the target object plus the distance to the horizon from the top of the CN tower must be greater than or equal to the distance between the objects.

H1 + H2 >= D

We already have H1, which Pongo calculated to be 45.65 miles. Since we're talking about a target object 100 miles away, the distance to the horizon from the target object is (generously) 54 miles (241,000 feet).

Let's reverse Pongo's equation to find how high this theoretical object would have to be.

241,000 = 20903520*ARCOS(20903520/(20903250+H))

Wolfram Alpha computes H as being equal to about 1400 feet. Thus, and unsurprisingly, this suggests that the target object would have to be much, much taller than the CN tower itself on a round Earth.

In conclusion:

ANOTHER VICTORY FOR FET!
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 05, 2015, 07:11:49 PM
That's how far the horizon is, not how far you can see.

I agree. Since the earth is flat, the distance to the horizon is irrelevant.

However, let's continue on with the delusion that globularism is a possible scenario here. In order to see something from the top of the CN tower, the distance to the horizon from the top of the target object plus the distance to the horizon from the top of the CN tower must be greater than or equal to the distance between the objects.

H1 + H2 >= D

We already have H1, which Pongo calculated to be 45.65 miles. Since we're talking about a target object 100 miles away, the distance to the horizon from the target object is (generously) 54 miles (241,000 feet).

Let's reverse Pongo's equation to find how high this theoretical object would have to be.

241,000 = 20903520*ARCOS(20903520/(20903250+H))

Wolfram Alpha computes H as being equal to about 1400 feet. Thus, and unsurprisingly, this suggests that the target object would have to be much, much taller than the CN tower itself on a round Earth.

Congratulations to FEers on correcting two errors explained by REers' critiques of this sophomoric, failed proof. I'm happy to see that Pongo and revised his proof to use the altitude, not the height above CN Tower's foundation, of the observer. That took only 5 pages to correct one error.

Now I see that you've decided to correct, though snidely, Pongo's error when he failed to consider that the formula Pongo used provides the distance to the horizon (inaccurately), not the distance the observer can see. Congratulations on that, though it took 5 pages too.

Now let's review the remaining errors in the FEer failed proof:
1) The altitude of the surface of Lake Ontario, not just the radius of the RE, should be used.
2) The result, once you correct for the above error, is not the height of the target, as you claim. It's the altitude. Pongo understands that on the Tower side now. Now you need to consider it of the target side.

Once you fix the remaining errors, do stop back. Thank you for trying harder.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 05, 2015, 07:34:48 PM
We are already bending over backwards to accommodate your nitpicking. Even if we conced every variable and give you benifit on each of them it won't account for the 55 more miles you need for round-earth theory to be true. Give it up, the proof is before you.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Gulliver on March 05, 2015, 07:51:03 PM
We are already bending over backwards to accommodate your nitpicking. Even if we conced every variable and give you benifit on each of them it won't account for the 55 more miles you need for round-earth theory to be true. Give it up, the proof is before you.
It's your faulty proof. Deal with the critiques or give up.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 05, 2015, 08:32:07 PM
Wolfram Alpha computes H as being equal to about 1400 feet. Thus, and unsurprisingly, this suggests that the target object would have to be much, much taller than the CN tower itself on a round Earth.
You forget that the 100 mile visibility claim is for ideal conditions.  Those ideal conditions could include any of several atmospheric refraction phenomena that cause objects that would normally be blocked by the horizon to be visible.  This is why I generally consider such convexity experiments to be largely inconclusive.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 05, 2015, 10:16:58 PM
Wolfram Alpha computes H as being equal to about 1400 feet. Thus, and unsurprisingly, this suggests that the target object would have to be much, much taller than the CN tower itself on a round Earth.
You forget that the 100 mile visibility claim is for ideal conditions.  Those ideal conditions could include any of several atmospheric refraction phenomena that cause objects that would normally be blocked by the horizon to be visible.  This is why I generally consider such convexity experiments to be largely inconclusive.

So the 100 mile visibility is only possible if conditions magically make the Earth look exactly as if it were flat, even though it's obviously round in reality. It's funny how often globularism falls back on that excuse.

