#### Tau

• Zetetic Council Member
• 911
• Magistrum Fallaciae
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #80 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!
« Last Edit: March 05, 2015, 06:12:39 PM by Tausami »
That's how far the horizon is, not how far you can see.

#### Gulliver

• 682
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #81 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.
Don't rely on FEers for history or physics.
[Hampton] never did [go to prison] and was never found guilty of libel.
The ISS doesn't accelerate.

#### Pongo

• Most Educated Flat-Earther
• Planar Moderator
• 759
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #82 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.

#### Gulliver

• 682
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #83 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.
Don't rely on FEers for history or physics.
[Hampton] never did [go to prison] and was never found guilty of libel.
The ISS doesn't accelerate.

#### markjo

• 7893
• Zetetic Council runner-up
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #84 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.
Abandon hope all ye who press enter here.

Science is what happens when preconception meets verification.

Ignorance more frequently begets confidence than does knowledge. -- Charles Darwin

If you can't demonstrate it, then you shouldn't believe it.

#### Tau

• Zetetic Council Member
• 911
• Magistrum Fallaciae
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #85 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.
« Last Edit: March 05, 2015, 10:19:05 PM by Tausami »
That's how far the horizon is, not how far you can see.

#### markjo

• 7893
• Zetetic Council runner-up
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #86 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
Abandon hope all ye who press enter here.

Science is what happens when preconception meets verification.

Ignorance more frequently begets confidence than does knowledge. -- Charles Darwin

If you can't demonstrate it, then you shouldn't believe it.

#### Pongo

• Most Educated Flat-Earther
• Planar Moderator
• 759
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #87 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.

#### markjo

• 7893
• Zetetic Council runner-up
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #88 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.
Abandon hope all ye who press enter here.

Science is what happens when preconception meets verification.

Ignorance more frequently begets confidence than does knowledge. -- Charles Darwin

If you can't demonstrate it, then you shouldn't believe it.

#### Tom Bishop

• Zetetic Council Member
• 10773
• Flat Earth Believer
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #89 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?

#### Tau

• Zetetic Council Member
• 911
• Magistrum Fallaciae
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #90 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.
That's how far the horizon is, not how far you can see.

#### markjo

• 7893
• Zetetic Council runner-up
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #91 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.
Abandon hope all ye who press enter here.

Science is what happens when preconception meets verification.

Ignorance more frequently begets confidence than does knowledge. -- Charles Darwin

If you can't demonstrate it, then you shouldn't believe it.

#### Rama Set

##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #92 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.

#### Pongo

• Most Educated Flat-Earther
• Planar Moderator
• 759
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #93 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!

#### Rama Set

##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #94 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.

#### Rama Set

##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #95 on: March 06, 2015, 05:13:22 PM »
So the tallest building in Niagara Falls, NY is the Seneca Niagara Casino Tower at 109 meters.  Niagara Fall, NY has an elevation of 51 meters.

The horizon is there for 45kms 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.  It appears that you can see well past the bulge between the two towers.

Does anyone agree or disagree with this?

#### Pongo

• Most Educated Flat-Earther
• Planar Moderator
• 759
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #96 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.

#### Rama Set

##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #97 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.

#### Pongo

• Most Educated Flat-Earther
• Planar Moderator
• 759
##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #98 on: March 06, 2015, 05:40:47 PM »
So the tallest building in Niagara Falls, NY is the Seneca Niagara Casino Tower at 109 meters.  Niagara Fall, NY has an elevation of 51 meters.

The horizon is there for 45kms 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.  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.

#### Rama Set

##### Re: Proving a Flat-Earth Using Round-Earth Maths
« Reply #99 on: March 06, 2015, 05:49:07 PM »
So the tallest building in Niagara Falls, NY is the Seneca Niagara Casino Tower at 109 meters.  Niagara Fall, NY has an elevation of 51 meters.

The horizon is there for 45kms 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.  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.