# The Flat Earth/Globe Earth Discussion Thread

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34 minutes ago, peter said:

So the sun light goes to the extremity of the green circle, I will just wait for your reply to confirm that before I continue

Since you haven't answered I will take that as a yes, and if so what stops the sun light at the edge of the circle , nothing, therefore the light should continue to the edges of the dome in all directions and as such there would be no night at all irrespective of where the sun or the moon is in relation to your explanation

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46 minutes ago, peter said:

Since you haven't answered I will take that as a yes, and if so what stops the sun light at the edge of the circle , nothing, therefore the light should continue to the edges of the dome in all directions and as such there would be no night at all irrespective of where the sun or the moon is in relation to your explanation

peter, you dont take anything as a yes or no unless specifically stated by myself,  can you please explain in spherist terms and logic where 100ft+ of curvature has disappeared to?

the question of 'wheres the curvature ?' was the original post and through 70 pages of divergence and return we now have multiple videos showing experiments over lakes and videos showing objects being viewed over long distances which, by sphere model, should be hidden behind hundreds of foot of land or water.

while you introduce questions and then questions upon answer upon answer the fact remains that the original issue is unanswered

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41 minutes ago, peter said:

therefore the light should continue to the edges of the dome in all directions

no thats wrong because you are assuming the suns size and distance from the land

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3 hours ago, zArk said:

no thats wrong because you are assuming the suns size and distance from the land

So what stops the light at the edge of the circle then another invisible barrier and how do you explain sun rise and sun set

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I was on a flight from London to Maine NY when it did an emergency landing in Goose Island Canada. This seems to support the fe map. Whilst I remain sceptical I do believe that there are unexplained anomalies with the sphere model. Not least the size of Africa for a start.

I am curious to learn what flat earthers think of the ionosphere and if they believe this is the true reason why short wave radio can be heard globally

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, Software Developer

Originally Answered: Chicago is 59 miles from the opposite shore of Lake Michigan.Given the earths curvature, it should be 2320 feet below the horizon. How can it be seen?

Well… Let’s see… We KNOW the rule is 8″ x miles squared, Right?

8*59²/12 = 2320.7 ft or 707.3 m

So this MUST be correct, right? It’s math, can’t be wrong!

Well…

IF you put your eye/camera lens at sea-level, so it is half-way under the dirt/water, exactly at 0 elevation, and you ignored refraction then yes — that is what you would expect.

Now ask yourself, are those EVER the ACTUAL viewing conditions?

No - no they are not.

Since none of those assumptions are correct let’s try that again using the correct mathematics for the real-life question.

In short, you are trying to measure (D) in the image below when it is clearly (B) that you should be asking about. But for some reason, Flat Earth people can never seem to grasp this difference. (B) applies whenever the observer has ANY height AT ALL — even 1 cm. And unless you’ve stuck your camera half-way under the water you aren’t at 0 elevation (you have to measure to center of the lens).

Tom Dreyfus gave an excellent answer on the math as well, I’m just expanding on it a bit using slightly different geometry here so we can avoid doing any trig, which I think is easier for a layperson to understand, and then I’ll do the actual math for Chicago - which puts this whole nonsense to rest for the rational and sane.

Here is the correct geometry to use for an observer who hasn’t (metaphorically or literally) buried their head in the sand…

We have two right triangles to solve. This geometry is obviously correct - right? If you don’t understand this diagram then you might as well stop here. It’s also the same as the geometry in Tom Dreyfus's answer to Chicago is 59 miles from the opposite shore of Lake Michigan. Given the earth’s curvature, it should be 2320 feet below the horizon. How can it be seen? We are solving the EXACT same equation here - I’m just doing it with simple algebra.

Inputs (the values we know, to some approximation)

h₀ = elevation of observer
d₀ = total distance to distant object
R = Earth Radius (we will calculate this based on our latitude rather than using an average value)

Outputs
d₁ = observer eye-line distance to Horizon
h₁ = height of object obscured by horizon (positive when √h₀ √[h₀ + 2 R] > d₀)

Equations

The first equation we will need is to figure out the value for d₁ which is the distance to our horizon point. We will need this value in the second equation.