I'd be interested in calculating whether or not even RE says it's possible for that much refraction to occur, but I honestly wouldn't know where to start.

That's how far the horizon is, not how far you can see.

I agree. Since the earth is flat, the distance to the horizon is irrelevant.

However, let's continue on with the delusion that globularism is a possible scenario here. In order to see something from the top of the CN tower, the distance to the horizon from the top of the target object plus the distance to the horizon from the top of the CN tower must be greater than or equal to the distance between the objects.

H1 + H2 >= D

We already have H1, which Pongo calculated to be 45.65 miles. Since we're talking about a target object 100 miles away, the distance to the horizon from the target object is (generously) 54 miles (241,000 feet).

Let's reverse Pongo's equation to find how high this theoretical object would have to be.

241,000 = 20903520*ARCOS(20903520/(20903250+H))

Wolfram Alpha computes H as being equal to about 1400 feet. Thus, and unsurprisingly, this suggests that the target object would have to be much, much taller than the CN tower itself on a round Earth.

Congratulations to FEers on correcting two errors explained by REers' critiques of this sophomoric, failed proof. I'm happy to see that Pongo and revised his proof to use the altitude, not the height above CN Tower's foundation, of the observer. That took only 5 pages to correct one error.

Now I see that you've decided to correct, though snidely, Pongo's error when he failed to consider that the formula Pongo used provides the distance to the horizon (inaccurately), not the distance the observer can see. Congratulations on that, though it took 5 pages too.

Now let's review the remaining errors in the FEer failed proof:
1) The altitude of the surface of Lake Ontario, not just the radius of the RE, should be used.
2) The result, once you correct for the above error, is not the height of the target, as you claim. It's the altitude. Pongo understands that on the Tower side now. Now you need to consider it of the target side.

Once you fix the remaining errors, do stop back. Thank you for trying harder.

The 'failures' you incessantly refer (and honestly,  You complain that I'm being snide?) to are actually simplifications, which are acceptable because a few dozen feet here or there don't really matter much when we're talking about thousands of feet. We're assuming a spherical cow, so to speak, because the true shape of the quadruped is a bit irrelevant. However, I'll sink to your level and address them anyway.

Pongo did use the altitude of Lake Ontario. He added that (inconsequential) value into his equation like, 2 pages ago or something. Anyway, so it's actually an 'altitude' of 1400 feet instead of a 'height' of 1400 feet. So what? It's still taller than the CN tower itself, even including the height above sea level. And believe it or not, I checked out the altitude of the Niagara Escarpment (since we're talking about Niagara) when I did the math. The exact altitude varied a lot and could be anywhere from 40 feet above sea level to 700 feet above sea level in a small area. Regardless, the building seen from the CN tower would need to be between 1400 and 700 feet tall, which would necessarily put it on the list of the tallest buildings in the US on Wikipedia. However, there are no such buildings in the area.

In conclusion, you cannot see 100 miles from the CN tower unless there's some magical RE refraction going on or the Earth is flat. Take your pick.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 06, 2015, 12:58:46 AM
Wolfram Alpha computes H as being equal to about 1400 feet. Thus, and unsurprisingly, this suggests that the target object would have to be much, much taller than the CN tower itself on a round Earth.
You forget that the 100 mile visibility claim is for ideal conditions.  Those ideal conditions could include any of several atmospheric refraction phenomena that cause objects that would normally be blocked by the horizon to be visible.  This is why I generally consider such convexity experiments to be largely inconclusive.

So the 100 mile visibility is only possible if conditions magically make the Earth look exactly as if it were flat, even though it's obviously round in reality. It's funny how often globularism falls back on that excuse.
Are you suggesting that refraction does not occur? ???

I'd be interested in calculating whether or not even RE says it's possible for that much refraction to occur, but I honestly wouldn't know where to start.
Well, you can try starting here: http://www.mike-willis.com/Tutorial/PF6.htm
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 06, 2015, 01:10:21 AM
Is this what the goal posts have been moved to now? Refraction? Can you let us know the next goal post move so when we shut this one down we have a head start on what's next?

