[Tau]

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?

[Gulliver]

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.

[Pongo]

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...

[Gulliver]

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.

[Tau]

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.

[Gulliver]

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

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.

[Pongo]

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.

[Gulliver]

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.

[Tau]

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

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.

[markjo]

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?

[Tau]

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.

[Gulliver]

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.

[markjo]

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.

I'm not asking about your argument, I'm asking about Pongo's premise that the horizon is 100 miles away from the top of the CN Tower.

[Pongo]

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?

[Gulliver]

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.

[Tom Bishop]

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.

[markjo]

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 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.).

[Gulliver]

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.

[Pongo]

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!

[Rama Set]

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