We can plug in these values into the Pythagorean theorem equations giving us:

(R+h₀)² = d₁² + R²

When we solve for d₁ we get (click on the equation to see Wolfram|Alpha solve it):

equation (1) d₁ = √h₀ √[h₀ + 2 R]

We can be very certain of this equation because I've provided the Wolfram|Alpha link which solves it for you.

And the FIRST thing we need to do is see if our observer can see GREATER THAN your total distance. If this horizon point (√h₀ √[h₀ + 2 R]) is GREATER THAN d₀ then your total calculation needs to be a NEGATIVE value, meaning. ONLY where √h₀ √[h₀ + 2 R] is LESS THAN d₀ then it will be positive h₁ value.

Next we just plug in the values for our second right triangle (on the right):

(R+h₁)² = d₂² + R²

And we solve for h₁, giving us (click on the equation to see Wolfram|Alpha solve it):

equation (2) h₁ = √[d₂² + R²] - R

We don't know d₂ directly but we do know that d₂ = (d₀ - d₁)

We can now combine these equations to give us our final formula (or you can calculate it yourself in the two separate steps since you need the first result anyway). You can simplify this equation further if you make a few assumptions but I don't want to do that.

h₁ = √[d₂² + R²] - R
h₁ = √[(d₀ - d₁)² + R²] - R
h₁ = √[(d₀ - [√h₀ √[h₀ + 2 R]])² + R²] - R

This is our complete formula!

Height of Distant Objects Obscured by Earth Curvature = h₁ = √[(d₀ - [√h₀ √[h₀ + 2 R]])² + R²] - R (when √h₀ √[h₀ + 2 R] is LESS THAN d₀)

Example for Chicago from Warren Dunes State Park

First we need to know what the Earth radius is at the point where we want to make our observation, we going to ESTIMATE it based on latitude for Chicago from Google Earth Pro:

There is a nifty calculator you can use here: Earth Radius by Latitude Calculator

Using 41.885367 and our elevation bias is 175m (the lowest elevation along our sightline which is the height of Lake Michigan, see more below) which gives us R of 6,368,824 meters.

The distance is ~53 miles or 85,350 meters (measured in Google Earth Pro from the shore of Warren Dunes State Park).

Observer height (182 - 175) = ~7m

Now we have to estimate the refraction… I’m going with 20% refraction as it was documented that this NOT the normal view and there was a thermal inversion over the lake. But even if you DO NOT take refraction into account (and we will do that also) you’ll just get a slightly larger value for the height obscured - this problem and the images of Chicago all still CLEARLY show curvature of the Earth! The thing you need to know about refraction here is that we can express in terms of Earth radius - so we just multiply R by our refraction factor (so 1.20 = +20%). But the person making the claim that “this is IMPOSSIBLE” needs to provide us with the EXACT value for refraction or else they cannot possibly know what is possible or not. So I’m left guessing some “reasonable” number — but since we doing upper and lower bounds here we can see any reasonable value gets us close enough to disprove a Flat Earth - the Spherical model is a better fit.

More details here: Derivation for Height of Distant Objects Obscured by Earth Curvature

Now we can plug these values into our equation:

√[(d - (√h √[h + 2 R]))² + R²] - R, d=85350, h=7, R=6368824*1.20

] = ~368 meters

And again WITHOUT refraction we get a max value of:

√[(d - (√h √[h + 2 R]))² + R²] - R, d=85350, h=7, R=6368824

] = ~452 meters

So we can ESTIMATE (Unless you know EVERY value to high precision you are ONLY making a rough estimate) that we would expect somewhere around 368 to 452 meters of Chicago [as measured from the water level of 175m] to be blocked at 53 miles away view from about 7m over the water, depending on the atmospheric conditions.