On a side note, it's an amazing bit of doublethink that round-earthers can both vehemently deny bendy light while fully supporting things like refraction.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 06, 2015, 02:34:43 AM
On a side note, it's an amazing bit of doublethink that round-earthers can both vehemently deny bendy light while fully supporting things like refraction.
I think that it's amazing that FE'ers think that bendy light is on par with refraction.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tom Bishop on March 06, 2015, 02:52:38 AM
Funny how refraction suspends the image hundreds of feet in the air, no higher and no lower, to the exact altitude it would need to be if the earth were flat.

Don't you see how astronomically unlikely that is?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 06, 2015, 04:41:01 AM
This thread is a golden example of RE stubbornness in face of the truth, as well as of the easily verifiable flatness of the Earth.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 06, 2015, 04:43:32 AM
Funny how refraction suspends the image hundreds of feet in the air, no higher and no lower, to the exact altitude it would need to be if the earth were flat.
In the case of ducting, refraction is not "suspending the image hundreds of feet in the air".  The image is following the curvature of the earth.  The conditions required, although not necessarily common, are well understood and completely plausible.

Don't you see how astronomically unlikely that is?
No, I don't.  Why don't you calculate the odds for us?  And please show your math.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 06, 2015, 06:48:50 AM
Funny how refraction suspends the image hundreds of feet in the air, no higher and no lower, to the exact altitude it would need to be if the earth were flat.

Don't you see how astronomically unlikely that is?

It doesn't look like the Earth is flat. You can't see the bottom of the building.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 06, 2015, 04:29:19 PM
It doesn't look like the Earth is flat. You can't see the bottom of the building.

According to Markjo, "The image is following the curvature of the earth."  So the image both follows the curvature of the earth AND cuts off at the bottom.  Round-Earth theory does all the things at all the times! DO NOT QUESTION IT!
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 06, 2015, 04:43:57 PM
It doesn't look like the Earth is flat. You can't see the bottom of the building.

According to Markjo, "The image is following the curvature of the earth."  So the image both follows the curvature of the earth AND cuts off at the bottom.  Round-Earth theory does all the things at all the times! DO NOT QUESTION IT!

What?  I am not really sure what you are referring to.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 06, 2015, 05:13:22 PM
So the tallest building in Niagara Falls, NY is the Seneca Niagara Casino Tower (http://en.wikipedia.org/wiki/List_of_tallest_buildings_in_Upstate_New_York) at 109 meters.  Niagara Fall, NY has an elevation of 51 meters.

The horizon is there for 45kms (http://www.ringbell.co.uk/info/hdist.htm) away from the top of the tower.

The CN Tower is roughly 101 kms away from the Seneca Niagara Casino Tower, as determined by a triangulation on Google Maps.  So in order to be able to see the Seneca Niagara Casino Tower from the CN Tower, you must have a horizon greater than (101-45)kms, or 56kms, away.  So do we?

Well if the CN Tower observation deck is roughly 450m tall then the distance to the horizon is 75kms (http://www.ringbell.co.uk/info/hdist.htm).  It appears that you can see well past the bulge between the two towers.

Does anyone agree or disagree with this?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 06, 2015, 05:30:23 PM
It doesn't look like the Earth is flat. You can't see the bottom of the building.

According to Markjo, "The image is following the curvature of the earth."  So the image both follows the curvature of the earth AND cuts off at the bottom.  Round-Earth theory does all the things at all the times! DO NOT QUESTION IT!

What?  I am not really sure what you are referring to.

It was two posts above mine that you quoted.  The last post my Markjo in this thread.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 06, 2015, 05:33:02 PM
It doesn't look like the Earth is flat. You can't see the bottom of the building.

According to Markjo, "The image is following the curvature of the earth."  So the image both follows the curvature of the earth AND cuts off at the bottom.  Round-Earth theory does all the things at all the times! DO NOT QUESTION IT!

What?  I am not really sure what you are referring to.

It was two posts above mine that you quoted.  The last post my Markjo in this thread.

Ah I see.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 06, 2015, 05:40:47 PM
So the tallest building in Niagara Falls, NY is the Seneca Niagara Casino Tower (http://en.wikipedia.org/wiki/List_of_tallest_buildings_in_Upstate_New_York) at 109 meters.  Niagara Fall, NY has an elevation of 51 meters.

The horizon is there for 45kms (http://www.ringbell.co.uk/info/hdist.htm) away from the top of the tower.

The CN Tower is roughly 101 kms away from the Seneca Niagara Casino Tower, as determined by a triangulation on Google Maps.  So in order to be able to see the Seneca Niagara Casino Tower from the CN Tower, you must have a horizon greater than (101-45)kms, or 56kms, away.  So do we?

Well if the CN Tower observation deck is roughly 450m tall then the distance to the horizon is 75kms (http://www.ringbell.co.uk/info/hdist.htm).  It appears that you can see well past the bulge between the two towers.

Does anyone agree or disagree with this?

101km is not >= 100mi as round-earthers boast that you can see from CN tower.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 06, 2015, 05:49:07 PM
So the tallest building in Niagara Falls, NY is the Seneca Niagara Casino Tower (http://en.wikipedia.org/wiki/List_of_tallest_buildings_in_Upstate_New_York) at 109 meters.  Niagara Fall, NY has an elevation of 51 meters.

The horizon is there for 45kms (http://www.ringbell.co.uk/info/hdist.htm) away from the top of the tower.

The CN Tower is roughly 101 kms away from the Seneca Niagara Casino Tower, as determined by a triangulation on Google Maps.  So in order to be able to see the Seneca Niagara Casino Tower from the CN Tower, you must have a horizon greater than (101-45)kms, or 56kms, away.  So do we?

Well if the CN Tower observation deck is roughly 450m tall then the distance to the horizon is 75kms (http://www.ringbell.co.uk/info/hdist.htm).  It appears that you can see well past the bulge between the two towers.

Does anyone agree or disagree with this?

101km is not >= 100mi as round-earthers boast that you can see from CN tower.

REers are also wrong sometimes, I should know.  That being said, the photo taken from the CN Tower towards Niagara Falls, NY appears to be completely possible on the RE model.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 06, 2015, 05:52:44 PM
It doesn't look like the Earth is flat. You can't see the bottom of the building.

According to Markjo, "The image is following the curvature of the earth."  So the image both follows the curvature of the earth AND cuts off at the bottom.  Round-Earth theory does all the things at all the times! DO NOT QUESTION IT!
No.  The refractive phenomena that I'm referring to only happens under certain conditions.  In fact, the 100 mile visibility claim from the CN Tower is admittedly under "ideal" conditions.  I'm just proposing that those "ideal" conditions could include some degree of atmospheric refraction that could extend the visibility beyond what might be normally expected on a round earth.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: model 29 on March 15, 2015, 05:41:05 PM
So are there any land-based objects that can actually be seen from the CN tower that are 100 miles away?

If not, then I guess Earth isn't flat.  I wonder if the writer of the 'facts' on that linked site had miles and kilometers confused, or simply wrote something they heard.

Using a couple different calculators, from a height of 1400 feet, an object 100 miles away would need to be 2000 feet tall to be seen on a round Earth over the horizon (based on sea-level elevations).

Here are some mountains viewed from about 100 miles away.  The peaks are 10,000 to 12,000ft. high.  With the camera 10-15ft high there should be about 6,000 feet hidden by the curvature.
(http://i42.tinypic.com/11kjn91.jpg)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Tau on March 15, 2015, 06:01:09 PM
It doesn't look like the Earth is flat. You can't see the bottom of the building.

According to Markjo, "The image is following the curvature of the earth."  So the image both follows the curvature of the earth AND cuts off at the bottom.  Round-Earth theory does all the things at all the times! DO NOT QUESTION IT!
No.  The refractive phenomena that I'm referring to only happens under certain conditions.  In fact, the 100 mile visibility claim from the CN Tower is admittedly under "ideal" conditions.  I'm just proposing that those "ideal" conditions could include some degree of atmospheric refraction that could extend the visibility beyond what might be normally expected on a round earth.

And of course, those certain conditions are "when it's convenient"
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 16, 2015, 01:39:59 AM
It doesn't look like the Earth is flat. You can't see the bottom of the building.