Since the Sears Tower is RIGHT at this limit, 442 meters tall (excluding the top spires), we would expect that under low refraction conditions we would JUST BARELY see the very tips of the tallest building (atmospheric conditions allowing) and during inversions we would see a significantly higher fraction - possibly 100+ meters.

Isn’t it AMAZING how, using the CORRECT mathematical equations and a deeper understanding of the phenomena actually being observed that we get results that align much more closely with our actual observations?

*NOTE: some of the images are from Grand Mere State Park which is 56 miles BUT you also have a higher observation point there of about +15 meters elevation (so you STILL get~368m obscured

). But, please DO THE MATH YOURSELF for the EXACT observation you are testing but make sure your input values MATCH the exact picture you are testing.

What Flat Earth model cannot do is explain where the bottom of the city goes or why our view varies so greatly. Can’t be refraction because that would allow us to see MORE of the city. Can’t be perspective because that makes a distant object appear proportionally smaller in angular size, it doesn’t hide one thing behind another - if the bottom of the tower were too small to see due to perspective then the top would be too small also.

Here is the awesome photo showing a very distant Chicago skyline with the Sears Tower (aka Willis Tower) (credit to Joshua Nowicki I believe) circled in red.

And we clearly see that a significant portion of the city is below the horizon line with only tall buildings jutting up. The only question here is how much curvature are we seeing.

And we can tell there is some serious refraction going on by looking at the Sears Tower inside the red outlined area. What do you suppose that building looks like? Does it have a tall thin section below the spires on top? No, it doesn’t. So we’re seeing the top of the building Stretched by differences in the refractive index along each sight-line.

This is unequivocal evidence for the refraction.

And where do you suppose the ENTIRE bottom of the city went exactly besides hiding behind the curvature of the Earth?

Here is the Google Earth Pro elevation profile between the two showing why I used 175m as my elevation bias (it’s the height of the water which is our low-point, AKA OUR horizon is biased up by 175 meters, the lake water is NOT at sea-level but ~577′ or 175m)

And here another, more typical view where the mirage effect is inferior.

~ Joshua Nowicki

Now tell me that isn’t refraction.

And another with OBVIOUS mirage effect right across the center of the ‘city’ which has significantly stretched all the buildings. This guy has some awesome images.

~ Joshua Nowicki

And a guy who did some side-by-sides comparing these photos:

)

Edited by peter
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1 hour ago, Diesel said:

I was on a flight from London to Maine NY when it did an emergency landing in Goose Island Canada. This seems to support the fe map.

Helloooo D! I love new faces on here, especially when they understand. Thank-you... uplifting!

The emergency landing will remain unexplained forever because the spherist can never explain them. If pilots (and ship's captains for that matter) navigated from a globe, they would remain lost forever. I'm not sure how much of the thread you have read, but point four in the OP discusses this point in depth. Six emergency landing s are presented and every single time we see that they only work on flat earth. Any honest poster will find this truth for themselves. There is no debunk. They are ignored and never spoken of to young children. Also, I recently posted three more. They, of course, were ignored as well.

1 hour ago, Diesel said:

Whilst I remain sceptical I do believe that there are unexplained anomalies with the sphere model. Not least the size of Africa for a start.

Or like when they tell us about their "total eclipses" with both the sun and moon clearly visible above the horizon? Pretty hysterical, yes?

1 hour ago, Diesel said:

I am curious to learn what flat earthers think of the ionosphere and if they believe this is the true reason why short wave radio can be heard globally

LOL, these are line of sight transmissions and since there is no geometric horizon, there simply exists no physical barrier. We have shown this multiple times now. Even with lasers. This how we know that earth's "radius" has been falsified, thus falsifying every bit of heliocentric idiocy that goes along with it.

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8 hours ago, peter said:
, Software Developer

Originally Answered: Chicago is 59 miles from the opposite shore of Lake Michigan.Given the earths curvature, it should be 2320 feet below the horizon. How can it be seen?

Well… Let’s see… We KNOW the rule is 8″ x miles squared, Right?

8*59²/12 = 2320.7 ft or 707.3 m

So this MUST be correct, right? It’s math, can’t be wrong!