According to Markjo, "The image is following the curvature of the earth."  So the image both follows the curvature of the earth AND cuts off at the bottom.  Round-Earth theory does all the things at all the times! DO NOT QUESTION IT!
No.  The refractive phenomena that I'm referring to only happens under certain conditions.  In fact, the 100 mile visibility claim from the CN Tower is admittedly under "ideal" conditions.  I'm just proposing that those "ideal" conditions could include some degree of atmospheric refraction that could extend the visibility beyond what might be normally expected on a round earth.

And of course, those certain conditions are "when it's convenient"
Convenient for whom?  I've been to the top of the CN Tower several times and the Toronto smog alone makes the conditions far from ideal.

Then again, refraction can be quite convenient for FE'ers when it makes a round earth appear flat.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 16, 2015, 01:57:50 PM
So are there any land-based objects that can actually be seen from the CN tower that are 100 miles away?

CN Tower's webpage says there are.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: mister bickles on March 18, 2015, 03:18:07 AM

Then again, refraction can be quite convenient for FE'ers when it makes a round earth appear flat.

i get the feeling that "refraction" is one of those convenient, throw-a-way "explanations" that TPTB trot out to try and dodge "the hard balls";

although "refraction" is a relatively simple concept/explanation, TPTB and their innumerable shills are quite content to spew out a barrage of meaningless scientific gobbly-de-gook to baffle and confuse the average punter...

most people haven't got the qualifications, the time or the inclination to penetrate these "smoke screens";

that's the way TPTB like it......keep every-one distracted with mindless entertainments and too busy working to earn a living to have any spare time to consider the fact that they're trapped in "the Matrix";

(the word "amuse", for instance,  actually means "don't think" or "anti-think"....'muse' being the Græco-Roman term for "thinking" or "pondering"/"reflecting on some-thing"....an "a" prefix is its opposite!)

as for refraction making a RE appear flat....well....where is the evidence for this?
where are the laboratory experiments (that have been double/triple checked and, then, peer-reviewed) to demonstrate this?

what are the refractive indices of air and water.....how do they compare and how can they distort an object's appearance....   ???

most people know that water has a certain refractive index that can distort objects under-water......
does air have the same sort of effect?

quite frankly, i doubt it!

if any-thing, it would be only of a minimal, marginal kind that would make no appreciable difference

Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: model 29 on March 19, 2015, 02:32:27 AM
as for refraction making a RE appear flat....well....where is the evidence for this?
[/color][/font]
A scope and a 20foot change in elevation will provide you with a nice example if the conditions are favorable.

The right half was taken from between 15-20 feet above the water.  The left half from about 6 inches.  I aligned the two shots using one of the taller trees (the hillside is 12 miles away).  The hillside and buildings have 'sunk' below the horizon/waterline in the left half, but refraction allows them to be visible, yet with compressed appearance.  The angled features on the hillside are mudslides, also noticeable compressed by the refraction.  The bridge is 9 miles away.
(http://i1368.photobucket.com/albums/ag167/jeffro556/bridge2_zps38b17185.jpg)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: mister bickles on March 19, 2015, 03:04:29 AM
less than convincing.....
all it looks like to me is distortion because of the close distance to the photographed object/small elevation above the water surface;

as per my previous post, i would want hard, objective scientific proof via laboratory experiments that have been double/triple (blind) checked and peer-reviewed before i'm convinced that 'refraction' (in the air) is playing any significant part at all in these distant "flat horizons";

for now and AFAIMC, its perfectly compatible with an FE model    ::)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: model 29 on March 19, 2015, 03:53:24 AM
less than convincing.....
all it looks like to me is distortion because of the close distance to the photographed object/small elevation above the water surface;

Can you explain how that distortion would result in what is seen in the pictures if it's something other than refraction and distant objects disappearing below Earth's curvature?

Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: mister bickles on March 19, 2015, 04:01:07 AM
less than convincing.....
all it looks like to me is distortion because of the close distance to the photographed object/small elevation above the water surface;

Can you explain how that distortion would result in what is seen in the pictures if it's something other than refraction and distant objects disappearing below Earth's curvature?

there are too many instances of these flat, distant horizons in photographs and every-day experience for 'refraction' to play any significant part.....many of the pix i'v seen are in perfect weather conditions, too!.....
(actually...i failed to mention that....the weather conditions in the bridge 'photo look less than ideal, too....the humidity may have played a factor in the refraction);

again: this needs to be proven by controlled, laboratory experiments!  :-B
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 19, 2015, 04:27:12 AM
most people know that water has a certain refractive index that can distort objects under-water......
does air have the same sort of effect?

quite frankly, i doubt it!

if any-thing, it would be only of a minimal, marginal kind that would make no appreciable difference
Have you never heard of a mirage? ???
(http://i.dailymail.co.uk/i/pix/2012/04/16/article-0-129FC49E000005DC-904_634x384.jpg)
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: mister bickles on March 19, 2015, 08:23:38 AM

Have you never heard of a mirage? ???

hardly the same thing;
"mirages" only happen under special atmospheric conditions and only in certain, peculiar geographical locations;
i'm struggling to believe that (air) "refraction" plays any significant part in these distant flat Earth horizon(s)/perspectives;
i can't entertain "mirages" for a micro-second!
sorry!  :(
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 19, 2015, 01:38:10 PM

Then again, refraction can be quite convenient for FE'ers when it makes a round earth appear flat.

i get the feeling that "refraction" is one of those convenient, throw-a-way "explanations" that TPTB trot out to try and dodge "the hard balls";

although "refraction" is a relatively simple concept/explanation, TPTB and their innumerable shills are quite content to spew out a barrage of meaningless scientific gobbly-de-gook to baffle and confuse the average punter...

most people haven't got the qualifications, the time or the inclination to penetrate these "smoke screens";

that's the way TPTB like it......keep every-one distracted with mindless entertainments and too busy working to earn a living to have any spare time to consider the fact that they're trapped in "the Matrix";

(the word "amuse", for instance,  actually means "don't think" or "anti-think"....'muse' being the Græco-Roman term for "thinking" or "pondering"/"reflecting on some-thing"....an "a" prefix is its opposite!)

as for refraction making a RE appear flat....well....where is the evidence for this?
where are the laboratory experiments (that have been double/triple checked and, then, peer-reviewed) to demonstrate this?

what are the refractive indices of air and water.....how do they compare and how can they distort an object's appearance....   ???

most people know that water has a certain refractive index that can distort objects under-water......
does air have the same sort of effect?

quite frankly, i doubt it!

if any-thing, it would be only of a minimal, marginal kind that would make no appreciable difference

What does TPTB stand for?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 19, 2015, 02:32:16 PM

Then again, refraction can be quite convenient for FE'ers when it makes a round earth appear flat.

i get the feeling that "refraction" is one of those convenient, throw-a-way "explanations" that TPTB trot out to try and dodge "the hard balls";

although "refraction" is a relatively simple concept/explanation, TPTB and their innumerable shills are quite content to spew out a barrage of meaningless scientific gobbly-de-gook to baffle and confuse the average punter...

most people haven't got the qualifications, the time or the inclination to penetrate these "smoke screens";

that's the way TPTB like it......keep every-one distracted with mindless entertainments and too busy working to earn a living to have any spare time to consider the fact that they're trapped in "the Matrix";

(the word "amuse", for instance,  actually means "don't think" or "anti-think"....'muse' being the Græco-Roman term for "thinking" or "pondering"/"reflecting on some-thing"....an "a" prefix is its opposite!)

as for refraction making a RE appear flat....well....where is the evidence for this?
where are the laboratory experiments (that have been double/triple checked and, then, peer-reviewed) to demonstrate this?

what are the refractive indices of air and water.....how do they compare and how can they distort an object's appearance....   ???

most people know that water has a certain refractive index that can distort objects under-water......
does air have the same sort of effect?

quite frankly, i doubt it!

if any-thing, it would be only of a minimal, marginal kind that would make no appreciable difference

What does TPTB stand for?

The Powers That Be
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Pongo on March 19, 2015, 02:33:13 PM
Thanks!
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Rama Set on March 19, 2015, 05:35:15 PM
Thanks!

Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: markjo on March 20, 2015, 01:49:44 AM

Have you never heard of a mirage? ???

hardly the same thing;
???  Are you saying that mirages are not caused by refraction?