Well…

IF you put your eye/camera lens at sea-level, so it is half-way under the dirt/water, exactly at 0 elevation, and you ignored refraction then yes — that is what you would expect.

Now ask yourself, are those EVER the ACTUAL viewing conditions?

No - no they are not.

Since none of those assumptions are correct let’s try that again using the correct mathematics for the real-life question.

In short, you are trying to measure (D) in the image below when it is clearly (B) that you should be asking about. But for some reason, Flat Earth people can never seem to grasp this difference. (B) applies whenever the observer has ANY height AT ALL — even 1 cm. And unless you’ve stuck your camera half-way under the water you aren’t at 0 elevation (you have to measure to center of the lens).

truly informative.

now back to the laser at the lakeside

after reading Tony Miller i am unsure if you are saying that the laser light is bent across the curvature of 13 / 16miles of lake OR the laser at the shoreline as observed by the camera is being refracted above the curvature /???

Edited by zArk

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1 hour ago, zArk said:

after reading Tony Miller i am unsure if you are saying that the laser light is bent across the curvature of 13 / 16miles of lake OR the laser at the shoreline as observed by the camera is being refracted above the curvature /???

If you read my post regarding the laser ,you can draw your own conclusions as I haven't said either, I was pointing out some, not all of the  variables that could influence the results over that distance

Also I see you only quoted the premise and not the math or conclusion, to my post ,rather selective don't you think as one of the pictures that bflat put up on on page 40 to further his cause was used in my last post as a rebuttal

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RE: EMERGENCY LANDINGS CANNOT BE DENIED

11 hours ago, bflat said:

They are ignored and never spoken of to young children. Also, I recently posted three more.

From the OP:

4)   Airlines use this map or one quite similar to navigate from:

EMERGENCY LANDINGS PROVE THIS BEYOND DOUBT:

From just a couple pages back:

Still confused?

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23 minutes ago, bflat said:

23 minutes ago, bflat said:

4)   Airlines use this map or one quite similar to navigate from:

Sounds great, THEY USE THIS MAP , or one very similar , how similar ,your going to have to be a bit more specific. Now look what is written on the map it's self ( scientifically and practically correct) which to me infers that it is not correct, then look at the patent date 1892.

Nowhere dose it say the earth is flat ,but what it dose say is it's a longitude and time calculator so I suspect that is what this is.

It's a brilliant heading  THE NEW STANDARD MAP OF THE WORLD, again sounds great but the key word here is MAP ,what is a map ? it is a two dimensional representation of 3 dimensional reality so of course it is going to be Flat, it doesn't mean the earth is flat.

Back in the day before L.E.D  screens and computers I will grant you that pilots used similar maps to navigate but only up to the point they are both drawn on paper as it would be a bit hard to balance a globe on your lap while trying to fly.

THEY USE THIS MAP , or one very similar , a fairly deceptive statement ,wouldn't you say

If you are wondering why the font is very small ,always check the fine print

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40 minutes ago, peter said:

Sounds great, THEY USE THIS MAP , or one very similar , how similar ,your going to have to be a bit more specific. Now look what is written on the map it's self ( scientifically and practically correct) which to me infers that it is not correct, then look at the patent date 1892.

Nowhere dose it say the earth is flat ,but what it dose say is it's a longitude and time calculator so I suspect that is what this is.

It's a brilliant heading  THE NEW STANDARD MAP OF THE WORLD, again sounds great but the key word here is MAP ,what is a map ? it is a two dimensional representation of 3 dimensional reality so of course it is going to be Flat, it doesn't mean the earth is flat.

Back in the day before L.E.D  screens and computers I will grant you that pilots used similar maps to navigate but only up to the point they are both drawn on paper as it would be a bit hard to balance a globe on your lap while trying to fly.

THEY USE THIS MAP , or one very similar , a fairly deceptive statement ,wouldn't you say

If you are wondering why the font is very small ,always check the fine print

The emergency landings prove the map. You still don't get it.