"mirages" only happen under special atmospheric conditions and only in certain, peculiar geographical locations;
i'm struggling to believe that (air) "refraction" plays any significant part in these distant flat Earth horizon(s)/perspectives;
i can't entertain "mirages" for a micro-second!
sorry!  :(
Perhaps you should look into air temperature gradients near the surface of water before you dismiss refraction so completely.
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: model 29 on March 21, 2015, 05:44:23 PM

Have you never heard of a mirage? ???

hardly the same thing;
"mirages" only happen under special atmospheric conditions and only in certain, peculiar geographical locations;
i'm struggling to believe that (air) "refraction" plays any significant part in these distant flat Earth horizon(s)/perspectives;
i can't entertain "mirages" for a micro-second!
sorry!  :(

Mirages occur quite often.  I guess you've never been on a road in hot weather.  There are several types of mirages though.  The shoreline buildings in my pictures are actually a type of superior mirage from what I remember, due to a layer of warmer air on top of the cooler air against the water's surface.

there are too many instances of these flat, distant horizons in photographs and every-day experience for 'refraction' to play any significant part...
Yes, sometimes there is little or no refraction or mirage.  Then we get to see a better example of objects disappearing from the bottom up beyond the curvature.

Quote
..many of the pix i'v seen are in perfect weather conditions, too!.....
And the temperature of the water and different air layers were...?

Quote
(actually...i failed to mention that....the weather conditions in the bridge 'photo look less than ideal, too....the humidity may have played a factor in the refraction);
A little hazy.  Not too bad though.
(http://i1368.photobucket.com/albums/ag167/jeffro556/narrows_zps6z066kkl.jpg)
Temperature has more to do with it, but are you now admitting that it does in fact happen?

Quote
again: this needs to be proven by controlled, laboratory experiments!  :-B[/color][/font]
And what would be the best way to go about simulating both both flat and slightly curved surfaces, consisting of distances of several miles, in a laboratory setting?
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: mister bickles on March 24, 2015, 02:01:07 AM

Quote
again: this needs to be proven by controlled, laboratory experiments!  :-B[/color][/font]
And what would be the best way to go about simulating both both flat and slightly curved surfaces, consisting of distances of several miles, in a laboratory setting?

doesn't necessarily have to be in a laboratory....but, rather, just controlled, laboratory-type conditions.....there are plenty of reliable, scientific experiments done out-side of an actual laboratory.....

that said: there are, also, plenty of laboratory experiments that successfully simulate real-world conditions...."wave tanks" and "wind tunnels", for instance......

again:
i'm not convinced that 'refraction' plays any significant part in this "flat Earth horizon" phenomenon....there are simply too many instances of it in too many different geographical locations with different weather conditions....
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Wes on May 18, 2015, 06:46:29 PM

Then again, refraction can be quite convenient for FE'ers when it makes a round earth appear flat.

i get the feeling that "refraction" is one of those convenient, throw-a-way "explanations" that TPTB trot out to try and dodge "the hard balls";

although "refraction" is a relatively simple concept/explanation, TPTB and their innumerable shills are quite content to spew out a barrage of meaningless scientific gobbly-de-gook to baffle and confuse the average punter...

most people haven't got the qualifications, the time or the inclination to penetrate these "smoke screens";

that's the way TPTB like it......keep every-one distracted with mindless entertainments and too busy working to earn a living to have any spare time to consider the fact that they're trapped in "the Matrix";

(the word "amuse", for instance,  actually means "don't think" or "anti-think"....'muse' being the Græco-Roman term for "thinking" or "pondering"/"reflecting on some-thing"....an "a" prefix is its opposite!)

as for refraction making a RE appear flat....well....where is the evidence for this?
where are the laboratory experiments (that have been double/triple checked and, then, peer-reviewed) to demonstrate this?

what are the refractive indices of air and water.....how do they compare and how can they distort an object's appearance....   ???

most people know that water has a certain refractive index that can distort objects under-water......
does air have the same sort of effect?

quite frankly, i doubt it!