Just like I have asked y'all to find the curve, do your own observations, and record the results, I'll ask you to finally watch the videos above AND THEN MAP THEM YOURSELF - BOTH ON ANY GLOBE OF YOUR CHOICE AND ALSO ON GLEASON'S MAP (or the freemason's UN flag!)

It's that simple!

Seriously... map it yourself... forget the rest!

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37 minutes ago, bflat said:

The emergency landings prove the map. You still don't get it.

Just like I have asked y'all to find the curve, do your own observations, and record the results, I'll ask you to finally watch the videos above AND THEN MAP THEM YOURSELF - BOTH ON ANY GLOBE OF YOUR CHOICE AND ALSO ON GLEASON'S MAP (or the freemason's UN flag!)

It's that simple!

Seriously... map it yourself... forget the rest!

And even if I went to all that trouble and didn't come to your conclusion , experience would tell me your reply would be one word "FAKE"

You are correct, there are a lot of things in this world I don't get ,but you old son are definitely not one of them

I'm still waiting for the answers to my questions by the way ,oh! and my shoe horn as well

Edited by peter
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6 hours ago, peter said:

If you read my post regarding the laser ,you can draw your own conclusions as I haven't said either, I was pointing out some, not all of the  variables that could influence the results over that distance

Also I see you only quoted the premise and not the math or conclusion, to my post ,rather selective don't you think as one of the pictures that bflat put up on on page 40 to further his cause was used in my last post as a rebuttal

ermm i am not drawing my own conclusions because i asked you

i only quoted the beginning because it was such a long post :S i didnt mean to offend you

some of the variables ... ok yes, i remember some of the variables. can you name any other variables?

i am intrigued how this observed refraction occurs with regard to the laser on the shore.

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25 minutes ago, zArk said:

i only quoted the beginning because it was such a long post :S i didnt mean to offend you

First of ,I'm not offended

34 minutes ago, zArk said:

i am intrigued how this observed refraction occurs with regard to the laser on the shore.

Lets be honest ,that I can't tell you,however the first question I would ask is, can the laser that was said to be used, able to travel that distance under those conditions and if so we only have the video makers word for it and since we are taking people at face value one would have to ask what if any financial benefits  do the people gain from making these videos  particularly the likes of Mark Sargent . I believe financial statements from U tube would be enlightening and very interesting,

As to other variables ,heat haze ,temperature differential between the water and surrounding air , temperature differential between the beam and surrounding air , there are probably more these and the ones in the other post  are just the ones I can think of off the top of my head

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13 minutes ago, peter said:

First of ,I'm not offended

Lets be honest ,that I can't tell you,however the first question I would ask is, can the laser that was said to be used, able to travel that distance under those conditions and if so we only have the video makers word for it and since we are taking people at face value one would have to ask what if any financial benefits  do the people gain from making these videos  particularly the likes of Mark Sargent . I believe financial statements from U tube would be enlightening and very interesting,

As to other variables ,heat haze ,temperature differential between the water and surrounding air , temperature differential between the beam and surrounding air , there are probably more these and the ones in the other post  are just the ones I can think of off the top of my head

oh! sorry i thought you were using Tony Millers calculations and his conclusion to apply somehow to the laser 13 mile and laser 16mile problem for spherists

do you agree with Tony Millers calcs, included science facts, diagrams, used photos, conclusion and applied flatearth problem???

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1 hour ago, zArk said:

oh! sorry i thought you were using Tony Millers calculations and his conclusion to apply somehow to the laser 13 mile and laser 16mile problem for spherists

do you agree with Tony Millers calcs, included science facts, diagrams, used photos, conclusion and applied flatearth problem???

Personally, what I would say is I couldn't give a shit one way or another if the world is round or flat but at the present time,in my opinion there is a more substantial amount of evidence coupled with actual observation to indicate the earth is a spheroid than there is to indicate that its flat,if that changes through scientific analysis ,fine but until that happens I find to base your theory on people that have an obvious vested interest in the perpetuation of this ,for what ever reason I believe is asking for problems.

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@peter I think that picture can easily be viewed , with refraction bringing up the sea level,  to contain the Chicago land behind the visual effect.

The apparent horizon

And we clearly see that a significant portion of the city is below the horizon line with only tall buildings jutting up. The only question here is how much curvature are we seeing

After looking at that picture, tony must mean the volume of buildings or the width of building as significant because I can see quite a lot of buildings across the picture and the whitish band plus the apparent horizon of sea level is covering the rest.

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He is also talking about the apparent increase in length of the tower spires due to refraction and puts in a comparison photo of the tower on the left. However I will grant you  the photo used as a comparison is taken at a different location and  therefore the perspective will be different so it can only be used as a rough indication as to the effect

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14 hours ago, zArk said:

oh! sorry i thought you were using Tony Millers calculations and his conclusion to apply somehow to the laser 13 mile and laser 16mile problem for spherists

do you agree with Tony Millers calcs, included science facts, diagrams, used photos, conclusion and applied flatearth problem???

Sorry I didn't answer your question before , lets just say that his explanation at present is feasible and is the best I have been able to find at present that takes into account effects you find in nature and not just perfect conditions in a laboratory. Is he 100% accurate? good question.

There are a lot of things I don't agree with in mainstream science ,however the earth being flat is not one of them .

If the flat earth crowd would come up with a theory that encompasses all the phenomena we see with regards to our physical world , well and good I would look at it , but the biggest problem I see at present  all you have is    (here is an 1892 time and longitude  calculator, look at this ,look at that, explain this, explain that ,here a bloke with a laser ,this plane made an emergency landing,here is some thumb nail sketches of 17th century sailing ships , here is a bloke with a globe and some string in his kitchen  ) do you see how this looks to the other camp.

Sorry but if you don't come up with a theory that dose explain all we see and experience with regards to the earth and moon and planets ,the average person will never take this subject seriously and the people with a vested interest will be reaping the rewards from U tube and the like for a good time to come yet

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6 hours ago, peter said:

,however the earth being flat is not one of them

well this i agree with in principal because in all social/worldly polarised debates i have found and still find that both camps are deliberately hiding facts that counter their positions

but this doesnt mean i will jump in bed with either as the best option

i dont think the earth is a sphere OR flat as per the FE model

i.e the solar eclipse argument

the moon has been spotted above the horizon during a solar eclipse

so FE say that there is another object in the sky that causes the eclipse

spherists scoff

i was on the side of spherists for solar eclipse but the whole situation was explained and the sphere model doesnt fit either

1 earth spins towards the east so the the sun is observed to rise in the east

2 the moon is observed to spin east to west but due to its velocity being slower than the earths spin

this is an observed motion of the moon and the sun --- fine, spherist model

but during an eclipse the effect was observed travelling in a west east motion

this is impossible with the stalwart sphere model of 1 and 2

the more i look into the arguments between the 2 the more i am left with nothing from both

Edited by zArk
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I could only put up the link as I was blocked from printing the article.

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42 minutes ago, peter said:

I could only put up the link as I was blocked from printing the article.

yes, and thats the problem aint it

spherist terms and argument

the earth observes the moon moving east to west HOWEVER

this is due to the speed of the moon rotating (west to east) around the earth visually /apparently slower than the rotation of the earth (west to east) on its axis

therefore the moon is observed to go backwards across the sky -- the east west movement

GREAT!!!

BUT.... alas the solar eclipse is observed , apparent, to move west to east

this is impossible using the spherist theory of the earths rotation on axis, its speed, the moons rotation , its speed and the suns position

the solar eclipse observed west to east does not fit with the spherists theory of sun/moon/earth and the kit kaboodle of all its other arguments

this is MATE, i wont say CHECK MATE because i am sure there will be a chalkboard soon by Professor 'dickface things can only get better' Cox

geddit??

well i am baffled at how the Spherists expect me, to turn a blind eye to this blatant conflict in theory but we see it and its fact

TIME STAMPED FOR RUN THROUGH

Edited by zArk

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