if any-thing, it would be only of a minimal, marginal kind that would make no appreciable difference

http://en.wikipedia.org/wiki/Atmospheric_refraction

In fact, not only does light refraction occur in laboratory settings, it is used by radio communications teams, from military applications, to commercial, to send high frequency radio signals greater distances than line of sight. The use of atmospheric refraction allows the signals to be sent at high angles as high as into the ionosphere, allowing the signal to refract back to the surface of the Earth beyond the horizon.

http://fas.org/spp/military/docops/afwa/U3.htm

Light refraction is a very real phenomenon. So to determine the distance one can see from a horizon, one necessarily must use control, as you suggest right? Wouldn't that mean controlling the test to eliminate for any possible variables that could interfere with accurately measuring the distances seen, such as the refraction of light, or at least factoring in a margin of error due to it?

*****************

Also, as Mr. Bickles raises the point of repeated controlled testing and peer-review, wouldn't such mathematical models serving as proof have to match replicable and observable tests?

Pongo, can you please explain if your model accounts for why we cannot see the lower portion of mountains in the photos, or the bottom sections of towers, if the Earth were flat? Wouldn't a linear line of sight produce the entire image that we are looking at? Also, if your model is to be used to demonstrate that a spherical Earth were not possible, then wouldn't the sight seen from the top of the CN tower also have to be visible from the base of the tower? Or are you suggesting that atmospheric conditions prevent us from seeing visible light beyond a set distance from the surface? If so, then why doesn't this match observation? Light waves have varying frequencies, and atmospheric distortion would fade out the shorter wavelengths first. Thus, if this were to be what caused our vision to perceive a horizon, then objects should appear more red the further out they are, just as the Sun does due to the angle as it sets. Another method of falsifying this claim could be tested with radiowaves, as they can travel further than visible light. If we were to set up radio signals, and monitor them, would the distance of line of sight be further for them?

Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: Cole smith on December 16, 2016, 04:32:23 PM
I'm truly impressed w the way you responded to several trolls. To take issue w a 10' oversight in your post is utterly mind boggling. Trolls and other g-earthers don't seem too concerned about the 55' discrepancy from Toronto's platform line of sight...can one of them explain airline travel from east to west or viceversa since aircraft seem to arrive either direction about the same time???
Title: Re: Proving a Flat-Earth Using Round-Earth Maths
Post by: rabinoz on December 17, 2016, 01:18:34 AM
I'm truly impressed w the way you responded to several trolls. To take issue w a 10' oversight in your post is utterly mind boggling. Trolls and other g-earthers don't seem too concerned about the 55' discrepancy from Toronto's platform line of sight...can one of them explain airline travel from east to west or viceversa since aircraft seem to arrive either direction about the same time???
I fail to see any problem with "airline travel from east to west or vice-versa since aircraft seem to arrive either direction about the same time".

Aircraft fly in the air. The air generally moves with the earth (we do have winds in various directions).
So, apart from any local winds the aircraft is flying relative to the moving earth.

Actually at high altitudes there are the jet-streams, which are high speed (to 200 km/hr or so) winds blowing from west to east and caused by the earth's rotation (among other things).
As a result, many west to east flights are significantly faster than the corresponding east to west flights.
For example:
New York City, USA (all airports) to London, United Kingdom (all airports): 26+ flights per day, 6h 55m duration  compare with
London, United Kingdom (all airports) to New York City, USA (all airports): 27+ flights per day, 8h 15m duration.

Hong Kong (HKG) to Los Angeles, USA (LAX): 5 flights per day, 12h 30m duration compare with
Los Angeles, USA (LAX) to Hong Kong (HKG): 5 flights per day, 15h 35m duration

Sydney (SYD) to Santiago, Chile (SCL): 5 flights per week, 12h 20m duration  compare with
Santiago, Chile (SCL) to Sydney (SYD): 5 flights per week, 14h 10m duration.

Shall I go on? Yes, the rotating Globe explains things just fine, thank you.

By the way, you should note that the Hong Kong to Los Angeles flight is at roughly the same latitude north as the Sydney to Santiago, Chile flight is south.
The times are very similar, indicating that the distance from Hong Kong to Los Angeles (actually 11,640 km) is similar to the distance from Sydney to Santiago, Chile (actually 11,340 km),
which is in complete disagreement with the "Ice Wall Map".